Recent content by Eigenentity

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    Hot Air Balloon Energy Estimate Comparison

    I'm not sure it's as simple as that. Sure, the work done by the burner goes into heat, but that heat is clearly doing mechanical work. Think of what happens if I run the burner a little longer: I lift the entire balloon and its payload off the ground. I'm definitely doing real work. One way of...
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    Hot Air Balloon Energy Estimate Comparison

    I'm playing around with the physics of hot air balloons, and I've come to two different estimates of the energy required to get a hot air balloon off the ground. The two estimates make wildly different predictions and I'm hoping to get some insight on which one is wrong. Okay, so first from...
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    Lagrangian: singularity in inverted pendulum EoMs?

    To anyone wondering, the "unphysical" singularity arises from the unphysical assumption of a light carriage. With a carriage of nonzero mass M the EoMs become: \ddot{x} = \frac{2 m \sin \theta \left(g \cos \theta - l \dot{\theta}^2\right)}{2 M + m (1 - \cos 2 \theta)} \ddot{\theta} =...
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    Lagrangian: singularity in inverted pendulum EoMs?

    Hi all, I'm doing some analysis of a bicycle mechanics problem and at one point the approximations I'm making mean that the problem reduces to the classic inverted pendulum. I'm very confused, as the equations I've worked out appear to have an unphysical singularity in them, and I can't see...
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    Density matrix and von Neumann entropy - why does basis matter?

    To anyone wondering. No, basis does not matter. The error in the above reasoning is the assumption that the logarithm of a matrix is performed elementwise. This is true for diagonal matrices, but not general ones. More concretely: \log_2 \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{pmatrix}...
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    Density matrix and von Neumann entropy - why does basis matter?

    Density matrix and von Neumann entropy -- why does basis matter? I'm very confused by why I'm unable to correctly compute the von Neumann entropy S = - \mathrm{Tr}(\rho \log_2{\rho}) for the pure state | \psi \rangle = \left(|0\rangle + |1\rangle\right)/ \sqrt 2 Now, clearly the simplest...
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