thermodynamics - How slow is a reversible adiabatic expansion of an ideal gas?

A truly reversible thermodynamic process needs to be infinitesimally displaced from equilibrium at all times and therefore takes infinite time to complete. However, if I execute the process slowly, I should be able to get close to reversibility. My question is, "What determines when something is slow?"

For definiteness, let's take an insulating cylindrical piston with cross-sectional area $A$ and original length $L_0$. There is an ideal gas inside with $n$ molecules of gas with mass-per-molecule $m$. The temperature is $T_0$, and the adiabatic index is $\gamma$.

I plan to expand the piston adiabatically to length $L_1$, taking a time $t$ to do so. If I take $t$ to be long enough, the process will be nearly-reversible. However, $t$ being long does not mean "one minute" or "one year". It means $t >> \tau$ for some

$$\tau = f(A, L_0, L, n, m, T, k_B, \gamma)$$

What is $\tau$?

From purely dimensional considerations, I guess the relationship is something like

$$\tau = \sqrt{\frac{mLL_0}{k_bT}}f(n, L/L_0, A/L^2, \gamma)$$,

but I don't have a strong physical explanation.

Edit A meaningful answer should let me do the following: I take a certain example piston and try expanding it a few times, putting it in a box so I can measure the heat released to the environment. I calculate the entropy change in the universe for the expansions. After doing several expansions, each slower than the last, I finally get $\Delta S$ for the universe down to a number that I think is sufficiently small. Next, I plan to repeat the experiment, but with a new piston that has different dimensions, different initial temperature, etc. Based on my results for the previous piston, how can I figure out how long I should take to expand the new one to achieve a similar degree of reversibility on the first try?

For reference, the pressure is

$$P = \frac{nk_BT}{V}$$

and the speed of sound is

$$v = \sqrt{\frac{\gamma k_b T}{m}}$$,

and I'm happy to have an answer in terms of these or other derived quantities. Formulas for entropy and thermodynamic potentials can be found in the Wikipedia article.

Answer

I am a student so please point out in gory detail anything I did wrong.

For a process to be quasistatic, the time scales of evolving the system should be larger than the relaxation time. Relaxation time is the time needed for the system to return to equilibrium.

We have an adiabatic process, so equilibrium must be preserved at each point, that is to say

(Working within the validity of the kinetic theory for ideal gases and ignoring friction)

$(A L(t))^\gamma P(t) = (A L_0)^\gamma P(t_0)$

Momentum gained by the piston:

$\Delta p = 2 m v_x$

A molecule would impact the piston every

$\delta t = \frac{2(L_0+ \delta x) }{v_x}$

The force exerted on the piston is $F =\frac{\Delta p}{\delta t} = \frac{m v_x^2}{L_0+\delta x}$ Pressure would be $P = \frac{P}{A}$ and for $N$ such molecules

$P = \frac{N m ^2}{A (L_0+\delta x)} = \frac{N m ^2}{3A (L_0+\delta x)}$

So at the instant $t=t'$ where the piston has been displaced by $\delta x$, we have

$(A L(t))^\gamma P(t) = \frac{N m ^2}{3A^{1-\gamma}} (L_0+\delta x)^{\gamma -1}$

Expanding in series

$ = \frac{N m ^2 L_0^{\gamma-1}}{3A^{1-\gamma}} (1 + \frac{(\gamma-1) \delta x}{L_0}+ O(\delta x^2) ) $

Substituing $\frac{\delta x}{L_0} = \frac{\delta t v_x}{2 L_0} -1$

$(A L_0)^\gamma P(t_0) = (A L_0)^\gamma P(t_0) (1 + (\gamma-1) (\frac{\delta t v_x}{2 L_0} -1))$

If we want our process to be reversibly adiabitic atleast to first order, we must have from above

$\delta t = \frac{2 L_0}{}$

Now, this is time until collision for the starting case. Investigating second orders

$ (A L_0)^\gamma P(t_0) = (A L_0)^\gamma P(t_0) (1 + \frac{(\gamma-1) \delta x}{L_0}+ \frac{1}{2} (\gamma-1)(\gamma-2) (\frac{\delta x}{L_0})^2 +O(\delta x^3) ) $

Looking at just the series terms

$ 1 + (\gamma-1)\frac{\delta x}{L_0} (1 +\frac{1}{2} (\gamma -2) \frac{\delta x}{L_0}) \approx 1$

This would be true for

$\delta t = \frac{4 L_0}{} (\frac{1}{2-\gamma} -1)$

Now, this is the "time until next collision" for a gas molecule hitting the piston. To maintain reversibility, at least to second order, the piston should be moved from $L_0$ to $L_0 + \delta x$ in time $\tau = \delta t$ so that the system variables follow the adiabatic curve.

The $$ can be calculated from the maxwell distribution

If I could stop ALL motion would I stop time?

I've never really understood time. Time is not a force. It isn't energy or matter. Time doesn't MAKE anything do anything. Time doesn't make a clock tick (motion or batteries do). Time doesn't make the earth revolve around the sun (motion/gravity does). I don't think time exists. People say time is used as a form of measurement but aren't we just measuring movement? Even atomic clocks measure the movement of electrons that orbit an atom's nucleus as they "jump" back and forth between energy states. So my question is if I were to stop ALL motion (even movement of electrons that orbit an atom's nucleus) would I stop time? Is time just masquerading as the measurement of motion?

EDIT: This question is not a duplicate of "Does time freeze at Absolute Zero?" Because I am NOT talking about temperature. I am speaking about the motion of ALL things. I am asking if ALL things (photons, energy, matter, forces) stop moving would time stand still? I would have to say yes it would. If ALL motion were to stop for a million years and then start again no one would even know. Atomic clocks wouldn't even miss a step.

field theory - Why do $psi_a$ and $bar{psi}_{dot{alpha}}$ represent two different degrees of freedom?

I am taking a course in QFT and I've been introduced to the concept of left-handed (undotted) and right-handed spinors (dotted).

I know that left-handed spinors are associated with the irreducible representation $(1/2,0)$ of the Lorentz group ($\mathcal{L}$) and the right-handed spinors with the representation $(0,1/2)$ and that these two representations are not equivalent.

My professor told us in class that $\psi_a$ and $\bar{\psi}_{\dot{\alpha}}$ represent two different degrees of freedom but he didn't say why. I've been trying to understand it myself but I haven't been able of convincing myself of why it must be so (I have very little knowledge of Lie algebras).

Why do these spinors represent two different degrees of freedom?

Answer

In fact these representations are related to each other by discrete transformations of the Lorentz group. But by the definition, dotted spinors are transformed as complex conjugated undotted spinors, and this provides understanding of the statement that $\psi_{a}$ and $\bar{\psi}_{\dot{a}}$ are different degrees of freedom: the situation is similar (but not equally coincide with) to the statement that complex scalar field $\varphi$ and conjugated one $\varphi^{*}$ are independent degrees of freedom.

Nuclear Salt Water Rockets: viability and follow-up

This is the original paper by R. Zubrin proposing the Nuclear Salt Water Rocket design.

Basically the design is that a capillar set of pipes store a uranium salt-water solution, inside a cadmium matrix, which helps to keep it below criticality. The fluid is simply flown at high speed into a combustion chamber where it reaches criticality, heats quickly and is expelled from some unspecified nozzle.

The paper is from 1991, and i'm sure that the design has been revisited many times, so there must be good critiques and discussion of the problems that would have to be addressed, but i could not find any serious follow-up literature about the viability of the design, just, you know, the usual blog and mailing list informal ranting that you can find from a simple google search.

Have there been any simulations or attempts to evaluate the design with more detail in the literature that i'm not aware? The original paper doesn't go into much discussion about the temperatures, which i presume are quite high, and most of the informal viability analysis i've read about focus on that aspect.

Answer

Although the paper was published in the early 1990s, I could identify two citations of it only:

• McNutt Jr., R.L., Andrews, G.B., McAdams, J., Gold, R.E., Santo, A., Oursler, D., Heeres, K., Fraeman, M. & Williams, B. 2003, Low-cost interstellar probe, Acta Astronautica, vol. 52, no. 2-6, pp. 267-279.


• R.L. McNutt Jr., G.B. Andrews, J.V. McAdams, R.E. Gold, A.G. Santo, Douglas A. Ousler, K.J. Heeres, M.E. Fraeman, B.D. Williams, A Realistic Interstellar Probe, In: Klaus Scherer, Horst Fichtner, Hans Jörg Fahr and Eckart Marsch, Editor(s), COSPAR Colloquia Series, Pergamon, 2001, vol. 11, pp. 431-434.

There are issues with citation databases and search engines, their are not conclusive, I know. But just two follow-up papers/abstracts/proceedings in 20 years is usually a sign, that there has not been any further investigation - at least in civilian public research.

electric circuits - How do surfaces charges accumulate?

When a voltage source is applied across an ideal wire, do the push of electric field of battery source causes the accumulation of charges on the surface of wire or the capacitance of wire causes it?

One more doubt regarding this is - if we have a straight ideal (no turns at all) wire, somehow if I generate electric field parallel to the wire through a battery source (so the circuit is closed) will there be accumulation of charges( along the wire length) to cancel electric field of battery in this case? If so how?

Answer

Ideal wire cannot have net electric field inside. Whatever electric field the source can create, is cancelled out in the ideal wire by charges around, mostly on the terminals and wire surface. These negative charges are always present there, but usually their field is cancelled out by equal amount of charge of positive sign. When you connect the voltage source, the negative charges move a little so their field is no longer completely cancelled out by the positive charges. This motion is due to electromotive force of the battery and mutual push of the negative charges on the surfaces.

Re the second question, yes, the fact that the wire is ideal means the charges on the wire will redistribute to cancel out the external field of the battery. This is only an approximation to what would happen in reality. In reality, the wire is not ideal and the voltage source is not ideal either. There would be large current, there would be non-zero electric field in the wire and the battery would heat up and its voltage would drop down. It would still be above zero, as long as the current is flowing. Which in case of shorted battery won't be long - don't try to short out any battery, you would destroy it and it may result in dangerous explosion.

quantum mechanics - Proof of de Broglie wavelength for electron

According to de Broglie's wave-particle duality, the relation between electron's wavelength and momentum is $\lambda =h/mv$.

The proof of this is given in my textbook as follows:

1. 
   

   De Broglie first used Einstein's famous equation relating matter and energy, $$E=mc^2,$$ where $E=$ energy, $m =$ mass, $c =$ speed of light.

   
   


2. 
   

   Using Planck's theory which states every quantum of a wave has a discrete amount of energy given by Planck's equation, $$E=h\nu,$$ where $E =$ energy, $h =$ Plank's constant ($6.62607 \times 10^{-34}\:\mathrm{ J\:s}$), $\nu =$ frequency.

   
   


3. 
   

   Since de Broglie believes particles and wave have the same traits, the two energies would be the same: $$mc^2=h\nu.$$

   
   



4. 
   

   Because real particles do not travel at the speed of light, de Broglie substituted $v$, velocity, for $c$, the speed of light: $$mv^2=h\nu.$$

   
   

I want a direct proof without substituting $v$ for $c$. Is it possible to prove directly $\lambda=h/mv$ without substituting $v$ for $c$ in the equation $\lambda =h/mc$?

electromagnetism - If the strong nuclear force is stronger than electrostatic repulsion, why don't nuclei collapse into a point?

Today in class we were discussing the strong nuclear force, and our teacher was explaining about how the strong nuclear force counteracts the repulsion force between protons in a nucleus.

When asked about the relative strength of the two forces in question, she said that "The strong nuclear force is the strongest force of nature, and is infinitely stronger than the repulsion force between the protons". Now, if that were true, how would the atom remain in equilibrium, because if I'm correct, Equilibrium is achieved when the net force on a body is zero. However, in this situation, that doesn't seem to be the case. Could someone elaborate on this apparent contradiction?

Answer

First, the strong force acts on scales where our classical idea of forces as something that obeys Newton's laws breaks down anyway. The proper description of the strong force is as a quantum field theory. On the level of quarks, this is a theory of gluons, but on scales of the nucleus, only a "residual strong force", the nuclear force remains, which can be thought of as being effectively mediated by pions.

Now, a force mediated by pions is very different from one mediated by photons, for the simple reason that pions are massive. Massive forces do not, in their classical limit, follow a pure inverse square law, but yield the more general Yukawa potential, which goes as $\propto \frac{\mathrm{e}^{-mr}}{r^2}$ where $m$ is the mass of the mediating particle. That is, massive forces fall off far faster than electromagnetism.

So this makes it already difficult to tell what the "strength" of a force exactly is - it depends on the scale you are looking at, as Wikipedia's table for the strengths of the fundamental forces rightly acknowledges. However, in no sense is the strong force "infinitely stronger" than the electromagnetic force - it is simply much stronger than it, sufficient to keep nuclei together against electromagnetic repulsion.

Now, the person who said that it is "infinitely stronger" might have had something different in mind which is not actually related to the strength of the force but to its fundamentally quantum mechanical nature: Confinement, the phenomenon that particles charged under the fundamental (not the residual) strong force cannot freely exist in nature. When you try - electromagnetically or otherwise - to separate two quarks bound by the strong force, then you will never get two free quarks. The force between these two quarks stays constant with increasing distance, it does not obey an inverse square law at all, and in particular the energy to being on of the two quarks to infinity is not finite. At some point, when you have invested enough energy, there will be a spontaneous creation of a new quark-antiquark pair and you will end up with two bound quark systems, but no free quark. In this sense, one might say that the strong force is "infinitely stronger", but crucially this is not the aspect of the strong force that keeps nuclei together; the theory of pions shows no confinement.