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seen Jul 7 '13 at 22:47

Jan
9
comment Equation for the trajectory of a frisbee?
You've picked a very challenging problem. Once you acquire your data, could you please share it with us? I see several articles on this topic (biosport.ucdavis.edu/research-projects/…, web.mit.edu/womens-ult/www/smite/frisbee_physics.pdf), do any of them roughly cover what you are interested in?
Dec
31
comment What is the difference between a photon and a phonon?
@Slaviks, yes, I was just using a periodic crystal as an example of something that doesn't even have continuous translational symmetry. Air and liquids has no problem carrying sound.
Dec
23
comment Can entropy be equal to zero?
I, for one, define temperature as $T = \partial U/\partial S$, so that a nonzero temperature always implies nonzero entropy.
Dec
23
comment Superposition of electromagnetic waves
I assume the question has a typo? Perhaps you mean: Is it possible to produce microwaves through the superposition of two optical beams? The answer is yes.
Dec
21
comment Simple explanation of Quantum Zeno Effect
This article (arxiv.org/abs/0809.4388) also has an elegant explanation.
Dec
21
comment Optical equivalent of a superconductor
If you could make a (very!) high temperature superconductor, where the pairing gap with in the eV range, then that would solve your problem. It's an outstanding scientific goal to make higher temperature superconductors.
Dec
21
comment Are all bose-einstein condensates superfluid?
A BEC without any interactions is not a superfluid, which is the example described in the above passage. However, I do not think that condensation without superfluidity has been observed. Exotic condensates, like photon condensates, exist for which the concept of superfluidity might not makes sense.
Dec
21
comment Counterpropagating beams in a ring cavity lasers
The whole point of a ring cavity laser is to eliminate the standing wave pattern. If only a single mode, e.g. CW, is present, there will be no standing wave, only a running wave.
Dec
20
comment Schrodinger's equation (explanation to non physicist)
@juanrga, thanks for cleaning up my answer and fixing the errors and text!
Dec
20
comment CERN projects on gravity
From what I've heard from ALPHA, the simplest/first gravity experiment they'd like to try is just to make antihydrogen fall towards Earth. Surprisingly, there aren't any experimental bounds on this! My understanding is that, beyond this test, they would measure local gravity with antihydrogen as a test of the equivalence principle (everything accelerates under gravity the same, regardless of mass, charge, etc.). A failure of antihydrogen to fall the same as hydrogen would, I presume, indicate beyond-GR physics.
Dec
19
comment What's the physical significance of the inner product of two wave functions in quantum region?
Let me elaborate on one point that hwlau made. Suppose I prepare a quantum system in state $|\beta\rangle$, and then send it through a filter that picks out state $|\alpha\rangle$. For instance, the states could refer to polarization of light, and the filter could be a polarizer. What's the probability by state will make it through the filter? Answer: $P = |\langle \beta | \alpha \rangle|^2$.
Dec
19
comment Would it be possible to have an electron-less solid?
Bose-Einstein condensates and Fermi gases of neutral particles (atoms) are regularly created in labs, including mine. There are may labs that trap ions (chargred plasmas) in electrical (Paul) or magnetic (Penning) traps and cool them to incredibly low temperatures, but even at T=0 they are not degenerate because of the enormous Coulomb repulsion. Instead, they form ion crystals. I agree with Argus, neutron stars should fit the bill, but (to my knowledge) we are still far from forming a trapped degenerate neutron sample in a laboratory.
Dec
19
comment Wigner's friend and quantum Zeno effect
I'm a sucker for experimental papers. Bernu et al. (arxiv.org/abs/0809.4388) has a very clear explanation (and data), take a look at the last paragraph on the first page. I suppose the point I want to make with this: any observation of a system requires coupling to that system, e.g. coupling an atom to a cavity and probing the cavity mode, so it's not surprising than an outside observer agrees that the evolution of the system is altered.
Dec
19
comment Who used the concept of symmetries first?
Noether's theorem probably marks our modern understanding of symmetry in physics, though features (such as linear and angular momentum) must have been understood earlier. It's hard to imagine a full appreciation of symmetry in physics without Lagrangian or Hamiltonian mechanics. My guess is that someone must have noticed that physics is (almost) parity symmetric much earlier.
Dec
18
comment Causality in a gedanken experiment on the hydrogen atom
A small detail: It's easy to get unphysical results when thinking experiments where charge can 'vanish'. In particular, you've just created monopole radiation.
Dec
18
comment Schrodinger's equation (explanation to non physicist)
@Temitope.A: Entanglement isn't obvious in anything here because I've only written single-particle wavefunctions. A two-particle wavefunction $\Psi(\vec r_1, \vec r_2)$ gives a probability $\int_{V_1}\int_{V_2}|\Psi|^2 d \vec r_1 d \vec r_2$ of detecting one particle in a region $V_1$ and a second particle in a region $V_2$. A simple solution for distinguishable particles is $\Psi(\vec r_1, \vec r_2) = \psi_1(\vec r_1) \psi_2(\vec r_2)$, and it can be shown that this satisfies all our conditions. An entangled state cannot be written so simply. (Indistinguishable particles take more care.)
Dec
18
comment Is a hard drive heavier when it is full?
Ron, I believe you're correct, and so my answer might be misleading. I do wonder whether information entropy can be thought of as a source of mass: to order bits, we must do work by removing entropy. Such an explanation would have an order of magnitude similar to my answer.
Dec
18
comment Amplitude of an electromagnetic wave containing a single photon
Anna V: the quantum electromagnetic field has operators analogous to the harmonic oscillator. Instead of the Hamiltonian $H = p^2 + x^2$ (I'm dropping the units here) and energy $(\frac{1}{2} + n) \hbar \omega$, we write a Hamiltonian for each frequency $\omega$: $H = \frac{\epsilon_0}{2} E^2 + \frac{1}{2\mu_0} B^2$, where $E$ and $B$ are treated as conjugate quantum operators (just like $x$ and $p$). We find creation and annihilation operators such that $E = a^\dagger + a$ and $B = a^\dagger - a$, so the energy must be $(\frac{1}{2} + n) \hbar \omega$. $\langle E \rangle$ is well-defined.
Dec
18
comment Schrodinger's equation (explanation to non physicist)
One type of measurement is strong measurement, where we the experimentalists, measure some differential operator $A$, and find some particular (real) number $a_i$, which is one of the eigenvalues of $A$. (Important detail: for $A$ to be measureable, it must have all real eigenvalues.) Then, we know the wavefunction "suddenly" turns into $\psi_i$, which is the eigenfunction of $A$ whose eigenvalue was that number $a_i$ we measured. The system has lost of knowledge of the original wavefunction $\psi$. The probability of measuring $a_i$ is $|<\psi_i | \psi>|^2$.