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Given R(x, y) dx ∧ dy, is there always a P dx + Q dy that gives R(x, y) = ∂Q ∂P − ? ∂x ∂y It is easy to see that there are lots of them. 1. Let ψ = xy sin(y) dx ∧ dy be a “spin field” on R2 . Is it derived from some 1-form ω = P dx + Q dy? Solution: Yes, put P = 0 and Q = x2 y sin(y)/2 Then ∂Q ∂P − = xy sin(y) ∂x ∂y All I did was to set P to zero and integrate Q with respect to x. This is a bit too easy to be interesting. It stops being so silly if we do it on R3 , as we shall see later. 2. It should be obvious that just as we had the derivative taking 0-forms to 1-forms, so we have a process for getting 2-forms from 1-forms.
We used an actual parametrisation only to make it easier to evaluate it. 1. For any continuous vector field F on Rn and any differentiable curve c, the value of the integral of F over c does not depend on the choice of parametrisation of c. I shall prove this soon. 1. For the vector field F on R2 given by F x y −y x evaluate the integral of F along the straight line joining 1 0 to 0 . 3. INDEPENDENCE OF PARAMETRISATION 37 Parametrisation 1 c : [0, 1] −→ R2 1 0 (1 − t) t 0 1 +t 1−t t = c = −1 1 Then the integral is −t 1−t t=1 t=0 q −1 1 dt 1 t + 1 − tdt = = 1dt = t]10 = 1 0 This looks reasonable enough.
X1 .. Vn . xn Rn to R, and P V= P V+ c 0 V P Where I have specified the endpoints only since V has the independence of path properly. For every point P ∈ D I define ϕ ( P ) to be above equation as P 0 V, and I can rewrite the P V ϕ(P ) = ϕ(P ) + P 1 = ϕ(P ) + t=0 V1 (c(t)) .. q . Vn (c(t)) 1 0 0 .. 5. CLOSED LOOPS AND CONSERVATISM where c(t) = x1 − a + t x2 .. 45 for 0 ≤ t ≤ a. xn Since the integration is just along the x1 line we can write x1 x2 x=x1 x3 ϕ(P ) = ϕ(P ) + V1 dx ..