# Newton's 2nd law for Massless Particles in SR

1. Jan 19, 2016

### andresB

I think I have never seen a force equation for massless particles, I wonder if such thing exist or if not why not?

2. Jan 20, 2016

### haushofer

Massless particles follow straight lines (i.e. Geodesics) in spacetime. An action can be derived using an auxiliary field (einbein); see e.g. Zee's book on GR or Green, Scwarz,Witten on string theory.

3. Jan 20, 2016

### pervect

Staff Emeritus
There's a differential equation that serves the same purpose as the Newtonian force equation, but it's called the geodesic equation. Massless particles follow what are called null geodesics.

Light is deflected not only by "forces", but by the curvature of space. The geodesic equation takes into account both the forces and the spatial curvature. In the language of GR, both forces and spatial curvature are aspects of curvature of space-time, and the geodesic equation describes the extremal path that light follows as a result of the curvature of space-time.

Last edited: Jan 20, 2016
4. Jan 20, 2016

### andresB

I'm aware about the geodesic equation in GR, but that is not what I'm wondering for (and to clarify I'm just talking about SR for simplicity).

Massless particles not always follows straight lines in flat space-time, for example photons are allowed to compton scatter electrons in a pure classical way. Are other interactions disallowed for some fundamental reasons?

5. Jan 20, 2016

### Staff: Mentor

All such interactions are best modeled as the incoming photon being absorbed in the interaction while a new photon is created. That way, the photons are always travelling on straight (geodesic) paths and we don't have to think about forces pushing them around and changing their trajectory.

The intuitive notion of a photon as a particle of light moving through space, with a trajectory determined by the forces applied to it, is very misleading. It may seem natural to think that a beam of light is a stream of photons passing by, the same way that a river is a stream of water molecules flowing by... But it's not.

6. Jan 20, 2016

### andresB

Well yes, but the same apply to electron and all other particles, yet there is a classical force equation for them.

It just seems curious to me that there is no such equation for massless particles.

7. Jan 20, 2016

### DrStupid

The classical equation F=dp/dt also works for massless particles in SR. But the point is: does it make sense? In some cases it might me a useful approximation.

8. Jan 20, 2016

### pervect

Staff Emeritus
Consider what happens in special relativity, (in GR terms, in flat space-time) first. A photon scatters off a particle. The details of the scattering are a quantum mechanical process, but classically we say that momentum is conserved, the momentum of the photon + particle before the collision is the same as the momentum afterwards.

The details of the force interaction are typically not important, the collision is regarded as taking a very short amount of time, so the forces are infinite. The momentum (unlike the "force") is always finite, though. It's rather to the situation of a bat hitting a baseball - the bat exchanges momentum with the ball over a short amount of time. The quantity of physical interest is usually the impulse, the amount of momentum exchanged, not the force. Under some circumstances the force might be of interest, for instance if one wants to determine whether the ball survives being hit with the bat or not. However, this requires rather detailed knowledge of the collision process, something that's not compatible with the model of a photon as a point particle. Using the point particle model of a photon (and ignoring it's wave aspect), we know that photons at normal energies don't explode when they bounce off something. If we had a high enough energy photon, we might get into exotic physics where the collision process generated other particles, but that's getting beyond the scope of the question, I think.

I see no particular reason to be especially interested in the force, considering that it is infinite and impulsive in the classic model, and the difficulties of trying to model the photon as some sort of classical extended object - which is rather likely to be wrong in principle, though the wrongness lies not in the classical mechanics, but as interpreting quantum objects as being classical objects.

In flat space-time, the equation for the momentum of a photon is $p = \frac{h \nu}{c}$. This can be derived from the quantum mechanical energy frequency relationship $E = h \nu$ and the speciall relativistic relation between energy, momentum, and mass valid for all particles (massless and not massless), $E^2 = (pc)^2 + (m\,c^2)^2$.

Occasionally we get people who are confused about photons having momentum without having mass - there are a number of FAQ's on the issue which may or may not help the confusion. Other than pointing out that photons do have momentum, and don't have mass, and giving the equations for the momentum of a photon (and the general relationship beteween energy, momentum, and mass), I am at a bit of a loss as to what else can be said coherently. If there's any questions about this, feel free to ask in another thread - but take a look at the FAQ's first to see if your questions have already been answered.

As far as photons bouncing off particles in curved space-time, the definition of global momentum turns out to be complicated in GR. Fortunately, one can separate the process out into a collision process, which one can regard as happening in a small, flat local frame of reference in which one can use the techniques of special relativity, and the propagation process (which we've already discussed) that describes how the photon transverses curved space-time when it's not colliding with anything.

So this has gotten long, so let me summarize. Photons have momentum, and they exchange momentum when they bump into particles. The conservation of momentum is what we need to understand the classical aspects of photon collisions in those cases where we can regard photons as being "point particles". We can separate the collision process of a photon with a particle in curved space-time as a collision process that happens in flat space-time plus a propagation process that hapens in curved space-time.

9. Jan 20, 2016

### Orodruin

Staff Emeritus
Can we please drop the talk about photons in the same framework as classical forces? It makes my brain cringe ... The classical concept of a force is not quantum mechanical and photons inherently are.

What might be discussed is whether or not it makes sense to talk about a 4-force acting on a massless particle (as of yet unspecified). However, this really also loses any significance as the 4-force is defined as $dP/d\tau$, where $P$ is the 4-momentum and $\tau$ the proper time (which is unspecified for a massless particle). You could of course differentiate with respect with some other affine curve parameter if you somehow manage to arrange for a force to act on the massless particle, but it would be inherently different from the 4-force. The geodesic equation is what would drop out when there is no external influence making the particle change direction. After all, this is what we have with normal classical mechanics as well - the force is essentially determines the geodesic deviation in the metric placed by the inertia tensor on the configuration space.

10. Jan 20, 2016

### Staff: Mentor

Electrons and other massive particles can be handled using the methods of ordinary quantum mechanics, where there is a position operator and we can think in terms of position (actually the expectation value of that position operator), velocity (the time derivative of that expectation value), and acceleration (the time derivative of the expectation value of the velocity). We can also expect by the correspondence principle that these quantities will behave like their classical counterparts under conditions where the quantum effects are can be ignored so that we can apply classical force equations.

With photons we have to use relativistic quantum electrodynamics; there is no position operator and no conditions under which the quantum effects can be ignored so none of the line of thought above works. It's not even wrong, just completely inapplicable.

11. Jan 20, 2016

### pervect

Staff Emeritus
I don't disagree with anything that is said, but - one might reasonably ask - does it make any sense to talk about the three-force on a massless particle?

The usual motivation for asking such a question would be that the student isn't familiar with 4-forces :(. But we can try to answer it anyway, though a textbook answer may be problematical as most textbooks will be talking about 4-forces. A serious student would really be better advised to learn about 4-forces just so they can understand their textbooks and read and and answer their own questions.

For the unprepared or casual student, though, my current thinking is that it doesn't make much sense to talk about the force (3-force) on a massless particle in general, except possibly under special circumstances. The special circumstances would be when dp/dt exists. In many of the scenarios mentioned in this thread, dp/dt doesn't exist either. It's already been noted that F=dp/d$\tau$ doesn't make sense for a massless particle.

What I would hope to get across to the casual student is that it is unwise to hold on over-strongly to F=ma, and much better to consider F=dp/dt instead. When p=mv, an m is constant, F=ma follows from dp/dt by the chain rule of calculus. However, p=mv is only true in non-relativistic Newtonian mechanics, it's not true in relativistic mechanics. In relativistic mechanics, $p = \gamma m v$ , $\gamma = 1/\sqrt{1-(v/c)^2}$.

The occasional students seem to grasp this point right away, other students don't seem to "get it", alas. I'm not positive why the students who don't get it don't get it, though my suspicion is that the chief obstacle is that understanding F=dp/dt requires familiarity with calculus and how to differentiate functions. So far I haven't thought of a good way of presenting F=dp/dt while assuming no familiarity with calculus. Trying to think about explaining physics without calculus makes me feel like I'm trying to play charades with both arms tied behind my back.

12. Jan 21, 2016

### DrStupid

That doesn't make much sense for massless objects. I would use p=E·v/c² instead.

13. Jan 22, 2016

### PAllen

Lightlike trajectories need not be geodesic at all, and there have been well worked out systems of kinematics and dynamics for classical massless particles in SR. Classically, you can posit a charged massless particle that follows a non-geodesic lightlike path in a Coulomb fields. As Orodruin noted, 4-force cannot be used, but 3-momentum and 3-force can be, and any frame choice produces equivalent results. You just lose the elegance of 4-vectors. I don't have time now, but this has been discussed here before, and a number of relevant papers were collected in those threads.