The chain rule and velocity transformation in relativity

Question

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 , $t'(t) equiv t' = gamma(t-xv/c^2)$.

Anon1759 · Accepted Answer

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?

Laurence Lurio · Answer

For your equation 2, the chain rule should be:

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

benrg · Answer

$t$ and $t'$ are just coordinates and you should treat them the same way as spatial coordinates. In Euclidean geometry you can write any curve parametrically as $x(s), y(s)$, but in special cases you may be able to write $y(x)$ or $x(y)$. Likewise you can write a curve in Minkowski space as $x(s), t(s)$ or in special cases as $x(t)$. In point of fact you can almost always write $x(t)$ since the curves are usually causal worldlines, but just because you can doesn't mean you should. It's usually less hassle to use a non-coordinate parameter (which may be proper time).
Given a curve $x(τ), t(τ)$, its Lorentz-boosted form is $$X(τ) = γ,x(τ)+γβ,t(τ),;T(τ) = γ,t(τ)+γβ,x(τ),$$ where I'm using upper case instead of primes to avoid confusion with the derivative. If you need $T$ as a function of $t$, it's $Tcirc t^{-1}$, where the $T$ and $t$ in that formula are treated as functions of $τ$. If you want the derivative of $T$ with respect to $t$, that's $T'(t) = T'(τ)/t'(τ)$, or $displaystyle frac{dT}{dt}=frac{dT/dτ}{dt/dτ}$ if you prefer that notation. All of this is exactly the same as the Euclidean case, and the fact that it's a time coordinate is not relevant.
In most cases you shouldn't need any of those things; you should just uniformly treat all of the coordinates as functions of proper time (or some other non-coordinate parameter).

The chain rule and velocity transformation in relativity

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