- #1
ArtVandolay
- 5
- 8
- TL;DR Summary
- Stuck on derivation in Griffiths' Electrodynamics
In deriving the work-energy theorem, Griffiths does the following:
##\frac{d\mathbf{p}}{dt}\cdot\mathbf{u} = \frac{d}{dt}\bigg(\frac{m\mathbf{u}}{\sqrt{1-u^2/c^2}}\bigg)\cdot\mathbf{u}=\frac{m\mathbf{u}}{(1-u^2/c^2)^{3/2}}\cdot\frac{d\mathbf{u}}{dt}##
I may have forgotten something essential from vector calculus, but for the life of me, I can't figure out how he goes from the second term to the third. Using ##\frac{d(u^2)}{dt}=\frac{d(\mathbf{u}\cdot\mathbf{u})}{dt}=2\mathbf{u}\cdot\frac{d\mathbf{u}}{dt}##, I'm getting:
$$\frac{d}{dt}\bigg(\frac{m\mathbf{u}}{\sqrt{1-u^2/c^2}}\bigg)\cdot\mathbf{u}=\bigg[\frac{m\frac{d\mathbf{u}}{dt}}{\sqrt{1-u^2/c^2}} - \frac{1}{2}\frac{m\mathbf{u}}{(1-u^2/c^2)^{3/2}}\bigg(-\frac{2\mathbf{u}}{c^2}\cdot\frac{d\mathbf{u}}{dt}\bigg)\bigg]\cdot\mathbf{u}$$
I can't see how to get this in the form shown above. What am I missing?
##\frac{d\mathbf{p}}{dt}\cdot\mathbf{u} = \frac{d}{dt}\bigg(\frac{m\mathbf{u}}{\sqrt{1-u^2/c^2}}\bigg)\cdot\mathbf{u}=\frac{m\mathbf{u}}{(1-u^2/c^2)^{3/2}}\cdot\frac{d\mathbf{u}}{dt}##
I may have forgotten something essential from vector calculus, but for the life of me, I can't figure out how he goes from the second term to the third. Using ##\frac{d(u^2)}{dt}=\frac{d(\mathbf{u}\cdot\mathbf{u})}{dt}=2\mathbf{u}\cdot\frac{d\mathbf{u}}{dt}##, I'm getting:
$$\frac{d}{dt}\bigg(\frac{m\mathbf{u}}{\sqrt{1-u^2/c^2}}\bigg)\cdot\mathbf{u}=\bigg[\frac{m\frac{d\mathbf{u}}{dt}}{\sqrt{1-u^2/c^2}} - \frac{1}{2}\frac{m\mathbf{u}}{(1-u^2/c^2)^{3/2}}\bigg(-\frac{2\mathbf{u}}{c^2}\cdot\frac{d\mathbf{u}}{dt}\bigg)\bigg]\cdot\mathbf{u}$$
I can't see how to get this in the form shown above. What am I missing?