# The Doppler effect from a uniformly moving mirror

Regarding this paper by Aleksandar Gjurchinovski,

The Doppler effect from a uniformly moving mirror

on page 4 he says that eq. (8) can be transformed into eq. (12) by the use of "some simple algebra". I've been trying to do this for several days without success. Could anybody please help me?

Here are the results of my fruitless attemps:

• Can you give some more details on what you have tried? – Munthe Apr 5 '18 at 0:59
• I tried to, I scanned the papers and tried to post them here, but they are in .pdf and the SE software didn't accept them. I don't know if there is a Linux program that converts .pdf to .jpg or something like that. – Luiz de Assis Netto Apr 6 '18 at 3:26
• I just did it. I converted the files and added to the post (see above). I just posted two pages but there's a lot more than that. – Luiz de Assis Netto Apr 6 '18 at 3:43
• Most pages are written by pencil, though. I'm afraid if I scan them and pass them on to you, you won't be able to read them. – Luiz de Assis Netto Apr 6 '18 at 6:33
• People here are much more likely to read your equations if you write them using MathJax. – PM 2Ring Apr 6 '18 at 17:18

By now I imagine you have either figured the math out or moved on to new problems, but just in case, here it is:

Gjurchinovski gives us in (8) that

$$\frac{\lambda}{\lambda_0} = \frac{c+v\frac{\cos(\alpha + \beta)}{\cos \alpha}}{c-\frac{v}{\cos \alpha}} = \frac{\cos \alpha +\frac{v}{c}\cos(\alpha + \beta)}{\cos \alpha - \frac{v}{c}} = \frac{\cos \alpha + w \cos(\alpha+\beta)}{\cos \alpha - w}$$

where we have defined $w = v/c$. He also gives in (11) that

$$\cos \beta = \frac{-2w+(1+w^2)\cos \alpha}{1-2w\cos \alpha + w^2} = \frac{-2w+(1+w^2)\cos \alpha}{D}$$

Where we, recognizing that the denominator in (11) is the same as the desired denominator of (12), have defined $$D = 1-2w\cos \alpha + w^2$$

Now look at $$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta = \cos \alpha \cos \beta - \sin \alpha \sqrt{1-\cos^2\beta}$$

Looking at the square root and substituting for $\cos \beta$ gives

$$\sqrt{1-\cos^2\beta} = \sqrt{1 - (\frac{-2w+(1+w^2)\cos \alpha}{D})^2} = \sqrt{\frac{D^2-(-2w+(1+w^2)\cos \alpha)^2}{D^2}}$$

Substituting for $D$ and crunching through the algebra, remembering that $\sin^2\theta + \cos^2\theta = 1$ gives

$$\sqrt{1-\cos^2\beta} = \sqrt{\bigg(\frac{\sin \alpha (1-w^2)}{D}\bigg)^2} = \frac{\sin \alpha (1-w^2)}{D}$$

where we have chosen $(1-w^2)$ rather than $(w^2-1)$ to ensure that the square root function gives the positive root, since $w = v/c < 1$ always.

Substituting our results for $\sqrt{1-\cos^2\beta}$ and $\cos \beta$ into the expression for $\cos (\alpha + \beta)$ gives

$$\cos(\alpha + \beta) = \cos \alpha \frac{-2w + (1 + w^2)\cos \alpha}{D} + \sin \alpha \frac{\sin \alpha (1-w^2)}{D} = \frac{-2w\cos\alpha + (1+w^2)\cos^2\alpha+(1-w^2)\sin^2\alpha}{D}$$

So for the complete expression we have $$\frac{\lambda}{\lambda_0} = \frac{D\cos \alpha+w\big(-2w\cos\alpha + (1+w^2)\cos^2\alpha+(1-w^2)\sin^2\alpha\big)}{(\cos \alpha -w)D}$$

Expand and simplify the numerator, substituting in the definition of D, and you find it is equal to $(\cos \alpha - w)(1-w^2)$, which gives $$\frac{\lambda}{\lambda_0} = \frac{1-w^2}{D} = \frac{1-(v/c)^2}{1-2(v/c)\cos \alpha + (v/c)^2}$$

as desired.