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Math Question

  1. Mar 4, 2005 #1

    DB

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    In this paper the author wrote:

    [tex]t'=\sqrt{\frac{4h^2}{c^2-v^2}}=\frac{2h}{\sqrt{c^2-v^2}}[/tex]

    [tex]t'=\frac{2h}{\sqrt{c^2}}\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

    Why so?

    Also, isn't the relativistic beta considered just v/c, not as the author stated (c/v)^2?
     
  2. jcsd
  3. Mar 4, 2005 #2
    The author factored out sqrt(c^2) in order to get the frequently occuring expression 1/sqrt(1-(v/c)^2), which is referred to by the symbol [tex]\gamma[/tex] in most texts. I'm not sure what you mean by beta, but the unitless replacement v := v/c is common.
     
  4. Mar 4, 2005 #3

    dextercioby

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    Customarily "c=1" and no such tricks are necessary...:wink:

    Daniel.
     
  5. Mar 5, 2005 #4

    DB

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    thanks guys, but I still can't see how
    [tex]\frac{2h}{\sqrt{c^2-v^2}}=\frac{2h}{\sqrt{c^2}}\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
    How did he factor out sqrt(c^2) to get 1/sqrt(1-(v/c)^2)?
     
  6. Mar 5, 2005 #5
    This is basic factoring:
    [tex]\frac{1}{\sqrt{c^2-v^2}} = \frac{1}{\sqrt{c^2(1-\frac{v^2}{c^2})}} = \frac{1}{\sqrt{c^2}\sqrt{1-\frac{v^2}{c^2}}} = \frac{1}{\sqrt{c^2}}\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
     
  7. Mar 5, 2005 #6

    selfAdjoint

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    Since [tex]c^2[/tex] is a constant, multiply and divide by it:[tex]c^2 (c^2 - v^2)/c^2 = c^2(\frac {c^2}{c^2} - \frac {v^2}{c^2}) = c^2 ( 1 - \frac{v^2}{c^2})[/tex], no?
     
    Last edited: Mar 5, 2005
  8. Mar 5, 2005 #7

    jtbell

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    Also, in general,

    [tex]\sqrt{ab} = \sqrt{a} \sqrt{b} [/tex]

    if that's what's bothering DB.
     
  9. Mar 5, 2005 #8

    DB

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    Thanks.
    Ok. I see the math there, but why do we multiply by c^2/c^2?
    (I know you've stated that its constant, but can you elaborate please?)
    Nawww, dont worry bout that.
     
  10. Mar 5, 2005 #9

    cepheid

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    c2 / c2 = 1, so you can multiply by it whenever you want without changing anything.

    selfAdjoint was just showing you the intermediate steps very explicitly, making it clear that having an 'extra' c2 multiplying the expression out front is fine as long as you divide both terms in the expression by c2 to compensate. But surely you can arrive at the end result straight away, just by thinking of it as "factoring out a c2" from both terms:

    [tex] (c^2 - v^2) = c^2(1- \frac{v^2}{c^2})[/tex]

    well that's exactly what hypermorphism and selfAdjoint already showed you.
     
  11. Mar 5, 2005 #10

    DB

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    Ahhh, I see it now, thanks guys.
     
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