Mass of a system of two photons

[Meir Achuz]

performing the transformation I arrive at the conclusion that the resultant rest mass of the system is equal to zero. Is that a paradox? How could I come out of it?

It is not a paradox. It would just be that you made a mistake in the transformation. If done correctly, M^2 is invariant.

 

[bernhard.rothenstein]

mass of two photons

It is not a paradox. It would just be that you made a mistake in the transformation. If done correctly, M^2 is invariant.

Thanks.
Let I be the inertial reference frame in which the frequencies of the two photons are equal to each other. Let I' be a reference frame which moves relative to K with speed V relative to K. In the case when the two photons move in the same direction theirs frequencies are affected in the same way by the Doppler Effect and so p'=0 and
m'(0)=0. If the photons move in opposite directions m'(0)=2E
In the case when the two photons move in opposite directions the frequencies of the two photons are differently affected by the Doppler Effect. With k(+)=sqrt((1+b)/(1-b) , k(-)=sqrt((1-b)/(1+b)) , b=V, c=1
m'(0)=sqrt{EE[k(+)+k(-)]-pp[k(+)-k(-)]}=0.
Please let me find out where is the error.

 

[pervect]

Thanks.
Let I be the inertial reference frame in which the frequencies of the two photons are equal to each other. Let I' be a reference frame which moves relative to K with speed V relative to K. In the case when the two photons move in the same direction theirs frequencies are affected in the same way by the Doppler Effect

yes

and so p'=0 and

No.

p = p', but with both photons moving in the same direction, the total momentum is p+p' (they add, not subtract). This means the total momentum is always greater than zero, just as it is for a single photon of nonzero frequency.

In fact, one photon of twice the frequency has the same energy and momentum of two photons of half the frequency moving in the same direction.

 

[bernhard.rothenstein]

mass of two photons

yes

No.

p = p', but with both photons moving in the same direction, the total momentum is p+p' (they add, not subtract). This means the total momentum is always greater than zero, just as it is for a single photon of nonzero frequency.

In fact, one photon of twice the frequency has the same energy and momentum of two photons of half the frequency moving in the same direction.

Thanks. It seems that I have discovered the error in my calculus.
Please tell me if you agree with the following statements;
1. If the two photons move in the same direction theirs resultant (rest) mass is equal to zero in all inertial reference frames in relative motion.
2. If the two photons move in opposite directions theirs resultant (rest) mass is equal to 2E/cc in all inertial reference frames in relative motion where E
represents the energy of one of the involved photons in the reference frame where the frequencies of the two photons equate each other and which can be arbitrarily choosen.
So there is no paradox and big Albert is right again.

 

[bernhard.rothenstein]

mass of two photons

I'm surprised there is so much confusion about this question.

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.

Following that hint the result is that the same results hold in all inertial reference frames in relative motion if E is the mass of a single photon in the reference frame where theirs enegies are equal to each other, Please confirm.

 

[bernhard.rothenstein]

mass of two photons

It is not a paradox. It would just be that you made a mistake in the transformation. If done correctly, M^2 is invariant.

Thanks. You are right. Did you see the problem treated somewhere in the literature (the problem of two photons detected from different inertial reference frames in relative motion)?

 

[masudr]

...the distinction between momentum of a system of two particles and the momenta of the individual particles.

I'm not sure I see the distinction; surely the 4-momentum of a system of particles is simply the sum of the 4-momenta of the particles that make up the system?

If this is the case, then what is the distinction between the two?

If this is not the case, then why isn't it? And what is incorrect?

 

[MeJennifer]

I'm not sure I see the distinction; surely the 4-momentum of a system of particles is simply the sum of the 4-momenta of the particles that make up the system?

If this is the case, then what is the distinction between the two?

If this is not the case, then why isn't it? And what is incorrect?

Addition of two 4-vectors in a Minkowski spacetime often produce an unintuitive result. Each separate photon has no mass, but in some cases a system of photons does have mass. Localizing that mass is an entirely different but very interesting question.

One thing to keep in mind is that relativity is a classical theory. I think relativity theory really does not model photons very well. This is even more obvious in general relativity. Einstein, unlike Weyl was never comfortable with modeling matter in terms of complex wave mechanics.

 

[masudr]

Addition of two 4-vectors in a Minkowski spacetime often produce an unintuitive result.

Is there or isn't there a distinction? And if so, then what is it?

 

[MeJennifer]

Is there or isn't there a distinction? And if so, then what is it?

They are different.

In case of two photons going in opposite directions each one has an individual mass of 0. But when you add both photons together you get something like 0 + 0 > 0. This is due to the non-Euclidean metric of spacetime.

 

[bernhard.rothenstein]

mass of two photons

They are different.

In case of two photons going in opposite directions each has one has an individual mass of 0. But when you add both photons together you get something like 0 + 0 > 0. The reason is related to the metric of spacetime.

Has that fact some influence on the discussion about the opportunity to use the concept of relativistic mass. That is only a question and not a statement! Thanks for the answer.

 

[MeJennifer]

Has that fact some influence on the discussion about the opportunity to use the concept of relativistic mass. That is only a question and not a statement! Thanks for the answer.

I would think so.

Personally I do not agree with this "grand objection" expressed by the prevailing winds here on PF against the usage of relativistic mass. In the end it is simply a matter of terminology. As long as one understands the difference between relativistic and proper mass I see no issue.

 

[jostpuur]

Addition of two 4-vectors in a Minkowski spacetime often produce an unintuitive result. Each separate photon has no mass, but in some cases a system of photons does have mass.

Energy of a system of particles is a sum of particles' energies, and the momentum is also a sum of the momentums. But the mass of the system is not a sum of the masses of the particles. Since $$m^2=p_\mu p^\mu$$, thinking with four-vectors, this is nearly equivalent to saying that a norm of a sum of vectors is not a sum of the norms. That is, $$||x+y||\neq ||x||+||y||$$ in general. I don't think this is unintuitive. Improve your intuition on the matter

The inequality $$0+0>0$$ doesn't seem very related to this matter. You are not supposed to add masses, or other norms of vectors, in the first place.

 

[MeJennifer]

Energy of a system of particles is a sum of particles' energies, and the momentum is also a sum of the momentums.

Note that energy and momentum are not Lorentz invariant properties, instead they are covariant.

 

[Chris Hillman]

Time to lock this thread?

I'm not sure I see the distinction; surely the 4-momentum of a system of particles is simply the sum of the 4-momenta of the particles that make up the system?

If this is the case, then what is the distinction between the two?

If this is not the case, then why isn't it? And what is incorrect?

This thread was always terribly confused, and unfortunately it seems that my half-hearted attempt to nudge in a more fruitful direction failed. In fact, although some participants are in my "ignore list", I have the impression it has descended even further into the murk since my own contribution. Masudr, you must have misunderstood what I tried to tell someone (I don't remember the circumstances of the post you quoted), but I don't think it would be worthwhile to pursue this, given the degree of confusion in this thread.

 

[pmb_phy]

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.

Yep. I seemed very clear to me that you recognize that fact. The sum of the invariant masses is not additive, only the sum of the inertial (aka relativistic mass) is additive (and conserved). This conservation can be proved with no use of the conservation of energy, All that is required is that momentum is conserved in all inertial frames.

Pete

 

[bernhard.rothenstein]

mass of two photons

Yep. I seemed very clear to me that you recognize that fact. The sum of the invariant masses is not additive, only the sum of the inertial (aka relativistic mass) is additive (and conserved). This conservation can be proved with no use of the conservation of energy, All that is required is that momentum is conserved in all inertial frames.

Pete

Please let me know how "all that is required is that momentum is conserved in all inertial frames." Your answer shows again that not using the concept of relativistic mass some confusion can arrise.
Thanks

Bernhard

 

[rbj]

Personally I do not agree with this "grand objection" expressed by the prevailing winds here on PF against the usage of relativistic mass. In the end it is simply a matter of terminology. As long as one understands the difference between relativistic and proper mass I see no issue.

reading this is refreshing (to me, at least).

 

[bernhard.rothenstein]

mass

reading this is refreshing (to me, at least).

for me too

 

[pmb_phy]

Please let me know how "all that is required is that momentum is conserved in all inertial frames." Your answer shows again that not using the concept of relativistic mass some confusion can arrise.
Thanks

Bernhard

you can find the derivation on y we site at

http://www.geocities.com/physics_world/sr/conservation_of_mass.htm

Best wishes Bernhard

Pete

 

[pmb_phy]

Note that energy and momentum are not Lorentz invariant properties, instead they are covariant.

Yay for MJ.

Not too many people know of that use of the term "covariant." Bravo to MJ

Pete

 

[pmb_phy]

reading this is refreshing (to me, at least).

Me too.

Pete

 

[masudr]

Note that energy and momentum are not Lorentz invariant properties, instead they are covariant.

Has anyone ever claimed anything contrary to that?