Constant Acceleration in Relativistic Mechanics

[kreil]

Homework Statement

Assume a particle moves 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 tau if the particle is at x0 at t=0 with no velocity.

The Attempt at a Solution

I am assuming the wording of the problem means

$$\frac{d^2 x}{d \tau^2}=g$$

which after some work gives the following results:

$$x(\tau) = \frac{1}{2}g \tau^2+x_0$$

$$\frac{dx}{d \tau} = \frac{dx}{dt} \frac{dt}{d \tau} = g \tau$$

$$\frac{dx}{dt} = \frac{g \tau}{\sqrt{1+(g \tau)^2}}$$

$$\frac{dt}{d \tau} = \sqrt{1+(g \tau)^2} \implies t(\tau) = \frac{g \tau \sqrt{1+(g \tau)^2)}+sinh^{-1}(g \tau)}{2g}$$

Is this correct? This expression for the time as a function of tau seems bizarre for such a simple situation, and it makes me wonder about whether my interpretation of the problem wording was correct.

 

[granpa]

look up rapidity

 

[vela]

Homework Statement

Assume a particle moves 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 tau if the particle is at x0 at t=0 with no velocity.

The Attempt at a Solution

I am assuming the wording of the problem means

$$\frac{d^2 x}{d \tau^2}=g$$

Is this correct? This expression for the time as a function of tau seems bizarre for such a simple situation, and it makes me wonder about whether my interpretation of the problem wording was correct.

No, it's not correct. The coordinates x and t are those of the lab frame. In the particle's frame, you'd have a different set of coordinates, say x' and t'. If the particle experiences an acceleration g in its frame, that means

$$\frac{d^2 x'}{dt'^2}=g$$

It's easiest to analyze this problem in terms of the rapidity θ, where the particle's velocity β is given by β=tanh θ.

 

[kreil]

we haven't covered rapidity yet. what would the steps be if i started from that equation? i.e.

$$\frac{d^2 x'}{dt'^2}=g$$

should i integrate to find x'(t'), then apply the LT to find x(t)? if so how would i find gamma?

 

[vela]

That equation only holds for an instant because a moment later, the particle will be in moving at a different speed, so the x' and t' coordinates no longer coincide with the coordinates of the particle's rest frame.

You can use the relativistic velocity addition formula. Say the particle is moving with speed β=v/c as observed in the lab frame. With respect to the particle's instantaneous rest frame, an infinitesimal (proper) time dτ later, the particle will be moving with speed g dτ. Use the velocity addition formula to calculate the corresponding increase in speed dβ as seen in the lab frame. You should be able to show, to first order, that

$$d\beta = (1-\beta^2) g\,d\tau$$

You can integrate this to find the velocity β as a function of proper time. (Use the substitution β=tanh θ to do the integral.)

 

[kreil]

The velocity addition formula I have is,

$$V^{x'}= \frac{V^x-v}{1-V^xv/c^2} \implies V^{x'}/c = \frac{\beta - v/c}{1-\beta v/c}$$

I don't see how I can get to

$$d\beta = (1-\beta^2) g\,d\tau$$

using that formula, since $V^x$ (lab observed speed of the particle) is not necessarily equal to v, the speed between the frames.

 

[vela]

If you solve for V^x, you get

$$V^x = \frac{v+V^{x'}}{1+vV^{x'}/c^2)} \rightarrow \beta^x = \frac{\beta+\beta^{x'}}{1+\beta\beta^{x'}}$$

where β^x=V^x/c, β^x'=V^x'/c, and β=v/c. (I prefer using β's just to get rid of those factors of c.) That's the form you want to use because you're interested in β^x, the speed of the particle in the lab frame, which I formerly called just β.

At proper time τ, the speed of the particle is β^x(τ). The second frame of reference, S', is the particle's instantaneous rest frame at proper time τ, so β=β^x(τ) and β^x'=0. If you plug those values in, you'll see the addition formula works out trivially for proper time τ.

At proper time τ+dτ, the particle will have sped up slightly. In the lab frame, its speed will be β^x+dβ^x, and in S', the particle will be moving with speed β^x'=g dτ. If you solve for dβ^x to first order, you should get the relationship I mentioned earlier.

 

[kreil]

thanks, that explanation makes sense

if you have time, I've got an open question on a momentum commutator relation which has been giving me a major headache and i would appreciate any help you could offer:

https://www.physicsforums.com/showthread.php?t=436473