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Derivation of the Equation for Relativistic Momentum 
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#1
Feb807, 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. 


#2
Feb807, 04:22 PM

P: 329

[tex]E=\gamma mc^2[/tex] 


#3
Feb807, 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 massenergy equivalence, one with and one without the gamma factor. What's the difference? 


#4
Feb807, 04:34 PM

P: 329

Derivation of the Equation for Relativistic Momentum



#5
Feb807, 04:42 PM

P: 392

You mentioned the following:



#6
Feb807, 05:53 PM

P: 329

"Relativistic mass" is an unfortunate misnomer that corresponds to the result of multiplying [tex]\gamma[/tex] by proper mass. It has no physical meaning. 


#7
Feb807, 06:03 PM

Sci Advisor
HW Helper
PF Gold
P: 4,137

Given the 4momentum vector [tex]\tilde P [/tex],
an inertial observer (with unit 4velocity [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]. 


#8
Feb807, 06:49 PM

Mentor
P: 11,782

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) 


#9
Feb807, 11:43 PM

P: 329




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