# Two different velocity variables in relativistic momentum?

1. Aug 16, 2011

### SamRoss

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?

2. Aug 16, 2011

### PAllen

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.

3. Aug 16, 2011

### SamRoss

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?

4. Aug 16, 2011

### bobc2

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

5. Aug 16, 2011

### jfy4

Hi,

The four velocity is the tangent to the world line, expressed
$$u^\alpha =\frac{d x^\alpha}{d\tau}$$
where $\tau$ is the proper time. This decomposes as
$$u^\alpha =(\gamma,\gamma \, \vec{v})$$
Momentum is defined as
$$p^\alpha =mu^\alpha$$
Here $m$ is the body's rest-mass.

6. Aug 17, 2011

### SamRoss

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.