Total energy in Schwarzschild spacetime from Killing vectors

[maverick280857]

Hi,

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

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?

 

[clamtrox]

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.

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.