# Constant Acceleration in Relativistic Mechanics

• kreil
In summary, In summary, if you have two frames of reference, each with a different momentum, then the addition of those moments will result in a change in momentum for the particle in the second frame.
kreil
Gold Member

## 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.

Last edited:
look up rapidity

kreil said:

## 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.

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?

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.)

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.

If you solve for Vx, 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 βx=Vx/c, βx'=Vx'/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 βx, 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 βx(τ). The second frame of reference, S', is the particle's instantaneous rest frame at proper time τ, so β=βx(τ) and βx'=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 βx+dβx, and in S', the particle will be moving with speed βx'=g dτ. If you solve for dβx to first order, you should get the relationship I mentioned earlier.

## What is "constant acceleration" in relativistic mechanics?

"Constant acceleration" refers to the uniform increase in velocity of an object over a given period of time. In relativistic mechanics, this concept is applied to objects moving at speeds close to the speed of light.

## How does constant acceleration affect an object's mass in relativistic mechanics?

In relativistic mechanics, as an object approaches the speed of light, its mass increases due to the effects of time dilation and length contraction. This means that the object's mass will continue to increase as it undergoes constant acceleration.

## What is the formula for calculating constant acceleration in relativistic mechanics?

The formula for calculating constant acceleration in relativistic mechanics is a = F/m, where "a" is the acceleration, "F" is the force acting on the object, and "m" is the object's mass.

## How does constant acceleration affect an object's velocity in relativistic mechanics?

In relativistic mechanics, constant acceleration leads to a continuous increase in the object's velocity. However, as the object approaches the speed of light, its velocity will continue to increase at a slower rate due to the effects of time dilation and length contraction.

## What are the practical applications of constant acceleration in relativistic mechanics?

Constant acceleration in relativistic mechanics is used in many practical applications, such as particle accelerators, space travel, and understanding the behavior of high-speed particles. It also plays a crucial role in the development of modern physics theories, such as special relativity and general relativity.

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