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Help with special relativity

  1. Jan 13, 2005 #1
    I am writing an essay on special relativity, but I am stuck on how to prove that mass increases with speed, e=mc squared and I do not know how to show how you add velocities at speeds approaching the speed of light. I would be very grateful for an explanation of these things.
     
  2. jcsd
  3. Jan 13, 2005 #2
    Proving mass increases with speed...well for a start:
    [tex]m = \frac {m_0} {\sqrt{1-v^2/c^2}}[/tex]

    Then you can look at the velocity of 2 objects travelling towards each other.
    Overall velocity is u, and the 2 velocities are v and u':
    [tex]u = \frac {v+u'} {1+ \frac {v u'} {c^2}}[/tex]

    Reverse sign of one of the velocities to get particles travelling towards each other.
    I believe it is called the Einstein addition velocity...not sure if that's right though, learnt it a long time ago.
     
  4. Jan 13, 2005 #3
    Yes but how do you derive those equations?
     
  5. Jan 13, 2005 #4

    jtbell

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    Staff: Mentor

    You can find derivations in many second-year college/university level "modern physics" textbooks. For example, Beiser, "Concepts of Modern Physics" (5th ed.) or Taylor/Zafiriatos/Dubson, "Modern Physics for Scientists and Engineers" (2nd ed.). Check your nearby libraries to see what they have. Or browse through some university Web sites. Surely someone has put lecture notes on line which give the derivations or at least sketch their outlines.
     
  6. Jan 13, 2005 #5
    Here's a derivation if that'll help.

    Derivation of E=MC^2
     
  7. Jan 13, 2005 #6

    jcsd

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    Science Advisor
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    Hmm I don't know if that's a great one as it starts off with the formula mass increase. Infact these days few talk about mass increase as the mass is usually defined as the rest mass which obviously does not change.
     
  8. Jan 13, 2005 #7
    1. I never implied that it was the best possible derivation. :approve:

    2. As a "Science Advisor" shouldn't you be actually pointing to a better derivation instead of merely posting criticisms? Are you sure you aren't just a "Science Critic" posing as an "Science Advisor"? :biggrin:
     
    Last edited: Jan 13, 2005
  9. Jan 14, 2005 #8
    well there is always the derivation from the metric:

    (Tau)^2=t^2-s^2
    (Tau)^2=t^2-(v^2)(t^2)
    (Tau)^2=(t^2)(1-(V^2))
    (Tau/t)^2=1-v^2
    (Tau/t)=(E/m)
    (E/m)^2=(1-v^2)
    E/m=(+ or -)(1-v^2)^(1/2)
    We then convert it to SI units
    (E/mc^2)=(+or-)(1-v^2)^(1/2)
    E=(+or-)(mc^2)(1-v^2)^(1/2)
    at rest a particle has 0 velocity so the equation reduces to:

    E=(+or-)mc^2

    sorry for the confusing notation, i dont know LaTex
     
  10. Jan 14, 2005 #9
    I cleaned up a little bit for you without using Latex. Hope I didn't make an error.
    I used an auto-replace macro and didn't check the results. :biggrin:

    You can use [ sup ]2[ /sup ] (no spaces) for exponents. Also [ sub ]n[ /sub ] works for subscripts.

    Hold down the alt key, type in 0177 then let up to produce ±
    Sorry, I couldn't figure out the code for the Tau symbol.

    Nice derivation by they way. If it's correct! I honestly don't know. :yuck:
     
    Last edited: Jan 14, 2005
  11. Jan 17, 2005 #10
    The modern interpretation discourages the idea of "mass increase" and favors the idea that momentum is not as simple as p = mv.

    The only thing SR does is tell us is that mathematically P = m*v*gamma, whether the gamma term associates with the mass term, or the velocity term, or both, is a philosophical difference of interpretation.
     
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