# The metric tensor

1. Jul 7, 2009

### dianaj

Hi all,
I have just started learning about general relativity. Unfortunately my book is very math-oriented which makes it a bit challenging to understand the content from a physical point of view. I hope you can help.

I am familiar with special relativity theory and Minkowsky Space. As I understand it the most simple version of the metric tensor, g, is used to describe this flat space.

If a pick a point in a gravitational field the space around it curves. Unlike flat space we can only consider a very small section around this point, in order to make the space around it seem flat, right? But if the small section is very-nearly-flat why not use the equation from Minkowsky Space to describe it?

I am referring to this equation:
$$(d\tau)^2 = c^2(dt)^2 + r^2$$

I mean, what is the point of zooming in to make it look flat if it is not treated as flat space anyway?

If space is curved the metric tensor becomes a lot more complex and the equation becomes:
$$(d\tau)^2 = -g_{\mu\nu}dx^{\mu}dx^{\nu}$$

Am I right when I say that the equation describes processes in a very small, free-falling nearly-flat-but-still-curving section around a point in a big(ger) gravitational field? That g is not simple anymore due to the curvature in this small section? And that the point of view is that of a free-falling observer in that exact point?

And here is a bonus question: if a free falling observer (or the very small section) does not feel gravity why does he still percieve the space as curved?

Diana

2. Jul 7, 2009

### George Jones

Staff Emeritus
Hi Diana; welcome to Physics Forums!

Which book are you using?
Typo? In a curved spacetime, coordinates can be set up such that the metric has Minkowski form at any single event. Even in these coordinates, the metric will look different away from the chosen event.
Then, things look close to the special relativistic case, and intuition built up using special relativity often (but not always) can be applied.
Any point of view is possible, freely falling, or non-freely falling.
Non-zero curvature manifests itself in the form of tidal effects in any extended region, no matter how small, of spacetime.

3. Jul 7, 2009

### Irrational

the minkowski metric is $$(-1,1,1,1)$$ (with c taken to be 1)

in general, the line element is $$d\tau^2 = -g_{\mu\nu}dx^{\mu}dx^{\nu}$$

so in minkowski space, the line element becomes $$d\tau^2 = -dt^2 + dx^2 + dy^2 + dz^2$$ (which is what i think you are referring to in the equation above involving r)

the 'complexity' of the line element depends really on the metric chosen.

not quite sure what the question is about zooming is but http://en.wikipedia.org/wiki/Minkowski_space#Locally_flat_spacetime" makes sense.

Last edited by a moderator: Apr 24, 2017
4. Jul 7, 2009

### BWV

ok here is another related question, if you plug the properly signed kronecker delta (sorry don't know the correct term) as the metric tensor into the GR equations you get Minkowski space, correct? Some curvilinear metric tensor is then what would differentiate general from special relativity?

5. Jul 7, 2009

### Irrational

perhaps i'm wrong but i'd imagine that if we include time, then

kronecker delta = (1,1,1,1)
mikowski metric = (-1,1,1,1)

so no, you don't end up with minkowski space.

if you take time out of the equaton, then yes, the two are the same but this is not really any help is it?

6. Jul 7, 2009

### Irrational

ah. you said 'properly signed kronecker delta'... i presume you mean the minkoswki metric? :)

yes, you would get minkowski space then alright.

7. Jul 7, 2009

### BWV

yes, thats what I meant with the -1,1,1,1, on the central diagonal of a 4x4 matrix

so another question, what determines a valid metric in curved space? because its a tensor by definition you could convert minkowski space in a covariant transformation to, say, 4-dimensional spherical coordinates and back again without moving outside of special relativity - analogous to the transformations in any introduction to tensor calculus and not have curved spacetime. Is it the fact that the covariant derivative would in that case be zero, wheras with a "real" curvature you would have to contend with the christofel symbols?

8. Jul 9, 2009

### dianaj

It's actually http://www.nbi.dk/~polesen/GRC.pdf" [Broken] (course starts in september but I can't wait ). If you could recommend a book to read on the side I would be very happy.

Ah, I think that's where I went wrong. I think I somehow got it into my head, that the equations only describe processes as a free-falling observer would see it. But this is not true.

I think what confused me was that the book started with small free-falling elevators, but I have realized that this was only to derive the more general equations.

Last edited by a moderator: May 4, 2017
9. Jul 9, 2009

### Naty1

10. Jul 9, 2009

### Phrak

In a sufficiently small region of space (the limit as space and time displacements go to zero) spacetime may appear flat. In a freely falling reference frame you can apply the Minkowski metric at some point. Visually, think of a flat 2D Minkowski spacetime tangentially intersecting a curved 2D spacetime surface at a single point.

The curvature of spacetime enters into relativity by considering first and second derivatives of the metric with respect to displacement. So, though the metric can be Minkowski, derivatives of the metric are not.

In your notes, dianaj, curvature starts-in at equation 1.74. To me, the text looks a little too fast--too much of an outline. In the defense of the author, I only took a very brief look, though.

You might want to take the bold step and change texts. This, I believe, is by far the best online text available:
http://xxx.lanl.gov/abs/gr-qc/9712019"

Last edited by a moderator: Apr 24, 2017
11. Jul 9, 2009

### George Jones

Staff Emeritus
In a curved spacetime, coordinates can be set up such that the metric has Minkowski form at any single event P, and such that at P, all the first derivatives of the metric with respect to coordinates are zero (page 50 of Carroll's on-line notes). Such coordinates are called local inertial coordinates.

12. Jul 9, 2009

### Phrak

I don't see anything in Sean Carroll's Lecture Notes on General Relativity (Dec 1997) page 50, or even pages 48-52 to indicate that the fist derivatives of the metric are zero in Riemann normal coordinates, or any other singled-out coordinate system. Can you give a specific equation number to better substantiate this? Otherwise I will be forever wondering how you came to this conclusion.

Last edited: Jul 9, 2009
13. Jul 10, 2009

### Daverz

It's in between eq. 2.34 and eq. 2.35.

14. Jul 10, 2009

### George Jones

Staff Emeritus
My personal favourite, comprehensive introductions to general relativity are given in