1. Jan 29, 2005

### anonymous299792458

If a charge is accelerating, but you are also accelerating WITH the charge, will you see any radiation? I.e. will the integral of the Poynting vector over a closed surface surrounding the charge be zero? It seems that it should be.

Now lets say YOU are accelerating, but the charge is NOT. You woudn't see radiation from the charge in this case, will you?

2. Jan 29, 2005

### vincentchan

for the first question, yes, you will see radiation
for the second question, No, you won't wee radiation

remember you are not in inertial reference fram in both case, you have no reason to expect the answer is same as you sitting in an inertial reference.... velocity is relative, but accelerating is absolute, which is independent to the observer's motion... the charge feels acceleration. and an accelerating charge emits radiation, period

it is a bit weird in the first case because you do not see the charge moving and it radiates, but again, you are not in the inertial frame, same as you see stuffs behave weird when were in an accelerating bus......

3. Jan 29, 2005

### anonymous299792458

Yes, yes, that's what I would have thought too (and you may very well be right). But lets say we have a charge sitting on a table in the earth's gravitational field. I don't think you'll measure any radiation, otherwise you'd have a source of "free" energy. Locally, this is equivalent to the charge being in an accelerating reference frame and you accelerating with the charge. So, I would assume that in the latter case the radiation you'd measure would also be zero.

4. Jan 29, 2005

### vincentchan

latter case? did u mean the former case?
if you have a charge on earth and you are rotating with it... you WILL see radiation... just the rotation of the earth is too slow (1 rotation / day) that the radiation is far too small for us to detect it.....
and NO, you won't get free energy... the rotation of earth will slow down by the charge when it radiate, you have a price to pay.... the day is longer (by a tiny amount). again, the effect will be too small for us to detect.....

5. Jan 29, 2005

### anonymous299792458

I wasn't talking about rotation. I was talking about the equivalence principle which states that a gravitational field is locally equivalent to an accelerating reference frame.

6. Jan 29, 2005

### pmb_phy

No.
The Poyting vector will be zero.

Pete

7. Jan 29, 2005

### anonymous299792458

Um, we've got two completely opposite answers here from pmb phy and vincenchan...

8. Jan 29, 2005

### pmb_phy

I'm going by results established in the physics journals. E.g. see -
http://www.geocities.com/physics_world/misc/falling_charge.htm

For example Radiation from an Accelerated Charge and the Principle of Equivalence, A. Kovetz and G.E. Tauber, Am. J. Phys., Vol. 37(4), April 1969
Pete

9. Jan 30, 2005

### pmb_phy

http://www.geocities.com/physics_world/misc/falling_charge.htm

Scroll down to the article and click the title. The title links to a PDF file containing the article.

Pete

10. Jan 30, 2005

### vincentchan

pmb_phy:
your article seems confirms what i said..... and it explain very clearly .....

11. Jan 30, 2005

### pmb_phy

I responded "No." As the article states in the abstract

Pete

12. Jan 30, 2005

### vincentchan

here is one of the newest research done by last year...
http://arxiv.org/PS_cache/gr-qc/pdf/9303/9303025.pdf [Broken]
go to page 15 and read the conclusion.....

edit: the 10th line..."in the accelerated frame...." don't wanna quote every thing, read...

Last edited by a moderator: May 1, 2017
13. Jan 30, 2005

### pmb_phy

You didn't explain how what I said contradicted the articles I quoted/referanced.

I've already read the article the you referred to. The author contradcits a lot of physics in that article. For example; the author assumes, using his own intuition or whatever, that some forms of the equivlance principle don't apply. Since the equivalence principle is a postulate then only experimentation can determine whether it is correct or not - not calculation.

The author asserts
This contradicts previous conclusions. For example; An extra force must be used to accelerate a charged particle over a non-charged particle. This extra force is known in the literature as the "self-force" acting on the particle. This also is known as the "radiation reaction force."

Details - The power radiated by an accelerating charge is given (in esu units) by P = (2/3)q2/c[sup3[/sup](a)2. To account for this energy loss modify Newton's equation by adding in an extra force Frad. Suppose the force on a non-charged particle of the same (rest) mass is (in the non-relativistic limit) Fext = ma. A particle with charge q and mass m is then given by Fext + Frad = ma. By demanding that energy be conserved this demands Frad = mTda /dt (where T is a constant and I'm too lazy to type it in :tongue: ). If the particle is uniformly accelerated then the radiative force is zero. Thus if you have a particle in a uniformly accelerating frame of reference and you're also at rest in that frame then the weight of the particle is independant of the charge. Hence the authors assertions are in contradiction with the assertions I've just stated.

Its also be shown that the weight of a particle depends on the spacetime curvature. That means that a charged particle at rest on, say, the earth will have a different weight (i.e. smaller required support force) than the same particle at rest in a uniform g-field.

The paper is newer. That doesn't mean that its better.

Pete

Last edited by a moderator: May 1, 2017
14. Jan 31, 2005

### da_willem

15. Jan 31, 2005

### da_willem

The article Vinchentan cited says:
Is this/can this/how can this be true?

16. Jan 31, 2005

### pmb_phy

Sorry. I screwed up the names. I'll correct it later.
In my opinion the author Vinchentan cited is wrong.

Pete