# Special relativity-analogy of rotation

1. May 28, 2017

### Karol

1. The problem statement, all variables and given/known data
Analogy to rotation:
$$(x')^2+(y')^2+(z')^2-c^2(t')^2=x^2+y^2+z^2-c^2t^2$$
It isn't

2. Relevant equations
Lorentz transformations:
$$x'=\frac{x-ut}{\sqrt{1-u^2/c^2}}$$
$$t'=\frac{t-ux/c^2}{\sqrt{1-u^2/c^2}}$$

3. The attempt at a solution
$~(x')^2-c^2(t')^2~$ must equal $~x^2-c^2t^2$ but it isn't so:
$$\frac{(x-ut)^2}{1-u^2/c^2}-\frac{c^2(t-\frac{ux}{c^2})}{1-u^2/c^2}\neq x^2-c^2t^2$$

2. May 28, 2017

### BvU

Perhaps if you restore the square in the numerator of t'2 you'll fare better $$\frac{(x-ut)^2}{1-u^2/c^2}-\frac{c^2(t-\frac{ux}{c^2})^2}{1-u^2/c^2}\ = x^2-c^2t^2 \ \ !$$

3. May 29, 2017

### Karol

$$\frac{(x-ut)^2}{1-u^2/c^2}-\frac{c^2(t-\frac{ux}{c^2})^2}{1-u^2/c^2}\ =\frac{x^2-2xut+u^2t^2-c^2t^2+2cxut-u^2x^2}{(c+u)(c-u)/c^2}$$
$$=\frac{(1-u^2)x^2+(c-1)2xut+(u^2-c^2)t^2}{(c+u)(c-u)/c^2}$$

4. May 29, 2017

### BvU

Nonsense. $c$ only occurs as $c^2$ so you can't have a $c^1$ in there. Another check you should do: dimensions: they don't fit !

5. May 29, 2017

### Karol

$$\frac{(x-ut)^2}{1-u^2/c^2}-\frac{c^2(t-\frac{ux}{c^2})^2}{1-u^2/c^2}\ =...=\frac{x^2+u^2t^2-c^2t^2-u^2x^2}{(c+u)(c-u)/c^2}$$
$$=\frac{(1-u^2)x^2+(u^2-c^2)t^2}{(c+u)(c-u)/c^2}$$

6. May 29, 2017

### BvU

Numerator dimensions aren't the same: can't have $x^2$ and $u^2 x^2$ side by side.

7. May 29, 2017

### Karol

$$\frac{(x-ut)^2}{1-u^2/c^2}-\frac{c^2(t-\frac{ux}{c^2})^2}{1-u^2/c^2}\ =...=\frac{x^2+u^2t^2-c^2t^2-\frac{u^2}{c^2}x^2}{(c+u)(c-u)/c^2}$$
$$=\frac{(1-\frac{u^2}{c^2})x^2+(u^2-c^2)t^2}{1-u^2/c^2}=x^2-c^2t^2$$

8. May 29, 2017

### BvU

Bingo !

9. May 29, 2017

### Karol

You are great, BvU