Total Energy in Schwarszchild Questions Answered

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Hi,

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

Carroll said:
If K^\mu is a Killing vector, we know that

K_\mu \frac{dx^\mu}{d\lambda} = \mbox{ constant }

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,

p^\mu = m \frac{dx^\mu}{d\tau}

What about a lightlike particle?

EDIT 2: On page 207, he says that for a lightlike particle, it is convenient to normalize \lambda in such a way that p^\mu = \frac{dx^\mu}{d\lambda} for a lightlike particle. How does one justify this?
 
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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.

maverick280857 said:
What about a lightlike particle?

EDIT 2: On page 207, he says that for a lightlike particle, it is convenient to normalize \lambda in such a way that p^\mu = \frac{dx^\mu}{d\lambda} for a lightlike particle. How does one justify this?

The same property holds for any geodesic.

The affine parameter is only defined up to some scaling factor \lambda \rightarrow a \lambda + b. The choice of a and b is arbitrary.
 
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