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I know there is a straightforward way to derive the law of velocity transformation in Special Relativity by just dividing $dx$ on $dt$. But I am interested here to apply the chain rule and see how to obtain the same formula. I just begin by writing the differential of $dx$ with respect to other coordinates; $$\frac{dx}{dt}=\frac{\partial x}{\partial x'}\frac{\partial x'}{\partial t}+\frac{\partial x}{\partial t'}\frac{\partial t'}{\partial t}$$ 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; $$u=\gamma u'\frac{1}{\gamma}+v\gamma\frac{1}{\gamma}=u'+v$$ However, this is non-relativistic transformation of the velocities. So probably, I missed something in the chain rule or in the substitution. Regards,

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  • $\begingroup$ I don't know the answer (yet), but I suspect it has something to do with mixing partial and total derivatives, and being careful about what's kept constant. I'll see if I can work it out. $\endgroup$
    – Javier
    Jul 16 '17 at 14:40
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$\newcommand{\t}[2]{\frac{\mathrm{d}#1}{\mathrm{d}#2}}$$\newcommand{\p}[2]{\frac{\partial #1}{\partial #2}}$Like I said in my comment, you have to be careful with the distinction between partial derivatives and total derivatives. The latter are evaluated along the worldline, which means that the pair of variables $(t,x)$ (or $(t',x')$) is not independent. On the other hand, partial derivatives are calculated for arbitrary variations of the variables, and that's where we use the Lorentz transformation.

The best way to start is to write

$$\t{x}{t} = \t{x}{t'} \t{t'}{t}.$$

This is simply an application of the chain rule for functions of one variable: on the worldline, $x$ can be considered a function of $t'$, and $t'$ a function of $t$.

Now we need to be careful. By the Lorentz transformation, $x$ is a function of $t'$ and $x'$, but $x'$ itself is a function of $t'$, so we can write $x = x(t', x'(t'))$, and

$$\t{x}{t'} = \p{x}{t'} + \p{x}{x'}\t{x'}{t'} = \gamma v + \gamma u'$$

$$\t{t'}{t} = \p{t'}{t} + \p{t'}{x} \t{x}{t} = \gamma - \gamma v u$$

Solving for $u'$ gives the correct relation

$$u' = \frac{u - v}{1 - uv}.$$


Edit: In response on the difference between my derivation and yours. Yours starts all right if you put total derivatives where they should be:

$$\begin{align} \t{x}{t} &= \p{x}{x'} \t{x'}{t} + \p{x}{t'} \t{t'}{t} \\ &= \p{x}{x'} \t{x'}{t'} \t{t'}{t} + \p{x}{t'} \t{t'}{t} \end{align}$$

Now you problem is that $\partial t' / \partial t$ is $\gamma$ but $\mathrm{d}t'/\mathrm{d}t$ is not; rather, it's what I wrote above. The difference is that $\partial t' / \partial t$ is taken with $x$ held constant, but you don't want that: you want $x$ to be a function of $t$, so you use a total derivative. If you do this you get the correct result; in fact, if you take out a common factor of $\mathrm{d}t'/\mathrm{d}t$, you get my first line.

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  • $\begingroup$ Great, I got the same result if $(\frac{\partial x}{\partial t})$=$(\frac{\partial x}{\partial x})(\frac{\partial x}{\partial t})(\frac{\partial t}{\partial t})$. Still I don`t get why the first equation should be in total derivative form. I mean this is just the second term of my original equation in partial derivative. $\endgroup$
    – Isaacadel
    Jul 16 '17 at 21:41
  • $\begingroup$ Great, I got the same result if dxdt=dxdx′dxdt′dtdt..Still I don`t get why the first equation should be in total derivative form. I mean this is just the second term of my original equation in partial derivative. $\endgroup$
    – Isaacadel
    Jul 16 '17 at 21:49
  • $\begingroup$ @Isaacadel see my edit, let me know if it answers your question. $\endgroup$
    – Javier
    Jul 17 '17 at 20:51
  • $\begingroup$ @Isaacadel If the answer answers your question then kindly consider marking it as the accepted answer. $\endgroup$
    – Dvij D.C.
    Jul 20 '17 at 11:51
  • $\begingroup$ Sure, but how to do that? $\endgroup$
    – Isaacadel
    Jul 20 '17 at 15:10

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