# Mach equation algebraic manipulation

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?

• 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. :)
– Blue
Commented May 18 at 11:31
• 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. Commented May 18 at 11:34
• Yes, I was missing a $\gamma$ term. I plugged it in the question just now. Thank you. Commented May 18 at 14:53

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