What is the displacement of an accelerated and relativistic object?

Question

Displacement in an accelerated classical object is: $$s=ut+\frac {at^2}{2}$$ What is the displacement of an accelerated relativistic object?

In Newtonian mechanics there are two types of displacement.

1. Displacement of an object with velocity constant: $$s=ut$$
2. Displacement of an accelerated object with acceleration constant: $$s=ut+\frac {1}{2}at^2$$

This is not completely clear, but I think the second displacement should be something different in relativity. Is that true?

What is the result of this integral in relativity? $$s=\int (u+at)dt$$ classicaly $$s=\int (u+at)dt=\int udt +a\int tdt=ut+a\frac {t^2}{2}+x_0$$

What is the relativistic one?

more details: $$v=u+at$$ $$s\to displacement$$ $$a\to acceleration$$ $$u\to initial \,velocity$$ $$v\to final \,velocity$$

Answer 1

It's important to note that the equation

$$ s = ut + \frac{at^2}{2} $$

is a mathematical equation, not a physical one. It is simply the equation for the integral of a velocity with constant acceleration:

$$ s = \int (u + at) dt $$

So, if you pick a particular frame and measure a particle's initial velocity to be $u$ and its constant acceleration to be $a$, then this equation for displacement will describe your relativistic particle. But there are a few differences that relativity does lead to. If you want to understand the force on a particle with mass $m$ necessary to give it a constant acceleration of $a$, you will have to use the relativistic version of Newton's law:

$$ F = \frac{d( \vec{p} )}{dt} = \frac{d( m\vec{v}\gamma )}{dt} $$

So, one will need to apply a non-constant force to achieve constant acceleration. In addition, if transform to a different frame, the observed displacement will change based on the relative velocity of the second frame to the first. This is described by a lorentz transformation.

Answer 2

Suppose a particle is moving at a velocity approaching $c$ where relativistic effects matter. This doesn't change the method you use to measure its physical properties; you still use the same space coordinates and clock that you would use if it was travelling at small velocities where relativity doesn't matter. Therefore, you define and measure velocity and acceleration in exactly the same way which means your relativistic displacement formulas are identical to the Newtonian ones.

Answer 3

I think that you are asking here about an object which has a constant proper acceleration, that is to say an object which has a constant acceleration in its instantaneous rest-frame. As previous replies have said, this is simply an exercise in applying Lorentz transformations. If we denote the proper acceleration by $\alpha$, and consider the object to start from rest at $t=0$, then the body's velocity and displacement measured by an observer at rest with respect to the body's initial position will be:

$ x(t) = (c^2/\alpha) ( \sqrt{1 + (\alpha t/c)^2} -1 ) $, and

$ v(t) = \alpha t / \sqrt{1 + (\alpha t/c)^2 } $.

Note how $\alpha t$ plays the role of $v$ in the Lorentz transformation. It is also interesting to see how $v$ asymptotically approaches $c$ in the limit of long times, exactly as one would expect.

When $\alpha t \ll c$, i.e. the non-relativistic limit, these expressions can be simplified by making a Taylor expansion of the square-roots to recover the familiar classical results you give. This is similar to a problem given at the end of chapter 5 of "Special Relativity" by A.P. French. Incidentally, French remarks (about the transformation of accelerations) "The results are somewhat complicated, and there is no point in trying to remember them unless relativistic kinematics is your livelihood". I think this is very true!