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I saw this video on Lorentz transformation and needed to refresh my memory a bit.

If

$$\frac{t}{t'}= \sqrt{1-\frac{v^2}{c^2}}$$

and

$$\gamma = \frac {t'}t $$

How do you make this equal ?

$$\gamma = \frac 1{\sqrt{1-\frac {v^2}{c^2}}}$$

Most of you will probably say "Oh If you can't do this you're way to ahead of yourself". You don't need to say that since I am aware of that and just took this as an example.

Daniel Fischer
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Tarmiac
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2 Answers2

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So we have an expression for $t/t'$, $$ \frac{t}{t'} = \sqrt{1-v^2/c^2},$$

but we need the reciprocal, $t'/t$, in order to know what $\gamma$ is. To do this we will write the right hand side of our equation as a fraction. This can be done by writing it "over one".

$$ \frac{\color{red}{t}}{\color{blue}{t'}} = \frac{\color{red}{\sqrt{1-v^2/c^2}}}{\color{blue}{1}}.$$

Now we can find the reciprocal by just flipping both sides upside down.

$$ \frac{\color{blue}{t'}}{\color{red}{t}} = \frac{\color{blue}{1}}{\color{red}{\sqrt{1-v^2/c^2}}},$$

Now that we know what $t'/t$ is we can replace it with $\gamma$ (because we know that $\gamma=t'/t$). This replacement gives us the desired equation.

$$ \gamma = \frac{1}{\sqrt{1-v^2/c^2}}.$$

Spencer
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  • man, how do you make colors in your equations ? – Fabrice NEYRET Oct 08 '15 at 19:29
  • @FabriceNEYRET, Here is an example "\color{blue}{equation}" => $\color{blue}{equation}$. – Spencer Oct 08 '15 at 19:30
  • $\color{red}G\color{green}r\color{blue}e\color{magenta}e\color{cyan}t \color{yellow}!$ :-) – Fabrice NEYRET Oct 08 '15 at 19:34
  • Compare the first equation in your answer and the first one in the question: they are not the same. Yours is the inverse of that of the question. It is $t'/t$ whereas you have $t/t'$ – null Oct 08 '15 at 19:34
  • @null, that wasn't what it looked like when I started anwering the question. Looks like the edit history is a bit fishy, the prime on the t keeps switching places. – Spencer Oct 08 '15 at 19:39
  • you are right, I didn't notice that there were so many edits. I only changed the $ to $$ in my edit to make this more readable. – null Oct 08 '15 at 19:48
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    @null In case you wonder, you started editing before the OP fixed that markup error. You submitted your suggestion a few seconds after the OP finished his fix, and when your suggestion was approved, the fix was undone. It's a relatively rare race condition. Doesn't happen too often, but every week or so, something like that occurs. – Daniel Fischer Oct 08 '15 at 20:40
  • @DanielFischer thank you for the clarification and fixing the problem. Good to know – null Oct 08 '15 at 20:47
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Let $\frac{x}{y} = z$ then:

$\frac{x}{y}*y = z*y$

$x = z*y$

$1 = z*\frac{y}{x}$

$\frac{1}{z} = \frac{y}{x}$

Basically $\frac{1}{\frac{x}{y}} = \frac{y}{x}$ is such a basic identity we tend to forget that it's not immediately obvious to everyone.

fleablood
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