# Relativistic Velocity

Which velocity can be said to be relativistic velocity? I mean is there any range beyond which velocities can be called as relativistic velocities?

Fredrik
Staff Emeritus
Gold Member
All velocities are relativistic, but relativistic effects will not be easily noticeable until the speed is large enough to make

$$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

(where v is the speed, and c is the speed of light)

significantly larger than 1.

An example: suppose that v=0.1c, i.e. 10% of the speed of light. Then gamma is about 1.005. Is that significantly larger than 1? That depends on that you're doing. Sometimes a 0.5% correction to the non-relativistic result isn't important, and sometimes it is.

Is same true for relativistic energies also?

The kinetic energy of a particle is $E_{kin.}=(\gamma -1)mc^2$ if this deviates substantially from classical kinetic energy you should use the relativistic result.

So you could use as a criterium: if $2 (\gamma -1)c^2/v^2$ is significantly larger than 1 you deal with relativistic velocities...

russ_watters
Mentor
Consider the orbital speed of a satellite or spacecraft: ~17,000mph. For an astronaut, this is not a relativistic velocity, meaning he won't notice the time dilation on his watch from takeoff to landing. For a GPS satellite, however, this is a relativistic velocity and must be accounted for in its functioning in order for GPS positions to be accurate.

Fredrik
Staff Emeritus
$$\frac{1}{\sqrt{1-x}}=1+\frac{1}{2}x+\frac{3}{8}x^2+\frac{5}{16}x^3+\dots$$
$$(\gamma-1)mc^2=\bigg(1+\frac{1}{2}\bigg(\frac{v^2}{c^2}\bigg)+\frac{3}{8}\bigg(\frac{v^2}{c^2}\bigg)^2+\frac{5}{16}\bigg(\frac{v^2}{c^2}\bigg)^3+\dots-1\bigg)mc^2=\frac12mv^2\bigg(1+\frac34\bigg(\frac{v^2}{c^2}\bigg)+\frac58\bigg(\frac{v^2}{c^2}\bigg)^2+\dots\bigg)$$