# Tensor pertubration: Hello to everyone

1. Apr 18, 2007

### darashayda

Hi

I just joined the forum and wanted to thank everyone about your informative posts.

I am currently studying the mathematical foundations of the general relativity and may need some help.

My first question:

Is there a way to perturb the fundamental tensor to obtain another metric tensor?

Or, as I guess, once fundamental tensor is found, it is very hard to make slightest changes!?

What I mean is can I perturb gij to say gij + e and still the new perturbation stays a metric?

Dara

2. Apr 18, 2007

### robphy

One often considers a one-parameter family of metrics.

http://link.aip.org/link/?JMAPAQ/31/2441/1 [Broken] (Damour and Schmidt) "Reliability of perturbation theory in general relativity"

(Wald) "General Relativity" chapter 7 has a discussion of perturbation.

Last edited by a moderator: May 2, 2017
3. Apr 18, 2007

### Chris Hillman

Perturb to a constant curvature metric, even

I guess it depends upon what you mean by "perturbation".

This is more of a research topic than a topic for newcomers to semi-Riemannian geometry, but FWIW there is a classic theme in Riemannian geometry ("bundle" positive definite bilinear forms) in which one shows how one can "perturb" any metric to some constant curvature metric. In fact one can "represent" any metric by systematically stipulating a perturbation (in a specific sense) to the reference metric. See Llosa and Soler, "On the degrees of freedom of a semi-Riemannian metric", Class. Quantum Grav. 22 (2005): 893-908 for some Lorentzian geometry ("bundle" indefinite but nondegenerate bilinear forms) analogues in low dimensions.

By the way, if you are interested in "counting" the "number of (semi)-Riemannian D-manifolds", the simplest approach I know is due to Einstein himself (possibly inspired by discussions with Hilbert/Noether about basically the same argument in what we now call commutative algebra). For that see S.T.C. Siklos, "Counting solutions of Einstein's equations", Class. Quantum Grav. 13 (1996): 1931-1948.

Last edited: Apr 18, 2007
4. Apr 18, 2007

### darashayda

Thanx

Ok this is what I need to find out.

If there is a fundamental tensor gij, for every infinitesimally small e can I find another fundamental tensor g'ij that satisfies: |gij - g'ij| < e at least for a local neighborhood if not for the entire space.

I will check the references as you suggested. But I appreciate the help though I am just a beginner :)

Dara

5. Apr 18, 2007

### darashayda

Dear robphy

I cannot get the files! No login/password and I could not find them in another place on the net.

Any suggestion or does anybody has a copy?

D

6. Apr 18, 2007

### robphy

Those are articles are from journals which allow online access to (individual and institutional) subscribers. You could try to google the titles and/or the individual authors.

7. Apr 18, 2007

### Stingray

You can write down just about anything you want and call it metric (though there are some minimal restrictions). Are you asking whether an initial solution of Einstein's equation can be perturbed in to another solution?

8. Apr 18, 2007

### Chris Hillman

Not sure I understand the question

I guess it depends upon what you mean by "find", but if you are asking about perturbation of a Lorentzian manifold to a "nearby" manifold, and if I understand the question correctly, then the answer is "in principle, sure, same as for any tensor field in a sufficiently small neighborhood". If you are asking about perturbation of vacuum solutions of the EFE to a "nearby" vacuum solution, the answer is "in principle, if you suitably constrain the perturbation, yes, sure".

9. Apr 18, 2007

### darashayda

Dear Chris and Stingray

What I mean by the metric is an invariant metric that satisfies the usual tensorial transformations. Not all functions can support that, only a few to my knowledge e.g. the euclidean length.

You may perturb a tensor to still act as a compliant vector, to be a solution to the equations, but the question is can you perturb the tensor ARBITRARILY close to its original value.

Best I can see you can perturb all the tensors but they are no longer a tensor they are a matrix of some kind and will not conform to the transformation rules nor to the invariance requirements.

In other words I can take the diff equation solutions; add a little hack function as in the case of the WEAK FORCE/ENERGY computations, and get myself an approximate solution.

But could I perturb the tensors, with arbitrarily close amounts to their original values, and satisfy the equations 100%.

Lets say I have a solution to Einstein equations, then add a little perturbation to the solutions and plug them back in and walla I get all the equalities satisfied without any error factor.

Lets say I did that, then can I do it again and again with perturbation closer and closer to the original solution? Again no approximate compliance but fully satisfying both sides of the equations.

Dara

10. Apr 19, 2007

### Chris Hillman

I cannot understand a word you are saying

Oh dear, I read this yesterday and it doesn't make any more sense to me today:

Invariant metric? Invariant with respect to what?

You can stipulate that an equation be "satisfied", but I've never seen anyone stipulate that a transformation be "satisfied". And what are these "usual tensorial transformations"?

A function can have a "support", namely the subset of the domain where it takes on nonzero values, but I have no idea you mean here.

Again, "compliant vector" is not a standard term in mathematical English, so I again I have no idea what you are talking about. Is this perhaps a loose translation of a term you saw in a book in French?

What equations?

I thought you were proposing to perturb something away from an initial state!

For perturbation of a mathematical object of type T to be well defined, it should result in a new object of type T, sure, but I don't see why you believe that the perturbations discussed in differential geometry are ill-defined.

What invariance requirements? Invariance with respect to what?

"hack" function?

There is a "weak-field approximation" to gtr, which you could reasonably call "low energy approximation" or "small curvature approximation", but there is no "weak force approximation".

Whereas equation (10) is only 54% satisfied?

(Sorry if that sounds a bit sarcastic, but I'm trying to convey to you the poor impression which your drastic and protracted abuse of language makes.)

At least now I can infer that you are not after all reading books in French :-/

Dara, seriously now:

For all I know it is possible that you are trying to ask a perfectly sensible question, but if that is the case, at this point I have grave doubts about your ability to make yourself understood until you have learned more of the standard language used in the gtr literature.

The only analogy I can use to convey what reading your questions above was like, would be hearing a sports announcer say "Well, the Babe really batted the old puck around the golf rink in the Superbow today, scoring billions and billions over pair, but we knew it was all over when Ali knocked down the warning flag with a wild drive to center field, leaving the Giants in command of the Cup".

11. Apr 19, 2007

### darashayda

Hello Chris

you can make fun of me as u like it wont reduce me sincerity and effort to learn GR. It only reflects upon your character.

INVARIANCE is with regards to COORDINATE TRANSFORMATIONS. This is a standard lingo u can find in all the books dealing with GR or related material.

> but I've never seen anyone stipulate that a transformation be "satisfied"

[url]http://www.iop.org/EJ/abstract/0305-4470/11/10/009[/url]
[url]http://press.princeton.edu/chapters/s8111.pdf[/url]

And many others, please do search the above files for "satisfying the transformation."

The term is also used satisfying coordinate transformation or coordinate systems or the transformations rules/laws. the lingo is clearly that of GR.

Compliant vectors: A'i = aij Aj normally called vector in most American texts.

[url]http://www3.interscience.wiley.com/cgi-bin/abstract/109573633/ABSTRACT?CRETRY=1&SRETRY=0[/url]
[url]http://arxiv.org/PS_cache/hep-th/pdf/0301/0301146v1.pdf[/url]
[url]http://www.iop.org/EJ/abstract/0022-3719/21/7/009[/url]
Search for 'tensorial transformations' or non-tensorial.

> There is a "weak-field approximation" to gtr, which you could reasonably call "low energy approximation" or "small curvature approximation", but there is no "weak force approximation".

yes my error in terminology sorry, u used the correct phrase.

My hack function was the perturbed gij + h I called it hack to indicate the h function used in the 'low energy' texts I am reading

> What equations?

Tensorial equations and coordinate transformation equations.

Ok my question again in more precise terms:

Can EVERY GR metric invariant under the coordinate transformations be perturbed in such a manner that a) it will still be an invariant metric under the coordinate transformations, b) perturbation can be arbitrarily close to the original metric i.e. for every e there is ds' that |ds - ds'| < e for some open region of the timespace.

Are there such examples for R4?

D

Last edited by a moderator: Apr 22, 2017
12. Apr 19, 2007

### MeJennifer

Good for you Darashayda!
Don't be discouraged, most people on this forum are actually quite nice, like to help and are not in the habit of showing off their knowledge by putting down others.

Last edited: Apr 19, 2007
13. Apr 19, 2007

### Stingray

I don't think this was Chris's intention at all. Your posts really are hard to understand. At the very least, please try to use real words (u = you and so on). I also have no idea what you mean by a "compliant" vector. How is it different from any other vector?

Regardless, your main misunderstanding appears to lie in the fact that invariance is not a property of any object's components in a single coordinate system. It has to do with the relations between these components in all possible coordinate systems. Any bounded 4 x 4 symmetric matrix with the appropriate signature can represent the metric components of some spacetime in some coordinate system. You'd then find the components in other coordinates using the standard transformation laws. But no matter what you start with, those laws always make sense.

It follows that metrics which are arbitrarily close always exist. Nothing changes if you require them to satisfy Einstein's equation. Almost any metric at all can be generated by an appropriate matter distribution (though it may be completely unphysical). Furthermore, two sets of metric components may even describe exactly the same spacetime while differing by an infinitesimal amount (try stretching each coordinate by a factor $1+\epsilon$).

Last edited: Apr 19, 2007
14. Apr 19, 2007

### darashayda

Hello Stingray

I believe in your latter statement about the generation of ANY metric from some matter distribution. I guess that is due to the fact that the matter distribution appears on one side of the equation and the curvature metric on the other side. So you can drive the metric by the matter distribution, and do so arbitrarily close. That was a good point and thank you.

I cant convince myself that any bounded 4x4 with appropriate signature is a metric. It may transform by the transformation functions/rules but it will not be invariant across the transformations.

D

15. Apr 19, 2007

### Stingray

It is probably more direct to pick a metric, compute the Einstein tensor, and then say that that's your stress-energy tensor (multiplied by $8 \pi$). Some people amuse themselves by "solving" Einstein's equation this way. Random guessing almost never yields reasonable types of matter, however.

I don't understand what you mean. The statement "invariant across transformations" means that the components transform according to the appropriate rules. It does not say that the the components don't change at all. If you required that, the 0 matrix would be the only "metric!"

16. Apr 19, 2007

### darashayda

I meant ds2 (squared) is invariant not the components. I just dont believe that almost all symmetric 4x4 with non-vanishing det do the job.

But again your point was really nice. you may suffer by seeing me asking more questions :)

D

17. Apr 19, 2007

### Stingray

Ah, ok. $ds^2$ is one of those things that shouldn't be taken too seriously. It's more of a "physicist's shorthand." Still, saying that it's invariant just means that distances between points do not depend on the coordinate system. This is actually equivalent to stating that a metric should obey the standard tensor transformation rules.

Writing down $ds^2$ implies the comparison of two neighboring points in a given spacetime. Fix some coordinate system in whichever spacetime you're interested in. Then say that point A has coordinates $x^{\mu}$. Let point B have coordinates $x^{\mu} + dx^{\mu}$. Then the squared distance between these points is
$$ds^2[A,B] = g_{\mu \nu}(x) dx^{\mu} dx^{\nu}$$
If you change coordinate systems, the left-hand side shouldn't change. The new separation "vector" is different, but so are the metric components. If the latter quantities transform according to the standard rules, the squared distance stays the same, as required.

If you change spacetimes, it is no longer meaningful to say that you're comparing the same two points. You certainly wouldn't expect that the distance between two (new) points that can have same coordinate representation to be the same. And indeed it isn't. $ds^2$ is not invariant under spacetime changes.

I hope that helps .

18. Apr 20, 2007

### darashayda

Dear Jennifer

Thanx for the encouragement, I hope to learn from you as well.

19. Apr 20, 2007

### darashayda

Dear Stingray

Right, what you wrote about ds2 is what I meant all along, I will try to learn how to use the equations in this forum instead of babbling :)

D

20. Apr 20, 2007

### Stingray

You can enter equations using Latex. Are you familiar with that? If so, inline equations can be written using the tags [ itex ] and [ /itex ]. Get rid of the i's for larger equations. There's a huge thread about this that you can probably dig up.