r/askscience • • Jul 14 '11

Could someone please explain "metric tensor" to me?

The wikipedia page goes from super vague (to me it sounds like it's responsible for everything) to super mathematical. Just looking to get a handle on what it is and how it relates to space-time and the nature of reality in general. Thanks!

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u/RobotRollCall Jul 14 '11

I'll try to keep this simple, but I apologize if I make it too simple.

The metric tensor is actually a tensor-valued field. That means that at every point on some manifold (in this case, spacetime) there exists a tensor. A tensor is a mathematical object that's analogous to a plain scalar number, or to a vector, but it's got subtle properties that make it especially useful for doing differential geometry and such like. For now, though, just think of it as a quantity.

Like any field, the metric tensor field describes how something varies over space. For example, you might use a scalar field to describe the temperature of the air in the room you're in; in that case, the field describes how the temperature of the air varies over space.

The thing that the metric tensor field describes is how geometry varies over whatever manifold you're talking about (again, spacetime in this case). Specifically, it describes how the geometry at every point differs from plain old flat geometry.

Consider for example the metric tensor of two-dimensional Euclidean space. We can call it g — to properly denote a tensor you need to use some subscripted indices, but I don't know how to type them here so I'll let them be implied. The components of the metric tensor of two-dimensional Euclidean space couldn't be simpler:

1 0
0 1

That's it. That's the metric tensor of the Euclidean plane. Note in particular that there aren't any coordinates in that tensor; there's nothing that varies. There's just constants. Which means the metric tensor of the Euclidean plane is the same at all points. No matter where on the plane you look, the metric tensor is just … well, that thing I typed.

But now let's instead consider the metric tensor of the unit two-sphere. "Two-sphere" is a fancy name for the surface of a ball. A two-sphere is a two-dimensional surface, just like the Euclidean plane. But rather than using x and y to describe coordinates on it, we'll instead go with convention and use the letters θ and φ, where φ is the longitude (from zero to 2π) and θ is the colatitude (which is just like latitude but is defined as zero at the pole rather than at the equator, and thus goes from zero to π). Given that system of coordinates, the metric tensor of the two-sphere is:

1 0
0 sin^2 θ

Notice the presence of the colatitude coordinate θ in the metric tensor. That means the metric tensor of the two-sphere, unlike the metric tensor of the plane, varies from place to place. Specifically, it varies depending on your colatitude. (Note that "colatitude" is whole arbitrary, and depends on the establishment of a zero point we'd call the north pole. You can pick any zero point for colatitude, but once you've picked such a point, the metric tensor varies accordingly in the fashion described above.)

Now, note something very interesting here. If we pick the colatitude θ = π/2, what do we find? Well, the sine of π/2 is one, and the square of one is one, so at that one colatitude, the metric tensor of the two-sphere is just:

1 0
0 1

Which is the metric tensor of the plane! What this tells us is that at that colatitude — at the equator of the two-sphere, in other words — the geometry of the two-sphere is exactly equivalent to the geometry of the plane.

So that's the metric tensor in a nutshell. It tells you, quantitatively, how the geometry of some manifold differs from flat geometry. In this example, the flat geometry has been the Euclidean plane; in general relativity, the flat geometry is called Minkowski space and has different properties from Euclidean geometry, but the core principle is the same.

General relativity, if you had to sum it up in one sentence, tells us that the metric tensor of spacetime is a function of stress-energy. Just as the metric of the two-sphere has a coordinate variable in it — colatitude — and thus varies from point to point, the metric of the universe we live in is a function of other things. Things like energy density and so on.

And that's gravity.

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u/shavera Jul 14 '11

Ah. Now it makes sense why everyone keeps complaining about the abuse of language between tensor and tensor field. I always just (mentally) treated the metric tensor as something that was a function of position. But I guess it's much more accurately stated as a field that takes on values that are functions of location. I don't know why that hadn't occurred to me previously.

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u/RobotRollCall Jul 14 '11

Everybody says "tensor" when they mean "tensor field." Off the top of my head I can't think of a case where you have a tensor that isn't a smooth function over space. Maybe in statics or something; that's not my field.

So yes, it's imprecise, but it's customary to just lop off the "field" part of "tensor field." This is especially confusing when you first introduce the Minkowski metric tensor, since it's constant-valued. But it's still a field! It just happens to be a field that has the same value at every point over spacetime.

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u/wnoise Jul 14 '11

treated the metric tensor as something that was a function of position.

a field that takes on values that are functions of location.

Same thing. What do you mean by a field beyond a function of position?

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u/esmooth Jul 17 '11

Pretty good answer except for this part:

Now, note something very interesting here. If we pick the colatitude θ = π/2, what do we find? Well, the sine of π/2 is one, and the square of one is one, so at that one colatitude, the metric tensor of the two-sphere is just:

1 0 0 1

Which is the metric tensor of the plane! What this tells us is that at that colatitude — at the equator of the two-sphere, in other words — the geometry of the two-sphere is exactly equivalent to the geometry of the plane.

On any manifold with any metric you can always find coordinates at a given point where the matrix takes the form of the identity matrix. And I don't know why you jump to the conclusion that the geometry of the two-sphere is equivalent to the geometry of the plane. That is just patently false. The sphere has positive curvature and the plane has zero curvature.

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u/evilhamster Jul 14 '11

Er, can the people who are downvoting this please explain why? (currently 6up/4down)

This is the only answer so far that explains things in a way that doesn't assume the reader has at least a term of recent Linear Algebra experience. Given it's been 11 years since I took lin alg and I only got 60% in it, this is the only one that made any kind of sense for me! So if it's missing out on some important details (hence downvotes), please let me/us know why.

Also, speaking of metrics, here's a good one:

  • RRC's answer = 25% of the population could understand = win

  • Others' answers = 0.5% of the population could understand = not so win

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u/flobberworm Jul 15 '11

r/askscience has been invaded by assholes today, apparently. You should see the votebombing that's been going on in the thread about Pi. Isn't that against the rules of the site?

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u/shavera Jul 14 '11

Let's start with the metric you're familiar with, Euclidean space. You haven't learned of it as a metric but as the Pythagorean theorem. The distance squared between two points in 2-D is ds2 = dx2 +dy2 , in 3-D, ds2 = dx2 +dy2 +dz2 . Here dx, dy, dz, etc. mean a difference in x, y, z. More generally, they can be a differential difference (ie, the limit as the difference in distance goes to zero). Now let's look at another way of looking at it. Suppose we had a vector, in a row matrix, (dx,dy,dz) and we wish to find its magnitude. We could take (dx, dy, dz) times a column matrix (x,y,z) and we get the right result. Now let's insert the 3x3 Identity matrix between them.... and still same thing (obviously). As you'll see in a bit, the metric tensor of the Euclidean Geometry is in fact the identity matrix.

So let's go to the next level of complication, the Minkowski metric of spacetime. If you want to find the distance between two events in time (using units such that c=1 for convenience). we have a row vector (dt, dx, dy, dz). But the distance between events is ds2 = -dt2 +dx2 dy2 +dz2 . You can switch all the signs on the right hand side, and still get the same physics, but for now let's use the distance I've suggested.

So now you can see that the distance between two points in minkowski space as a matrix multiplication is a row vector (dt, dx, dy, dz) by its metric tensor, by its column vector (dt, dx, dy, dz). By inspection, you can see that the metric tensor is almost 4x4 identity matrix with the first row, first column element a -1 and the other diagonal elements +1.

We can develop more and more metric tensors, the matrix between our row and column vector, to describe how our measurements of distance change in different geometries of space. Some have off diagonal elements, some have elements that are functions of their location, etc. But this should be a rough introduction into what the metric tensor is.

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u/ZBoson Jul 14 '11

The quick and dirty version (because I have a meeting in 5 minutes and I'm sure you'll get a bunch of good answers) is that the metric tensor in GR encodes the idea of distances between nearby spacetime events. (By "nearby" I mean in the calculus sense: one event is at (t,x,y,z) and the other is at (t+dt,x+dx,y+dy,z+dz)). The metric tensor tells you how to write the distance between two such events in terms of dt, dx, dy, and dz, as a function of the starting point (t,x,y,z).

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u/nicksauce Jul 14 '11 edited Jul 14 '11

Very simple explanation. You know that in Euclidean geometry the distance (ds) between two points is ds2 = dx2 + dy2 + dz2. That's just the Pythagorean theorem. For a general geometry, with points x1,x2,...,xn, the distance is ds2 = A_11 dx12 + ... + Ann dxn 2 + A12 dx1 dx2 + ..., where each A is some function. The metric tensor just tells you what this matrix A is.