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 Exercise [21.06] 
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Joined: 26 Mar 2010, 04:39
Posts: 109
Post Exercise [21.06]
There are a couple of errors in the book which make this question harder than it should have been. Namely, Penrose got his signs mixed up in the definition of the free falling frame, and the expression for the wave function is slightly wrong.

I have explained and corrected these errors in my attached solution.

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27 Jul 2010, 00:09

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [21.06]
Robin, you are correct about the sign mixup in the definition of the freely-falling frame. The exercise should actually say:

Z = z + \tfrac{1}{2}t^2g

(Possibly Penrose forgot which set of coordinates were which: If you switch them around, the sign is correct as written - i.e. changing the sign actually gives you the inverse coordinate transformation. However, the given expression for the wavefunction does indeed require the lower-case coordinates to be the original (non-falling) ones in order to make sense.)

You are also correct that there is a minor error in the expression for the wavefunction, however it isn't at all what your solution indicates. The error is just a missing \frac{1}{\hbar} in the exponent of the e. The equation in the exercise statement should actually look like this:

\Psi = e^{\frac{i}{\hbar}\left(\frac{1}{6}mt^3g^2+mtzg\right)}\psi


The method you used in your posted solution is a good one, however your result is incorrect because you made an error expanding out the term {\partial_z}^2\left(e^\phi\psi\right). It should be:

{\partial_z}^2\left(e^\phi\psi\right) = e^\phi\left(\left[(\phi_z)^2+\phi_{zz}\right]\psi + 2\phi_z\psi_z + \psi_{zz}\right)

But you omitted the (\phi_z)^2\psi term. I re-did your calculations with this term included, and you do indeed get:

\phi = \tfrac{i}{\hbar}\left(\tfrac{1}{6}mg^2t^3+mgtz\right)

Which differs from the book only by the missing factor of \frac{1}{\hbar}.

27 Aug 2012, 08:35
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