# Velocity transformation using the chain rule

1. Jul 15, 2017

1. The problem statement, all variables and given/known data
How to obtain the famous formula of velocity transformation using a chain rule.
I know that there is a straightforward way by dividing $dx$ as a function of $dx$ and $dt$ on $dt$ which is also a function of them. But I would rather try using the chain rule.
2. Relevant equations
$x=\gamma(x+vt)$
$t=\gamma(t+\frac{v}{c^2}x)$

3. The attempt at a solution
I tried the following chain rule $\frac{dx}{dt}=\frac{dx}{dx}\frac{dx}{dt}\frac{dt}{dt}$ so $u=\frac{dx}{dx}\frac{dt}{dt}u$
The first term $\frac{dx}{dx}=\gamma(1+\frac{v}{u})$
The second term $\frac{dt}{dt}$ requires me to take a derivative of $t$ with respect to $t$. Here the equation, $t=\gamma(t+\frac{v}{c^2}x)$ will not help because I have to differentiate $x$ with respect to $t$.

2. Jul 15, 2017

### Orodruin

Staff Emeritus
Your chain rules ignores the fact that it is a differentiation wrt many variables. You need the full chain rule with all partial derivatives. An important thing to keep in mind is when t and t' are used in their capacity as coordinates in the Lorentz transformation and when they are used to refer to acurve parameter.

3. Jul 16, 2017

Here is another trial;
$\frac{dx}{dt}=\frac{\partial x}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial x}{\partial t}\frac{\partial t}{\partial t}$
But $\frac{\partial x}{\partial t}=\frac{\partial x}{\partial t}\frac{\partial t}{\partial t}$
So, $\frac{dx}{dt}=\frac{\partial x}{\partial x}\frac{\partial x}{\partial t}\frac{\partial t}{\partial t}+\frac{\partial x}{\partial t}\frac{\partial t}{\partial t}$
Substitute $u$ for $\frac{dx}{dt}$ and $u$ for $\frac{\partial x}{\partial t}$
Also, from the two equation of LT $\frac{\partial x}{\partial x}=\gamma$, $\frac{\partial t}{\partial t}=\frac{1}{\gamma}$, $\frac{\partial x}{\partial t}=v\gamma$ and $\frac{\partial t}{\partial t}=\frac{1}{\gamma}$ (here, I did not use the inverse LT to substitute for $\frac{\partial t}{\partial t}$, instead I used the main transformation)
yields; $u=\gamma u\frac{1}{\gamma}+v\gamma\frac{1}{\gamma}=u+v$ This is non-relativistic transformation.

Last edited: Jul 16, 2017
4. Jul 17, 2017

$\frac{dx}{dt}=\frac{dx}{dx}\frac{dx}{dt}\frac{d t}{dt}$
Now $\frac{dx}{dx}$ and $\frac{d t}{dt}$ are expressed in term of partial derivatives;
$\frac{dx}{dx}=\frac{\partial x}{\partial x}\frac{\partial x}{\partial x}+\frac{\partial x}{\partial t}\frac{\partial t}{\partial x}=\frac{\partial x}{\partial x}+\frac{\partial x}{\partial t}\frac{\partial t}{\partial x}$
$\frac{dt}{dt}=\frac{\partial t}{\partial t}\frac{\partial t}{\partial t}+\frac{\partial t}{\partial x}\frac{\partial x}{\partial t}=\frac{\partial t}{\partial t}+\frac{\partial t}{\partial x}\frac{\partial x}{\partial t}$
Replacing; $\frac{dx}{dt}$ and $\frac{dx}{dt}$ by $u$ and $u`$, and the other partial derivatives from Lorentz Transformation yields the proper result.