I am struggling to differentiate relativistic momentum formula.
Considering special relativity,
$$ \vec{F}=\frac{d\vec{P}}{dt}=\frac{d}{dt}\frac{m\vec{v}}{\sqrt{1-v^2/c^2}}$$
which I understand.
The textbook proceeds, "when the net force and velocity are both along the x-axis,"
$$ F=\frac{m}{(1-v^2/c^2)^{3/2}}a $$
This is where I am stuck.
I am not sure how to compute the derivative $$F=\frac{d}{dt}\frac{m\vec{v}}{\sqrt{1-v^2/c^2}}$$ to get $$ F=\frac{m}{(1-v^2/c^2)^{3/2}}a $$
Best Answer
If net force and velocity are both along the $x$-axis, $F_y,F_z,v_y,v_z$ are 0 and there remains $$ F = F_x = \partial_t \frac{m v_x}{\sqrt{1 - v^2/c^2}}~, $$ where also $v = v_x$. This means $$ F = \partial_t \frac{m v}{\sqrt{1 - v^2/c^2}} = \partial_t \frac{m}{\sqrt{\frac 1 {v^2} - \frac 1 {c^2}}} \overset{\text{chain rule}}= - \frac{m}{2\sqrt{\frac 1 {v^2} - \frac 1 {c^2}}^3} \partial_t \left( \frac{1}{v^2} - \frac{1}{c^2} \right) \overset{\text{again chain rule}}= - \frac{m}{2\sqrt{\frac 1 {v^2} - \frac 1 {c^2}}^3} \left( \frac{-2}{v^3} \right) \underbrace{\partial_t v}_{=a}~. $$ Multiplying the $v^3$ back into the root and canceling out the signs and the twos yields the result from the textbook. The trick here is to initially multiply both the denominater and the numerator with $1/v$, which leaves the fraction unchanged but reduces the number of occurances of $v$.