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

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2 Answers 2

up vote 1 down vote accepted

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.

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-1 "is a mathematical equation, not a physical one" acceleration, velocity, time, displacement are physical quantities in that equation. –  Physiks lover Apr 20 '12 at 12:18
    
You should also mention that acceleration as defined in classical mechanics is hardly useful in relativity. Motion of constant proper accerelation is more useful as a model. –  C.R. Apr 20 '12 at 15:02
    
"acceleration, velocity, time, displacement are physical quantities in that equation. – Physiks lover" "Motion of constant proper accerelation is more useful as a model. – Karsus Ren" agree –  user8784 Apr 20 '12 at 20:28

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.

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"Suppose a particle is moving at a velocity approaching where relativistic effects matter. This doesn't change the method you use to measure its physical properties;" light is constant but displacement have effect on light properties –  user8784 Apr 20 '12 at 20:34
    
@BadBoy could you give me an example where displacement has an effect on light properties? –  Physiks lover Apr 20 '12 at 21:27
    
when light moves which means light have displacement. please do not ask question inside question!. i want my answer –  user8784 Apr 20 '12 at 21:41
    
@bad boy I queried the comment you made to my answer which is perfectly ok to do. You still use the same ruler to measure the displacement of light as you would a car moving at low velocities. It's still displacement. –  Physiks lover Apr 21 '12 at 15:14

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