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I'm trying to go from:

$$ M^2 = \frac{M_S^2 + \frac{2}{\gamma-1}}{\frac{2M_S^2 \gamma}{\gamma-1} - 1}$$

to:

$$ M^2 =1 - \left[\frac{\gamma +1}{2 \gamma} \right] \left[ \frac{M_S^2-1}{(M_S^2-1) + \frac{\gamma +1}{2 \gamma}} \right]$$

but I'm finding it difficult. So far, I got:

$$ M^2 = \frac{M_S^2 + \frac{2}{\gamma-1}}{\frac{2M_S^2 \gamma}{\gamma-1} - 1} = \frac{\gamma -1}{2 \gamma} \left[ \frac{M_S^2 + \frac{2}{\gamma-1}}{M_S^2 \gamma - \frac{\gamma-1}{2 \gamma}} \right] = \frac{\gamma -1}{2 \gamma} \left[ \frac{M_S^2 + \frac{2}{\gamma-1}}{M_S^2 \gamma -1 + \frac{\gamma+1}{2 \gamma}} \right]$$

but was unsuccessful from here on out. What should I do after this?

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  • $\begingroup$ As written (and barring extra relations between the values), the expressions are not equal. Looking at the denominators ... In the first expression, $2M_S^2$ doesn't (and won't) get multiplied by $\gamma$; but the second expression's $M_S^2$ gets multiplied by $2\gamma$. ... Changing the first denominator's $2M_S^2$ to $2\gamma M_S^2$ happens to fix the problem, but I don't know if that's valid. :) $\endgroup$
    – Blue
    Commented May 18 at 11:31
  • $\begingroup$ If you substitute some values for $\gamma$ and $M_S$ into the first and second expressions you will see that the expressions are not equivalent, so there must be some mistake with your question as posed. $\endgroup$ Commented May 18 at 11:34
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    $\begingroup$ Yes, I was missing a $\gamma$ term. I plugged it in the question just now. Thank you. $\endgroup$ Commented May 18 at 14:53

1 Answer 1

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Do it reverse:

\begin{align*} M^2 & =1 - \left[\frac{\gamma +1}{2 \gamma} \right] \left[ \frac{M_S^2-1}{(M_S^2-1) + \frac{\gamma +1}{2 \gamma}} \right] \\ & =1 - \frac{M_S^2-1}{\frac{2\gamma(M_S^2-1)}{\gamma +1} + 1} \\ & =1 - \frac{(M_S^2-1)(\gamma + 1)}{2\gamma(M_S^2-1)+(\gamma +1)} \\ & = \frac{2\gamma M_S^2-2\gamma + \gamma + 1 -\gamma M_S^2 - M_S^2 + \gamma +1}{2\gamma(M_S^2-1)+(\gamma +1)} \\ & = \frac{M_S^2 (\gamma - 1) + 2}{2\gamma M_S^2-(\gamma -1)} \\ & = \frac{M_S^2 + \frac{2}{\gamma-1}}{\frac{2\gamma M_S^2}{\gamma-1} - 1} \end{align*}

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  • $\begingroup$ You have an extra $\gamma$ in your final denominator. (Or, perhaps, OP is missing a $\gamma$ in their first denominator.) $\endgroup$
    – Blue
    Commented May 18 at 11:43
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    $\begingroup$ I think OP is missing a $\gamma$. $\endgroup$
    – user326159
    Commented May 18 at 11:43
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    $\begingroup$ That is correct, I was, in fact, missing a $\gamma$ term in the denominator. Thanks $\endgroup$ Commented May 18 at 14:52

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