Register to reply

Derivation of the Equation for Relativistic Momentum

by NanakiXIII
Tags: derivation, equation, momentum, relativistic
Share this thread:
NanakiXIII
#1
Feb8-07, 04:16 PM
P: 392
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.
Phys.Org News Partner Science news on Phys.org
Law changed to allow 'unlocking' cellphones
Microsoft sues Samsung alleging contract breach
Best evidence yet for coronal heating theory detected by NASA sounding rocket
nakurusil
#2
Feb8-07, 04:22 PM
P: 329
Quote Quote by NanakiXIII View Post
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]
NanakiXIII
#3
Feb8-07, 04:26 PM
P: 392
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?

nakurusil
#4
Feb8-07, 04:34 PM
P: 329
Derivation of the Equation for Relativistic Momentum

Quote Quote by NanakiXIII View Post
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.
NanakiXIII
#5
Feb8-07, 04:42 PM
P: 392
You mentioned the following:

Quote Quote by nakurusil View Post
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?".
nakurusil
#6
Feb8-07, 05:53 PM
P: 329
Quote Quote by NanakiXIII View Post
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
#7
Feb8-07, 06:03 PM
Sci Advisor
HW Helper
PF Gold
robphy's Avatar
P: 4,120
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
#8
Feb8-07, 06:49 PM
Mentor
jtbell's Avatar
P: 11,632
Quote Quote by NanakiXIII View Post
I can't seem to dig up any derivation for the equation for relativistic momentum:

[tex]p=\gamma mv[/tex]
There are some leads in this thread from last year:

http://www.physicsforums.com/showthread.php?t=107361

(which I should have thought of looking for when I posted in your other thread about relativistic mass)
nakurusil
#9
Feb8-07, 11:43 PM
P: 329
Quote Quote by jtbell View Post
There are some leads in this thread from last year:

http://www.physicsforums.com/showthread.php?t=107361

(which I should have thought of looking for when I posted in your other thread about relativistic mass)
There is an even better one here.


Register to reply

Related Discussions
Derivation of the Equation for Relativistic Mass Special & General Relativity 78
Derivation of relativistic acceleration and momentum Special & General Relativity 13
Relativistic Doppler Shift Equation - Derivation Introductory Physics Homework 7
Non-relativistic derivation of Schwarzschild's equation Special & General Relativity 8
Derivation of Momentum Equation (Eulers Equation) Differential Equations 0