General Relativity Homework #7 problems 2, 3 and 4

Using the GREAT package which can be found at:
http: //library.wolfram.com/infocenter/MathSource/4781/

Problem 2

First define an n-coordinate vector:

x={ψ,θ,ϕ};

And metric:

(metric={{1,0,0},{0,((Sin[ψ])^2),0},{0,0,((Sin[ψ])^2)((Sin[θ])^2)} })//MatrixForm

With these inputs, the GREAT package will calculate everything else.

(a)

The Christoffel connection coefficients are:

Christoffel[metric,x]

You should read this output as: {{{G111,G112,G113},{G121,G122,G123},{G131,G132,G133}} , {{G211,G212,G213},{G221,G222,G223},{G231,G232,G233}} , {{G311,G312,G313},{G321,G322,G323},{G331,G332,G333}} 
where these are Gijk where i is upper and  j,k are lower.  For example G133 = - Cos[ψ] ((Sin[θ])^2) Sin[ψ].  Also note that every component is displayed even though many of them are related through the symmetry under exchange of the last two indices, e.g. G212=G221.   This way of expressing the components can get cumbersome, but usually we don’t use these directly.

(b)

The Riemann tensor, Ricci tensor, and the Ricci scalar are:

Riemann[metric,x]

Again you read this output as {{{{R1111,R1112,R1113},{R1121,R1122,R1123},{R1131,R1132,R1133},etc.}}}   This can get really confusing, but again we usually don’t use these directly.   Again note that many components are related through the expected index symmetries if the Riemann tensor.  In this case the nonzero independent components are:
 R1211=((Sin[ψ])^2)
 R1221=-((Sin[ψ])^2)
 R1313=((Sin[θ])^2) ((Sin[ψ])^2)
 R2112= -1
 R2323=((Sin[θ])^2) ((Sin[ψ])^2)
 R3113=-1
 R3223=-((Sin[ψ])^2)
 Note that since the first index is upper, we do not have the explicit antisymmetry under exchange of the first two indices, e.g. R2323 is not equal to R3223.
 
 
The next command calculates the components of the Riemann tensor with all of the indices lowered.  This expression will be useful for seeing the full index symmetry of the Riemann tensor and of course will be needed for the last part of this  homework.  This is NOT part of the GREAT package, so I had to define the operation explicitly.

Riemannlower[metric_,x_]:=
  Block[ {Dim, Riem, Riemannlower, a, b, i, k, j},
          Dim = Length[x];
          Riem = Riemann[metric,x];
          Riemannlower = Table[ Sum[metric[[b,a]]*Riem[[a,i,j,k]], {a,Dim}],
                         {b,Dim},{i,Dim}, {j,Dim} , {k,Dim}];
          (* Return the Riemann tensor (four lower indices b,i,j,k): *)
          Simplify[Riemannlower] ]

Riemannlower[metric, x]

The nonzero terms are:
R1211=((Sin[ψ])^2)
 R1221=-((Sin[ψ])^2)
 R1313=((Sin[θ])^2) ((Sin[ψ])^2)
 R2112= -((Sin[ψ])^2)=R1221
 R2323=((Sin[θ])^2) ((Sin[ψ])^2)
 R3113=-((Sin[θ])^2) ((Sin[ψ])^2)=-R1313
 R3223=-((Sin[θ])^2) ((Sin[ψ])^4)=-R2323
We can now see the full index symmetry including the first index.

Ricci[metric,x]

These components can be usefully represented by a matric since this is a two-index tensor.  To do this we can use:

Ricci[metric,x]//MatrixForm

Note the expected index symmetry of the Ricci tensor.

SCurvature[metric,x]

Note that in this case the Ricci scalar is a constant, though this need not be true for a general space.

(c)

We want to show that:
(Subscript[R, ρσμν])=(R/(n(n-1)))((Subscript[g, ρμ])(Subscript[g, σν])-(Subscript[g, ρν])(Subscript[g, σμ]))

We have already evaluated the components on the LHS.  The RHS can be evaluated using:

RHS[metric_,x_]:=
  Block[{Dim, RHS, S, b, i, j, k},
         Dim = Length[x];
         S = SCurvature[metric,x];
         RHS =
           Table[ S*(metric[[b,j]]*metric[[i,k]]
                - metric[[b,k]]*metric[[i,j]])/(Dim*(Dim-1)),
          	   {b,Dim}, {i,Dim}, {j,Dim},{k,Dim} ];
          	 (* Calculates the RHS of the equation for arbitrary R and n with lower indices b,i,j,k *)
         Simplify[RHS] ]

RHS[metric,x]

To see that the LHS and RHS are equal, we can just subtract them.   We expect to get zeros everywhere.

Riemannlower[metric,x]-RHS[metric,x]

{{{{0,0,0},{0,0,0},{0,0,0}},{{0,0,0},{0,0,0},{0,0,0}},{{0,0,0},{0,0,0},{0,0,0}}},{{{0,0,0},{0,0,0},{0,0,0}},{{0,0,0},{0,0,0},{0,0,0}},{{0,0,0},{0,0,0},{0,0,0}}},{{{0,0,0},{0,0,0},{0,0,0}},{{0,0,0},{0,0,0},{0,0,0}},{{0,0,0},{0,0,0},{0,0,0}}}}

Since the difference is zero for each element, this proves the equality of the LHS and RHS of the expression.

Problem 3
(a) A doughnut in 3D

Clear[x,metric]

To determine if the doughnut's surface is curved, we want to compute the Riemann curvature tensor for the 2D surface of the doughnut.  In order to do this we first need to determine the metric of the doughnut's surface.

We start with the embedding map that expresses the 2D surface of the doughnut (with coordinates {u,v}) to the coordinates of the 3D embedding space {x,y,z}:
   x = (R1 + R2 Sin[v]) Cos[u]
   y = (R1 + R2 Sin[v]) Sin[u]
   z = R2 Cos[v]

Since the embedding space is flat, we know the line elements squared is given by: 
(ds^2)=(dx^2)+(dy^2)+(dz^2)

To determine the "induced" metric on the doughnut, we simply compute the coordinate differentials {dx,dy,dz} in terms of {u,v} using the embedding map, and then put these expressions into the line element above:

dx= -R1 Sin[u]du+R2(-Sin[v] Sin[u] du + Cos[u] Cos[v] dv);
dy=R1 Cos[u]du+R2(Sin[v] Cos[u] du + Sin[u] Cos[v] dv);
dz=-R2 Sin[v]dv ;

FullSimplify[ (dx^2)+(dy^2)+(dz^2)]

This expression gives the line element on the surface of the 2D doughnut.  We can use this to infer the matrix form of the induced metric on the doughnut.  Noting that it is still diagonal, we simply read off the coefficients of the squared terms in the line element.

(metric={{(R1^2)+((R2^2)/2)-(1/2)(R2^2) Cos[2 v]+2R1 R2 Sin [v],0},{0,(R2^2)} })//MatrixForm

Too proceed we define our coordinate vector and then use the GREAT package to calculate the Riemann tensor:

x={u,v};

Finally, we can find the Riemann curvature tensor :

Riemann[metric,x]

Since the Riemann tensor is not zero, we conclude that the surface of the 2D doughnut in 3D space is NOT flat.

(b) A torus in 4D

We can repeat the above analysis in 4D using the following embedding map : 
x=R1 Sin[u]
y=R1 Cos[u]
z=R2 Sin[v]
w=R2 Cos[v]

With this embedding, we can proceed to find the induced metric on the surface of the torus.

Clear[dx,dy,dz,dsds,metric]

dx=R1 Cos[u] du;
dy=-R1 Sin[u] du;
dz=R2 Cos[v] dv;
dw=-R2 Sin[v]dv;

dsds=Simplify[(dx^2)+(dy^2)+(dz^2)+(dw^2)]

(metric={{(R1^2),0},{0,(R2^2)} })//MatrixForm

From the form of this metric it should be clear that the surface in this case is actually flat.  Just because the diagonal elements are not 1 does not change this result.  What is important is that the metric is diagonal and constant.  Just to see that this is in fact the case though, we can go ahead and compute the Riemann curvature tensor for this case.

Riemann[metric,x]

As expected all components of the Riemann tensor vanish indicating that the 2D torus in 4D space is truly flat.

Problem 4

We have an 8D spacetime coordinatized by:

x={t,ψ,χ,δ,η,θ ,ϕ,β};

with metric:

(metric={{-((Cosh[ψ])^2),0,0,0,0,0,0,0},{0,1,0,0,0,0,0,0},{0,0,((Sinh[ψ])^2),0,0,0,0,0 },{0,0,0,((Sinh[ψ])^2)((Sin[χ])^2),0,0,0,0 },{0,0,0,0,1 ,0,0,0},{0,0,0,0,0,((Sin[η])^2),0,0 },{0,0,0,0,0,0,((Sin[η])^2)((Sin[θ])^2),0 },{0,0,0,0,0,0,0,((Sin[η])^2)((Sin[θ])^2)((Sin[ϕ])^2)}})//MatrixForm

To determine the energy-momentum tensor (Subscript[T, μν]) that generates this geometry, we can use Einstein’s equation (Subscript[G, μν])=(Subscript[R, μν])-1/2(Subscript[g, μν])R=8(Subscript[πGT, μν]) and evaluate the left hand side with the metric above and divide by 8πG:

1/{8πG}{EinsteinTensor[metric,x]//MatrixForm}

To determine the trace of the energy-momentum tensor we can just use the relation between the trace of (Subscript[R, μν]) and the trace of (Subscript[T, μν]) that can be obtained by tracing Einstein’s equation, i.e. in 8D T=-3R/8πG.  If we evaluate R:

SCurvature[metric,x]

Since R=0 then we know T=0 as well.