0
$\begingroup$

Context

I am studying special relativity. In [1], Gray derives an equation that is written in terms of three differentials. Namely, $$\left(d\tau\right)^2 = \left(dt\right)^2 - \frac{\left|d\mathbf{r}\right|^2}{c^2}\tag{1}.$$

Gray then wishes to derive that $$ \frac{dt}{d\tau} = \frac{1}{\sqrt{1-\frac{1}{c^2}\frac{\left|\mathbf{dr}\right|^2 }{\left(dt\right)^2}}}. $$

Gray derives this as follows: \begin{align} \left( \frac{d\tau}{dt}\right)^2 &= \frac{\left(d\tau\right)^2}{\left(dt\right)^2} &&\text{derivative is a ratio!?!} \\ &= \frac{dt^2 - \frac{1}{c^2}\left|d\mathbf{r}\right|^2}{\left(dt\right)^2} &&\text{substitution} \\ &= \frac{dt^2}{dt^2} - \frac{1}{c^2}\frac{\left|d\mathbf{r}\right|^2}{\left(dt\right)^2} &&\text{association} \\ &= 1 - \frac{1}{c^2}\frac{\left|d\mathbf{r}\right|^2}{\left(dt\right)^2} &&\text{ratios} \end{align} After taking the reciprocal of the square root, Gray shows what he wishes to show.

I do not like Gray's approach. This is because Gray treats the derivative as a ratio of differentials; yet the derivative is not the ratio of differentials (cf., [2,3]).

This brings me to three questions...

Questions

(1) Can we obtain $dt/d\tau$ by taking the square root of (1)?

(2) Can we use implicit differentiation on (1) to obtain the result?

(3) Though no one can speak for Gray, perhaps what is really wanted here $\frac{\partial{t}}{\partial{\tau}}$?

Bibliography

[1] N. Gray, A Student's Guide to Special Relativity, Cambridge University Press, 2022, p. 55.

[2] Is $\frac{\textrm{d}y}{\textrm{d}x}$ not a ratio?

[3] Derivative $\frac{d\left\{B(A_1, \ldots, A_M)\right\}}{d\left\{C(A_1, \ldots, A_M)\right\}}$ equals what?

$\endgroup$
3
  • $\begingroup$ Why do you have an issue with “ratio of differentials”? For instance, if y=x^2, dy/dx = 2x and dx/dy = 1/2x. Hopefully an example like this can help to give a geometric understanding of “ratios of differentials”. $\endgroup$
    – Johnson
    Jul 27 at 22:56
  • $\begingroup$ @Johnson. There is a very full discussion of the matter in [2], which is linked from my problem statement. $\endgroup$ Jul 29 at 23:26
  • $\begingroup$ Ahh yes, thank you. $\endgroup$
    – Johnson
    Jul 31 at 13:01

2 Answers 2

2
$\begingroup$

Derivatives are not ratios but they behave similar to ratios because of theorems like inverse function theorem and FTC (although in some cases you'll have to be careful). To make your argument more sound mathematically, starting from the below expression which is just the same expression as $(1)$ in your post:

$$\Big(\dfrac{d\tau(t)}{dt}\Big)^2=1-\beta(t)^2$$

Where

$$\beta(t)^2=\Big(\dfrac{d\vec{r}(t)}{dt}\Big)^2\dfrac{1}{c^2}$$

Note that $1-\beta(t)^2>0$ since no massive particle can reach the speed of light, hence by the inverse function theorem, $t$ is invertible as a function of $\tau$ locally. And the derivative of $t(\tau)$ is the inverse of the derivative of $\tau(t)$. More specificlaly we have:

$$\Big(\dfrac{dt(\tau)}{d\tau}\Big)^2=\dfrac{1}{1-\beta(t(\tau))^2}$$

Which is what is meant by your final expression. So as you can see, no infinitesimal arguments are required and the statement follows by the inverse function theorem in one step.

$\endgroup$
0
$\begingroup$

I found it best to avoid the differentials all together. I went back to the relation
\begin{align} \tau = \gamma \left[ t - \frac{v_1}{c^2} x ^{(1)} - \frac{v_2}{c^2} x ^{(2)} - \frac{v_3}{c^2} x ^{(3)} \right], \end{align} which I obtain from the Lorentz transform for proper time time $\tau$. Here $$\gamma=\frac{1}{\sqrt{ 1- \frac{v_1^2}{c^2} - \frac{v_2^2}{c^2} - \frac{v_3^2}{c^2} }},$$ and $v_1,v_2.v_3$ are constants with respect to time. Then assuming that $\frac{\left\|\mathbf{v}\right\|^2}{c^2}\neq 1$, it is strait forward to write $d\tau /dt$.

\begin{align} \frac{d\tau}{dt} &= \gamma \left[ \frac{dt}{dt} - \frac{v_1}{c^2} \frac{dx ^{(1)} }{dt} - \frac{v_2}{c^2} \frac{dx ^{(2)} }{dt} - \frac{v_3}{c^2} \frac{dx ^{(3)} }{dt} \right], \\ &= \gamma \left[ 1- \frac{v_1^2}{c^2} - \frac{v_2^2}{c^2} - \frac{v_3^2}{c^2} \right], \\ &= \gamma \frac{1}{\gamma ^2}, \\ &= \frac{1}{\gamma }. \end{align}

Then, as @Leonid indicates, we can, indeed, use the inverse function theorem [1]. In particular, assume that $\frac{\left\|\mathbf{v}\right\|^2}{c^2} \neq 1$. Then, adapting from [1], I write as follows. I have a function $\tau$ of a single variable $t$. The inverse function theorem states that since $\tau$ is a continuously differentiable function with nonzero derivative at the point $t_a$; then $\tau$ is injective in a neighborhood of $t_a$, the inverse is continuously differentiable near $\tau_b=\tau(t_a)$, and the derivative of the inverse function at $\tau_b$ is the reciprocal of the derivative of $\tau$ at $t_a$. In other words, $$\bigl(\tau^{-1}\bigr)'(\tau_b) = \frac{1}{\tau'(t_a)} = \frac{1}{\frac{1}{\gamma }} = \gamma.$$

[1] https://en.wikipedia.org/wiki/Inverse_function_theorem

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.