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

How does one get time dilation, length contraction, and E=mc^2 from spacetime metric?

  1. Oct 20, 2014 #1
    How does one get time dilation, length contraction, and E=mc^2 from the spacetime metric?

    Suppose all that you are given is x12 + x22 + x32 - c2t2 = s2

    How do you derive time dilation, length contraction, and E=mc^2 from this?

    What is the most direct way to do this?
  2. jcsd
  3. Oct 20, 2014 #2

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2017 Award

    You should take a look at Spacetime Physics, by Taylor and Wheeler, where this is all covered. It's only about 100 pages.
  4. Oct 20, 2014 #3
    Thanks! Is there a quick answer of any resources on the web? Thanks! :)
  5. Oct 20, 2014 #4
  6. Oct 20, 2014 #5


    User Avatar

    Staff: Mentor

    Start with time dilation. The squared distance between (0,0,0,0) and (T,vT,0,0) must equal the squared distance between (0,0,0,0) and (T',0,0,0) because it's the same interval between the same two points, namely the endpoints of a journey taken at speed ##v## as viewed by the traveller (primed coordinates) and an observer moving at speed ##v## relative to the traveller (unprimed coordinates). Use the metric to calculate the squared distances, equate them, and solve for the ratio of T' to T.
  7. Oct 20, 2014 #6


    User Avatar
    Science Advisor
    Gold Member

    This is not a simple. I don't think it can really be derived from the metric alone. You need to bring in, for example, energy and momentum conservation. Then, what you end up deriving from this plus the metric is E2 - p2c2 = m2c4 (once you have the right prior motivation, this just says that the 4-norm of m * 4-velocity unit vector is m, which is trivially true). When momentum is zero, you have the special case E=mc2.
    Last edited: Oct 20, 2014
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook