Brownian motion and the resultant of forces

[theegyptian]

TL;DR Summary

Why would the instant forces resultant on the Brownian motion be not equal to zero

Hello,

Why isn't the forces resultant on a "large" molecule (by small molecule: water for example) zero? The reason for this Brownian motion is the thermal agitation of the water molecule. If we talk about white and Gaussian noise in electronics (due to the thermal agitation of the electrons) the AVERAGE of this white noise is zero (even if the power is not). So I would expect the force resultant to be zero. At each moment (time), there are a very large number of forces on the large molecule so I think about a gaussian law of mean/average equal to zero because intuitively there is absolutely no reason why the large molecule will go to a specific direction.

Sorry for my english, its not my first language.

Thanks

 

[hutchphd]

There is no reason to predict a particular direction, but one would predict a meandering wander. Particles that are macroscopic indeed do not exhibit Brownian Motion, so the supposition that they are always overwhelmed by coincidnt countervailing strikes is evidently wrong as thingd get small. Einstein used the observed motion of pollen to calculate the size of atoms so you may need to rethink your pontification Look up the paper and tell us why you are not correct. The fluctuations clearly win for some sized particulatesx. Folks liked the Einstein result pretty well

 

[theegyptian]

Particles that are macroscopic indeed do not exhibit Brownian Motion, so the supposition that they are always overwhelmed by coincidnt countervailing strikes is evidently wrong as thingd get small. Einstein used the observed motion of pollen to calculate the size of atoms so you may need to rethink your pontification Look up the paper and tell us why you are not correct. The fluctuations clearly win for some sized particulatesx. Folks liked the Einstein result pretty well

- Can you elaborate on what you call macroscopic ? (compared to water ?) can you take examples ?
- How small the approximation won't be right ?
- I never said I am not correct !
- Win over what ?
- I believe (if I understood you well) Pollen molecule is considered as a small one (compared to water molecule ?) ?

Can't we just apply (whatever the size of the molecule that we introduce in water) the central limit theorem and say that at each instant the resultant is a random variable that follow Gaussian law with an average of 0 ?

Thanks

 

[hutchphd]

The standard historical particles are grains of pollen which are macroscopic but not baseballs or automobiles. I will give this thought at a more reasonable hour.

 

[PeroK]

TL;DR Summary: Why would the instant forces resultant on the Brownian motion be not equal to zero

Hello,

Why isn't the forces resultant on a "large" molecule (by small molecule: water for example) zero? The reason for this Brownian motion is the thermal agitation of the water molecule. If we talk about white and Gaussian noise in electronics (due to the thermal agitation of the electrons) the AVERAGE of this white noise is zero (even if the power is not). So I would expect the force resultant to be zero. At each moment (time), there are a very large number of forces on the large molecule so I think about a gaussian law of mean/average equal to zero because intuitively there is absolutely no reason why the large molecule will go to a specific direction.

Sorry for my english, its not my first language.

Thanks

The average force is zero, but the variance is not zero. Imagine tossing one million coins. The average is an equal number of heads and tails, but there would almost certainly not be exactly an equal number. In fact, as the number of coin tosses increases, the expected difference in the absolute value of the number of heads minus the number of tails increases. For a million coins, you would expect perhaps up to 1000 more heads or up to 1000 more tails.

It's the same with molecular collisions in random directions. To get the expected average of zero is rare. Usually, there are more in one direction than another. Statistically this is called variance.

 

[theegyptian]

The average force is zero, but the variance is not zero. Imagine tossing one million coins. The average is an equal number of heads and tails, but there would almost certainly not be exactly an equal number. In fact, as the number of coin tosses increases, the expected difference in the absolute value of the number of heads minus the number of tails increases. For a million coins, you would expect perhaps up to 1000 more heads or up to 1000 more tails.

It's the same with molecular collisions in random directions. To get the expected average of zero is rare. Usually, there are more in one direction than another. Statistically this is called variance.

I believe you meant decrease in bold.



So what you say is that the larger the number of water molecule the closer to zero is the average force ? I agree with that but I tend to think that the number of molecule is so large that we are not supposed to see (by naked eyes) the Brownian motion... but we see it in the case of dye in water for example

 

[PeroK]

I believe you meant decrease in bold.

No. It increases. The percentage of heads and tails tend to 50%. But, the expected value of $|H-T|$ increases the more times you toss a coin.

Try running a computer simulation.

I believe you meant decrease in bold.

So what you say is that the larger the number of water molecule the closer to zero is the average force ? I agree with that but I tend to think that the number of molecule is so large that we are not supposed to see (by naked eyes) the Brownian motion... but we see it

No. The number of molecules is so large that we see a net force, as a result of the statistical variance in random collisions. It's a delicate calculation, but it earned Einstein his PhD.

 

[theegyptian]

No. It increases. The percentage of heads and tails tend to 50%. But, the expected value of $|H-T|$ increases the more times you toss a coin.

I didn't get that but I will give it a thought.

Try running a computer simulation.


No. The number of molecules is so large that we see a net force, as a result of the statistical variance in random collisions. It's a delicate calculation, but it earned Einstein his PhD.

I will try to make a computer simulation. The bold sentence drive me crazy. You mean that if the number were not that high there wouldn't be a net force ? The larger number of molecules, the more we tend to a Gaussian law (central limit theorem) which has an average of 0 ? (I agree there is a variance). I don't get that there is a net force BECAUSE the number of molecule is that large.

I see a temporal average and another average (on the number of molecule). And I thought that the Gaussian average is the temporal average but things are messy in my mind

 

[theegyptian]

No. It increases. The percentage of heads and tails tend to 50%. But, the expected value of |H−T| increases.

I think you mean that the variance increase with the number of molecule !!!!

Do you think that we can get the parameter of the Gaussian (average and variance) with a type of binomial law for which N tend to infinite ? where N is the number of molecules !!!

 

[theegyptian]

No. It increases. The percentage of heads and tails tend to 50%. But, the expected value of $|H-T|$ increases the more times you toss a coin.

Try running a computer simulation.


No. The number of molecules is so large that we see a net force, as a result of the statistical variance in random collisions. It's a delicate calculation, but it earned Einstein his PhD.

I GOT IT !!! Thank you very much. There is only a single average which is coming from the number of water molecules !!! the evolution of the motion of dye over TIME is actually the throwing a second, or third or 4th etc time the coins to see if we get tail or head. The average motion of Dye over TIME is 0. The instant net force value at a specific time depends on the variance !!!

It is really counter intuitive that the variance increase with the increase of the molecule number !! I really think I got it with your help but if you think I am bullsh*** let me know ! Thanks !!!

 

[hutchphd]

TL;DR Summary: Why would the instant forces resultant on the Brownian motion be not equal to zero

At each moment (time), there are a very large number of forces on the large molecule so I think about a gaussian law of mean/average equal to zero because intuitively there is absolutely no reason why the large molecule will go to a specific direction.

Can't we just apply (whatever the size of the molecule that we introduce in water) the central limit theorem and say that at each instant the resultant is a random variable that follow Gaussian law with an average of 0 ?

You need to study the random walk in 3 dimensions. I will, out of personal laziness, direct you to this very good Youtube discussion of the Chi- squared distribtion. The derivation of this is not difficult, just slightly tedious to LaTeX.
Now you see that the distributon of r will follow the $${\chi}^2 (3)$$ the chi-squared distribution with 3 degrees of freedom

Notice that for dimensions one and two, the likely value for r is in fact zero as your intuition seemingly demands But in dimensions 3 and larger, the likely displacement grows like N √3 where N is the number of steps of unit size.
With apologies for my LaTeX sloth....

 

[theegyptian]

You need to study the random walk in 3 dimensions. I will, out of personal laziness, direct you to this very good Youtube discussion of the Chi- squared distribtion. The derivation of this is not difficult, just slightly tedious to LaTeX.
Now you see that the distributon of r will follow the $${\chi}^2 (3)$$ the chi-squared distribution with 3 degrees of freedom

View attachment 340049

Notice that for dimensions one and two, the likely value for r is in fact zero as your intuition seemingly demands But in dimensions 3 and larger, the likely displacement grows like N √3 where N is the number of steps of unit size.
With apologies for my LaTeX sloth....

Are we saying that r is the average ?N is the number of molecule ? Not really clear but I will study this distribution. thanks again. The average should really not get more far to 0 as the number of molecule increase but I will see if that is what you meant...

 

[theegyptian]

No. The number of molecules is so large that we see a net force, as a result of the statistical variance in random collisions. It's a delicate calculation, but it earned Einstein his PhD.

I understood the video and what you said.

I would just say that the reason of this net force is not only the variance but even the average is not zero because the freedom degree is above 3. For me the number of freedom degrees is the number of water molecule that apply forces to the pollen molecule. I am not sure if you agree...

 

[hutchphd]

For me the number of freedom degrees is the number of water molecule that apply forces to the pollen molecule. I am not sure if you agree...ould model the molecules

The distribution for the distance traveled grow with the number of interactions (steps in the random walk). The degrees of freedom comes from the dimensionality of space. The step size will get smaller for bigger particles (but the frequency of the steps will get bigger) Ya need to do the complete threory to figure out the actual result....but it makes diffusion work as well as viscous drag. The fluctuation-dissipation theorem is also a good thing to look at.

 

[PeroK]

I understood the video and what you said.

I would just say that the reason of this net force is not only the variance but even the average is not zero because the freedom degree is above 3. For me the number of freedom degrees is the number of water molecule that apply forces to the pollen molecule. I am not sure if you agree...

I think this is the same confusion between an average displacement, which is total displacement divided by the number of steps. This will tend to zero as the number of steps gets large. And the absolute displacement that may not tend to zero as the number of steps gets large.

This is the same concept as I tried to explain earlier regarding coin tosses.

 

[PeroK]

I understood the video and what you said.

I would just say that the reason of this net force is not only the variance but even the average is not zero because the freedom degree is above 3. For me the number of freedom degrees is the number of water molecule that apply forces to the pollen molecule. I am not sure if you agree...

Here's the output from a computer simulation of a random walk in one dimension. At each stage the number -1 or +1 is chosen at random with equal probability, and the current position changed by -1 or +1. The starting position is 0. After so many steps (10, 100, 1,000 ...), the position is reported, as well as the average step (position/number of steps). You can see that the average step goes to zero. But, the position itself tends to get further away from 0 as the number of steps increases. Here's a typical output (up to 10 million steps):

Steps = 10; Position = 4; Average = 0.4
Steps = 100; Position = 16; Average = 0.16
Steps = 1000; Position = 46; Average = 0.046
Steps = 10000; Position = 176; Average = 0.0176
Steps = 100000; Position = 424; Average = 0.00424
Steps = 1000000; Position = -930; Average = -0.00093
Steps = 10000000; Position = 2240; Average = 0.000224

At the end, after 10 million steps, the position was +2240 steps away from the starting position. This still represents a very small average step away from the starting position.

 

[theegyptian]

I think this is the same confusion between an average displacement, which is total displacement divided by the number of steps. This will tend to zero as the number of steps gets large. And the absolute displacement that may not tend to zero as the number of steps gets large.

This is the same concept as I tried to explain earlier regarding coin tosses.

I understand what you are thinking but this was not my confusion. I am not that beginner :).

The problem was my understanding of central limit theorem and how to apply it. I thought there are 2 averages. One average over the number of molecules, and another average over the time. The average over the number of molecule would give me the net force that is random and the other average (over time) would give me the average of the Gaussian.

I understand now my mistake. There is no average over the molecule. the summation of many random variable following the same law (individual force of each molecule of water on the pollen) "create" a variable following a gaussian law. This average (in bold above) does not exist and do not give us the net force. it is actually a gaussian variable and as with any variable there is the frequency and the probability (average). If I throw a coin twice I won't necessarily have 1 face then the other face which is obvious.

Now what would be interesting is studing the condition(s) for which we can apply the central limit theorem and for which we have a gaussian. An important question is the size of the molecule (here pollen) compared to the size of the pollen molecule. Thinking about size makes me think about the average cinetique energy of the water molecule in regard with the pollen mollecule mass because utimately that would dictate the amount of displacement of the pollen molecule (non elastic collision)

 

[hutchphd]

At the end, after 10 million steps, the position was +2240 steps away from the starting position. This still represents a very small average step away from the starting position.

I fear you are obscuring the point. In 3 dimensions, with independence of x,y and z random walks, the particle wanders away from zero, "~never" to return to the origin. For equal step size l, the distance r will most likely be $l\sqrt 3 ~~$ Of course for this problem the step size is not fixed so the counting is more interesting but the "phase space available" argument is the key. Apologies if I misunderstand.

 

[PeroK]

I fear you are obscuring the point. In 3 dimensions, with independence of x,y and z random walks, the particle wanders away from zero, "~never" to return to the origin. For equal step size l, the distance r will most likely be $l\sqrt 3 ~~$ Of course for this problem the step size is not fixed so the counting is more interesting but the "phase space available" argument is the key. Apologies if I misunderstand.

For a random walk in 3D there is a probability of about 1/3 of returning to the origin.

 

[theegyptian]

I fear you are obscuring the point. In 3 dimensions, with independence of x,y and z random walks, the particle wanders away from zero, "~never" to return to the origin. For equal step size l, the distance r will most likely be $l\sqrt 3 ~~$ Of course for this problem the step size is not fixed so the counting is more interesting but the "phase space available" argument is the key. Apologies if I misunderstand.

Let me ask you a question regarding the X^2 video that you told me to watch. I understand it is a test and it helps take a decision with a confidence level/pourcentage. I am just thinking that the size of the samples play a role and obscure a bit my understanding. In the video example, there are 6 categories and 600 datas (in such a way that each categorie is 100 to ease the calculation). If I increase the number of data then the number obtained (X^2) might be different and I might take another decision. Then is confusing because with different sample size I take 2 potential different decision (if we keep everything else the same for the point of comparison). If you ask me which decision I would trust I would say the one with more data (higher than 600) because more data means more trustable.

BTW I understood your point of 3D displacement has a different behavior the 2 D displacement (average not zero).

 

[theegyptian]

For a random walk in 3D there is a probability of about 1/3 of returning to the origin.

so the probability of returning to the origin is 1/n with n the spatial dimension. In one dimension we have one value (not a distribution). But for 2 and 3 dimensions the probability of returning to origin is 1/2 or 1/3 right ? So how is the distribution ?

Because ultimately what is interesting is the average not the probability of retrurning to origin. Obviously in one dimension probability of returning to origin is 1 so the origin IS the average...

Can we say that in one dimension it is a succession of Bernoulli because if you don't go left then you obviously go right but in other dimensions (2 and 3) then Bernoulli does not apply any more because it is not binary anymore ?

 

[PeroK]

so the probability of returning to the origin is 1/n with n the spatial dimension. In one dimension we have one value (not a distribution). But for 2 and 3 dimensions the probability of returning to origin is 1/2 or 1/3 right ? So how is the distribution ?

Because ultimately what is interesting is the average not the probability of retrurning to origin. Obviously in one dimension probability of returning to origin is 1 so the origin IS the average...

Can we say that in one dimension it is a succession of Bernoulli because if you don't go left then you obviously go right but in other dimensions (2 and 3) then Bernoulli does not apply any more because it is not binary anymore ?

I don't understand what you are asking.

 

[hutchphd]

For a random walk in 3D there is a probability of about 1/3 of returning to the origin.

This is the overall probability as N gets large of ever returning to the origin at any step. It is Polya's number I believe. I think this is a different animal than you think.....Sorry gotta go beat on my car while the sun shines.