- #1

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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