Is there a Space-Time Curvature equation?

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I've been wondering if there is a relationship between the mass of an object and how much it "curves" space-time. I can't seem to find an equation or connection, I have looked at four-momentum but am not certain what it actually calculates. Maybe I just don't know it and there is yet an equation for this.
 

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  • #3
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Alright, so what it is basically saying is that space-time curvature = the Ricci tensor - half the metric tensor * scalar curvature..the only problem I am having is that the Ricci tensor, metric tensor and scalar curvature all break down like the einstein tensor and those break downs break down even more and more to the point where it is so much work that it is almost not even worth the calculation. Do you know of a way of either simplifying this or maybe an alternate way? If it comes down to it I may just need to find a "Einstein tensor: for Dummies" book
 
  • #4
Ich
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In certain symmetric conditions, the field equations are a lot simpler. Check out http://math.ucr.edu/home/baez/einstein/einstein.html" [Broken].
 
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George Jones
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I've been wondering if there is a relationship between the mass of an object and how much it "curves" space-time. I can't seem to find an equation or connection, I have looked at four-momentum but am not certain what it actually calculates. Maybe I just don't know it and there is yet an equation for this.
The link that FunkyDwarf tell only half the story, the geometry and curvature part. The other half of the story is the energy/mass/momentum part, and this, too is described by a tensor, the energy-momentum (sometime called stress-energy tensor).

Einstein's equation of general relativity sets theses two tensors "equal", so that curvature and geometry equals distribution of energy/mass/momentum. Since this is an equation, it is impossible to change just one side: change the geometry side and this means that distribution of energy/mass/momentum must salso be changed; change the distribution of energy/mass/momentum, and this means geometry must also be changed.

This is the equation that Ich's link discusses; see also

http://en.wikipedia.org/wiki/Einstein_field_equations.
 
  • #6
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Ok, so I now have the equation, thank you guys very much for that by the way, but now can anyone tell me how to apply it? It would seem simple enough as it is looks to serve like a function almost, but for every constant there are numerous other constants involved. What I am asking here is if I MUST go down the chain of constants in order to arrive upon my final answer
 
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  • #7
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As Ich mentioned, in highly symmetric situations you can get much simpler solutions, but otherwise it usually requires numerical solutions. Also, the EFE is non-linear which adds all sorts of other problems.
 
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By "other problems" do you mean other equations or complications? My guess is probably both, and do you know if there is an example or if you can give me one of a highly symmetric situation? Thanks for the help
 
  • #9
Ich
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do you know if there is an example or if you can give me one of a highly symmetric situation?
I gave you a link. I don't think you'll find anything more appropriate for your purposes.
 
  • #10
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By "other problems" do you mean other equations or complications? My guess is probably both, and do you know if there is an example or if you can give me one of a highly symmetric situation? Thanks for the help
The main problem with non-linear equations is that you cannot just add up two solutions to get a third solution. For example, the Schwarzschild solution is a good description of the spacetime around the earth and it is also a good description of the spacetime around the moon, but if you want a the spacetime around the earth-moon system you cannot simply add the two Schwarzschild solutions to get the right answer.

For the most important examples of highly symmetric solutions look up: Schwarzschild metric, Kerr metric, and FLRW metric.
 
  • #11
Fredrik
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By "other problems" do you mean other equations or complications? My guess is probably both, and do you know if there is an example or if you can give me one of a highly symmetric situation? Thanks for the help
The equation is extremely difficult to solve. For example, the Kerr metric, which describes a universe that's completely empty except for a single rotating star or black hole with no electric charge, wasn't found until 1963. In other words, it took 48 years to find that solution, even though it describes one of those "highly symmetric situations".

The first solutions that were found are even simpler. The Schwarzschild metric was found in 1915 and describes a universe that's completely empty except for a single non-rotating star or black hole with no electric charge. That's probably the simplest solution, but it still isn't really easy to find. Another one of the simplest solutions was found in 1922. It's called the FLRW metric. It's the most general solution that's consistent with the assumption that spacetime can be "sliced" into a one-parameter family of spacelike hypersurfaces that are homogeneous and isotropic (in a certain technical sense). You can think of the parameter as "time" and the hypersurfaces as "space, at different times".

Some of the FLRW solutions are very accurate descriptions of the large-scale behavior of the universe. That last claim is often called "the big bang theory", because all the FLRW solutions have a property called an "initial singularity" or a "big bang".
 
  • #12
George Jones
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Just reinforcing what others have said ...

The Einstein equation is a single tensor equation that is equivalent to a set of ten coupled non-linear partial differential equations. Without symmetry, this is as bad (or worse!) than it sounds.
 
  • #14
If all you want is a means of solving the equations, then unless you have a nice computer package, GOOD LUCK!

However, for a good explanation of the theoretical underpinning, check out Leonard Susskind's lectures on GR at YouTube or iTunes
 
  • #15
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tgramling...If you are looking for a simple expression for the space curvature - take a look at Feynman in vol II
(42-6). The excess radius is MG/3c^2 ... so for example if you wanted to know how much the mass of the earth curves space you can make a quick calculation and see that its in the range of a one millimeter
 
  • #16
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Thanks guys for the help, sorry for such a late response, I've been at the lake and studying for finals the past week. yogi, I'm just going to ask the straight up question, what is the excess radius? I have tried to look into but just cant find out what it is! It would seem like there is a simple denotation or connotation for this.
 

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