Two different velocity variables in relativistic momentum?

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 4K views
SamRoss
Gold Member
Messages
256
Reaction score
36
For the relativistic momentum

p=(1/[1-u^2/c^2])mv

does u always equal v or does u refer to the speed of the reference frame and v refer to the speed of the object?
 
Physics news on Phys.org
SamRoss said:
For the relativistic momentum

p=(1/[1-u^2/c^2])mv

does u always equal v or does u refer to the speed of the reference frame and v refer to the speed of the object?

Where does this come from? There is only one velocity involved in this formula. I could guess and say some author is using u for speed (magnitude of velocity) and v for the velocity vector. If so, it is correct and self explanatory.
 
The separation of the single velocity variable into two variables came from me trying to fill in some logical gaps in some derivations I was looking at and I guess I didn't do such a great job.

My real problem is this. I want to find a simple derivation of relativistic momentum that does NOT involve the ambiguous relativistic mass quantity. Every supposed derivation that I look at turns out really to be a derivation of relativistic mass and I have come to understand that this is a dangerous concept. Do you know of any simple derivations of relativistic momentum using only invariant mass?
 
samross said:
the separation of the single velocity variable into two variables came from me trying to fill in some logical gaps in some derivations i was looking at and i guess i didn't do such a great job.

My real problem is this. I want to find a simple derivation of relativistic momentum that does not involve the ambiguous relativistic mass quantity. Every supposed derivation that i look at turns out really to be a derivation of relativistic mass and i have come to understand that this is a dangerous concept. Do you know of any simple derivations of relativistic momentum using only invariant mass?

samross, I've attached a couple of Microsoft Word files (the second is a continuation of the first). The idea is to give you a presentation that includes space-time diagram graphics to help keep track of the vector components. In this display I've used two observers moving with the same speed in opposite directions with respect to a rest system. By doing this the distances can be directly compared without worrying about hyperbolic proper time curves.

View attachment Four_Momentum1_Forum.doc

View attachment Four_Momentum2_Forum.doc
 
Hi,

The four velocity is the tangent to the world line, expressed
[tex] u^\alpha =\frac{d x^\alpha}{d\tau}[/tex]
where [itex]\tau[/itex] is the proper time. This decomposes as
[tex] u^\alpha =(\gamma,\gamma \, \vec{v})[/tex]
Momentum is defined as
[tex] p^\alpha =mu^\alpha[/tex]
Here [itex]m[/itex] is the body's rest-mass.
 
Thanks for the quick reply. The Minkowski method has always been a bit less intuitive for me than the kinematic equations themselves but I am trying to understand the document. In the meantime I've posted a related topic titled Help with Relativistic Collision. I feel that somewhere in those collision thought experiments is a line of reasoning that's at my level. Thanks again.