From October 24, 2018:
Consider a closed system at thermal and mechanical equilibrium with \(PV\)-work only, and where the temperature and pressure of the system are constant. Then, for an arbitrary process, we have \(\text{d}U = dq + dw \leq T\text{d}S-P\text{d}V\), or, equivalently, \(\text{d}G = \text{d}(U-TS+PV) \leq -S\text{d}T+V\text{d}P\,(= \text{d}G_\text{rev}) = 0\), where the expression in parentheses follows from the differential of \(G\). But \(G\) is a state function, so how can equality not hold for \(\text{d}G \leq \text{d}G_\text{rev}\)?
From June 20, 2018:
What is the expected number of hits taken for a monkey to type out the word ABRACADABRA on a typewriter containing only the 26 letters A-Z, given that a random letter is depressed on each hit? (Adapted from Williams' Probability with Martingales.) Answer.
From September 19, 2017:
A block is placed on a plane inclined at an angle \(\theta\). The coefficient of friction between the block and the plane is \(\mu = \tan\theta\). The block initially moves horizontally along the plane at a speed \(V\). In the long-time limit, what is the speed of the block? (Introduction to Classical Mechanics, Morin, 3.7) Answer.
Find the maximum work that can be extracted from \(N\) identical objects with initial temperatures \(T_1, \dotsc, T_N\). Each object has constant heat capacity \(C_V = C\). Answer.
From November 26, 2016:
If \(f(x)\) and \(g(x)\) are linearly independent functions on an interval \(I\), they are linearly independent on any interval \(J\) contained inside \(I\).
If \(f(x)\) and \(g(x)\) are linearly dependent functions on an interval \(I\), they are linearly dependent on any interval \(J\) contained inside \(I\).
If \(f(x)\) and \(g(x)\) are linearly independent solutions of \(L(y)=0\) [\(L\) a differential operator] on an interval \(I\), they are linearly independent on any interval \(J\) contained inside \(I\).
If \(f(x)\) and \(g(x)\) are linearly dependent solutions of \(L(y)=0\) [\(L\) a differential operator] on an interval \(I\), they are linearly dependent on any interval \(J\) contained inside \(I\).
From September 20, 2016:
The dot product of any vector with itself is simply the squared norm of the vector: \(\mathbf{a \cdot a = ||a||}^2\). Hence differentiating with respect to time, we obtain \(\mathbf{a \cdot \dot{a}} = 0\) -- but this implies that every infinitesimal displacement is perpendicular to \(\mathbf{a}\), which is manifestly false. What goes wrong?
From September 3, 2016:
Let \(\langle u_A|\) denote \(\langle u|A\), and let \(|v_A\rangle\) denote \(A|v\rangle\). Then \(A^\dagger|u\rangle = (\langle u|A)^\dagger = \langle u_A|^\dagger = |u_A\rangle\) and \(\langle v|A^\dagger = (A|v\rangle)^\dagger = |v_A\rangle^\dagger = \langle v_A|\), so that \(\langle v_A|u\rangle = \langle v|A^\dagger|u\rangle = \langle v|u_A\rangle\), and \(\langle v_A|u\rangle = \langle v|A|u\rangle = \langle v|u_A\rangle\), where left and right equalities denote the action of \(A\) or \(A^\dagger\) on a bra or ket respectively. But this implies \(A = A^\dagger\) for all \(A\)! What goes wrong?
From August 13, 2016:
A mechanical system consists of only one point. Show that its acceleration in an inertial coordinate system is equal to zero ("Newton's first law").
A mechanical system consists of two points. If at the initial moment their velocities in some inertial coordinate system are equal to zero, show that the points will stay on the line that connected them at the initial moment.
A mechanical system consists of three points. If at the initial moment their velocities in some inertial coordinate system are equal to zero, show that the points will remain in the plane that contained them at the initial moment.
A mechanical system consists of two points. For any initial conditions, show that there exists an inertial coordinate system in which the two points remain in a fixed plane. (Mathematical Methods of Classical Mechanics, Arnol'd, pg. 10)
From August 5, 2016:
Suppose that there are two operators \(\textbf{A}\) and \(\textbf{B}\) such that \([\textbf{A}, \textbf{B}] = c\textbf{I}\), where \(c\) is a constant [and \(\textbf{I}\) is the identity operator]. Show that the vector space in which such operators are defined cannot be finite-dimensional. Conclude that the position and momentum operators of quantum mechanics can be defined only in infinite dimensions. (Mathematical Physics, Hassani, 5.45)