# Relativistic Energy Derivation math problem

1. Jul 10, 2013

### michael154

Hey,

In a derivation of relativistic energy (in Physics for Scientists and Engineers, 5th edition, Serway and Beichner) they use a method of integration by substitution:

Given that

$F=\frac{dp}{dt}$

and relativistic momentum is given by

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

$W=∫F dx=∫\frac{dp}{dt} dx$

which is fine then they say

$\frac{dp}{dt}=\frac{d}{dt}\frac{mv}{\sqrt(1−(\frac{(v^2)}{c^2}))}=\frac{m\frac{dv}{dt}}{(1−(\frac{v^2}{c^2}))^(\frac{3}{2})}$

(This is about halfway through the derivation.)
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 like that? Any help would be appreciated.

Last edited: Jul 10, 2013
2. Jul 10, 2013

### robphy

is there a missing square-root symbol?

3. Jul 10, 2013

### Bill_K

(d/dt)(mv(1-v2)-1/2) = m(1-v2)-1/2dv/dt -1/2(-2v)(mv)(1-v2)-3/2)dv/dt
[first term is from d/dt'ing the v in the numerator, second is from d/dt'ing the denominator]

= m(1-v2 + v2)(1-v2)-3/2dv/dt = m(1-v2)-3/2dv/dt

4. Jul 10, 2013

### michael154

Sorry but that's not very clear. I have trouble reading non formatted math like that. As a second year physics student in uni, should I be able to do that differentiation by hand?

5. Jul 10, 2013

### pervect

Staff Emeritus
Try looking up "chain rule". Wiki has a definition, it might not be the easiest to read. Note that you want to find (d/dt) p(t) which is equal to (d/dt) p(v(t)), and that m is assumed to be constant (not a function of time).

6. Jul 10, 2013

### Raze

Sorry I use u instead of v for velocity

In plain English the reason you get an exponent of -3/2 is because γ(u) is an exponent of -1/2, and what's the derivative of x-1/2? It's (-1/2)x-3/2.

I actually just posted a similar question regarding this topic, but with the time component rather than space.

Last edited: Jul 10, 2013