Velocity transformation using the chain rule

[Adel Makram]

Homework Statement

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.

Homework Equations

$x=\gamma(x`+vt`)$
$t=\gamma(t`+\frac{v}{c^2}x`)$

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 $t$.

 

[Orodruin]

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.

 

[Adel Makram]

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.

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.

 

[Adel Makram]

From some help, I found a way out.
First the velocity should be represented by the total derivatives not partial derivatives.
$\frac{dx}{dt}=\frac{dx}{dx`}\frac{dx`}{dt`}\frac{d t`}{dt}$
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.