0
$\begingroup$

Here is the whole Box 2.2, at Page 55

Gravitation Box 2.2

The dot behind the second $-p^2$ seems to be a "planck mass" (sarcasm, flea egg) or just the book's style to use Dot behind the equations. So the Equation is basically;

$-p^2=m=E^2-p^2$ which can be written

$-p^2=E^2-p^2$ and if we add on both sides $+p^2$ we have

$0 = E^2$

My question is "What is wrong here?" as here is no logic.

The same problem is already written above as $p^2=-E^2+p^2$ which is again nothing else than
$0=-E^2$

In my World I see it so; $m^2=0$
thus $m^2=E^2-p^2$ is $0=E^2-p^2$ and $E^2=p^2$

$\endgroup$
  • 3
    $\begingroup$ The "dot" behind the second p is the period that ends the sentence (note the use of a semicolon on the penultimate block-set equation). Most, but not all typographers recommend that, though many suggest more space setting it off than appears here. And as you can see in ACourisousMind's answer, typography matter, so pay attention. Most books define their style conventions somewhere. $\endgroup$ – dmckee --- ex-moderator kitten Mar 5 '16 at 16:45
  • 1
    $\begingroup$ Looks to me that there's two different fonts here, a serif font ($\mathsf{p}$) to denote the four-vector, and a Roman font ($p$) to denote the three-vector. $\endgroup$ – Kyle Kanos Mar 5 '16 at 17:31
5
$\begingroup$

It's a bit hard to see in this typography, but the two p are supposed to be different. The p on the l.h.s. is the four-momentum $p = (p^0,p^1,p^2,p^3)^T$, the one on the r.h.s is the three-momentum $\vec p = (p^1,p^2,p^3)^T$, and then $$ p^2 = E^2 - \vec p ^2$$ for $p^0 = E$ is tautologically true just from the definition of $p^2$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks. Also for the comments. This makes the the stuff pretty clear. $\endgroup$ – Jokela Mar 6 '16 at 10:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.