Recent content by SunGod87

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    Do Angular Momentum Components Commute with Nabla Squared and r Squared?

    Homework Statement Show the three components of angular momentum: L_x, L_y and L_z commute with nabla^2 and r^2 = x^2 + y^2 = z^2Homework Equations [A, B] = AB - BA For example: [L_x, \nabla^2] = L_x \nabla^2 - \nabla^2 L_x The Attempt at a Solution L_x \nabla^2 =...
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    Derivative of a Tensor: Solving for d_a A^b

    That's what I thought, but the lecturer is saying otherwise. I'm tempted to think he just misunderstood my question, since "How does d_a A^b transform under coordinate transformations" doesn't actually have a question mark after it, maybe it's just a sentance to describe the rest of the...
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    Derivative of a Tensor: Solving for d_a A^b

    My question is, I can only answer "How does d_a A^b transform under coordinate transformations" after having done: "compute d'_a A'^b", right? I've also revised my solution to the second part here http://img88.imageshack.us/img88/3954/38821206vp5.jpg This is what you meant? Thanks for the...
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    Integration Questions: Check Answers and Simplify Solutions | Homework Help

    You forgot to increase the exponent by one before dividing!
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    Derivative of a Tensor: Solving for d_a A^b

    Thing is, I've asked my lecturer about this now. He says we should answer that part first? Surely he's either made a typo or he's wrong? You HAVE to do the second part to know the answer to the first, right? In fact, while I have your attention... can you check my working, please...
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    Derivative of a Tensor: Solving for d_a A^b

    Sure, I've done that for the next part: "compute d'_a A'^b" and found it isn't a tensor so the partial derivative operator isn't a good operator in tensor analysis and a good operator should return a tensor. But what about the first part? I can't answer it without having attempted the next parts...
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    Derivative of a Tensor: Solving for d_a A^b

    Homework Statement http://img522.imageshack.us/img522/3511/80377551yt7.jpg Homework Equations None... I think. Seems like something I should just know rather than have to work out? The Attempt at a Solution I can do everything in this problem apart from the very first part. I...
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    Finding Eigenstates of Spin Operators in Quantum Mechanics

    So I should have (on the RHS) -hbar/2 [c_1 | 1/2 > + c_2 | -1/2 >] When I'm done, right? Or should I have: -hbar/2 [ | 1/2 > + | -1/2 >] I'm pretty sure it's the first one, right? Edit: Maybe not, I'm confusing myself with random coefficients multiplied for our cause and normalisation...
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    Finding Eigenstates of Spin Operators in Quantum Mechanics

    [SOLVED] Quantum Mechanics - Spin Homework Statement Problem is attached. Homework Equations The Attempt at a Solution The first part is seemingly straight forward. Measurements are +/- hbar/2, both with probability (1/sqrt[2])^2 = 1/2 of being observed. For the next part I...
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    Proving 2nd Order Differential Eqns

    y = 3x*e^(-2x) y' = 3e^(-2x) - 6xe^(-2x) y'' = -6e^(-2x) - 6e^(-2x) + 12xe^(-2x) = 0 -12e^(-2x) + 12xe^(-2x) = 0 12xe^(-2x) = 12e^(-2x) x = 1 Differentiated using the chain rule (for exponentials) and product rule.
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    Maximising area for constant perimeter

    Ah I can just write it like that. It makes sense but I was worried perhaps the lecturer was looking for something more. Thanks very much!
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    Maximising area for constant perimeter

    I tried another way and got the equation of a circle! But I'm still stuck! I have (x + B)^2 + (y + D)^2 = C^2 y(a) = y(-a) = 0 (a + B)^2 + (y + D)^2 = C^2 (-a + B)^2 + (y + D)^2 = C^2 so: (a + B)^2 = (-a + B)^2 and B = 0 x^2 + (y + D)^2 = C^2 Using the constraints: y(a) = y(-a) = 0 INT[-a,a]...
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    Maximising area for constant perimeter

    Are you sure it shouldn't be lambda*(1 + (dy/dx)^2)^(1/2) + y ?
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    Maximising area for constant perimeter

    (1 + (dy/dx)^2)^(1/2) + lambda*y = F + lambda*G lambda = C dy/dx = y' No explicit dependence on x so: (1 + (y')^2)^(1/2) + C*y - (y')^2 (1 + (y')^2)^(-1/2) = A (constant) (1 + (y')^2) + C*y*(1 + (y')^2)^(1/2) - (y')^2 = A*(1 + (y')^2)^(1/2) 1 + C*y*(1 + (y')^2)^(1/2) = A*(1 + (y')^2)^(1/2) 1 =...
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