Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Math Question

  1. Mar 4, 2005 #1


    User Avatar

    In http://www.freewebs.com/mouldy-fart/Space,%20Time%20and%20SR.pdf [Broken] paper the author wrote:



    Why so?

    Also, isn't the relativistic beta considered just v/c, not as the author stated (c/v)^2?
    Last edited by a moderator: May 1, 2017
  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.
    Last edited by a moderator: May 1, 2017
  4. Mar 4, 2005 #3


    User Avatar
    Science Advisor
    Homework Helper

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

  5. Mar 5, 2005 #4


    User Avatar

    thanks guys, but I still can't see how
    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


    User Avatar
    Staff Emeritus
    Gold Member
    Dearly Missed

    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


    User Avatar

    Staff: Mentor

    Also, in general,

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

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


    User Avatar

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar

    Ahhh, I see it now, thanks guys.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook