Recent content by mef51

Ohh haha the lightbulb just clicked. Perhaps the unambiguous way of phrasing it would be "Show that if both processes start at an initial pressure ##P_i## and end at a final pressure ##P_f##, then the difference in temperature obtained by the second method is always higher."

Indeed

I dunno! On a PV diagram an adiabat with constant pressure doesn't really make sense either. Let's say that we don't have this condition that the initial and final pressures are the same. Then we'd have that the temperature difference for each method is $$...

So the pressure starts and ends at the same point but changes in between?

The initial and the final pressures are the same. How do you know that? I could equivalently say that the initial temperatures are the same for both processes and the final temperatures are different. Further, it's not even necessary that any of the temperatures are the same. We just want to...

Hm okay well for adiabatic throttling I know that the Joule-Thomson coefficient ##\mu## is positive when a gas is cooling so we know that ##\mu \equiv \Big( \frac{\partial T}{\partial P} \Big)_H > 0##. Using the fact that the adiabatic throttling is a constant-enthalpy process you can show that...

Homework Statement One method for cooling a gas is adiabatic throttling (Joule-Thomson Experiment). Another method is a reversible adiabatic expansion. Show that if the initial and final pressures are the same, the difference in temperature obtained by the second method is always higher. Hint...

It's not a homework problem. I had an attempt at it and this is what I got: (I used ##\alpha## as the polytropic index instead of ##n##) Using: ##\delta Q = nC_\alpha dT## and ##dU = nC_vdT## $$ \delta Q = dU + \delta W \\ nC_\alpha dT = nC_vdT + PdV \\ nC_\alpha \triangle T = nC_v\triangle T +...

Consider an ideal gas. For a polytropic process we have ##PV^n = const##. Different values of ##n## will represent different processes; for example isobaric (##n=0##), isothermal (##n=1##), and isochoric (##n=\infty##). The Wikipedia article on polytropic processes states that the specific heat...

Okokok. Since ##Q(s)## is differentiable, it must be continuous. ##det(Q)## must also be continuous. Since ##det(Q_0) = 1##, if ##det(Q)## were ever -1, then by the Intermediate Value Theorem there should be an ##s## such that ##det(Q(s)) = 0 ## (or any other value between 1 and -1). But...

Since ##Q^TQ## is constant ##\forall s##, ##Q^TQ = {Q_0}^T Q_0 = I_{33}## since ##Q_0## is in ##SO(3)##. Ok. I now need to show that ##det(Q) = 1 \quad \forall s##, knowing that ##det(Q_0) = 1## I know that ##det(Q^TQ) = det(I_{33}) = 1## So using properties of determinants ##1 = det(Q^TQ) =...

Alright let's take the derivative of ##Q^TQ##. If this ends up being zero, then ##Q^TQ## is constant and could be equal to the identity ##I_{33}##. This would prove one of the properties we want ##Q## to have to be in ##SO(3)##. We need to use ##\dot{Q} = AQ## and ##A^T=-A## $$ (Q^TQ)' =...

I made a mistake in my attempt: Since ##Q_0## is constant, taking the derivative of ##Q_0## is just 0. Should perhaps take the derivative of ##A(s)Q(s)## instead

Homework Statement Suppose that ##s \to A(s) \subset \mathbb{M}_{33}(\mathbb{R})## is smooth and that ##A(s)## is antisymmetric for all ##s##. If ##Q_0 \in SO(3)##, show that the unique solution (which you may assume exists) to $$\dot{Q}(s) = A(s)Q(s), \quad Q(0) = Q_0$$ satisfies ##Q(s) \in...

You're right. My prof has a convention where you only write the primes on the object instead of the indices (so ##T_{\mu\nu}'## instead of ##T_{\mu'\nu'}## and he also defined the transformation ##\Lambda## so that ##\Lambda## has a prime on the top and ##\Lambda^{-1}## has a prime on the...