# Relativistic mechanics Taylor expansion

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1. Oct 16, 2014

### infinitylord

1. The problem statement, all variables and given/known data
For a particle traveling near the speed of light, find the first non-vanishing term in the expansion of the relative difference between the speed of the particle and the speed of light, (c-v)/c, in the limit of very large momentum p>>mc. Hint: Use (mc/p) as a small parameter for your expansion.

3. The attempt at a solution
I know the answer is (1/2)(mc/p)^2, but I'm not sure where this answer comes from. I understand the basic idea of expansions, however, I'm not sure as to what "in the limit of very large momentum" means. Can someone guide me through this problem?

2. Oct 16, 2014

### Orodruin

Staff Emeritus
"In the limit of large momentum" means that the momentum is very large in comparison to the mass, i.e., exactly what the problem statement tells you. It means that the mass can be considered a small parameter in which you can perform a series expansion.

3. Oct 16, 2014

### infinitylord

I understand that it means that momentum is very large compared to mass. However, I'm still not really sure where to begin with the actual expansion. Am I expanding the (c-v)/c?

4. Oct 16, 2014

### vela

Staff Emeritus
Since you want the momentum to appear in your answer, try writing v/c in terms of momentum and energy.

5. Oct 27, 2014

### infinitylord

Thanks for the help and sorry it has taken me so long to respond.
Okay, here's where I got with this.
v = (p/m)/(1+(p/mc)2)1/2
multiply the top and bottom by mc/p to get
c/(1+(mc/p)2)1/2
Then I tried using the known expansion (1+x)n = 1+nx+1/2(n-1)nx^2... where n=-1/2 and x=(mc/p)2 (and c will just be multiplied by each term).
So this all should yield:
c - 1/2*c*(mc/p)2 + 3/8*c*(mc/p)4.

This is not the correct answer so I'm still not quite sure what's going wrong with this.

6. Oct 27, 2014

### Orodruin

Staff Emeritus
This is your expression for v, not for (c-v)/c. Try expressing (c-v)/c using what you already obtained for v. (Also, neglect the term proportional to m^4 c^4/p^4 ...)

7. Oct 27, 2014

### infinitylord

Oh so the c's cancel out leaving the desired first term! Thanks for the help