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Jeff Knapowsk
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Homework Statement
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.
Homework 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))
The Attempt at a Solution
Assume k*tau << 1 and let's 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?