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I'm trying to prove an identity from physics. I have the following two equations ($M$ is a constant):

$$e^2 = \left(1-\frac{2M}{r}\right)\left(1+\frac{l^2}{r^2}\right)$$ and $$r = \frac{l^2}{2M}\left[1 - \sqrt{1 - 12\left(\frac{M}{l}\right)^2}\right]$$ and I need to show that $$\frac{l}{e} = \sqrt{Mr}\left(1 - \frac{2M}{r}\right)^{-1}$$

I've tried rearranging the second eqation to get $l^2$ on its own and then dividing this by $e^2$ from equation 1 to get an expression for $l^2/e^2$ but this is quite messy and I can't see a clear way to simplify it. Can anyone see a straightforward way of doing this?

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  • $\begingroup$ Should that "12" inside the radical be a "2"? $\endgroup$ – colormegone May 15 '13 at 0:32
  • $\begingroup$ I do believe it's supposed to be a 12. $\endgroup$ – saurs May 15 '13 at 1:00
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let $x=\dfrac{l^2}{M^2},a=\dfrac{2r}{M} \to a=x-\sqrt{x^2-12x} \to x^2-12x=(x-a)^2 \to x=\dfrac{a^2}{2a-12}$

$\sqrt{1-\dfrac{12}{x}}=\sqrt{\dfrac{a^2-12*2a+12*12}{a^2}}=\dfrac{|a-12|}{a}$

now you can simplify more easy.

ya,if 12 is wrong, you give a right number but the middle result is same.

But it seems the last one is not correct,there should a factor of $\sqrt{\left(1 - \dfrac{2M}{r}\right)^{-1}}$

edit: I am interesting to go further after seeing the source poster.

$l^2=\dfrac{(aM)^2}{2a-12}=\dfrac{4r^2}{\dfrac{4r}{M}-12}=\dfrac{r^2M}{r-3M}=\dfrac{rM}{1-\dfrac{3M}{r}}$ if $a>12$, we have :

$r=\dfrac{l^2}{2M}\left[1-\dfrac{a-12}{a} \right]=\dfrac{6l^2}{aM}=\dfrac{3l^2}{r} \to \dfrac{l^2}{r^2}=\dfrac{1}{3}\to e^2 = \left(1-\dfrac{2M}{r}\right)\left(1+\dfrac{1}{3}\right)=\left(1-\dfrac{2M}{r}\right)\dfrac{4}{3}$

$\dfrac{l^2}{e^2}=\dfrac{3l^2}{4}\left(1 - \dfrac{2M}{r}\right)^{-1}=\dfrac{3rM}{4}\left(1 - \dfrac{2M}{r}\right)^{-1}\left(1 - \dfrac{3M}{r}\right)^{-1}$

if $a<12$ ,we have :

$r=\dfrac{l^2}{2M}\left[1-\dfrac{12-a}{a} \right]=\dfrac{l^2}{2M}\dfrac{2a-12}{a}=\dfrac{a-6}{2r} \to \dfrac{l^2}{r^2}=\dfrac{2}{a-6}=\dfrac{M}{r-3M} \to$ $1+\dfrac{l^2}{r^2}=\dfrac{r-2M}{r-3M} \to e^2=\left(1 - \dfrac{2M}{r}\right)\dfrac{r-2M}{r-3M} \to $

$\dfrac{l^2}{e^2}=\dfrac{r-3M}{r-2M}*rM*\left(1 - \dfrac{2M}{r}\right)^{-1}\left(1 - \dfrac{3M}{r}\right)^{-1}=rM\left(1 - \dfrac{2M}{r}\right)^{-2}$

so that is the answer!

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  • $\begingroup$ my calculation shows 12 should 4 and the final result is $mr\sqrt{(1-(\frac{2M}{r})^{-1}}$ $\endgroup$ – chenbai May 15 '13 at 1:34
  • $\begingroup$ Yes, the equation shown for $ \ l/e \ $ is not correct dimensionally: $l$ has the units of $r$ , and $e$ is proportional to $\ (1 - \frac{2M}{R})^{1/2} \ $ , since $ \ (1 + \frac{l^2}{r^2} ) \ $ is dimensionless. So $ \ l/e \ $ should be proportional to $r$ and contain a factor of $\ (1 - \frac{2M}{R})^{-1/2} \ $. $\endgroup$ – colormegone May 15 '13 at 2:31
  • $\begingroup$ (missed the 5-minute deadline) That last line in the OP would work if the radical were placed on the other factor. Is there a typo? $\endgroup$ – colormegone May 15 '13 at 2:38
  • $\begingroup$ @RecklessReckoner Well this identity is from the book by Hartle called "Gravity: An Introduction to Einstein's General Relativity". I have copied it down exactly as it's written in the book. See eqn 9.45 here: gyazo.com/6c762b2b4ec530242945c4d54fc92604 (the eqn 9.34 that he refers to is the second equation in my original post) $\endgroup$ – saurs May 15 '13 at 3:24
  • $\begingroup$ As it should happen, I have a library copy right here (I thought this looked like orbital mechanics)... I suspect there is a typo where that second factor should have been under the square root also. As for the "12" , Equation 9.34 may only be applicable at the pericenter and apocenter of the orbit. $\endgroup$ – colormegone May 15 '13 at 3:40
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I'd let $\alpha =l/e$ so $ e=l/\alpha$ can be substituted in. Then one has $\alpha,l$ as coordinates. Eliminate $l$ and one is left with an equation for just $\alpha$. There might be neater ways of doing it, but this sort of method is at least free of guesswork.

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  • $\begingroup$ When you say "eliminate $l$", do you mean use the second equation to write $l$ in terms of $r$ then substitute into the first equation? I can see how that works in principle but it looks like it would be (very) messy in this case. $\endgroup$ – saurs May 15 '13 at 1:16
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I concur with chenbai about the "12" being a "4" (I think it's a "2" inside the squared ratio). Here's how I worked this out.

Rearrange the second equation to deal with the radical:

$$2Mr \ = \ l^2 \ - \ l^2 \cdot \sqrt{1 - (\frac{2M}{l})^2} \ \Rightarrow \ l^2 \cdot \sqrt{1 - (\frac{2M}{l})^2} \ = \ l^2 \ - \ 2Mr $$

$$\Rightarrow \ l^4 \cdot (1 - \frac{4M^2}{l^2}) \ = \ l^4 \ - \ 4Mrl^2 \ + \ 4M^2r^2 $$

$$ \Rightarrow \ 4Mrl^2 \ = \ 4M^2r^2 \ + \ 4M^2l^2 \ = \ 4M^2 \cdot (r^2 + l^2) \ , \ \ \ [1] $$

after squaring to eliminate the radical and making some further algebraic rearrangement.

Now,

$$\frac{l^2}{e^2} \ = \ \frac{4M^2 \cdot (r^2 + l^2)}{4Mr} \ \cdot \ (1 - \frac{2M}{r})^{-1} \ \cdot \ (\frac{r^2}{r^2 + l^2}) \ \ , $$

[using equation [1] and the equation for $e^2$ ]

$$\Rightarrow \ \frac{l^2}{e^2} \ = \ Mr \cdot \ (1 - \frac{2M}{r})^{-1} \ \Rightarrow \ \frac{l}{e} \ = \ \sqrt{Mr \cdot \ (1 - \frac{2M}{r})^{-1}} . $$

OK, so I was partly wrong: the "Mr" is under the radical...

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  • $\begingroup$ Thanks for your help. The equation for $r$ again comes from Hartle: see gyazo.com/971e3a9d424815bc1419601a1738c50c. This comes from finding the extrema of a potential function $V(r)$. Perhaps it is a typo in Hartle, as your working seems to make sense. $\endgroup$ – saurs May 15 '13 at 3:45
  • $\begingroup$ I'll have to look at the passage further to see whether 9.34 for the extrema is intended to be used here, or in its stead, a similar expression over the orbit in general. I was a bit concerned about the "12" because I can't see how to make a factor of 3 disappear for the final expression... $\endgroup$ – colormegone May 15 '13 at 3:49

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