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In this question-

Is proper time equal to the Invariant Interval or the time elapsed in the Rest Frame?

@Dale in the comments says-

no $dt$ is never physical time. It is always coordinate time. The equation $d\tau=dt’$ needs to be interpreted carefully. It is not a general fact, but instead it is only true along the worldline of the clock and only in the (primed) rest frame of said clock. But even then, the physical time is $d\tau$ and the coordinate time $dt’$ is designed to match it. That is what makes the primed coordinates the rest frame of the clock [...]

What is coordinate time?

What does one mean by coordinate time? Isn't it a time measured on the clock? @Dale says no. Then what is meant coordinate time?

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  • $\begingroup$ Time is measured on a clock. But whose clock? $\endgroup$ – R. Emery Dec 26 '20 at 22:33
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In relativity we deal with spacetime which is a 4-dimensional structure called a (pseudo-Riemannian) manifold. It includes space and time together. Meaning that time is just another direction in spacetime, perpendicular to the three dimensions of space.

In the manifold we have a metric which describes all of the geometric properties of spacetime. It describes physically measured distances and durations and angles and speeds. You can use it to identify straight or curved lines and so forth. Straight lines in this sense are technically called geodesics.

A point particle forms a line in the spacetime manifold, called its worldline. An extended object forms a worldsheet or a worldtube, but we will stick with worldlines for now. A physically measured duration is called proper time, $d\tau$, and is measured by a physical clock on a specified worldline. An inertial object (in free-fall, no real forces acting on it) moves in a straight line at constant speed according to Newton’s first law. In other words, an inertial object’s worldline is straight.

Now, spacetime does not come with any physical labels or grid lines on it, but often we find it convenient to add labels/gridlines. These are called coordinates. Coordinates are ordered set of four numbers that label each event in spacetime. They must have some nice mathematical properties like being smooth and invertible. Everything else is a matter of convenience.

One approach that is often convenient is to use the first coordinate for time and the last three coordinates for space $(t,x,y,z)$. A further convenience is to use straight lines and planes. A further convenience is to choose an inertial object as a reference and orient those lines so that the time coordinate axis is parallel with the reference object’s worldline and so that the space axes remain fixed and don’t rotate. A final convenience is to scale the coordinates so that on the reference object’s worldline the physically measured proper time matches the coordinate time and so that coordinate distances match measured physical distances. Such coordinates are called the inertial reference object’s rest frame. In the absence of gravity such coordinates can extend to cover all of spacetime, but with gravity they can only be local, near the path of the inertial object.

Importantly, in these coordinates the metric can be written as $ds^2 = -c^2 d\tau^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2$. The physical time measured on any clock can be obtained by integrating the above expression for $d\tau$ along the worldline of that clock. The coordinate time is merely a convention for assigning the labels as described above. This convention is chosen to match the physical time on the reference object’s clock, but not necessarily on other clocks. Elsewhere, $dt$ is merely a label.

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  • $\begingroup$ If I have understood correctly then in the Schwarzschild metric case -$(1-R/r) dt$ ( not just dt but it's coefficient included) measures the physical time and similarly $(1-R/r) ^{-1}dr$ measures the physical distance just like in the Minkowski metric case $cdt$ would give the exact physical time and $dx$ or $r^d\theta$ would give the physical distance. Is that right. Have what I understood now correct. That means when the $g_{00} =1$ then $dt$ alone would represent physical time but when $g_{00}\neq 1$ then the entire $g_{00}dt$ ( dt together with the coefficient) represents the physical time $\endgroup$ – Shashaank Dec 26 '20 at 12:38
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    $\begingroup$ @Shashaank for 1) yes, although it is easier to write it $d\tau$. 2) no, time dilation is the relationship between proper time and coordinate time at the same location. So $\gamma = dt/d\tau$ $\endgroup$ – Dale Dec 26 '20 at 13:27
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    $\begingroup$ @Shashaank comments are for improving the question, not for discussion. You really need to go to a site that is more conducive for discussion. I agree that what you describe is more physically meaningful, but it is called redshift, not time dilation. Redshift is directly measurable as the change in the frequency of light emitted by one object and received at another. If redshift is measured between two stationary objects then it is called gravitational redshift. If the objects are moving then it includes Doppler redshift. $\endgroup$ – Dale Dec 26 '20 at 14:21
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    $\begingroup$ @Zorawar “all objects in GR move along straight (geodesic) paths” this is incorrect. Only inertial (free falling) objects move on geodesics. Non inertial objects do not travel on geodesics. You are currently probably not in free fall and so not moving on a straight path in GR. See Carroll’s Lecture Notes on p 70: arxiv.org/abs/gr-qc/9712019 “The primary usefulness of geodesics in general relativity is that they are the paths fol- lowed by unaccelerated particles. In fact, the geodesic equation can be thought of as the generalization of Newton’s law f = ma for the case f = 0” $\endgroup$ – Dale Dec 26 '20 at 15:27
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    $\begingroup$ @Zorawar I am sorry but I do not like your proposed terminology at all. It surprises me that you think it is acceptable to write “all paths” and mean “only pure-gravitational paths” and yet object to “perpendicular” which is a perfectly valid way to describe a tetrad. I also do not think the word “Euclidean” belongs anywhere in a discussion about pseudo-Riemannian manifolds (except maybe “not Euclidean”). Your terminology does not at all seem to be an improvement. I have no intention to alter my answer as suggested. The only suggestion that seems palatable is “locally”. I will consider that. $\endgroup$ – Dale Dec 26 '20 at 22:56
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In special relativity one traditionally imagines that spacetime is filled with a network of clocks that are Einstein synchronized. The clocks define an inertial reference frame, and the time coordinate of an event is the reading on the nearest clock.

When dealing with curved spacetime you usually can't Einstein synchronize clocks, and it's usually not convenient to use clock readings (however synchronized) directly as a coordinate, so coordinates are treated more abstractly. The same is true when dealing with ordinary curved surfaces like the surface of the Earth. We use latitude and longitude to measure global position, but there are no rulers that measure latitude and longitude; measurements are done locally with metersticks. In general relativity, the metric tells you how to convert between local measurements of clocks and metersticks and the global coordinates that you're using.

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  • $\begingroup$ If I have understood correctly then in the Schwarzschild metric case -$(1-R/r) dt$ ( not just dt but it's coefficient included) measures the physical time and similarly $(1-R/r) ^{-1}dr$ measures the physical distance just like in the Minkowski metric case $cdt$ would give the exact physical time and $dx$ or $r^d\theta$ would give the physical distance. Is that right. Have what I understood now correct. That means when the $g_{00} =1$ then $dt$ alone would represent physical time but when $g_{00}\neq 1$ then the entire $g_{00}dt$ ( dt together with the coefficient) represents the physical time $\endgroup$ – Shashaank Dec 26 '20 at 11:56
  • $\begingroup$ This looks to me by far the best explained answer. Please just confirm the above comment I have put. Also it would be really helpful if you could please elaborate the last sentence in your answer mathematically i.e please elaborate mathematically a bit on "the metric tells you how........ the global coordinates that you're using." I think it will be perfect then. Thanks $\endgroup$ – Shashaank Dec 26 '20 at 11:59
  • $\begingroup$ Could you please confirm the above point. $\endgroup$ – Shashaank Dec 27 '20 at 20:34
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Mathematically Space-Time is a 4-dimensional Lorentzian manifold $L$. Coordinates of an event, i.e. $(x_{\mu})=(ct, x, y, z)$ only make sense when you define an Atlas $\mathcal{A}={(\mathcal{U}_{\alpha}, \phi_{\alpha})}$ on $L$ where $\mathcal{U}_{\alpha}\subset L$ and $\phi_{\alpha}:L\rightarrow\mathbb{R}^4$ is a homeomorphism (i.e. continus functions between the topological Spaces $M$ and $\mathbb{R}^4$). The coordinates of a point $p\in U_{\alpha}\subset L$ are then given by $\phi_{\alpha}(p)=(x_1, x_2, x_3, x_4)$.

There are a few restrictions on the $\phi_{\alpha}$ in order for the math to make sense, but other than that you're more or less free to choose any parametrization you want. The main idea is now, that physical quantities and laws should in a sense be independent of this choice. $\tau$ is such a quantity, the coordinates $x_i=\phi_{\alpha}(p)_i$ are obviously not.

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$d\tau$ is an invariant of the metric. $dt$ is the time measured by a clock in a particular frame, as $dx$ is the length measured in this frame.

It is like saying that in 2D spatial cartesian coordinates, $ds^2 = dx^2 + dy^2$. $ds$ here is invariant by rotation. And we can also say that for any orientation of $ds$, there is a particular frame so that $ds^2 = dx^2$ (or $ds^2 = dy^2$).

But for 2D polar coordinates it is different. $ds^2 = dr^2 + r^2 d\theta^2$. $ds$ is also invariant by rotation. But there is no angle of rotation that leads to $ds^2 = d\theta^2$. It only happens in the case of $r = 1$. It corresponds to the example of $d\tau = dt$ for a static observer in the Schwartschild metric, that is only valid at $r = \infty$.

The Minkowski metric can be compared to the example of cartesian coordinates, and $d\tau$ can be understood as the time for a clock at rest of an inertial frame.

But in GR, similar to the example of polar coordinates, it is not generally true.

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  • $\begingroup$ If I have understood correctly then in the Schwarzschild metric case -$(1-R/r) dt$ ( not just dt but it's coefficient included) measures the physical time and similarly $(1-R/r) ^{-1}dr$ measures the physical distance just like in the Minkowski metric case $cdt$ would give the exact physical time and $dx$ or $r^d\theta$ would give the physical distance. Is that right. Have what I understood now correct. That means ever the $g_{00} =1$ then $dt$ alone would represent physical time but when $g_{00}\neq 1$ then the entire $g_{00}dt$ ( dt together with the coefficient) represents the physical time $\endgroup$ – Shashaank Dec 26 '20 at 11:56
  • $\begingroup$ Could you please confirm the above point. $\endgroup$ – Shashaank Dec 27 '20 at 20:35
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    $\begingroup$ Yes, as already answered by Dale. It is the same principle for measuring a distance between 2 points in an open landscape. It is possible to use some topographic device and make a local direct measurement. Or use GPS to get the ccordinates of each point, and make a calculation to get the distance. The spacetime metric is the rule to calculate not only distance between points, but time between events. And that must coincide with a local measurement by metersticks or clocks. $\endgroup$ – Claudio Saspinski Dec 27 '20 at 20:57
  • $\begingroup$ Thanks I understand now. But your last line in the above comment is a bit confusing to me. I am a bit confused what do you mean by the last and the 2nd last line. Could you please add a mathematical equation regarding your last line "And that must coincide.......... metersticks or clocks". It will be perfect then, I guess. Thanks again $\endgroup$ – Shashaank Dec 28 '20 at 7:22
  • $\begingroup$ @ClaudioSaspinski Sorry for nitpicking, but distances and times are defined along some paths, not between two events. $\endgroup$ – Umaxo Dec 28 '20 at 10:47

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