Relativistic energy is sometimes derived in the following manner and I'm having trouble deciphering one mathematical step. So it goes like this:

Given that $$F=\frac{ dp }{ dt }$$ and reletavistic momentum is given by
p=\frac{ mv }{ sqrt(1-\frac{v^2}{c^2}) }

W=int_{x _{1}}^{x_{2}} F dx=int_{x _{1}}^{x_{2}} \frac{dp}{dt}dx

which is fine
then they say
\frac{dp}{dt}=\frac{d}{dt}\frac{mv}{sqrt(1-(\frac{v^2}{c^2}))}=\frac{m(dv/dt)}{(1-(\frac{v^2}{c^2}))^\frac{3}{2}}

its that last step that i dont understand. why does the denominator term now have a 3/2 index? and why is du/dt given in such an odd way?

Question

Relativistic energy is sometimes derived in the following manner and I'm having trouble deciphering one mathematical step.  So it goes like this:

Given that $$F=\frac{ dp }{ dt }$$ and reletavistic momentum is given by
 p=\frac{ mv }{ sqrt(1-\frac{v^2}{c^2}) }

W=int_{x _{1}}^{x_{2}} F  dx=int_{x _{1}}^{x_{2}} \frac{dp}{dt}dx

which is fine
then they say 
\frac{dp}{dt}=\frac{d}{dt}\frac{mv}{sqrt(1-(\frac{v^2}{c^2}))}=\frac{m(dv/dt)}{(1-(\frac{v^2}{c^2}))^\frac{3}{2}}

its that last step that i dont understand. why does the denominator term now have a 3/2 index? and why is du/dt given in such an odd way?

theeric · Answer

Hi! I don't know if I'll be able to help or not, but I want to repost your question here with the equations in brackets. I don't see them appearing right now.





"Relativistic energy is sometimes derived in the following manner and I'm having trouble deciphering one mathematical step.  So it goes like this:

Given that $$\large F=\frac{dp}{ dt}$$   
 and reletavistic momentum is given by
 $$\large p=\frac{ mv }{\sqrt{1-\frac{v^2}{c^2}} }$$

 $$\large W=\int_{x _{1}}^{x_{2}} F\  dx=\int_{x _{1}}^{x_{2}} \frac{dp}{dt}\ dx$$

which is fine
then they say 
$$\large\frac{dp}{dt}=\frac{d}{dt}\frac{mv}{\sqrt{1-(\frac{v^2}{c^2})}}=\frac{m(dv/dt)}{(1-(\frac{v^2}{c^2}))^\frac{3}{2}}$$

its that last step that i dont understand. why does the denominator term now have a 3/2 index? and why is du/dt given in such an odd way?"

anonymous · Answer

Thanks didnt know why they weren't showing up right.

theeric · Answer

Yeah! You typed it in well! And I'm wondering the same thing as you. I'm not a math buff, but it looks like all substitution until that last step, then they multiply the equation by $\gamma^2$ and turn $v$ into $\large\frac{dv}{dt}$, which is acceleration, I think.

So I don't know how they got that, at the moment. I didn't use that work equation at all, maybe that has something to do with it?

theeric · Answer

Well, I guess what they did in that last part is take the derivative with respect to $t$. I missed that. I'm not sure if I can check that out with my own math!

When my class had these equations and math processes, some of us just watched in awe and confusion... Others were better!

theeric · Answer

Did they take the derivative? Wow, I need to polish my math skills...

anonymous · Answer

It is integration by substitution. They then take the new expression for dp/dt and substitute it into the into to evaluate it.

theeric · Answer

$$\large\frac{d}{dt}\frac{mv}{\left(1-\left(\frac{v^2}{c^2}\right)\right)^\frac{1}{2}}=\frac{m\left(dv/dt\right)}{\left(1-\left(\frac{v^2}{c^2}\right)\right)^\frac{3}{2}}$$

Does this mean the $v$ gets differentiated with respect to $t$? Or does that come from somewhere else?
$$\large\frac{d}{dt}\frac{m\boldsymbol v}{\left(1-\left(\frac{v^2}{c^2}\right)\right)^\frac{1}{2}}=\frac{m\boldsymbol{\left(dv/dt\right)}}{\left(1-\left(\frac{v^2}{c^2}\right)\right)^\frac{3}{2}}$$
I'm sorry I can't be of more help. Good luck!

egenriether · Answer

Hey, there was too much writing to put this in the post so I attached the answer to a handwritten page I scanned in.  Let me know if you need any clarification.

anonymous · Answer

Thanks