I What does equiprobable mean in the context of thermal motion?

[Mike_bb]

Really? What is the specific direction in the figure in post 29?

Does $<V_x>=<V_y>=<V_z>=0$ hold in your case in post 29?

 

[Dale]

Does $<V_x>=<V_y>=<V_z>=0$ hold in your case in post 29?

Yes. The velocity in each direction is given by a standard normal distribution, so it has mean = 0 and standard deviation = 1.

 

[weirdoguy]

I depicted 12 particles and their velocities.

Since when is 12 "extremely large"? You are not listening to what people are saying. All the book is saying is true:
- on average
- when we have around $10^{23}$ molecules

12 molecules is not a case for statistical physics.

 

[Mike_bb]

Yes. The velocity in each direction is given by a standard normal distribution, so it has mean = 0 and standard deviation = 1.

Thanks! I understand how it works.

But why does my book say another thing?

 

[Dale]

But why does my book say another thing?

From what I can tell it is not a very high quality book on this specific topic. I think it might be suitable as a reference for someone who already understands the material, but it doesn’t seem like a good source to learn it.

 

[Mike_bb]

Dale, could you explain what is idea of this proof? I can't understand it. Thanks.

 

[Herman Trivilino]

In the book

Which book? Are you still reading that advanced book?

there was following expression:

$N\frac{mc_x^2}{2} = \frac{N}{3} \frac{mc^2}{2}$
$c^2 = 3с_x^2$


Explanation: 1/3 of all particles moves along one axis (in both directions).

Your explanation isn't correct. What they're saying is that the square of the magnitude of $\vec{c}$ equals three times the square of the x-component of $\vec{c}$.

The Pythagorean theorem tells us that $c^2=c_x^2+c_y^2+c_z^2$. And since $c_x^2=c_y^2=c_z^2$, we have $c^2=3c_x^2$.

I depicted 12 particles and their velocities. But I don't understand which 4 molecules move along one axis.

View attachment 364673

Could anyone explain it?

Yes, your conclusion is based on an incorrect assumption.

As I've tried to tell you before, if I had barely completed the university-level introductory course, I would not be able to understand a book written for either upper-level undergraduate or graduate-level physics majors. You need the understanding of vector calculus, differential equations, thermodynamics, and statistics that can only be gained by taking university-level courses in those topics.

Again, as I told you before, get a college-level introductory physics textbook and read the section where the ideal gas law is derived. A review of vectors in the earlier chapters would also be helpful.

 

[Herman Trivilino]

could you explain what is idea of this proof?

It's the first sentence in the passage you quoted:

"The main assumption here is that the direction of the velocity is distributed uniformly."

 

[weirdoguy]

You need the understanding of vector calculus, differential equations, thermodynamics, and statistics that can only be gained by taking university-level courses in those topics.

I think OP has even more basic problem - understanding what this:

"The main assumption here is that the direction of the velocity is distributed uniformly."

means. And thinking in terms of mean behavior of large amount of particles. I think so especially after the 12-particle drawing...

 

[jbriggs444]

Explanation: 1/3 of all particles moves along one axis (in both directions).

Is this explanation verbatim from the book? Or is it your own interpretation? Either way, It is utterly incorrect.

If you total ${v_x}^2$ across all 12 particles you will get a result.
If you total ${v_y}^2$ across all 12 particles you will get another result.
If you total ${v_z}^2$ across all 12 particles you will get a third result.

Obviously, the sum of the three sub-totals is exactly equal to the total squared velocity across the 12 particles.

For a sample this small, the three sub-totals will be slightly different. Before looking at the velocities, there is no reason to expect any particular axis to have a greater sub-total than any other. On average (over a large number of 12 particle samples) we expect the total squared velocity to be spread evenly across the three sub-totals. Accordingly, we expect that each sub-total will be 1/3 of the overall total. On average.

For larger sample sizes, the percentage variation between the three sub-totals will become smaller and smaller. [The expected percentage variation will scale approximately with the inverse square root of the sample size.] For a sample size of $10^{23}$, we would expect variation on the rough order of $10^{-10}$ percent.

There is no particular axis along which 1/3 of the particle trajectories. will align.

 

[Mike_bb]

It's the first sentence in the passage you quoted:

"The main assumption here is that the direction of the velocity is distributed uniformly."

It's not enough.

 

[Herman Trivilino]

It's not enough.

You asked for "idea of proof", and that's the idea: To show that the direction of the velocity is uniformly distributed.

@jbriggs444 gave you a thorough explanation of what that means in Post #30.

If you understand vector components and how the ideal gas law is derived, you can understand that "proof", which is not really a proof, but an explanation.

 

[Steve4Physics]

@Mike_bb, I’m coming into this thread rather late but would like to add some random thoughts which might help.

It looks like you are dealing with an ideal gas made of identical particles (i.e. with equal masses).

(Note that if you have a mixture of different gases, e.g. hydrogen and oxygen, then at a given temperature, the average speed of the hydrogen molecules is much more than the average speed of the oxygen molecules because of their different masses; this would complicate things.)

I’m guessing you are looking at something like the derivation of gas pressure at an introductory level, e.g. to derive $PV = \frac 13 Nm c_{rms}^2$ or similar.

It is important to distinguish between speed (a scalar) and velocity (a vector). We should use the words ‘speed’ and ‘velocity’ carefully – they are not interchangeable.

For a stationary (zero momentum) container of gas, the average particle velocity is zero (or there would be net momentum). This applies to components of momentum in a given direction. And note that we don’t need equal and opposite pairs, e.g. the average of 1,2,3 and -6 is zero.

The kinetic energy of a particle (of mass $m$ and speed $c$) is $\frac 12 mc^2$. If you want the average kinetic energy of the particles you need the average value of speed-squared which we can write as <c^2>; this is is called the mean square speed. Its square-root is called the root mean square speed and is given a symbol such as as $c_{rms}$. So $c_{rms} = \sqrt {<c^2>}$. The value of $c_{rms}$ is similar to (but not the same as) the average speed.

The randomly moving gas particles have kinetic energy. The equipartition theorem in this situation tells us that:
$\frac 13$ of the total kinetic energy is due to components of motion in the x-direction;
$\frac 13$ of the total kinetic energy is due to components of motion in the y-direction;
$\frac 13$ of the total kinetic energy is due to components of motion in the z-direction.
Since a particle’s kinetic energy is $\frac 12 mc^2$ this leads to:
$\frac 13 c_{rms}^2 = <c_x^2> = <c_y^2> = <c_z^2>$
where, for example, $<c_x^2>$ is the average of the squares of the x-components of velocity (always positive because of the squaring).

Plenty to think about!

 

[Gavran]

Hi all!

In the book there was following expression:

$N\frac{mc_x^2}{2} = \frac{N}{3} \frac{mc^2}{2}$
$c^2 = 3с_x^2$


Explanation: 1/3 of all particles moves along one axis (in both directions).

I depicted 12 particles and their velocities. But I don't understand which 4 molecules move along one axis.

View attachment 364673

Could anyone explain it?
Thanks.

It seems to me that you are trying to make a deterministic model of a stochastic process. What you write here is not in the context of randomness.

 

[Mike_bb]

Herman Trivilino,

$<V_x^2> = \frac{1^2+3^2+2^2+7^2+...}{N}$
$<V_y^2> = \frac{(-3)^2+(-1)^2+2^2+7^2+...}{N}$

Is this true?

 

[jbriggs444]

$<V_x^2> = \frac{1^2+3^2+2^2+7^2+...}{N}$
$<V_y^2> = \frac{(-3)^2+(-1)^2+2^2+7^2+...}{N}$

Looking at your drawing here, those numbers are quite wrong. Where are you getting them?

As I read the drawing, $v_{1x}$ is negative, $v_{2x}$ is positive, $v_{3x}$ is positive and $v_{4x}$ is negative. $N$ is 12 for that situation.

 

[Mike_bb]

Looking at your drawing here, those numbers are quite wrong. Where are you getting them?

As I read the drawing, $v_{1x}$ is negative, $v_{2x}$ is positive, $v_{3x}$ is positive and $v_{4x}$ is negative. $N$ is 12 for that situation.

My post #65 is not about this situation.

 

[weirdoguy]

So what is it about?

 

[Mike_bb]

So what is it about?

For this proof:

 

[jbriggs444]

My post #65 is not about this situation.

So, again, where did those specific numbers come from?

 

[Mike_bb]

So, again, where did those specific numbers come from?

These numbers are random but from expressions for $<V_x^2>$ and $<V_y^2>$ we can see that X-axis and Y-axis are equiprobable. (gas is isotropic)

 

[jbriggs444]

View attachment 364722

Is this true?

Technically the quoted passage is false. It is an argument, not a valid proof.

The first error is minor. The author says that if the mean of ${v_x}^2$ is greater than the mean of ${v_y}^2$ that means that the x component of the velocities is greater in absolute value than the y component. As a handwave, that claim is arguably true. However, as a statement of mathematical fact, it is wrong. It is wrong because taking the absolute value is not the same thing as taking the square. The two functions are shaped similarly, but not identically. One can exploit this to construct counter-examples.

The second error is more fundamental. Let us reduce the problem to two dimensions as the author has done. We begin with an array of velocities that are uniformly distributed in direction. Suppose that these velocities are also uniformly^(*) distributed in magnitude. There is no general direction such that velocities in that direction have a larger magnitude than tuples in the orthogonal direction.

Now let us modify this array of velocities in the following way...

For every velocity that is generally in the $x$ direction we double its magnitude. For every velocity that is generally in the $y$ direction we halve its magnitude.

I claim that the distribution of directions in the resulting array is still uniform. Yet the average squared velocity in the $x$ direction will be larger than the average squared velocity in the $y$ direction.

This illustrates the point that having a uniform distribution of directions is not sufficient to make a velocity distribution isotropic.

(*) There is no such thing as a uniform distribution over the range $[ 0, \infty )$ so technically my objection is not entirely rigorous. But, like the author's passage, it is close enough to provide the flavor for a technically correct argument.

 

[Herman Trivilino]

This illustrates the point that having a uniform distribution of directions is not sufficient to make a velocity distribution isotropic.

But isn't the author arguing that having a uniform distribution of velocities is sufficient to make a uniform distribution of directions?

 

[Mike_bb]

Herman Trivilino,

Please, answer to my post #65. Thanks.

 

[Herman Trivilino]

Herman Trivilino,

Please, answer to my post #65. Thanks.

I've nothing to add to the responses given in Posts #66 through #73.

 

[Mike_bb]

Herman Trivilino,

$<V_x^2> = \frac{1^2+3^2+2^2+7^2+...}{N}$
$<V_y^2> = \frac{(-3)^2+(-1)^2+2^2+7^2+...}{N}$

It's my example with random components but uniformly distributed. Is it true? I want to understand.

 

[jbriggs444]

Herman Trivilino,

$<V_x^2> = \frac{1^2+3^2+2^2+7^2+...}{N}$
$<V_y^2> = \frac{(-3)^2+(-1)^2+2^2+7^2+...}{N}$

It's my example with random components but uniformly distributed. Is it true? I want to understand.

How can we infer a uniform distribution from a pair of samples that are not?

 

[jbriggs444]

But isn't the author arguing that having a uniform distribution of velocities is sufficient to make a uniform distribution of directions?

That may be the idea that he is pursuing. But the argument he made does not use the premise that the velocities are uniformly distributed. Only that their directions are uniformly distributed. That premise is insufficient for the desired conclusion.

 

[Mike_bb]

How can we infer a uniform distribution from a pair of samples that are not?

Ok. How can we infer that $<V_x^2> = <V_y^2>$ if we don't know all components of velocities?

 

[jbriggs444]

Ok. How can we infer that $<V_x^2> = <V_y^2>$ if we don't know all components of velocities?

From the premise that the velocity distribution is isotropic.

 

[Mike_bb]

From the premise that the velocity distribution is isotropic.

Ok. But I don't understand again. I think that each direction is not specific and particles move randomly in any direction. But how components of velocity vectors depend on it?

I can't imagine it.

 

[jbriggs444]

Ok. But I don't understand again. I think that each direction is not specific and particles move randomly in any direction. But how components of velocity vectors depend on it?

I can't imagine it.

If you have one particle, the components of the velocity vector at some chosen time are not isotropic. The particle is [almost certainly] moving in some direction. Its [squared] velocity in that direction is non-zero. Its [squared] velocity in the two orthogonal directions is zero. That is not isotropic.

[To be fair, if we are working in the center of momentum frame, the one particle case is rather degenerate. The particle has zero velocity. There is no randomness in play at all]

If you have two particles, you again cannot have isotropy. There is a some plane within which the two velocity vectors will lie. The sum of the squared velocities in the direction orthogonal to the plane will be zero. The sum of the squared velocities in the two orthogonal directions will be non-zero. That is not isotropic.

[Again, if we were playing fair, using the center of momentum frame and demanding zero total angular momentum then the two particles would be moving toward or away from each other on a common line rather than a common plane]

If you have three particles, you again cannot have isotropy. Pick almost any orthogonal triple of directions. The sum of the squared velocities in the three directions will differ from one another due to random variation.

If you have a large number of particles, the law of large numbers makes itself evident. The sum of squared velocities in any three orthogonal directions will be in remarkably good agreement with each other.

How large is "large" depends on how small you want the variation to be.

It is, perhaps, worth pointing out the distinction between a "distribution" and a "sample" from that distribution. The distribution may be isotropic even though no individual sample is. If the samples are large and we do not look too closely, it is easy to overlook that distinction.

 

[Herman Trivilino]

Herman Trivilino,

$<V_x^2> = \frac{1^2+3^2+2^2+7^2+...}{N}$
$<V_y^2> = \frac{(-3)^2+(-1)^2+2^2+7^2+...}{N}$

It's my example with random components but uniformly distributed. Is it true? I want to understand.

When the ellipses (...) is used to continue a sequence, the author's intent is obvious because the terms in the sequence form a pattern, thus making it obvious what the next term is, and the terms after that.

Here is an example. $1+3+5, ...$ . Here's it's clear how to continue the sequence, the next terms to be added are $7, 9, 11,$ etc.

Here's another example. $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+ ... = 1$. Here it's clear that the next term is $\frac{1}{16}$, and so on, with the sequence summing to $1$.

Thus I cannot answer your question because there is no pattern to the sequence that I can see.

This is probably why @jbriggs444 asked you where those numbers came from.

If you eliminated the ellipses, and you explained that the numbers in the sequences were the x- and y-components of the velocities, then yes, it would be true.

Again, it's not clear to me why you're asking these questions. If I knew that I would be better able to answer them. I know you answered before that you want to understand what's in that book, but why? What is the concept in that book that you're trying to understand? In other words, why that book and not some other book? For example, a book about the structure of pine needles!

 

[Mike_bb]

the x- and y-components of the velocities, then yes, it would be true.

Yes! I meant that these numbers are x- and y-components.

 

[jbriggs444]

Yes! I meant that these numbers are x- and y-components.

The x and y components of some set of velocities, yes?

What set of velocities? You have claimed that the components are "uniformly distributed". But we cannot infer a distribution from a small sample. And there is no such thing as a uniform distribution over an infinite range. So what actual distribution are you proposing?

In a statistics course, we might be able to do a bit more than throw up our hands. A finite sample can allow us to "estimate" one or more parameters that characterize a matching distribution. We could look for something like a "maximum likelihood estimator" or an "unbiased estimator". We could look for an estimator that matches the distribution mean to the sample mean and the distribution variance to the sample variance. Regardless, it is best to start by at least guessing at the general form of the distribution before trying to estimate its parameters.

 

[Mike_bb]

The x and y components of some set of velocities, yes?

What set of velocities?

For example, $N=10^{23}$

 

[jbriggs444]

For example, $N=10^{23}$

That tells us the sample size. It does not tell us the distribution from which the samples are drawn.

Google ("what is the distribution of velocities in an ideal gas") suggests https://en.wikipedia.org/wiki/Maxwell–Boltzmann_distribution

 

[Mike_bb]

Herman Trivilino,

I read now on Wiki how to derive expression for $<V_x^2>$ using integral. But I'm confused because this integral is taking from $-\infty$ to $+\infty$. We have already found out that it can be possible that symmetry of components isn't needed. As is mentioned above: for example, in X-axis in negative direction we can have -6 and in positive direction we can have 1,2,3 (average is 0). But if we use integral from $-\infty$ to $+\infty$ then we obtain that every positive component in X-axis has opposite (negative) component (1,2,3) and (-1,-2,-3). How is it possible? Thanks.

 

[jbriggs444]

Herman Trivilino,

I read now on Wiki how to derive expression for $<V_x^2>$ using integral.

But what are you integrating over?

Are you "integrating" over a small finite sample (e.g. [-6, 1, 2, 3])

That would be $\frac{(-6)^2 + (1)^2 + (2)^2 + (3)^2}{4} = \frac{36 + 1 + 4 + 9}{4} = \frac{50}{4} = 12.5$

Or are you integrating over a larger finite sample where $N \approx 10^{23}$? This margin is too small to write that sum in term by term fashion.

Or are you integrating over some particular Maxwell Boltzmann distribution? [erroneous -- we want the distribution for one component only]

Or are you integrating over some particular normal distribution? [correct -- and what you appear to be doing].

But I'm confused because this integral is taking from $-\infty$ to $+\infty$.

If you are integrating between these limits, you are probably integrating over a distribution rather than over a finite sample. The general form would be $\int_{-\infty}^{+\infty}f(x) \text{pdf}(x) dx$ where we have arranged that $\text{pdf}(x)$ is a proper probability density function, i.e. $\int_{-\infty}^{+\infty} \text{pdf}(x) dx = 1$.

But this has nothing whatsoever to do with any finite sample except that the sample measures can be expected to closely match the distribution measures when the sample size is very large.

I do not believe that [-6, 1, 2, 3] were actually drawn at random from a normal distribution.

 

[Herman Trivilino]

read now on Wiki how to derive expression for <Vx2> using integral. But I'm confused because this integral is taking from −∞ to +∞.

Stop reading that stuff and read a university-level introductory physics textbook. That's the best advice I can give you given that you haven't answered my questions about what you're doing and why. I have to glean from your posts that you are trying to understand something about the distribution of velocities of gas molecules. A university-level introductory physics textbook is the best place for you to do that.

 

[Steve4Physics]

... As is mentioned above: for example, in X-axis in negative direction we can have -6 and in positive direction we can have 1,2,3 (average is 0). But if we use integral from $-\infty$ to $+\infty$ then we obtain that every positive component in X-axis has opposite (negative) component (1,2,3) and (-1,-2,-3). How is it possible? Thanks.

View attachment 364741

@Mike_bb, can I throw this in...

Suppose you have ten 6-sided dice, each marked with the values:
-3, -2, -1, +1, +2, +3.

You throw them.

A = the number of -3s
B = the number of +3s.

You will probably find $\frac AB$ is not 1.

But if you have $10^{23}$ dice you will find that $\frac AB$ is almost exactly equal to 1. It’s a result of the statistical behaviour when the number of dice is very large.

The same idea applies to x-components of velocity for particles in a gas.

If you haven’t yet learned about and understood continuous probability distributions, you should probably (IMO) avoid the calculus-based analysis.

Edit - typo's.

 

[Mike_bb]

But what are you integrating over?

I'm integrating over continuous distribution. (components of velocities on X-axis)
The result is $\frac{kT}{m}$. This result coincide with $<V_x^2>$ (finite large sample). How is it possible?

jbriggs444,

Am I right that for every particle with positive component on X-axis we can find particle with opposite (negative) component? If it's so then I really understand how it works.

Google AI:

 

[jbriggs444]

I'm integrating over continuous distribution. (components of velocities on X-axis)
The result is $\frac{kT}{m}$. This result coincide with $<V_x^2>$ (finite large sample). How is it possible?

How is it possible? Why should it not be possible? Why the surprise?

It is very difficult to respond to an open ended question like this. You need to be specific about your concerns. Or, as has been aptly suggested, use a textbook and take a course.

jbriggs444,

Am I right that for every particle with positive component on X-axis we can find particle with opposite (negative) component? If it's so then I really understand how it works.

No.

If we choose a frame of reference where the total momentum is zero (i.e. the wind is not blowing) and assume a gas where all the particles are identical (so that momentum matches velocity), all we get is that the sum of the velocities in any direction is zero. Not that the velocities are paired up evenly.

You've given an example yourself: { -6, 1, 2, 3 }.

Google AI:

Is not a valid reference here.

For a very large sample from a normal distribution with mean zero, we do get that the number of particles in a small velocity range: $(-v \pm \Delta v)$ is approximately equal to the number of particles in the symmetric velocity range: $(+v \pm \Delta v)$.

We do not get a guarantee of exact equality. Nor do we need one for practical purposes.

Of course, the normal distribution is not special in this respect. The same property will hold for any symmetric distribution.

I would suggest a course in statistics. You need help nailing down concepts like "distribution", "sample", "mean" and "variance".

 

[Steve4Physics]

I'd like to add this to what @jbriggs444 has said.

Am I right that for every particle with positive component on X-axis we can find particle with opposite (negative) component?

No. You are wrong! Try working through the following carefully.

Step-1

Question: what is the probability of a gas particle having a speed of exactly 50.1 m/s?

You might say I haven’t given you enough information – but I have!

The answer is zero!

That’s because ‘exactly 50.1 m/s’ means
50.100000000000000000000000000… (to infinitely many decimal places)

The probability of a speed being exactly equal to this value (to the infinite-th decimal place) is zero.

That’s maths, not physics. It’s an issue for any continuous variable, not just speed.
______________________
Step-2

How do we deal with this? Answer: we break-down our continuous variable into convenient intervals. We can then ask: what is probability of a gas particle having a speed in some interval, e.g. $v$ to $v+\delta v$.

E.g. Using m/s for speed, our intervals could be:
.
$49 \le v \lt 50$
$50 \le v \lt 51$
$51 \le v \lt 52$
.
Here $\delta v = 1$.

We can how ask what is the probability of a particle's speed being in the range 50. to 51. The answer will be some sensible value.
_________________

Step-3 (which may answer your question)

Say:
P gas particles have x-components of velocity in the range $50 \le v_ x \lt 51$
Q gas particles have x-components of velocity in the range$-51 \le v_ x\lt -50$

Does P = Q? In fact it’s mathematically better to ask: does $\frac PQ = 1$?

For a small number of particles, it is unlikely that $\frac PQ = 1$.

But for a large number of particles (e.g. $10^{23}$) $\frac PQ$ is almost exactly equal to 1.

This is the result of the statistical behaviour when the number of particles is very large.

This is a bit of an oversimplification but it illustrates the principle.

 

[Mike_bb]

jbriggs444,

Ok. Is it right that if we have particle with X-component 4 then we can find particle with Y-component 4? (Google AI says it's true)

You wrote:
"If we choose a frame of reference where the total momentum is zero (i.e. the wind is not blowing) and assume a gas where all the particles are identical (so that momentum matches velocity), all we get is that the sum of the velocities in any direction is zero. Not that the velocities are paired up evenly."

I didn't say that sum of the velocities in any direction is zero. For example, two vectors: $V_1(1,5)$ and $V_2(-1,7)$ Why not?

 

[jbriggs444]

jbriggs444,

Ok. Is it right that if we have particle with X-component 4 then we can find particle with Y-component 4? (Google AI says it's true)

1. Google AI is not an acceptable reference. [See the rules here. Search for "ChatGPT and AI-generated text"]
2. If Google AI says this then Google AI is wrong. AI is frequently wrong but plausible. Hence the rule above.
3. I do not believe that Google AI says this. However since Google AI is not a valid reference here, that matter is not available for discussion.
4. [For the pedants out there] An implication that starts with a false premise will always be logically true. If false then anything. As has been pointed out, the likelihood of a particle with an X component of 4 being selected from a normal distribution is zero. We can find a mathematical formula for this:$$\int_4^4 \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} dx = 0$$
If you can find a valid reference that agrees with what you are saying, please link to it.

You wrote:
"If we choose a frame of reference where the total momentum is zero (i.e. the wind is not blowing) and assume a gas where all the particles are identical (so that momentum matches velocity), all we get is that the sum of the velocities in any direction is zero. Not that the velocities are paired up evenly."

I didn't say that sum of the velocities in any direction is zero. For example, two vectors: $V_1(1,5)$ and $V_2(-1,7)$ Why not?

I am not sure what point you are trying to make here.

Edit. Maybe I get it. You are suggesting that the velocity in the x direction can be independent of a wind blowing in the y direction?

1. Two [two dimensional] velocities that you have made up in your head are the same thing as four random samples from a normal distribution.
2. For any actual random sample, the mean for the sample will usually be different from the mean of the distribution.
3. If we are playing fair, we are working with a gas at rest rather than a gas in a wind tunnel. So we are arranging [within experimental limits] for the distribution to have a mean of zero in all directions. Or, equivalently, we are arranging for the total momentum to be zero.
4. [For the pedants out there] If we imagine an ideal gas of $10^{23}$ particles with a total momentum of zero and imagine that it is modelled with perfect mathematical accuracy as [per dimension] $10^{23}$ random samples from a normal distribution, we will be off by one. There are only $10^{23}-1$ degrees of freedom available. When we specified a total momentum of zero, that cost us one degree of freedom [per dimension]. We can have at most $10^{23}-1$ independent samples. Good luck measuring this discrepancy with a physical experiment.

 

[Mike_bb]

jbriggs444,

I opened another russian book and found there (translation):

"Strictly speaking, the average value of any velocity component is zero, since it can be positive or negative with equal probability."

It strange to me because I found in all sources that there are particle with positive and particle with negative component.

Could anyone provide source where it's written the opposite?

 

[jbriggs444]

jbriggs444,

I opened another russian book and found there (translation):

"Strictly speaking, the average value of any velocity component is zero, since it can be positive or negative with equal probability."

I read this as "given any distribution which is symmetric about zero, the distribution mean will be zero".

That is a true statement about the distribution. Nonetheless, the following statement is false:

"Any finite sample drawn from a symmetric distribution will have a sample mean of zero"

An example is a fair coin with +1 stamped into one face and -1 stamped into the other. We can immediately see that the distribution mean is zero since the positive and negative results are equiprobable. However, if we flip this coin 100 times, the total over those throws is unlikely to be zero. [Google says roughly 8% probability of 50 heads and 50 tails]

It strange to me because I found in all sources that there are particle with positive and particle with negative component.

I see no strangeness. Yes, any large sample from any distribution which is symmetric about zero is overwhelmingly likely to have both positive and negative elements. The probability that all elements share the same sign is one in $2^{N-1}$. [Pedant point - provided that p(0) = 0].

For a one element sample, it is 100% certain that the single element will share the same sign with every other element.

For a two element sample, it is 50% certain.

For a three element sample, it is 25% certain.

For a $10^{23}$ element sample, it is practically impossible.

So what?

 

[Mike_bb]

jbriggs444,

I agree with this statement:

"Yes, any large sample from any distribution which is symmetric about zero is overwhelmingly likely to have both positive and negative elements.

 

[Mike_bb]

jbriggs444,

So what?

Your statement "Yes, any large sample from any distribution which is symmetric about zero is overwhelmingly likely to have both positive and negative elements." contradicts with your following statements:

"The probability that all elements share the same sign is one in $2^{N−1}$"