- #1
bernhard.rothenstein
- 991
- 1
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.
Thanks in advance
Thanks in advance
It's a Very good question!bernhard.rothenstein said: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.
Thanks in advance
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.Chris Hillman said:Don't forget that momentum is a vector. But what does this really tell you? (Hint: do these "photons" neccessarily have the same momentum?)
jtbell said:What is your definition of "mass" for this situation?
lightarrow said: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?
lightarrow said: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?
pervect said: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
[tex]\sqrt{(2E)^2 - (2|p|)^2} = 2 \sqrt{E^2 - |p|^2} = 0[/tex]
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
[tex]\sqrt{(2E)^2 - 0} = 2E[/tex]
This sort of problem can be found in several textbooks, including Taylor & Wheeler's "Spacetime physics.
bernhard.rothenstein said: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!
I am not, it is not at all that simple.pervect said:I'm surprised there is so much confusion about this question.
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.Chris Hillman said: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.
I try to restate my question:pervect said:I don't quite understand the question - what do your variable names signify?
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.Ich said: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.
I'm not sure I get the point. E² = m² + p² is valid anyway, as are the transformation formulas.The question is why.
bernhard.rothenstein said: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?
Thanking in advance for your answer which is of imporrtancce for me.
jtbell said:I agree that the quantity
[tex]\sqrt {E_{total}^2 - (|{\vec p}_{total}| c)^2}[/tex]
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.
I try to state my problem. We speek about tardyons and photons as particles characterized by energy and momentum. We derive transformation equations for energy and momentum in the case of a tardyon and in the case of a photon as well.pervect said:I think you're asking: giving a 1+1 energy-momentum vector (E,p) we have (in geometric units):
[tex]
E = \gamma(E' + v p') \hspace{.5 in}
p = \gamma(p' + v E')
[/tex]
where [tex]\gamma = \sqrt{1-v^2}[/tex]
The answer to your question is: just square and add
[tex]
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
[/tex]
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.
I disagree on that point. Suppose you take a closed system of particles and define the "mass of the system", M, as the magnitude of the systems 4-momentum. A similar procedure is done with E = energy of the system, i.e. solve for E = sqrt[E^2-(pc)^2]. In such a case you can then determine the mass, M, by first determining the energy/mass of each particle. Then there are two possibilities. Either M' is zero or it is non-zero. If M is zero then the mass of the system is zero. If M is none-zero then the mass of then system has a mass of M' = gamma(v) M where gamma is the determined from the solving for gamma in M^2c^4 = sqrt[E^2-(pc)^2 for p. In the other case you would get the exact same but expressed as E = gama^2.anantchowdhary said:if it is rest mass...its ZERO
This is very interesting. You made me think about it but.bernhard.rothenstein said:2. I try since long time to explain for myself why starting with the transformation equation for the tardyon presented as
E=g(V)[E'+Vp']=g(V)E'[1+Vu'/cc] (u' the velocity of the tardyon in I')
leads for u'=c to
E=E'g(V)(1+V/c)
the last accounting for the transformation of the energy of a photon. Has that fact an explanation? Could we say that special relativity theory ensures a smooth transition from tardyon to photon?
I consider your hint. Consider an inertial reference frame in which two identical photons move i opposite directions. It is clear that the resultant momentum is equal to zero the total rest mass being different from zero.pervect said: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
[tex]\sqrt{(2E)^2 - (2|p|)^2} = 2 \sqrt{E^2 - |p|^2} = 0[/tex]
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
[tex]\sqrt{(2E)^2 - 0} = 2E[/tex]
This sort of problem can be found in several textbooks, including Taylor & Wheeler's "Spacetime physics.
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 said: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?
Thanks.Meir Achuz said:It is not a paradox. It would just be that you made a mistake in the transformation. If done correctly, M^2 is invariant.
yesbernhard.rothenstein said: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
No.and so p'=0 and
Thanks. It seems that I have discovered the error in my calculus.pervect said: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.
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.pervect said: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
[tex]\sqrt{(2E)^2 - (2|p|)^2} = 2 \sqrt{E^2 - |p|^2} = 0[/tex]
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
[tex]\sqrt{(2E)^2 - 0} = 2E[/tex]
This sort of problem can be found in several textbooks, including Taylor & Wheeler's "Spacetime physics.