
#37
Feb1312, 12:05 AM

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I would neither use the spherical charge (b/c then we would have an external electric field which modifies the equations of motion) nor the Faraday cage (b/c we don't need it in a universe w/o any external electric field  and we cannot measure the electromagnetic farfield generated by the charge).
We take a spaceship in free fall; we place a proton and a neutron at exactly the same point (at rest w.r.t. the spaceship); we switch of the strong interaction between them and look what happens. We are not interested in the difference due to external fields. The modified geodesic equation [tex]\frac{d^2 x^\mu}{d \tau^2} + \Gamma^{\mu}_{\nu \rho} \frac{d x^\nu}{d\tau} \frac{d x^\rho}{d \tau} = \frac{q}{m} F^\mu{}_\nu \frac{d x^\nu}{d\tau}[/tex] reduces to [tex]\frac{d^2 x^\mu}{d \tau^2} + \Gamma^{\mu}_{\nu \rho} \frac{d x^\nu}{d\tau} \frac{d x^\rho}{d \tau} = 0[/tex] as long as we neglect backreaction of the electromagnetic field generated by the charged particle itself. Now we can safely separate two effects: a) a deviation of the worldline of the proton from the geodesic motion b) the electromagnetic farfield generated by the proton which we can observe in different (free falling) inertial frames Any deviation from geodesic motion can be observed locally w/o ever referring to the electromagnetic field. If we observe a deviation, then we can try to study the effect for the farfield. If we don't observe a deviation, then every effect in the farfield is due to different inertial (inertial) frames. Of course a deviation from geodesic motion would indicate a violation of the equivalence principle. And b/c we start with a geodesic equation for the charge (neglecting backreaction) it should be clear where we have to look for a correction of the equation of motion. Hope this makes clear what I have in mind. (I have to admit that I haven't checked the references; I'll do that asap) 



#38
Feb1312, 12:22 AM

P: 296

Because we have to account for the energy dissipated due to radiation, now this can only mean that object is not freefalling but experiencing a net force(radiation reaction force). 



#39
Feb1312, 12:37 AM

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That's why I would like to get rid of the energy and the farfield. 



#40
Feb1312, 01:08 AM

P: 296





#41
Feb1312, 01:16 AM

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P: 5,307

The problem may be due to a missing understanding of the selfenergy. But even w/o this problem we would not be able to define energy (as a volume integral) in arbitrary spacetimes; this has nothing to do with the selfenergy problem but simply with the geometry of spacetime. (I changed my post #39)




#42
Feb1312, 01:49 AM

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#43
Feb1312, 02:00 AM

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First there is an electric field inside the shell, so the particle at the center feels an additional force (this applies to the shell as well).
Second the shell is not pointlike, so it does not follow a single geodesic but a family of geodesics and is subject to deformation. I do not say that the problem is uninteresting or wrong, but it's much more complicated. 



#44
Feb1312, 02:27 AM

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P: 1,732

Tom,
Regarding (approximate) motion of test particles in GR (possibly spinning, possibly charged), R.P. Kerr worked out a reasonable amount of this. R. P. Kerr, "The lorentzcovariant approximation method in general relativity.", [Series of three papers]: Il Nuovo Cimento (1959) Volume 13, Number 3, 469491 Volume 13, Number 3, 492502, Volume 13, Number 4, 673689 It's an improved (i.e., lorentzcovariant) version of the old EinsteinInfeldHoffman method which showed (by solving the field equations) that test particles do indeed follow geodesics, (and hence this need not be postulated as a separate hypothesis). The Lorentzcovariant method reveals extra terms if the test particle has intrinsic spin. The final paper above extends the treatment to a charged test particle in an EinsteinMaxwell context. 



#45
Feb1312, 03:45 AM

P: 2,900

It is also discouraging that there seems not to exist a practical way to decide this experimentally. 



#46
Feb1412, 12:51 AM

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P: 5,307

I was thinking about this post and #24 and I have to say that I don't understand your objections.
In #24 you are writing [perhaps there's the possibility to calculate some kind of energy loss which could be defined w.r.t. a small spherical shell, but I doubt that is is possible in general] Your reasoning here is a bit strange. It's not that the question if freefall corresponds to geodesic motion determines whether one can neglect backreaction, but that backreaction may cause deviations from freefall i.e. from geodesic motion. 



#47
Feb1412, 01:07 AM

P: 1,583

Regarding experimental confirmation, it should be noted that the idea of radiation from charges accelerating with respect to you is not limited to classical physics; it remains in quantum field theory on curved spacetime, where it gives rise to socalled Unruh radiation: people on earth should be able to detect blackbody radiation from charges in outer space that are more or less in geodesic motion. (I take the point many in this thread have raised that backreaction may lead to slight deviations from freefall.) I think to date there has only been one experiment claiming to detect the Unruh radiation, and even that is disputed.




#48
Feb1412, 01:29 AM

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P: 5,307

But Unruh radiation is not the effect of a freefalling, notsofreefalling or accelerated charge, but of an accelerated observer. Inertial observers will not detect Unruh radiation; noninertial obversevers will detect Unruh radiation, but this is caused by vacuum in the absence of charges.
So I don't understand how this is related. (I agree that backreaction could be studied using QFT in curved spacetime) 



#49
Feb1412, 05:20 AM

P: 2,900

I find it difficult to understand. 



#50
Feb1412, 06:00 AM

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P: 5,307

no, proper acceleration is of course not observerdependent;
but if a charge is freefalling it is by definition not accelerated (w.r.t. a local inertial observer), nevertheless we discuss radiation; for a distant observer proper acceleration of the charge becomes meaningless  and in addition the radiation he observes may be due to his own motion i.e. observer dependent that's why I want to get rid of the observer and the farfield with its 1/r behavior at all 



#51
Feb1412, 06:23 AM

P: 3,015

Someone had pointed out that the static field of a point charge may become inhomogeneous and/or timedependent in a grav field.
I remember that a curved metric plays an analogous role to a(n inhomogeneous) dielectric constant. This means that the speed of propagation of emwaves is less than c. But, then, there arises the possibility of wakefields being created by charged particles traveling with speed higher than the corresponding phase velocity? Is my conclusion correct? 



#52
Jul2212, 09:02 AM

P: 205

"I think one can safely neglect selfcoupling of the charge with its own electromagnetic field)"
In some places that is a good approximation, but this is definitely a place where you should not use such an approximation. When a particle radiates energy, energy goes into the electromagnetic field. Unless there is a radiation reaction force on the particle, the field gets energy for "free", and energy will not be conserved. 



#53
Jul2212, 10:50 AM

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ApplePion, you're posting in a thread that's been inactive since February.
There is a huge literature on this subject, dating back to the 60's. Here are some references: C. MoretteDeWitt and B.S. DeWitt, "Falling Charges," Physics, 1,320 (1964), http://www.scribd.com/doc/100745033/Dewitt1964 http://arxiv.org/abs/grqc/9303025  Parrott, 1993 http://arxiv.org/abs/physics/9910019  Harpaz and Soker, 1999 http://arxiv.org/abs/0905.2391  Gralla, Harte, and Wald, 2009 http://arxiv.org/abs/0806.0464  Grøn and Næss, 2008 http://arxiv.org/abs/grqc/0006037 http://arxiv.org/abs/grqc/9909035 Some expositions: http://www.itp.unihannover.de/~giul...Talks/Lyle.pdf http://www.mathpages.com/home/kmath528/kmath528.htm 



#54
Jul2212, 10:52 AM

P: 205

"ApplePion, you're posting in a thread that's been inactive since February."
So pretty much are you! Best wishes. 


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