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Kiwithepike
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Homework Statement
Consider a particle in one-dimensional so called hyperbolic motion
x(t)=[itex]\sqrt{b^{2}+t^{2}}[/itex]
where b is a constant.
a) Find[itex]\gamma[/itex](t).
b) Find the proper time [itex]\tau[/itex](t). (assume that [itex]\tau[/itex]=0 when t = 0
c) Find x and v[itex]_x[/itex] as functions of the propertime [itex]\tau[/itex].
d) FInd the 4-velocity u[itex]^{\mu}[/itex].
The Attempt at a Solution
A) ok to begin I took the derivative of x(t) to get velocity. tuned out to be t(b[itex]^{2}[/itex]+t[itex]^{2}[/itex])[itex]^{-1/2}[/itex].
soo therefor [itex]\gamma[/itex](t) = [itex]\frac{[itex]\sqrt{b^{2}+t^{2}}[/itex]}{[itex]\sqrt{1-\frac{t^{2}}{\sqrt{b^{2}+t^{2}}}}[/itex]}[/itex]
b) so now [itex]\tau[/itex](0) = [itex]\sqrt{t^{2}-(b^{2}+t^{2}}[/itex]
[itex]\tau[/itex](0) = [itex]\sqrt{0^{2}-(b^{2}+t^{0}}[/itex] = 0
[itex]\tau[/itex](0) = [itex]\sqrt{-b^{2}}[/itex] = 0
so would b = 0?
this is where I'm getting lost.
c) x as a function of \tau would be [itex]\sqrt{t^{2}-\tau^{2}}[/itex]=x?
where does v[itex]_x[/itex] come in? would i solve v(t) for t^2?
d) I know the 4 vector for u[itex]^{\mu}[/itex] is (u^0,u^1,u^2,u^3) and the roattional lorrentz for hyperbolic is
|t'| = |cosh[itex]\varphi[/itex] -sinh[itex]\varphi[/itex] |
|x'| |-sinh[itex]\varphi[/itex] cosh[itex]\varphi[/itex] |
where tanh[itex]\varphi[/itex]=v
where cosh[itex]\varphi[/itex]= [itex]\gamma[/itex]
where do i go from here? Thanks for all the help.
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