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

Total energy in the Schwarszchild Solution

  1. Apr 27, 2014 #1

    I am working through Carrol's book on General Relativity. On page 206, he makes the following statement:

    How does this follow from the definition of a Killing vector?

    He uses this equation to determine the total energy (equation 5.61).

    Thanks in advance.

    EDIT: This makes sense for a timelike particle for which,

    [tex]p^\mu = m \frac{dx^\mu}{d\tau}[/tex]

    What about a lightlike particle?

    EDIT 2: On page 207, he says that for a lightlike particle, it is convenient to normalize [itex]\lambda[/itex] in such a way that [itex]p^\mu = \frac{dx^\mu}{d\lambda}[/itex] for a lightlike particle. How does one justify this?
    Last edited: Apr 27, 2014
  2. jcsd
  3. Apr 27, 2014 #2
    I don't have Carrolls book, but I think he explains this very well in his lecture notes. See page 140, equation (5.43) and the text around that equation.

    The same property holds for any geodesic.

    The affine parameter is only defined up to some scaling factor [itex] \lambda \rightarrow a \lambda + b [/itex]. The choice of a and b is arbitrary.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Total energy in the Schwarszchild Solution