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This type of question might be voted down or frowned upon, but if anyone could point me in the right direction I would be very grateful. I have worked through a question getting the following result:

$$ \frac{\nu{'}}{\nu}=\frac{(\hat{\Omega}\cdot{\beta}-1)}{(\hat{\Omega{'}}\cdot{\beta}-1)+\frac{h\nu}{\gamma{m_e}c^2}(\hat{\Omega}\cdot\hat{\Omega}'-1)} $$

Now, the problem is: I think I have done this question correctly. That is my assumption. However, the question asks to show that:

$$ \frac{\nu{'}}{\nu}=\frac{(1-\hat{\Omega}\cdot{\beta})}{(1-\hat{\Omega{'}}\cdot{\beta})+\frac{h\nu}{\gamma{m_e}c^2}(1-\hat{\Omega}\cdot\hat{\Omega}')} $$

So my question is this: have I gone wrong or am I able to re-arrange my expression into the expression I need to show...?

Michael Roberts
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    You're right there, just multiply by $-1$ on top and bottom... – abiessu Nov 03 '15 at 16:24
  • Could you show me please...I'm struggling to see this. – Michael Roberts Nov 03 '15 at 16:24
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    Note that we have (for instance) $$-1\cdot (\hat{\Omega}\cdot{\beta} - 1) = 1-\hat{\Omega}\cdot{\beta}$$Also, as with any fraction, we're allowed to multiply the numerator and denominator both by $-1$. – Arthur Nov 03 '15 at 16:30

1 Answers1

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To get

$$\frac{\nu{'}}{\nu}=\frac{(\hat{\Omega}\cdot{\beta}-1)}{(\hat{\Omega{'}}\cdot{\beta}-1)+\frac{h\nu}{\gamma{m_e}c^2}(\hat{\Omega}\cdot\hat{\Omega}'-1)}$$

changed into the correct form, we have $(\hat{\Omega}\cdot{\beta}-1)=-(1-\hat{\Omega}\cdot{\beta})$ and $(\hat{\Omega{'}}\cdot{\beta}-1)+\frac{h\nu}{\gamma{m_e}c^2}(\hat{\Omega}\cdot\hat{\Omega}'-1)=-(1-\hat{\Omega{'}}\cdot{\beta})-\frac{h\nu}{\gamma{m_e}c^2}(1-\hat{\Omega}\cdot\hat{\Omega}')$, and so we have

$$\frac{\nu{'}}{\nu}=\frac{-(1-\hat{\Omega}\cdot{\beta})}{-(1-\hat{\Omega{'}}\cdot{\beta})-\frac{h\nu}{\gamma{m_e}c^2}(1-\hat{\Omega}\cdot\hat{\Omega}')}=\frac{(1-\hat{\Omega}\cdot{\beta})}{(1-\hat{\Omega{'}}\cdot{\beta})+\frac{h\nu}{\gamma{m_e}c^2}(1-\hat{\Omega}\cdot\hat{\Omega}')}$$

where we have used the distributive property to extract $-1$ in the denominator.

abiessu
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