Relativistic Velocity

  • Thread starter nelufar
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  • #1
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Main Question or Discussion Point

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

Answers and Replies

  • #2
Fredrik
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All velocities are relativistic, but relativistic effects will not be easily noticeable until the speed is large enough to make

[tex]\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

(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.
 
  • #3
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Is same true for relativistic energies also?
 
  • #4
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The kinetic energy of a particle is [itex]E_{kin.}=(\gamma -1)mc^2 [/itex] if this deviates substantially from classical kinetic energy you should use the relativistic result.

So you could use as a criterium: if [itex]2 (\gamma -1)c^2/v^2[/itex] is significantly larger than 1 you deal with relativistic velocities...
 
  • #5
russ_watters
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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.
 
  • #6
Fredrik
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nelufar said:
Is same true for relativistic energies also?
To see the answer to your question it helps to use the series expansion

[tex]\frac{1}{\sqrt{1-x}}=1+\frac{1}{2}x+\frac{3}{8}x^2+\frac{5}{16}x^3+\dots[/tex]

The kinetic energy is

[tex](\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)[/tex]

Note that the first term after the last equality sign is just the non-relativistic kinetic energy. The other terms are relativistic corrections to the non-relativistic result. If the speed is large enough these terms can't be ignored.
 

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