I Help Understanding Metric Tensor

dsaun777

I am trying to get an intuition of what a metric is. I understand the metric tensor has many functions and is fundamental to Relativity. I can understand the meaning of the flat space Minkowski metric ημν, but gμν isn't clear to me. The Minkowski metric has a trace -1,1,1,1 with the rest being zero but how do I "find" what the metric is in General Relativity? I know that the metric is spatially dependent and locally flat, but how exactly does it depend on spacetime? I heard sources say that the metric is simply given or stated for the type of curved coordinates that you happen to be working in. From the metric derivatives you can get the Christoffels, from the derivatives of the Christoffels you can get the Riemann and from the Riemann you can get he Ricci curvature...etc. but how do you get the metric to begin with? I know its related to the Newtonian potential for the 00 components 1/2g00⇒Φ I am sort of lost when it comes to understanding the other components.

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kent davidge

I am trying to get an intuition of what a metric is. I understand the metric tensor...
It's what you said, a tensor.
I can understand the meaning of the flat space Minkowski metric ημν, but gμν isn't clear to me
The difference is that the Minkowski metric $\eta_{\mu \nu}$ might be used to describe Minkowski spacetime, while a general metric $g_{\mu \nu}$ is used to describe a more general spacetime.
how do I "find" what the metric is in General Relativity?
This is something tensors are good at. You already know the Minkowski metric. On a general spacetime, "general" coordinates must be used. If you know how they are related to the "local" coordinates, you will automatically know the general metric, via $$g_{\mu \nu} = \frac{\partial \xi^\alpha}{\partial x^\mu} \frac{\partial \xi^\beta}{\partial x^\nu} \eta_{\alpha \beta}$$
You can always do that because of the Equivalence Principle.
I know that the metric is spatially dependent and locally flat, but how exactly does it depend on spacetime?
The metric is locally flat by the Equivalence Principle. The explicit form of its dependence on the coordinates will depend on your choice of coordinate system, which in turn is dictated by the spacetime you are working with. For example, Minkowski spacetime allows you to choose coordinates such that the metric doesn't depend at all on either space or time coordinates.
but how do you get the metric to begin with?
I know its related to the Newtonian potential for the 00 components 1/2g00⇒Φ I am sort of lost when it comes to understanding the other components.
This is valid only for the situation where you assume three things:
- that the gravitational field is weak
- that matter is moving in non relativistic velocities
- that the gravitational field is static

I believe that the other (non-vanishing) components of the metric $g_{ik}$ are not completely fixed by this approach because you can always perform Lorentz rotations or boosts which will still keep the three assumptions invariant.

Nugatory

Mentor
The metric is locally flat by the Equivalence Principle.
It's the other way around. Because the spacetime is locally flat, the equivalence principle works.

kent davidge

By the way, don't let yourself get confused by symbols like $g_{\mu \nu}$ and $\eta_{\alpha \beta}$ for the metric coefficients, or $x^\mu$ and $\xi^\alpha$ for the coordinates. When I started studying Relativity, I had a hard time understanding the concepts just because of this notational mess. It doesn't matter at all what symbols you use as long as you keep track of what you are doing, you are free to use whatever symbols you want.

kent davidge

It's the other way around. Because the spacetime is locally flat, the equivalence principle works.
I like to take the Equivalence Principle as the dictator of how spacetime should behave. Do you think there's a problem with that?

dsaun777

It's what you said, a tensor.

The difference is that the Minkowski metric $\eta_{\mu \nu}$ might be used to describe Minkowski spacetime, while a general metric $g_{\mu \nu}$ is used to describe a more general spacetime.

This is something tensors are good at. You already know the Minkowski metric. On a general spacetime, "general" coordinates must be used. If you know how they are related to the "local" coordinates, you will automatically know the general metric, via $$g_{\mu \nu} = \frac{\partial \xi^\alpha}{\partial x^\mu} \frac{\partial \xi^\beta}{\partial x^\nu} \eta_{\alpha \beta}$$
You can always do that because of the Equivalence Principle.

The metric is locally flat by the Equivalence Principle. The explicit form of its dependence on the coordinates will depend on your choice of coordinate system, which in turn is dictated by the spacetime you are working with. For example, Minkowski spacetime allows you to choose coordinates such that the metric doesn't depend at all on either space or time coordinates.

This is valid only for the situation where you assume three things:
- that the gravitational field is weak
- that matter is moving in non relativistic velocities
- that the gravitational field is static

I believe that the other (non-vanishing) components of the metric $g_{ik}$ are not completely fixed by this approach because you can always perform Lorentz rotations or boosts which will still keep the three assumptions invariant.
So the metric in flat space can be transformed into the metric in GR using the property of tensor transformation? Another way to view the metric is just that gij=eiej where ei and ej are the basis vectors of your coordinate system correct? Pardon my symbolic illiteracy I'm fairly new. So in your equation you're changing reference frame with respect to another reference frame. Can you choose any coordinate basis you want as long as they transform properly?

Nugatory

Mentor
So the metric in flat space can be transformed into the metric in GR using the property of tensor transformation?
No. The metric of flat space can be transformed from one coordinate system to another (for example, from Cartesian coordinates to polar coordinates) but it's still the same metric tensor describing the same flat spacetime, just written in a different way. (It's also just as much "the metric of GR" as is the metric for the curved spacetime around a star; both are solutions to the Einstein field equations, just for different configurations of matter).
The metric tensor takes as input the coordinates of two nearby points and outputs the distance between them. Every possible shape of spacetime has a particular metric tensor associated with it; in fact, the metric tensor defines the spacetime and vice versa. Coordinate transformations don't change the spacetime or the relationship between points in it, they just change the way we label these points.
Can you choose any coordinate basis you want as long as they transform properly?
Yes, and that's the power and beauty of tensor methods: they don't care what coordinates we use. Just be careful not to confuse the same metric written in different coordinates with two different metrics. For example, I can write the metric of a flat two-dimensional plane using either polar or Cartesian coordinates; but no coordinate transformation will turn that metric into the fundamentally different metric of the two-dimensional surface of a sphere.

Nugatory

Mentor
I like to take the Equivalence Principle as the dictator of how spacetime should behave. Do you think there's a problem with that?
No inherent problem.... but the equivalence principle is an approximation that will always fail if examined closely enough so is a poor candidate for dictator. We've already started with the postulate that spacetime is a pseudo-Riemannian manifold; this implies local flatness, which in turn justifies the equivalence principle as an approximation. There's no natural way of running the logic in the other direction.

Of course the equivalence principle is a powerful motivation for choosing the postulate in the first place, and that's how the historical development went.

PeroK

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I am trying to get an intuition of what a metric is. I understand the metric tensor has many functions and is fundamental to Relativity. I can understand the meaning of the flat space Minkowski metric ημν, but gμν isn't clear to me. The Minkowski metric has a trace -1,1,1,1 with the rest being zero but how do I "find" what the metric is in General Relativity? I know that the metric is spatially dependent and locally flat, but how exactly does it depend on spacetime? I heard sources say that the metric is simply given or stated for the type of curved coordinates that you happen to be working in. From the metric derivatives you can get the Christoffels, from the derivatives of the Christoffels you can get the Riemann and from the Riemann you can get he Ricci curvature...etc. but how do you get the metric to begin with? I know its related to the Newtonian potential for the 00 components 1/2g00⇒Φ I am sort of lost when it comes to understanding the other components.
Are you studying GR in your master's? That information would help a lot in terms of the replies you might get.

A metric is a "distance function". It tells you the distance between any two points by specifying the distance for infinitesimal changes in the coordinates. You can get the distance between any two points, along a given path, by integrating along this path.

In this way the metric actually defines the geometry. This is the beauty of differential geometry: all the information you need about the geometry of whatever you are studying - in this case spacetime - is encapsulated in that one "differential" equation.

Now, for every choice of coordinates your metric will take a different form. But, it's still the same metric, just expressed in a different form dependent on your choice of coordinates. Take any two points in spacetime and the same path between them and the distance must be the same regardless of the coordinates you chose.

The metric is a tensor: it exists independent of any coordinate system and its components transform accordingly when you move from one coordinate system to another.

Note that it is not always easy to tell when two metrics are the same, i.e. describe the same spacetime geometry and are related by a coordinate transformation. You can have a very unflat looking metric that, with a change of coordinates, transforms into the usual metric for flat spacetime. The analogy is that Euclidean 3D space is the same space in Cartesian and spherical coordinates, even though the metric looks very different in each case.

If you have a region of spacetime, how do you determine the metric? You could go out and measure everything. You label enough points in your spacetime and measure the distance between any two neighbouring points. That would directly determine your metric.

(There is a story that Gauss actually wanted to try this: to measure the distance bewteen mountain tops to see whether space is actually Euclidean. Obviously he would have only got one measurement and he would have had no idea that the time coordinate might be relevant.)

Alternatively, you could study the stress-energy sources in your spacetime and write down the Einstein Field equations (EFE) and solve them!

That's what Schwarzschild did in 1916 for the simple case of a spherically symmetric mass surrounded by vacuum.

Because the EFE are non-linear they are very difficult to solve. In this way GR is much harder than Electromagnetism in that there is not a wide range of scenarios available to be studied. As the equations admit far fewer available solutions. That's why you are often simply given a metric to work with.

pervect

Staff Emeritus
I am trying to get an intuition of what a metric is. I understand the metric tensor has many functions and is fundamental to Relativity. I can understand the meaning of the flat space Minkowski metric ημν, but gμν isn't clear to me. The Minkowski metric has a trace -1,1,1,1 with the rest being zero but how do I "find" what the metric is in General Relativity? I know that the metric is spatially dependent and locally flat, but how exactly does it depend on spacetime? I heard sources say that the metric is simply given or stated for the type of curved coordinates that you happen to be working in. From the metric derivatives you can get the Christoffels, from the derivatives of the Christoffels you can get the Riemann and from the Riemann you can get he Ricci curvature...etc. but how do you get the metric to begin with? I know its related to the Newtonian potential for the 00 components 1/2g00⇒Φ I am sort of lost when it comes to understanding the other components.
Let's talk about a 2 dimensional flat space for the moment, a plane.

Then the metric (more precisely the line element) in cartesian coordinates (x,y) is $dx^2 + dy^2$.

And the metric in polar coordinates $r, \theta$ is $dr^2 + r^2\,d\theta^2$.

The general philosophy is that a plane is a plane, a geometric entity regardless of whether one uses cartesian coordinates or polar coordinates. We can go further, and say that the coordinates are just lablels, the labels for cartesian coordinates $x,y$ are just different labes than the labels for polar coordinates $r, \theta$. Sometimes I get pushback on this simple statement. I regard it as pretty basic, but I suppose if there is pushback on the idea, it's better if you (the original poster or reader) have some doubts, it's best if you think about it, and if you still don't agree, then post something.

I can move on a bit to something that is not a plane, but the two-dimensional surface of a sphere, which is a different geometric enity. But I'm out of time, and it's probably best not to try and do to much.

Adding more dimensions, including time, is another topic I'll skip for the moment that might be of interest.

dsaun777

Are you studying GR in your master's? That information would help a lot in terms of the replies you might get.

A metric is a "distance function". It tells you the distance between any two points by specifying the distance for infinitesimal changes in the coordinates. You can get the distance between any two points, along a given path, by integrating along this path.

In this way the metric actually defines the geometry. This is the beauty of differential geometry: all the information you need about the geometry of whatever you are studying - in this case spacetime - is encapsulated in that one "differential" equation.

Now, for every choice of coordinates your metric will take a different form. But, it's still the same metric, just expressed in a different form dependent on your choice of coordinates. Take any two points in spacetime and the same path between them and the distance must be the same regardless of the coordinates you chose.

The metric is a tensor: it exists independent of any coordinate system and its components transform accordingly when you move from one coordinate system to another.

Note that it is not always easy to tell when two metrics are the same, i.e. describe the same spacetime geometry and are related by a coordinate transformation. You can have a very unflat looking metric that, with a change of coordinates, transforms into the usual metric for flat spacetime. The analogy is that Euclidean 3D space is the same space in Cartesian and spherical coordinates, even though the metric looks very different in each case.

If you have a region of spacetime, how do you determine the metric? You could go out and measure everything. You label enough points in your spacetime and measure the distance between any two neighbouring points. That would directly determine your metric.

(There is a story that Gauss actually wanted to try this: to measure the distance bewteen mountain tops to see whether space is actually Euclidean. Obviously he would have only got one measurement and he would have had no idea that the time coordinate might be relevant.)

Alternatively, you could study the stress-energy sources in your spacetime and write down the Einstein Field equations (EFE) and solve them!

That's what Schwarzschild did in 1916 for the simple case of a spherically symmetric mass surrounded by vacuum.

Because the EFE are non-linear they are very difficult to solve. In this way GR is much harder than Electromagnetism in that there is not a wide range of scenarios available to be studied. As the equations admit far fewer available solutions. That's why you are often simply given a metric to work with.
I have BS in mathematics and I am studying relativity independently from Wisner and Thorne's massive book Gravitation. I took up to Special relativity at university So GR is pushing my comfort level of physics and mathematics, which is always a good thing!
So I could actually go out and with high precision instruments, say a gyroscopic light measuring device, and measure "points" in a given spacetime and construct a suitable metric?

Nugatory

Mentor
So I could actually go out and with high precision instruments, say a gyroscopic light measuring device, and measure "points" in a given spacetime?
You cannot measure points, but you can measure the distance between them, and that's basically what the metric is telling you. So in principle you could check your measurements against calculations using the metric to see whether you in fact have the right metric for the spacetime we live in. For example, very precise measurements could in principle tell you that the distance between the point "10 meters east of this stake in the ground" and "ten meters north of this stake in the ground" is not quite exactly $10\sqrt{2}$ meters (and the interior angles of the triangle do not add up to exactly 180 degrees) and therefore the surface of the earth is not a flat Euclidean plane - $ds^2=dx^2+dy^2$ is not giving exactly the right relationship between the lengths.

An added complication with spacetime is that time is also a coordinate, so a measurement like the one I described above requires that the distance measurements are between the same points at the same time. However, "at the same time" just means "has the same time coordinate" so any attempt to analyze the spatial geometry carries an implicit assumption about how we're assigning time coordinates; that's what @pervect was getting at when he said that Gauss wasn't aware of this pitfall. (In the example above I glossed over this pitfall by choosing a coordinate system in which the spatial coordinates of my three points (stake, ten meters north, ten meters east) do not change with time).

Do I have to make my metric dimensionless?
You can save yourself much grief by setting $c=1$. This is tantamount to measuring time in seconds and distances in light-seconds, or measuring distance in meters and time in units of the amount of time it takes for light to move one meter.

I am studying relativity independently from Wisner and Thorne's massive book Gravitation.
That's an excellent text, but the initial learning curve is a bit steep. You might want to take a quick swing through Sean Carroll's no-nonsense intro and then go back to MTW; it's a lot easier to follow MTW after you've seen the quick overview.

pervect

Staff Emeritus
I have BS in mathematics and I am studying relativity independently from Wisner and Thorne's massive book Gravitation. I took up to Special relativity at university So GR is pushing my comfort level of physics and mathematics, which is always a good thing!
So I could actually go out and with high precision instruments, say a gyroscopic light measuring device, and measure "points" in a given spacetime and construct a suitable metric?
For a spatial metric, all you need to be able to do to construct a metric is to be able to measure the distance between any two nearby points, with some sort of mechanism that I'll call "a ruler". You don't need a gyroscope, as we are not yet dealing with time - you just some way of measuring distances. You need to have an appropriate set of labels for these points, and they have to be "smooth". With a BS in math, you may (or may not?) appreciate that you can define a 1:1 invertible mapping between points on a line and points on a plane. This sort of labelling is not "smooth" in the needed sense, you can't construct a metric if you use this sort of non-smooth labelling of points. Point set topology introduces the necessary ideas to understand "smoothness". (This is different from other sorts of topics that are still called topology).

But that's more abstract than you really need to get, though if your focus is on the math and not the physics, you might find it interesting. MTW (the book you are studying) won't cover this sort of math, though, it's a physics textbook and not a math textbook. Walds text, "General Relativity", touches a bit on the math, but it is not as interesting a read as MTW.

Once you have an appropriate smooth labelling of your set of points on the geometric surface you are studying, you can construct the metric as long as you have some way of measuring the distances between nearby points.

The space-time metric is slightly more abstract. Rather than measuring the distances between points, you measure the Lorentz intervals between events. And events are just things that happen in a specific place at a specific time, since we are studying space-time. The coordinate labels we assign to events in space-time are similar to the labels we use to describe spatial geometry, but they include an additional label for "time". The time label is arbitrary, just as the spatial labels are - the tensor approach allows this sort of arbitrary labelling.

Hopefully the Lorentz interval is familiar to you from your study of special relativity. If your treatment of SR didn't stress the Lorentz interval (many don't), you might try "Space-time Physics" by Taylor and Wheeler, in addition to MTW.

You still don't need a gyroscope to find the space-time metric. With the introduction of time, one introduces the possibility of rotation. It turns out one can duplicate the functions of a gyroscope by measure the Lorentz interval between all points. Getting into the details would take us off-topic, but I will say that one can construct a mathematical model of a ring laser gyroscope, and this is my justification for saying that one doesn't need to postulate the existence of gyroscopes, one simply needs to postulate the existence of the Lorentz interval as a quantity that is independent of the observer and represents the physics one is studying.

The space-time metric of flat, empty space is the Minkowskii metric, usually written as

$$ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$$

You can see that there is a minus sign in front of the "time" term. This makes the metric not positive definite - it's called a Lorentzian metric.

Light beams in a vacuum are a convenient physical representation of a curve with a zero-length Lorentz interval.

It may take some thought to figure out how one can use knowledge of the Lorentz interval to measure other things. It turns out that the key instrument for measuring Lorentz intervals is the clock, which measures something called proper time, which is the timelike interval along the worldline of the clock. This is a different sort of time than the coordinate label that we assigned to time earlier was that was arbitrary. The proper time is something that can be measured with a clock (often called wristwatch time), and it's not a matter of convention, it's a physical quantity that can be measured.

Having a clock and using the property that light in a vacuum travels a path of zero Lorentz interval, it's possible to measure various observer-dependent notion of "length" or "spatial distance" between points. It's an observer dependent notion, because of so-called "Lorentz contraction", so we expect different observers to have different notions of "spatial distance".

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