# Photon mass

1. Feb 19, 2005

### gonzo

So, I'm having trouble with the concept of the mass of a "system" being greater than the sum of the masses of the components of that system. Specifically in the case of photons.

A system consisting of 1 photon has no mass.
A system consisting of 2 photons in the same direction has no mass.
However a system consisting of 2 photons moving on different direction has mass, even if neither of the photons do.

Just like the idea that heat has mass and can be weighed with precise enough instruments in the future (as in the case of how a system that involves the inelastic collision of two objects seems to end up with more mass after the collision than it had before, which is supposed to come from the heat gained by the objects), does this mean you can theoretically weight two phtons moving in different directions, just not one or the other individual photon?

2. Feb 19, 2005

### dextercioby

Please explain this assertion:"However a system consisting of 2 photons moving on different direction has mass, even if neither of the photons do."What' the theory behind it...?QFT or GRT,or something else...?

Daniel.

3. Feb 19, 2005

### gonzo

Sorry, this is just 4-vector momentum-energy in SR.

4. Feb 19, 2005

### jcsd

I think you're a little confused:

Photons in SR don't have rest mass, but they do have something called relativstic mass which is really just there energy.

The systems you are decsribing are for GR and whethr or not two phtons will interact gravitationallly.

5. Feb 19, 2005

### polyb

Dext, it's from GRT and is a consequence of the what Wheeler called the 'momenergy' () or Einstein's mass-energy relation. So it does drop out of the math but only when you have 2 or more photons, or so I learned from a class on GR. Here is one small write up that gives a breif discourse:

Does light have mass?

I would reccomend also to look into John Wheeler's book on Spacetime Physics where he gives his justifications for the idea of photon mass. It is pretty straight forward and simple. Here is a write up for the theretical consideratioins on this topic. Here is a small write up from aps. And just to make things really interesting, Here is one from the arxive that involves brane theory. Enjoy!

6. Feb 19, 2005

### gonzo

I read the book, in fact that's why I'm confused. He talks about it, and talks about how "systems" can have mass that isn't accounted for by their parts. But I'm still having trouble with it. And it's SR not GR from my understanding, although GR may deal with it too, but I haven't read that much on GR yet and so don't know.

JCSD ... I am not talking about their gravitational attraction or their relativistic mass. This the total mass of a system, which is also the magnitude of the momentum-energy vector.

I just wonder if it means you can theoretically weight a system of photons and detect the system mass that way. Just like you can in theory for heat.

7. Feb 19, 2005

### polyb

Ooh, my bad, your right it is SR, not so much GR. Read the first couple of links I made, hopefully that may clarify some of your confusion. Other than that, it is a mathematical relic IMHO. Of course I could be wrong there as well.

Conceptually mass is usually the measure of the resistance to the change in momentum. How about considering this: when light passes from one medium to another, say like from a vacuum into the air, what happens?

8. Feb 19, 2005

### jcsd

oh right actually if I think about it:

If two photons with energy momentum p and q having compoents in some frmae of: $(p_t, p_x, p_y, p_z)$ and $(q_t, q_x, q_y, q_z)$.

We know that $p_x^2 + p_y^2 + p_z^2 - p_t^2 = m^2 = 0$ and $q_x^2 + q_y^2 + q_z^2 - q_t^2 = m^2 = 0$, but $(p_x + q_x)^2 + (p_y + q_y)^2 + (p_z + q_z)^2 - (p_t + q_t)^2 = 2(p_x + p_y + p_z)(q_x + q_y + q_z) - 2p_tq_t$
which is equal to zero only when p = Cq where C is a scalar constant (i.e. when they are parallel).

Last edited: Feb 19, 2005
9. Feb 19, 2005

### pervect

Staff Emeritus
A system with one photon can have mass, but you have to imagine the photon being in a perfectly mirrored box. The direction of travel of the photon isn't really relevant or not to whehter the system has mass. Putting the photon in a mirrored box is highly relevant, however. I believe it's impossible for the mirrors to have zero mass, BTW, so your system really is a system, it's not just a photon.

Probably the best thing to do is to think about measuring the momentum of the box as a function of its velocity.

When the photon bounces of a mirrored wall, the total momentum of the box+photon is conserved. If we actually have only one photon in the box, the walls of the box must be vibrating back and forth (the center of mass of the box will remain fixed). This is speaking classically, in spite of the fact that we are talking only about one photon - I don't want to introduce the quantum complications into the problem.

In the rest frame of the box, the photon has the same frequency in both directions, hence it has the same momentum in both directions.

In a frame moving with respect to the box, the photon is blue shifted when moving in one direction, and red shifted in the other. There are several ways to see this must be true, one of the simplest is to know that the photon has the same frequency in both directions when the box is stationary, and to apply the Lorentz transforms.

The average momentum of the box is the average of the momentum of the walls of the box and the photon. Because the frequency of the photon depends on its direction, the average momentum of the photon will not be zero, hence it contributes to the momentum of the box.

It's seems slightly more difficult to work out what happens when you accelerate the box, but if you do, you'll find that f=ma, and that a (on the average) includes a contribution from the photon. It's easy to see that the two approaches must be the same by considering the equation f = dp/dt.

10. Feb 19, 2005

### gonzo

I don't know if we are talking about the same thing here, pervect. I'm talking about invariant mass, which is also the magnitude of 4-vector momentum-energy (or momenergy).

As I understand it as explained by Wheeler in "Spacetime Physics", one photon has 0 invariant mass. Two photons moving in the same direction also have 0 invariant mass, and this can be understood by realizing that two photons heading the same direction are equivalent to one photon heading in that direction with more energy. However, two photons heading in different directions taken together have positive invariant mass, without the need for anything else as part of the "system".

11. Feb 19, 2005

### jcsd

It is the same thing, it just demonstrates physicallywhat it means for a system of photons to have a non-zero rest mass.

12. Feb 19, 2005

### gonzo

I guess I'm just confused then. From what pervect wrote it seems to me he is talking about the momentum of a photon, which, although related, is a bit different than what I mean.

Judgeing from Wheeler's discussion of weighing heat when talking about the increase in mass in certain systems from heat, in the same section and under the same topic as he discusses the mass of multi-photon systems, it seems to be implied that you should be able to, in theory *weigh* two photons moving in different directions in a way that would give 0 weight for 1 photon or two photons moving in the same direction.

Pervect mentions that you could use one photon, which even further makes it seem that we are talking about different though possibly related things.

How you would even theoretically weigh two photons moving in different directions in some way that compares it to two photons moving in the same direction (or I assume a million photons moving in the same or different directions) I have no idea, which is one of the reasons I started the thread.

I guess you can't really ever have a system consisting ONLY of photons moving in the same direction and actually do any experiments on it ... so maybe it's all moot. Unless there is some theoretical way to do this?

13. Feb 19, 2005

### pervect

Staff Emeritus
I think I've probably confused a perfectly good textbook explanation of system mass - mea culpa.

I don't have Taylor & Wheeler's textbook (which has a very good reputation), but the basic idea behind their defintion of system mass as you describe it is very simple

You look at the total energy of the system E, in any inertial frame, and you look at the total momentum p, of the system, in any inertial frame, and you take the quantity (E/c^2)^2 - (p/c)^2 = m^2 as the mass of the system.

If you apply this simple definition to a pair of photons, you get their result very directly.

If you apply their result to a more complex system (my favorite is still the photon in a box problem), you will find that the system of a mirrored box plus a photon has a higher mass than just the mirrored box, and that the force required to accelerate the box of photons is just the invariant mass of the box + photons.

The problem with my idea is that you can't really imagine "pushing" photons around, unless they are interacting with matter (such as a mirror). Taylor & Wheeler's definition can be applied without the necessity of pushing anything - one simply looks at the system's total energy and momentum from several viewpoints, and computes an invariant quantity.

14. Feb 19, 2005

### Haelfix

People get so confused about this. Photons traveling antiparrelel to one another *attract* one another at least in the weak field limit. You can see this from either field theory, or simply writing down the worldline equations in linearized gravity. This is called the Tolman-Ehrenfest-Podolsky effect and its a bit puzzling if you study it in depth. The so called gravity of light indeed comes with a caveat, photons travelling parrelel to one another do not feel this effect at all. This is actually a statement of the nonlinearity in the Einstein field equations, and I think there is still some open debate about some of the technicalities.

The thing is it does makes sense in say a Scattering experiment, viewed by a stationary observer and I believe it has been observed.

15. Feb 20, 2005

### hellfire

Two parallel photon currents do "feel" a gravitomagnetic effect. Why do not two single parallel photons?

16. Feb 20, 2005

### pmb_phy

If you choose to define rest mass as "energy in zero momentum frame"/c2. For two photons moving parallel to each other in the same direction then there is no rest frame. However if the two photons move in opposite directions then there exists a frame in which the total energy in that frame is zero and you can then apply the definition of rest mass above.

As far as weight - I don't see how you could weigh a photon. But any photon will have inertial mass, passive gravitational mass and active gravitational mass. Each of these are equal to each other so its often referred to as simply "mass."

A single photon carries momentum, can be deflected by a gravitational field and it can generate a gravitational field.

Pete

17. Feb 20, 2005

### jcsd

I made a school boy error in my claculation earlier, it should of been:

$$(p_x + q_x)^2 + (p_y + q_y)^2 + (p_z + q_z)^2 - (p_t + q_t)^2 = 2p_xq_x + 2p_yq_y + 2p_zq_z - 2p_tq_t = 2(\vec{p}\cdot\vec{q})$$

i.e. the sum of two null vector is equal to twice their scalar product which is zero when they are 'parallel' due to the fact they are orthogonal with themsleves!!

Last edited: Feb 20, 2005