I'm still struggling with Carroll's discussion of the Newtonian Limit. I'm hoping no one will mind if I ask several questions here as they all relate to the same section (pages 105-106) of his “Lecture notes on general relativity” and I'm trying to get this thing straight in my mind.

He states that “moving slowly” means $\frac{dx^{i}}{d\tau}\ll\frac{dt}{d\tau} $ , and therefore the geodesic equation becomes $\frac{d^{2}x^{\mu}}{d\tau^{2}}=\Gamma_{00}^{\mu}+\left(\frac{dt}{d\tau}\right)^{2}=0 $

Q1. Why does $\frac{dx^{i}}{d\tau}\ll\frac{dt}{d\tau}$ ? What's the justification in assuming that the rate of change of coordinate time wrt proper time is much larger than the rate of change of distance wrt proper time? If I can understand this I (I think I) can understand why he only considers the $\Gamma_{00}^{\mu}$ Christoffel symbols.

Because the field is static, he changes the geodesic equation $\Gamma_{00}^{\mu}=-\frac{1}{2}g^{\mu\lambda}\partial_{\lambda}g_{00}$ to $\Gamma_{00}^{\mu}=-\frac{1}{2}\eta^{\mu\lambda}\partial_{\lambda}h_{00} $

Q2. So, he's gone from $g^{\mu\lambda}$ to $\eta^{\mu\lambda}$ . Is that because he ignores the $h^{\mu\lambda}$ bit of $g^{\mu\lambda}=\eta^{\mu\lambda} -h^{\mu\lambda}$ because it's negligible?

Q3. He's also gone from $g_{00}$ to $h_{00}$ . Is that because the rate of change of $g_{00}=\eta_{00}+h_{00}$ wrt $\lambda$ (I know this is a clumsy way of stating it - sorry) equals $h_{00}$ because $\eta_{00}$ doesn't change wrt to anything?

Thank you.

  • $\begingroup$ I've corrected my original mess up with various indexes. Thank you Nikolaj. $\endgroup$
    – Peter4075
    Sep 12, 2011 at 14:31

1 Answer 1

  1. Think in units of $c=1$. If $\frac{∂x}{∂t}=\frac{∂t}{∂t}$, then you are moving at the speed of light. The faster you are, the more your world line is tangent to the light cone. The condition just says that you're moving at non-relativistic speed, i.e. about tangent to the time axis in a Minkowski-diagram.

  2. The term is $g∂g=(\eta+h)∂(\eta+h)=(\eta+h)∂h==\eta∂h+O(h∂h)$. In Minkowski coordinates, you have $∂\eta=0$ in any case. And he probably says somewhere, that the pertubation $h$ doesn't vary too much, so that $∂h$ is also small and the second order expression $h·∂h$ clearly doesn't contribute.

You also messed up an index in "$-\frac{1}{2}\eta^{\mu\lambda}\partial_{\gamma}h_{00}$".

  • $\begingroup$ Q1 - Understood. I was confused at first by $\frac{\partial t}{\partial t'}$ but realised the prime was a comma! In other word because the particles are moving slowly the time-component (ie the 0th component of the particle's four-velocity) dominates the other (spatial) components. Q2 & 3 - Understood after a night's sleep. Thanks $\endgroup$
    – Peter4075
    Sep 13, 2011 at 8:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.