What does equiprobable mean in the context of thermal motion?

[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.