Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relativistic Velocity

  1. Oct 26, 2004 #1
    Which velocity can be said to be relativistic velocity? I mean is there any range beyond which velocities can be called as relativistic velocities?
     
  2. jcsd
  3. Oct 26, 2004 #2

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
  4. Oct 27, 2004 #3
    Is same true for relativistic energies also?
     
  5. Oct 27, 2004 #4
    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...
     
  6. Oct 27, 2004 #5

    russ_watters

    User Avatar

    Staff: 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.
     
  7. Oct 28, 2004 #6

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Relativistic Velocity
  1. Relativistic velocity (Replies: 4)

  2. Relativistic Velocity (Replies: 9)

Loading...