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Wald: Ch. 2, Problem 8.b |
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| Apr8-11, 01:14 PM | #1 |
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Wald: Ch. 2, Problem 8.b
Find the coefficients in coordinate bases, of the metric tensors for Minkowski space, for "rotating coordinates" defined by
[tex]t' = t,[/tex] [tex]x' = (x^2+y^2)^{-1/2} \cos(\phi - \omega t),[/tex] [tex]y' = (x^2+y^2)^{-1/2} \sin(\phi - \omega t),[/tex] [tex]z' = z.[/tex] I think the inverse coordinate transformation should be [tex]t = t',[/tex] [tex]x = (x'^2+y'^2)^{-1/2} \cos(-\phi + \omega t),[/tex] [tex]y = (x^2+y^2)^{-1/2} \sin(-\phi + \omega t),[/tex] [tex]z = z'.[/tex] (EDIT: Insert prime symbols on x and y inside the brackets in the 3rd line of the inverse transformation.) In Mathematica, I calculated the Jacobian matrix of this inverse transformation, using doubled letters for primed ones: In[1]:= q = {tt, Sqrt[xx^2 + yy^2]*Cos[-phi + omega*tt], Sqrt[xx^2 + yy^2]*Sin[-phi + omega*tt], zz}; J = D[q, {{tt, xx, yy, zz}}] Out[1]:= {{1, 0, 0, 0}, {omega Sqrt[xx^2 + yy^2] Sin[phi - omega tt], ( xx Cos[phi - omega tt])/Sqrt[xx^2 + yy^2], (yy Cos[phi - omega tt])/ Sqrt[xx^2 + yy^2], 0}, {omega Sqrt[xx^2 + yy^2] Cos[phi - omega tt], -(( xx Sin[phi - omega tt])/Sqrt[xx^2 + yy^2]), -(( yy Sin[phi - omega tt])/Sqrt[xx^2 + yy^2]), 0}, {0, 0, 0, 1}} Then I calculated the new coefficients of the metric tensors thus: In[2]:= g = DiagonalMatrix[{-1, 1, 1, 1}]; gg = Transpose[J].g.J Out[2]: {{-1 + omega^2 (xx^2 + yy^2) Cos[phi - omega tt]^2 + omega^2 (xx^2 + yy^2) Sin[phi - omega tt]^2, 0, 0, 0}, {0, (xx^2 Cos[phi - omega tt]^2)/(xx^2 + yy^2) + ( xx^2 Sin[phi - omega tt]^2)/(xx^2 + yy^2), ( xx yy Cos[phi - omega tt]^2)/(xx^2 + yy^2) + ( xx yy Sin[phi - omega tt]^2)/(xx^2 + yy^2), 0}, {0, (xx yy Cos[phi - omega tt]^2)/(xx^2 + yy^2) + ( xx yy Sin[phi - omega tt]^2)/(xx^2 + yy^2), ( yy^2 Cos[phi - omega tt]^2)/(xx^2 + yy^2) + ( yy^2 Sin[phi - omega tt]^2)/(xx^2 + yy^2), 0}, {0, 0, 0, 1}} The result is symmetric, but both the Jacobian matrix and the new coefficient matrix of the metric tensor field have determinant zero. I guess this means I'm doing something wrong, since the determinant of the latter matrix is used to measure spacetime volumes, but volume wouldn't be well defined if a particular volume could be zero when measured in one chart, and nonzero in another. Any suggestions? |
| Apr8-11, 01:58 PM | #2 |
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It took me a moment to notice this issue, but there is something fishy about the transform. For given z and t, any x, y on the circle (x^2+Y^2)=k produce the same x', y' values. Thus, a circle in (x,y,z,t) gets mapped to a point in (x',y',z',t'). That's not a valid coordinate transform.
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| Apr8-11, 02:32 PM | #3 |
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Ah, thanks, I see what's amiss now. I omitted his final condition: tan(phi) = y/x. Setting phi = ArcTan[y/x] gives a Jacobian matrix with determinant 1, and a metric matrix with determinant -1, as expected.
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| Apr8-11, 04:30 PM | #4 |
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Wald: Ch. 2, Problem 8.b
Several typos in your transformation as given. (x2 + y2) should have an exponent + 1/2. Also in the sin, cos argument you want to change the sign of just t, not both φ and t. In other words, you want the (x,y) coords to rotate in the opposite (time-reversed) sense wrt (x',y'). Since cos is an even function and sin is odd, what you have written amounts to a reflection y' = - y.
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| Apr8-11, 06:06 PM | #5 |
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Thanks, Bill. I never spotted the minus sign in the exponent in the LaTeX till you pointed it out; that was just a typo, and didn't enter my calculations. I did evenually realise my mistake with inserting a minus sign before the phi. Also, the argument of sine and cosine in the inverse transformation should be phi'+omega*t', where phi' = arctan(y'/x'), shouldn't it?
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