Perturbational Curvilinear Dustortions

[Mystery111]

Einstein derived what is called the Einstein tensor which is:

 

[math]\nabla_{\mu} G^{\mu \nu}=0[/math]

 

and [math]G^{\mu \nu}[/math] is just the Ricci tensor minus [math]\frac{1}{2}g^{\mu \nu}R[/math]. These equations express the local continuity of energy and momentum.

 

As a side note, it could be imagined for a universe to be devoid of mass, so this indiates the right hand side of

 

[math]\nabla_{\mu} R^{\mu \nu} = \frac{1}{2 \nabla_{\mu}g^{\mu \nu} R[/math]

 

But does the absence of matter imply zero curvature for a metric? The answer is no. Gravitational waves are not trivial, where the Ricci tensor is zero everywhere, but the Reimann tensor is not.

 

 

Let us now concentrate on a specific wave equation, quite a famous one:

 

[math]\frac{\partial^2 \phi}{\partial t^2} = c^2\frac{\partial^2 \phi]{\partial x^2}[/math]

 

This describes two kind of waves, one wave which moves to the left, another to the right. From now on, we will use natural units. To express this equation in three dimensions

 

[math]\frac{\partial^2 \phi}{\partial t^2} = [\frac{\partial^2 \phi]{\partial x^2} + \frac{\partial^2 \phi]{\partial y^2} + \frac{\partial^2 \phi]{\partial x^2}[/math]

 

We can rewrite this as

 

[tex]\eta^{\mu \nu} \frac{\partial^2 \phi}{\partial X^{\mu} \partial X^{\nu}=0[/tex]

 

To make it into a tensorial equation, we can take [tex]\eta^{\mu \nu}[/tex] to be [tex]g^{\mu \nu}[/tex] thus we can state that [tex]g^{\mu \nu} \frac{\partial \phi}{\partial X^{\nu}[/tex] and differentiate as:

 

[math]\frac{\partial}{\partial x^{\mu} g^{\mu \nu} \frac{\partial \phi}{\partial X^{\mu}[/math]

 

We need to bring in the Covariant derivative, and make:

 

[math]\frac{\partial}{\partial x^{\mu} g^{\mu \nu} \frac{\partial \phi}{\partial X^{\nu}+ \Gamma_{\mu \alpha}^{\mu} g^{\nu \beta} \frac{\partial \phi}{\partial X^{\beta}}=0[/math]

 

[math] \nabla g^{\mu \nu} \frac{\partial \phi}{\partial x^{\nu}}[/math]

 

To work out the covariant derivative involves Christoffel Symbols. This is by definition, the wave equation in curved coordinates. The equations coeffients depend on position, and describes a wave in curvilinear coordinates.

 

Curvature varies throughout space, and understanding the relativity of photons, for instance, you disover that photons couple to spacetime - the intrinsic relationship between curvature, matter and energy are numerous.

 

Of course, you wind the clock back, then the curvature becomes stronger and stronger, it should get to a point where there is a perfect curvature, i.e (an object of infinite curvature, with infinite mass and energy). The parameters are defined in a negative spacetime region. Could it be possible, to have a lot of curvature which typically emanates mass from it's geometry?

 

The answer might be yes, because you can deal with

 

[math]\nabla_{\mu} R^{\mu \nu} = \frac{1}{2 \nabla_{\mu}g^{\mu \nu} R[/math]

 

Where you can remove all the mass from the universe, but still have a non-zero curvature.

 

 

Part Two

 

[math]R_{\mu \nu} - 1/2g_{\mu \nu}R + \Lambda g_{\mu \nu} = kT_{\mu \nu}[/math]

 

The Newtonian analogue is:

 

[math]\nabla^2 \phi + \Lambda = 4 \pi G \rho[/math]

 

where [math]\Lambda[math] is our energy density, the cosmological constant. I ran into a problem with this, because the 00-component tells me that the [math]g_{\mu \nu}[/math] part is close to the limits of special relativity, and as we know, special relativity is a flat spacetime. The idea that this describes our early universe very well seems bleak, but let's work with it for now.

 

Suppose this is taken as a uniform energy density, then we would have

 

[math]\nabla \phi + \Lambda = 0[/math]

[math]\nabla \phi = \Lambda[/math]

 

we must assume that we cannot deal with [tex]\phi[/tex] properly as we did if we have the metric of spacetime any significant time past the first instant of the universe. We would differentiate [tex]\phi[/tex] with the function [math]x^2 + y^2 + z^2[/math] but that is an solution for three dimensional spacetime - this kind of representation for [tex]\phi[/tex] must become obsolete. Of course, normally you do this by saying the del of our function is 6, since differentiating [math]x[/math] twice gives you [math]2[/math]], same with the [math]y[/math] and [math]z[/math] components; you simply end up with

 

[math]\phi = \frac{\Lambda}{6}[x^2+y^2+z^2][/math]

 

But of course, the spatial dimensions must vanish to leave a point on the metric - we would also find [math]\Lambda|_{\infty}[/math] since the singularity can permit an infinite volume of energy. This must mean that the curvature is also infinite.

 

Then again, if we wanted to draw out the use of this energy density when the universe becomes sufficiently large enough, then you simply differentiate the gravitational potential to find the force, and these equations let you retrieve the important terms involving the cosmological constant as a driving force for continued expansion.

 

Using the cartesian coordinates, we can understand for instance that the x-component of force is given as:

 

[math]\frac{\partial \phi}{\partial x} = \frac{\Lambda}{6}2x = \frac{\Lambda x}{3}[/math]

 

This tells us there is a component of force along the x-direction. This is just basic relativity, but in order to understand my theory, we need a new understanding of how to treat [math]\phi[/math] as a potential in zero dimensions, with an infinitely large [math]\Lambda[/math] which would be related to the identity of either the stress energy or the curvature; you can swap the constant term around to either the right or the left of Einsteins field equation describing how space tells matter how to move, and how matter tells space how to bend, by understanding that the constant can be seen in terms of either it's geometry or by the energy-distribution of spacetime - of course, I have surmized it is more rewarding to think of the geometry in this case, where it contributes to the potential of new particles which are created as you highly stress spacetim.

[imatfaal]

dftt!

 

http://www.sciforums.com/showpost.php?p=2800681&postcount=22

[Mystery111]

...messed the latex here.

Edited September 28, 2011 by Mystery111