Is there an easy way to show that x^2−t^2=1/g^2 for a (relativistic) body undergoing acceleration g?

Awainaideannagi

Awainaideannagi

Answered question

2022-07-17

Is there an easy way to show that x 2 t 2 = 1 / g 2 for a (relativistic) body undergoing acceleration g?

Answer & Explanation

Cael Cox

Cael Cox

Beginner2022-07-18Added 11 answers

Consider the particle's view from its own (accelerating) reference frame. Since 𝑔 is a proper acceleration, the particle always measures a constant acceleration for itself. Combined with the fact that by definition, its velocity is zero in that reference frame, you know that the particle's trajectory has to be invariant under a combined time translation and Lorentz boost. (In other words, the trajectory has to locally look the same after some time has elapsed and the particle's coordinate velocity has changed accordingly.) The only trajectory that satisfies that condition is a hyperbola.
Mathematically, you can come up with this by repeatedly applying a Lorentz transformation. The transformation can be represented as
( c t x ) ( γ β γ β γ γ ) ( c t x )
Suppose the particle starts with x ( 0 ) = 0, which corresponds to setting up a coordinate system in which it starts out at rest. Using regular old kinematics, you can figure out that in the vicinity of t = 0, its path should follow x ( t ) = 1 2 g t 2 + O ( t 3 ) (after all, this should match the nonrelativistic result over short time scales). So an infinitesimal time δ t after t = 0, its speed is β c = g δ t, and the corresponding Lorentz transformation is
( 1 g δ t / c g δ t / c 1 )
Now, if the particle's path is invariant under the time translation with corresponding boost, then if you want to advance δ t further forward in time, you just do the same thing again, and again, and again. The only catch is that this increment δ t is proper time, because you're always measuring it in the particle's own rest frame.
In general, to figure out the particle's path after a finite proper time τ, you just apply the transformation N times, using the limit N with N δ t = τ:
( 1 g δ t / c g δ t / c 1 ) N = [ 1 2 + ( 0 g / c g / c 0 ) δ t ] ( τ / δ t ) = exp [ ( 0 g / c g / c 0 ) τ ] = ( cosh g τ c sinh g τ c sinh g τ c cosh g τ c )
Apply this to the starting position four-vector ( 0 x ( 0 ) ) and you have your coordinate time and coordinate position,
( c t ( τ ) x ( τ ) ) = ( x ( 0 ) sinh g τ c x ( 0 ) cosh g τ c )
and if you solve this to actually put x in terms of t, you get
x ( 0 ) 2 cosh 2 g τ c x ( 0 ) 2 sinh 2 g τ c = x ( 0 ) 2 x ( t ) 2 t 2 = x ( 0 ) 2
To figure out what x ( 0 ) is, you can take the second derivative with respect to proper time,
d 2 d τ 2 x ( τ ) = x ( 0 ) g 2 c 2 cosh g τ c = g x ( 0 ) = c 2 g
suchonosdy

suchonosdy

Beginner2022-07-19Added 3 answers

Your expression is a solution to the integration of the transformation, g = γ 3 a , for acceleration along V between the lab and proper frames with initial conditions x = c 2 / g at t = 0. I'll give a derivation for this standard transformation.
Let frames S and S be in standard configuration with S moving with velocity V along x . Differentiating the velocity transformation for u wrt t gives the transformation of acceleration along V
a = d d t ( u + V 1 + u V c 2 ) = d u d τ d τ d t d d u ( u + V 1 + u V c 2 ) = a 1 γ ( 1 + u V c 2 ) 1 γ 2 ( 1 + u V c 2 ) 2 = a 1 γ 3 ( 1 + u V c 2 ) 3
For completion, the same can be done for the acceleration perpendicular to V , a
a = 1 γ 3 ( 1 + v u c 2 ) 3 a a = 1 γ 2 ( 1 + v u c 2 ) 3 ( a + v c 2 × ( a × u ) )
which reduce to the important standard expressions for a proper frame
γ 3 a = a γ 2 a = a

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