Derivation of the Equation for Relativistic Momentum

  • Thread starter NanakiXIII
  • Start date
392
0

Main Question or Discussion Point

I asked a quite similar question about relativistic mass and the reason for this question is identical: I can't seem to dig up any derivation for the equation for relativistic momentum:

[tex]p=\gamma mv[/tex]

If anyone could point me in the right direction, I'd much appreciate it.
 

Answers and Replies

329
0
I asked a quite similar question about relativistic mass and the reason for this question is identical: I can't seem to dig up any derivation for the equation for relativistic momentum:

[tex]p=\gamma mv[/tex]

If anyone could point me in the right direction, I'd much appreciate it.
Because this is a definition. And so is the relativistic total energy:


[tex]E=\gamma mc^2[/tex]
 
392
0
I see. Why is it defined as such, exactly? There must be a reason for adding the so frequently seen gamma, no?

Another question on the side: I've seen two different versions of the equation for the mass-energy equivalence, one with and one without the gamma factor. What's the difference?
 
329
0
I see. Why is it defined as such, exactly? There must be a reason for adding the so frequently seen gamma, no?
I explained that to you in the other thread (relativisic mass) a few minutes ago.


Another question on the side: I've seen two different versions of the equation for the mass-energy equivalence, one with and one without the gamma factor. What's the difference?
Same reason as momentum, I also explained that to you in the other thread.
 
392
0
You mentioned the following:

The whole darned thing was introduced in order to reconcile the relativistic momentum/energy:

[tex]p=\gamma m(0)v[/tex] (1)
[tex]E=\gamma m(0)c^2[/tex]

with the Newtonian counterpart:


[tex]p=mv[/tex] (2)

So the best thing is to tell your teacher that your proof is you grouped together [tex]\gamma[/tex] and proper mass m(0) into [tex]\gamma m(0)[/tex] and you assigned that quantity to m
You mean the reason for the gamma is because of the use of relativistic mass? And yet in the other thread you tell me to base relativistic mass on relativistic momentum. That's circular. Are you saying that relativistic mass is a definition and thus relativistic momentum is as well? That would still leave the question of "why the gamma?".
 
329
0
You mentioned the following:



You mean the reason for the gamma is because of the use of relativistic mass? And yet in the other thread you tell me to base relativistic mass on relativistic momentum. That's circular.
"Relativistic mass" is just a misnomer, ok? Try to learn to forget about it, it has no meaning. Tolman's derivation from the other thread has no "relativistic momentum" in it, ok?


Are you saying that relativistic mass is a definition and thus relativistic momentum is as well? That would still leave the question of "why the gamma?".
Can't you read ? Relativistic momentum is a definition.
"Relativistic mass" is an unfortunate misnomer that corresponds to the result of multiplying [tex]\gamma[/tex] by proper mass. It has no physical meaning.
 
robphy
Science Advisor
Homework Helper
Insights Author
Gold Member
5,399
676
Given the 4-momentum vector [tex]\tilde P [/tex],
an inertial observer (with unit 4-velocity [tex]\tilde t[/tex]
can write that vector as the sum of two vectors, a "temporal" one parallel to [tex]\tilde t[/tex] and a "spatial" one perpendicular to [tex]\tilde t[/tex].
[tex]\gamma[/tex] is [tex]\cosh{\theta}[/tex], the analogue of cosine(angle) , and [tex]\gamma\beta[/tex] is [tex]\sinh{\theta}[/tex], the analogue of sine(angle), where the [Minkowski-]angle [tex]\theta[/tex] (called the rapidity) is between [tex]\tilde P [/tex] and [tex]\tilde t [/tex].
 
jtbell
Mentor
15,402
3,190

Related Threads for: Derivation of the Equation for Relativistic Momentum

  • Last Post
Replies
5
Views
3K
Replies
78
Views
55K
  • Last Post
Replies
6
Views
8K
Replies
5
Views
3K
Replies
13
Views
15K
Replies
4
Views
2K
  • Last Post
Replies
6
Views
2K
Top