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Relativistic corrections to classical physics formulae

  1. Jul 22, 2013 #1
    How are classical formulas in physics (such as p = mv, or kinetic energy, or maxwell distribution of speeds) treated with the appropriate relativistic correction/modification? Is it done by using the Lorentz transformation equations? Could anyone give me a few examples of relativistic corrections to classical formulae? Thanks.
     
    Last edited: Jul 22, 2013
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  3. Jul 22, 2013 #2

    ghwellsjr

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    Yes. The formulas must remain unchanged when subjected to the Lorentz transformation process. (But don't ask me how they actually do it.)
     
  4. Jul 22, 2013 #3
    Thanks! :) yeah sometimes I get confused when looking at the classical formulas, and the relativistic version, and try to see how its done using the Lorentz stuff.
     
  5. Jul 22, 2013 #4

    PAllen

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    IMO, the best approach is to learn about 4-vectors. Typically, the relativistic formulas look natural as 4-vectors, while sometimes looking 'unnatural' in 3-vector notation.

    For example, starting with 4-velocity as the derivative of (t,x,y,z) by proper time ([itex]\tau)[/itex], denoted U, you have:


    p = m U ; m is (rest) mass, p momentum.

    A = proper acceleration = what is measured by an accelerometer = dU/d[itex]\tau[/itex]

    F = dp/[itex]\tau[/itex] = m A

    (in the above, I assume a particle whose rest mass does not change).

    Note, this approach explains the disfavor of relativistic mass: there is no relativistic mass in any of the above forumulas. The factor γ is buried within U (and in a more complex way, within A).
     
    Last edited: Jul 22, 2013
  6. Jul 24, 2013 #5
    The relativistic corrections to the first:

    p = γ m0 v, with m0 = "rest mass" (the Newtonian mass concept which assumes that inertial effects are independent of speed had to be abandoned). As PAllen mentioned relativistic mass, it is easy to see where the concept of "relativistic mass" came from: one can bundle γm0 together as m = "relativistic mass", so that one gets again p = m v.

    Further, F = dp / dt remains unchanged.

    However, the relationship between force F and acceleration a - coordinate acceleration of an object as measured in an inertial system - is much more complex; that's a neat textbook exercise. :tongue:
    See: http://en.wikipedia.org/wiki/Force#Special_relativity
     
    Last edited: Jul 24, 2013
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