Middle East Technical University Fall 1997 Math 457 Calculus on Manifolds

FINAL

JANUARY 10, TIME: 120 MIN

Problem 1. (20 pts) Electro-magnetic field is described by the tensor

T = (E^{1}dy ∧ dz + E^{2}dz ∧ dx + E^{3}dx ∧ dy) + (H^{1}dx + H^{2}dy + H^{3}dz) ∧ dt

in the space-time ℝ^{4}, where E^{i} = E^{i}(x, y, z, t), H^{i} = H^{i}(x, y, z, t).

(1) Determine dT.

(2) Express the condition dT = 0 (the Maxwell equation) in terms of the electric vector field E = (E^{1}, E^{2}, E^{3}) and the magnetic vector field H = (H^{1}, H^{2}, H^{3}) (using the operators div, curl).

Solution. (1)

dT = (E_{x}^{1} + E_{y}^{2} + E_{z}^{3}) dx ∧ dy ∧ dz + (E_{t}^{1}dt ∧ dy ∧ dz + E_{t}^{2}dt ∧ dz ∧ dx + E_{t}^{3}dt ∧ dx ∧ dy)

+ (−H_{y}^{1} + H_{x}^{2}) dx ∧ dy ∧ dt + (−H_{z}^{1} + H_{x}^{3}) dx ∧ dz ∧ dt + (−H_{z}^{2} + H_{y}^{3})dy ∧ dz ∧ dt

= (divE) dx ∧ dy ∧ dz + (E_{t}^{1} − H_{z}^{2} + H_{y}^{3}) dy ∧ dz ∧ dt

+ (−E_{t}^{2} − H_{z}^{1} + H_{x}^{3}) dx ∧ dz ∧ dt + (E_{t}^{3} − H_{y}^{1} + H_{x}^{2}) dy ∧ dz ∧ dt

(2) By the above dT = 0 means that divE = 0, E_{t}^{1} = H_{z}^{2} − H_{y}^{3}, E_{t}^{2} = −H_{z}^{1} +

H_{x}^{3}, E_{t}^{3} = H_{y}^{1} − H_{x}^{2}. The latter three identities are read in vector form as \frac{dE}{dt} = curlH.

Problem 2. (25 pts) Consider the tensor field g = y dx ⊗ dx − x dy ⊗ dy

(1) Is g a symmetric tensor vector field ? Is it alternating ? Explain your answer.

(2) Evaluate g(v, w), where v = 2\frac{∂}{∂x} − \frac{∂}{∂y} and w = \frac{∂}{∂x} + 3\frac{∂}{∂y} are vectors located at point p = (3, 4).

(3) Find Alt(g) at point q = (1, 1).

(4) Write down g in the polar coordinate system.

Solution. (1) g is symmetric, because it is defined by a symmetric matrix

g = (dx dy) ⊗ ( \begin{matrix}y & 0 \\ 0 & −x \end{matrix} ) ( \begin{matrix}dx \\ dy \end{matrix} )

g is not alternating tensor field, because Alt(dx⊗dx) = Alt(dy⊗dy) = 0 and thus Alt(g) = 0, i.e., Alt(g) ≠ g at every point (x, y), except (0, 0).

(2) g(v, w) = ydx(v)dx(w) − xdy(v)dy(w) = (y)(2)(1) − (x)(−1)(3) = 2y + 3x = 8 + 9 = 17

(3) Alt(g) = 0 at any point, as is shown in (1).

(4) dx = cos θ dr − r sin θ dθ, dy = sin θ dr + r cos θ dθ,

g = r sin θ(cos^{2} θ dr ⊗ dr − r cos θ sin θ (dr ⊗ dθ + dθ ⊗ dr) + r^{2} sin^{2} θ dθ ⊗ dθ)

− r cos θ(sin^{2} θ dr ⊗ dr + r cos θ sin θ(dr ⊗ dθ + dθ ⊗ dr) + r^{2} cos^{2} θ dθ ⊗ dθ)

= r sin θ cos θ(cos θ − sin θ) dr ⊗ dr − r^{2} sin θ cos θ(cos θ + sin θ)(dr ⊗ dθ + dθ ⊗ dr)

+ r^{3}(sin^{3} θ − cos^{3} θ) dθ ⊗ dθ Problem 3. (25 pts) Consider R = {(x, y, z, t) ∈ ℝ^{4} | z = x^{2} + y^{2}, x + y + z − t = 0, z ≤ 4}.

(1) Verify that R is a manifold. What is its dimension ? What is its boundary ∂R ?

(2) Parametrize ∂R and evaluate ∫_{∂R} ω, where ω = x dz + z dt.

(3) Parametrize R and express ∫_{R} dω as a double integral (do not evaluate the integral).

(4) Check if the orientations of R and ∂R defined by your parametrizations agree. Taking this into account write down Stokes’ theorem and apply it to evaluate the double integral.

Solution. (1) R = F^{−1}(0, 0) ∩ {(x, y, z, t) ∈ ℝ^{4} | z ≤ 4}, where F: ℝ^{4} → ℝ^{2}, F (x, y, z, t) = x^{2} + y^{2} − z, x + y + z − t). Since

F′ = ( \begin{matrix}2x & 2y & −1 & 0 \\ 1 & 1 & 1 & −1 \end{matrix} )

has rank 2 for any x, y, F^{−1}(0, 0) is a manifold of dimension 4 − 2 = 2. Its part lying in the half-space {(x, y, z, t) ∈ ℝ^{4} | z ≤ 4} is a manifold with the boundary L = F^{−1}(0, 0) ∩

{(x, y, z, t) ∈ ℝ^{4} | z = 4}, because (0, 0, 4) is a regular value of the mapping G(x, y, z, t) = x^{2} + y^{2} − z, x + y + z − t, z) and L = G^{−1}(0, 0, 4).

Another explaination: R is a manifold since it can be parametrized by the disc D = {(x, y) ∈

ℝ^{2} | x^{2} + y^{2} ≤ 4}, as the graph of the mapping f: D → ℝ^{2}, f(x, y) = (x^{2} + y^{2}, x + y + (x^{2} + y^{2})).

(2) Since ∂R = f(∂D), we can parametrize ∂R as follows: x = 2 cos θ,y = 2 sin θ, z = 4 and t = 2 cos θ + 2 sin θ + 4, θ ∈ [0, 2π]. Furthermore, since z = 4 is a constant, we have

dz = 0 and

∫_{∂R} ω = ∫_{0}^{2π}(0 + 4(−2 sin θ + 2 cos θ)) dθ = 0.

(3) R is parametrised in (1) above, (x, y) ∈ D, z = x^{2} + y^{2}, t = x + y + x^{2} + y^{2} and

dω = dx ∧ dz + dz ∧ dt, so,

∫_{R} dω = ∫_{D} dx ∧ (2x dx + 2y dy) + (2x dx + 2y dy) ∧ (dx + dy + 2x dx + 2y dy)

= ∫_{D}(2y + 2x(1 + 2y) − 2y(1 + 2x)) dx ∧ dy = ∫_{D} 2x dx dy

(4) The orientation of ∂D induced from D agrees with the parametrization x = 2 cos u, y = 2 sin u, of ∂D; therefore, the orientations of R and ∂R defined by our parametrizations also agree, and we can use the Stokes’ theorem ∫_{R} dω = ∫_{∂R} ω to deduce that ∫_{D} 2x dx dy = 0 (which can also be easily found by explicit calculation).