elfmotat

You're in my thread. Maybe other people appreciate whatever shtick you think you're doing right now, but I don't. Cut it out or start your own thread.

I can't tell whether or not you're serious right now. Because if so, you're being extremely obnoxious.

[math]\frac{1}{2 \sqrt{2}}[/math] units. See for example: http://www.seas.harvard.edu/softmat/downloads/2011-10.pdf

I have this theory that the moon is made of cheese...

Apologies for the sloppiness, all I had to write on was my desk! So, if you can read my drawing well enough, you should be able to see that we have: [math]tan \alpha = \frac{PQ}{OQ}[/math] [math]tan \beta = \frac{PQ}{QC}[/math] [math]tan \gamma = \frac{PQ}{QI}[/math] If we expand out tanx we have: [math]tan x=\frac{sinx}{cosx} = \frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-...}{1-\frac{x^2}{2}+\frac{x^4}{4!}-...}[/math] Now if we let x be very small, then all powers of x will be even smaller. So if we discard all higher order terms, we can approximate tanx with: [math]tan x \approx \frac{x}{1}=x[/math] So when the angles in the picture are small, we have: [math]\alpha \approx \frac{PQ}{OQ}[/math] [math]\beta \approx \frac{PQ}{QC}[/math] [math]\gamma \approx \frac{PQ}{QI}[/math]

Yes. I remember going over this quickly when I first started studying GR. I've also looked up the question on a few other science boards and nearly everyone seems to agree with me. Plus, I just stumbled across this 1964 paper by DeWitt, which also agrees with me: https://www.scribd.com/doc/100745033/Dewitt-1964 . Of course we could all be missing something fundamental, but I doubt it. The topic is apparently a lot more subtle than I originally thought! Yes, I believe it should classically still radiate. It is different from the particle sitting on the ground because the particle on the ground has proper acceleration (i.e. a net force acting on it) due to the ground! Unless you mean the particle w.r.t. the Earth-Sun system as JonathanApps suggested. It should also classically radiate while it's stuck to the Earth orbiting the Sun. Indeed. The observation of EM radiation is frame dependent, and accelerated observers co-moving with a charge will not detect any! But this isn't because it's not emitting any radiation, it's because the radiation lies beyond the Rindler horizon. See for example here: http://arxiv.org/pdf/physics/0506049%E2%80%8E . (Of course this opens the philosophical question: if I can't measure it, is it really there? Which makes this a problem of definitions as well: what constitutes radiation? Does it need to be observable for all observers?) To add to the previous comments about the testability of all this, Jackson does a calculation in section 14.2 for the energy per meter that would need to be supplied to an electron in a linear accelerator to get significant radiation loss, and it's ~1014 MeV/meter, which is far beyond what current linear accelerators are capable of (he gives the figure 50 MeV/meter).

No, it is precisely because spacetime is curved that the particle radiates. The charge generates an EM field, which affects the motion of the charge, which affects the EM field, which affection the motion of the charge, etc. I'm not sure why you think negligible quantities must necessarily add up into some measurable effect. That's certainly not true. I'm fairly certain I'm correct that falling charges will indeed radiate as a prediction of GR.

I agree, they seem to be completely unrelated. The latter case can be calculated with Maxwell's equations in flat spacetime. The former case, you need to solve Maxwell's equations on a curved background, while simultaneously solving the equation of motion for the particle. I'm pretty sure there are no exact solutions for such a scenario.

I suspect it would be many, many orders of magnitude too small to detect.

I've thought this over and realize it's incorrect. The EM field of a charge in an accelerating rocket will locally look like the EM field of that charge sitting on Earth, but globally they are not the same due to curvature, i.e. tidal forces. Some of the field will be inside of the Earth, and some of it will be on the other side of the Earth. Obviously this is different from the case with the rocket. In fact, a charged particle sitting on Earth will NOT radiate. Otherwise we could plug in a Van de Graaf generator and expect a lot of radiation (and free energy)! The upward force on the particle exerted by the ground acts through zero distance, so no work is being done on the particle. That means no energy is being put into it, so it cannot radiate. So then the question becomes, if gravity isn't a force (and therefore can't do any work) then why would a free-falling charge radiate? The answer is slightly more complicated, but it has to do with the fact that charged particles make bad test particles. A charged particle will emit its own EM field, but the value of the EM field is dependent on the gravitational field. The gravitational field will have the effect of forcing the particle to interact with its own EM field. In other words, it will feel a Lorentz force! (For anyone interested in QFT, a way to visualize this would be virtual photons scattering off of virtual gravitons, such that the photons come back and interact with the charge.) Or, to put it more simply, charged particles *CAN'T* freefall because they will always be subject to a Lorentz force "pushback" from the EM field in curved spacetime. So there we have it: a charge sitting on the ground WILL NOT radiate. A charge in a uniformly accelerating rocket WILL radiate. If you drop a charge into a gravitational field it WILL radiate, but it will not freefall.

That sounds tedious. Try making your equations at the link imatfaal suggested (http://www.codecogs.com/latex/eqneditor.php), then copy and paste the code between [*math][/math] tags (without the *).

It's an onion site, you need tor to browse it.

[math]\int \frac{a+bx}{cx} dx = \int \frac{a}{cx} dx + \int \frac{bx}{cx} dx[/math] Typing into the normal reply box works just fine for me. I'm not sure what your problem is imatfaal.

I've noticed that I get errors when I have double spaces in my code. For example "a=b+c " will not render due to the two spaces at the end.

You really need to pick up an intro physics textbook, because this is unbearable. It's like the Hydra of misconceptions: you correct one and two more pop up. But worst part of it is the sheer arrogance you display in your posts. You refuse to listen to what others are saying, which makes discussion with you pointless.

No problem. By the way, the site will render your LaTeX for you if you wrap it in [*math][*/math] tags (without the *s).

They're called "seagull graphs." See, for example, section 6.5.1 in David Tong's lecture notes: http://www.damtp.cam.ac.uk/user/tong/qft/six.pdf .

I'll drop the coefficient to simplify things: [math]\psi^* \psi =(e^{ikx} e^{-\alpha x^2})(e^{-ikx} e^{-\alpha x^2})=e^{ikx-ikx-\alpha x^2-\alpha x^2}=e^{-2 \alpha x^2}[/math]

Well, I feel like an idiot. I was integrating the wavefunction instead of the square of the wavefunction. The square of it is just [math]\sqrt{\frac{2 \alpha}{\pi}} e^{-2\alpha x^2}[/math], which you can use u-substitution to get it into the form of integral (2). Everything works out, there are no extra factors.

My hint was meant to suggest that you can use u-substitution to solve the integral. Let [math]u= \sqrt{\alpha} x [/math]. It doesn't have any physical meaning. Indeed. It seems the function isn't actually normalized.

For the wave function they give you, see what happens if you change variables: [math]x \to x / \sqrt{\alpha}[/math]. Then compare the result to the third integral they give you.

In the context of relativity, which is what we're discussing, "invariant" is synonymous with "the same in all reference frames." That equation is in my signature, it has nothing to do with this thread.

"Invariant" means "the same value in all reference frames." The number of dust particles on the floor actually is an invariant quantity. The energy of a particle is not. Energies depend on your reference frame.

I'm not sure where your confusion is coming from. An electron in a rocket accelerating at 1g will radiate the same way that an electron sitting on earth will radiate.

What is the rocket doing in the "rocket experiment?" If it's moving uniformly then masses are weightless! If it's sitting on a launchpad on Earth then it weighs the same as if it were accelerating at 1g in deep space.