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

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...
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 reallife application I'd go with MAPLE all the way
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.
Would it be possible to work out the nonvanishing 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.
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
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 nonvanishing 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 KerrNewman metric.
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 2020. But Einstein's field equations are a set of 10 simultaneous nonlinear equations and nonlinear diffeqs are inherently difficult to solve.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 nonvanishing stressenergymomentum 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.
As one scientist states  Solutions of the Einstein Equations
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.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, nonlinear, partial differential equations; we may take that as precise shorthand for "very difficult to solve"!)
Last edited by pmb; January 2nd, 2013 at 01:05 AM.
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
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 stressenergy 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 spacetime, the resulting set of equations becomes extremely difficult to handle. So I agree with you, pmb !
By calculating all of them. This is something I would use a CAM package for, like Maple.For example, how do you know the others vanish?
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.There are 41 independant Christioffel symbols to find and it can be a pain going through each derivation.
Yes, I have seen this done for the basic gravitational wave solution to the field equations.So you take the tensor for the matter and then use the linear appoximation to derive the metric.
I always wanted to see the proof of
Start with the momentum 4vector in Minkowski space :
Now calculate the norm ( length ) of the vector :
...and rearrange this to isolate E...
And you're done. The above is the generalized energymomentum relation; for particles at rest ( p=0 ) this gives E=mc^{2}, 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 massenergy relation to know that P = (E/c,p) is a 4vector.You also had to know the massenergy relation to know what the magnitude of P is.
I've written down two of Einstein's massenergy relation and placed it on my website at
Mass Energy Equivalence
Einstein
The last one is my modified version of Einstein's first derivation.
Wise Man, the "M" and "C" should be lowercase. Apparently it's quite crucial to the meaning of the equation
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).Originally Posted by pmb
So can be rewritten as , which indicates a Pythagorean relationship. Pretty cool I think, since the equivalence can be visualized as (the alwaysintuitive) right triangle.
Hmm... I'm going to take a look at pmb's proof, it seems more promising.
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 handwave 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......
.
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?
Einstein came up with the equation itself, and Schwarzschild provided a ( not "the" ) solution. The equations have very many solution for different physical scenarios.
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.did he not solve them?
Which one are you referring to ? There are several animations there.Could you explain the meaning of the animation at Wiki's Gravitational wave  Wikipedia, the free encyclopedia, is it real or just a guess?
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.
Yes, a system of particles, when impacted by a gravitational wave, would oscillate exactly as shown in the animation.if it is real, does a particle oscillate as the animation shows, or is it pushed inside only during the contraction, or what?
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?
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.
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.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.
In general terms the exterior Schwarzschild solution as derived on this thread tells us the metric of spacetime outside a static, spherical mass distribution without charge or angular momentum. It is a vacuum solution.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)
There is also an interior Schwarzschild solution, which would give the metric of spacetime 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 ).
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 arebut how do the equations apply to the whole universe,
One example of a cosmological solution would be the FriedmanLemaitreRobertsonWalker ( FLRW ) metric.
The universe does not have a centre.where there is no matter at the center of the universe
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?
Yes, it could be observed, though the effect is extremely small and thus very hard to detect.
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.By measurements you mean the real measures or the atoms?what if the object is brittle?
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.
The ball follows the curvature of spacetime; since the curvature is oriented towards the centre of mass, that is where the ball will tend to move towards.What makes the ball fall? the curvature they show in pictures shows space i bent in the opposite direction
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.So all the analogies with a balloon they make in cosmology are fake?
There is no such analogy that I am aware of.then the analogy between the Universe and a star/ black hole does not apply
It is not the same metric. The former is described by the Schwarzschild metric, whereas the latter is described by the RobertsonWalker 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.How can the same metric apply to a sphere with matter inside an a univerce with no matter at the center?
Hi logic. You have asked some interesting questions.
Regarding this one:
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.
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/
Last edited by Strange; January 12th, 2013 at 08:33 AM.
Show me just one prediction that sgr makes in the universe, just a single one!
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.
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.
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.
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.
Wasn't it Hawking who made Einstein's theory more relevant.
Nor was Einstein a dropout. SR and GR are complete within their realm of applicability.
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.
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.
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 Riccitensor..
You are right, this is a typo in the indices. Well spotted, and thanks for pointing it out.
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 ?Second of all, some of the ones you've written do in fact vanish (i think...)..
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!
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 Mapleskills or spend the rest of the day in hell calculating tensors manually
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?
A year.What am I missing?
Correct. R is called the Ricci scalar, and is the contraction of the Ricci tensor across both indices.
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 energymomentum tensor to each point in spacetime. 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.2. I never quite understood why making produces the "vacuum solution".
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 spacetime. The distinction is irrelevant in the case of Schwarzschild, but becomes important for other solutions.
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 ).The gravitating mass is always present, so .
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.What am I missing?
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.
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.
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.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.
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.
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