Mass of a system of two photons

[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.
Thanks in advance

 

[jtbell]

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

 

[anantchowdhary]

if it is rest mass...its ZERO

 

[lightarrow]

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

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.

 

[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?)

 

[lightarrow]

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

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.

 

[bernhard.rothenstein]

mass of a system of photons

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

Rest mass!

 

[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.

 

[pervect]

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.

 

[Chris Hillman]

'Nuff 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?

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

 

[pervect]

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?

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.

 

[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.

 

[bernhard.rothenstein]

mass of a system of 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.

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!

 

[pervect]

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

 

[MeJennifer]

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

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.

 

[blumfeld0]

i learned a lot by reading this
http://www.mathpages.com/home/kmath232/kmath232.htm

5th paragraph from the bottom onwards

 

[lightarrow]

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.

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.

 

[bernhard.rothenstein]

system of two photons

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

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.

 

[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.

 

[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.

 

[jtbell]

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.

 

[bernhard.rothenstein]

two photons

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.

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'.
For u'=c it leads to
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

 

[Ich]

The question is why.

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

 

[pervect]

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.

I think you're asking: giving a 1+1 energy-momentum vector (E,p) we have (in geometric units):<br /> E = \gamma(E&#039; + v p&#039;) \hspace{.5 in}<br /> p = \gamma(p&#039; + v E&#039;)<br />

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

The answer to your question is: just square and add

<br /> E^2 - p^2 = \gamma^2(E&#039; + v p&#039;)^2 - \gamma^2(p&#039; + v E&#039;)^2 = \gamma^2 (E&#039;^2 + v^2 p&#039;^2 - p&#039;^2 - v^2 E&#039;^2) = \gamma^2 (1-v^2) (E&#039;^2 - p&#039;^2) = E&#039;^2 - p&#039;^2<br />

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.

 

[pervect]

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.

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.

 

[bernhard.rothenstein]

two photons

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


<br /> E = \gamma(E&#039; + v p&#039;) \hspace{.5 in}<br /> p = \gamma(p&#039; + v E&#039;)<br />

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

The answer to your question is: just square and add

<br /> E^2 - p^2 = \gamma^2(E&#039; + v p&#039;)^2 - \gamma^2(p&#039; + v E&#039;)^2 = \gamma^2 (E&#039;^2 + v^2 p&#039;^2 - p&#039;^2 - v^2 E&#039;^2) = \gamma^2 (1-v^2) (E&#039;^2 - p&#039;^2) = E&#039;^2 - p&#039;^2<br />

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 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.
1. Do the formulas we have for the tardyon hold in the case of a photon?
If yes, how could we explain that?
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?
Please consider that all above are questions (facts) and not statements.
g(V)=1/sqrt(1-VV/cc)

 

[pmb_phy]

if it is rest mass...its ZERO

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.


The web page I created here http://www.geocities.com/physics_world/sr/mass_energy_equiv.htm
demonstrates all the math involved.


Pete

 

[lightarrow]

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?

This is very interesting. You made me think about it but.
Edit: probably you have "rediscovered" one of the facts that De Broglie used to justify the introduction of matter waves.
The only thing it came to my mind (but I think there must be much more than just this) is:
E = hf
p = h/l
with f = frequency; l = wavelength
are true for matter particles as well as for photons
So, E/p = f*l and the last one is the phase velocity of the wave.
On the other hand, E^2 = (cp)^2 + (mc^2)^2 which is true for every system. If we put m = 0 in the last equation, we have: E = cp --> E/p = c.
This means that the phase velocity must be = c.
So, if we assume (from QM) that tardions have an associated wave, we can conclude that, in the limit for m --> 0, the wave must move with phase velocity = c.
What does it also imply?

 

[country boy]

Try working in terms of the frequencies of the two photons. In the case of oppositely moving photons, apply a doppler shift to get the frame in which the two frequencies are the same. The momenta are then equal and opposite and cancel. This is therefore the rest frame of system. The sum of the energies in this frame gives the rest mass.

This cannot be done with photons moving in the same direction, because you can't find a frame in which the two have the same frequency.

 

[bernhard.rothenstein]

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.

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.
Consider that scenario from another inertial reference frame. The frequencies of the two photons are differently affected by the Doppler Effect and 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?

 

[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&gt;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