Is Proper Time Only Perceived by External Observers?

[Kairos]

I am confused about the notion of proper time because it is defined as one's own time measured by one's own clock, but any given person is in all circumstances naturally at rest with respect to himself and therefore his "own proper time" is the coordinate t; am I wrong?
So contrary to this definition, are the proper times (tau, different from time coordinates t) in fact perceived only by external observers, like Doppler effects which are only seen by observers external to the frame of reference of the source?

 

[PeroK]

I am confused about the notion of proper time because it is defined as one's own time measured by one's own clock,

What could be simpler than that? There's no need to consider any coordinate system, as the reading on a clock at any point in spacetime is a single, invariant number - independent of any coordinate system.

A clocking reading $t_0$ is an event, and the spacetime coordinates of that event depend on the coordiante system. The clock reading is its proper time.

 

[Ibix]

but any given person is in all circumstances naturally at rest with respect to himself and therefore his "own proper time" is the coordinate t; am I wrong?

You have it back to front. The time coordinate at an observer's location in their own (possibly non-inertial) reference frame can be defined as the reading on their clock. That is, you can use someone's proper time reading as the beginnings of a coordinate system. In fact, one way to define an inertial frame is to imagine a flock of synchronised clocks at mutual rest and use their proper times to coordinate time.

Proper times are the fundamental physical things. Coordinate times are a convenient abstract structure built on them. Just as, in spacen theodolite readings are fundamental physical things and map references are convenient abstract structure built on them.

So contrary to this definition, are the proper times (tau, different from time coordinates t) in fact perceived only by external observers, like Doppler effects which are only seen by observers external to the frame of reference of the source?

No. The proper time on some clock is the only thing you can measure. Coordinates aren't physical things that can be measured. It's the same as coordinates on Earth. You cannot measure your latitude and longitude - you need to measure altitudes of stars and compare the sun to a clock to deduce your latitude and longitude.

 

[Kairos]

The time coordinate at an observer's location in their own (possibly non-inertial) reference frame can be defined as the reading on their clock.

but this is also the definition of the proper time, right?

 

[etotheipi]

Not the definition, no. The proper time $\tau$ along a timelike worldline $\mathscr{L}$ of an observer $\mathscr{O}$ is just a parameterisation of the curve. You can, however, use this parameter to assign a time co-ordinate to arbitrary events. Consider that $\mathscr{O}$ emits a photon at an event $A_1 \in \mathscr{L}$, which reaches an event $M$ and is immediately deflected back so that it reaches $\mathscr{O}$ again at an event $A_2 \in \mathscr{L}$. Then, using the definition of Einstein-Poincaré simultaneity, the event $M$ is said to be simultaneous to an event $A \in \mathscr{L}$ on $\mathscr{O}$'s worldline iff $\tau(A) = \frac{1}{2}[\tau(A_1) + \tau(A_2)]$.

If the curvature of $\mathscr{L}$ is zero, or can be neglected in some neighbourhood close to the worldline, then the simultaneity hyperplane of $\mathscr{O}$ to an event $A$ can simply be identified with the set of events in orthogonal directions (with respect to the metric) to his worldline. Then, the proper time $\tau$ along with three spatial co-ordinates $(x^1, x^2, x^3)$ such that $\overrightarrow{AM} = x^i \mathbf{e}_i$ define the co-ordinates with respect to his so-called local frame $\{ \mathbf{u}(\tau), \mathbf{e}_i(\tau)\}$.

This is analogous, but distinct in concept, to an affine co-ordinate system $(O, \mathbf{e}_{\mu})$, which is defined simply by picking a point $O \in \mathbf{R}^4$ and letting $(x^0, x^1, x^2, x^3)$ be the numbers such that, for any $M \in \mathbf{R}^4$, you have $\overrightarrow{OM} = x^{\mu} \mathbf{e}_{\mu}$.

 

[Ibix]

but this is also the definition of the proper time, right?

No. Your proper time is the reading on your clock (give or take where you set the zero). Full stop.

Your time coordinate may or may not be the same as your proper time, depending on what coordinate system you are using. In SR in a coordinate system where you are at rest then your time coordinate will be the same as your proper time, yes. However, the time coordinate is also defined off your worldline. Furthermore, in GR your proper time does not necessarily advance at the same rate as coordinate time. So there are important reasons to keep in mind that coordinate time and proper time are distinct concepts, even if they coincide in specific circumstances.

 

[Kairos]

Not the definition, no. The proper time $\tau$ along a timelike worldline $\mathscr{L}$ of an observer $\mathscr{O}$ is just a parameterisation of the curve. You can, however, use this parameter to assign a time co-ordinate to arbitrary events. Consider that $\mathscr{O}$ emits a photon at an event $A_1 \in \mathscr{L}$, which reaches an event $M$ and is immediately deflected back so that it reaches $\mathscr{O}$ again at an event $A_2 \in \mathscr{L}$. Then, using the definition of Einstein-Poincaré simultaneity, the event $M$ is said to be simultaneous to an event $A \in \mathscr{L}$ on $\mathscr{O}$'s worldline iff $\tau(A) = \frac{1}{2}[\tau(A_1) + \tau(A_2)]$.

If the curvature of $\mathscr{L}$ is zero, or can be neglected in some neighbourhood close to the worldline, then the simultaneity hyperplane of $\mathscr{O}$ to an event $A$ can simply be identified with the set of events in orthogonal directions (with respect to the metric) to his worldline. Then, the proper time $\tau$ along with three spatial co-ordinates $(x^1, x^2, x^3)$ such that $\overrightarrow{AM} = x^i \mathbf{e}_i$ define the co-ordinates with respect to his so-called local frame $\{ \mathbf{u}(\tau), \mathbf{e}_i(\tau)\}$.

This is analogous, but distinct in concept, to an affine co-ordinate system $(O, \mathbf{e}_{\mu})$, which is defined simply by picking a point $O \in \mathbf{R}^4$ and letting $(x^0, x^1, x^2, x^3)$ be the numbers such that, for any $M \in \mathbf{R}^4$, you have $\overrightarrow{OM} = x^{\mu} \mathbf{e}_{\mu}$.

Did I understand: the proper time of an observer A can be determined by A directly by reading his watch, but can also be calculated by an observer B from another frame using the definition of simultaneity?

 

[Kairos]

No. Your proper time is the reading on your clock (give or take where you set the zero). Full stop.

but that's what I wrote at the beginning?

 

[PeroK]

Did I understand: the proper time of an observer A can be determined by A directly by reading his watch, but can also be calculated by an observer B from another frame using the definition of simultaneity?

Let me draw an analogy. The distance traveled by a car is the reading on the odometer. That is a simple definition. If you want to know the distance a car has travelled, you look at the odometer reading.

Now, of course, you can calculate that reading by whatever method you like from observations of the car's position at various times. But, that doesn't change the definition - nor does it need to be part of the definition.

The important point that I believe you are missing is that the definition of proper time is coordinate independent. You seem to want to find some way to give a coordinate dependent definition. Which is a backwards step, IMO.

 

[Ibix]

but that's what I wrote at the beginning?

Yes. My criticism was that you seemed to be regarding someone's proper time as derived from their time coordinate in their rest frame. That's backwards - proper time is the direct observable and coordinate time is the derived quantity.

Did I understand: the proper time of an observer A can be determined by A directly by reading his watch, but can also be calculated by an observer B from another frame using the definition of simultaneity?

Your proper time is the "distance" along your worldline. This can be calculated by anyone who has a history of your coordinates in some frame and a calculator.

 

[Kairos]

Furthermore, in GR your proper time does not necessarily advance at the same rate as coordinate time

even in GR your proper time and coordinate time remain the same as long as you are in inertial motion?

 

[PeroK]

even in GR your proper time ... is the time shown on your clock.

This is getting silly!

 

[Kairos]

This is getting silly!

in other words for PeroK : even in GR the time shown on your clock and the coordinate time remain the same as long as you are in inertial motion?

 

[PeroK]

in other words for PeroK : even in GR the time shown on your clock and the coordinate time remain the same as long as you are in inertial motion?

Proper time has nothing to do with inertial motion. You keep adding extraenous elements that are irrelevant to the defintion.

 

[Kairos]

Proper time has nothing to do with inertial motion. You keep adding extraenous elements that are irrelevant to the defintion.

I didn't say that proper time has something to do with the inertial motion, but I just said, following a post that mentioned GR, that the time coordinate and the proper time are the same as long as you are in inertial motion... and I'm not sure, this is a question!

 

[DrGreg]

even in GR your proper time and coordinate time remain the same as long as you are in inertial motion?

In GR as well as SR, it is always possible to choose a coordinate system in which your own spatial coordinates are a constant $(0,0,0)$ and your own time coordinate is equal to your proper time.

However, other coordinate systems are available.

 

[Dale]

therefore his "own proper time" is the coordinate t; am I wrong?

Yes, you are wrong. Even in an inertial observer’s rest frame proper time is not the same as coordinate time. They are equal where they are both defined, but the proper time is only defined along the worldline of the observer while coordinate time is defined throughout spacetime.

even in GR your proper time and coordinate time remain the same as long as you are in inertial motion?

No. As I said above they are defined in different domains. For two functions to be the same they must share the same domain as well as have the same value everywhere on that domain.

Also, in GR there is no way to extend an inertial observer’s coordinates in curved spacetime to get a global inertial frame.

 

[Kairos]

Even in an inertial observer’s rest frame proper time is not the same as coordinate time. They are equal where they are both defined, but the proper time is only defined along the worldline of the observer while coordinate time is defined throughout spacetime.

No. As I said above they are defined in different domains. For two functions to be the same they must share the same domain as well as have the same value everywhere on that domain.

Also, in GR there is no way to extend an inertial observer’s coordinates in curved spacetime to get a global inertial frame.

thank you for your corrections. I had (misleadingly) assumed that as a clock in free fall does not feel gravity, in its rest frame its proper time was the coordinate time.

I have another question: Is the coordinate time unique and is it the one used when talking about the age of the universe as a whole?

 

[etotheipi]

That's cosmic time, to define it you first need to introduce a set of so-called isotropic observers comoving with the substratum. These define a congruence of timelike worldlines filling the entire spacetime. In a homogeneous spacetime there is a one-parameter family of spacelike hypersurfaces $\Sigma_t$ orthogonal to these worldlines, which foliate the spacetime; you can construct a convenient time co-ordinate for these hypersurfaces by labelling each with the proper time of a clock carried by one of the isotropic observers.

 

[PeroK]

Is the coordinate time unique and is it the one used when talking about the age of the universe as a whole?

Coordinates are not unique: you always have a choice. There are useful coordinates (e.g. the heliocentric model of the solar system) and comoving coordinates in cosmology.

 

[vanhees71]

in other words for PeroK : even in GR the time shown on your clock and the coordinate time remain the same as long as you are in inertial motion?

No, in GR the proper time is defined analogously to how it's defined in SR, i.e.,
$$\tau=\frac{1}{c} \int \mathrm{d} \lambda \sqrt{g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}},$$
where the dot means derivation wrt. the arbitrary world-line parameter $\lambda$. Note that this expression does neither depend on the choice of coordinates nor on the choice of the parametrization. It's an invariant and thus an observable quantity. Neither $\lambda$ nor the coordinates have a priori some physical meaning. Their choice is arbitrary, but no really physical quantity depends on it.

It's again analogous to usual Euclidean analytical geometry: Changing the coordinates used to describe a geometrical object doesn't change any true geometric property of this object, e.g., the length of a curve or the angle between two intersecting lines, the area of a surface or the volume of some body.

 

[Dale]

Is the coordinate time unique and is it the one used when talking about the age of the universe as a whole?

Coordinate time is not unique.

The age of the universe at any event can be defined as the proper time for a comoving worldline from that event back to the Big Bang. That is unique for each event.

 

[Dale]

the arbitrary world-line parameter λ.

Doesn’t it need to be an affine parameter?

 

[Kairos]

Coordinate time is not unique.

Thank you for this clear answer! So it is impossible, for comparison purpose, to superimpose the graduations of the time coordinate axis of the different references frames of SR

The age of the universe at any event can be defined as the proper time for a comoving worldline from that event back to the Big Bang. That is unique for each event.

I don't see how to do this in practice and believed that the age of the universe was determined through the expansion rate. At the scale of the universe, all the observers are approximately comoving, aren't they?

 

[Mister T]

Proper time is the time that elapses between two events that occur at the same place. It's a relativistic invariant, meaning all observers will agree on its value.

 

[etotheipi]

Proper time is the time that elapses between two events that occur at the same place. It's a relativistic invariant, meaning all observers will agree on its value.

That's not a good definition. Firstly, because between any two timelike-separated events there are infinitely many timelike curves with generally different elapsed proper times; so you also need to specify the curve. And secondly because the events need not occur at the same spatial position in any given frame in order to define proper time; indeed, the notion of proper time along a curve is frame-independent.

 

[vanhees71]

Doesn’t it need to be an affine parameter?

no, why? The trick is that $\mathrm{d} s=\mathrm{d} \lambda \sqrt{\dot{x}^{\mu} \dot{x}^{\nu} g_{\mu \nu}}$ is parametrization independent. If $\lambda$ is an affine parameter, of course you get $\mathrm{d} s=C \mathrm{d} \lambda$, where $C=\text{const}$, which makes life easier.

 

[Dale]

no, why? The trick is that $\mathrm{d} s=\mathrm{d} \lambda \sqrt{\dot{x}^{\mu} \dot{x}^{\nu} g_{\mu \nu}}$ is parametrization independent. If $\lambda$ is an affine parameter, of course you get $\mathrm{d} s=C \mathrm{d} \lambda$, where $C=\text{const}$, which makes life easier.

Excellent, thanks. I guess I just got in the habit of always getting an affine parameter such that I have completely forgotten when it is necessary.

 

[Vanadium 50]

I guess I just get in the habit of always getting an affine parameter that I have completely forgotten when it is necessary.

It's always necessary in a B-level thread.

 

[stevendaryl]

Speaking of non-affine parameters,

In one sense, General Relativity is a natural generalization of Special Relativity. General Relativity is to Special Relativity as Riemannian geometry is to Euclidean geometry.

In another sense, a lot of the issues that are so important in motivating Special Relativity are tossed out the window when it comes to General Relativity. In teaching Special Relativity, a lot of time (no pun intended) is spent on such things as clock synchronization and setting up coordinate systems and reference frames, and defining proper time along a path as the time shown by a standard clock traveling along that path. The distinction between inertial and noninertial reference frames is really important, and a lot of conceptual errors trace back to not making such a distinction.

But that's mostly tossed out when it comes to General Relativity. Don't worry about clock synchronization. Just pick any old way to assign times to spacetime events (preferably so that a signal that is traveling at lightspeed or slower arrives at a later time than it left). Don't worry about how to operationally set up a coordinate system. Just pick any 4 independent scalar fields $X, Y, Z, T$ so long as distinct events have distinct coordinates. Don't worry about the distinction between inertial and noninertial coordinates, because any coordinate system is as good as any other. In describing parametrized paths, don't worry about whether the parameter is proper time, or not. If it's not, it makes some equations a little more complicated, but no big deal.

 

[vanhees71]

Excellent, thanks. I guess I just got in the habit of always getting an affine parameter such that I have completely forgotten when it is necessary.

Of course, working with affine parameters is of great advantage. That's why it's better to use the "square form" of the Lagrangian for the geodesics, i.e.,
$$L=-\frac{1}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu},$$
where the dot again means the derivative wrt. an arbitrary parameter $\lambda$, but thanks to Noether's theorem, since the Lagrangian is quadratic in the $\dot{x}$ and since it doesn't explicitly depend on $\lambda$ the "Hamilton-like" conserved quantity $H=L$ is conserved along the solutions of the equations of motion (which are just the geodesic equation). This means that for the solution $\lambda$ is automatically an affine parameter along the trajectories of the particle.

Another advantage is that this works without trouble for both light-like as well as time-like geodesics. In the latter case you simply choose the conserved quantity $g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=c^2$. Then you have $\lambda=\tau$, with $\tau$ the proper time along the geodesic. For the light-like case you have to set $g_{\mu \nu} =g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=0$, and of course there's no proper time, but $\lambda$ is still some arbitrary affine parameter. The physics of course doesn't depend on the choice of this parameter.

 

[Vanadium 50]

Of course, working with affine parameters is of great advantage.

Not so much in a B-level thread.

 

[vanhees71]

Particularly in a B-level thread, because it simplifies the task to solve the equations of motion ;-)).