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I'm working in Weinberg's Gravitation and Cosmology and I'm having trouble seeing how he reduced a certain equation. He starts a derivation on p.185 going from a general metric in the standard form $$d\tau^2=B(r)dt^2-A(r)dr^2-r^2d\theta^2-r^2\sin^2\theta d\varphi^2 \tag{8.4.1}.$$

From the geodesic (free-fall) equation $$\frac{d^2x^{\mu}}{dp^2}+\Gamma^\mu_{\nu\lambda}\frac{dx^\nu}{dp}\frac{dx^\lambda}{dp}=0$$ he works out 4 differential equations using a parameter p and the non-vanishing terms of the affine connection.

Then finds $$\frac{dt}{dp}=\frac{1}{B(r)} \tag{8.4.10},$$ angular momentum $$r^2\frac{d\varphi}{dp}=J\tag{8.4.11},$$ and energy $$A(r)(\frac{dr}{dp})^2+\frac{J^2}{r^2} - \frac{1}{B(r)}=-E \tag{8.4.13}.$$

Then eliminates the parameter p to get equations of motion back in terms of $t$.

He finds that: $$r^2\frac{d\varphi}{dt}=JB(r)\tag{8.4.18}$$ $$\frac{A(r)}{B^2(r)}(\frac{dr}{dt})^2+\frac{J^2}{r^2 }-\frac{1}{B(r)}=-E\tag{8.4.19}$$ $$d\tau^2=EB^2(r)dt^2\tag{8.4.20}$$

Then he says:

For a slowly moving particle in a weak field $J^2/r^2$, $(dr/dt)^2$, $A-1$, and $B-1\approx2\phi$ will all be small, and to first order in these equations the above equations of motion become $$r^2\frac{d\varphi}{dt}\approx J$$ $$\frac{1}{2}(\frac{dr}{dt})^2+\frac{J^2}{2r^2}+\phi\approx \frac{1-E}{2}$$

I'm sure this is just a question of algebra but for the life of me I can't see how eqn. 8.4.19 reduces to the last equation. I've wasted a ton of scrap paper trying to juggle the terms in 8.4.19 and letting $B^2=(1+2\phi)^2\approx(1+4\phi)$ but in never works out quite right. Maybe I'm missing something about how the approximations for slow motion and weak field interact with 8.4.19?

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If $B \approx 1 + 2\phi$ then $\frac{1}{B} \approx 1 - 2\phi$, hence the term before the equal sign becomes $(1 - 2\phi)$, hence

$$\frac{A}{B^2}\left(\frac{dr}{dt}\right)^2 + \frac{J^2}{r^2} - (1 - 2\phi) = -E$$

$$\frac{A}{B^2}\left(\frac{dr}{dt}\right)^2 + \frac{J^2}{r^2} + 2\phi = -E + 1$$

$$\frac{A}{2B^2}\left(\frac{dr}{dt}\right)^2 + \frac{J^2}{2r^2} + \phi = \frac{-E + 1}{2}$$

It is also reasonable to assert that $\frac{A}{B^2} \approx 1$, which eventually brings you to

$$\frac{1}{2}\left(\frac{dr}{dt}\right)^2 + \frac{J^2}{2r^2} + \phi = \frac{1 - E}{2}$$

More on the last assertion

$$\frac{A}{B^2} = \frac{A-1+1}{B^2} = \frac{1}{B^2}$$

Since $A-1$ is small. The rest is in the comments.

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    $\begingroup$ Can you explain that last estimation? I would think that $B^2-1=(B-1)(B+1) = 2\phi(2+2\phi)=4\phi+4\phi^2$. So where did the 1st order term go in your estimation? $\endgroup$ – Iron Charioteer Mar 17 '18 at 18:01
  • $\begingroup$ Yes, that was an error of mine, sorry. We can use a very better and dirtier trick for that: $$\frac{1}{B^2} = \frac{1}{B}\cdot\frac{1}{B}$$ Since $\frac{1}{B} \approx 1-2\phi$ we have $$\frac{1-2\phi}{B} = \frac{1-2\phi}{1 + 2\phi} \approx 1$$ for small $\phi$. $\endgroup$ – Von Neumann Mar 17 '18 at 18:15
  • $\begingroup$ Notice that with the use of the previous trick, we don't need to write $B^2 + 1 - 1$ anymore. Just for $A$. $\endgroup$ – Von Neumann Mar 17 '18 at 18:16
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    $\begingroup$ Ah ha! Yes, I like that method much better :) $\endgroup$ – Iron Charioteer Mar 17 '18 at 18:24

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