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Regarding Special Relativity

  1. Dec 10, 2004 #1
    Deriving SR Formula

    edit 3:
    How does one go about deriving the formula:[tex]mr =\frac{m0}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex]?
    Last edited by a moderator: Dec 10, 2004
  2. jcsd
  3. Dec 10, 2004 #2


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    Mass in your equation does not change with velocity.

    Some people call [itex]\gamma m[/itex] the relatvistic mass and this cleraly does increase with velocity, but this defintion is not widely used and it's not a very good defintion anyway.
  4. Dec 10, 2004 #3
    But [itex]\gamma m[/itex] isn't mass itself; it is the multiple of mass by a factor of [itex]\gamma[/itex]. Or is this the "wrong" way of looking at the situation?
  5. Dec 10, 2004 #4
    You're right... According to my equation, mass will decrease. If velocity is say, 0.7c, [tex]\sqrt{1-\beta^2[/tex] will equal 0.714 (as compared to the a value of 1 which it would be if velocity was much smaller than c). Therefore, mass would decrease.

    I don't understand how this is so.
  6. Dec 10, 2004 #5


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    Almost universally m is regarded as the mass aka the rest mass, but a few people regard [itex]\gamma m[/itex] as the mass aka the relatvistic or transverse mass.

    Defining m as the mass is useful as it doesn't change from frame to frame. Defining [itex]\gamma m[/itex] as the mass means that it does change from frame to frame and you find for conssitency you have to define a second kind of mass [itex]\gamma^3 m[/itex] called the longitudinal of mass.

    The reason for the two definitions is due to how the concepts of mass and force are related, if you think that Netwon's second law is [itex]\vec{F} = m\vec{a}[/itex] then you might argue that the concept of relatvistic mass is just sticking with this concpet, but most physcists like to think of Netwon's second law as [itex] \vec{F} =\frac{d\vec{p}}{dt}[/itex] anyway.
    Last edited: Dec 10, 2004
  7. Dec 10, 2004 #6
    Ah. I have confused several concepts in SR. I now understand the fact that:

    [tex]mr =\frac{m0}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex]

  8. Dec 10, 2004 #7
    edit 3: Topic changed.
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