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Phys 514 - Assignment 6 Solutions Phys 514 - Assignment 6 Solutions Prepared by Bryce Cyr April 3, 2018 1. The metric for the three-sphere in coordinates ( ; θ; φ) is ds2 = d 2 + sin2( )(dθ2 + sin2(θ)dφ2) (1) a) Calculate the Christoffel symbols b) Calculate the Riemann tensor, Ricci tensor, and Ricci scalar. Solution a) Recall that 1 Γσ = gσρ(@ g + @ g − @ g ) µν 2 µ νρ ν ρµ ρ µν And with our metric, the nonzero metric elements are 2 2 2 g = 1 gθθ = sin gφφ = sin sin θ We will approach this systematically, by check the upper index of the connection coefficients first. 1 Γ = g (@ g + @ g − @ g ) µν 2 µ ν ν µ µν The first two terms are 0 since derivatives of g vanish. This gives us two choices for the last term, µ = ν = θ and µ = ν = φ. These yield the coefficients Γθθ = − sin cos 2 Γφφ = − sin cos sin θ That's it for the upper index. Lets move onto θ. We have 1 Γθ = gθθ(@ g + @ g − @ g ) µν 2 µ νθ ν θµ θ µν Lets start with the last term. This is only nonzero if µ = ν = φ. This yields the symbol θ Γφφ = − sin θ cos θ 1 This takes care of the φ indices. The only other way to get a nonzero symbol is by µ = θ or ν = θ. In the first instance, we get 1 Γθ = gθθ(@ g + @ g − @ g ) θν 2 θ νθ ν θθ θ θν The first term and the final term are 0 by inspection. The middle term is nonzero if ν = . By the symmetry of the symbols we get θ θ Γθ = Γ θ = cot Finally, lets look at the last upper index. It is 1 Γφ = gφφ(@ g + @ g − @ g ) µν 2 µ νφ ν φµ φ µν The final term is always 0 since no part of the metric is φ dependent. Now, either µ or ν must be φ to get a nonzero symbol. The derivative will alternate between the symmetric indices, and so a straightforward computation yields the final four symbols φ φ Γ φ = Γφψ = cot φ φ Γθφ = Γφθ = cot θ To recap, the nonzero Christoffel symbols here are Γθθ = − sin cos 2 Γφφ = − sin cos sin θ θ Γφφ = − sin θ cos θ θ θ Γθ = Γ θ = cot φ φ Γ φ = Γφψ = cot φ φ Γθφ = Γφθ = cot θ b) The Riemann tensor is defined as ρ ρ ρ ρ λ ρ λ Rσµν = @µΓνσ − @ν Γµσ + ΓµλΓνσ − ΓνλΓµσ (2) The techniques to find the Riemann tensor were illustrated in one of the solutions for assignment 4, and so we will compute as many as my energy will allow for these Christoffel symbols. We recall the useful expression λ Rρσµν = gλρRσµν Which allows us to fully exploit the symmetries Rρσµν = −Rρσνµ Rρσµν = −Rσρµν Rρσµν = Rµνρσ Rρ[σµν] = 0 2 With this in mind, lets start by computing the upper index Riemann tensor indices. Since g = 1, we don't have to do any conversion between the two forms. We get λ λ R σµν = @µΓνσ − @ν Γµσ + ΓµλΓνσ − ΓνλΓµσ Now, to keep the first term nonzero, we note that we have three options. First, we can set ν = σ = θ and µ = to find λ λ R θ θ = @ (− sin cos ) − @θΓ θ + Γ λΓθθ − ΓθλΓ θ = sin2 − cos2 + cos2 = sin2 The middle two terms are 0 in the top line. By the symmetries, we can write a few more terms 2 2 R θ θ = sin Rθ θ = R θθ = − sin R θθ = 0 As our second check, we note that choosing ν = σ = φ and µ = yield a nonzero first term as well. This expression is 2 λ λ R φψφ = @ (− sin cos sin θ) − @φΓ φ + Γ λΓφφ − ΓφλΓ φ = sin2 θ(sin2 − cos2 ) + cos2 sin2 θ = sin2 θ sin2 The symmetries then yield 2 2 2 2 R φψφ = sin θ sin Rφψ φ = R φφψ = − sin θ sin R φφ = 0 Finally, the last option to keep the first term nonzero is ν = σ = φ with µ = θ This yields 2 λ λ R φθφ = @θ(− sin cos sin θ) − @φΓθφ + ΓθλΓφφ − ΓφλΓθφ = −2 sin cos sin θ cos θ + sin cos sin θ cos θ + sin cos sin θ cos θ = 0 So this one vanishes, as well as all its forms related by symmetries. Now we move to the second term, @ν Γµσ. We have multiple choices again, so first, set µ = σ = θ and ν = . This means we would be computing R θθ . We already know what this is from above! Its simply − sin2 θ sin2 , thus no computation is necessary. Moving on, we can choose µ = σ = φ and ν = ρ. This is the computation of R φφψ which is also done. Finally, the last thing that keeps the second term nonzero is µ = σ = φ with ν = θ. This computes R φφθ, which is 0 by an antisymmetric transformation of our previous result. Thus, we didn't have to perform any explicit calculation for this term! Moving now to the third term, λ θ φ θ φ ΓµλΓνσ = ΓµψΓνσ + ΓµθΓνσ + ΓµφΓνσ = 0 + ΓµθΓνσ + ΓµφΓνσ 3 We have options here, first, we can set µ = θ and ν = σ = φ. This will mean we compute R φθφ We have computed this already, and its 0. Next option, set µ = θ, ν = θ, and σ = . This computation is of R θθ, 2 2 also 0. Our next choice is µ = φ, ν = , σ = φ. This is computing R φφψ = − sin θ sin from symmetry. Finally, we can also make the choice µ = φ, ν = θ, σ = φ. This yields R φφθ, which is 0 once again by symmetry. Lets move onto the final term, expanded it is λ θ φ θ φ ΓνλΓµσ = Γν Γµσ + ΓνθΓµσ + ΓνφΓµσ = 0 + ΓνθΓµσ + ΓνφΓµσ We have four choices again, so lets speed through them. First, choose ν = θ, µ = σ = φ. This corresponds to computing R φφθ, again 0. Next choose ν = θ, µ = θ, σ = . This computes R θθ = 0. Next choice would be µ = φ, ν = , σ = φ. This computes R φφψ, already computed above. Finally, choose ν = φ, µ = θ, σ = φ. This computes R φθφ = 0. This concludes the computation of the Riemann symbols for the upper index of . As you can see, (and I'm sure you know after doing it yourself) it is a bit of a slow process. I will quote the results of the other nonzero Riemann tensor elements here, but rest assured they come from the exact same process as above but for upper indices θ and φ instead. Our nonzero elements are 2 2 R θ θ = Rθ θ = sin Rθ θ = R θθ = − sin 2 2 2 2 R φψφ = Rφψφψ = sin θ sin Rφψ φ = R φφψ = − sin θ sin 2 4 2 4 Rθφθφ = Rφθφθ = sin θ sin Rφθθφ = Rθφφθ = − sin θ sin ρ ρλ To get back to the proper form, we take the transformation Rσµν = g Rλσµν , where we have 1 1 g = 1 gθθ = gφφ = sin2 sin2 sin2 θ With these transforms, we get our final Riemann tensor elements. 2 2 2 2 2 2 Rθ θ = sin Rθθ = − sin Rφψφ = sin θ sin Rφφψ = − sin θ sin θ θ θ 2 2 θ 2 2 R θ = 1 R θ = −1 Rφθφ = sin θ sin Rφφθ = − sin θ sin φ φ φ 2 φ 2 R φψ = 1 R φ = −1 Rθφθ = sin Rθθφ = − sin λ The Ricci tensor is determined by a contraction of the Riemann tensor, Rµν = Rµλν Since we know this is a symmetric tensor in three dimensions, we can calculate the six independent components individually without much work. θ φ R = R + R θ + R φψ = 0 + 1 + 1 = 2 θ φ Rθθ = Rθ θ + Rθθθ + Rθφθ = sin2 + 0 + sin2 = 2 sin2 4 θ φ Rφφ = Rφψφ + Rφθφ + Rφφφ = sin2 sin2 θ + sin2 sin2 θ = 2 sin2 sin2 θ θ φ R θ = R θ + R θθ + R φθ = 0 + 0 + 0 = 0 θ φ R φ = R φ + R θφ + R φφ = 0 + 0 + 0 = 0 θ φ Rθφ = Rθ φ + Rθθφ + Rθφφ = 0 + 0 + 0 = 0 So the independent components of our Ricci tensor are R = 1 2 Rθθ = 2 sin 2 2 Rφφ = 2 sin sin θ R θ = R φ = Rθφ = 0 µν Our Ricci scalar is defined as R = g Rµν . Using the inverse metric written above, this yields θθ φφ R = g R + g Rθθ + g Rφφ = 2 + 2 + 2 = 6 5 2. Do the same calculation using the tetrad basis. Solution Recall our line element from the previous problem ds2 = d 2 + sin2 dθ + sin2 sin2 θdφ2 To make use of the tetrad formalism, we would like to have a line element that looks like 2 a b ds e e δab Where a; b run from 1 to 3. This allows us to define our tetrad basis. It will be e = d eθ = sin dθ eφ = sin sin θdφ Note that we are now in a noncoordinate basis. The appendix J of the book provides a good background on the application of the tetrad formalism, so refer to it if you are having any confusion. We wish to compute the Riemann tensor, which by equation J:29 is a a a c Rb = d!b + !c ^ !b (3) If we expect to find this, we had better start by first computing the spin connection, !. Note that the a a µ ν Riemann tensor in the above expression has been expressed in a basis of one forms, so Rb = Rbµν dx dx .
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