# Derivation of the momentum-energy relation

1. Feb 23, 2005

### misogynisticfeminist

In deriving the relation $$E^2=(pc)^2+(mc^2)^2$$, I have used the relations $$p=\gamma mu$$ and $$E=\gamma m_0c^2$$. I am currently stuck until a certain step and would appreciate it if someone could show its derivation, thanks alot...

: )

2. Feb 23, 2005

### dextercioby

Okay,here goes:

$$p^{2}=\gamma^{2}m^{2}v^{2}=\frac{c^{2}m^{2}v^{2}}{c^{2}-v^{2}}$$

Add what u need,but only after multiplying by $c^{2}$:

$$p^{2}c^{2}+m^{2}c^{4}=m^{2}c^{4}(\frac{v^{2}}{c^{2}-v^{2}}+1)= m^{2}c^{4}\frac{c^{2}}{c^{2}-v^{2}}=(\gamma m c^{2})^{2}=E^{2}$$

,where i made use of Einstein's formula...

Daniel.

$$p 3. Feb 23, 2005 ### robphy Just to encourage more use of the rapidity [tex]\theta$$ (and one's trigonometric intuition):

$$u=c\tanh\theta$$ and $$\gamma=\cosh\theta$$ (So, $$\gamma u=c\sinh\theta$$)

So, start with $$p=m_0c\sinh\theta$$ and $$E=m_0c^2\cosh\theta$$.

Since \begin{align*} \cosh^2\theta - \sinh^2\theta&\equiv1\\ \left(\frac{E}{m_0 c^2}\right)^2 - \left( \frac{p}{m_0c}\right)^2&=1\\ E^2 - (pc)^2&=(m_0c^2)^2\\ \end{align*}

4. Feb 23, 2005

### dextercioby

Though not taught in HS,your method is elegant.Mine is simply lacking in inspiration.:yuck:

Daniel.

5. Feb 23, 2005

### pmb_phy

6. Feb 23, 2005

Staff Emeritus
IMHO, there is no reason why hyperbolic trig functions should not be taught in HS trig (in 1950 we got them in "College Math" which I took Junior year in HS). Nor is there that rapidity should not be taught in intro to relativity for math-enabled students.

7. Feb 23, 2005

### dextercioby

I didn't mean hyperbolic trig functions,but special relativity & SLT using them...Maybe they're taught somewhere,but i would find that level (of teaching physics) too high...

Daniel.

8. Feb 23, 2005

### misogynisticfeminist

hey, thanks for the help ! and i'll definitely check out that trig derivation once I've started on hyperbolic functions though...

: )

9. Feb 23, 2005

### DB

Umm, not to bother you guys but, how does
$$p^{2}=\gamma^{2}m^{2}v^{2}=\frac{c^{2}m^{2}v^{2}}{ c^{2}-v^{2}}$$?
What I get is:

$$(\frac{1}{\sqrt{1-v^2/c^2}})^2=\frac{1}{1-v^2/c^2} *m^2v^2$$

$$\frac{m^2v^2}{1-v^2/c^2}$$

Now I'm stuck....

10. Feb 23, 2005

### robphy

multiply by $$\frac{c^2}{c^2}$$

11. Feb 24, 2005

### DB

oo, ya now I got it, thanks. Is that what Dexter meant when he said:
"Add what u need,but only after multiplying by $c^2$:"?