# Minkowski Metric

1. May 6, 2013

### shounakbhatta

Hello,

I don't know whether I have mentioned the subject line properly. Many times while reading over General Relativity I come across the following equation:

ds^2=dx1^2+dx2^2+dx3^2+dx4^2

=dx^2+dy^2+dz^2-c^2dt^2.

Now, my question from the above equation is:

(a) Are we putting the Pythagorean theorem into Minkowski Metric?
(b) dx1,dx2......are they four dimensions like, length, breadth, height and time?
(c) c is it the speed of light?
(d) How the factor -c^2dt^2 comes?
(e) What is the use of converting this to Minkowski metric?

Kindly explain me in step-by-step as I am new to this world.

Thanks

2. May 6, 2013

### yenchin

A metric simply describes how to measure distance. In 2D space, the Pythagoras theorem says
$s^2=x^2 + y^2$. In 3D space, it says $s^2=x^2+y^2+z^2$, but you know that already I assume. The infinitesimal version is simply $ds^2=dx^2+dy^2+dz^2$.

In Special Relativity, we are interested in both space and time {t,x,y,z}. But they don't have the same unit since t is time and the rest are spatial dimensions. Thus one uses {ct,x,y,z} instead. Note that ct has dimension of length. The metric is *defined* to be $ds^2=-c^2dt^2+ dx^2+dy^2+dz^2$. This has a few nice properties: The metric gives zero for light ray, since light travels at speed c, i.e. $c^2 dt^2 = dx^2 + dy^2 + dz^2$ is precisely the distance traveled by light in time interval dt. I hope this provides a nice picture of what is going on.

Note that in many relativity texts, one chooses the unit where $c=1$ so that the metric is simply written $ds^2=-dt^2+ dx^2+dy^2+dz^2$. There are also convention in which one writes $ds^2=dt^2- dx^2-dy^2-dz^2$.

Does that help?

3. May 6, 2013

### shounakbhatta

Hello Yenchin,

That was really a great reply. Actually, it cleared out lot of doubts that I have been carrying for a long time. Now, I still have certain questions. I would be thankful if you can please answer me:

(1) As you have told that metric is basically how to measure the distance. Now, in General Relativity we have different metrics like Schwarzschild metric and other metric. Does that mean that we have different type of metrics to measure different things in different co-ordinates?

(2) As you have mentioned the metric has been **defined**. Does that mean ds^2=-c^2dt^2 is not a derivation but an assumption?

It would be very kind of you if you plz.explain a bit.

Thanks

4. May 6, 2013

### WannabeNewton

The metric tensor is not a coordinate dependent quantity. It is something that assigns an inner product to each fiber of the tangent bundle of space-time. We express the metric in certain bases such as the coordinate basis when doing actual computations but the metric tensor itself is a purely geometric quantity. The different metrics you mentioned are different solutions to Einstein's equations and describe different space-time geometries.

5. May 7, 2013

### yenchin

You can assign many types of metric on a given manifold [if you are not familiar with this notion, think of it as an abstract mathematical "space"]. A manifold a priori does not come with metric. For example the four dimensional space $\Bbb{R}^4$. You can equip it with the usual Euclidean metric $d = \sqrt{dx^2 + dy^2 + dz^2 + dw^2}$, or you can equip it other the various other metrics, including the Minkowski metric, to define your geometry. It is not a derivation, but it is not just an "assumption", it really depends on what do you want to do; in relativity the theory is a theory of spacetime and we use a pseudo-Riemannian manifold to model the physics.

6. May 7, 2013

### shounakbhatta

Hello,

Thanks for the reply. It cleared up certain concepts. I just want to ask you few more questions. I hope it is ok with you, as more discussions clear certain concepts and crops up new things:

(a) The other types of metric like Schwarzschild metric and others, are they for measuring distances at different co-ordinate system?

(b) Do we have only three types of metric, the Euclidean, the Lorenzian and Schwarzschild metric?

(c) The x,y,z,t (4 dimension) is measured in Lorentzian metric, right?

(d) The Schwarzschild metric helps us to measure around a spherically symmetric body, such as a planet, or a black hole etc, right?

Now, I have one confusion. The Lorentzian metric g, when taken to General Relativity, does that take a general form and becomes Schwarzschild metric?

Thanks.

7. May 7, 2013

### WannabeNewton

I would like to stress again that the metric tensor is an object that is by itself coordinate independent. We can express it in any allowed coordinate basis but the value given by the action of the metric tensor on elements of each fiber of the tangent bundle is independent of the coordinates; you are confusing the metric tensor with coordinate systems, the latter of which have no real physical meaning. However, coordinate transformations of the metric tensor are important in the sense that since it is coordinate independent, we can always pick a convenient or more revealing coordinate system to express our metric in, knowing we will get the same value regardless for the "infinitesimal displacements".

Here's a relatively interesting example for anyone interested in a neat coordinate transformation (this is problem 3.2 in Wald): consider a 2 dimensional manifold $M$ with metric $g_{ab}$ and associated derivative operator $\nabla_{a}$. Let $\alpha$ be a scalar field on $M$ that is harmonic meaning it satisfies $\nabla^{a}\nabla_{a}\alpha = 0$ and let $\epsilon_{ab}$ be a 2-form (i.e. a tensor field such that $\epsilon_{ab} = -\epsilon_{ba}$) satisfying $\epsilon_{ab}\epsilon^{ab} = 2(-1)^{s}$ where $s = 0$ if $g_{ab}$ is a riemannian metric and $s = 1$ if $g_{ab}$ is a lorentzian metric. $\epsilon_{ab}$ is also called the natural volume element on $M$; note that the above definition implies $\epsilon_{ij}\epsilon^{ab} = (-1)^{s}2\delta^{[a}_{i}\delta^{b]}_{j}$. Finally, consider another scalar field $\beta$ given by the equation $\nabla_{a}\beta = \epsilon_{ab}\nabla^{b}\alpha$.

We first want to show that locally there always exists a $\beta$ that actually satisfies the above equation. That is, we must show that given the 1-form $\omega_{a} = \epsilon_{ab}\nabla^{b}\alpha$, there always exists locally a scalar field $\beta$ such that $\nabla_{a}\beta = \omega_{a}$. The most straightforward way to do this would be to show that $\omega_{a}$ is closed, meaning $\nabla_{[a}\omega_{b]} = 0$ because then the Poincare Lemma guarantees that locally $\omega_{a}$ is exact, that is locally there exists a scalar field $\beta$ such that $\nabla_{a}\beta = \omega_{a}$, which is exactly what we want.

Note that if we can show $\epsilon^{ab}\nabla_{a}\omega_{b} = 0$ then we are done since this would imply $\epsilon_{ij}\epsilon^{ab}\nabla_{a}\omega_{b} = \nabla_{[i}\omega_{j]} = 0$. Also note that since $\epsilon_{ab}\epsilon^{ab} = 2(-1)^{s}$, we have that $\epsilon^{ab}\nabla_{c}\epsilon_{ab} = 0$ implying $\epsilon_{ij}\epsilon^{ab}\nabla_{c}\epsilon_{ab} = \nabla_{c}\epsilon_{ij} = 0$ identically. Proceeding with this, we have $\epsilon^{ab}\nabla_{a}\omega_{b} = \epsilon^{ab}\nabla_{a}(\epsilon_{bc}\nabla^{c}\alpha) = \nabla_{a}(\epsilon^{ab}\epsilon_{bc}\nabla^{c}\alpha)$ and this reduces to $\epsilon^{ab}\nabla_{a}\omega_{b} = (-1)^{s + 1}\nabla_{a}(\delta_{c}^{a}\nabla^{c}\alpha) = (-1)^{s + 1}\nabla_{a}\nabla^{a}\alpha = 0$ hence $\nabla_{[a}\omega_{b]} = 0$ as desired so locally there always exists a $\beta$ such that $\nabla_{a}\beta = \epsilon_{ab}\nabla^{b}\alpha$ as desired. Finally, note that $\nabla^{a}\nabla_{a}\beta = \epsilon_{ab}\nabla^{a}\nabla^{b}\alpha = 0$ because $\nabla_{a}$ commutes on scalar fields (torsion-free) so the second covariant derivative is symmetric in the two indices whereas $\epsilon_{ab}$ is anti-symmetric in the two indices so the contraction vanishes. Thus $\beta$ itself is harmonic.

Now, what does all this have to do with coordinate transformations? Let's say we had originally expressed our metric in some coordinates $(x^1,x^2)$ so that $ds^{2} = g_{\mu\nu}dx^{\mu}dx^{\nu}$ where $\mu,\nu = 1,2$. Let's choose the scalar fields $\alpha(x^{1},x^{2}), \beta(x^{1},x^{2})$ as coordinates. The coordinate transformation law for the metric tensor tells us that $g_{\alpha\alpha} = \nabla^{\mu}\alpha \nabla^{\nu}\alpha g_{\mu\nu} = \nabla_{\mu}\alpha \nabla^{\mu}\alpha$ and similarly $g_{\beta\beta} = \nabla_{\mu}\beta\nabla^{\mu}\beta = \epsilon_{\mu\nu}\epsilon^{\mu\gamma}\nabla^{\nu}\alpha \nabla_{\gamma} \alpha = (-1)^{s}\nabla_{\mu}\alpha\nabla^{\mu}\alpha$. Finally, $g_{\alpha\beta} = \nabla_{\mu}\beta \nabla^{\mu}\alpha = \epsilon_{\mu\nu}\nabla^{\nu}\alpha \nabla^{\mu}\alpha = 0$ again because we are contracting symmetric indices with anti-symmetric ones. Letting $\Omega^{2} = \nabla_{\mu}\alpha \nabla^{\mu}\alpha$, under the coordinate transformation our line element takes the form $ds^{2} = \pm \Omega^{2}(d\alpha^{2} + (-1)^{s}d\beta^{2})$. We call these coordinates harmonic coordinates. So given any lorentzian metric $g_{ab}$ on a 2-manifold, we can always locally put the metric in the form $ds^{2} = \Omega^{2}(-dt^{2} + dx^{2})$ i.e. we can always make a coordinate transformation to locally make the metric equivalent to the flat 1 + 1 lorentzian metric up to a conformal factor. I thought this was a cool result

The Schwarzschild metric is a Lorentzian metric. A Lorentzian metric on space-time is a pseudo-Riemannian metric of signature (-,+,+,+).

Yes.
It is, to be more specific, for a spherically symmetric static body.

8. May 7, 2013

### pervect

Staff Emeritus
There are as many metrics as there are ways of giving events coordinates, i.e. an infinite number.

It might be useful to talk about metrics in a more familar context - the surface of the Earth. It is convenient to represent locations on the surface of a spherical Earth by two coordinates which we call lattitude φ and longitude $\lambda$. This is not the only possible approach, but it's a very common one.

Suppose you have two points and we want to know the great circle distance (the distance on the Earth's surface) between them. You can look up an exact function for the "great circle" distance between two points on the Earth's surface, this goes like

http://en.wikipedia.org/wiki/Great-circle_distance

$s = 2\,\arcsin \left( \sqrt {\sin^2\frac{d\phi}{2} + \cos \phi_1 \cos \phi_2 \sin^2 \frac{d\lambda}{2} } \right)$

This is rather complicated and doesn't illustrate the concept of a metric yet. The metric comes into play when we approximate the above, for "nearby" points. If you're familiar with taylor series, we can just use a multi-variable taylor series. The result then becomes the much simpler formula

$ds^2 = dφ^2 + cos^2 φ \, d \lambda^2$

If you compare this to the "Schwarzschild " metric, you'll see that it's rather similar, it's a quadratic equation, or bilinear form. Symbolically

$ds^2 = g_{φφ} d φ^2 + g_{\lambda\lambda} d \lambda^2 \quad g_{φφ}=1 \quad g_{\lambda\lambda} = \cos^2 φ$

But there is no time involved, it's a "distance" metric, not a space-time metric. A little more on the time part later.

So, our metric is just a function that converts changes in coordinates to changes in distances - for nearby points.

You can think of a metric as also representing how distorted a map made using your coordinates will be, if we draw it on a flat sheet of paper.

In the example above, we can see that near the equator, the metric is diagonal, indicating no distortion. Near the poles, $g_{\lambda\lambda}$ goes to zero, indicating large distortions.

We could choose different coordinates (a polar projection) to make the poles appear less distorted - but then the equator would be distorted. We will never be able to represent a curved surface undistorted on a flat piece of paper.

Note that if we choose different coordinates, we get different values of the metric components. The values of the metric components tell us how distorted are maps will be, but they aren't anything physical in and of themselves, by choosing different coordinates we will have different metric component values.

The above is how spatial metrics work. Lets go back to the space-time metrics.

One of the things one learns in special relativity is that time, and distance, are not fundamental. What's fundamental is the "Lorentz interval" ds, which is giving by the Minkowskii metric.

It's tough to say a lot about GR without learning more about SR first. One thing we can say is that the metric coefficients for space-time tell us how distorted our representation is, just as they did in our simpler example on the Earth.

Metrics that approach the Minkowskii metric (or the various spherical forms of it) represent undistorted space-time.

i.e
$ds^2 = dx^2 + dy^2 + dz^2 -c^2 dt^2$ is a cartesian form of the Minkowskii metric
$ds^2 = dr^2 + r^2 d \theta^2 + r^2 cos^2 \theta d \phi^2 - c^2 dt^2$ is a spherical form of the same metric.

As the metric coefficients get to be significantly different from their Minkowskii values, our representation of space-time is becoming more distorted. Unless we know more , we can't say for sure whether this distortion represents an odd choice of coordinates, or whether it reflects effects that are due to actual space time curvature. We can always change coordiantes to make the distortion go away at any one point, but in the presence of actual curvature, we can't choose coordinates to make it go away everywhere.

If you compare the Schwarzschild metric to the spherical form of the Minkowkii metric, you can see that it's similar, but become very distorted near r=2m, where some of the components vanish, and others become infinite, indicating infinite distortion.

It turns out that this is a coordinate distortion, similar to the way that we had vanishing metric coefficients near the poles of the Earth with our lattitude-longitude coordinates. But it takes quite a bit of math to demonstrate this.

9. May 7, 2013

### shounakbhatta

Hello Pervect,

Thank you very much for the detailed reply. I can understand the following things. Please correct me if I am wrong:

(a) A metric measures distortion over a surface, a manifold.
(b) ds is the Lorentz interval
(c) The line element in Euclidean space is: ds^2=g11 (dx1)^2+g22(dx2)^2, where g is the metric tensor.

Now, I want to ask you the Euclidean metric, in the matrix format gives (1 0 0,0 1 0,0 0 1). I want to know, are there any simultaneous equations which give rise to the values mentioned in the matrix?

I can well understand the example that you have provided in the cartesian form of Minkowski metric and the spherical form of Minkowski metric.

10. May 7, 2013

### WannabeNewton

11. May 7, 2013

### pervect

Staff Emeritus
I should clarify a few things of what I said, though you quoted me correctly.

a) The distortion isn't the distortion of the mainfold, per se. The distortion is in the map you draw. Making the map is technically a projection operation. The metric gives you the information on how badly the physical shape of the actual surface is distorted when you project it on to a flat piece of paper so that the coordinates form a cartesian grid.

c) ds^2=g11 (dx1)^2+g22(dx2)^2 [+ g12 dx1 dx2] is the general line element for a 2 dimensional Euclidean space.

If we consider the metric:
$$\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$$

if your coordinates are x,y,z, the line element corresponding to this metric would be dx^2 + dy^2 + dz^2. It would be a flat three dimensional Euclidean space. It's not a set of simultaneous equations, it's the matrix representation of a "quadratic form"

In matrix notation, it looks like this:

$$\left( \begin{array}{c} dx & dy & dz \end{array} \right) \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{ccc} dx\\ dy\\ dz \end{array} \right) = \left( \begin{array}{c} dx & dy & dz \end{array} \right) \left( \begin{array}{ccc} dx\\ dy\\ dz \end{array} \right) = dx^2 + dy^2 + dz^2$$

12. May 7, 2013

### yenchin

By the way, what is your background in mathematics? Are you reading certain GR book? We may be able to suggest suitable texts for further reading. I think you need to firmly grasp the idea of metric spaces and differential geometry of surfaces before discussing general relativity.

13. May 7, 2013

### WannabeNewton

I agree with this whole-heatedly. It is very hard to give rigorous accounts of something if the background is lacking and it is also very hard to hand wave something as important as metric tensors; "intuitive" but highly non-rigorous, non-mathematical explanations will only take you so far.

14. May 7, 2013

### shounakbhatta

Hello,

To be very honest with you all, my background of mathematics is not very strong. I am doing a self-learning from Special Relativity, graduating to learn GR.

Thanks for the help.

15. May 7, 2013

### yenchin

That does not help us to suggest a suitable text... could you be more specific about your current level of mathematics, i.e., what have you already learned? Calculus? Differential equations? Linear Algebra?

16. May 7, 2013

### shounakbhatta

Yes. Calculus, differential eqn.and linear algebra I have done but not into differential geometry or topology.

17. May 8, 2013

### yenchin

OK. I think if you are ambitious, you are quite ready to learn some differential geometry.

Some people prefer to directly learn abstract manifold theory, but I think it is good to get some geometric intuitions by first learning 3-dimensional differential geometry of surfaces. A standard text of this is Do Carmo's "Differential Geometry of Curves and Surfaces ". Depending on your interest, you don't really have to go over the whole book though. You can then pick up Riemannian geometry. Do Carmo has a book on that too. But my favorite differential geometry text is John M. Lee's "Riemannian Manifolds: An Introduction to Curvature". His book on smooth manifold is also good. Of course GR is really about semi-Riemannian geometry [though physicists like to refer to it as merely 'Riemannian'], so you would want to also refer to texts like O'Neill's Semi-Riemannian Geometry With Applications to Relativity.

For GR, again, there are many choices. If you prefer to see the "physics first" approach, Hartle is a popular choice. If you are more mathematically inclined, go with Carroll. I personally don't like Hartle's book, and first learned GR via the little book of Foster & Nightingale. You should of course explore many texts and see which one suits you the best.

By the way if you are more mathematically-minded, you may want to check out the following book for the structure of special relativity [it won't teach you much physics, but it is a superb math book]: The Geometry of Minkowski Spacetime: An Introduction to the Mathematics of the Special Theory of Relativity .

18. May 8, 2013

### VantagePoint72

I realize you prefaced by saying it would just help develop intuition, but I think it should be emphasized that none of this is really necessary. Any standard intro GR text, of which you mentioned a couple, will teach all the necessary differential geometry in the first few chapters. Nothing more than some very basic multivariable calculus and a bit of linear algebra are necessary to get started. More important than prior experience with differential geometry is, I think, a solid foundation in special relativity, going as far as the covariant formulation of electrodynamics. It is the simplest example of a classical field theory and I think it's much more valuable for developing the sort of intuition needed to get through GR. Unfortunately, I don't have a good SR book to recommend since when I was an undergraduate and took relativistic electrodynamics the professor wrote his own course notes, guided by Landau and Lifgarbagez (which is not an acceptable stand-alone text for a self-guided student). For GR, though, I especially like Carroll's book.

19. May 8, 2013

### Fredrik

Staff Emeritus
I just want to add that the curvature book is not an introductory text on differential geometry. The other one is, but since Lee wrote the curvature book before the introductory book, he didn't include anything about curvature in the introductory book. Both books are great. Everyone who wants to learn differential geometry for GR should get both.

20. May 8, 2013

### WannabeNewton

Learning math from a physics book is a terrible idea no matter what physics subject it is, period.