A question on proper time in special relativity

[Sagittarius A-Star]

I know, but every time I have to remember that even though it's the same symbol at the denominator of a second derivative, for instance, in the SR context has another meaning.

I think, the sign of $ds^2$ is not a property of nature, because it is an artifact of the convention, which of the two signatures (+---) or (-+++) you have selected for your given scenario. That means, if you calculate a realistic scenario (for example without tachyons) and calculate only measurable physical quantities, like temporal interval or spatial distance, then the $ds^2$, which might be negative, will automatically disappear from your equations before the end of your calculation. See for example the calculation in posting #3.

 

[PeterDonis]

I think, the sign of $ds^2$ is not a property of nature

More precisely, which kind of interval (spacelike or timelike) you pick to have a positive $ds^2$ and which one you pick to have a negative $ds^2$ is not a property of nature, it's a matter of which signature convention you choose. But the fact that there are intervals with both positive and negative $ds^2$ is a property of nature; there is no way to have only positive (or only negative) $ds^2$ in spacetime by any choice of convention.

 

[PeterDonis]

if you calculate a realistic scenario (for example without tachyons) and calculate only measurable physical quantities, like temporal interval or spatial distance, then the $ds^2$, which might be negative, will automatically disappear from your equations before the end of your calculation.

More precisely, if you know what kind of interval you are dealing with (spacelike or timelike), you can always adjust things during the calculation to avoid having to take the square root of a negative number. However, that doesn't change the fact that spacelike and timelike intervals inherently have opposite signs for $ds^2$. Those opposite signs reflect a fundamental physical difference between those types of intervals.

 

[Sagittarius A-Star]

Ok, Here's something a little more recent

Richard Feynman proposed, that the spacetime interval can be imaginary ...

the interval squared would be negative and we would have an imaginary interval, the square root of a negative number. Intervals can be either real or imaginary in the theory.

Source:
https://www.feynmanlectures.caltech.edu/I_17.html

... and Wolfgang Rindler proposed the opposite:

We shall regard it, as we justify below, as the square of a vector – the "squared displacement" – rather than as the square of a possibly complex number.

Source:
http://www.scholarpedia.org/article/Special_relativity:_mechanics

 

[vanhees71]

H. Minkowski also proposed to shift the "i" issue from the spacetime interval to the unit system by defining the ratio between the unit of distance and the unit of time to be imaginary:
$$3 \cdot 10^5 \ km = \sqrt{-1} \ seconds$$
Source: same link, page 50, PDF page 60:
https://www.minkowskiinstitute.org/mip/MinkowskiFreemiumMIP2012.pdf#page=60

But this proposal is not accepted today. Maybe, because it would only shift the "i" issue from the spacetime interval (were it does not harm) to everywere else in physics.

I hope, we don't need to discuss the infamous $\mathrm{i} c t$ convention. It's a disease ;-)).

 

[cianfa72]

A frame is an infinite family of observers, one at every place in space, all following parallel worldlines, all carrying a three number spatial coordinate, and all agreeing a mechanism for when they should zero their watches. Then the proper time since zero of one of this family is this frame's coordinate time at their location.

In this definition of frame, the agreed 'mechanism' you were talking about for zeroing their watches, is actually the Einstein synchronization procedure in the rest frame of that infinite family of observers ?

Are other mechanisms possibile in principle to do that ?

 

[jbriggs444]

In this definition of frame, the agreed 'mechanism' you were talking about for zeroing their watches, is actually the Einstein synchronization procedure in the rest frame of that infinite family of observers ?

Are other mechanisms possibile in principle to do that ?

Send machine gun bullets between the clocks in the grid at a velocity that is verifiably consistent in both the "forward" and "reverse" directions. (Measure momentum at launch and arrival, for instance).

Measure round trip time. Synchronize clocks by zeroing the shooter's clock when a shot is made and setting the target's clock to RTT/2 upon arrival.Or, slow clock transport. Synchronize one grid clock with the wrist watch of a traveler. Send the traveler to the next grid clock. Synchronize the next grid clock with the traveler's watch. In the limit of slower and slower traveler's, you get perfect synchronization.

For speedier convergence, send the traveler and his watch at an arbitrarily great fixed speed and synchronize in both directions, using a round trip. Measure the amount by which the original clock was adjusted after the return and only adjust it by half that much.

 

[cianfa72]

Send machine gun bullets between the clocks in the grid at a velocity that is verifiably consistent in both the "forward" and "reverse" directions. (Measure momentum at launch and arrival, for instance).

Measure round trip time. Synchronize clocks by zeroing the shooter's clock when a shot is made and setting the target's clock to RTT/2 upon arrival.

ok, assume the mechanism of machine gun bullets to synchronize clocks in the grid the way you said. It defines the coordinate time $t$ for (I would say one of) the frame (aka coordinate chart) in which clocks in the grid are at rest.

Now I think there is actually no logical reason why the one-way speed of light should result c in that frame. In other words there is no logical reason why the metric in that frame should be (two spatial dimension and speed of light normalized to 1) $$ds^2=dx^2 + dy^2 - dt^2$$

 

[jbriggs444]

ok, assume the mechanics of machine gun bullets to synchronize clocks in the grid the way you said. It defines the coordinate time $t$ for (I would say one of) the frame (aka coordinate chart) in which clocks in the grid are at rest.

Now I think there is actually no logical reason why the one-way speed of light should result c in that frame. In other words there is no logical reason why the metric in that frame should be (two spatial dimension and speed of light normalized to 1) $$ds^2=dx^2 + dy^2 - dt^2$$

The speed of light could be less than c. Yes. Light pulses could behave like bullets with source-relative speeds rather than always traveling at the invariant speed.

One would have to re-derive special relativity without reference to light. But that has been done anyway.

 

[cianfa72]

One would have to re-derive special relativity without reference to light. But that has been done anyway.

ok, but in that case (light did not travel at invariant speed) which procedure we would use to physically synchronize clocks in the grid in order to define the rest frame coordinate time?

 

[Dale]

ok, but in that case (light did not travel at invariant speed) which procedure we would use to physically synchronize clocks in the grid in order to define the rest frame coordinate time?

Slow clock transport would be the easiest.

 

[cianfa72]

Slow clock transport would be the easiest.

So using it we would be able to define the coordinate time $t$ for a frame (coordinate chart) in which clocks are at rest and the metric is in the standard Minkowski form, though. All that even if light nor other physical process had the property to travel at the invariant speed.

 

[Dale]

So using it we would be able to define the coordinate time $t$ for a frame (coordinate chart) in which clocks are at rest and the metric is in the standard Minkowski form, though. All that even if light nor other physical process had the property to travel at the invariant speed.

Yes. It isn’t the only way, but the easiest

 

[Sagittarius A-Star]

ok, assume the mechanism of machine gun bullets to synchronize clocks in the grid the way you said. It defines the coordinate time $t$ for (I would say one of) the frame (aka coordinate chart) in which clocks in the grid are at rest.

Now I think there is actually no logical reason why the one-way speed of light should result c in that frame. In other words there is no logical reason why the metric in that frame should be (two spatial dimension and speed of light normalized to 1) $$ds^2=dx^2 + dy^2 - dt^2$$

There is a logical reason, why the bullet clock synchronization leads to the same time-coordinate definition as the Einstein synchronization with light:

The basic principle of clock synchronization is to ensure that the coordinate description of physics is as symmetric as the physics itself. For example, bullets shot off by the same gun at any point and in any direction should always have the same coordinate velocity dr/dt .

Source:
http://www.scholarpedia.org/article...nematics#Galilean_and_Lorentz_transformations

 

[cianfa72]

There is a logical reason, why the bullet clock synchronization leads to the same time-coordinate definition as the Einstein synchronization with light.

With light-based Einstein synchronization procedure applied to a grid of clocks mutually at rest, by very definition the one-way speed of light is c (provided that its two-way speed is actually c). For the same reason if the speed of bullets on a closed path was costant (say b) then the one-way speed of bullets would be b.

However I do not understand your logic about why using bullets clock synchronization the one-way speed of light should be c in that frame.

 

[PAllen]

With light-based Einstein synchronization procedure applied to a grid of clocks mutually at rest, by very definition the one-way speed of light is c (provided that its two-way speed is actually c). For the same reason if the speed of bullets on a closed path was costant (say b) then the one-way speed of bullets would be b.

However I do not understand your logic why using bullets clock synchronization the one-way speed of light should be c in that frame.

The logical reason is assuming isotropy in your synchronization. This is the only free choice - isotopropy or a very special type of one way anisotropy that preserves two way isotropy of lightspeed. Once you adopt isotropy for any synchronization method, then constant one way speed of light is forced. For example, if you simply assume that if two identical colocated clocks are moved apart with identical acceleration profile, then they are syncchronized, that alone is sufficient for one way lightspeed isotropy to follow.

 

[cianfa72]

The logical reason is assuming isotropy in you synchronization. This is the only free choice - isotopropy or a very special type of one way anisotropy that preserves two way isotropy of lightspeed. Once you adopt isotropy for any synchronization method, then constant one way speed of light is forced.

Do you mean isotropy of the synchronization method chosen or isotropy of lightspeed ?

if you simply assume that if two identical colocated clocks are moved apart with identical acceleration profile, then they are synchronized, that alone is sufficient for one way lightspeed isotropy to follow.

Is it just an assumption or it can be checked in some way ?

 

[PAllen]

Do you mean isotropy of the synchronization method chosen or isotropy of lightspeed ?Is just an assumption or it can be checked in some way ?

1) of synchorinization method

2) Yes, it can be checked. It is a perfectly feasible experiment (in principle).

 

[Sagittarius A-Star]

With light-based Einstein synchronization procedure applied to a grid of clocks mutually at rest, by very definition the one-way speed of light is c (provided that its two-way speed is actually c). For the same reason if the speed of bullets on a closed path was costant (say b) then the one-way speed of bullets would be b.

However I do not understand your logic why using bullets clock synchronization the one-way speed of light should be c in that frame.

With Einstein-synchronized clocks we have Reichenbach's ε = 1/2. Then the one-way-bullet-speed is measured as isotropic. The same is valid backwards.

 

[cianfa72]

1) of synchronization method

But...if the synchronization method (e.g bullets synchronization method) was isotropic but light propagation process was not, it did not follow the constancy of one-way speed of light, I believe.

 

[PAllen]

But...if the synchronization method (e.g bullets synchronization method) is isotropic and light propagation process was non isotropic, it did not follow the constancy of one-way speed of light, I believe.

Wrong. There is no logically consistent model in which identically launched bullets can be assumed to have isotropic one way speed while light does not. The epsilon parameter describing the only allowed form of anisotropy applies to all physics. That is, any clock synchronization technique in SR amounts to a choice of epsilon. Any that assumes full isotropy for any synchronization method is choosing epsilon of 1/2.

 

[cianfa72]

Then the one-way-bullet-speed is measured as isotropic.

...only if the process of sending bullets is actually isotropic.

 

[PAllen]

...only if the process of sending bullets is actually isotropic.

is assumed to be isotropic for a fully specified launch method.

 

[cianfa72]

Wrong. There is no logically consistent model in which identically launched bullets can be assumed to have isotropic one way speed while light does not. The epsilon parameter describing the only allowed form of anisotropy applies to all physics. That is, any clock synchronization technique in SR amounts to a choice of epsilon. Any that assumes full isotropy for any synchronization method is choosing epsilon of 1/2.

So in this case at hand the point is that if we assume full isotropy (epsilon of 1/2) it applies to identically launched bullets as well to one-way light speed

 

[PAllen]

So in this case the point is that if we assume full isotropy (epsilon of 1/2) it applies to identically launched bullets as well to one-way light speed

Yes.

 

[Dale]

But...if the synchronization method (e.g bullets synchronization method) was isotropic but light propagation process was not, it did not follow the constancy of one-way speed of light, I believe.

It does follow, actually. That is part of what Reichenbach showed. If the identical bullets are isotropic then the identical lasers are isotropic.

 

[cianfa72]

It does follow, actually. That is part of what Reichenbach showed. If the identical bullets are isotropic then the identical lasers are isotropic.

So choosing epsilon=1/2 two spatially separated events are said simultaneus if and only if they are simultaneous accoring any specific "isotropic" synchronization method chosen.

 

[Dale]

So choosing epsilon=1/2 two events are said simultaneus if and only if they are simultaneous accoring any specific "isotropic" synchronization method chosen.

Yes, that is correct

 

[cianfa72]

Yes, that is correct

Furthermore it follows that whatever "isotropic" method used -- assuming the constancy of two-way speed of light on closed paths equals to c -- then the one-way lightspeed has to be c (or just 1 when normalized).