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<p>I understand that the black-and-white images you see looking through a household telescope are only like that due to the intensity of the light that reaches us, and that most of the astronomy images we find online have some color modification. But if we were to eventually travel between galaxies, how would those galaxies appear to the naked eye? How close to the false color images would they be, and how accurately do we know this?</p> | 1,217 |
<p>I understand the mathematical derivation of the Tully Fisher relation from basic physics formulas, as shown on <a href="http://www.jca.umbc.edu/~george/html/courses/glossary/tully_fisher.html" rel="nofollow">this site</a>. However, after using the physics equations, it seems that several assumptions are made from this point on.</p>
<p>First are statistical assumptions. There are statistical errors because the observable, luminous mass of the galaxy is less than the actual mass of the galaxy, and the mass of the galaxy is assumed to be only the observed mass, not the actual mass. Second, this relationship seems to assume that all galaxies are perfectly circular (with negligible thickness).</p>
<p>So, with all these assumptions necessary for the Tully-Fisher relation, how is the relationship derived, how can corrections be made for the assumptions when attempting calculations using this relationship, and why is the Tully-Fisher relation generally accepted by the astrophysics community?</p> | 1,218 |
<p>How can I prove Thevenin's and Norton's theorem? <a href="http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem" rel="nofollow">Thévenin's theorem</a> can be used to convert any circuit's sources and impedances to a Thévenin equivalent.</p> | 1,219 |
<p>I'm doing simulations of copper, where the temperature can reach up to ~1300 K.</p>
<p>Some calculations depend on the mean free path (MFP) of copper. The only value I've found for it is 39nm and it's always accompanied with the clause of "at room temperature".</p>
<p>Can anybody tell me (or provide some references), how does the MFP depend on temperature?</p>
<p>Copper's resistivity dependence on temperature is known to me, can that be used to derive the mean free path dependence in the following way?</p>
<p>So we got the conductivity $\sigma(T)$ and that can be related to the relaxation time as</p>
<p>$\sigma = \frac{n e^{2}}{m} \tau$</p>
<p>And the relation between the mean free path and relaxation time is $\Lambda = v_F \tau$, where $v_F$ is the Fermi velocity (is this an approximation?).</p>
<p>Thus we get</p>
<p>$\Lambda(T) = \frac{v_F m}{n e^{2}} \sigma(T)$</p>
<p>Is this a valid way of finding the dependence of the mean free path of copper on temperature?</p> | 1,220 |
<ol>
<li><p>What does a <a href="http://en.wikipedia.org/wiki/Quantum_state" rel="nofollow">"STATE"</a> exactly mean in <a href="http://en.wikipedia.org/wiki/Quantum_mechanics" rel="nofollow">quantum mechanics</a>? </p></li>
<li><p>What is the equivalence of "STATE" in classical mechanics?</p></li>
<li><p>If we have a wave function $\Psi$ , its absolute square $|\Psi|^2$ is the probability density of finding the particle somewhere in the space, and I know it can be written as $\langle \Psi |\Psi \rangle$. But what is the physical meaning of $\langle \Phi | \Psi \rangle$, where $\Phi$ and $\Psi$ are two different wave functions. Why we need to take inner product of two different wave functions. </p></li>
</ol> | 1,221 |
<p>If I drop a ball in a train moving at a constant speed, will it land on the spot I aimed it at or a little away as the train has moved while it was in air? If it lands away, will the observer not know that he is in a moving frame of reference? If it lands on the intended spot, how did the ball know it is inside a train? </p> | 1,222 |
<p>I'm struggling with the notion of an inertial frame of reference. I suspect my difficulty lies with the difference between Newtonian and relativistic inertial frames, but I can't see it. </p>
<p>I've read that Newton's laws apply in any non accelerating frame of reference, which are called inertial frames. So, if I play pool on a train moving with uniform velocity, the balls behave in the same way as if I were playing pool in a pool hall. So the train is (to a good approximation, ignoring forces caused by the Earth's rotation and movement around the Sun etc) an inertial frame. </p>
<p>I've also read that in an inertial frame, Newton's first law is satisfied. So, if I slide a rock on a sheet of ice and I could somehow eliminate the frictional forces between the rock and the ice, the rock would carry on sliding for ever, as predicted by Newton's first law. </p>
<p>Question 1 - is the sheet of ice also therefore a good approximation to an inertial frame?</p>
<p>Question 2 - are these two definitions of inertial frames saying the same thing in different ways?</p>
<p>Both these examples occur in gravitation fields, which doesn't seem to matter as both the train and the ice sheet are presumably good approximations to inertial frames.</p>
<p>Question 3 - is gravity irrelevant when defining these two inertial frames?</p>
<p>In special relativity, I've read (Foster and Nightingale) that an inertial frame is also one where Newton's first law holds. But as it's special relativity there cannot be an inertial frame if there's a gravitational field, so I'm assuming the above two examples are not inertial frames in special relativity.</p>
<p>Question 4 - how come you can use the same definition of an inertial frame (satisfied Newton's first law) but in the case of special relativity, the train and the ice sheet aren't inertial frames. Is this to do with not being able to synchronize clocks in a gravitational field?</p>
<p>In general relativity, I've read (Schutz) that a freely falling frame is (on Earth) the only possible (and local) inertial frame. So again the first two examples would not be inertial frames in general relativity.</p>
<p>Question 5 - am I right to think that my two examples aren't even approximate approximations to an inertial frame in relativity?</p>
<p>Question 6 - have I missed anything else that might be useful?</p>
<p>Many thanks</p>
<p>Edit. On reflection, am I right in thinking that because Newtonian mechanics assumes universal, absolute time we don't need to worry about synchronizing clocks in a Newtonian inertial frame. Therefore we don't need to worry about gravity in a Newtonian inertial frame, because in such a frame gravity does not affect time.</p>
<p>This is not the case in spacetime, as here gravity does affect time and the only way to have synchronized clocks in an inertial frame in spacetime is:
1. have no gravitational field, or
2. use a local, freely falling frame.</p>
<p>Am I on the right track here? </p> | 285 |
<p>Example of pendulum is inertial frame of reference or non inertial frame of reference?
because if pendulum starts moving its continuously moves without changing there period of time but is changes its velocity with there motion... </p> | 1,223 |
<p>Let $\rho = \begin{bmatrix}\ 1&0 \\ 0&0 \end{bmatrix}$, $\rho' = \begin{bmatrix}\ 0&0 \\ 0&1 \end{bmatrix}$, $\rho'' = \dfrac{1}{2}\begin{bmatrix}\ 1&1 \\ 1&1 \end{bmatrix}$ (all density operators).</p>
<p>Consider a physical operation $\phi$ such that $\phi(\rho) = \rho$, $\phi(\rho') = \rho'$, $\phi(\rho'') = \dfrac{1}{5}\begin{bmatrix}\ 4&2 \\ 2&1 \end{bmatrix}$.</p>
<p>Why is $\phi$ not a realisable physical operation? It certainly preserves trace and positivity...</p> | 1,224 |
<p>I remember considering in class in college, the case of a photon gas trapped in a d-dimensional box as a subject of interest, whose energy distribution, heat capacity, etc. should be calculated. </p>
<p>This momentum/energy distribution was then related to that emitted by a black body - that up to a constant factor, they are one in the same.</p>
<p>I could understand the argument that making a small hole in such a box would constitute a "black body", as the hole can't reflect back light impinging on it, but I can't see how to make the conclusion that this is the exact same distribution. Shouldn't the atomic structure of matter be brought into account as well, or is it considered "hollow" on sufficiently large scales? </p> | 1,225 |
<p>There are some definitions and properties for <a href="http://en.wikipedia.org/wiki/Pauli_matrices" rel="nofollow">Pauli matrices</a> and their combinations:</p>
<p>$$
\varepsilon^{\alpha \beta } = \varepsilon^{\dot {\alpha} \dot {\beta} } = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}_{\alpha \beta }, \quad \varepsilon_{\dot {\alpha} \dot {\beta}} = \varepsilon_{\alpha \beta} = -\varepsilon^{\alpha \beta },
$$
$$
(\sigma^{\mu})_{\alpha \dot {\alpha} } = (\hat {\mathbf E} , \hat {\mathbf \sigma})^{\mu}_{\alpha \dot {\alpha}}, \quad (\tilde {\sigma}^{\mu})^{\dot {\beta } \beta } = \varepsilon^{\alpha \beta}\varepsilon^{\dot {\alpha }\dot {\beta }}(\sigma^{\mu})_{\alpha \dot {\alpha}} = (\hat {\mathbf E}, -\hat {\mathbf \sigma})^{\mu , \dot {\beta }\beta },
$$
$$
(\sigma^{\mu \nu})_{\alpha \beta} = -\frac{1}{4}\left( (\sigma^{\mu} \tilde {\sigma}^{\nu})_{\alpha \beta } - (\sigma^{\nu} \tilde {\sigma}^{\mu})_{\alpha \beta }\right), \quad (\tilde {\sigma}^{\mu \nu})_{\dot {\alpha }\dot {\beta }} = -\frac{1}{4}\left( (\tilde{\sigma}^{\mu} \sigma^{\nu})_{\dot {\alpha} \dot {\beta} } - (\tilde {\sigma}^{\nu} \sigma^{\mu})_{\dot {\alpha} \dot {\beta} }\right),
$$
$$
(\tilde {\sigma}^{\mu})^{\dot {\alpha }\alpha}(\sigma_{\mu})_{\beta \dot {\beta }} = 2\delta^{\dot {\alpha}}_{\dot {\beta}}\delta^{\alpha}_{\beta}.
$$
How to show, that
$$
(\sigma^{\alpha \beta})_{a b }(\tilde {\sigma}^{\mu \nu})_{\dot {c} \dot {d}}g_{\alpha \mu} = 0?
$$
(It helps to show, that spinor irreducible representation of the generators of Lorentz group expands on two spinor subgroups).</p>
<p><strong>My attempt</strong>.</p>
<p>I tried to show this, but only got following.
$$
(\sigma^{\alpha \beta})_{a b }(\tilde {\sigma}^{\mu \nu})_{\dot {c} \dot {d}}g_{\alpha \mu} = \frac{1}{16}\left( (\sigma^{\alpha})_{a \dot {n}}(\tilde {\sigma}^{\beta })^{ \dot {n}}_{\quad b}(\tilde {\sigma}^{\mu})_{\dot {c}}^{\quad m }(\sigma^{\nu })_{m \dot {d}} - (\sigma^{\alpha})_{a \dot {n}}(\tilde {\sigma}^{\beta })^{ \dot {n}}_{\quad b}(\tilde {\sigma}^{\nu})_{\dot {c}}^{ \quad m}(\sigma^{\mu })_{m \dot {d}}\right)
$$
$$
+ \frac{1}{16}\left(-(\sigma^{\beta })_{a \dot {n}}(\tilde {\sigma}^{\alpha })^{ \dot {n}}_{\quad b}(\tilde {\sigma}^{\mu})_{\dot {c}}^{\quad m}(\sigma^{\nu })_{m \dot {d}} + (\sigma^{\beta })_{a \dot {n}}(\tilde {\sigma}^{\alpha })^{ \dot {n}}_{\quad b}(\tilde {\sigma}^{\nu})_{\dot {c}}^{\quad m }(\sigma^{\mu })_{m \dot {d}} \right)g_{\alpha \mu}
$$
After that I transformed each summand like
$$
(\sigma^{\alpha})_{a \dot {n}}(\tilde {\sigma}^{\beta })^{ \dot {n}}_{\quad b}(\tilde {\sigma}^{\mu})_{\dot {c}}^{\quad m }(\sigma^{\nu })_{m \dot {d}}g_{\alpha \mu} = (\sigma^{\alpha})_{a \dot {n}}(\tilde {\sigma}_{\alpha})_{\dot {c}}^{\quad m }(\tilde {\sigma}^{\beta })^{ \dot {n}}_{\quad b}(\sigma^{\nu })_{m \dot {d}} =
$$
$$
= \varepsilon_{\dot {c}\dot {\gamma}}(\sigma^{\alpha})_{a \dot {n}}(\tilde {\sigma}_{\alpha})^{\dot {\gamma} m }(\tilde {\sigma}^{\beta })^{ \dot {n}}_{\quad b}(\sigma^{\nu })_{m \dot {d}} = 2\varepsilon_{\dot {c}\dot {\gamma }}\delta^{\dot {\gamma}}_{\dot {m}}\delta^{n}_{a}(\tilde {\sigma}^{\beta })^{ \dot {n}}_{\quad b}(\sigma^{\nu })_{m \dot {d}} =
$$
$$
2\varepsilon_{\dot {c}\dot {m}}(\tilde {\sigma}^{\beta })^{ \dot {m}}_{\quad b}(\sigma^{\nu })_{a \dot {d}} = 2(\sigma^{\beta } )_{b \dot {c}}(\sigma^{\nu })_{a \dot {d}}.
$$
Finally, I got
$$
(\sigma^{\alpha \beta})_{a b }(\tilde {\sigma}^{\mu \nu})_{\dot {c} \dot {d}}g_{\alpha \mu} = \frac{1}{8}\left( (\sigma^{\beta})_{b\dot {c}}(\sigma^{\nu})_{a\dot {d}} + (\sigma^{\beta })_{b \dot {d}}(\sigma^{\nu})_{a \dot {c}} + (\sigma^{\beta })_{a \dot {c}}(\sigma^{\nu})_{b \dot {d}} + (\sigma^{\beta})_{a \dot {d}}(\sigma^{\nu})_{b \dot {c}}\right).
$$
What to do next?</p> | 1,226 |
<p>I'm curious if you have a radiator or say a block of metal (lets say it's copper since it has the highest thermal conductivity) and on one side is a processor producing heat.</p>
<p>At idle the processor heats the block to 50°C, and the block is 120mm x 120mm. The ambient room temp is 22.2°C. You put a fan on the block (120mm x 120mm @ 1000RPM) and it cools the block to 40°C. Now drop the ambient temp 2 degrees C to 20.2°C.</p>
<p>Would the full change in ambient temp take effect upon the aluminum block? Would it have less effect due to the blocks efficiency? The processor's heat is constant and the fan blowing on the block is constant.</p> | 1,227 |
<p>If I have two balls with masses and charges $m_1, q_1^{+}$, $m_2, q_2^{+}$, initially held at distance $d$, and then released, how can I know the kinetic energies of each of the balls at infinite distance between them? I'm quite stuck on that, because they both have the same potential energy at the beginning, and it decreases not in the same pattern, as if one of the balls was stationary. So it not only falls like $1/R$, because at the same time, the other ball that is causing this potential energy, is also being repelled. So how can I really find out the energies? I tried to apply the conservation of energy law, because I know that at infinite distance from each other they'll have zero potential energy, thus all the initial was transformed into kinetic form, however I'm stuck with the initial potential energy (they both have it, so should I put $2U_p$?), and even so, I can't find their kinetic energies separately, without having another equation.</p> | 1,228 |
<p>Or: can it be proved that anti-matter definitely is nót matter going backwards in time?</p>
<p><a href="http://en.wikipedia.org/wiki/Antimatter" rel="nofollow">From wikipedia</a>:</p>
<blockquote>
<p><em>There is considerable speculation as to why the observable universe is apparently almost entirely matter [...] the apparent asymmetry of matter and antimatter in the visible universe is one of the greatest unsolved problems in physics.</em></p>
</blockquote>
<p>Couldn't the concept of reversal in time contribute to the explanation of this problem?</p>
<p>Measurements that reveal reversal in time must be rare and exceptional, just like anti-matter.</p> | 420 |
<p><strong>Short question</strong></p>
<p>Given any Lagrangian density of fields one could possibly conceive, is it the case that after one has performed a Legendre transformation, if the Hamiltonian is then expressed in terms of the original fields, will it contain all of the terms originally in the Lagrangian but with the signs of the potentials 'flipped'? Or are there cases when terms will be dropped in the transformation? Or is my statement altogether wrong. This question is inspired by the long version of my question below, relating to a problem I am currently working on.</p>
<p><strong>Long, specific question</strong></p>
<p>Given the Lagrangian density</p>
<p><img src="http://i.stack.imgur.com/bIVKZ.png" alt="enter image description here"></p>
<p>I have derived (several times to check for errors) using a Legendre transformation, the Hamiltonian density:</p>
<p><img src="http://i.stack.imgur.com/pZJVT.png" alt="enter image description here"></p>
<p>which is basically the Lagrangian density with a few sign flips <em>apart</em> from that it is missing the two terms outside the brackets in the Lagrangian density $\frac{1}{4\pi c}\frac{\partial\vec{A}}{\partial t}\cdot \nabla\phi + \frac{1}{c}\vec{A}\cdot\frac{\partial \vec{P}}{\partial t}$.</p>
<p>This worries me, as naively I expected to obtain a Hamiltonian very similar looking to the Lagrangian with all the same terms but some of the signs flipped. However I noticed that the 'missing' terms are the only ones which contain non-squared time derivatives so thought that might have something to do with it.</p> | 1,229 |
<p>I have a set of trajectories of three particles and their instantaneous velocities. I would like to compute the 3 components of the angular velocity pseudovector of the fictive triangle formed by these three particles. How can I proceed for that?</p> | 1,230 |
<p>Several different sources online state that the average temperature of interstellar space (or the universe in general) is around 2-3K.</p>
<p>I learned that temperature is basically the wiggling of matter, and I find it somewhat counterintuitive that the wiggling of so few particles can cause a temperature of 2-3K. Is there a (order-of-magnitude) calculation which can show that this average temperature estimation is correct, using an estimation of the average density of interstellar space (or the universe in general)?</p> | 1,231 |
<p>Is it possible to write down a partition function for a <a href="http://en.wikipedia.org/wiki/Microcanonical_ensemble" rel="nofollow">microcanonical ensemble</a>?</p> | 1,232 |
<p>I am very curious if an easy calculable formula for the bremstrahlung radiation of deeply relativistic, charged particles exists, if they are moving on circular orbit:</p>
<p>$P(E,m_0,Q,r)=?$</p>
<p>...where</p>
<ul>
<li>$P$ is the power of the Bremstrahlung radiation;</li>
<li>$E$ is the total kinetic energy of the particles (we are in deeply relativistic case, thus $E\gg{m_0}c^2$);</li>
<li>$m_0$ is the total rest mass of the particles;</li>
<li>$Q$ is the total charge of the particles;</li>
<li>and $r$ is the radius of the orbit.</li>
</ul>
<p>If a such clean, trivial formula doesn't exist, a link were also okay, where it can be found.</p> | 1,233 |
<p>I came across the notion of complex energy while studying instanton method to study the unstable state. Unstable states are those which have energy with an imaginary part. But as we know Hamiltonian is a Hermitian operator. So how can the energy be imaginary?</p> | 1,234 |
<p>Where do the terms <a href="http://en.wikipedia.org/wiki/Microcanonical_ensemble" rel="nofollow">microcanonical</a>, <a href="http://en.wikipedia.org/wiki/Canonical_ensemble" rel="nofollow">canonical</a> and <a href="http://en.wikipedia.org/wiki/Grand_canonical_ensemble" rel="nofollow">grand canonical</a> (ensemble) come from?</p>
<p>When were they coined and by whom? Is there any reason for the names or are they historical accidents?</p> | 1,235 |
<p>Trying to understand <a href="https://en.wikipedia.org/wiki/Limits_to_computation" rel="nofollow">the physical limits to computation</a>, I notice that among these we have two types of limits that constrain the minimum allowable energy for a computation.</p>
<ul>
<li>Limits that constrain <strong>the product of energy and time taken</strong>, which include <a href="https://en.wikipedia.org/wiki/Bremermann%27s_limit" rel="nofollow">Bremermann's limit</a> and <a href="https://en.wikipedia.org/wiki/Margolus%E2%80%93Levitin_theorem" rel="nofollow">Margolus-Levitin</a>. These two laws pretty much state the same thing as far as I can tell, with the only difference being a constant which doesn't concern me since only order of magnitude matters for my own purposes.</li>
<li>Limits that constrain <strong>only energy</strong> required for computation, notably <a href="https://en.wikipedia.org/wiki/Landauer%27s_principle" rel="nofollow">Landauer's principle</a>.</li>
</ul>
<p>So let's just write these. I'll utilize the Bremermann's limit since it seems to be more commonly referenced.</p>
<p>$$ E \ge \frac{ 2 \pi \hbar }{ \Delta t } $$</p>
<p>$$ E \ge k T \ln{(2)} $$</p>
<p>If you believe what these two equations are telling you literally, then obviously there is a certain time range where the energy-time limit will be more restrictive and a range where the energy-alone limit will be more restrictive. With trivial algebra, we set the RHS equal to each other and find the time for computation where the crossover happens. I used the temperature of space here. For room temperature, it is much smaller/faster.</p>
<p>$$ \Delta t = \frac{ 2 \pi \hbar }{ k T \ln{(2)} } \approx 2 \times 10^{-13} s $$</p>
<p>If we're considering serial computations, this would leave us in the $\text{THz}$ range. It's not anywhere near Planck time or anything like that, but it also doesn't seem like much of a practical limitation.</p>
<p>Is the energy-time limitation only discussed for academic purposes, with the understanding that it will be swamped by the more restrictive limit? Or is there some deeper reason why it should matter?</p> | 1,236 |
<p>By reading <a href="http://www.sciencedirect.com/science/article/pii/0550321388904555" rel="nofollow">an article</a>, I found a partition function that, according to the author, describes an interacting with random variables as coupling constant.</p>
<p>$$Z =\int \mathrm{d} \lambda_i e^{i(K^{ij}\lambda_i\lambda_j + V^{ijk}\lambda_i\lambda_j\lambda_k)}\mathrm{exp}(e^{iS_{eff}(\lambda)})$$</p>
<p>This expression is totally unfamiliar to me. Could someone show me how to derive that, providing a reference (online course, textbook, etc.) if necessary?</p> | 1,237 |
<p>I'm studying optics at the moment and I am wondering how the refractive index changes when a liquid is in super critical state?</p>
<p>Thank you.</p> | 1,238 |
<p>I recently learnt that the conservation laws are a consequence of the symmetries of space and time (the Lagrangian in Newton mechanics). Since space-time change in a black hole wouldn't quantum mechanics also change?</p>
<p>Let me give an example by what I mean: <br>
In classical space: <br>
$$x' = x + e \\
L(x') = L(x)$$</p>
<p>But the above transformation is not true in relativistic space. Hence, the conservation of momentum changes in special relativity (i.e momentum is not $mv$). So similarly shouldn't all the conservation laws change for a black hole and therefore it's action? </p>
<p>Side note: the reason I was thinking of a black-hole was that the incoming particle of mass will not drastically affect the space-time (due to the heavy centre) </p> | 1,239 |
<p>I'm trying to calculate the spin and color averaged gg->gg cross section, and I am stumbling upon gauge invariance:</p>
<p>Must the amplitude not be invariant under replacements $\epsilon_i \to \epsilon_i + \kappa_i p_i$ where $p_i\cdot\epsilon_i=0$ and $\kappa_i$ is arbitrary? For me it is not :/</p>
<p>I have summed the 4-gluon diagram and s,t,u-channel diagrams with two 3-gluon vertices each.</p>
<p>When I square the amplitude and carry out the spin averaged polarization sum in axial gauge, the extra "n"-vector does not vanish.</p>
<p>Am I missing something trivial? Can someone point me to some resource? I do not want to calculate the amplitude with spinor helicity formalism. I am thankful for any hints because I've been sitting on this problem with a collegue for quite a while.</p>
<p>Thanks, Tobias</p> | 1,240 |
<p>Squeezed electromagnetic vacuum does have a renormalized energy density smaller than the vacuum. So it makes it in my opinion a inconspicuous candidate for a dark energy carrier.</p>
<p>Are there observatories at the moment attempting to detect squeezed radiation from astrophysical and cosmic background sources? If not, What sort of equipment do you need to measure squeezed vacuum radiation?</p> | 1,241 |
<p>If the atmosphere is filled with electromagnetic waves all oscillating at different wavelengths and speeds how is it that they don’t all interfere with each other? For example turning on your light seems to have no effect on the sound coming from your radio. </p> | 1,242 |
<p>Does time exist in space or does space exist in time?</p>
<p>Alternatively, perhaps both are independent existences?</p>
<p>Another possibility is that this question is beyond the capacities of the human mind.</p> | 11 |
<p>So here is an interesting question about inductors or coils in general. </p>
<p>Suppose you had a inductor which was 12 inch's in diameter and say 12 inch's in length and has 100 turns of wire. The total length of that wire would be 12(diam) x 3.14(pie) x 100(turns) = 3,768 inches/314 ft in length. </p>
<p>So if you apply a voltage across the coil, I assume the time needed for the current to start moving in the whole coil would about the time needed for the signal to traverse the length of the wire which would take about 314 nanoseconds at 1 ft per nanosecond. Otherwise if the electric field from the first wire just permeates out from wire to wire, then it would take a max of 1 nanosecond before the current in every wire was moving.</p>
<p>So which is it? Does the electric field need to move through all of the 314 ft of wire before the current is moving at the end of the coil or does electric field move through the length of the wire and ultimately cause the current in the end of the coil to start moving after 1 nanosecond?</p> | 1,243 |
<p>For projector $p$, in finite dimension say, some unitaries $u, v$ does $upu^\dagger = vpv^\dagger$ implies $u = v$ ?
Intuitively, can we not say that a unitary is matrix permuting the basis and since $p$ is diagonal then obviously $u$ is $v$ ?
But for an exact proof ?</p>
<p>what if further, $p = upu^\dagger = vpv^\dagger$ ?</p> | 1,244 |
<p>I know the main problem with a quantum computer is that you can't read and write data from it until after it has run though all of its calculations. But would reading and writing quantum data to particles <i>without collapsing their probability fields</i> make this possible?</p> | 1,245 |
<p>Interacting fermionic SPT phases in 1d and 3d with $\mathbb{Z}_2^T$ symmetry are classified by $\mathbb{Z}_8$ and $\mathbb{Z}_{16}$ respectively, as shown in the paper by Fidkowski and Kitaev <a href="http://arxiv.org/abs/1008.4138" rel="nofollow">http://arxiv.org/abs/1008.4138</a>, and Wang and Senthil <a href="http://arxiv.org/abs/1401.1142" rel="nofollow">http://arxiv.org/abs/1401.1142</a>. I'd like to know the classification for the 2d case. Can anyone suggest some reference on that?</p> | 1,246 |
<p>This is a problem that has been periodically bugging me, so I finally decided to work on it. I haven't done any physics since high school, so I'm a bit out of practice:</p>
<p>Consider a doorway with two hinged doors, one being the mirror of the other. Someone opens one of the doors (which open outward) and lets go, allowing the door to begin shutting. You want to walk though the doorway, but you need one of the doors to be fully open. When is it more optimal to open the closed door completely, as opposed to pulling the closing door back open? By optimal, I refer to the torque over time I would have to exert to open the door fully in time $t_{open}$</p>
<p><strong>Specifications</strong></p>
<ul>
<li>The doors use 2 manual door closers (a spring that is stretched when the door is open and pulls the door back) which are connected the top and bottom of the door (for simplicity) in such a way that they always pull perpendicular to the door frame</li>
<li>The doors have two sets of hinges equidistant from the center of the door (again, for simplicity)</li>
<li>The doors are fully open when they reach an angle $\theta_{open} < {\frac{\pi}{2}}$ from the frame.</li>
</ul>
<p><strong>Assumptions:</strong></p>
<ul>
<li>Ignoring friction/air resistance/gravity entirely</li>
<li>ignoring dampening mechanisms of door closer (I have no idea how they work)</li>
<li>Treating the door as though it had no width</li>
</ul>
<p><strong>Attempt</strong></p>
<p>So, we have a door of mass $m$ with length across $l_d$, and a handle offset some distance $l_h$ from the hinges of the door. There is are two springs connected to the inner side of the door some distance $l_{s}$ from the hinge edge.</p>
<p>Let the two springs each have a spring constant $\frac{k}{2}$. When offset some distance $x$, they pull on the door with force of magnitude $kx$, giving a torque on the door of $l_{s}kx\cos{\theta}$. The offset $x$ is related to $\theta$ by $l_{d}\sin{\theta} = x$, so torque is equal to $l_{d}l_{s}k\sin{\theta}\cos{\theta}$, and torque over time is equivalent. Since we can treat the door as a thin rod, moment of inertia $I$ of the door as $\frac{ml_{d}^{2}}{3}$, we know the angular acceleration $\alpha$ of the door: $$\alpha(t) = \frac{3l_{s}k\sin{\theta}\cos{\theta}}{ml_{d}}$$ Then, the angular velocity $\omega$ is : $$\omega(t) = \frac{3l_{s}k\sin{\theta}\cos{\theta}}{ml_{d}}t $$</p>
<p>and $\theta$ is: $$\theta(t) = \frac{3}{2}\frac{l_{s}k\sin{\theta}\cos{\theta}}{ml_{d}}t^2 $$</p>
<p>At this point, I pretty sure I've taken a wrong turn somewhere, probably involving conversion of angular acceleration to velocity, since theta is a function of time as well...</p>
<p>I feel like this is a problem I should be able to do relatively simply. Anyone want to help me along? Also, it seems like I'm probably over-complicating the situation...</p> | 1,247 |
<p>I find the profound asymmetry in the sensitivity of left and right chiral particles to be one of the most remarkable analytical observations captured in the Standard Model. Yet for some, I've not found much in the way of discussions that worry about <em>why</em> of such as truly remarkable fact is true. I can't help but be reminded a wee bit of views on the motions of planets before Newton... you know, "it be Angels that do push them around, ask ye not why!"</p>
<p>Seriously, I know the Standard Model has a lot of givens in it... but surely <em>someone</em> has mulled over why the universe might exhibit such a non-intuitive and thus <em>interesting</em> asymmetry? And perhaps even developed some solid speculations or full theories on why such in-your-face chiral asymmetries exist in nature?</p>
<p>Do such theories exist, or is this asymmetry truly just a "given" and nothing more?</p> | 1,248 |
<p>Since the Higgs mechanism is so intimately tied to binding together massless chiral fermions, does it happen to have anything to say about the spin statistics issue?</p>
<p>I'm actually assuming the answer is no, but perhaps there is a deeper tie there that I'm missing completely. So, just curious...</p> | 1,249 |
<p>According to equation (6) on the first page of <a href="http://www.nhn.ou.edu/~gut/notes/cm/lect_14.pdf" rel="nofollow">some lecture notes online</a>, the above equation is used to prove the virial theorem. For rectangular coordinates, the relation</p>
<p>$$
2T~=~\sum_i p_{i}\dot{q}^{i}
$$</p>
<p>is obvious. How would I show it holds for generalized coordinates $q^{i}$?</p> | 1,250 |
<p>My question is about <code>photo electric</code> but it could be applied to other daily routine phenomenon. As we know rest mass of photon is zero. When a photon strikes the metal surface it transfers its energy to the electrons. Whether electron will be emitted or not, it depends upon work function. But my question is about that photon which was hit on metal surface. Does that photon vanished? Does that photon turned into nothing? where does it go after scattering? When I study this I only find the story about the emitted electrons but not about photons after collision. Am i missing some basic concept?</p> | 1,251 |
<p>Suppose, in whatever dimension and theory, the action $S$ is invariant for a global symmetry with a continuous parameter $\epsilon$.</p>
<p>The trick to get the Noether current consists in making the variation local: the standard argument, which does't convince me and for which I'd like a more formal explanation, is that, since the global symmetry is in force, the <em>only</em> term appearing in the variation will be proportional to <em>derivatives</em> of $\epsilon,$ and thus the involved current $J^\mu$ will be conserved on shell:
$$ \delta S = \int \mathrm{d}^n x \ J^\mu \partial_\mu \epsilon $$</p>
<p>This is stated, e.g., in <em>Superstring Theory: Volume 1</em> by Green Schwarz Witten on page 69 and <em>The Quantum Theory of Fields, Volume 1</em> by Weinberg on page 307.</p>
<p>In other words, why a term $ \int \mathrm{d}^n x \ K(x) \epsilon(x)$ is forbidden?</p>
<hr>
<p>Taking from the answer below, I believe two nice references are</p>
<ol>
<li><a href="http://arxiv.org/abs/hep-th/0002245" rel="nofollow">theorem 4.1</a></li>
<li><a href="http://arxiv.org/abs/math/0108160" rel="nofollow">example 2.2.5</a></li>
</ol> | 1,252 |
<p>This is a theoretical problem. Suppose we have two ropes of the same material and diameter, but differing in length. We excert force on both ropes until they break. Let rope A be the longer one. What can be said about the stress in rope A compared to the stress in rope B at the moment of breaking, taking in account the deformation required to keep a constant volume?</p>
<p>You would expect them to be the same as the stress-strain diagram is derived for a specific material. But at the same strain of both ropes, the stress in rope A would be higher as it crosssection decreases more.</p>
<p><img src="http://i.stack.imgur.com/aj7XJ.jpg" alt="!"></p>
<p>Stress-strain diagram from <a href="http://cdn.transtutors.com/" rel="nofollow">http://cdn.transtutors.com/</a></p>
<p><a href="http://i.stack.imgur.com/aj7XJ.jpg" rel="nofollow">1</a></p> | 1,253 |
<p>All particles exhibit wave-particle duality. And I have a strange question. </p>
<p>Why does a larger system, liken an atom that is just a set of smaller systems, itself exhibit wave-particle duality?</p>
<p>In principle all large systems can be defined as a set of smaller systems. An atom is a set of nucleus (a smaller set of up and down quarks held together by the strong interaction) and some electron(s) bound to the set of quarks by the electromagnetic force.</p>
<p>How is this set of smaller systems being able exhibit wave-particle duality as a whole as if it is a particle itself?</p>
<p>Does this imply that electron and all other elementary particles are indeed just another set of smaller sets and any sets can exhibit wave-particle duality?</p>
<p>And here comes the question: how do you define a set? We naturally define composition such as atom, molecule or "elementary particle" like electron as a set. Can the composition of electrons and up quarks be defined as a set (a system with wavefunction) and the down quarks that present in the atom be defined as a separated set (another system)?</p> | 1,254 |
<p>I saw a diagram of the photosensitivity (Current per Power) of a photodiode.
<img src="http://i.stack.imgur.com/44MQw.png" alt="Photodiode sensitivity"></p>
<p>So there is this diagonal stating the 100% quantum efficiency.
I wondered why the sensitivity for bluer light lower than that of redder light. </p>
<p>Is it because a blue photon has a higher energy? Thus fewer blue photons are needed to achieve the same power as with red photons?
The photocurrent is proportional to the number of converted photons. So a 1 watt beam in red has more photons, which generate more photo-electrons, thus more current per watt. </p>
<p>Is this reasoning correct and explains the curve in the diagram?</p> | 1,255 |
<p>What is a practical way in an engineering or physics laboratory to create bubbles of a specified size in water within plastic tubing? The tubing is a few mm inner diameter. We'd like to make bubbles smaller in size than the diameter, for example 1/2 the diameter. We'd like to have control over the size, especially volume, and rate of bubble generation. </p>
<p>We have tried a metal hypodermic needle connected to a low-power air pump, but air accumulates over time to make one big bubble, which eventually breaks away. Maybe there is a way to knock the bubble off when it's the desired small size, instead of waiting for it to do as it pleases? </p>
<p>The water is at roughly 1 to 1.3 atm pressure, at room temperature. </p>
<p>We imagine possibilities of electrolysis, or acoustics, but haven't found any written experience through google searches.</p> | 1,256 |
<p>Why is it that in stars undergoing gravitational collapse <a href="http://en.wikipedia.org/wiki/Electron_degeneracy_pressure" rel="nofollow">electron degeneracy</a> kicks in? Why couldn't the electrons form <a href="http://en.wikipedia.org/wiki/Energy_bands" rel="nofollow">energy bands</a> like in semiconductors? </p> | 1,257 |
<p>For my bachelor's thesis, I am investigating <a href="http://en.wikipedia.org/wiki/Divergent_series" rel="nofollow">Divergent Series</a>. Apart from the mathematical theory behind them (which I find fascinating), I am also interested in their applications in physics. </p>
<p>Currently, I am studying the divergent series that arise when considering the Riemann zeta function at negative arguments. The Riemann zeta function can be analytically continued. By doing this, finite constants can be assigned to the divergent series. For $n \geq 1$, we have the formula: </p>
<p>$$ \zeta(-n) = - \frac{B_{n+1}}{n+1} . $$</p>
<p>This formula can be used to find: </p>
<ul>
<li>$\zeta(-1) = \sum_{n=1}^{\infty} n = - \frac{1}{12} . $ This formula is used in Bosonic String Theory to find the so-called "critical dimension" $d = 26$. For more info, one can consult the relevant wikipedia <a href="http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%C2%B7_%C2%B7_%C2%B7" rel="nofollow">page</a>. </li>
<li>$\zeta(-3) = \sum_{n=1}^{\infty} n^3 = - \frac{1}{120} $ . This identity is used in the calculation of the energy per area between metallic plates that arises in the <a href="http://en.wikipedia.org/wiki/Casimir_effect" rel="nofollow">Casimir Effect</a>. </li>
</ul>
<p>My <strong>first question</strong> is: do more of these values of the Riemann zeta function at negative arguments arise in physics? If so: which ones, and in what context? </p>
<p>Furthermore, I consider summing powers of the Riemann zeta function at negative arguments. I try to do this by means of
<a href="http://mathworld.wolfram.com/FaulhabersFormula.html" rel="nofollow">Faulhaber's formula</a>. Let's say, for example, we want to compute the sum of $$p = \Big( \sum_{k=1}^{\infty} k \Big)^3 . $$
If we set $a = 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2} $, then from Faulhaber's formula we find that $$\frac{4a^3 - a^2}{3} = 1^5 + 2^5 + 3^5 + \dots + n^5 , $$ from which we can deduce that $$ p = a^3 = \frac{ 3 \cdot \sum_{k=1}^{\infty} k^5 + a^2 }{4} .$$
Since we can also sum the divergent series arising from the Riemann zeta function at negative arguments by means of <a href="http://en.wikipedia.org/wiki/Ramanujan_summation" rel="nofollow">Ramanujan Summation</a> (which produces that same results as analytic continuation) and the Ramanujan Summation method is linear, we find that the Ramanujan ($R$) or regularised sum of $p$ amounts to $$R(p) = R(a^3) = \frac{3}{4} R\Big(\sum_{k=1}^{\infty} k^5\Big) + \frac{1}{4} R(a^2) . $$
Again, we know from Faulhaber's Formula that $a^2 = \sum_{k=1}^{\infty} k^3 $ , so $R(a^2) = R(\zeta(-3)) = - \frac{1}{120} $, so $$R(p) = \frac{3}{4} \Big(- \frac{1}{252} \Big) + \frac{ ( \frac{1}{120} )} {4} = - \frac{1}{1120} . $$</p>
<p>My <strong>second</strong> (bunch of) <strong>question(s)</strong> is: Do powers of these zeta values at negative arguments arise in physics? If so, how? Are they summed in a manner similar to process I just described, or in a different manner? Of the latter is the case, which other summation method is used? Do powers of divergent series arise in physics in general? If so: which ones, and in what context?</p>
<p>My <strong>third and last</strong> (bunch of) <strong>question(s)</strong> is: which other divergent series arise in physics (not just considering (powers of) the Riemann zeta function at negative arguments) ?
I know there are whole books on renormalisation and/or regularisation in physics. However, for the sake of my bachelor's thesis I would like to know some concrete examples of divergent series that arise in physics which I can study. It would also be nice if you could mention some divergent series which have defied summation by any summation method that physicists (or mathematicians) currently employ. Please also indicate as to how these divergent series arise in physics. </p>
<p>I also posted a somewhat more general and improved version of this question on <a href="http://mathoverflow.net/questions/161872/what-are-some-geometric-physical-probabilistic-interpretations-of-the-rieman">MO</a>.</p> | 1,258 |
<p>Let's say I am flowing a fluid through a pipe that has a constant surface temperature. Is it possible to determine the heat flux on the inner surface without knowing the pipe's length when the only two temperatures I know are the pipe surface temperature and the inlet flow temperature?</p>
<p>I am able to assume the flow is thermally and hydrodynamically developed. A fully developed pipe with constant surface temperature has a Nusselt number of 3.6568 which allows me to then determine the convection coefficient. I was going to calculate the heat flux using $$q''=\bar h\Delta T_{lm}$$</p>
<p>The log mean temperature difference would need to use the outlet temperature found by $$\frac{T_{s}-T(L)}{T_{s}-T_{in}}=exp(\frac{-\pi DL}{\dot m c_{p}}\bar h)$$</p>
<p>The problem lies in that I don't have the pipe's length so I can't calculate the outlet temperature or the LMTD. The only other method I could think of is using $$q''=h(T_{s}-T_{in})$$ but that will greatly underestimate the actual heat flux. </p> | 1,259 |
<p>The $\mathrm{U(1)}$ QED case has two physical degrees of freedom, which is easy to understand because the free electromagnetic field must be transverse to the direction of propagation. But what are the physical degrees of freedom the $\mathrm{SU(N)}$ Yang-Mills theory? </p>
<p>I believe that the $\mathrm{SU(N)}$ Y.M. theory has four degrees of freedom, and it is easy to see that by gauge fixing (such as the Lorentz condition or the Coulomb condition) that we can always remove one redundant degree of freedom. But then we still haven't completely fixed the redundant degrees of freedom. Thus my question is:</p>
<blockquote>
<p>How many physical degrees and how many are redundant degrees of freedom does the Y.M. theory have? </p>
</blockquote>
<p>I would be interested to understand how we can determine this mathematically, and also to understand what the physical intuition behind this is?</p> | 192 |
<p>I'm currently studying at the spectra of some supersymmetric models, and would like to know whether the parameter points I'm looking at are ruled out due to excessive CP violation.</p>
<p>I am using <a href="http://www.physik.uni-wuerzburg.de/~porod/SPheno.html" rel="nofollow">SPheno</a>, which allows me to test my spectra against various other experimental bounds. It provides a <code>Block SPhenoLowEnergy</code> in its SLHA output that has, for example, the predicted branching ratio $B_s \rightarrow \mu\mu$, or the anomalous muon $g-2$.</p>
<p>What's the variable to look for when one colloquially says "ruled out because of CP violation"? How to get it in SPheno (or with any other spectrum generator / event generator, I can switch)? I feel like the answer is obvious but I'm missing the forest for the trees.</p> | 1,260 |
<p>Is it true:From equation $C=Q/V$</p>
<p>1)If Potential difference is kept
<strong>constant</strong>
$C \propto Q$</p>
<p>2)If charge is kept <strong>constant</strong>
$C \propto 1/V$
Then if 2nd is true I do have a question:
Don't we say that capacitance is the ability of a body to store charge,so my question goes this way,if the charge is kept constant and potential difference is decrease,then by the above assumption the capacitance should increase(and which is true).But we have said that charge is kept constant,then how would the capacitance increase?
I need an answer</p> | 1,261 |
<p>I know QFT at graduate level but I'll like to master the skill of working with Feynman diagrams. I'm looking for a book of solved exercises on this topic. </p>
<p>Specifically, I'm looking for the kind of repetitive exercises, just like in elementary school you have to do e.g. 10 multiplications, 10 divisions, etc. I haven't found anything like that.</p>
<p>The topics I'm interested in are, for example, calculating invariant amplitudes, cross sections, etc. from Feynman diagrams, computing the Feynman rules from some Lagrangian theory and calculating the lower order tree expansion, etc.</p> | 12 |
<p>These are some further important queries regarding the question here <a href="http://physics.stackexchange.com/questions/102910/why-would-spacetime-curvature-cause-gravity">Why would spacetime curvature cause gravity?</a></p>
<p>Q1. Explain the statement “Everything in spacetime moves at the speed of light”. Is this statement even true or false? My naïve understanding is as follows: If an object is stationary in space (relative to an observer?) then it will be moving in time, so, it will, say, move from (x=0,t=0) to (x=0,t=10). The velocity is the space interval / time interval, which from above coordinates is still zero. So it is moving through spacetime, but at zero velocity. So velocity will just determine the angle of its path to the time axis. If this angle is zero, ie it is moving parallel to the time axis, then velocity = zero, and it can really move in time from one time point to the other at zero velocity. Where am I wrong, and what is the real explanation?</p>
<p>Q2. Suppose there are two objects (see fig), an apple A above the earth at x=0, and the earth E at x=10 as shown in the fig. If there is no earth and just the apple in open space, then the spacetime is not curved due to the gravity of the earth, and the apple stays at x=0 but moves in time from A to B (fig a). Now the earth is at x=10 and presumably the spacetime curves and the axes tilt as in fig b. Then the apple moves from A to C, just following the geodesic. But if we assume this, then the apple has not moved at all, because, due to the tilt of the axis, point C is also at x=0. So fig b cannot be the correct situation, otherwise the apple will arrive at point C which is still on the x=0 axis.
So I assume that in fact the apple has not moved from A to C in fig b, but has moved from A to D in fig. a, where D is really at x=10. But if I assume this, then it is not spacetime itself that has curved. The spacetime is still straight, but the apple has moved from (x=0,t=0) to (x=10,t=10). Again, where am I wrong, and what is the correct explanation? How exactly has the spacetime curved or tilted for the apple at A due to the presence of the Earth at E, and how does the apple move to the earth by following the geodesic in free fall? (assume only one spatial dimention)</p>
<p><img src="http://i.stack.imgur.com/9QQXe.png" alt="spacetime curved by gravity?"></p>
<p>Q3: We say that gravity is not just a “force of attraction” between two pieces of mass, and it does not “pull” the two pieces of mass towards each other. Instead, acceleration is manifested because the two pieces of mass are simply following their now curved geodesics. But it can be shown that there really is an attractive force due to gravity. E.g. Take the apple in the earth’s gravity and suspend it on a spring balance. The spring will extend. Yes, you can say that since it is suspended motionless, so it is no longer in free fall, and since the spring balance exerts an external force upwards on the apple, so it no longer follows its free fall geodesic. But it is not only the spring balance which is exerting an upward force on the apple. The apple is also exerting a downward force on the spring, which causes the spring to extend. Surely it is too naïve to say that the downward force which the earth exerts on the apple and in turn the apple exerts on the spring suddenly vanishes if the apple is released into free fall. </p>
<p>If two masses were remotely attracting each other with a force, why would they cease to attract each other with a force when in free fall? It is logically a much more satisfying explanation that the gravitational attraction force which was there at rest remains in free fall too and it is this gravitational attraction force which causes the freefall acceleration by F=ma, just like any force on any object would cause an acceleration. Also, it is the net force on the object which causes the acceleration. When the object was suspended, the net force was zero, so acceleration was zero. When in free fall, the net force IS THE GRAVITATIONAL ATTRACTION FORCE, so acceleration is accordingly. This seems totally opposite to the statement that there is no gravitational force in free fall. </p>
<p>The reason why the spring of the spring balance does not extend when in free fall is that there IS a gravitational force downwards, but there is no force exerted by the spring upwards, so the net force on the object is downwards due to gravity, which causes the acceleration F=ma. Where am I wrong? What is the correct explanation?</p> | 1,262 |
<p>This is just a quick question. I would figure this out myself if I wouldn't have an exam about this tomorrow.</p>
<p>I am working on the non-relativistic approximation of the Dirac equation for an electron in an EM field. On one point, I need the following relation:</p>
<p>$$
\epsilon^{klm} \sigma^{m} = \sigma^m \epsilon^{mkl}
$$
where $\sigma^m$ denotes the $m$th Pauli matrix and $\epsilon^{klm}$ denotes the Levi-Civita symbol and the Einstein summation convention is used.</p>
<p>The question is: does this relation hold in general for the Levi-Civita symbol or is this specific for the Pauli matrices?</p>
<p>TIA, Eric.</p> | 1,263 |
<p>I'm slightly befuddled by is what it means when I'm asked to </p>
<blockquote>
<p>Draw the Feynman diagram in momentum space for the two point function of $\frac{\lambda}{3!}\phi^3$ theory for order $O(\lambda^2).$</p>
</blockquote>
<p>I can draw Feynman diagrams, and I thought two-point function meant</p>
<p>$$\langle0\|\phi(x)\phi(y)\|0\rangle$$</p>
<p>and what I know about $ O(\lambda^2)$ is that it will have more diagrams than $ O(\lambda).$</p>
<p>Other than that, I'm a bit lost. I mean, I'm not even sure if this is a really simple calculation or quite a long one. </p>
<p>Apologies to myself if anything I've written above is embarrassing. </p> | 1,264 |
<p>I try to understand why Algodoo software (freeware) give an efficiency greater than unity. I tested several scenes and at each time I can look the software give energy, I don't say it's all the time the case, a lot of scene have energy constant or decrease but with particular forces Algodoo give energy. </p>
<p>The energy is not very high compare to sum of energy, I only have 50 J each second (the system give 50 W) with a total energy of 10000 J, it's small only 0.5 % of the total energy. But I can test until 12000 Hz for some scenes and the energy don't decrease. The system can be a 2 objects or better 3 objects in a free rotation, there is no axis. I compute with a large scale of frequencies from 100 Hz to 12000 Hz. I can't increase more frequency because forces become unstable: forces are unstable. </p>
<p>It seems 1200 Hz give good results in a lot of scene and for a lot of scenes results don't change from 800 Hz to 1200 Hz. I compute datas with a spreadsheet for test the result of energy, when Algodoo give energy (efficiency >1) datas give energy. I looked at center of gravity and with good frequency it don't move, sometimes like 0.1 mm for an object of 4 m * 4 m for the main object and 0.3 m for the second object (for example). No friction and no gravity in this study. Friction and restitution is at 0.5 for all material. I changed only density for have best result.</p>
<p><strong>Example of a scene: (not the best error but easy)</strong></p>
<p><img src="http://i.stack.imgur.com/6MjbY.png" alt="enter image description here"></p>
<p><strong>Forces:</strong></p>
<p><img src="http://i.stack.imgur.com/zOnz2.png" alt="enter image description here"></p>
<p><strong>Energy and informations:</strong></p>
<p><img src="http://i.stack.imgur.com/KeCbC.png" alt="enter image description here"></p>
<p><strong>Position of center of gravity:</strong></p>
<p><img src="http://i.stack.imgur.com/pMt5N.png" alt="enter image description here"></p>
<p>I'm not sure, but with some tests (not all), I find a phase angle between 2 objects (or 3 with 3 free objects in rotation). Like sum of force from one object is the same (value) than other, if there is a phase angle between two center of gravity, there is a torque from these center of gravity. Or at least one center works in positive and other in negative. Like centers of gravity don't move in the same distance, the work is not the same. </p>
<p><strong>Angle: 6°</strong></p>
<p><img src="http://i.stack.imgur.com/9y1BB.png" alt="enter image description here"></p>
<p><strong>A link for download an example:</strong></p>
<p><a href="https://drive.google.com/file/d/0B63Jbse1IMAkUUo2Sm0tcnI5ak0/edit?usp=sharing" rel="nofollow">https://drive.google.com/file/d/0B63Jbse1IMAkUUo2Sm0tcnI5ak0/edit?usp=sharing</a></p>
<p><strong>I have several questions:</strong></p>
<p>1/ For you, the error come from this phase angle ? If not, where the error come from ? </p>
<p>2/ If there is a phase angle, there a torque ? </p>
<p>3/ Is it possible in reality that centers of gravity are in phase angle ? I think it's not possible but I asked the question to be sure. </p>
<p>4/ Why Algodoo find a phase angle ? Is there a physical explanation ?</p>
<p>Maybe the error don't come from the phase angle but when I compute sum of energy, center of gravity of red object give energy but in the same time green object don't consume this energy. I made a lot of tests and at each time it's the energy come from the small object. If you have an idea for find the error ? Or a test I can do for find the error ? If necessary I can give another scenes with more power with 3 objects.</p>
<p><strong>For example, this 3 objects:</strong></p>
<p><img src="http://i.stack.imgur.com/t7v7X.png" alt="enter image description here"></p>
<p><strong>The energy increase like:</strong></p>
<p><img src="http://i.stack.imgur.com/u68hf.png" alt="enter image description here"></p>
<p><strong>I thought 4800 Hz is a good frequency, but some seconds after the sum is:</strong></p>
<p><img src="http://i.stack.imgur.com/GPfWF.png" alt="enter image description here"></p>
<p><strong>I study the system and find something: the energy increase only when forces are like :</strong></p>
<p><img src="http://i.stack.imgur.com/GYihS.png" alt="enter image description here"></p>
<p><strong>Details of forces:</strong></p>
<p><img src="http://i.stack.imgur.com/UCtRp.png" alt="enter image description here"></p>
<p><strong>The energy increase more if red and orange objects have forces like before (but in opposite direction):</strong></p>
<p><img src="http://i.stack.imgur.com/cw6ZH.png" alt="enter image description here"></p>
<p><strong>When I compute energy I have more than Algodoo:</strong></p>
<p><img src="http://i.stack.imgur.com/bb0g6.png" alt="enter image description here"></p>
<p><strong>A link for download the scene:</strong></p>
<p><a href="https://drive.google.com/file/d/0B63Jbse1IMAkTUUzTzNlc1VORXM/edit?usp=sharing" rel="nofollow">https://drive.google.com/file/d/0B63Jbse1IMAkTUUzTzNlc1VORXM/edit?usp=sharing</a></p>
<p>In this scene, I don't find always a phase angle or small. But each time forces are like image shows, the energy increase and if I compute datas with a spreadsheet, it's red and orange objects (small objects) that give energy. In the contrary, green object don't consume energy, the sum is near zero.</p>
<hr>
<p>It's not possible it's the moment of inertia the problem ? If one part of object give a torque from friction it receive a torque in the same time, but the work from a torque is torque by angle of rotation. If inertia is not the same, angle of rotation is not the same too even the torque is equal in value, no ? I don't find the error from Algodoo and it's the only idea I have found. At least, if someone can say if it's a possibility or not.</p> | 1,265 |
<p>I'm fine with $U(1)$ symmetry and <a href="http://en.wikipedia.org/wiki/Noether%27s_theorem" rel="nofollow">Noether's Theorem</a>, but struggling with the translations of the field; namely</p>
<p>$$\phi'(x^{\mu})=\phi(x^{\mu}-a^{\mu}),$$ where $a^{\mu}$ constant four-vector</p>
<p>$$x^{\mu}=x^{\mu}+a^{\mu},$$</p>
<p>and the Lagrangian density</p>
<p>$${\cal L}=\frac{1}{2}\partial_{\mu}\phi^*\partial^{\mu}\phi-V(\phi^*\phi).$$</p>
<p>So a few questions:</p>
<ol>
<li><p>I can't show the Lagrangian is invariant under this transformation. Is it just a case that as $a^{\mu}$ is constant then the first term in the Lagrangian will obviously stay the same? But what about $V$? How I can show that's invariant?</p></li>
<li><p>Infinitesimally, is the transformation $\phi'(x^{\mu})=\phi(x^{\mu})-a^{\mu}\partial_{\mu}\phi(x^{\mu})?$ </p></li>
<li><p>If I'm right in point 2., how can I apply Noether's Theorem to this?</p></li>
</ol> | 1,266 |
<p>As I understand it, <a href="http://en.wikipedia.org/wiki/Paramagnetism" rel="nofollow">paramagnetism</a> is similar in its short-term effect to <a href="http://en.wikipedia.org/wiki/Ferromagnetism" rel="nofollow">ferromagnetism</a> (spins of the electrons line up with the magnetic field, etc.), though apparently the effect is weaker. What is it exactly that determines whether an atom or molecule is ferromagnetic versus paramagnetic, and why is the paramagnetic effect weaker?</p> | 1,267 |
<p>I can conceive of a particle existing in empty spacetime, but not a wave. A wave appears to me at least, to insist upon a medium for its very definition. </p>
<p>I understand that the 19C physicists postulated the lumineferous aether for this very reason, but abandoned it in favour of the electromagnetic field.</p>
<p>In the textbooks I've looked at, its always said that the physicists eventually understood that a wave does not necessarily require a medium. But it seems to me tht they abandoned the idea of a medium mechanically conceived (I assume because of the high prestige of newtonian mechanics, its status as a defining, fundamental paradigm which privileges our naive & intuitive understanding of mechanical phenonema) for a medium conceived in a a more general manner.</p>
<p>Afterall, this field permeates spacetime, and its state evolves through time. This is how I would naively conceive of a medium in a general sense.</p>
<p>I should add this is maybe more of a history of physics question, and I've added a tag as such to stop people getting upset.</p> | 1,268 |
<p>As a fun project, I would like to roughly simulate the suspension operation of a full-suspension mountain bike. This is <em>not</em> another one of those "How does a bicycle stay upright?" questions.</p>
<p>Follow some metrics of interest.</p>
<ul>
<li>Pedal bob - the amount of movement in the rear triangle upon pedaling up an incline. An estimate of the lost energy due to this.</li>
<li>Sharp obstacle transient response - for example how does the bike move, when hitting a curb at high speed.</li>
<li>Large obstacle transient response - for example, how does the bike move when it falls (along with the rider) 1 meter.</li>
</ul>
<p>I am somewhat familiar with python, Mathematica, <a href="http://www.scilab.org/" rel="nofollow">ScyLab</a>, Matlab, Ansys. I guess an analytical solution would be too difficult.</p>
<p><em>Which approach will give quickly a parameterizable model and analyze it for the abovementioned information?</em></p>
<p><a href="http://www.bikeradar.com/gear/article/buyers-guide-to-mountain-bike-suspension-part-2-28438/" rel="nofollow">Here</a> several common frame variations are mentioned.</p>
<p>EDIT:</p>
<p>More concretely:</p>
<ul>
<li>What model to use? Is there a widespread "bicycle" model for dynamic analysis?</li>
<li>What software to utilize, in order to help me solve the problem?</li>
</ul> | 1,269 |
<p>In order to determine the relative motion between the Earth and the ether (the medium through which light supposedly propagated. It has zero density and complete transparency), scientists used the concept of the <a href="http://en.wikipedia.org/wiki/Aberration_of_light" rel="nofollow">aberration of starlight</a>. If the Earth frame of reference was not <a href="http://en.wikipedia.org/wiki/Aether_drag_hypothesis" rel="nofollow">dragging the ether frame of reference</a>, then the direction of starlight as seen from both frames of reference separately would be different. Whereas if the Earth frame of reference was indeed dragging the ether frame of reference, then the starlight observed from both frames of reference separately would appear to come from the same direction. This is all intuitive. But how do we prove whether either is the case, when we are unable to identify the ether frame of reference itself, and conduct experiments within this frame of reference to determine the observed direction of starlight from within this frame of reference?</p> | 1,270 |
<p>Context: My room is being painted, and i sit and study in a corner of the room, surrounded by walls on 2 sides, such that i am facing the wall. A tube light is at my 4-5 o'clock, and a door(transparent) at my 7 o'clock. These are the only two sources of light. There is a TV mounted just in front of my desk. </p>
<p>Problem: I want to paint only one wall with a dark color and the rest with a light yellowish green color. Now, which wall should be painted with the dark color so as to maximize light in the corner?? IE, the one facing me, on which the TV is mounted, or any other?</p> | 1,271 |
<p>When writing a method for an experiment, does it always have to be set out in orderly numbered steps? Can it not also be a paragraph of text that outlines the method?</p>
<p><strong>A mundane example</strong>:</p>
<ol>
<li>Place a tripod on a table, and the wire gause mesh ontop </li>
<li>Following that, place a busen burner underneath it, and hook it up to the gas</li>
<li>Fill a beaker with 100ml of water</li>
<li>Place the beaker on the wire gause, turn on the gas and lit the Busen burner</li>
<li><p>Heat the water until it boils</p>
<p><strong>...</strong></p></li>
</ol>
<p><strong>VERSUS</strong> </p>
<p>Firstly, place a tripod, ensuring that you place a wire gause on on top. Following that step, ...</p>
<p>Thanks.</p> | 1,272 |
<p>Does the curvature of space-time cause objects to look smaller than they really are? What is the relationship between the optical distortion and the mass of the objects?</p> | 1,273 |
<p>Are virtual particles only popping in and out of existence where the local energy density is below a certain point? What I wonder is, does any kind of matter prevent the pairs from appearing? Is there a shell surrounding an atom or maybe I should call it a boundary beyond which particle pair production occurs, and within the boundary it does not, I have wondered if the different orbitals around an atom are affected (set)by the influence of the virtual particles.</p> | 1,274 |
<p>In a wave guide, graphics of propagation of Transversal Magnetic modes show closed field lines for the electric field. </p>
<p>For example, for a rectangular guide:</p>
<p>$E_x (x,y,z) = \frac {-j\beta m \pi}{a k^2_c} B_{mn}\cos\frac{m\pi x}{a}\sin\frac{n\pi y}{b}e^{-j(\beta z + \omega t)}$</p>
<p>$E_y (x,y,z) = \frac {-j\beta n \pi}{b k^2_c} B_{mn}\sin\frac{m\pi x}{a}\cos\frac{n\pi y}{b}e^{-j(\beta z + \omega t)}$</p>
<p>$E_z (x,y,z) = B_{mn}\sin \frac{m\pi x}{a}\sin\frac{n\pi y}{b}e^{-j(\beta z + \omega t)}$</p>
<p>Is it possible to have closed lines for the electric field?</p> | 1,275 |
<p>Recently I have started to study the classical theory of gravity. In Landau, Classical Theory of Field, paragraph 84 ("Distances and time intervals") , it is written</p>
<p>We also state that the determinanats $g$ and $\gamma$, formed respectively from the quantities $g_{ik}$ and $\gamma_{\alpha\beta}$ are related to one another by $$-g=g_{00}\gamma $$
I don't understand how this relation between the determinants of the metric tensors can be obtained. Could someone explain, or make some hint, or give a direction? </p>
<p>In this formulas $g_{ik}$ is the metric tensor of the four-dimensional space-time and $\gamma_{\alpha\beta}$ is the corresponding three-dimensional metric tensor of the space. These tensors are related to one another by the following formulas $$\gamma_{\alpha\beta}=(-g_{\alpha\beta}+\frac{g_{0\alpha}g_{0\beta}}{g_{00}})$$ $$\gamma^{\alpha\beta}=-g^{\alpha\beta}$$</p>
<p>Thanks a lot.</p> | 1,276 |
<p>Suppose I am trying to determine the value of the acceleration due to gravity by simply dropping objects from a ledge. If I did not account for air resistance, would the acceleration due to gravity measured be less than if I did account for air resistance?</p>
<p>This is what a thought: Suppose I had an object of mass $m$ that was being dropped and let $F_{drag}$ be the force due to air resistance. Then </p>
<p>$mg - F_{drag} = ma$. Then with some algebra I get $a = g - \frac{F_{drag}}{m}$. If I do not account for air resistance, I get $a = g$. But if I do account for air resistance $a < g$. So I get that the value of $g$ is greater if I do not account for air resistance. </p> | 1,277 |
<p>The question goes along the lines:</p>
<blockquote>
<p>Uranium-$236$ fissions when it absorbs a slow-moving neutron. The two fission fragments can be almost any two nuclei whose charges $Q_1$ and $Q_2$ add up to $92e$ ($e$ is the charge of a proton, $1.6 \times 10^{-19}C$), and whose nucleons add up to $236$ protons and neutrons. One of the possible fission modes involves nearly equal fragments of the two palladium nuclei with $Q_1 = Q_2 = 46e$. The rest masses of the two palladium nuclei add up to less than the rest of the mass of the original nucleus. Make the assumption that there are no free neutrons, jus the palladium nuclei. The rest mass of the U-$236$ is 235.996 u (unified atomic mass units), and the rest mass of each $Pd-118$ nucleus is 117.894 u, where $1$ $u = 1.7 \times 10^{-27} kg$.</p>
<p>(a) Calculate the final speed $v$, when the Pd nuclei have moved far apart (due to the mutual electric repulsion). Is this speed small enough that $\frac{p^2}{2m}$ is an adequate ($p$ is momentum) approximation for the kinetic energy of one of the palladium nuclei? (make the non relativistic assumption first, then compare $v$ is indeed small enough to $c$)</p>
<p>(b) Using energy considerations, calculate the distance between centers of the palladium nuclei just after fission, when they are starting from rest.</p>
</blockquote>
<p>So the beginning step is to chose the system, for which I include both the Pd nuclei. So the energy principle is now: $$\Delta E_{sys} = W_{ext}$$ We know that $W_{ext} = 0$ since my system is defined not including any external work done on the system. So now instead the system is $\Delta E_{sys} = 0 = \Delta K_{m_1} + \Delta K_{m_2} + \Delta U_{elec_1} + \Delta U_{elec_2}$ We can ignore the earth and the gravitational potential energy sine the masses are very small. If we expand the equation, we get:
$$K_{f_1} - K_{i_1} + K_{f_2} + K_{i_2} + U_{f_1} - U_{i_1} + U_{f_2} - U_{i_2} = 0$$
getting rid of terms that are $0$, we get:
$$K_{f_1} + K_{f_2} - U_{i_1} - U_{i_2} = 0$$
furthermore:
$$K_{f_1} + K_{f_2} = U_{i_1} + U_{i_2}$$
We know that $K = \frac{1}{2}mv^2$ and that $K_{f_1} = K_{f_2}$ are equal since the masses are the same. So we can simplify that to $2K_f = mv^2$. Same for $U = \frac{1}{2 \pi \epsilon_0} \frac{Q_1 Q_2}{r}$. So $U_{i_1} = U_{i_2} = 2U_i = \frac{2}{2 \pi \epsilon_0} \frac{Q_1 Q_2}{r}$. So we can essentially write:
$$mv^2 = 2 \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r}$$
$$v = \sqrt{\frac{2}{m} \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r}}$$
I then get lost n finding $r$ since it is not given, furthermore, (b) asks us to find the distance before they move apart. So I am approaching the problem incorrectly. I would need to figure out the error here and how to commence part b. </p> | 1,278 |
<p>I realize I'm trying to get a simple answer on a complicated subject, but here goes anyway. I've done some research and I understand (well, mostly) about how Vc is the amount of energy required to get two nucleons past the electric repulsive force so the strong force can take over. I know about the Maxwell-Boltzmann distribution, quantum tunneling, and the Gamow peak.</p>
<p>But all that is about how the Coulomb barrier is overcome by ambient temperatures in stars. The texts always go from how Vc is calculated, to how hot a star would have to be to overcome it, and how they aren't actually that hot, and why quantum tunneling comes to the rescue, yada yada, yada.</p>
<p>I want a more concrete understanding of what Vc really represents. If I had two protons and was using my patented Nucleon Energizer 3000, does Vc represent the amount of energy I'd have to set the MeV dial to so that I could force the two protons together past the Coulomb barrier and get them to fuse? And how much would my electric bill be afterwards?</p> | 1,279 |
<p>There is a very fundamental flaw in the common explanation given of the space-time curvature due to massive objects. It is said that a massive object curves space time just like a bowling ball on a rubber sheet, and another object near the massive object simply rolls down-hill on the rubber sheet, that is the reason why we observe that the other object is experiencing gravitational attraction to the massive object which curved the space-time. </p>
<p>The flaw is this:</p>
<p>Even if the massive object did bend the spacetime downwards, and there was a second object in the vicinity, this second object just would not roll down-hill automatically, unless there was a SECOND EXTERNAL source gravity, other than these two objects pulling everything downwards. In other words, even if the space time is curved, there is no particular reason for the second object to move closer to or be attracted by the first, because there is no other external force that would force it to move down-hill.</p>
<p>As a visualization, consider the bowling ball on a rubber sheet and another object on the same rubber sheet in outer space where there is no other source of gravity. Now, neither will the bowling ball curve the rubber sheet downwards (because it is not being pulled downwards), nor will the other object move down-hill on the rubber sheet even if the rubber sheet was depressed downwards, because there is no downward gravitational force from earth or another external source which would cause either of these two effects. So, there would be no reason for the other object to be attracted towards the first object. This is inconsistent with observation because objects do really experience gravitational attraction towards each other. But it certainly can't be explained by the bowlig ball and rubber sheet example.
What is the explanation for all this?</p>
<p>Flaw 2:
If there are 2 massive objects on the rubber sheet, they will both bend spacetime around themselves, and they will both remain trapped in their own depressions, and would never move towards or be attracted to each other. This is again contrary to observation. Since both masses are in reality attracted to each other, so they can't possibly bend spacetime and can't be trapped in their own depressions. Why?</p>
<p>Flaw 3: All pictures of bent spacetime on the web show spacetime being bent downwards. Why? In open space, with no other object nearby, all directions are equivalent to all other directions, and "downward" is undefined. So if spacetime is really bent, in which direction is it really bent?</p> | 13 |
<p>When I was young I read one book in which is written that you get more tan on the morning than on evening even light angle from Sun to Earth is the same. Don't remember exact reason, I think because ultraviolet is more absorbed on the evening because air is more humid or something like this. Is it true?</p> | 1,280 |
<p>Condensed matter physicists have shown using quantum information that in many condensed matter systems, entanglement entropy only scales as the area of the boundary, and not the volume. This is the basis for the density matrix renormalization group and Projected Entangled Pair States (PEPS). Does this also explain the holographic principle in quantum gravity?</p> | 1,281 |
<p>I just thought of an interesting question.Is the sun actually moving?
I have learnt that the way the lunar landers and the space ships of the Mercury and Apallo missions moved and controlled their positions through the use of thrusters.NoW apply that same concept to the sun are the constant explosions on the surface of the sun enough to move it?</p> | 1,282 |
<p>Our calculus book, Stewart, has a problem where they claim that for a metal cable (inner radius $r$) encased in insulation (outer radius $R$), the speed of an electrical impulse is given by</p>
<p>$$v = - k \left(\frac{r}{R}\right)^2 \ln \left(\frac{r}{R}\right)$$</p>
<p>where $k$ is a positive constant.</p>
<p><strong>My question</strong>
What I would like to know is the physical justification for their claim.</p>
<p><strong>My thoughts</strong>
There claim is somewhat surprising, since for sufficiently high insulation R, with r fixed, the speed of the impulse <em>decreases</em> (by L'Hospital) with <em>more</em> insulation.</p>
<hr>
<p><em>EDIT</em>: I received this email after contacting Brooks/Cole, the publisher of the textbook. The response didn't really help unfortunately.</p>
<blockquote>
<p>Hi Professor ...,</p>
<p>I just heard back from the author regarding your query: “I can understand why Professor
... thinks this equation is counterintuitive, but it is in fact correct. I have been >trying to track down the source that I used in devising this problem, but unfortunately I >can’t seem to find it right now.” I will certainly let you know if he is able to track >down the source information. I’m sorry I can’t give you a more concrete answer at this >time. Best, ...</p>
<p>[JIRA] (KYTS-1199) Content Feedback from Instructor for ISBN: 0495014281 Essential >Calculus: Early Transcendentals 1st edition.</p>
</blockquote> | 1,283 |
<p>I'm a maths undergrad taking a course on electromagnetism, I've drawn a diagram to represent this following question, but I'm having a bit of trouble approaching it:</p>
<p>"Two tiny balls of mass m = 0:1 g and charge q are suspended on silk threads of
length l = 30 cm (see picture below). Their centers are separated by D = 6 cm.
Find the charge q." </p> | 14 |
<p>If an astronomer moves at relativistic speed, the stars and constellations are distorted. He sees the stars towards which he is moving blue shifted, while the ones he's moving away from are red shifted. In addition, the apparent direction of distant stars is modified. I think that the coolest way of representing this is the "relativistic ellipse".</p>
<p>Think of a spherical shell with the inner surface reflective. Imagine a flashbulb going off at the center. The light is reflected and returns to the source.</p>
<p>Now think of the same situation observed from a moving frame of reference. The light is still emitted and returns to the center at the same time, but that point has now moved. Since the speed of light is the same in all frames of reference, the distance traveled by each light ray has to be the same. So the paths of the light rays (and the boundary of the spherical shell) describes an ellipse, the <a href="http://www.relativitybook.com/resources/relativistic_ellipse.html" rel="nofollow">"relativistic ellipse"</a>: </p>
<p><img src="http://i.stack.imgur.com/9PoOJ.png" alt="RelativisticEllipse"></p>
<p>The above drawing is from the book "Relativity in Curved Spacetime" which I highly recommend and can be purchased <a href="http://rads.stackoverflow.com/amzn/click/0955706807" rel="nofollow">here</a> or <a href="http://www.relativitybook.com/book_cover.html" rel="nofollow">here</a>.</p>
<p>Perhaps when my copy arrives it will have the equation. Should be possible to work it out using Lorentz invariance or contraction or something similar.</p>
<hr>
<p>Per the answer by <a href="http://physics.stackexchange.com/users/1257/helder-velez">Helder Velez</a>, <a href="http://physics-quest.org/Book_Chapter_Non_Simultaneity.pdf" rel="nofollow">Chapter 4 of Hans de Vries's book</a> has the following useful diagram:<br>
<img src="http://i.stack.imgur.com/gC4Gu.jpg" alt="HansEllipse"></p> | 1,284 |
<p>It would be nice to have a cute method that uses Lorentz transformations of basis vectors by exponential transformation using gamma matrices. To avoid confusion, let's assume -+++ signature. Given $\gamma_\mu$ as gamma matrices that satisfy ${\rm tr}(\gamma_\mu\gamma_\nu) = 4\eta_{\mu\nu}$, then we have ${\rm tr}(\gamma'_\mu\gamma'_\nu) = 4\eta_{\mu\nu}$ if we put:<br>
$\gamma'_\mu = \exp(-A)\gamma_\mu\exp(+A)$<br>
where $A$ is any matrix.</p>
<p>Boosts and rotations use $A$ as a bivector. For example, with $\alpha$ a real number, $A=\alpha\gamma_0\gamma_3$ boosts in the z direction while $A=\alpha\gamma_1\gamma_2$ gives a rotation around the z axis.</p>
<p>Solving for the value of $\alpha$ that gives a boost with velocity $\beta = v/c$ appears to be straightforward. But how to do the rotations?</p>
<p>And by the way, what happens when you generalize $A$ to be something other than bivectors?</p> | 1,285 |
<p>You may have noticed over the last few years that Moore's law is no longer applying to the real world.
This observation states that over the history of computing hardware, the number of transistors on integrated circuits doubles approximately every two years.</p>
<p>However, as microprocessors in computers have continued to become smaller, the architecture size has become incredibly small:- the new Skylake architecture to be released by Intel will use 14nm semiconductors.</p>
<p>We are beginning to reach the limit in size where the information being passed is still completely reliable.</p>
<p>Would it be reasonable to ask how much quantum tunneling would occur in a semiconductor 5nm in length? We can assume that in this CPU</p>
<p>Voltage = 1.2V
Amperage = 63A</p>
<p>I enjoy physics, but I'm not that great at it; I'd love to hear the results! Even if massive approximations!</p> | 1,286 |
<ol>
<li><p>Where the gravitational pull of Earth exist up to? </p></li>
<li><p>What distance from Earth it will be zero?</p></li>
<li><p>How do the skydivers fly at a same altitude? </p></li>
<li><p>Won't they feel gravitational pull? </p></li>
<li><p>What is the Earth's gravitational pull?</p></li>
</ol> | 1,287 |
<p>What are the units of the damping constant from the following equation by dimensional analysis?</p>
<p>$$\zeta = \frac{c}{2\sqrt{mk}}$$</p>
<p>I'm assuming the units have to be s^-1, as the damping constant is present in the exponential equation which plots damping of y=Ae^kt (which plots amplitude vs time). Is that a correct assumption?</p>
<p>If somebody could do a quick dimensional analysis to confirm it would be great.</p> | 1,288 |
<p>Let us say a block of mass is placed on the surface of earth. Then while drawing the forces on that body, we say:</p>
<ol>
<li>Force $F = mg$ acting towards the center of Earth.</li>
<li>Normal reaction $N$ offered by the surface of Earth.</li>
</ol>
<p>If no other forces are acting, we say $F=N$. But what about the centrifugal force $m\omega^2R$ . Why don't we ever bring that into picture? What am I missing?</p> | 18 |
<p>I am working on a variational problem involving elastic stability of a beam.</p>
<p>The deformation of the beam is given by six functions of the material coordinate along the beams longitudinal axis. The governing functional looks like
$$
\int_{s_1}^{s_2} F \left(
u(s),v(s),w(s),\phi(s),\chi(s),\psi(s) \right) ds.
$$
(It represents the beam's strain energy.)</p>
<p>Now I want to impose the condition that transverse shear deformation and elongation are zero. The consequence is that the 6 degrees of freedom of the beam are reduced to essentially 3. It can be shown that the three relations
$$
\frac{du}{ds} = \sin(\chi), \quad
\frac{dv}{ds} = -\sin(\phi)\cos(\chi), \quad
\frac{dw}{ds} = \cos(\phi)\cos(\chi) -1
\qquad (*)
$$
between the coordinates are valid in case transverse shear deformation and elongation are neglected.</p>
<p>My understanding is that there are now two ways to incorporate this constraint condition:</p>
<ol>
<li>eliminate variables in the original functional by substitution </li>
<li>include equations (*) by means of Lagrange multipliers</li>
</ol>
<p>I would like to use the latter method. The book <em>The variational principles of mechanics</em> by Lanczos explains this for both holonomic and non-holonomic constraints (and isoperimetric ones as well). Though I can see that equations (*) are clearly not some algebraic relation
$$
f_i(u,v,w,\phi,\chi,\psi)=0, \quad i=1\dots3
$$
between the coordinates, does this means the constraint qualifies as non-holonomic? I am confused because it seems integrable. Or does it then become an isoperimetric constraint?</p>
<h2>Edit</h2>
<p>I am getting more confused; can we only call constraints non-holonomic if derivatives with respect to <em>time</em> are involved? (Meaning that all equations (*) above just represent holonomic constraints, since it is all static?)</p>
<p>In <em>The enigma of nonholonomic constraints</em> by Flannery (2005), it is explained that for the typical action
$$
S = \int_{t_1}^{t_2} L({q},\dot{q},t) dt
$$
the Lagrange multiplier rule can only be used for holonomic and semi-holonomic (exact linear) conditions. Here, the independent variable clearly is time, but in my case it is some spatial coordinate.</p>
<p>If we assume that the independent variables space and time are interchangeable, my (*) equations appear non-holonomic (so, not to be incorporated with Lagrange multipliers), as they cannot be expressed as exact differential.. But, is this correct reasoning?</p> | 1,289 |
<p>Consider a thick wire that narrows for some part of its length, connected to the terminals of a battery, hence carrying some current.</p>
<p>Presumably the electrons have to speed up as the wire gets thinner - if the current is constant everywhere, then if the wire cross-section is less, they have to go through faster in order for the same number of electrons to pass any point in a given time.</p>
<p>What accelerates them as the wire narrows?</p>
<p>And what decelerates them again as the wire widens out again after the narrow bit?</p> | 1,290 |
<p>I tried to calculate earth's orbital period using <a href="https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law" rel="nofollow">Kepler's third law</a>, but I found 365.2075 days for the orbital period instead of 365.256363004 which is the correct value. I checked everything, and I couldn't find what's the problem.
I used these values for my calculation:</p>
<ul>
<li>Semi-major axis, a: 149,598,261 km</li>
<li>Gravitational constant, G: 6.67*10<sup>-11</sup> N·(m/kg)<sup>2</sup></li>
<li>Solar mass, M: 1.9891*10<sup>30</sup> kg</li>
</ul> | 1,291 |
<p>for any real and positive 's' and in the sense of semiclassical approximation </p>
<p>is this valid ??</p>
<p>$ \sum_{n}exp(-sE_{n})\sim \iint_{C}dxdpe^{-s(p^{2}+V(x))}$ valid for every 's'</p>
<p>here simply both the sum and integral are convergent, so in this case the approximation can be seen as the replacement of a sum by an integral.</p> | 1,292 |
<p>Can someone answer and explain it please.
Here are the options:</p>
<ul>
<li>Moving downwards and decelerating.</li>
<li>Moving downwards with a constant velocity.</li>
<li>Moving upwards and decelerating.</li>
<li>Moving upwards with a constant velocity.</li>
</ul> | 1,293 |
<p>I suggest the following thought experiment that describes a machine which makes everybody happy.</p>
<p>Suppose a lottery is conducted. The winner is awarded a billion dollars plus the title of eternal Earth dictator.</p>
<p>The winner is determined in the following way.</p>
<ul>
<li><p>There is a thorough isolated black box inside which a pretender is placed.</p></li>
<li><p>Inside the box a quantum experiment is conducted with equal probability of the two outcomes: 0 or 1 (it may be based on an atom decay)</p></li>
<li><p>The pretender measures the result with an apparatus. </p></li>
<li><p>The black box is connected by a quantum link (in form of an optical fiber) to an external quantum computer</p></li>
<li><p>After the measurement by the pretender is done the result is automatically transmitted by the optical link to the quantum computer</p></li>
<li><p>The transmitted qubit is used by quantum computer as an initial state of the quantum register instead of usual Hadamard transform before performing a Shor factorization algorithm.</p></li>
<li><p>After the algorithm is performed, the result is checked whether it is correct.</p></li>
<li><p>If it is correct the pretender is declared the looser. If the factorization returned a wrong result, the pretender is declared the winner.</p></li>
</ul>
<p>What happens.</p>
<p>Any average person placed in the black box will measure the result of the atom decay and similarly to a Schroedinger cat will enter the superposition of two states: the one which saw 0 and the one which saw 1.</p>
<p>Similarly the measuring apparatus will enter the superposition. So the qubit sent via the optical fiber will be a superposition of two states:</p>
<p>$$\frac{1}{\sqrt{2}}|0\rangle+\frac{1}{\sqrt{2}}|1\rangle$$</p>
<p>As this value is exactly the same what Hadamard gate returns after being applied to an empty qubit, this will allow the quantum computer to perform the factorization as usual and return the correct result, dispite that the initial step of applying the Hadamard gate has been replaced by loading the qubit obtained from the black box.</p>
<p>Now suppose the observer himself is placed in the black box. After the atom decay he will perform measurement which will collapse the wave function. The observer will be in either state where he saw 0 or where he saw 1. So either basis state 0 or basis state 1 is sent via the cable to the quantum computer.</p>
<p>Since this is different from the usual value which should be loaded in the register before performing the factorization, the factorization will return a wrong result. So the pretender will be declared the winner.</p>
<hr>
<p>It follows that any person, when he or she is observing the lottery from outside will see all other pretenders to loose.</p>
<p>At the same time any participant that tries the lottery himself will find himself a winner.</p>
<p>This is a fascinating lottery which will make all the participation pretenders the winners in their own eyes. Each person who goes through this machine will see other people to congratulate them, to proclaim them the eternal emperor and so on.</p>
<p>I would like to see whether there any mistake in the reasoning.</p> | 1,294 |
<p>I have the following reaction: $^{10}_5\mathrm{Be} +\space ^2_1\mathrm{H} \rightarrow \space^{11}_5\mathrm{B} + \space ^1_1\mathrm{H}$ </p>
<p>And I know that I have to use the formula: $E = \Delta m\cdot c^2 = \Delta m \cdot \frac{931,5MeV}{u}$.</p>
<p>So I just need $\Delta m$ which is equal to:
$\Delta m = m_b - m_a$ where $m_b$ represents the mass "before the reaction" and $m_a$ the mass "after the reaction" so we have:<br>
$m_b = m(^{10}_5\mathrm{Be}) + m(^2_1\mathrm{H})$<br>
$m_a = m(^{11}_5\mathrm{B}) + m(^1_1\mathrm{H})$ </p>
<p>The book which contains this problem contains the following table:
<a href="http://i.imgur.com/esoGDVf.png" rel="nofollow">http://i.imgur.com/esoGDVf.png</a> but from this table, I only know $ m(^1_1\mathrm{H})$ and $m(^2_1\mathrm{H})$ i.e.<br>
$m_b = m(^{10}_5\mathrm{Be}) + 2.01410u$<br>
$m_a = m(^{11}_5\mathrm{B}) + 1.00783u$ </p>
<p>How do I calculate $m(^{10}_5\mathrm{Be})$ and $m(^{11}_5\mathrm{B})$ ?</p>
<p>P.S. I don't know if the tag is correct. The chapter in the book where I found this exercise is called "Basics of nuclear physics".</p> | 1,295 |
<p>Consider a $\varphi^3$ theory:
$$
Z_1(J) \propto \exp\left[\frac{i}{6} Z_g g\int \mathrm{d}^4 x \left(\frac{1}{i}\frac{\delta}{\delta J}\right)^3\right] Z_0(J),
$$
where
$$
Z_0(J) = \exp\left[\frac{i}{2} \int \mathrm{d}^4 x \mathrm{d}^4 x' J(x)\Delta(x-x')J(x')\right].
$$
That is
$$
Z_1(J) \propto \sum_{V=0}^\infty \frac{1}{V!}\left[\frac{i}{6} Z_g g \int \mathrm{d}^4 x \, \left(\frac{1}{i}\frac{\delta}{\delta J}\right)^3\right]^V \times \sum_{P=0}^\infty \frac{1}{P!}\left[\frac{i}{2} \int \mathrm{d}^4 y \, \,\mathrm{d}^4 z\, J(y)\Delta(y-z)J(z)\right]^P.
$$
In particular, we can consider the term when $V=2, P=3$. Calculation shows that
\begin{equation}
- i \frac{1}{2!}\frac{1}{3!} \frac{(Z_g g)^2}{6^2*2^3} \left[\int \mathrm{d}^4\, x_1 \, \mathrm{d}^4 x_2 \, \left(\frac{\delta}{\delta J(x_1)}\right)^3 \left(\frac{\delta}{\delta J(x_2)}\right)^3\right] \left[\int \mathrm{d}^4 y \, \mathrm{d}^4 z \, J(y)\Delta(y-z)J(z)\right]^3\\=
- i \frac{1}{2!}\frac{1}{3!} \frac{(Z_g g)^2}{6^2*2^3} \int \mathrm{d}^4 x_1 \, \mathrm{d}^4 x_2 \, \left[3^3*2^4 \Delta(x_1-x_1)\Delta(x_1-x_2)\Delta(x_2-x_2) + \\ 3^2\times 2^5\Delta(x_1-x_2)\Delta(x_1-x_2)\Delta(x_1-x_2)\right]\\=
-i(Z_g g)^2 \int \mathrm{d}^4 x_1 \, \mathrm{d}^4 x_2\, \times \left[\frac{1}{2^3} \Delta(x_1-x_1)\Delta(x_1-x_2)\Delta(x_2-x_2) + \\ \frac{1}{2\times 3!}\Delta(x_1-x_2)\Delta(x_1-x_2)\Delta(x_1-x_2)\right],
\end{equation}
where $\Delta(x_1-x_1)\Delta(x_1-x_2)\Delta(x_2-x_2)$ and $\Delta(x_1-x_2)\Delta(x_1-x_2)\Delta(x_1-x_2)$ correspond to their Feynman diagram. Then the question is why $\frac{1}{2^3}$ and $\frac{1}{2\times 3!}$ are just the reciprocal of symmetry factors of the corresponding Feynman diagram respectively? </p>
<p>In the general case of $V, P$, why the coefficients of the terms in the result of calculation are just the reciprocal of symmetry factors of the corresponding Feynman diagram respectively? </p> | 1,296 |
<p>I have the following equation:</p>
<p>$$\frac{\partial U}{\partial t}=k\frac{\partial^2 U}{\partial x^2}-v_{0}\frac{\partial U}{\partial x}, x>0$$</p>
<p>with initial conditions:</p>
<p>$$U(0,t)=0$$</p>
<p>$$U(x,0)=f(x)$$</p>
<p>In the problem is requested to give an interpretation of each of the terms in the above equation, and noting that such systems can model, besides solving by Fourier Transform.
The Fourier Transform solution is quite simple to do; however, I can not give a physical interpretation of the terms of the equation not to mention a system that can model it. So I wanted to ask your help to answer this question. Thank you very much for your help and attention.</p> | 1,297 |
<p>$$y(x,t)=2A\sin(Kx)\cos(\omega t)$$
$A$ and $x$ are in metre, $\omega$ is angular frequency.
Then find dimensions of $A$ and $K$.</p>
<p>In this equation how can I find the dimension of $K$?</p> | 1,298 |
<p>I have been taught that space-time should be viewed as a fabric and that objects with a large gravitational influence indent that fabric. My question is, if the singularity of a black-hole punctures space-time, how is this accomplished if the universe is 3D? Can an object move completely around the black-hole in all directions? Would you be able to travel "below" a black-hole?</p> | 13 |
<p>We know that the CMB is isotropic when viewed outside of the spinning and revolving earth.</p>
<p>Is it homogeneous?</p>
<p>Can we relate the CMB to an inertial frame in the Newtonian sense (in which space and time are homogeneous and isotropic)? Or can it just provide an idea to build upon a new theory in which global (privileged) inertial frames exist?</p>
<p>Einstein's General Theory of Relativity could be useful in an answer, but some outside-the-box thinking would be appreciated; feel free to question GR.</p> | 1,299 |
<blockquote>
<p>Bodies $A$ and $B$ are moving in the same direction in a straight line with a constant velocities on a frictionless surface. The mass and the velocity of $A$ are $2 \text{kg}$ and $10 \text{m/s}$. The mass and the velocity of $B$ are $6 \text{kg}$ and $4 \text{m/s}$. A spring is connected to the back of the body $B$ and its rate is $800 \text{N/m}$.</p>
<ul>
<li><p>What is the velocity of $A$ relatively to $B$ before the collision and after the collision?</p>
<p>(Answer: $6 \text{m/s}$ and $-6 \text{m/s}$)</p></li>
<li><p>What is the relative velocity between the objects when the spring achieves its maximum contraction? </p>
<p>(Answer: $0$)</p></li>
</ul>
</blockquote>
<p><img src="http://i.stack.imgur.com/lB3mI.png" alt="Illustration"></p>
<p>I had no problem with the first question. We know that in elastic collision the coefficient of restitution is $1$, therefore the relative velocities before and after the collision are equal but are opposite in sign. The relatve velocity before the collision is $10-4=6\text{m/s}$ .</p>
<p>My trouble is with the second question. I though a lot about it and came to a contradiction in my thoughts (of course it is because I'm missing something):</p>
<p>On the one hand, it seems to me that when the contraction of the spring is maximal, all the kinetic energy of $A$ was transformed to a potential elastic energy and to some portion of kinetic energy of $B$ (it will move faster). So it means that $A$ has no velocity at that moment, but $B$ has, so it's impossible that their relative velocity is zero. Intuitively, it means that $A$ has some kinetic energy. So it means that $A$ will be slowed down by the spring, until its velocity is zero, and the spring can finally return to its normal state. So it seems that the spring will be contracted maximally for some relatively short period of time until the kinetic energy of $A$ becomes zero. But something tells me that I'm wrong. Anyway, I've tried to play with the law of conservation of energy and to wrap it around my thoughts, but it didn't quite work well. So what am I missing? I will appreciate any help.</p>
<hr>
<p>Proposed solution:</p>
<p><strong>Relative speed during the maximal contraction</strong></p>
<p>The moment $A$ reaches the spring, the spring will push both of the bodies as a reaction to the $A$'s push. Therefore, the $A$ will start to decelerate while $B$ will start to gain more speed. Although $A$ is losing its kinetic energy, its velocity won't be zero at the moment of the maximal contraction. This is because $A$ will continue to contract the spring as long as its speed is higher than the speed of $B$ (and therefore the speed of the spring itself). The moment their velocities are the same, $A$ will no longer be "overtaking" the spring, therefore its contraction will be maximal at that point. Therefore, the relative speed is zero.</p>
<p><strong>Finding the maximal contraction:</strong></p>
<p>The net momentum is conserved at any point in the time, so:</p>
<p>$m_A v_A + m_B v_B = (m_A + m_B) V$</p>
<p>In this particular question, $V$ would be: $5.5 \text{m/s}$</p>
<p>We also know that the total mechanical energy is conserved (the collision is perfectly elastic), therefore, the mechanical energy before the collision (which only consists of kinetic energies) will be equal to the total mechanical energy during the max contract. Therefore:</p>
<p>$\frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = \frac{1}{2} (m_A + m_B) V^2 + \frac{1}{2} k (\Delta x_{max})^2$</p>
<p>The energy when the contraction is maximal consists of kinetic energies of $A$ and $B$ (which are moving with the same velocity) and of elastic potential energy due to the fact that the spring is contracted.</p>
<p>In this particular question $\Delta x_{max}$ would be:</p>
<p>$\Delta x_{max} = \sqrt{\frac{27}{400}} \approx 26 \text{cm}$</p> | 1,300 |
<p>This question is trying to see if anyone has some simple notation (or tricks) for dealing with operators acting on coherent states in a Fock space. I use bosons for concreteness; what I'm interested in might not be applicable to fermions.</p>
<p>If I have a multiparticle state defined by</p>
<p>$$|\phi\rangle=|n_1 n_2 ...n_k\rangle=\frac{(a_1^\dagger)^{n_1}}{\sqrt{n_1 !}}\frac{(a_2^\dagger)^{n_2}}{\sqrt{n_2 !}}...\frac{(a_k^\dagger)^{n_k}}{\sqrt{n_k !}}|0\rangle$$</p>
<p>I can write this compactly as</p>
<p>$$|\phi\rangle=\prod_{i=1}^{k}\frac{(a_i^\dagger)^{n_i}}{\sqrt{n_i !}}|0\rangle.$$</p>
<p>Now, I would like to act on this state with some operator consisting of creation and annihilation operators; this could be quite complicated, like</p>
<p>$$a_ja_ka^\dagger_l.$$</p>
<p>Now, IF the above product was a sum, I could find the answer very easily using commutation relations and delta functions:</p>
<p>\begin{align}
a_ja_ka^\dagger_l|\phi\rangle&=a^\dagger_la_j\sum_i\frac{(\delta_{ki}+a^\dagger_{i}a_k)}{\sqrt{n_i!}}|0\rangle\\
&=a^\dagger_la_j\left(\frac{1}{\sqrt{n_k!}}|0\rangle \right)
\end{align}</p>
<p>...continue until you get it to normal form, and you are done. Of course, this is a different problem but I am just illustrating how neat this notation is.</p>
<p>You can't just throw around delta functions in the product above, since they would make the entire product vanish. What I want is a clean way to denote "I have passed $a_k$ through all $i\neq k$, used the commutation relations to get $1+a^\dagger_ia_k$, and am now ready to hit this with another creation opertaor $a_l$."</p>
<p>I think it's fairly clear what I am looking for; anyone have any notation or tricks to make calculations using arbitrary operators like the one I have above easier?</p> | 1,301 |
<p>See Griffiths Quantum Mechanics, eq. 11.21. Evidently,
$$\psi(r,\theta,\phi)=Ae^{ikz}+A\sum\limits_{l,m}^{\infty}C_{l,m}h_{l}(kr)Y_{l}^{m}(\theta,\phi).$$
But I don't see why the $l$th Hankel function necessarily needs to multiply the $l$th spherical harmonic. Seeing as we arrived at this equation through separation of variables, my feeling is that the equation should look something like:
$$\psi(r,\theta,\phi)=Ae^{ikz}+A\sum\limits_{l,m,n}^{\infty}C_{l,m,n}h_{n}(kr)Y_{l}^{m}(\theta,\phi).$$</p> | 1,302 |
<ul>
<li><p>How does unitarity require that every scalar operator in a $2+1$ SCFT will have to have a scaling dimension $\geq \frac{1}{2}$ ? </p></li>
<li><p>Why is an operator with scaling dimension exactly equal to $\frac{1}{2}$ said to be "free (i.e decoupled from the rest of the theory)" ? </p></li>
<li><p>Let us say that we know in some such theory the R-charge is monotonically decreasing with increasing coupling constant (say $\lambda$). Then define $\lambda _n ^f$ to be that value of the coupling constant at which the R-charge of the operator $Tr[\phi ^n]$ becomes $= \frac{1}{2n}$. Let $\lambda _n ^m$ be that that value of the coupling at which the R-charge of the same operator becomes $= \frac{2}{n}$ (..marginal ?..) Then clearly $\lambda _n ^m < \lambda_n ^f$. </p></li>
</ul>
<p>How does this imply that there has to exist a $\lambda _c \leq \lambda_2^f$ where the theory might undergo a phase transition? </p> | 1,303 |
<p>I have read an answer on this site regarding the change of laws over time . However a physisct told me that the laws did evolve at planck era and then stopped evolving after it , is that true even in case of a TOE ( namingly as the best candidate string theory )</p> | 1,304 |
<p>A clock near the surface of the earth will run slower than one on the top of the mountain. If the equivalence principal tells us that being at rest in a gravitational field is equivalent to being in an accelerated frame of reference in free space, shouldn't the clock near the earth run <em>increasingly</em> slower than the other clock over time? If those two clocks are considered to be in two different frames of acceleration, the one near the earth will have a greater acceleration than the one on the mountain, and thus over time, their relative velocity will increase over time which will increase the time dilation. Or is this not a proper use of the equivalence principle?</p> | 1,305 |
<p>I have heard that area is a vector quantity in 3 dimensions, e.g. <a href="http://physics.stackexchange.com/q/14165/">this</a> Phys.SE post, what about the length/distance? Since area is the product of two lengths, does this mean that length is also a vector quantity, and why?</p> | 1,306 |
<p>One usually hears about graphene as a good thermal conductor, and good light absorber due to its tunable bandgap properties. But i haven't heard about its aplicability as an optical mirror. In fact, mostly the opposite is true: the optical transmissivity of monolayer graphene is very high ($\approx$ 98%)</p>
<p>Since there are material engineering tricks to tune the separation between two or more graphene layers, i would expect that multi-layer destructive interference can be achieved with optical wavelengths, enhancing reflectivity</p>
<p>I know that probably graphene is not the best material to do this, but allow me to insist. Why? Well, graphene mantains it's properties well over 3000K, and a couple of layers of it can be quite strong and very light. All these properties make it the ideal material for a laser-pushed sail. </p>
<p>By the way, i'm asking about optical and UV frequencies, i know that monolayer graphene is reflective on the microwave region, and i'm not asking about that. Thanks</p> | 1,307 |
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