# Question about electromagnetic force and mass

1. Aug 15, 2008

### Mårten

According to Coulomb's law we have that the force between two charged particles could be described by

$$(1)~~~F = k \frac{Q_1 Q_2}{r^2}.$$

Now, a force is according to most sources something that causes a mass to accelerate (not a charge to accelerate), so the force in (1) above will solely act on charged bodys and charged particles with mass. So if we have a charged particle P_1 with charge Q_1 and without mass in (1) (and of course another particle P_2 with charge Q_2), P_1 will not be affected by the force resulting from (1), and therefore it will not start to move.

Am I right in this? That is, electromagnetic forces stemming from Coulomb's law, is like all other forces, they're acting on masses and not on charges?

I was thinking of this, when wondering about how electrostatic induction works. When a positive charged object is close to a neutral object, electrons in the neutral object will move to the side facing the positive charged object. And then I was thinking that if the electrons had no mass, they wouldn't move here, because a force is something which acts on masses not on charges. Or could it be that the EM force actually acts on the particular charge, and not on the mass of the particle holding the charge?

2. Aug 15, 2008

### tiny-tim

Hi Mårten!

Electric forces only act on charges, and they act per mass.

The smaller the mass, the larger the acceleration (because a = F/m).

For example, an electron has the same charge (well, opposite) as a proton, but a much smaller mass. So the same electric field will accelerate the electron much more than the proton (and in the opposite direction).

Last edited: Aug 15, 2008
3. Aug 16, 2008

### h0dgey84bc

Are there any massless particles with charge? If you had a massless particle with the same charge as an electron, the electrostatic force would be the same on this particle, but the acceleration a=F/m (with m=0) would seem to be infinite.

With a photon F=ma doesn't apply, and you have to invoke gen rel to find out how the gravitational field acts on it remembering (remembering the equivalence of energy and mass, and that E=pc).

What would be the analogue of this for the E-field acting on a massless charged particle, to avoid the apparent problem of infinite acceleration?

4. Aug 16, 2008

### tiny-tim

Hi h0dgey84bc!

The m in F = ma is the relativistic mass, in other words the energy.

All particles (including photons) have energy, so there are no massless particles, and the problem of infinite acceleration doesn't apply.

5. Aug 16, 2008

### h0dgey84bc

Hi Tiny-tim,

I was referring to zero "rest mass" particles that have charge.

F=ma doesn't apply when considering the *gravitational* force on a massless particle like a photon. The momentum-energy tensor is responsible for bending space-time, and the photon just follows the appropriate geodesic.

I really don't see why F=ma should automatically apply in the case of an ultrarelatvisitc charged particle in the EM case (since if it has zero rest mass, this hypothesised charged particle would always be moving at c, meaning Newtonian laws are in accurate)

6. Aug 16, 2008

### Redbelly98

Staff Emeritus
The only entities with zero rest mass (that I'm aware of) are photons and gravitons. And maybe one of the neutrinos. All of these have zero charge, however.

Last edited: Aug 16, 2008
7. Aug 16, 2008

### h0dgey84bc

Yep, sounds about right, I can't think of any others either that have zero rest mass.

Let's say there existed such a particle however, it has a zero rest mass and the charge of an electron. Obviously F=kQe/r^2, but how would one find the acceleration on this particle? surely the Newtonian F=ma, will not apply since the particle is moving at c.

Infact F=kQe/r^2 probably doesn't apply either, since this is an electrostatic equation (unless the test charge Q was also one of the hypothesised rest mass zero charged particles, also moving in the same frame at c)

8. Aug 16, 2008

### tiny-tim

Hi h0dgey84bc!

Gravity isn't the only force.

And, as you say, the momentum-energy tensor is responsible for bending space-time, and it is that, rather than F = ma, that governs the "gravitational" effect on all particles, not just photons.
EDIT: F=kQe/r^2 doesn't apply to a moving charge anyway:
we need the Lorentz force, F = q(E + v x B), or its relativity equivalent.

And the Newtonian formula, technically, isn't F=ma, it's F = d(mv)/dt, which works fine when translated into relativity.

From example, an "acceleration" in the same direction as a photon simply changes its frequency.

Last edited: Aug 16, 2008
9. Aug 16, 2008

### Redbelly98

Staff Emeritus
For this hypothetical particle,

F = q ( E + VxB ) = dp / dt

where p is momentum.

Such a particle could be made to alter its direction, but not its speed. One could also change its de Broglie wavelength by changing the momentum's magnitude.

10. Aug 16, 2008

### h0dgey84bc

Of course, but what I was trying to illustrate is that F=ma doesn't apply to a photon (or any general rest-mass zero particle) when we are considering gravities action upon it. We can't just say a=F/m, and that m is the relativistic mass in the case when F is a gravitational force, thusly I don't see why we would if instead F was an electric force.

I was not implying that the momentum-energy tensor was the source of gravitational force for just photons (or rest mass zero particles in general), merely that it was the momentum-energy tensor NOT F=ma we must invoke when considering the gravititational force on a particle (in this case a rest mass zero particle).

My implication was that by analogy, why would F=ma work on something intrinsically relativistic (a zero rest mass particle moving a speed c) in the EM case.

Even in special rel the expression is:
$\vec{f}= \gamma m \vec{a} + \gamma^3 m \frac{\vec{v} \cdot \vec{a}}{c^2} \vec{v}$

But here m is the rest mass something a photon doesnt have.
In the past I have only ever dealt with a gravitational force on a zero rest mass particle (since this is the only force in reality that acts on them) and thus have just used Gen rel and the momentum energy tensor, geodesics etc to work out its effect on them.

But how would a "charged photon" accelerate in an E-field?

Seems that F=dP/dt always holds, but p=E/c, which just leads to F=1/c dE/dt

hmm

Last edited: Aug 16, 2008
11. Aug 16, 2008

### h0dgey84bc

very true, F=dP/dt is the general formula that works even with special relativity, but all the derivations from this I have seen have usually involved rest mass. If you can show me it for a photon I will be happy

12. Aug 16, 2008

### tiny-tim

Hi h0dgey84bc!

For something at the speed of light (and therefore with zero rest-mass), (m,mv)/√(1 - v2/c2) is replaced by (|p|,p).

13. Aug 16, 2008

### h0dgey84bc

Are you referring to the four momentum $$<BR> p^{\mu} </BR>$$ ?

Also why is my tex not working

Last edited: Aug 16, 2008
14. Aug 16, 2008

### tiny-tim

testing: $$A = B$$

(it's not you … it isn't working on other threads, either )

yes, that's the 4-momentum for speed-of-light particles …

it doesn't need an m, only a p and an E = |p|.

15. Aug 16, 2008

### h0dgey84bc

yes, that makes sense then, so you end up with F_mu=d/dt (P_mu)

the latin spatial comps of the four force: F_i=d/dt p_i....so I guess that the problem simply reduces to equating this with the i'th comp of the (4 vec version) Lorentz force, and solving this to find p as a function of time.

wonder why there are no massless particles with charge/colour etc

16. Aug 16, 2008

### Redbelly98

Staff Emeritus
Perhaps there is something pathological about the electric field of a charge moving at c, which prevents it from existing. If I weren't 20+ years removed from my last electromagnetism course, I might even attempt the calculation

17. Aug 16, 2008

### h0dgey84bc

I think you may be on to something perhaps, def got me thinking anyway. It's kind of interesting that the E field only propagates its affects at c, so if the generating particle is also moving at c in the opposite direction: relativistic velocity addition formula v=w-u/(1-wu/c^2), if w and u are both c you get something undefined.
time to try and find my electrodyn notes maybe...

Last edited: Aug 16, 2008
18. Aug 16, 2008

### h0dgey84bc

hmm sorry to be a stickler on this problem, but I've been trying to derive the acceleration a massless particle would undergo in an E field (or for any force field except gravitational) and I can't derive a single relation without using rest mass somewhere, four vectors methods or otherwise. Even if you get the momentum as a function of time, I cant seem to relate this back to velocity/acceleration for a massless particle
....this problem is really bugging me now, haha

19. Aug 16, 2008

### Mårten

Hi tiny-tim and others!

Thanks for thorough replies!
Okey, that was also my theory, even though I didn't formulate it so well... Thanks for confirming this!
But maybe, if the mass equals zero, then maybe NewtonII won't apply. Otherwise we would have packages of vacuum, with zero mass flying around us with infinite acceleration all the time. Maybe we have, we would never notice them...

20. Aug 17, 2008

### tiny-tim

Hi Mårten!

If … if … if …

NetwonII is basic … if you're going to throw it out, anything could happen …

massless pigs could fly!

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