# I Equalization of velocity components and momentum conservation

#### PainterGuy

Summary
I have been trying to understand this problem that how redistribution and equalization of velocity components take place and how momentum is conserved in a particular direction.
Hi,

I understand and I'm sorry that there are going to be many loopholes in what I'm trying to put together and that too without any mathematical formulation but I don't even know where and what to start with.

Suppose we have a finite length insulated hollow cylinder filled with with air at 1 atm. The cylinder is lying in horizontal position. Let's say it's sitting along the x-axis where x-axis runs horizontally parallel to your computer screen, vertical of your screen represents y-axis, and the z-axis is vertical to your screen.

The cylinder's right end is fitted with a pressure sensor. The other end, the left, is fitted with a movable sealed piston which could move at very fast rate like an audio speaker.

If the piston moves forward a small distance at an extremely fast rate, it would give the air molecules a positive momentum toward the right of screen. In other words, the x-component of velocity of molecules would increase and this increase would flow toward the sensor at the speed of sound, the sensor would register an increase in pressure. After a while, all the three velocity components need to equalize because pressure needs to be same in all directions.

I understand that the temperature would also rise and there would more number of molecules per unit volume.

My question is once the velocity components have equalized, wouldn't the sensor reading drop?
If there is a drop, wouldn't it signal a loss of momentum in the x-direction? The momentum should have been (and should be!) conserved because the applied momentum was in the x-direction.
If there is no drop, it'd mean that the x-component of velocity doesn't change after a while but then how come other velocity components increase and at what expense?

Thank you for your help.

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#### andrewkirk

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I wonder whether you mean speed rather than velocity. The average velocity of particles in the cylinder is zero, because the negatives in any direction approximately cancel out the positives.

Immediately after the impulse given by the piston, the average velocity in the x direction will become positive, but it will rapidly return to zero as the system returns to a new, higher-pressure, higher-temperature, equilibrium, in which the average speed of molecules is greater than it was before the impulse. It follows by symmetry arguments that the average x-component of velocity of particles striking the sensor will be higher than before.

The equilibration occurs firstly via collisions of the molecules pushed by the piston with other molecules. For those that make it to the end of the cylinder, the collision with the end will help equilibrate.

The key is to note that, while the energy inside the cylinder is almost conserved, given good enough insulation, the momentum of the air in the cylinder is not conserved. It is transferred to the cylinder walls and, through those, to whatever is holding the cylinder. So the nonzero momentum imparted by the piston is fairly rapidly transferred to the cylinder walls and then to its support.

• PainterGuy

#### PainterGuy

Thanks a lot!

I wonder whether you mean speed rather than velocity. The average velocity of particles in the cylinder is zero, because the negatives in any direction approximately cancel out the positives.
Yes, you are right. I should have used the word 'speed' instead.

It follows by symmetry arguments that the average x-component of velocity of particles striking the sensor will be higher than before.
I understand that the sensor reading will be higher than before the piston was moved but let's split it into two separate phases - phase #1: almost immediately after the piston is moved, phase #2: once the system has reached equilibrium. Wouldn't the reading be higher for phase #1? I think so because right after the piston is moved, the motion of molecules is directed in only directed but after a while the system starts going toward equilibrium.

The key is to note that, while the energy inside the cylinder is almost conserved, given good enough insulation, the momentum of the air in the cylinder is not conserved. It is transferred to the cylinder walls and, through those, to whatever is holding the cylinder. So the nonzero momentum imparted by the piston is fairly rapidly transferred to the cylinder walls and then to its support.
I thought that I should clarify few points.

So, the energy transferred by piston is conserved inside the cylinder. If the cylinder is well insulted, the energy conservation would include the air and cylinder walls because the entire system should be at the same temperature at equilibrium.

Let's suppose the cylinder is fixed to the ground. So, would the momentum get transferred to the ground?

Thank you for your help!

#### A.T.

Science Advisor
Let's suppose the cylinder is fixed to the ground. So, would the momentum get transferred to the ground?
Yes, and there are simpler examples of this: Throwing a ball against the wall that is fixed to the ground.

• PainterGuy

#### andrewkirk

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Let's suppose the cylinder is fixed to the ground. So, would the momentum get transferred to the ground?
Yes. and it should exactly cancel out the extra momentum in the opposite direction from the feet of the person that pushed the piston. So the net effect on the ground should be zero.
Wouldn't the reading be higher for phase #1? I think so because right after the piston is moved, the motion of molecules is directed in only directed but after a while the system starts going toward equilibrium.
Yes. A pulse of thicker air is sent towards the right side and when it hits there is a short higher reading, like how your windows shake when there is a nearby boom. If the gas is sparse and the impulse is very intense it may even bounce from one end to the other a few times, constantly getting weaker before dissipating to become undetectable. The weakening comes from collisions between molecules in the impulse 'wave' and other molecules, via which the original x-direction momentum is transferred to other directions.

• PainterGuy

#### PainterGuy

Thanks a lot for your help.

So the energy is conserved within the system. Temperature rises because molecules have more average speed than before and the pressure is increased because there are more molecules per unit volume which means more number of collisions with the walls per unit time.

As in my initial question statement, it was assumed that the piston can move at a very fast rate like an audio speaker. Suppose that the piston is moving back and forth at a rate of 40 Hz and generating a pressure of ±0.2 atm. I'm using "±" because when the piston pushes forward, it generates positive pressure, compression, and when the piston moves backwards it results in negative pressure, i.e. rarefaction.

Suppose the process has been going on for some time and the system has reached a steady state. What should the pressure sensor on the other end read? Would it read |0.2| atm assuming it gives only absolute value? Or. would it read something lower?

Thank you for your time and help!

#### jbriggs444

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Suppose the process has been going on for some time and the system has reached a steady state. What should the pressure sensor on the other end read? Would it read |0.2| atm assuming it gives only absolute value? Or. would it read something lower?
Obviously, the system cannot reach a steady state with vibrations continually being introduced. But we could expect it to settle down toward a repeating, periodic state. Assume that one has a cooling mechanism to drain the injected energy.

Gauge pressure is not limited to positive values. Why would you expect a non-zero average gauge pressure?

• PainterGuy

#### PainterGuy

Thank you!

But we could expect it to settle down toward a repeating, periodic state.
This is what I should have said.

Gauge pressure is not limited to positive values. Why would you expect a non-zero average gauge pressure?
Let me rephrase my original question from the previous posting.

It was assumed that the piston can move at a very fast rate like an audio speaker. Suppose that the piston is moving back and forth at a rate of 40 Hz in a sinusoidal fashion and generating a peak pressure of ±0.2 atm. The piston acts a sinusoidal device to generate pressure at 40 Hz with an amplitude of 0.2 atm. It could be given as 0.2*sin(2*π*40*t) and its RMS value is 0.2/√2=0.14142 atm.

Suppose the process has been going on for some time and the system has settled down toward a repeating, periodic state. What should the pressure sensor on the other end read? Would it read the same RMS value, i.e. 0.14142 atm or some value lower than that?

Your help is appreciated!

#### andrewkirk

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As JBriggs444 pointed out, some assumption needs to be made about the dissipation of the energy injected by the piston, otherwise the energy will build up without limit. With such an assumption the cylinder is no longer adiabatic. We could assume that energy is lost to the surroundings at the same rate that it is injected by the piston. If we make that assumption then the time-average of pressure on the sensor at the far end will not change as a result of the piston's movement.

Under that assumption, the pressure on the sensor will vary over time, with the average being the original pressure. The pattern of variation will depend in complex ways on many details including the size and shape of the cylinder and the nature of the gas. If the details meet certain tight constraints, a standing wave may be generated, under which the amplitude of the pressure variation on the sensor is greater than that applied by the piston, because of reinforcement from reflections (echos). If the cylinder is very long and a standing wave is not generated, the amplitude of the variation may be small enough to be disregarded.

• PainterGuy and jbriggs444

#### PainterGuy

Thank you!

I'm sorry for the late follow-up but I have been busy with something.

As JBriggs444 pointed out, some assumption needs to be made about the dissipation of the energy injected by the piston, otherwise the energy will build up without limit. With such an assumption the cylinder is no longer adiabatic. We could assume that energy is lost to the surroundings at the same rate that it is injected by the piston. If we make that assumption then the time-average of pressure on the sensor at the far end will not change as a result of the piston's movement.
So, it's a kind of Boyle's law situation where temperature is kept constant.

From my previous post where it was said that the piston exerts an RMS pressure of 0.14142 atm = 14 kPa = 14329.4 N/m². I think we can convert 14 kPa into 14 kJ/m³. So, the piston adds 14 kJ/m³ per cycle.

Do I have it correct?

Under that assumption, the pressure on the sensor will vary over time, with the average being the original pressure. The pattern of variation will depend in complex ways on many details including the size and shape of the cylinder and the nature of the gas. If the details meet certain tight constraints, a standing wave may be generated, under which the amplitude of the pressure variation on the sensor is greater than that applied by the piston, because of reinforcement from reflections (echos). If the cylinder is very long and a standing wave is not generated, the amplitude of the variation may be small enough to be disregarded.
So, the RMS pressure on sensor would be same as the RMS pressure exerted by piston.

We can assume that the cylinder is long enough to ignore standing waves build up.

I have a related question but first let me repeat something so there is no confusion. It was assumed that the piston can move at a very fast rate like an audio speaker. Suppose that the piston is moving back and forth at a rate of 40 Hz in a sinusoidal fashion and generating a peak pressure of ±0.2 atm. The piston acts a sinusoidal device to generate pressure at 40 Hz with an amplitude of 0.2 atm. It could be given as 0.2*sin(2*π*40*t) and its RMS value is 0.2/√2=0.14142 atm.

During the positive cycle of piston, the gas in cylinder is compressed and this compression results in rise of temperature. During negative cycle the piston moves backward from its mean position, it results into expansion of gas. This expansion would result into loss of temperature but that loss won't be equal to the temperature increase during the positive cycle of piston because it's not the gas which pushes piston back rather piston moves back on its own using energy from some outside source.

Do I have it correct?

Suppose the volume of cylinder is 1 m³. In my first post it was said that initially the cylinder is filled with with air at 1 atm. We have P₁V₁=P₂V₂ assuming temperature is kept constant therefore (1)(1)=(1+0.14142)V₂ ⇒V₂=0.876 m³. I hope the calculation is correct.

Thanks a lot for your help!

#### andrewkirk

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From my previous post where it was said that the piston exerts an RMS pressure of 0.14142 atm = 14 kPa = 14329.4 N/m². I think we can convert 14 kPa into 14 kJ/m³. So, the piston adds 14 kJ/m³ per cycle.
I am not sure that RMS is the correct measure here. It is different from AC current, where RMS is an appropriate measure for energy calculations. Certainly when the piston is moving inwards it is adding energy to the gas, but not when it is moving outwards. An RMS approach assumes that just as much energy is injected during an expansion as during a compression. I doubt that is the case. For a start there will be a slower collision rate between molecules and the piston during expansion, and the average speed of collision will be lower.

I don't know what the correct calculation is, but I'm fairly confident it's not RMS-based.

Perhaps a conservative approach (ie underestimating the energy injection rate) would be to assume no energy is added or withdrawn during expansion, and focus only on the compression phases.

• PainterGuy

#### PainterGuy

Thanks a lot for your help!

My original question was as follows.

It was assumed that the piston can move at a very fast rate like an audio speaker. Suppose that the piston is moving back and forth at a rate of 40 Hz and generating a pressure of ±0.2 atm. I'm using "±" because when the piston pushes forward, it generates positive pressure, compression, and when the piston moves backwards it results in negative pressure, i.e. rarefaction.

Suppose the process has been going on for some time and the system has settled down toward a repeating, periodic state. What should the pressure sensor on the other end read? Would the time-average of pressure on sensor be the same as time-average of pressure on piston's end?

I believe that we have already concluded that the time average of piston's pressure and the sensor's readings would be equal. I was after finding this answer.

I am not sure that RMS is the correct measure here. It is different from AC current, where RMS is an appropriate measure for energy calculations. Certainly when the piston is moving inwards it is adding energy to the gas, but not when it is moving outwards. An RMS approach assumes that just as much energy is injected during an expansion as during a compression. I doubt that is the case.
I agree with you but at the same time I'm little confused. Let me explain.

The piston moves back and forth around its mean position in a sinusoidal fashion as shown below. We have established that the temperature stays constant which means all the added energy is taken out which implies some kind of a heat sink but let's ignore it.

From A to B, as the piston moves inward, it collides with the gas molecules and the intensity it pushes the molecules with depends upon the rate of change, i.e. derivative of its movement. This movement generates a series of compressions and it results in rise in temperature.

From B to C, it starts going back toward its mean position, it generates a series of rarefactions and it results in decrease in temperature. The same thing happens from C to D.

From D to E, a series of compressions is generated and it results in rise of temperature.

If we follow what I'm saying then it'd mean that over one complete cycle of the piston, there shouldn't be any change in temperature. But I'm sure this is wrong because the piston does add energy to the gas which results in rise of temperature. Let's try to find the answer. I think that the following picture from the top right corner of this webpage, https://www.acs.psu.edu/drussell/Demos/waves/wavemotion.html, is helpful here. The webpage uses 'gif' image which shows the movement. We can say that the picture represents piston movement from A→B→C.

From A→B piston runs directly and forcefully into the molecules and pushes them back, i.e. compressions. As the piston runs into them, it comes into contact with lots of atoms/molecules. When the piston is moving back, i.e. B→C, molecules would come into contact with piston but not as many as during A→B, and also the piston is no longer running into the molecules. During B→C some of molecule would loose some of their speed, which results in decrease in temperature and hence loss of energy. But the main point is that energy supplied during A→B phase is way more than energy lost by gas molecules during B→C

In case of A→B the bat is fast moving toward ball and at the same time ball is rushing toward the bat. In case of B→C the bat is moving away from the ball. From C→D the gas molecules keep on loosing some their energy but from D→E the piston once again starts hitting the molecules very forcefully which means supplying them with a rather large amount of energy. During D→E the gas is at lower pressure compared to A→B therefore energy added would be a little less than compared to A→B.

Therefore, we can see that overall one complete cycle of piston adds energy to the system.

Perhaps a conservative approach (ie underestimating the energy injection rate) would be to assume no energy is added or withdrawn during expansion, and focus only on the compression phases.
In my opinion, compression phases are D→E and A→B, and energy added, as I visualize it, is going to be quite larger compared to expansions phases, i.e. B→C→D. If we ignore expansion phases and assume energy added during D→E and A→B is same then the energy input follow the pattern shown below. Do I make any sense at all?

Thank you for your patience and time!

Helpful link:
1: https://hemantmore.org.in/science/physics/sound-waves/3257/

#### andrewkirk

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If we follow what I'm saying then it'd mean that over one complete cycle of the piston, there shouldn't be any change in temperature.
No it won't average out. What will be zero by averaging out is the average over time of the excess speed of collisions ('excess' is measured relative to the average over time of collision speed given a stationary piston). But that doesn't make the time average of the injected energy zero, for two reasons:

1. In a collision between an object O1 of tiny mass m moving left at speed u and a massive object moving right at speed v, the leftwards component of velocity of the tiny object changes from u to -(u+v). So the kinetic energy increases from $\frac12 mu^2$ to $\frac12 m(u+v)^2$ so that the injected energy is $\frac12 mv(v+2u)$. This is dominated by the square of the piston's speed v. The average of the square of a random variable is greater than the square of its average.

2. What matters is not the raw average over time, but the average weighted by collision rate. Collision rate will be higher when the piston is moving in from what it is when it's moving out.

We can write an expression for the injected energy as follows:

Let the density of the gas at time t be $\rho_t$.

Let the distribution density function of leftwards velocities of particles at time t be $f_t$. Consider the energy injected in very short time interval $(t,t+dt]$ by collisions between the piston and particles with leftwards velocities in the range $(u,u+du]$.

The leftwards velocity of those particles is approximately u and the rightwards velocity of the piston is $v = A\sin 2\pi f t$ where A is the maximum velocity of the piston in its cycle and f is the frequency of the piston's cycle.

The density of particles with leftwards speed component in the prescribed range is $\rho f_t(u) du$ and the closing speed of the piston with those particles is approx $A\sin 2\pi f t + u$. So the mass of particles colliding with leftwards velocity component in the prescribed range during the time interval is $(\rho f_t(u) du) (A\sin 2\pi f t + u) dt$

So the energy injected is:

$$\frac12 (\rho f_t(u) du) ( (A\sin 2\pi f t + u) dt) (A\sin 2\pi f t) ((A\sin 2\pi f t) + u) = \frac12 \rho A f_t(u) (A\sin 2\pi f t + u)^2 \sin 2\pi f t\ du \ dt$$

A double integral of this expression over the range of possible values of u and t in a single cycle will give us the energy injected over that cycle.

The integration limits for t will be 0 to $1/f$. For u the upper limit will be $\infty$ and the lower limit will be the speed that is the negative of the piston's speed, as a leftward speed any smaller than that results in no collisions. So the lower limit for u is $-A\sin 2\pi f t$.

The time dependency of the velocity distribution $f_t$ makes things more complex but, if we assume a reasonably high frequency f, we can assume the distribution remains constant over a single cycle, and thereby regard $f_t(u)$ as constant over the range $(t,t+dt]$.

I'll leave it to you to work out the integral. I'm not sure if an analytic solution is possible. The changing lower limit of the u integration may be an obstacle to that. If an analytic solution is not possible, assume some values for the key parameters and do a numerical integration.

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• PainterGuy

#### PainterGuy

I genuinely appreciate your help!

I will take some time to get back to this. I'm working on few things and would like to get those done first. Thanks for the understanding.

#### PainterGuy

Hi again,

I was thinking of calculating the integral tomorrow but had few questions about the derivation. Could you please help me with those?

This is dominated by the square of the piston's speed v. The average of the square of a random variable is greater than the square of its average.
Why are you mentioning random variable? Is it because of "u", which is the speed of leftward moving molecule or particle? Yes, "u" is a random variable in the given context of distribution of speeds, but I'd say that this is true even when the variable is not random, so I'd say, "average of the square of a set of numbers is greater than the square of their average".

Do I make sense?

The density of particles with leftwards speed component in the prescribed range is ρft(u)duρft(u)du\rho f_t(u) du and the closing speed of the piston with those particles is approx Asin2πft+uAsin⁡2πft+uA\sin 2\pi f t + u. So the mass of particles colliding with leftwards velocity component in the prescribed range during the time interval is (ρft(u)du)(Asin2πft+u)dt(ρft(u)du)(Asin⁡2πft+u)dt(\rho f_t(u) du) (A\sin 2\pi f t + u) dt
I was trying to apply dimensional analysis but the units don't match up.

I believe the units for should be kg/s or informally mass/s.

You can see below that I'm ending up with mass/ms which is wrong. Where am I going wrong?

You also say that closing speed of piston with those particles is . I understand that the magnitude of the velocity difference at impact is called the closing speed.

If piston is moving is moving toward right, let's say this is positive velocity. Then the particle would be moving toward left, it'd be negative velocity. So, we can write the expression which is same as your expression. But there is a problem. There could be molecules/particles which are also moving rightward as the piston with speed less than that of the piston so the collision would take place. How do we take care of such particles? Does the expression include such cases? Can we ignore them?

Thank you!

#### andrewkirk

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Do I make sense?
Yes. Forget my reference to random variables. It was unnecessary.

Re dimensional analysis, the units of the three factors in the expression $(\rho f_t(u)du)(A\sin 2\pi ft+u)dt$
are $(\textit{kg}\,m^{-3})(m\,s^{-1})s$, which simplifies to $\textit{kg}\,m^{-2}$, which are the correct units for the mass colliding with the piston per square metre of piston surface area, in the time interval $(t,t+dt]$. If we multiply that by the surface area of the piston, which has units $m^2$, we get the kg of mass colliding in that time interval.
How do we take care of such particles? Does the expression include such cases?
They are included by our using a lower limit of integration that depends on the velocity of the piston. That lower limit ensures that we include all particles that are moving rightwards no faster than the piston. See the third last para of post #13.

• PainterGuy

#### PainterGuy

Thanks a lot for the clarification.

I was trying to do the numerical integration but came across few problems.

1. In a collision between an object O1 of tiny mass m moving left at speed u and a massive object moving right at speed v, the leftwards component of velocity of the tiny object changes from u to -(u+v). So the kinetic energy increases from $\frac12 mu^2$ to $\frac12 m(u+v)^2$ so that the injected energy is $\frac12 mv(v+2u)$. This is dominated by the square of the piston's speed v. The average of the square of a random variable is greater than the square of its average.
I'm sorry but it's not sitting well with me. I'm familiar with the two equations below about two masses colliding with each other. The density of particles with leftwards speed component in the prescribed range is $\rho f_t(u) du$ and the closing speed of the piston with those particles is approx $A\sin 2\pi f t + u$. So the mass of particles colliding with leftwards velocity component in the prescribed range during the time interval is $(\rho f_t(u) du) (A\sin 2\pi f t + u) dt$

So the energy injected is:

$$\frac12 (\rho f_t(u) du) ( (A\sin 2\pi f t + u) dt) (A\sin 2\pi f t) ((A\sin 2\pi f t) + u) = \frac12 \rho A f_t(u) (A\sin 2\pi f t + u)^2 \sin 2\pi f t\ du \ dt$$
$$ I understand that one can say that if the surface area is taken to be unity then Sp could be ignored but it won't be right, in my humble opinion. so that the injected energy is $\frac12 mv(v+2u)$. The final equation below is an equivalent of the quoted equation above. Right?$$
\frac12
(\rho f_t(u) du) ( (A\sin 2\pi f t + u) dt)
(A\sin 2\pi f t) ((A\sin 2\pi f t) + u)

Aren't we missing a factor of "2"? I tried to write a MATLAB code. There is something wrong with it because the answer was way too large. For "u", instead of infinity I used "10000" which is quite large. Anyway, I will try to fix it later.

Matlab:
% file name "Equalization of velocity components and momentum conservation"

close all;
clear all;
clc;

f = 100; % T = 1/f
t = linspace(0,1/100,50);
rho = 4;
A = 2;
f_u = 0.43;
u_l = -A.*sin(2.*pi.*f.*t);

f = @(u,t)(1/2).*rho.*A.*f_u.*(A.*sin(2.*pi.*f.*t)+u).^2.*sin(2.*pi.*f.*t);

for i=1:50

e=integral2(f,u_l(i),10000,0,5);

end

TE = sum(e) % total energy injected over one cycle
Code:
TE =

1.2451e+09
Thank you for the help!

Note to self:
I was getting confused over . Let's suppose rightward velocity is considered positive and leftward velocity is considered negative. The piston which is moving rightward has velocity v and the molecules moving leftward have velocity u=-v_mol. So, when we say that lower limit for u=-Asin(2πft), it also means -v_mol=-Asin(2πft) => v_mol=Asin(2πft). Anyway, we don't need to do and can just use "u" and the math takes care of it.

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