0
$\begingroup$

I am trying to figure out some notation issues (or at least that what I assume this is).

I assume the "proper time" could refer to a "proper time functional" of a timelike path $P(\lambda)$ with coordinates $x^\alpha(\lambda)$ $$\tau[P(\lambda)] = \int_{\lambda_A}^{\lambda_B}\sqrt{(- g_{\alpha \beta} (x^\gamma)\frac{dx^\alpha}{d \lambda}\frac{dx^\beta}{d \lambda})} d \lambda $$ and is a separate mathematical object than when we use proper time as simply the parameter along the worldline $P(\tau)$?

I assume you could also write $$ \tau[P(\tau')] = \int_{\tau'_A}^{\tau'_B} d \tau'$$ and the dummy variable $\tau'$ would be defined as the proper time parameter, $P(\tau')$, but I haven't seen any textbooks make this distinction (I assume this might just because it's mathematical/notational pedantry, but perhaps it's incorrect).

Edit to add: I suppose this is sometimes distinguished by referring to $\tau[P(\tau')] $ as the proper time interval with notation $\Delta \tau$.

$\endgroup$
5
  • $\begingroup$ It is simply not very fun to be working with physically meaningless/arbitrary path parameters $\lambda$ when we have the physically meaningful proper time $\tau=\tau[P(\lambda)]$, which can be seen as a relationship between $\tau$ and $\lambda$. Now, the path parametrisation $\lambda$ should be one-to-one and monotonic, and it should then be obvious that $\tau$ is too; and that they are in one-to-one correspondence with each other. Then you can just invert the relationship and use $\tau$ as the parametrisation by considering $P[\lambda(\tau)]$ $\endgroup$ Commented Aug 17, 2023 at 5:12
  • $\begingroup$ It seems like there is overloading of notation when you write $\tau = \tau[P(\lambda)]$; e.g., I'm not sure it makes sense to extremise a variable $\tau$ (or perhaps it does?), but it makes sense to extremise a function(al) $\tau[P(\lambda)]$. I'm not sure I understand this change when you invert the relationship to get $\tau$ as the parametrisation. $\endgroup$
    – qwerty
    Commented Aug 17, 2023 at 5:30
  • $\begingroup$ You are correct that, before the extremisation of the functional $\tau[P(\lambda)]$, even the numerical value of the extremised function itself is not defined. It is only after the extremisation that, everywhere along the extremised path, do you have a well-defined proper time to numerically associate with each position in spacetime. Only then can you invert the relationship, only for the extremised path. What you are arguing is indeed very pedantic, but it will turn up later when thinking of quantum theory, because the non-extremal paths also contribute. i.e. you are onto something. $\endgroup$ Commented Aug 17, 2023 at 5:39
  • $\begingroup$ How are $P$ and $x$ related? $\endgroup$
    – Qmechanic
    Commented Aug 17, 2023 at 6:19
  • $\begingroup$ @Qmechanic, I should have specified x refers to the coordinates of the path P in a chosen basis. $\endgroup$
    – qwerty
    Commented Aug 17, 2023 at 6:22

1 Answer 1

1
$\begingroup$

When people wish to refer to the proper time of a curve as a functional, they will use whatever local equivalent they wish to call the length functional for a timelike curve :

\begin{equation} l_\gamma = \int_{t_1}^{t_2} \sqrt{g_{\gamma(t)}(\dot{\gamma}(t), \dot{\gamma}(t))} dt \end{equation}

up to an appropriate sign under the square root to make the square root make sense. It is also called the Lorentzian length of a curve, and in the case of a geodesic is sometimes called the geodesic interval. It also relates to some other common functions in relativity such as Synge's world function.

$\endgroup$
1
  • $\begingroup$ Is "t" the time coordinate of the usual 4 coordinates? Is it not an affine parameter of the world line? $\endgroup$ Commented Jan 1 at 22:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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