Mass of a system of two photons

1. May 22, 2007

bernhard.rothenstein

Consider two photons moving in the same direction and in opposite ones. Please let me know what is the mass of the system of two photons in the two mentioned cases.

2. May 22, 2007

Staff: Mentor

What is your definition of "mass" for this situation?

3. May 22, 2007

anantchowdhary

if it is rest mass.....its ZERO

4. May 22, 2007

lightarrow

It's a Very good question!
It would seem that when they travel in opposite directions, the system's mass is not zero:
E^2 = (cp)^2 + m^2
but p = 0 so m =/= 0.

Last edited: May 22, 2007
5. May 22, 2007

Chris Hillman

Don't forget that momentum is a vector. But what does this really tell you? (Hint: do these "photons" neccessarily have the same momentum?)

6. May 22, 2007

lightarrow

Yes, this depends on the ref. frame in which we observe the two photons. In a ref. frame, p = 0, in another one, p can have every value we want.
However I can't understand; the mass m would seem to be = 0 only in the ref. frame where p = 0, but m = 0 in all the others. What does it mean?

Edit. I intended: "the mass m would seem to be =/= 0 only in the ref. frame where p (of the system) = 0; of course it means p1 = - p2.

Last edited: May 23, 2007
7. May 22, 2007

bernhard.rothenstein

mass of a system of photons

Rest mass!

8. May 22, 2007

smallphi

P1 = 4-momentum of first photon = (c|p1|, 3vector p1)
P2 = 4-momentum of second photon = (c|p2|, 3vector p2)

Total 4-momentum is P = P1 + P2 = (c(|p1|+|p2|), 3vector p1+p2).

By definition, the total mass^2 of the system is

M^2 = P.P = c^2 (|p1|+|p2|)^2 - 3vector(p1+p2) . 3vector(p1+p2)
(scalar product for the 3vector)

M is the same in any inertial coordinate system, i.e. invariant.

9. May 22, 2007

pervect

Staff Emeritus

A little bit of context is needed. If two photons are moving in opposite directions, I assume we are supposed to assume that the total momentum is zero, and that if they are moving in the same direction, that the total momentum is equal to twice the momentum of a single photon.

This assumption is true in SR, for instance. Since that's where the term 'rest mass' is defined, I assume that's the case you are considering.

So, making this clarification, it should be quite clear that for the system where the photons are moving in the same direction, E^2 - |p|^2 = 0 for each individual photon, and therefore the rest mass (invariant mass) of the system , which in geometric units is simply
$$\sqrt{(2E)^2 - (2|p|)^2} = 2 \sqrt{E^2 - |p|^2} = 0$$

because the total energy of the system is twice the energy of each photon, and the magnitude of the total momentum of the system is also twice that of an individual photon.

Similarly, when the photons are moving in the opposite direction, the total momentum of the system is zero, so the mass of the system is 2E (in geometric units), or 2E/c^2 (in standard units), because

$$\sqrt{(2E)^2 - 0} = 2E$$

This sort of problem can be found in several textbooks, including Taylor & Wheeler's "Spacetime physics.

Last edited: May 22, 2007
10. May 22, 2007

Chris Hillman

'Nuff said

I was trying to get you to discover for yourself what pervect just told you. I think this topic is now exhausted!

11. May 22, 2007

pervect

Staff Emeritus
There will never be a case where p=0 for two photons moving in the same direction.

Therefore I assume you're talking about the case where the photons move in the opposite direction.

If you actually perform the calculation, you'll find that the total mass of the system of two photons is an invariant. Thus, for two photons moving in opposite directions, it never goes to zero, and is in fact always 2E_0, where E_0 is the energy of the photons in the particular frame where the momentum is zero.

If you consider a different frame, you'll get something (using geometric units) like:

E = k E_0 + (1/k) E_0

p = K p_0 - (1/k) p_0

where E, p are the energy and momentum of the system in the new frame. The factor 'k' here is the relativistic doppler shift factor. When k=1, p=0, but when k is not 1, p is not zero.

Note that it's always true that |E| = |p| for a photon, so E_0 = p_0.

If we calculate E^2 - p^2 from the above formula we get

(k^2 + 2 + 1/k^2)E_0^2 - (k^2 -2 + 1/k^2)p_0^2

(k^2 + 1/k^2) (E_0^2 - p_0^2) + 2 (E_0^2 + p_0^2)

Because E_0 = p_0, this simplifies to 4 E_0^2, so the square root is 2E_0, as before, and independent of the doppler shift factor k.

12. May 22, 2007

Chris Hillman

Actually, it was never clear to me whether or not Lightarrow recognizes the distinction between momentum of a system of two particles and the momenta of the individual particles. That might be part of the problem.

13. May 22, 2007

bernhard.rothenstein

mass of a system of photons

Thank you for the two illuminating answers. Please give me a good reason for the fact that the starting equation holds in the case of a photon. Can I say that E=g(V)[E'+Vp'] holds in the case of a photon?
That is a question not a statement!

14. May 22, 2007

pervect

Staff Emeritus
I don't quite understand the question - what do your variable names signify?

15. May 23, 2007

MeJennifer

I am not, it is not at all that simple.

While this is might be as easy as apple pie for you, I can imagine that some might have trouble in understanding that while the rest mass of a system of two photons moving in the same direction is zero the rest mass of a system of two photons moving away from each other is not!

Since the two photons have energy they do form a combined gravitational field. Wheeler's geons come to mind here.

By the way, it is really better to use 4-vectors to describe these systems, as soon as we have more than two components who move at different angles it is ludicrous to keep working with 3-vectors, as precessions make in increasingly complicated.

Last edited: May 23, 2007
16. May 23, 2007

blumfeld0

17. May 23, 2007

lightarrow

Of course I recognize it! My problem was I didn't recognize that mass IS NOT additive, that is, m1 + m2 IS NOT equal to the mass of the system.

Last edited: May 23, 2007
18. May 23, 2007

bernhard.rothenstein

system of two photons

I try to restate my question:
Give me please a reason for the fact that the formula
E=csqrt(pp+mmcc)
holds in the case of a photon as well.
Does the formulae (energy and momentum transformations)
E=g(V)[E'+Vp']
and
p=g(V)[p'+VE'/cc]
hold in the case of a photon?

19. May 23, 2007

Ich

E=cp can be derived from classical electrodynamics; you don't have to derive it from Lorentz-transformations with v=c, if that is what bothers you.
The transformation equations hold for photons as well.

20. May 23, 2007

Voltage

Bernard, perhaps I can rephrase the answer another way. Apologies if this is under your bar:

A photon is considered to have zero mass, because mass is commonly understood to mean "inertial mass". However it does have energy/momentum, and energy is commonly understood to relate to "relativistic mass".

Both energy and mass are scalar quantities, whilst momentum is a vector quantity. Your measure of momentum takes account of its direction. This isn't true of energy or mass. If a photon changes direction, your measurement of its momentum will change, but your measurement of its energy will not.

Hence when you're looking at system comprising two photons, the energy content and the "relativistic mass" is not affected by direction of travel. This also applies to inertial mass.

To consider inertial mass, you need to look at the system as a whole. You can do this by enclosing your photons in a mirrored box. The photons bounce back and forth from one side to another. Because the photons have a relativistic mass, the inertial mass of the box is increased by the presence of the photons. But it does not oscillate when one photon bounces of the inside of the box and now starts heading back towards the other photon.

See Navier-Stokes for further information.

Last edited: May 23, 2007
21. May 23, 2007

Staff: Mentor

I agree that the quantity

$$\sqrt {E_{total}^2 - (|{\vec p}_{total}| c)^2}$$

is nonzero for two photons, in general (unless they are moving in the same direction). But is there any situation in which this quantity is useful for describing the behavior of an unbound collection of particles? Operationally, how does one "weigh" such a collection, or measure this quantity, besides calculating it from the individual energies and momenta as shown above?

Experimental high-energy particle physicists often calculate the "invariant mass" of a collection of particles emerging from an interaction, to determine whether they might have resulted from the decay of some other particle. But this calculation is always aimed at determining the properties of the initial particle. It doesn't have any significance for the outgoing particles viewed in isolation, as far as I know.

22. May 23, 2007

bernhard.rothenstein

two photons

Thanks for your answer. What bothers me is a good explanation for the fact that an equation that accounts for the behaviour of a tardyon accounts for the behaviour of a photon as well.
Consider please the equation which accounts for the transformation of the energy of a tardyon presented as
E=g(V)[E'+Vp']=g(V)E'[1+Vu'/cc]
where u' represents the O'X' component of the speed of the tardyon in I'.
E=g(V)E'(1+Vc]
that accounts for the behaviour of a photon. The question is why.
Please consider that what I say is a question and not a statement!
Hochachtungsvoll
Bernhard

23. May 23, 2007

Ich

I'm not sure I get the point. E² = m² + p² is valid anyway, as are the transformation formulas.
E is always defined, even for v=c. You simply plug in the value you like. It's only gm that is not defined. As long as you don't use it, there will be no problem.
So by defining p=uE/c² instead of p=gmv, you only rely on defined (known) parameters.
Maybe there are some philosophical (or even physical) implications that I'm not aware of.

Ergebenst
Ich

24. May 23, 2007

pervect

Staff Emeritus
I think you're asking: giving a 1+1 energy-momentum vector (E,p) we have (in geometric units):

$$E = \gamma(E' + v p') \hspace{.5 in} p = \gamma(p' + v E')$$

where $$\gamma = \sqrt{1-v^2}$$

$$E^2 - p^2 = \gamma^2(E' + v p')^2 - \gamma^2(p' + v E')^2 = \gamma^2 (E'^2 + v^2 p'^2 - p'^2 - v^2 E'^2) = \gamma^2 (1-v^2) (E'^2 - p'^2) = E'^2 - p'^2$$

You can also work it out including the factors of c if you must use standard units.

This is what is meant by saying that mass is the invariant of the energy-momentum 4 vector. We've only used 1 space + 1 time dimension in this example, so our 4-vector is actually a 2 vector as we have suppressed the other two spatial dimensions.

Last edited: May 23, 2007
25. May 23, 2007

pervect

Staff Emeritus
I think that the application you do mention, finding the mass of particles from their byproducts, is enough to give the idea an operational meaning and usefulness.

For instance, it shows that the mass of a positron and an electron, considered as a system, doesn't change when the positron annihilates the electron to form a pair of photons.