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Inverse Funcation of Relativistic mass

  1. Sep 24, 2009 #1
    The relativistic mass equation shows the relationship between relativistic mass and speed. Basically showing that the closer u get to light speed the less relativistic mass you have, but ur rest mass will be the same.

    I was asked by my friend what the inverse of that formula means.

    Thus I took the original equation and came up with the inverse: m = sqrt(c^2-(m0c^2/v)^2)

    I also looked at the graph of the original and the inverse would just be a reflection about y=x.

    I still can't interpret it. Any hints/ help?
     
  2. jcsd
  3. Sep 24, 2009 #2

    jtbell

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    Staff: Mentor

    Do you mean

    [tex]m = \frac{m_0}{\sqrt{1 - v^2 / c^2}}[/tex]

    Don't you mean "more relativistic mass" (not "less")?

    You must have made a mistake in your algebra somewhere. The units of the various quantities don't work out consistently in that equation. Also, if you started with the equation I gave above, an "inverse" equation would not start with "m =" but with "v =".
     
  4. Sep 24, 2009 #3
    I switched m for v and isolated for m
     
  5. Sep 25, 2009 #4

    Fredrik

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    We were supposed to guess that the relativistic mass in your equation was actually a velocity!? :confused: It's always a bad idea to change the meaning of the symbols without telling anyone.

    If you solve the equation in jtbell's post for v, you get

    [tex]v=c\sqrt{1-\frac{m_0^2}{m^2}}=c\sqrt{1-\frac{m_0^2c^4}{E^2}}[/tex]

    where E=mc2 is the energy. Note that v→c when E→∞.
     
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