# A Roller Coaster Ride through Relativity/Appendix F

## Relativistic Kinetic Energy

Suppose a mass of rest-mass M0 is accelerated from rest by a constant force F for a time t.

An accelerating firework

Under Newtonian mechanics, the acceleration of the mass a will be

${\displaystyle a={f \over M_{0}}}$

the distance travelled s will be

${\displaystyle s={\tfrac {1}{2}}at^{2}={FT^{2} \over 2M_{0}}}$

and the final speed reached will be

${\displaystyle v=at={Ft \over M_{0}}}$

from which we deduce that

${\displaystyle Ft=M_{0}v}$

Now the Kinetic Energy KE acquired by the mass will be equal to the work done by the force which is, of course, force × distance or Fs. Hence:

${\displaystyle KE=Fs={F^{2}t^{2} \over 2M_{0}}={M_{0}^{2}v^{2} \over 2M_{0}}}$
${\displaystyle KE={\tfrac {1}{2}}M_{0}v^{2}}$

Now in order to carry out the same kind of analysis using Special Relativity, we need to use the results obtained during the analysis of the constantly accelerated rocket, namely:

${\displaystyle s={c^{2} \over a}\left(\cosh {at \over c}-1\right)}$

and

${\displaystyle v=c\tanh {a \over c}t}$

What we need to do is eliminate t from these two equations which can be done using the following standard relation:

${\displaystyle {1 \over \cosh ^{2}\theta }+\tanh ^{2}\theta =1}$

The algebra is a bit messy but it is not difficult and reduces to

${\displaystyle as={c^{2} \over {\sqrt {1-v^{2}/c^{2}}}}-c^{2}=c^{2}(\gamma -1)}$

Now we need to think carefully what we have found out. This equation gives us a relation between the speed reached and the distance travelled for a rocket undergoing a constant acceleration of a. But from whose point of view? The person on the rocket or the person who remains at rest?

Well, there is no problem about the speed v. As we have seen, both observers agree about the relative speed of any object. But in any case, it is the speed as seen by the stationary observer which we are talking about.

The distance s is the distance travelled as seen by the stationary observer as well (remember it is the actual distance travelled to the star, not the length contracted distance).

What about a the acceleration? This is the acceleration as experienced by the occupants of the rocket. You will remember that the rocket is supposed to accelerate in such a way that the occupants of the rocket experience a constant 'artificial gravity'. As far as the astronauts are concerned, the rocket motors produce a constant thrust of F and the rocket has a constant mass of M0 so we can still assume that

${\displaystyle a={F \over M_{0}}}$

hence

${\displaystyle Fs=c^{2}(\gamma -1)}$

Normally we would just write

${\displaystyle KE_{r}=Fs}$

but this requires a bit of justification. You see, s is the distance travelled as seen by the stationary observer, while F is the thrust as measured by the astronauts and it is not immediately obvious that you can multiply these two quantities together.

Suppose that instead of accelerating a rocket with on board rocket motors, we consider accelerating an electron using a (stationary) electric field E.

An electron undergoing acceleration by an electric field E

The question is; is the force as experienced by the electron the same as the force exerted on the electron by the accelerator? In a sense, there is no answer to this question because it all depends on what you mean by force. What we are effectively doing is constructing a relativistic definition of force which is as close to the Newtonian definition as possible. One of the things which we would like our relativistic force to do is to obey Newton's third law which can be stated as action and reaction are equal and opposite; so if we assume that this is true we can go ahead and complete the proof. ie:

${\displaystyle KE_{r}=M_{0}c^{2}(\gamma -1)}$

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