Derivation of the momentum-energy relation

[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 a lot...

: )

 

[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 $c^{2}$:

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

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

Daniel.



$$p$$

 

[robphy]

Just to encourage more use of the rapidity $$\theta$$ (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 $$\begin{align*}<br /> \cosh^2\theta - \sinh^2\theta&\equiv1\\<br /> \left(\frac{E}{m_0 c^2}\right)^2 - \left( \frac{p}{m_0c}\right)^2&=1\\<br /> E^2 - (pc)^2&=(m_0c^2)^2\\<br /> \end{align*}$$

 

[dextercioby]

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

Daniel.

 

[pmb_phy]

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

For those of you who have not yet followed the derivation for the mass-energy relation please see
http://www.geocities.com/physics_world/sr/mass_energy_equiv.htm

Pete

 

[selfAdjoint]

Though not taught in HS

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.

 

[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.

 

[misogynisticfeminist]

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

: )

 

[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...

 

[robphy]

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

 

[DB]

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