# Relativity: position/acceleration 4-vectors to newton's laws

1. Jul 2, 2008

### Jeff Knapowsk

1. The problem statement, all variables and given/known data
Given the binomial expansions for sinh and cosh, so that the acceleration and position 4-vectors reduce to x=1/2 a t^2 for small t.

2. Relevant equations

Okay, so we have cosh (x) = 1 + x^2/2 + ..... and
sinh(x) = x + x^3/3 ......

We have the position (sigma) 4-vector = (B*sinh(k*tau): B*(cosh(k*tau) -1) and
the acceleration (alpha) 4-vector = (-B*k^2*sinh(k*tau): -B*k^2*(cosh(k*tau))

3. The attempt at a solution

Assume k*tau << 1 and lets ignore higher order terms above x^2.

so, cosh(k*tau) = 1+ 1/2 (k*tau)^2 and sinh(k*tau) = k*tau

So, know I'm somewhat stumped. What are we setting equal to what? What is the small in both of these equations? t? or tau?

I know that the first term in the position 4-vector is c*t. So I could say:

B*k*tau = ct. And the last term in the acceleration 4-vector (alpha) is a. So I could say

-B*k^2*(1 + (k*tau)^2/2) = a. But that gets me nowhere.

Anyone got a thought? What's the direction here?

I read in another thread that we should try to understand the physical meaning of the first term of the acceleration or velocity 4-vectors. It doesn't have a clear explanation. It just sort of "is". Right?

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