# Transition from Flat Minkowski to GR

1. Feb 28, 2014

### clerk

Hello friends ,
I have some conceptual problems in understanding the difference between Minkowski spacetime and the spacetime of general relativity. The general spacetime of GR is defined as a smooth manifold which is locally like Minkowski spacetime . What does this statement mean ?

Does it mean that at a point I can always find a co-ordinate chart such that the associated basis of tangent vectors have an inner product which is (1,1,1,-1)? If that is the case , then I can always use such charts at all points of the smooth manifold and make it locally flat everywhere ? What happens if I had used some different chart ? Will the inner products at the point change to some crazy numbers like (2,2,7,4 ) ?

Infact , lets take a simple example of a 2-sphere of unit radius.. the natural metric is (1, sin^2 theta) ..how do I mathematically show that it is locally flat ie. have a metric (1,1) ?
Thanks for any help ...

2. Feb 28, 2014

### tiny-tim

hello clerk!
every space is locally flat (except at a singularity) …

tangent spaces are flat, and every space at a particular point is locally like its tangent space at that point, ie locally flat

(the (1,1) part of your question, i don't understand … isn't it obvious that it's locally (1,1) ?)
yes!
as i said, every space is locally flat everywhere

as for crazy numbers, you can always change the coordinates locally to make the metric whatever you like, provided you don't change the signs

3. Feb 28, 2014

### clerk

Thanks a lot , Tiny Tim !!
In the 2 -sphere example , I wanted to understand the required co-ordinate transformation that would reduce the metric (1, sin^2 theta) to (1,1) ..if it is locally flat , I should be able to explicitly show that , right ?
Also please will you see if my following statements are correct .. I want to ensure that I understand the things correctly..
There exists a co -ordinate chart for every point of spacetime such that the associated basis of tangent vectors have an inner product which is (1,1,1,-1). Had I used a different chart , I would have got a different metric but with the same index. So in the GR texts where we see that the authors make a co-ordinate redefinition to give the metric a different form , he/she is just using a different co-ordinate chart to describe the same geometry..this new co-ordinate chart may not cover the whole spacetime because if it did then we could have made a co-ord transformation to make the metric flat everywhere globally ..this inability to use a single co-ordinate chart to cover the full spacetime is due to curvature of spacetime - which is gravity ..?

4. Mar 1, 2014

### pervect

Staff Emeritus
You can transform the coordinates with $\theta' = k \theta$ where k is some constant. But - you might want to read up on non-coordinate basis. (I assume you have a text that you can look it up in). Basis vectors are not necessarily tied to coordinates.

5. Mar 1, 2014

### tiny-tim

hi clerk!
correct
no, there are no flat coordinate charts!

"locally flat" means that in the limit it is flat, ie the smaller the chart region gets, the nearer it is to flat

the only coordinate charts that are flat are of spaces which are flat

6. Mar 1, 2014

### Naty1

I happend to be reading about the same subject recently trying to verify terminology. I found some nice clarifications here:

http://en.wikipedia.org/wiki/Tangent_space

I concluded tangent vectors live in tangent spaces. So a tangent space is a flat coordinate chart in tiny tim's explanation. {I hope!}

7. Mar 1, 2014

### tiny-tim

no, as i understand it, a chart is a coordinate map of a region of the space

eg you can cover the earth's surface with two charts, one of the whole surface minus the north pole, and one of the whole surface minus the south pole (with any sensible projection you choose)

(and yes, a tangent vector is an element of a tangent space …

if the original space can be embedded in a higher-dimensional space, then the tangent space at a point is homomorphic to the tangent plane at that point in that higher-dimensional space)

8. Mar 1, 2014

9. Mar 1, 2014

### clerk

@all - thanks a lot for the stimulating discussions ..
Just one more clarification - in the popular QFT text Anthony Zee makes the comment that under suitable restrictions , the effect of a gravitational field is equivalent to a co-ordinate transformation - how exactly do I see that ? (Reference chapter : Field theory in curved spacetime , 1st paragraph)

10. Mar 1, 2014

### WannabeNewton

1. The term "locally flat coordinates" is a horrible misnomer. Nothing is flat in a set of "locally flat coordinates". A better term would be "locally inertial coordinates". Note that we can always diagonalize the metric tensor at any given point-this is just part of the finite dimensional spectral theorem. The really important property of "locally inertial coordinates" is the vanishing of the Christoffel symbols at the origin of the coordinate system.

2. Basis vectors in general have nothing to do with coordinate systems. Only the holonomic bases do but we can easily work with non-holonomic bases and in fact do so all the time because it is the latter which allow us to properly describe physics relative to different observers. Non-holonomic bases are what we use for local Lorentz frames.

3. The effect of a gravitational field is definitely not equivalent to a coordinate transformation. Actually what in the world does that even mean? He (Zee) says this is the principle of general covariance but that's certainly not true. At best he could be talking about the fact that in a locally inertial coordinate system the first order effects of a gravitational field vanish, such as gravitational redshift.

It's almost sacrilegious to equate the gravitational field, which is a purely geometric object, with coordinate systems.

EDIT: I checked out the Weinberg reference in Zee and nowhere does Weinberg say that the effect of a gravitational field is equivalent to a coordinate transformation (whatever that means). All he says is, if a law of physics is generally covariant and holds in the absence of gravitation then it will hold in the presence of gravitation as well. Actually this is not precisely true as stated-it is only true if the law is only zeroth order or first order in $\nabla_{\mu}$. But the basic idea is if we go to a locally inertial coordinate system then $\nabla_{\mu} \rightarrow \partial_{\mu}$. If we then write down a law of physics in generally covariant form using $\partial_{\mu}$ in this locally inertial system then it will also hold under $\nabla_{\mu}$ i.e. even when gravitation is present.

Last edited: Mar 1, 2014
11. Mar 1, 2014

### TrickyDicky

If more of Zee's statements in his nutshell series are like this one it should send up a red flag about those books.

12. Mar 1, 2014

### clerk

WannabeNewton : I agree with you .. I was never able to digest that statement.. but at the same time felt that may be I am missing some deep secret of GR , feel more relaxed now.. :)

13. Mar 2, 2014

### stevendaryl

Staff Emeritus
Maybe he's talking about parallel transport? If you take a ball and let it drift in freefall, it might start with a velocity 4-vector in which the spatial components are all zero. But if you wait a second, it will acquire a nonzero spatial component to its velocity 4-vector. That sort of looks like there is a boost operator relating 4-vectors in one region of spacetime to 4-vectors in a neighboring region of spacetime.

Mathematically, if $X^\mu$ is the displacement vector connecting two neighboring regions of spacetime, and $U^\mu$ is a velocity 4-vector for an object in freefall that drifts from one region to the other, then in the second region, it will have a velocity 4-vector given approximately by:

$\tilde{U}^\mu = L^\mu_\nu U^\nu$

where

$L^\mu_\nu = \delta^\mu_\nu + \Gamma^\mu_{\nu \lambda} X^\lambda$

It looks a little like an infinitesimal coordinate change.

14. Mar 2, 2014

### pervect

Staff Emeritus
A tangent space is indeed flat - the usual example is if you have a manifold which is the surface of a sphere, the tangent space is a plane that's tangent to the sphere.

The tangent space (the plane) isn't the manifold (the sphere). But there's a mapping from the tangent space (the plane) to the manifold (the sphere) called the exponential map that can map vectors from the flat tangent space to displacements on the not-necessarily-flat manifold.

15. Mar 2, 2014

### Staff: Mentor

MTW say something similar in Chapter 7 (basically that the effects of a gravitational field are equivalent to a uniform acceleration of the coordinates), but they say it's the equivalence principle, not the principle of general covariance.

I think part of the problem is the term "gravitational field", which is ambiguous, as has been discussed in a number of previous threads on PF. As far as I can tell, Zee here is using it to mean "coordinate acceleration due to gravity", which is similar to MTW's usage in the passage I just mentioned. That is not a geometric object, and can, of course, be transformed away by a change of coordinates. (Of course, if that is indeed what Zee meant, he could have just said so directly, instead of using the ambiguous term "gravitational field".)

The geometric object is *tidal* gravity, which the term "gravitational field" is sometimes used to refer to, but I don't think Zee means it that way here.

You also have to include $\sqrt{-g}$ in the spacetime integration measure (which Zee notes). The rest of the discussion in the chapter of Zee in question basically says the same thing Weinberg is saying (though he talks more about replacing $\eta_{\mu \nu}$ with $g_{\mu \nu}$, which also needs to be done to correctly generalize to curved spacetime). I agree that Zee's remark about a "gravitational field" being equivalent to a coordinate transformation isn't helpful; it could have been left out altogether without any loss of understanding, since the rest of his discussion covers the necessary points.

16. Mar 2, 2014

### Staff: Mentor

I've only read QFT in a Nutshell (the book under discussion here), and it has helped me better understand quite a few aspects of QFT, so I don't think a red flag is warranted for that book. I would be interested in any information about the second "nutshell" book that specifically covers GR.

17. Mar 2, 2014

### WannabeNewton

But that isn't a coordinate transformation. Parallel transport moves a vector along a curve from one tangent space to another. A coordinate transformation simply changes the components of a vector in the same tangent space by changing the coordinate basis.

18. Mar 2, 2014

### WannabeNewton

Right if it's with regards to the equivalence principle then I can agree if, as you say below, we interpret "gravitational field" to mean the coordinate acceleration due to gravity or, as is made apparent in the Newtonian limit, the Christoffel symbols.

It's very run of the mill so you aren't going to get any insights out of it that you don't already possess. I'm not a fan of the book simply because I much prefer books that stick to coordinate-free (abstract index or index-free) calculations whenever possible and only use coordinates when absolutely necessary, such as Wald. Zee's GR book abuses coordinates at every bend. It's quite reminiscent of Weinberg's book which I also dislike.

19. Mar 2, 2014

### Staff: Mentor

That was kind of the impression I was getting from reading reviews of it. Oh, well.

I think part of this is the particle physicist viewpoint as contrasted with the relativist's viewpoint. Particle physicists seem to want to think of everything in terms of coordinates.

Although, as a counterpoint to this, Zee's QFT book does a fair bit of analysis (for example, in the section on non-Abelian gauge theory) in terms of differential forms, with nary a coordinate in sight. So whatever is going on is probably more complicated than just "particle physicists prefer coordinates".

20. Mar 2, 2014

### WannabeNewton

It might be that the GR book in question was meant to be accessible to undergrads, in which case coordinates would be much friendlier than the sea of indices in coordinate-free calculations.

I haven't had much experience with Zee's QFT book but the writing in the book is absolutely phenomenal. It's the only physics book apart from Griffiths' electrodynamics that I can read as if I were reading a novel.

21. Mar 3, 2014

### clerk

@all : I also enjoyed Zee's QFT book a lot .. infact it is the one from which I got a very nice picture of what QFT is ..
Sorry I keep asking for clarifications , I will be grateful if anybody explains a bit more about what "co-ordinate acceleration due to gravity " is .. As Wannabenewton correctly points out , it must the christoffel connections in the Newtonian limit ..but how is it different from using the term "gravitational field " ? Is it correct to say that the acceleration due to gravity is something co-ordinate dependent but the curvature associated with such connection is a geometric object ?

22. Mar 3, 2014

### WannabeNewton

Let me just clarify that this would be true in a non-rotating coordinate system. In a rotating coordinate system the Christoffel symbols constitute the acceleration due to gravity as well as centrifugal and Coriolis forces. The term "gravitational field" is just ambiguous in GR is all-there is no actual quantity we can unequivocally claim is the "gravitational field" unless we are in the Newtonian limit.

Yes that's correct.

23. Mar 3, 2014

### Naty1

Hey clerk: [Sorry this got so long but I decided to summarize for myself maybe 20 pages of my notes.]

I have greatly benefitted from prior descriptions in these forums from these same posters ...and others. Here are some of the mostly physical attributes I found insightful.

The 'Newtonian 'gravitational field" finds its more general counterpart in GR as geometric curvature. The question is which 'curvature', and there is no simple answer. There is no single space-time curvature measurement that captures the GR concept of 'gravity'. The elements of gravitational curvature are those of the EFE. When physical curvature becomes part of an observation, the measured "c" varies. [edit: Such gravitational curvatures are captured in measures like Ricci, Riemann,scalar curvatures and Christoffel symbols...but none are individually 'gravity' nor overall gravitational curvature. Seems most the direct way to distinguish betwee flat and geometrically curved spacetime is via tidal measures.]

One great tip I came across is that the source of gravity in GR, the Stress Energy Tensor [SET] is calculated in the frame of the source. So it's invarient and so is the associated gravitational curvature. To visualize this, consider some object moving along: In it's own frame where SET GRAVITY is computed, no matter how fast it goes the gravitational description remains the same. So no matter how fast an object goes, it's gravity can't collapse it to a black hole. It's kinetic energy of linear motion is irrelevant. But it's internal component motion, say as captured by temperture, pressure, momentum flow and so forth is represented by the SET.

So how does a high speed particle follow a different trajectory than a slow one? The forces associated with speeds ARE coordinate based effects. Different observers, for example, at different intertial speeds see different trajectories. Different observers see different worldlines. So do observers in accelerating frames; They see curves inertial observers don't. But these types of coordinate based 'curves' are not aspects of GRAVITATIONAL curvature which is invariant.

Questions about a rapidly moving massive body's effect on a stationary test body can be transformed to an equivalent question about the interaction between a rapidly moving test body and a stationary massive body. The results are identical for any invariant. Thus all observables relating to a rapidly moving massive body can be answered as if the body is stationary and the observer so moving. You can always transform away coordinate acceleration due to gravity.

PeterDonis:
Consider a flat graph paper and rectilinear shapes and curves which may be drawn upon it representing overall space and time, including that observed by both inertial and non inertial observers.

From DrGreg:

Gravitational curvature of spacetime is an intrinsic property of spacetime and does not depend on the observer. In the absence of gravity, spacetime [graph paper] is always "flat" whether you are an inertial observer or not; non-inertial observers draw a curved grid on flat graph paper. With gravity, the graph paper itself must be curved in a complex way so that it cannot be flattened without wrinkling. So a cylindrical curvature, for example, is not that kind of curvature

Spatial curvature is a co-ordinate dependent property of a given space-time. GSC is not.

More here:
Space time curvature caused by fast electron

BenCrowell:

A simplified version of the equivalence principle is embodied in Einstein's elevator experiment:

.... there is no observable distinction between local inertial motion and free fall motion under the influence of a [uniform] gravitational field. This suggests the definition of a new class of inertial motion, namely that of objects in free fall under the influence of gravity. This new class of preferred motions defines a geometry of space and time—in mathematical terms, it is the geodesic motion associated with a specific connection which depends on the gradient of the gravitational potential.

Pallen:
Pervect:
Wannabe:

Last edited by a moderator: May 6, 2017
24. Mar 4, 2014

### clerk

Hey thanks a lot !! That was so nice :)