Dr Phill’s Science Made Simple

Schwarzschild Outer Metric

In 1915, the same year as the publication of the Field Equations, the German physicist Karl Schwarzschild produced the first exact solution. He did this while in hospital having developed a fatal skin condition while serving in the German army on the Russian front in World War I.

The Schwarzschild Outer Metric is an exact solution to the Field Equations for the vacuum of space around a static and symmetric body. The metric must be independent of time and the spherical θ and ϕ coordinates.

Define the metric

The metric can be written as:

ds2=A(r)dt2B(r)dr2r2dθ2r2sin2θdϕ2

Where A and B are unknown functions of r. The solution must become the Minkowski metric as r and as M0.

The metric and its inverse are now defined to be:

gαβ=[A0000B0000r20000r2sin2θ] gαβ=[1A00001B00001r200001r2sin2θ]

Calculate the Christoffel Symbols

Using the comma notation, where a comma subscript means a partial derivative of that variable, the Christoffel symbols are:

Γμνλ=12gλρ(gμρ,ν+gνρ,μgμν,ρ)

There are four sets of Christoffel symbols, one set for each of t,r,θ,ϕ.

For the t component, λ=t, and for all non-zero symbols ρ=t.

Γμνt=12gtt(gμt,ν+gνt,μgμν,t)

The term gμν,t=0 as nothing is time dependent. Substitute the value of gtt, gives:

Γμνt=12A(gμt,ν+gνt,μ)

This leaves two symbols:

Γrtt=Γtrt=A2A

For the r component, λ=r, and for all non-zero symbols ρ=r.

Γμνr=12grr(gμr,ν+gνr,μgμν,r)

This gives four symbols:

Γttr=12grrgtt,r=A2BΓrrr=12grrgrr,r=B2BΓθθr=12grrgθθ,r=rBΓϕϕr=12grrgϕϕ,r=rsin2θB

For the θ component, λ=θ, and for all non-zero symbols ρ=θ.

Γμνθ=12gθθ(gμθ,ν+gνθ,μgμν,θ)

This gives three symbols:

Γθrθ=Γrθθ=12gθθgθθ,r=12r22r=1rΓϕϕθ=12gθθgϕϕ,θ=12r22r2sinθcosθ=sinθcosθ

For the ϕ component, λ=ϕ, and for all non-zero symbols ρ=ϕ.

Γμνϕ=12gϕϕ(gμϕ,ν+gνϕ,μgμν,ϕ)

This gives four symbols:

Γϕrϕ=Γrϕϕ=12gϕϕgϕϕ,r=12r2sin2θ2rsin2θ=1rΓθϕϕ=Γϕθϕ=12gϕϕgϕϕ,θ=12r2sin2θ2r2sinθcosθ=cotθ

So, the Christoffel symbols are:

Γrtt=Γtrt=A2AΓttr=A2BΓrrr=B2BΓθθr=rBΓϕϕr=rsin2θBΓθrθ=Γrθθ=1rΓϕϕθ=sinθcosθΓϕrϕ=Γrϕϕ=1rΓθϕϕ=Γϕθϕ=cotθ

Construct the Ricci Tensor

The Ricci tensor is defined to be:

Rμν=Γμν,λλΓλμ,νλ+ΓληλΓμνηΓμηλΓλνη

Due to symmetry the Ricci tensor has just the four trace emements:

Rμμ=Γμμ,λλΓλμ,μλ+ΓληλΓμμηΓμηλΓλμη

To calculate Rtt, μ=t expand each term and eliminate zero symbols. Only derivatives of r exist.

Rtt=Γtt,λλΓtλ,tλ+ΓληλΓttηΓtηλΓλtη=Γtt,rr+(Γtrt+Γrrr+Γθrθ+Γϕrϕ)Γttr2ΓtrtΓttr=(A2B)+(A2A+B2B+1r+1r)A2B2A2AA2B=A2BAB2B2+A24AB+AB4B2+ArBA22AB=A2BA4B(AA+BB)+ArB

To calculate Rrr, μ=r expand each term and eliminate zero symbols. Only derivatives of r exist.

Rrr=Γrr,λλΓrλ,rλ+ΓληλΓrrηΓrηλΓλrη=Γrr,rr(Γrt,rt+Γrr,rr+Γrθ,rθ+Γrϕ,rϕ)+(ΓtrtΓrrr+ΓrrrΓrrr+ΓθrθΓrrr+ΓϕrϕΓrrr)(ΓrttΓrtt+ΓrrrΓrrr+ΓrθθΓrθθ+ΓrϕϕΓϕrϕ)=(Γrt,rt+Γrθ,rθ+Γrϕ,rϕ)+(ΓtrtΓrrr+ΓθrθΓrrr+ΓϕrϕΓrrr)(ΓrttΓrtt+ΓrθθΓrθθ+ΓrϕϕΓϕrϕ)=(A2A)+2r2+AB4AB+BrB(A2A)22r2=A2A+A22A2A24A2+AB4AB+BrB=A2A+A4A(AA+BB)+BrB

To calculate Rθθ, μ=θ expand each term and eliminate zero symbols. Only derivatives of r and θ exist.

Rθθ=Γθθ,λλΓθλ,θλ+ΓληλΓθθηΓθηλΓλθη=Γθθ,rrΓθϕ,θϕ+(Γtrt+Γrrr+Γθrθ+Γϕrϕ)Γθθr2ΓθrθΓθθrΓθϕϕΓθϕϕ=(rB)+csc2θrB(A2A+B2B+1r+1r)+21rrBcot2θ=1B+rBB2+2Br2B(AA+BB)2B+1=1Br2B(AABB)+1

To calculate Rϕϕ, μ=ϕ expand each term and eliminate zero symbols. Only derivatives of r and θ exist.

Rϕϕ=Γϕϕ,λλΓϕλ,ϕλ+ΓληλΓϕϕηΓϕηλΓλϕη=Γϕϕ,rr+Γϕϕ,θθ+(Γtrt+Γrrr+Γθrθ+Γϕrϕ)Γϕϕr2ΓϕrϕΓϕϕrΓϕθϕΓϕϕθ=sin2θB+rsin2θBB2cos2θ+sin2θrsin2θB(A2A+B2B+1r+1r)+21rrsin2θB+cos2θ=sin2θ(1B+rBB2r2B(AABB)+1)=sin2θRθθ

The Ricci tensor is:

Rtt=A2BA4B(AA+BB)+ArBRrr=A2A+A4A(AA+BB)+BrBRθθ=1Br2B(AABB)+1Rϕϕ=sin2θRθθ

Construct the Ricci Scalar

The Ricci scalar is:

R=gμνRμν

R=RttARrrBRθθr2Rϕϕr2sin2θ=AABA2AB(AA+BB)+2r2B(1+rAArBBB)

Construct the Einstein Tensor

The Einstein Tensor can now be constructed.

Gμν=Rμν12Rgμν

Gtt=Ar2B(1rBBB)Grr=1r2(1+rAAB)Gθθ=r22B(AA+ArABrBA(AB)2A2B)Gϕϕ=Gθθsin2θ

Construct the Field Equations

The Einstein tensor is zero because the stress energy tensor Tμν=0 in a vacuum.

The field equations are:

Ar2B(1rBBB)=01r2(1+rAAB)=0r22B(AA+ArABrBA(AB)2A2B)=0Gθθsin2θ=0

Solve the Equations

The first two field equations reduce to:

1rBBB=01+rAAB=0

Subtracting gives:

AA+BB=BA+ABAB=(AB)AB=0

Integrating gives:

AB=k1

Where k1 is a constant of integration. As the metric must conform to the Minkowski metric k1=c2, AB=c2.

The first equation can be written as:

1BrBB2=ddr(rB)=1

Integrating gives:

rB=r+k2

Where k2 is a constant of integration.

This gives.

B=(1+k2r)1

Therefore.

A=c2(1+k2r)

As the solution must become the Minkowski metric as M the constant must be a multiple of the mass k2=kM. The A and B terms are in fact the Lorentz transformations.

1+kMr=1v2c2

kMr=v2c2

This must agree with Newton’s law of gravity for a small object of mass m falling from an infinite distance. In this case the kinetic energy of the object is equal to the gravitational potential energy.

12mv2=GMmr

v2=2GMr

Now substitute the value of v2.

kMr=2GMc2r

k=2Gc2

The Schwarzschild radius is:

rs=2GMc2,c2rs=2GM

The Christoffel Symbols

Now, the Christoffel symbols are:

Γrtt=Γtrt=rs2r(rrs)Γttr=GM(rrs)r3Γrrr=rs2r(rrs)Γθθr=(rrs)Γϕϕr=sin2θ(rrs)Γθrθ=Γrθθ=1rΓϕϕθ=sinθcosθΓϕrϕ=Γrϕϕ=1rΓθϕϕ=Γϕθϕ=cotθ

The Schwarzschild Solution

The Schwarzschild outer solution becomes:

ds2=c2dτ2=(12GMc2r)c2dt2(12GMc2r)1dr2r2dθ2r2sin2θdϕ2 ds2=c2dτ2=(1rsr)c2dt2(1rsr)1dr2r2dθ2r2sin2θdϕ2

The solution has two singularities. The singularity ar r=0 is not an issue as the solution is only valid in the vacuum of space around a massive object.

The singularity at r=rs has interesting consequences. The radius rs is normally well inside the object and hence out of scope of the solution. If however, all of the mass is compressed into a volume less than the Schwarzschild radius, the Schwarzschild radius defines the event horizon of a black hole.