The chain rule and velocity transformation in relativity

From elementary calculus, we have that the chain rule occurs when we differentiate a function like $$f(y(x)) \equiv f(x)$$:

$$\frac{\mathrm{d}}{\mathrm{dx}}[f] = \frac{\mathrm{d}}{\mathrm{dx}}[f(y(x))] = \frac{df}{dy}\frac{dy}{dx} \tag{1}$$

But, consider now a well-known [1] expression about velocity transformation in elemetary Special Relativity context:

$$\frac{\mathrm{d}}{\mathrm{dt'}}[x'] = \frac{dx'}{dt}\frac{dt}{dt'} \tag{2}$$

So, I'm struggling to understand this expression because suppose the situation: I'm a observer in reference frame $$S'$$, so I would construct quantities like velocity "with my proper primed coordinates", ($$x'$$, $$t'$$). Hence, the spatial coordinates will be parametrized by "my proper primed coordinate $$t'$$"

$$U' =: \frac{\mathrm{d}}{\mathrm{dt'}}[x'(t')] \tag{3}$$

Of course that if I want to know the velocity in a reference $$S$$, we have to transform $$(1)$$ accordingly Lorentz Transformations (LT). My problem isn't acctualy to derive the velocity transformations between $$S'$$ and $$S$$ (vice-versa), but to deal properly with the chain rule. Note that if $$(1)$$ is the chain rule, and $$(2)$$ is a valid expression, then the function $$x'$$ must have the form (I guess) of:

$$x' \equiv x'[t(t')] \tag{4}$$

And my question arise here in $$(4)$$: What suppose to mean, physically, $$t(t')$$? . Because it's occur naturally that if we perform a LT like $$S \to S'$$ time will have the dependency like $$t'(t)$$ because , $$t'(t) \equiv t' = \gamma(t-xv/c^2)$$.

2 Answers

I'm not sure exactly what your question is, but perhaps we can start with the following, and work from there....

From the Lorentz transform we know x' = F(x, t). Specifically:

$$x' = \gamma \left(x - vt \right)$$.

We also know

$$t' = \gamma \left(t - \frac{v x}{c^2}\right)$$

So,

$$v' = \frac{\partial x'}{\partial t'} = \frac{\partial x'}{\partial t}\frac{\partial t}{\partial t'} + \frac{\partial x'}{\partial x}\frac{\partial x}{\partial t'}$$

So we arrive at

$$v' = \frac{\partial x'}{\partial t'} = F(x', t') = G(x, t)$$

Specifically, $$x'$$ is a function of both $$t$$ and $$x$$, which are independent variables. In the same way $$t'$$ is also a function of $$x$$ and $$t$$, again, two independent variables. Does this help at all?

• Thank you for your help. But I need a explicit form of dependency, something like: $x'[x'(x,t), t'(x,t)]$ – M.N.Raia Mar 29 '19 at 21:28
• Well I guess you could write v' = v'(x',t') = v'(x'(x,t), t'(x,t)). Is that the sort of thing you're looking for? – Anon1759 Mar 30 '19 at 19:21

For your equation 2, the chain rule should be:

$$\frac{dx'}{dt'} = \frac{\partial x'}{\partial t}\frac{\partial t}{\partial t'} + \frac{\partial x'}{\partial x}\frac{\partial x}{\partial t'}$$

• I'm sorry, but this expression means nothing to me. How could have unprimed coordinates been given in terms of primed coordinates, if the primed coordinates are those we want to know? – M.N.Raia Mar 29 '19 at 17:37
• I apologize if this was confusing. The point was that $x'$ depends on both $t$ and $x$, not just $t$. I rewrote my answer focussing on your equation [2] rather than equation [4]. – Laurence Lurio Mar 29 '19 at 19:29