Why is time dilation expressed as t' in special relativity?

[DB]

In http://www.freewebs.com/mouldy-fart/Space,%20Time%20and%20SR.pdf paper the author wrote:

$$t'=\sqrt{\frac{4h^2}{c^2-v^2}}=\frac{2h}{\sqrt{c^2-v^2}}$$

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

Why so?

Also, isn't the relativistic beta considered just v/c, not as the author stated (c/v)^2?

 

[hypermorphism]

In http://www.freewebs.com/mouldy-fart/Space,%20Time%20and%20SR.pdf paper the author wrote:

$$t'=\sqrt{\frac{4h^2}{c^2-v^2}}=\frac{2h}{\sqrt{c^2-v^2}}$$

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

Why so?

Also, isn't the relativistic beta considered just v/c, not as the author stated (c/v)^2?

The author factored out sqrt(c^2) in order to get the frequently occurring expression 1/sqrt(1-(v/c)^2), which is referred to by the symbol $$\gamma$$ in most texts. I'm not sure what you mean by beta, but the unitless replacement v := v/c is common.

 

[dextercioby]

Customarily "c=1" and no such tricks are necessary...

Daniel.

 

[DB]

thanks guys, but I still can't see how
$$\frac{2h}{\sqrt{c^2-v^2}}=\frac{2h}{\sqrt{c^2}}\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$
How did he factor out sqrt(c^2) to get 1/sqrt(1-(v/c)^2)?

 

[hypermorphism]

thanks guys, but I still can't see how
$$\frac{2h}{\sqrt{c^2-v^2}}=\frac{2h}{\sqrt{c^2}}\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$
How did he factor out sqrt(c^2) to get 1/sqrt(1-(v/c)^2)?

This is basic factoring:
$$\frac{1}{\sqrt{c^2-v^2}} = \frac{1}{\sqrt{c^2(1-\frac{v^2}{c^2})}} = \frac{1}{\sqrt{c^2}\sqrt{1-\frac{v^2}{c^2}}} = \frac{1}{\sqrt{c^2}}\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

 

[selfAdjoint]

thanks guys, but I still can't see how
$$\frac{2h}{\sqrt{c^2-v^2}}=\frac{2h}{\sqrt{c^2}}\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$
How did he factor out sqrt(c^2) to get 1/sqrt(1-(v/c)^2)?

Since $$c^2$$ is a constant, multiply and divide by it:$$c^2 (c^2 - v^2)/c^2 = c^2(\frac {c^2}{c^2} - \frac {v^2}{c^2}) = c^2 ( 1 - \frac{v^2}{c^2})$$, no?

 

[jtbell]

Also, in general,

$$\sqrt{ab} = \sqrt{a} \sqrt{b}$$

if that's what's bothering DB.

 

[DB]

Since $$c^2$$ is a constant, multiply and divide by it:$$c^2 (c^2 - v^2)/c^2 = c^2(\frac {c^2}{c^2} - \frac {v^2}{c^2}) = c^2 ( 1 - \frac{v^2}{c^2})$$, no?

Thanks.
Ok. I see the math there, but why do we multiply by c^2/c^2?
(I know you've stated that its constant, but can you elaborate please?)

Also, in general,

$$\sqrt{ab} = \sqrt{a} \sqrt{b}$$

if that's what's bothering DB.

Nawww, don't worry bout that.

 

[cepheid]

Thanks.
Ok. I see the math there, but why do we multiply by c^2/c^2?
(I know you've stated that its constant, but can you elaborate please?)

c² / c² = 1, so you can multiply by it whenever you want without changing anything.

selfAdjoint was just showing you the intermediate steps very explicitly, making it clear that having an 'extra' c² multiplying the expression out front is fine as long as you divide both terms in the expression by c² to compensate. But surely you can arrive at the end result straight away, just by thinking of it as "factoring out a c²" from both terms:

$$(c^2 - v^2) = c^2(1- \frac{v^2}{c^2})$$

well that's exactly what hypermorphism and selfAdjoint already showed you.

 

[DB]

Ahhh, I see it now, thanks guys.