1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lorentz Transformation

  1. Jan 27, 2009 #1
    1. Start with the expression x'^2 + y'^2 _ z'^2 -c^2t'^2 and show, with the aid of the Lorentz transofmration, that this quantity is equal to x^2 + y^2 + z^2 -c^2t^2. This result establishes the invariance of s^2 defined by s^2 = x^2 + y^2 + z^2 -c^2t^2

    2. s^2 = x^2 + y^2 + z^2 -c^2t^2

    3. So far I have y = y' and z = z' because of the transformation law. basically there is no lorentz contraction perpendicular to the motion of X so y = y' and z = z'

    how do i really solve this problem? where should I start?
  2. jcsd
  3. Jan 28, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper

    So what does the Lorentz transformation look like for x' and t', in terms of x and t?
    I mean, what is the correct full expression for

    t' = ... t + ... x
    x' = ... x + ... t
    y' = y
    z' = z
  4. Jan 28, 2009 #3

    Attached Files:

  5. Jan 28, 2009 #4


    User Avatar
    Science Advisor
    Homework Helper

    Very nice. It's straightforward algebra then: plug them into
    [itex]x^2 - c^2 t^2[/itex]
    and show that you can simplify it to
    [itex](x')^2 - c^2 (t')^2[/itex].
  6. Jan 28, 2009 #5
    so I would have
    (x'+vt'/sqrt(1-v^2/c^2)^2 - c^2(t' + vx'/c^2 / sqrt(1-v^2/c^2) and simplify it algebraically to x'^2 - c^2t'^2 ?
  7. Jan 28, 2009 #6


    User Avatar
    Science Advisor
    Homework Helper

    Yes, the expression is a bit unclear, but it looks like what I got,
    [tex]\frac{(x' + v t')^2 - (t' + v x' / c^2)^2}{(\sqrt{1 - v^2/c^2)^2}[/tex]
    Just work out the multiplication (don't forget the cross-term). Then you'll see some cancellation taking place and you can factor the rest into something[itex]\,{}\times (x'^2 - c^2t'^2)[/itex].
  8. Jan 28, 2009 #7
    thanks a lot, i will get on this now :D
  9. Jan 28, 2009 #8
    I'm sort of stuck, my algebra isn't as fresh as I thought it would be. But am I on the right track though?

    http://img264.imageshack.us/img264/1081/1233176495203dl2.jpg [Broken]

    ^ My work
    Last edited by a moderator: May 3, 2017
  10. Jan 28, 2009 #9


    User Avatar
    Science Advisor
    Homework Helper

    You are on the right track, I see two terms cancelling.
    Watch the squares however, I see c^2 t'^2 instead of (c^2 t')^2 and v x'^2 instead of (v x')^2...

    Also it may help to write everything in one fraction.
  11. Jan 28, 2009 #10
    I need a common denominator to start canceling things out right, or can I do it before? This is the part where my algebra isn't as fresh as it should be. To get the common denominator, I have to multiply the left equation by c^4 (top and bottom) and multiple the right equation by 1 (top and bottom) correct?
  12. Jan 28, 2009 #11
    Okay I did what I said on the previous post and so far I have:

    c^4[x'^2 + (vt')^2] - (c^2t')^2 - (vx'^2) / c^4(1-v^2/c^2)

    What can I do now to move forward? I don't see anything else cancel out.
  13. Jan 28, 2009 #12
    Okay my way isn't right, still stuck on this :|
  14. Jan 29, 2009 #13


    User Avatar
    Science Advisor
    Homework Helper

    It seems you're having a little problem with the algebra. Let me get you started.

    Since y' = y and z' = z, it suffices to show that x'^2 - c^2 t'^2 = x^2 - c^2 t^2.
    Plugging it in:

    [tex]x^2 - c^2 t^2 = \left( \frac{ x' + v t' }{ \sqrt{1 - v^2 / c^2} } \right)^2 - c^2 \left( \frac{ t' + v x' / c^2 }{ \sqrt{1 - v^2 / c^2} } \right)^2.[/tex]
    Let me take the c^2 in the second term inside the brackets (c^2 a^2 = (c a)^2) and take the square root outside ( (a/b)^2 = a^2 / b^2, sqrt(a)^2 = a):
    [tex]x^2 - c^2 t^2 = \frac{ \left( x' + v t' \right)^2 - \left( c t' + v x' / c \right)^2 }{ 1 - v^2 / c^2 }.[/tex]
    Open up the brackets:
    [tex]x^2 - c^2 t^2 = \frac{ x'^2 + 2 x' v t' + v^2 t'^2 - c^2 t'^2 - 2 t' v x' - v^2 x'^2 / c^2 }{ 1 - v^2 / c^2 }.[/tex]

    Two terms cancel, after tracing the above steps carefully (make sure you understand which manipulations I did and that I didn't make any sign errors :tongue:) collect the terms in (x')^2 and (t')^2, i.e. write
    [tex]x^2 - c^2 t^2 = \frac{ A x'^2 + B t'^2 }{ 1 - v^2 / c^2 }.[/tex]
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Lorentz Transformation
  1. Lorentz Transformation (Replies: 3)

  2. Lorentz Transformation (Replies: 6)