Why is $\frac{d}{dt}(\frac{m \vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}) \cdot \vec{u} =\frac{m \vec u}{(1-\frac{u^2}{c^2})^{3/2}} \cdot \frac{d \vec u}{dt}$?

Question

Given that $\vec u(t)$, $u=|\vec u(t) |$ and $c$, $m$ are constants, how does one get from the LHS of the following equation to the RHS?

$$\frac{d}{dt}\left(\frac{m \vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}\right) \cdot \vec{u} =\frac{m \vec u}{\left(1-\frac{u^2}{c^2}\right)^{3/2}} \cdot \frac{d \vec u}{dt}$$

What is the differentiation rule used?

Answer 1

Using the product rule reveals

$$\begin{align} \frac{d}{dt}\left(\frac{m\vec u}{\sqrt{1-u^2/c^2}}\right)&=m\vec u \frac{d}{dt}\left(\frac{1}{\sqrt{1-u^2/c^2}}\right)+\frac{1}{\sqrt{1-u^2/c^2}}\frac{d}{dt}\left(m\vec u\right)\\\\ &=m\vec u\left(-\frac12(1-u^2/c^2)^{-3/2}\left(\frac{-2u}{c^2}\right)\,\frac{du}{dt}\right)\\&+(1-u^2/c^2)^{-3/2}(1-u^2/c^2)\frac{d}{dt}\left(m\vec u\right)\\\\ &=\frac{m}{c^2}(1-u^2/c^2)^{-3/2}\left(\vec uu\frac{du}{dt}+(c^2-u^2)\frac{d\vec u}{dt}\right)\tag1 \end{align}$$

Forming the inner product of the right-hand side of $(1)$ with $\vec u$, we find that

$$\begin{align} \frac{d}{dt}\left(\frac{m\vec u}{\sqrt{1-u^2/c^2}}\right)&=\frac{m}{c^2}(1-u^2/c^2)^{-3/2}\left(u^3\frac{du}{dt}+(c^2-u^2)\vec u\cdot \frac{d\vec u}{dt}\right)\\\\ &=\frac{m}{c^2}(1-u^2/c^2)^{-3/2}\left(u^2 \frac12\frac{du^2}{dt}+(c^2-u^2)\vec u\cdot \frac{d\vec u}{dt}\right)\\\\ &=\frac{m}{c^2}(1-u^2/c^2)^{-3/2}\left(u^2 \frac12 \frac{d(\vec u\cdot \vec u)}{dt}+(c^2-u^2)\vec u\cdot \frac{d\vec u}{dt}\right)\\\\ &=\frac{m}{c^2}(1-u^2/c^2)^{-3/2}\left(u^2 \vec u\cdot \frac{d\vec u}{dt}+(c^2-u^2)\vec u\cdot \frac{d\vec u}{dt}\right)\\\\ &=m(1-u^2/c^2)^{-3/2}\vec u\cdot \frac{d\vec u}{dt} \end{align}$$

as was to be shown!

Answer 2

It's just the usual product and chain rules, but it helps to work with indices and sum implicitly when they repeat. I'll set $c=1$ to simplify notation:$$\frac{d}{dt}\left((1-u_ju_j)^{-1/2}mu_i\right)u_i=\left(u_k\dot{u}_k(1-u_ju_j)^{-3/2}mu_i+(1-u_ju_j)^{-1/2}m\dot{u}_i\right)u_i\\=(1-u_ju_j)^{-3/2}mu_i\dot{u}_i\left(u_iu_i+1-u_iu_i\right)=(1-u_ju_j)^{-3/2}mu_i\dot{u}_i.$$

Answer 3

The usual one. Try this: $u^2=(u,u)$ and then $\dfrac{d}{dt}u^2= 2(u',u).$

Answer 4

Leave $m$ out and start with,

$$\frac{\vec{u}\cdot\vec{u}} {\sqrt{1-\frac{u^2}{c^2}}}=\frac{u^2} {\sqrt{1-\frac{u^2}{c^2}}}$$

Take derivatives on both sides,

$$\frac{d}{dt}\left(\frac{\vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}\right) \cdot \vec{u} +\frac{\vec u}{\sqrt{1-\frac{u^2}{c^2}}} \cdot \frac{d \vec u}{dt} =\left[ \frac{1}{\sqrt{1-\frac{u^2}{c^2}}} + \frac 12 \frac{\frac{u^2}{c^2}}{\left(1-\frac{u^2}{c^2}\right)^{3/2}} \right]\frac{du^2}{dt}$$

Since $ du^2/dt = 2\vec{u}\cdot d\vec{u}/dt$, rearrange the RHS as,

$$ RHS = \frac{\vec u}{\sqrt{1-\frac{u^2}{c^2}}} \cdot \frac{d \vec u}{dt}+\left[ \frac{1}{\sqrt{1-\frac{u^2}{c^2}}} + \frac{\frac{u^2}{c^2}}{\left(1-\frac{u^2}{c^2}\right)^{3/2}} \right]\vec{u}\cdot \frac{d\vec{u}}{st}$$

$$= \frac{\vec u}{\sqrt{1-\frac{u^2}{c^2}}} \cdot \frac{d \vec u}{dt}+ \frac{\vec{u}}{\left(1-\frac{u^2}{c^2}\right)^{3/2}}\cdot \frac{d\vec{u}}{st}$$

Compare with the LHS to cancel the first term. Thus,

$$\frac{d}{dt}\left(\frac{\vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}\right) \cdot \vec{u} =\frac{\vec u}{\left(1-\frac{u^2}{c^2}\right)^{3/2}} \cdot \frac{d \vec u}{dt}$$