Physics/Astronomy 226, Problem set 2, Due 1/27 Solutions

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1 Phsics/Astronom 226, Problem set 2, Due 1/27 Solutions 1. (Repost from PS1) Take a tensor X µν and vector V µ with components (ν labels columns and µ labels rows): X µν = , V µ = (1, 2, 0, 1) Let the metric be η µν, and consider each of the following. For each valid tensor equation, evaluate the l.h.s. For each invalid equation, state wh it is invalid. (a) Y = X µ µ (b) Z = X µµ (c) V = V µ V µ (d) T ν = X µν V µ (e) Q µν = X µ αx αν + X (µν) (f) G µναβ = X µν + V α X βδ V δ (g) R [µν] = X [µν] V [µ V ν] (a) Y = η µν X µν, which is perfectl OK. To compute it, ou know that ou will onl get a contribution to the sum then µ = ν, so ou can write it as: Y = η 00 X 00 + η 11 X 11 + η 22 X 22 + η 33 X 33 = = 1. (b) This is not OK as we are contracting over two lower indices. In this case the result will depend on the coordinate sstem (can ou see wh? the coordinate transformations will not form δ µν when multiplied), whereas the l.h.s. will not. (c) V = η µν V µ V ν = = 4. (d) This is a valid equation but also a good reminder that we have to be careful in converting to matrix notation in order to do the actual computation. First, we need to lower the index of V, giving V u = ( 1, 2, 0, 1). Then, we must multipl in b our matrix so that for each ν, we multipl V u b the column of X µν, i.e. T 0 = = 1. We then end up with T ν = (1, 2, 1, 4). (e) This is also fine. If we write the first term as X µα η αβ X βν, it is straightforward matrix multiplication XηX (considered as matrices). The second term is a smmetrization. We thus get: 1

2 Q µν = = (f) This is no good as it equates a (0 4) tensor to a (malformed) sum of (0 2) tensors. (g) The second term vanishes because V µ V ν is smmetric, but then we anti-smmetrize over it. So we just have to anti-smmetrize X µν to get: R [µν] = 1 2 [Xµν X νµ ] = Note that we don t actuall know what R µν is. 2. (a) Show that the equation is equivalent to the Maxwell equations ɛ βαµν α F µν = 0 i B i = 0 and ɛ ijk j E k + 0 B i = 0. (b) Show that it is also equivalent to the two alternative forms [α F µν] = 0 or α F µν + µ F να + ν F αµ = 0. (a) We can have either β = 0 or β = i where i = 1, 2, 3. Beginning with β = 0, we notice that ɛ 0αµν = ɛ ijk, where i, j, k are spatial dimensions. The equation above then simplifies to Now setting β = i, we have ɛ ijk i F jk = i ( ɛ ijk F jk ) = 2 i B i = i B i = 0. ɛ iαµν α F µν = ɛ i0µν 0 F µν ɛ ijµν j F µν. (1) On the first term on the RHS, we get contributions onl from µ, ν = j, k (i.e. spatial indices), so that ɛ i0µν 0 F µν = ɛ 0ijk 0 F jk = 0 (ɛ ijk F jk ) = 2 0 B i. In the second term on the RHS of Eq. 1, we must perform the analsis with µ and ν set to zero in separate cases and add the resulting two terms, but since both the Levi-Civita smbol and field-strength tensor are anti-smmetric, we can just 2

3 calculate it in one case and multipl b two to get the final answer. Proceeding with µ = 0, ɛ ijµν j F µν ɛ ij0k j F 0k = ɛ ijk j F 0 k = ɛ ijk j E k. Thus, combining these (and remembering the our original equation equals zero so we can drop both the factors of two): ɛ iαµν α F µν = 0 B i + ɛ ijk j E k = 0. (b) First, we ll show that the two equations listed above are the same: [α F µν] = 1 6 ( αf µν µ F αν + µ F να α F νµ + ν F αµ ν F µα ) = 1 3 ( αf µν + µ F να + ν F αµ ) = 0. where we have exploited the anti-smmetric propert of the field-strength tensor. Obviousl the factor of 1/3 is irrelevant since the whole thing is equal to zero. With this done, we just have to show that one of these two equations equals the equation given. This is easier to see with the more compact version. Since we began with ɛ βαµν α F µν, the effect of the Levi-Civita smbol is to anti-smmetrize the indices that it contracts with. Thus, another wa to write this equation showing the anti-smmetric relations of the α, µ, and ν indices without explicitl showing the LC smbol is to simpl write it as [α F µν]. 3. Calculate the nonzero components of the energ-momentum tensor T µν in cartesian coordinates in an inertial frame in which there is a flat disk of radius r 0 composed of N particles of mass m, rotating counterclockwise in the x plane about some fixed point, with fixed (radius-independent) angular velocit ω. Assume that the thickness of the disk is r 0, and that N is large enough that one can treat the particles as continuousl distributed with fixed number densit in the rest frame of the disk. (Notes: (i) Don t worr about what is keeping the particles rotating like this. (ii) Nor should ou worr about the effect of their mass on the spacetime assume it is Minkowski. (iii) Also, ou can express our answer using phrases like inside the disk and outside the disk. (iv) Assume that the particle number densit n is uniform in the rest frame of the disk. (v) Express ou answer in Cartesian coordinates.) Now suppose there is another such disk present with the same radius and center-ofrotation but with angular velocit ω, and that the particles do not collide or interact in an wa. What is T µν in this case? Assume the particles are dust-like, in which case the energ-momentum tensor is given b T µν = p µ N ν = mnu µ U ν where m is the mass of each particle, n is the number densit in the rest frame of the particles, and U µ is the four velocit vector field. Situate the disk so that it is centered 3

4 at x=0, =0 in the x- plane. Since geometr is normal, the disk will have a particle number densit given b n = if we let a be the thickness of the disk. The four velocit of a particle is given b U µ = χµ τ N πr 2 oa, = χµ t We are restricted to the x- plane, so U 3 = dz = 0. The time component, U 0, is given dτ b γ, with the 3-velocit magnitude equal to ωr for a particle at radius r. The x and t τ. components, U 1 and U 2, are given b γ dx dt Since this is circular motion, x and can be parametrized as x = r cos(ωt) = r sin(ωt) which can be differentiated to ield the velocities v x = rω sin(ωt) v = rω cos(ωt). and γ d dt, respectivel. Since U µ is a velocit field, we need to rewrite the velocit components as a function of position, not time. There is an inherent degenerac in using sine and cosine alone, so we will use the tangent and solve for t: x = sin(ωt) cos(ωt) = tan(ωt) t = 1 ( ) ω arctan x We can then rewrite the velocit equations as ( v x = rω sin(arctan ) ( v = rω cos(arctan ). We can simplif this further b drawing a right triangle with sides x and and hpotenuse r, and noticing that ( θ = arctan ( sin θ = sin(arctan ) = ( r cos θ = cos(arctan ) = x r. 4

5 Thus, our velocities reduce to and our four velocit field is given b v x = ω v = ωx U µ = γ(1, ω, ωx, 0). Multipling out the energ-momentum tensor (T 00 = mnu 0 U 0, T 01 = mnu 0 U 1, etc.), we arrive at 1 ω ωx 0 T µν = mnγ2 ω ω 2 2 ω 2 x 0 πr o a 2 ωx ω 2 x ω 2 x 2 0 Now we add another disk, with the same number densit n, same radius r o, and opposite angular velocit ω. The position and velocit of a particle for this ring will be given b x = r cos( ωt) = r cos(ωt) = r sin( ωt) = r sin(ωt) v x = rω sin(ωt) v = rω cos(ωt). We can solve once again for the time: x = tan( ωt) t = 1 ( ) ω arctan x which makes the velocities become ( v x = rω sin( arctan ) = ω ( v = rω cos( arctan ) = ωx. (We could have obtained this triviall b changing ω to ω, but it s nice to see that it works out.) The energ-momentum tensor for this ring is then given b 1 ω ωx 0 T µν = mnγ2 ω ω 2 2 ω 2 x 0 πroa 2 ωx ω 2 x ω 2 x 2 0. The total energ-momentum tensor for the sstem is simpl the sum of the energmomentum tensors for each ring: T µν = T µν (ω) + T µν ( ω) = 2mNγ2 0 ω 2 2 ω 2 x 0 πroa 2 0 ω 2 x ω 2 x 2 0. Note that ou can see that no energ is flowing anwhere, but that there is (anisotropic) pressure and nonzero shear. 5