# Force on a massless classical particle

1. Jul 14, 2011

### atyy

Can a massless classical particle experience a nonzero Newton's second law force?

Dickfore produced a very interesting formula in https://www.physicsforums.com/showpost.php?p=3333233&postcount=52 .

Is this generally accepted? Are there other expressions that work? Or are all acceptable expressions equivalent?

2. Jul 14, 2011

### PAllen

After posting doubts about how this would work in that other thread, I realized I thought it might work, but didn't have time to work out details. This post by Dickfore is very interesting. I was thinking that the key thing to work around, given preference to us 4 vectors is the following:

- we have a well defined 4-momentum for a massless particle (happens to have null norm)
- a null path can definitely change direction and not be geodesic (I was simply confused
for not realizing this).

but you need a consistent approach for not being able to take covariant derivatives by proper time along a path. 4-force is normally covariant derivative by proper time. What definition do you use for a lightlike path? I would hope there is something more elegant than Dickfore's post. So, for me, the key to making all this work is substituting something for covariant derivative by tau in the 4-vector formulation of relativistic kinematics.

3. Jul 14, 2011

### atyy

Dickfore's equation says the force sees the energy of the particle, which reminds one a bit of gravity (as a field in flat spacetime). How about http://arxiv.org/abs/gr-qc/0405030, Eq 12? Can it hold for null 4-velocities?

Another useful reference for particle equations of motion in gravity is http://arxiv.org/abs/gr-qc/0611100.

I would also like to know if Gralla and Wald's http://arxiv.org/abs/0905.2391 derivation of the electromagnetic self-force of point charge as a certain limit applies to massless classical point charges.

Last edited: Jul 14, 2011
4. Jul 14, 2011

### bcrowell

Staff Emeritus
I don't think it's absolutely necessary to start off by defining the four-force as $dp/d\tau$. For a massless particle, a finite three-force acting on it becomes an infinite four-force, so the three-force is well defined but the four-force isn't.

But if we don't use $dp/d\tau$ as our definition of force in relativity, and use the three-force instead, then we have to deal with the question of what is the fundamental definition of the three-force in relativity. I think the history is that very early on (maybe ca. 1920), Einstein decided that force was not a useful concept in relativity, and stopped referring to it. Newton's laws aren't valid, so it's not necessarily easy to define what is meant by the three-force, if we don't start with the four-force.

A related issue that has always bugged me, and about which I've never been able to get a satisfactory answer, is that a lot of treatments of SR derive the equation for relativistic kinetic energy by using the work-kinetic energy theorem with three-vectors, but I've never seen any justification for the assumption that the work-KE theorem should hold.

5. Jul 14, 2011

### PAllen

At least for massive particles, I've always been satisfied with the approach that takes 4-momentum as a definition; energy is its timelike component; rest energy its norm; and kinetic energy simply total energy - rest energy.

6. Jul 14, 2011

### bcrowell

Staff Emeritus
I don't have any objection to any of those statements. What bothers me is the link to the 3-force, and the approach in which the work-KE theorem is used as a fundamental assumption in order to derive the relativistic relations involving energy and momentum.

7. Jul 14, 2011

### PAllen

After reviewing my books and finding nothing of use on this question, I searched again online and found the following:

http://arxiv.org/abs/hep-th/9508081

The first part of this paper covers exactly this topic! (A consistent approach to the dynamics of massless particles in SR).

8. Jul 14, 2011

### atyy

Thank you! Looks very interesting.

9. Jul 15, 2011

### Dickfore

10. Jul 15, 2011

### bcrowell

Staff Emeritus
I got lost at this point:
What does it mean for a certain four-vector to "represent" a certain three-vector?

This also bugged me:
I don't see how you can have instantaneous action at a distance in relativity. He says it's only instantaneous in the rest frame of a certain object...so it's not instantaneous in the rest frames of other objects...?? This doesn't make a lot of sense to me. If object A acts on object B, then how do you decide whose frame it's instantaneous in?

11. Jul 15, 2011

### PAllen

Hmm, I just skimmed the claimed results. Hopefully, this weekend I can spend some time reading it.

12. Jul 15, 2011

### atyy

Very interesting! If you Lorentz transform, do you use cXB for the magnetic force, and do you get the same trajectory?