Question

A particle is moving along the x axis. It is uniformly accelerated in the sense that the acceleration measured in its instantaneous rest frame is always g, a constant. Find x and t as functions of the proper time assuming the particle passes through 2o at time t=0 with zero velocity. Hint: Consider the velocity u and acceleration a in an intertial rest frame. What are the values of a . a, au and u. u? Use these to find the particle's 4-velocity and integrate to find position. Draw the worldline of the particle on a spacetime diagram. Be careful to indicate correctly the asymptotic behaviour T → too.

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Answer 1

Answer: Page 0 As we are speaking about space time here we should consider 4 - velocity, that is velocity which represents the relatisvitic component of velocily along with its 3- dimensional vector space. Now Considering the Normalization condition for four velocity u-u=-1 - 0 also we find that a. u= 0 - ③ (that is derivate of u (four- acceleration about his perpendicular to u) In the rest frame following given information in question a- (0,8) Conly t and a components) Now the invariant a.a is a. a = gr -

In general frame the equations 9 and ③ page 2 become, -Cut) + (U?) % - -1 - ① -Catut) + (a’u") = 0 — ☺ -(at)t(ax)" =97 - © lets solve the above 96 4 6 equations for at fan in terms of g and u. That is at gua - dut - . dr a = gut- dut -o Now look at the ② & ③ equations, they are coupled for ut, un with their differential form. let's eliminate ut from 6 & 6 for find equation in terms of u² and g, dar = gut - 6

The solution where page 3 the uk is o at p =0 is, U² = Asinh(97) - 16 Where A is some constant Similarly, this leads to ut= A cash (gr) But consider equation 4 with its normalization condition, which implies A=1 Thus, u² = dx = sinhor) - ① ut - dt - cosh (gr) - (12) - dr Now integrating above equation abng with given boundary conditions, x(%) = x + + [cosh (9%) -1] t(P) = 16 [sinh (9%)]

Now can from a and t equations above are - Page 4 determine word line that is t² = (x-20 ) + ( 2 ) (x-20) Ly which is hyperbola.