# I Cox and Forshaw: Why E=mc^2

1. Jun 29, 2016

### EveningLight

Great book. Reading it for the second time. There are several "jumps" that are made in their lines of reasoning that certainly make the explanations easier to follow, but also left me wondering "Why or where did that assumption or statement come from?"

On page 77 (in my 2009 Da Capo Press version) when working towards setting up the spacetime model, they state: "The length of the hypotenuse must be either s^2 = (ct)^2+x^2 or s^2 = (ct)^2-x^2. There is no other option."

So why is the negative version even considered as a option? They try to address this in the paragraph following this, but really don't. They just say it is the only other option.

Can anyone explain to me why that is the only other option? Why is not s = ct + x not considered an option? Or any other s = function(ct,x)?

2. Jun 29, 2016

### Andrew Mason

Welcome to PF EveningLight! I can't really comment on the book without seeing it. Can you post the relevant pages?

From the assumptions of special relativity, it can be shown that for two events separated by x and t in a "rest frame" and x' and t' in a frame moving at speed v relative to the rest frame, $c^2t^2 - x^2 = c^2t'^2 - x'^2$. This is easier to see where x' = 0 (the two events occur at the same place in the moving frame) i.e.: $c^2t^2 - x^2 = c^2t'^2$. Since this equality $c^2t^2 - x^2 = c^2t'^2 - x'^2$ does not depend on v, the relative speed of the two observers, the relationship must be true for all observers. The spacetime interval between two events is defined as $s^2 = x^2 - c^2t^2$ precisely because it is invariant for observers in all frames.

AM

Last edited: Jun 30, 2016
3. Jun 30, 2016

### EveningLight

Thanks.

"From the assumptions of special relativity, it can be shown that for two events separated by x and t in a "rest frame" and x' and t' in a frame moving at speed v relative to the rest frame, c 2 t 2 − x 2 = c 2 t ′ 2 − x ′ 2"

So I guess I need the bit about "It can be shown that". I probably need someone to look at the book and understand the authors' progression of their presentation to help me out here.

4. Jun 30, 2016

### EveningLight

Would probably been good for me to also read another source about this topic. Can anyone else suggest other laymens' sources to this topic?

Thanks.

5. Jul 1, 2016

### Andrew Mason

There are any number of good physics texts on SR. My old https://www.amazon.com/Special-Relativity-M-I-T-Introductory-Physics/dp/0393097935 is easy to follow.

It is simply a matter of applying the Lorentz transformations to $c^2t'^2 - x'^2$. Texts usually skip the detailed algebra steps but I have put them in for you, one at a time:

$c^2t'^2 - x'^2 = c^2(\gamma(t-vx/c^2))^2 - (\gamma(x -vt))^2$
$= c^2\gamma^2(t^2 - 2vtx/c^2 + v^2x^2/c^4) - \gamma^2(x^2 - 2vtx + v^2t^2)$
$= \gamma^2(c^2t^2 - 2vtx + v^2x^2/c^2 - x^2 + 2vtx - v^2t^2)$
$= \gamma^2(c^2t^2 + v^2x^2/c^2 - x^2 - v^2t^2)$
$= \gamma^2(c^2t^2 - x^2 + (v^2x^2/c^2 - v^2t^2)$
$= \gamma^2(c^2t^2 - x^2 + v^2(x^2 - c^2t^2)/c^2$
$= \gamma^2(c^2t^2 - x^2 - v^2(c^2t^2 - x^2)/c^2$
$= \gamma^2(c^2t^2 - x^2)(1 - v^2/c^2)$
$= \frac{1}{(1 - v^2/c^2)}(c^2t^2 - x^2)(1 - v^2/c^2)$
$= c^2t^2 - x^2$

AM

Last edited by a moderator: May 8, 2017