Special relativity-analogy of rotation

[Karol]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

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

 

[BvU]

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

 

[Karol]

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

 

[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 !

 

[Karol]

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

 

[BvU]

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

 

[Karol]

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

 

[BvU]

Bingo !

 

[Karol]

You are great, BvU