# Thread: SPLIT : Discussion of Solving Einstein Equations Sticky

1. As a neat side note, looking at the metric shows that something interesting happens with and Both of those should look quite familiar.

2.

3. Originally Posted by river_rat
As a neat side note, looking at the metric shows that something interesting happens with and Both of those should look quite familiar.
Of course river_rat...the first one is a physical singularity, the second one a coordinate singularity, which, as it so happens, is also the Schwarzschild Radius.
One can distinguish between the two types of singularities by way of curvature invariants, like the Kretschmann invariant.

This brings a question to mind, and perhaps you can help with this river_rat : do you have a simple way to explain the Hodge Dual operator ?? Never managed to get my head around that one...

4. On a side note; use maple, or mathematica. Nobody wants to start with a metric, and calculate it's Einstein tensor starting of from scratch by hand. Trust me, you want to make a code!

5. Originally Posted by Kerling
On a side note; use maple, or mathematica. Nobody wants to start with a metric, and calculate it's Einstein tensor starting of from scratch by hand. Trust me, you want to make a code!
Ha ha, I hear you my friend, and wholeheartedly agree
On this occasion however my intention was to show interested readers how it is actually done, so I went through the trouble of doing it all manually ( with the help of the textbook referenced in the PDF ). In a real-life application I'd go with MAPLE all the way

6. Originally Posted by Markus Hanke
Originally Posted by Kerling
On a side note; use maple, or mathematica. Nobody wants to start with a metric, and calculate it's Einstein tensor starting of from scratch by hand. Trust me, you want to make a code!
Ha ha, I hear you my friend, and wholeheartedly agree
On this occasion however my intention was to show interested readers how it is actually done, so I went through the trouble of doing it all manually ( with the help of the textbook referenced in the PDF ). In a real-life application I'd go with MAPLE all the way
I remember that 5 years ago my professor told us to derive the FLRW metric. We could use the maple code, or do it by hand. He didn't care.
It was 120 pages
output...

7. Originally Posted by Kerling
I remember that 5 years ago my professor told us to derive the FLRW metric. We could use the maple code, or do it by hand. He didn't care.
It was 120 pages
output...
Cool...did you print them out and hand it in ?
The FLRW is quite tedious to calculate, and I really don't know if I could do it without help. Having vacuum equations is hard enough, even without making them inhomogeneous.

8. Originally Posted by Markus Hanke
Cool...did you print them out and hand it in ?
The FLRW is quite tedious to calculate, and I really don't know if I could do it without help. Having vacuum equations is hard enough, even without making them inhomogeneous.
No, I mailed him the pdf. I was appearently the only one that bothered to change the code to change the integratand with the appropiate constant of c^2 though. But I love Maple, I just don't like it that it uses 3 Gb of RAM

9. Originally Posted by Kerling
But I love Maple, I just don't like it that it uses 3 Gb of RAM
It's resource intensive alright. I am lucky that I run it on a fairly modern iMac, so it's not an issue for me - it's lightning fast, and appears to be much more stable than under Windows also.

10. Well, our IT department discourages apple usage. As they all break in their first year and have high maintenance costs. We use Linux. Besides we have some sort of super computer with 50 processors and 394 gb of RAM. Luckily my work is analytical.

11. Originally Posted by Markus Hanke

...

The non-vanishing elements of the Ricci tensor are thus

Would it be possible to work out the non-vanishing elements for the full metric

in that same manner while also including C(r) for an arbitrary coordinate system? No solutions are needed, no coordinate choices made, just those same equations including C(r) if possible.

12. for those such as myself who have yet to officially study these equations can someone please explain how ricci curvature tensor relates to field equations, i have an inkling however not sure on how they connect, mathematically

13. Originally Posted by grav-universe
Would it be possible to work out the non-vanishing elements for the full metric

in that same manner while also including C(r) for an arbitrary coordinate system? No solutions are needed, no coordinate choices made, just those same equations including C(r) if possible.
Yes, this would be a more general ansatz for a solution to the field equations, but definitely permissible. Physically, if C(r) differs from 1, such a metric would describe a gravitational field which is not spherically symmetric; for C(r)=1 this then reduces to the normal Schwarzschild metric as I have derived it in my calculation. You could of course attempt to solve the field equations for this ansatz, but the issue is that you get more non-vanishing elements of the Ricci tensor, and that those elements will be much more complicated algebraically. You would also need an additional set of boundary conditions to compensate for the emergence of additional integration constants, and also your A(r) and B(r) would likely take on a different form in the final solution.
I haven't done any maths, but I can already tell you that this would be very complicated and tedious to do. The general form of the above metric is roughly reminiscent of vacuum solutions to the field equations around bodies with angular momentum and electric charge, e.g. the Kerr-Newman metric.

14. Originally Posted by jshelley
for those such as myself who have yet to officially study these equations can someone please explain how ricci curvature tensor relates to field equations, i have an inkling however not sure on how they connect, mathematically
If you look in my initial post #1, equations (1)-(4) you can see how the Ricci tensor relates to the stress-energy tensor. In fact, the Einstein field equations are in essence that very relation between curvature and energy.

15. As has been shown, solving these equations is conceptually straightforward as one only needs to follow the prescribed steps; however, the actual algebra and analysis is tedious and time consuming, and increasingly so once one abandons some of the presumed symmetries, thereby complicating the ansatz.
The next step up from here would be to perform the same calculation for non-vanishing stress-energy-momentum tensors, i.e. for the interior of a mass. Finally, one can allow a cosmological constant into the field equations. Each of these steps will rapidly complicate the resulting differential equations, which is why in many cases the solutions are possible only numerically, but not in closed algebraic form.
Yipes! I strongly disagree. It's quite well known that Einstein’s equations are very very hard to solve. What can be misleading is to know of a particular solution and how its solved and then believe that these equations aren't that hard. The problem with that is that following such a derivation can be relativily easy but doing it from scratch when nobody has ever done it before can be a nightmare. But when someone shows you how it was done it only appears tractable. But that’s only because hindsight is 20-20. But Einstein's field equations are a set of 10 simultaneous non-linear equations and non-linear diff-eqs are inherently difficult to solve.

As one scientist states - Solutions of the Einstein Equations
The quantitative description of the Universe by General Relativity requires obtaining solutions to what are termed the Einstein field equations. These are 10 equations that must be solved simultaneously; they are notoriously difficult, and only a few solutions are known. (Technically, the equations to be solved are known to mathematicians as coupled, non-linear, partial differential equations; we may take that as precise shorthand for "very difficult to solve"!)
For example; you wrote The elements of the Christoffel symbols which do not vanish are.. its misleading to show that without knowing why. For example, how do you know the others vanish? Sometimes you can see by looking why. Othertimes to have to calulate them. There are 41 independant Christioffel symbols to find and it can be a pain going through each derivation.

16. There are other ways to approach this situation too. One isn't always given the form of the metric and then fill in the blanks. Sometimes you're given a distribution of matter doing it' thing, like a long rod which is moving in the direction of its axis. So you take the tensor for the matter and then use the linear appoximation to derive the metric. I worked this one out at
Gravitational Field of a Long Moving Rod

17. Originally Posted by pmb
Yipes! I strongly disagree. It's quite well known that Einstein’s equations are very very hard to solve.
Yes, I totally agree with that, there is absolutely no question whatsoever that these are very hard. It wasn't my intention to state otherwise, so my apologies if I came across that way. I was merely trying to say that, given an ansatz to a metric, the overall steps which lead to a possible solution are conceptually straightforward :

1. Write down the equations
2. Calculate the Christoffel symbols
3. Calculate the Ricci tensor
4. Calculate the elements of the stress-energy tensor ( if applicable )
5. Solve the resulting set of differential equations

The devil is then of course in the detail, because for anything even slightly less symmetric then a Schwarzschild space-time, the resulting set of equations becomes extremely difficult to handle. So I agree with you, pmb !

For example, how do you know the others vanish?
By calculating all of them. This is something I would use a CAM package for, like Maple.

There are 41 independant Christioffel symbols to find and it can be a pain going through each derivation.
Agreed, see above. I would utilize computer power to calculate all of them, and see which ones are zero. There are probably other ways to do that for someone who has much deeper insight into differential geometry than I do - I bet just by looking at the symmetries of the metric ansatz one could conceivable tell which elements must vanish, but I don't know how to do that.

So you take the tensor for the matter and then use the linear appoximation to derive the metric.
Yes, I have seen this done for the basic gravitational wave solution to the field equations.

18. Originally Posted by Markus Hanke
Yes, I totally agree with that, there is absolutely no question whatsoever that these are very hard. It wasn't my intention to state otherwise, so my apologies if I came across that way. I was merely trying to say that, given an ansatz to a metric, the overall steps which lead to a possible solution are conceptually straightforward :
Ah! Then I'm glad I mentioned it.

As one becomes used to doing this stuff it becomes easier to do.

19. I always wanted to see the proof of

20. Originally Posted by Wise Man
I always wanted to see the proof of

Now calculate the norm ( length ) of the vector :

...and rearrange this to isolate E...

And you're done. The above is the generalized energy-momentum relation; for particles at rest ( p=0 ) this gives E=mc2​, as expected, and for massless particles ( m=0 ) you get E=pc, as is the case for photons.

21. Originally Posted by Markus Hanke
Originally Posted by Wise Man
I always wanted to see the proof of

Now calculate the norm ( length ) of the vector :

...and rearrange this to isolate E...

And you're done. The above is the generalized energy-momentum relation; for particles at rest ( p=0 ) this gives E=mc2​, as expected, and for massless particles ( m=0 ) you get E=pc, as is the case for photons.
This really is't a derivation because you had to start with the mass-energy relation to know that P = (E/c,p) is a 4-vector.You also had to know the mass-energy relation to know what the magnitude of P is.

I've written down two of Einstein's mass-energy relation and placed it on my website at
Mass Energy Equivalence
Einstein

The last one is my modified version of Einstein's first derivation.

22. Originally Posted by pmb
This really is't a derivation
Maybe not, but it is a fast way to show why the relation looks as it does

23. Originally Posted by Markus Hanke

...and rearrange this to isolate E...

.
Is the minus not supposed to be a plus, or am I thinking of something else?

24. Originally Posted by epidecus
Originally Posted by Markus Hanke

...and rearrange this to isolate E...

.
Is the minus not supposed to be a plus, or am I thinking of something else?
Yes. You're correct. Good catch.

25. Originally Posted by Wise Man
I always wanted to see the proof of
Wise Man, the "M" and "C" should be lowercase. Apparently it's quite crucial to the meaning of the equation

Originally Posted by pmb
Is the minus not supposed to be a plus, or am I thinking of something else?
Yes. You're correct. Good catch.
Thanks! I was going post the following, but the typo made me hesitate. Still a good post though Markus; I'm always learning (despite my vast ignorance of physics notation and mathematics).

So can be rewritten as , which indicates a Pythagorean relationship. Pretty cool I think, since the equivalence can be visualized as (the always-intuitive) right triangle.

26. Originally Posted by epidecus
Is the minus not supposed to be a plus, or am I thinking of something else?
Gosh, you are right. Bad mistake on my part. Thanks for pointing that out, and my apologies

27. Hmm... I'm going to take a look at pmb's proof, it seems more promising.

28. Well, before you do that - and I haven't - you might want to look at Einstein's own argument, Which I paraphrase below. But note that, at least in my opinion, Einstein's brilliance is to bolster his lack of mathematical rigour with his physical intuition......

So here's Einstein (hugely paraphrased):

Consider a material body B with energy content . Let B emit a plane wave of light for some fixed period of time . One easily sees that the energy content of B is reduced by , which depends only on .

Let i.e.the light energy "withdrawn" from B.

Now, says Einstein, consider the situation from the perspective some body B' moving uniformly at velocity with respect to B. Then, evidently, by Lorentz time dilation in SR, depends only on , which is .

Therefore, the difference between and is simply . By expanding as a Taylor series, and dropping terms of order higher than 2 in , he finds that

.

With a flourishing hand-wave Einstein now says something like this: the above is an equation for the differential energy of bodies in relative motion; but then so is , the equation for kinetic energy - and these can only differ by an irrelevant additive constant, so set

and then .

But, says he, is simply a "quantity" of energy, light in this case, that now depends only on and so......

.

29. Originally Posted by Markus Hanke
The solution is called the Schwarzschild Metric.
Hi Markus,, this is a great thread !, I hope you can answer a few questions:

everybody says that Schwarzschild provided the exact solution, did Einstein only frame the equations? did he not solve them?

Could you explain the meaning of the animation at Wiki's Gravitational wave - Wikipedia, the free encyclopedia, is it real or just a guess?
if it is real, does a particle oscillate as the animation shows, or is it pushed inside only during the contraction, or what?

30. Originally Posted by Guitarist
but then so is , the equation for kinetic energy - and these can only differ by an irrelevant additive constant, so set
and then .
But, says he, is simply a "quantity" of energy, light in this case, that now depends only on and so.......
Although irrelevant, why drop the constant?

31. Originally Posted by logic
everybody says that Schwarzschild provided the exact solution, did Einstein only frame the equations?
Einstein came up with the equation itself, and Schwarzschild provided a ( not "the" ) solution. The equations have very many solution for different physical scenarios.

did he not solve them?
This is a good question. From what I know Einstein himself obtained only an approximate solution from which he derived the light ray deflection near the sun, and the Mercury perihelion precession. Schwarzschild was the first to obtain an exact solution to the equations.

Could you explain the meaning of the animation at Wiki's Gravitational wave - Wikipedia, the free encyclopedia, is it real or just a guess?
Which one are you referring to ? There are several animations there.
As to whether gravitational waves are real or not - they are an integral part of General Relativity, and there is plenty of indirect observational evidence for their existence. However, to date no such waves have been directly detected, which is because such detections are extremely difficult to perform.

if it is real, does a particle oscillate as the animation shows, or is it pushed inside only during the contraction, or what?
Yes, a system of particles, when impacted by a gravitational wave, would oscillate exactly as shown in the animation.

32. Originally Posted by Markus Hanke
Which one are you referring to ?
As to whether gravitational waves are real or not - they are an integral part
Yes, a system of particles, when impacted by a gravitational wave, would oscillate exactly as shown in the animation.
Thanks, Markus, I suppose you find the animation at "effects of..."
No I am not asking if the waves are "real" (I just read some posts in the other thread), if they are real or not, what counts are its effects.
I am asking if a ball in the air oscillates like a cork on a pond, like the animation suggests, or if it is pushed down a little bit by each "wave", else I do not see how it falls.

As to the solutions, I know there are a few, but they all seem to refer to the pressure a star or a body will stand.
If I got it right,Schwarzchild solution tells when a sphere of MATTER (star of neutrons or else) will collapse into a BH (i.e. when the PE is rougly 10^20)

but how do the equations apply to the whole universe, where there is no matter at the center of the universe , which they say is expanding like a ballon, and gravity runs on the curved space of its surface?

33. Originally Posted by logic
I am asking if a ball in the air oscillates like a cork on a pond, like the animation suggests, or if it is pushed down a little bit by each "wave", else I do not see how it falls.
I am not really sure if I understand your question correctly.
There would be no net change in relative position of such a ball, what would happen is that the ball's measurements would oscillate as shown in the animation.

As to the solutions, I know there are a few, but they all seem to refer to the pressure a star or a body will stand.
There are actually infinitely many solutions, but only a few are exact in the sense that they can be expressed in terms of analytical functions. Most solutions can only be done numerically and/or via approximation methods.

If I got it right,Schwarzchild solution tells when a sphere of MATTER (star of neutrons or else) will collapse into a BH (i.e. when the PE is rougly 10^20)
In general terms the exterior Schwarzschild solution as derived on this thread tells us the metric of space-time outside a static, spherical mass distribution without charge or angular momentum. It is a vacuum solution.
There is also an interior Schwarzschild solution, which would give the metric of space-time inside a mass distribution such as a star.
Both of these also tell us what happens when mass is contained in a volume of radius equal to or smaller than the Schwarzschild radius ( a black hole ).

but how do the equations apply to the whole universe,
One can make an ansatz for the universe as a whole, these are called cosmological solutions to the field equations. In that case it is necessary to use the complete, generalised Einstein field equations with cosmological constant, which are

One example of a cosmological solution would be the Friedman-Lemaitre-Robertson-Walker ( FLRW ) metric.

where there is no matter at the center of the universe
The universe does not have a centre.

34. Originally Posted by Markus Hanke
, what would happen is that the ball's measurements would oscillate as shown in the animation.
....
2) The universe does not have a centre.
When I asked it the animation is real I meant if it has/can be observed
By measurements you mean the real measures or the atoms?what if the object is brittle?
What makes the ball fall? the curvature they show in pictures shows space i bent in the opposite direction

2) So all the analogies with a balloon they make in cosmology are fake? then the analogy between the Universe and a star/ black hole does not apply. How can the same metric apply to a sphere with matter inside an a univerce with no matter at the center?

35. Originally Posted by logic
When I asked it the animation is real I meant if it has/can be observed
Yes, it could be observed, though the effect is extremely small and thus very hard to detect.

By measurements you mean the real measures or the atoms?what if the object is brittle?
I mean the measured dimensions of the object itself, so for example the radius of your ball along one direction, or the spacing between the point masses in the animation.
I have never really thought about it, but I suppose if you had a gravitational wave with a very large amplitude it is conceivably possible for an object's structural integrity to be compromised.

What makes the ball fall? the curvature they show in pictures shows space i bent in the opposite direction
The ball follows the curvature of space-time; since the curvature is oriented towards the centre of mass, that is where the ball will tend to move towards.

So all the analogies with a balloon they make in cosmology are fake?
No, they are pretty good analogies. Bear in mind that the analogy only extends to the surface of the balloon, not its interior; once this is understood it is immediately obvious that such a surface does not have a centre point.

then the analogy between the Universe and a star/ black hole does not apply
There is no such analogy that I am aware of.

How can the same metric apply to a sphere with matter inside an a univerce with no matter at the center?
It is not the same metric. The former is described by the Schwarzschild metric, whereas the latter is described by the Robertson-Walker metric. Most cosmological solutions are based on a (3+1)-dimensional manifold which contains a matter distribution, like for example dust. That is not the same as a star.

36. Hi logic. You have asked some interesting questions.

Regarding this one:

Originally Posted by logic
No I am not asking if the waves are "real" (I just read some posts in the other thread), if they are real or not, what counts are its effects.
I am asking if a ball in the air oscillates like a cork on a pond, like the animation suggests, or if it is pushed down a little bit by each "wave", else I do not see how it falls.
Are you thinking that the waves are responsible for gravitational attraction? If so, that is not the case. A planet, for example, has a static gravitational field that causes things to fall towards it. Gravitational waves are "disturbances in the force" caused by dynamic systems like two orbiting neutrons stars, etc. There would still be an overall attractive force but with "ripples" superimposed on it.

37. Originally Posted by logic
2) So all the analogies with a balloon they make in cosmology are fake?
Just to add to Markus's explanation, it is worth noting that all analogies are, well, analogies. They are approximations; just "stories" to try and help understand the more complex reality. They are always limited in their scope and, in general, cannot be used to derive further results.

It is worth noting that pretty much everything you read in popular science books and articles is an analogy. That includes descriptions of big bang cosmology and the nature of black holes.

Edit: http://xkcd.com/895/

38. Show me just one prediction that sgr makes in the universe, just a single one!

39. Originally Posted by TheVoiceInside
Show me just one prediction that sgr makes in the universe, just a single one!
What is "sgr"?

Tests of general relativity - Wikipedia, the free encyclopedia

40. god save him! special general relativity ! sgr!

41. Originally Posted by TheVoiceInside
god save him! special general relativity ! sgr!
Which is an abbreviation that is not used outside of your posts. You may verify that this is the case by launching a google search with search terms "sgr relativity -a*" (the latter exclusion eliminates many spurious hits).

The absence of hits is not surprising: There is no such thing as "special general relativity." There's general relativity (abbreviated GR) and special relativity (SR). I've never encountered "sgr" in any textbook or paper. You're making this up, showing once again your lack of knowledge of the subject.

42. Showing the lack of illusion of knowledge, the lack of delusion. INDEED!!!

43. Originally Posted by TheVoiceInside
Showing the lack of illusion of knowledge, the lack of delusion. INDEED!!!
I go with data. The data says that you don't know what you're talking about. People who actually use SR and GR in their daily work do not ever use "SGR" as an abbreviation. Your use of it, and your prickly response to being questioned about it, simply reveals you to be an empty poser.

44. It truly is amazing that Einstein was a drop out and his theory of relativity was incomplete. Physicists today find themselves trying to perfect it and prove it but they only scientist who has been remotely close was Hawking.

45. Originally Posted by sciencematters28
Physicists today find themselves trying to perfect it and prove it but they only scientist who has been remotely close was Hawking.
Tests of general relativity - Wikipedia, the free encyclopedia
Hawking is good at what he does, but most certainly nowhere alone in working with S.R. and G.R.

46. Wasn't it Hawking who made Einstein's theory more relevant.

47. Nor was Einstein a dropout. SR and GR are complete within their realm of applicability.

48. Originally Posted by sciencematters28
Wasn't it Hawking who made Einstein's theory more relevant.
Hawking can be credited for popularizing certain aspects of theoretical physics. Although he works with Relativity in his own work, he's not expanded on Relativity. He's offered insight into Black Holes and Q.M. And he is one of very, very many that understand and utilize relativity.

49. Originally Posted by Neverfly
Hawking is good at what he does, but most certainly nowhere alone in working with S.R. and G.R.
Agreed.
Hawking's reputation is largely built on his popular science books.
I.e. because he's had a book that is "recognised" by the public then the public recognise his name.
And go on to make the assumption that because they know his name and not those of others in the field then he must be the "top guy".
AFAIK Hawking is regarded as "good, but certainly not the best, working today" by other physicists.

Not knocking the guy, just pointing out a "common fallacy" - as apparently promoted by sciencematters28.

50. Hi,

I'm trying to follow your derivation of the solution. There seems to be a problem with your Cristoffel symbols..

[QUOTE=Markus Hanke;347669]

Calculating the Christoffel Symbols

The elements of the Christoffel symbols which do not vanish are

First of all, you have both and ...? I get the latter when calculating it.. Second of all, some of the ones you've written do in fact vanish (i think...)..

I don't know if this is affecting the solution, but I can't seem to reproduce your results since I get different elements of the Ricci-tensor..

51. Originally Posted by tp16
First of all, you have both and .....
You are right, this is a typo in the indices. Well spotted, and thanks for pointing it out.

Second of all, some of the ones you've written do in fact vanish (i think...)..
I am actually pretty sure that my calculations are correct ( simply because the end result comes out right ), but I will run it through MAPLE when I get home today, just to be 100% sure. It's easy to make mistakes with this stuff. Which specific ones do you think should vanish ?

52. I think the only difference is minor typos. I get the same as you except for

However, I still can't seem to get the same equations of the Ricci tensor.. But it might just be my Maple implementation that messes with me. I would very much like your calculations to be correct as the equations you end up with are way nicer than mine!

53. I double checked my Maple code and it seems that my implementation of the sums in the Ricci tensor was wrong since I now get the same equation as you for when doing it by hand(!) So it looks like I'll have to brush up on my Maple-skills or spend the rest of the day in hell calculating tensors manually

54. I have two questions:

1. I presume that in your equation R represents the trace of , correct?
2. I never quite understood why making produces the "vacuum solution". The gravitating mass is always present, so . So, the condition doesn't seem to be physical. What am I missing?

55. What am I missing?
A year.

56. Originally Posted by AlexG
What am I missing?
A year.

57. Originally Posted by xyzt
1. I presume that in your equation R represents the trace of , correct?
Correct. R is called the Ricci scalar, and is the contraction of the Ricci tensor across both indices.

2. I never quite understood why making produces the "vacuum solution".
Tensors are purely local objects, i.e. they are geometric objects evaluated at a specific point. The is in actual fact a tensor field, i.e. a field that assigns an energy-momentum tensor to each point in space-time. For those points which are in a vacuum exterior to your mass, the tensor must therefore vanish; if it doesn't vanish, the point in question is in the interior of a mass/energy distribution.

I should also note that the parameter M that appears in the Schwarzschild metric technically does not represent the central mass, but the total mass content of the entire space-time. The distinction is irrelevant in the case of Schwarzschild, but becomes important for other solutions.

The gravitating mass is always present, so .
You are right, it is always present, but see above - tensors are local. Hence we must choose if we want to consider the interior of the mass, or the exterior vacuum, i.e. we must choose where we want to evaluate the field equations. The Einstein equations will yield different solutions for both cases ( though the exterior vacuum solution is just a special case of the full interior metric with vanishing mass density ).

What am I missing?
Only that tensors are purely local objects. Once you understand that, it should become clear. Note that even though we just call the various objects in the field equations "tensors", they are in fact tensor fields - we just don't normally write , as a way to shorten the notation.

58. Originally Posted by Markus Hanke
Originally Posted by xyzt
1. I presume that in your equation R represents the trace of , correct?
Correct. R is called the Ricci scalar, and is the contraction of the Ricci tensor across both indices.

2. I never quite understood why making produces the "vacuum solution".
Tensors are purely local objects, i.e. they are geometric objects evaluated at a specific point. The is in actual fact a tensor field, i.e. a field that assigns an energy-momentum tensor to each point in space-time. For those points which are in a vacuum exterior to your mass, the tensor must therefore vanish; if it doesn't vanish, the point in question is in the interior of a mass/energy distribution.

I should also note that the parameter M that appears in the Schwarzschild metric technically does not represent the central mass, but the total mass content of the entire space-time. The distinction is irrelevant in the case of Schwarzschild, but becomes important for other solutions.

The gravitating mass is always present, so .
You are right, it is always present, but see above - tensors are local. Hence we must choose if we want to consider the interior of the mass, or the exterior vacuum, i.e. we must choose where we want to evaluate the field equations. The Einstein equations will yield different solutions for both cases ( though the exterior vacuum solution is just a special case of the full interior metric with vanishing mass density ).

What am I missing?
Only that tensors are purely local objects. Once you understand that, it should become clear. Note that even though we just call the various objects in the field equations "tensors", they are in fact tensor fields - we just don't normally write , as a way to shorten the notation.
Thank you, got it. This explains why , in the case of the Schwarzschild interior solution, .
I still can't get "Like" to work, what is going on?

59. Originally Posted by xyzt
Thank you, got it. This explains why , in the case of the Schwarzschild interior solution, .
Exactly

I still can't get "Like" to work, what is going on?

60. The above explains the difficulty of solving the EFE's. You are faced with finding out both the components of the Ricci tensor () AND the ones of the metric (). So, you start with an educated guess ("ansatz") of the metric in parametric format (the "solution") and you plug it into the equation and you nail down the different parameters. This gives you the solution.
It goes : "guessed" metric->Christoffel symbols->Rieman tensor->Ricci tensor->Ricci curvature->plug back into the EFE's and verify that the solution is good.

61. Originally Posted by xyzt
You are faced with finding out both the components of the Ricci tensor () AND the ones of the metric ().
Well, the Ricci tensor is just a function of the metric and its derivatives, but yes - since it is the Ricci tensor that appears in the field equations, you have little option but to calculate its components first.

It goes : "guessed" metric->Christoffel symbols->Rieman tensor->Ricci tensor->Ricci curvature->plug back into the EFE's and verify that the solution is good.
You don't need the Riemann tensor if you're doing it this way - there is a general formula to calculate the Ricci tensor directly from the metric.
There is another way to go about this - instead of the Ricci tensor you can calculate the Riemann tensor first; there is then a way to directly obtain the Einstein tensor ( turns out this is just the contracted double dual of the Riemann tensor ) from this, which you can then plug into the field equations. It's a little more work, but can be easier in some circumstances.

62. Originally Posted by Markus Hanke
Originally Posted by xyzt
You are faced with finding out both the components of the Ricci tensor () AND the ones of the metric ().
Well, the Ricci tensor is just a function of the metric and its derivatives, but yes - since it is the Ricci tensor that appears in the field equations, you have little option but to calculate its components first.

It goes : "guessed" metric->Christoffel symbols->Rieman tensor->Ricci tensor->Ricci curvature->plug back into the EFE's and verify that the solution is good.
You don't need the Riemann tensor if you're doing it this way - there is a general formula to calculate the Ricci tensor directly from the metric.
There is another way to go about this - instead of the Ricci tensor you can calculate the Riemann tensor first; there is then a way to directly obtain the Einstein tensor ( turns out this is just the contracted double dual of the Riemann tensor ) from this, which you can then plug into the field equations. It's a little more work, but can be easier in some circumstances.
Thank you.

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