I Relativistic Mass Calculation with 4th Dimensional Vector

[afeser]

I have a sample problem :
Photon, with energy E, is absorbed by m. Then, mass m becomes m' (i think it's relativistic mass). Calculate the new mass.

My teacher solved the problem as follows;

P₁=( E/c, P_3dphoton); // momentum of the photon
P₂=(mc, 0); // momentum of the stationary mass
Length=((P₁+P₂)|(P₁+P₂)); // which is to be conserved

=(P₁|P₁)+2*(P₁|P₂)+(P₂|P₂)
=0+2*E*m+m²c²; // (P₁|P₁) is 0
=m'²*c'²; // since it is invariant
Therefore:
m'=m*(1+2E/mc²)^1/2
​

But my solution is a bit different because I think we should only sum up the 3 dimensional values, and the relativistic mass will be γ*m where γ is 1/(1-(v/c)²)^1/2

P_initial=(m*c, 0, 0, 0)
P_final=(γ*m*c, P_3dphoton)
and (P_initial|P_initial)=(P_final|P_final) because it's invariant
Also, since (P_photon|P_photon)=0,
we have (P_3dphoton|P_3dphoton)=(E/c)²,
which implies γ=(1+E²/(m*c²)²)
then we have m'=m*γ=m*(1+E²/(m*c²)²)^1/2 which is obviously different from the first solution.

​

Which solution is correct? I think the first, of course, because we conserve the 4th dimensional vector, however claiming (m'c)² is invariant makes me uncomfortable because that means P_final=(γ*m'*c, P_x, P_y, P_z), where γ*m'*c factor should be E/c (as I learned from Wikipedia), but it's not.(m'=γ*m -> γ*m'*c=γ²*m'*c, which is not E/c=γ*m*c)

Please help me! What am I missing?

Thanks a lot...

 

[Orodruin]

(i think it's relativistic mass)

No it isn't. Relativistic mass is not a very useful concept, it hangs around because of historical context and popular scientific renditions. When learning relativity you would be better off forgetting you ever saw it. It essentially is just another name for total energy (after letting c = 1).

m'2*c'2; // since it is invariant

Obviously therefore referring to the invariant mass.

 

[Dale]

Then, mass m becomes m' (i think it's relativistic mass)

This is incorrect. In this problem m is the invariant mass, not the relativistic mass. You are getting the wrong answer because you misunderstood the question. Work the question where m is the invariant mass and you should get the correct answer.

 

[afeser]

No it isn't. Relativistic mass is not a very useful concept, it hangs around because of historical context and popular scientific renditions. When learning relativity you would be better off forgetting you ever saw it. It essentially is just another name for total energy (after letting c = 1).Obviously therefore referring to the invariant mass.

Thank you, I got it but now I have one more question.
We said absorbing a photon increases the mass, okey... but if electromagnetic interaction changes the mass(as in centre of mass experiment), then accelerating a particle also changes it since photons are the carrier of the electromagnetic interaction(the force we use). Then, how do we approach to the invariant mass if it varies every time we give an energy?

Also, in the example above why did we use P_final=(γ*m'*c, Px, Py, Pz) instead of P_final=(γ*m*c, Px, Py, Pz). Didn't we count Lorentz factor 2 times(first when calculating m' and second in the equation)?

 

[Orodruin]

then accelerating a particle also changes it

No it doesn't. It is unclear to me how you reached this conclusion.

Also, in the example above why did we use Pfinal=(γ*m'*c, Px, Py, Pz) instead of Pfinal=(γ*m*c, Px, Py, Pz). Didn't we count Lorentz factor 2 times(first when calculating m' and second in the equation)?

Again, $m'$ is not the relativistic mass. It is the invariant mass after the photon absorption.

 

[afeser]

No it doesn't. It is unclear to me how you reached this conclusion.

Because, as far as I know, photons are force carriers, they are sent between two charged particles when they're very close(such as electrons). That is, pushing something means electromagnetic interaction which requires photon transfer.

If photon-mass interaction creates mass, then why electromagnetic interaction does not?

 

[Orodruin]

Because, as far as I know, photons are force carriers, they are sent between two charged particles when they're very close(such as electrons). That is, pushing something means electromagnetic interaction which requires photon transfer.

If photon-mass interaction creates mass, then why electromagnetic interaction does not?

This is a grossly over-popularised argumentation. In general, popular science is only good for you to learn about physics, not to learn actual physics and basing any kind of conclusion on this sort of argumentation is just bound to get you into trouble.

In quantum field theory, electromagnetism may be seen as being mediated by virtual photons, which do not generally obey the on-shell relation between energy and momentum. However, even in that picture, you can never talk about actual photons being exchanged. The real process will be a linear combination of all possible processes that give you the same final state, the situation is not as if the two charged particles were throwing small balls between them.

What you are talking about when you are talking about absorption of a photon is the absorption of a real photon. To be honest, until you are ready to learn the full quantum field theory treatment, you are better off thinking about electromagnetic interaction in terms of its classical theory. Being a gauge theory, quantising electromagnetism is significantly more involved than basic quantum physics.

 

[Dale]

if electromagnetic interaction changes the mass(as in centre of mass experiment), then accelerating a particle also changes it since photons are the carrier of the electromagnetic interaction(the force we use).

This is not generally true. If a photon scatters off an atom without changing its internal state, then the atom will accelerate with no change in invariant mass. If a photon is absorbed then the internal structure of the atom will change (an atomic electron is raised to an excited state) and there will be a change in mass.

This is one reason why an isolated fundamental particle, like an electron, cannot absorb a photon only scatter it.

 

[Herman Trivilino]

Then, how do we approach to the invariant mass if it varies every time we give an energy?

Invariant doesn't mean "unable to vary". It means that the value is the same for all observers.

Photon, with energy E, is absorbed by m. Then, mass m becomes m'.

I suppose that it's clear from the context that there is some object of mass $m$ that absorbs a photon and afterwards has a mass $m'$. This object might be an atom. It's a bad habit to refer to an object as being $m$ because the value of $m$ is a property of the object, it's not the object itself.

My point is that all observers will agree on the values of $m$ and $m'$ regardless of how fast they are moving relative to the object. In other words, their value doesn't vary from frame to frame. But the object's mass does vary from $m$ to $m'$.

 

[afeser]

Thank you all, finally understand.

Just realized i have a lot to learn :)