By Edwin Bidwell Wilson
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Extra resources for A Statistical Discussion of Sets of Precise Astronomical Measurements IV
Starting with Gauss’ theorem written as B · dσ = ∇ · B dτ, V ∂V substitute B = a × P, where a is a constant vector and P is arbitrary. The left-hand integrand then becomes (a × P) · dσ = (P × dσ) · a. The right-hand integrand expands into P · (∇ × a) − a · (∇ × P), the first term of which vanishes because a is a constant vector. Our Gauss’ theorem equation can then be written a· P × dσ = −a · ∂V ∇ × P dτ . V Rearranging to a· P×σ+ ∂V ∇ × P dτ = 0, V we note that because the constant direction of a is arbitrary the quantity in square brackets must vanish; its vanishing is equivalent to the relation to be proved.
A · ((dσ × ∇) × P) = ∂S S Bringing a outside the integrals and rearranging, we reach (dσ × ∇) × P − a· S dr × P = 0 . ∂S Since the direction of a is arbitrary, the quantity within the square brackets vanishes, thereby confirming the desired relation. 1. The solution is given in the text. 2. ϕ(r) = ϕ(r) = Q , 4π ε0 r a ≤ r < ∞, Q 3 1 r2 − , 4π ε0 a 2 2 a2 0 ≤ r ≤ a. 3. The gravitational acceleration in the z-direction relative to the Earth’s surface is GM GM GM − + 2 ∼ 2z 3 for 0 ≤ z R. 2 (R + z) R R Thus, Fz = 2z Fx = −x GmM , and R3 GmM GmM ∼ −x , 3 (R + x) R3 Fy = −y GmM GmM ∼ −y .
Then make a cyclic permutation if needed to reach CBA. 17. Taking the trace, we find from [Mi , Mj ] = iMk that i trace(Mk ) = trace(Mi Mj − Mj Mi ) = trace(Mi Mj ) − trace(Mi Mj ) = 0. 18. Taking the trace of A(BA) = −A2 B = −B yields −tr(B) = tr(A(BA)) = tr(A2 B) = tr(B). 19. (a) Starting from AB = −BA, multiply on the left by B−1 and take the trace. After simplification, we get trace B = −trace B, so trace B = 0. 20. This is proved in the text. 21. (a) A unit matrix except that Mii = k, (b) A unit matrix except that Mim = −K, (c) A unit matrix except that Mii = Mmm = 0 and Mmi − Mim = 1.
A Statistical Discussion of Sets of Precise Astronomical Measurements IV by Edwin Bidwell Wilson