0
$\begingroup$

What is proper time? Is it a part of Minkowski space (that is a mere spacetime interval)? Or is it an intrinsic characteristic of massive particles (a sort of "aging")? Example: In the following Minkowski diagram,

  • A is the worldline of a particle moving with the observer's reference frame.

  • B is the worldline of another particle with proper time between two events = 4 (space distance x= 3, time interval t = 5, proper time τ = 4, according to the equation $$ \tau^2 = \delta t^2 - \delta x^2 $$)

  • C is just 2 events without any worldline between, and if a particle would travel through both events, its proper time would be 4

enter image description here

Is there proper time in C even if no particle is travelling there? Or is in C only hypothetic proper time in case a particle is travelling this path?

|cite|improve this question
$\endgroup$
  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/122327/2451 $\endgroup$ – Qmechanic♦ Jul 2 '14 at 12:50
  • $\begingroup$ Could you define what you mean by "There is proper time in C" as opposed to "There is only hypothetical proper time in C"? $\endgroup$ – ACuriousMind♦ Jul 2 '14 at 13:31
  • $\begingroup$ @ACuriousMind: "hypothetical proper time" = no proper time at all. - However obviously anybody can calculate what would be the proper time of a particle which would be travelling through both events. $\endgroup$ – Moonraker Jul 2 '14 at 13:41
  • 1
    $\begingroup$ Proper time is a property of paths in space, not of particles. It is, essentially, their length in the given metric. Proper time has no existence, it is not an object, so saying "There is proper time" or "There is no proper time" is as meaningful as saying "There is height". Would you call it a meaningful question to ask "Does the point $1m$ above my desk have height even if there is nothing there?" $\endgroup$ – ACuriousMind♦ Jul 2 '14 at 13:45
  • $\begingroup$ @ACuriousMind: At my information, proper time = aging. Am I wrong? $\endgroup$ – Moonraker Jul 2 '14 at 13:52
3
$\begingroup$

Ok, before we fill up the comment section with this, I will write this as an answer:

Proper time $\tau$ along a path $\gamma$ is

$$ \tau := \int_\gamma \sqrt{\mathrm{d}x^\mu\mathrm{d}x_\mu}$$

and a clock moving along $\gamma$ will have $\tau$ as its elapsed time at the end of the path. Yet, the definition of proper time $\tau$ involves such clocks not more than the definition of voltage involves voltmeters. You would not say "There is no voltage" because you have nothing to measure it with, and you would not say "There are no lengths" just because you have no ruler. And just as well you would not say "There is no proper time" just because there is no clock travelling along that path.

|cite|improve this answer
$\endgroup$
0
$\begingroup$

As stated by Moonraker your point C is the same as saying there is no space at all thereby any definition of 2 points makes no sense at all.

Now to the question "what is proper time?"

"It is the time which is measured in the rest frame of an observer passing through two events in space-time"

This will depend on the kind of movement the observer experiences thereby you cannot just take the direct distance in space-time as done in B but you have to integrate over all small space-time segments. You can do that because the space-time is an invariant under lorentz transformations so that you can always transform into a system in which your you are at rest so that you only move along the time axis.(Only allowed locally if you take acceleration into account)

|cite|improve this answer
$\endgroup$
0
$\begingroup$

The proper time between two space-time points is very, very like the ordinary distance between two points in ordinary space. In fact, in the case that the two points are at the same time in a given frame the proper time between them is just (minus) the distance between them divided by c, the speed of light. It does not require a clock to be real, any more than distance requires a ruler

|cite|improve this answer
$\endgroup$

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.