Einstein's Field Equation: 4D Spacetime & Cosmological Constant

[wam_mi]

I read the following on wiki

"The Einstein's field equation is a tensor equation relating a set of symmetric 4 x 4 tensors. It is written here using the abstract index notation. Each tensor has 10 independent components. Given the freedom of choice of the four spacetime coordinates, the independent equations reduce to 6 in number."

So is it true to say that the Einstein tensor is a symmetric 4 x 4 tensor which consists of 10 independent highly non-linear equations. Does this imply we are talking about a general 4 dimensions of space-time? But what I don't understand is the last bit. It says given the freedom of choice of the 4 space-time coordinates, the number of independent equations are now down to 6. Could someone explain this bit to me please? I just can't see how it works.

Finally, assuming that the cosmological constant in Einstein's equation is not zero, could one still replace the metric tensor by the flat metric to solve Einstein's equation?

Thanks

 

[Nabeshin]

So is it true to say that the Einstein tensor is a symmetric 4 x 4 tensor which consists of 10 independent highly non-linear equations.

The Einstein tensor is just a tensor. You need to form the Einstein equations from it to get the 10 non-linear equations.

It says given the freedom of choice of the 4 space-time coordinates, the number of independent equations are now down to 6. Could someone explain this bit to me please? I just can't see how it works.

So as you know, in GR any set of coordinates is just as good as any other. So in 4-d space time, there are four coordinates we can choose: the three spatial and the one time-like coordinates. Now, given that we have 10 equations, essentially we can eliminate four or them by making a clever choice of coordinates. Perhaps it would be good to consider a simpler example. Imagine I have a planet in an orbit around a star, such that the orbit is tilted with respect to the rotation of the star. see: http://www.aerospaceweb.org/question/astronomy/pluto/orbit.jpg for a good picture. Now, if I choose spherical coordinates with the star's rotation, I have to worry about all three degrees of freedom (the radial component and two angles). However, if I instead change coordinates to a system inclined into the plane of the planet's orbit, I now only worry about two (radial and one angle), since the object is now constrained to move in the plane. In essence, I have used my coordinate freedom to reduce the number of equations from 3 to 2.

The situation is analogous in the field equations, only it's much more difficult to think of good coordinate transformations.

Finally, assuming that the cosmological constant in Einstein's equation is not zero, could one still replace the metric tensor by the flat metric to solve Einstein's equation?

I'm not sure what you're getting at. Sure, why couldn't you have a flat universe with cosmological constant?

 

[Ich]

Sure, why couldn't you have a flat universe with cosmological constant?

"Flat universe" is not equivalent to "flat metric". The metric with cosmological constant is curved. See http://en.wikipedia.org/wiki/De_Sitter_space" .

 

[Altabeh]

I read the following on wiki

"The Einstein's field equation is a tensor equation relating a set of symmetric 4 x 4 tensors. It is written here using the abstract index notation. Each tensor has 10 independent components. Given the freedom of choice of the four spacetime coordinates, the independent equations reduce to 6 in number."

Actually this is not like that we just deal straightly with the four spacetime coordinates as being in charge of the reduction of number of the algebraically independent field equations. You have to be carefull here. It is the four equations of conservation laws of GR that do make the 10 independent equations reduce to 6 because this results in the appearece of metric components together in four equations; thus all metric components get involved together which does not bring any more independence than 6 equations to the set of 10 equations we already had.

Here since we can relate the components of the former metric to a new one, say, $$\bar{g}_{\mu\nu}$$ by four coordinates $$\bar{x}^{\alpha}$$ where latin indices run over 0 to 3, and the new metric is also a solution to the field equations as is the former, we can say that the new coordinates are free degrees of freedom of the 10 field equations because they can be chosen freely.

So is it true to say that the Einstein tensor is a symmetric 4 x 4 tensor which consists of 10 independent highly non-linear equations.

As Nabeshin said, this is just a tensor not a set of equations. Your argument is like we omit the right hand sides of a set of linear equations and say the reduced set represents again a set of linear equations while it never can. Einstein's field equations consist of two tensors and each by itself cannot show the equations at all. But yes these equations are highly nonlinear and they include in 10 independent equations if we for one second neglect the Bianchi identities.

Does this imply we are talking about a general 4 dimensions of space-time?

Yes, by looking at the number of independent components of the metric tensor, we can conclude that the number of dimensions of the spacetime is 4.

But what I don't understand is the last bit. It says given the freedom of choice of the 4 space-time coordinates, the number of independent equations are now down to 6. Could someone explain this bit to me please? I just can't see how it works.

The number of free values to be adjusted by hand in any set of equations (here the new four coordinates $$\bar{x}^{\alpha}$$ make the four relations between the new metric, involving linear or nonlinear functions, and the former metric arbitrary to be chosen. This process can be followed below:

According to the transformation law of metric tensor,

$$\bar{g}_{\mu\nu}=\frac{\partial x^{\alpha}}{\partial \bar{x}^{\mu}}\frac{\partial {x}^{\beta}}{\partial\bar{x}^{\nu}}g_{\alpha\beta}$$

if we have four free coordinates $$\bar{x}^{\alpha}$$ there can just be four equations involving 4 freely-chosen Jaccobi factors $$\frac{\partial x^{\alpha}}{\partial \bar{x}^{\mu}}$$ (Remember that the initial coordinates are not arbitrarly chosen), and thus there we have 10 components of $$\bar{g}_{\mu\nu}$$ to satisfy these equations including 4 freely-chosen components according to what conditions we like to impose on the set of the new coordinates. Hence we are left with the total of 10-4=6 independent components of the metric tensor and 4 out of 10 being arbitrary.

All we do to not hit this snag is to take into account the harmonic coordinates, as one alternative, to reduce the number of free values to zero by creating more independent equations.

Finally, assuming that the cosmological constant in Einstein's equation is not zero, could one still replace the metric tensor by the flat metric to solve Einstein's equation?

The existence of the cosmological constant in the field equations is by itself inconsistant with the metric being flat because the condition $$g_{\mu\nu}\rightarrow \eta_{\mu\nu}$$ so that $$T_{\mu\nu}\rightarrow 0$$ will no longer hold. This can also be understood from the property of the non-conservation of the matter tensor if the cosmological constant is neither constant nor covariantly constant.

AB

 

[wam_mi]

The existence of the cosmological constant in the field equations is by itself inconsistant with the metric being flat because the condition $$g_{\mu\nu}\rightarrow \eta_{\mu\nu}$$ so that $$T_{\mu\nu}\rightarrow 0$$ will no longer hold. This can also be understood from the property of the non-conservation of the matter tensor if the cosmological constant is neither constant nor covariantly constant.

AB[/QUOTE]

Sorry, I don't really understand what you mean by the inconsistancy between the existence of the cosmological constant and the flat metric. Why is it that the stress energy tensor is no longer equal to zero if the cosmological constant is neither constant nor covariantly constant?

Thanks

 

[haushofer]

Take the covariant derivative of the full Einstein equation with cosmological constant :) If Lambda is not constant, then the covariant derivative of the stress-tensor will equal the (covariant) derivative of Lambda.

 

[Nabeshin]

"Flat universe" is not equivalent to "flat metric". The metric with cosmological constant is curved. See http://en.wikipedia.org/wiki/De_Sitter_space" .

Ah right of course, thanks for the correction.