On the width of the kinetic energy distribution of a gas

In summary, the lecture notes on statistical mechanics discuss the distribution of kinetic energy of gas particles, represented by a graph with the mean value of energy, ##U##, and the width of the distribution, ##\sigma_{E}##. The author approximates the width of the distribution to be approximately ##\frac{\sigma_{E}}{\langle E\rangle}=\frac{\langle (E-U)^{2} \rangle ^{1/2}}{\langle E\rangle}## and explains the concept of the thermodynamic limit where the mean and standard deviation of the total energy approach infinity as the number of particles increases. However, the notation that the width of the peak is proportional to ##\frac{1}{\sqrt
  • #1
Aaron121
15
1
In these lecture notes about statistical mechanics, page ##10##, we can see the graph below.

1578510366663.png


It represents the distribution (probability density function) of the kinetic energy ##E## (a random variable) of all the gas particles (i.e., ##E=\sum_{i}^{N} E_{i}##, where ##E_{i}## (also a random variable) is the kinetic energy of gas molecule ##i##, and ##N## is the total number of gas molecules). ##U## above is the mean value of ##E## (##U=\langle E\rangle##).

The author claims that the width of the distribution above is approximately ##\frac{\sigma _{E}}{\langle E\rangle}=\frac{\langle (E-\textrm {U})^{2} \rangle ^{1/2}}{\langle E\rangle}##. Where does this approximation come from? Shouldn't it be just ##\sigma _{E}=\langle (E-\textrm {U})^{2} \rangle ^{1/2}##?

The exact quote reads,
the distribution of the system's total energy ##E## (which is a random variable because particle velocities are random variables) is very sharply peaked around its mean ##U=\langle E\rangle##: the width of this peak is ##\sim \Delta E_{rms}/U \sim \frac{1}{\sqrt{N}}## . This is called the thermodynamic limit— the statement that mean quantities for systems of very many particles approximate extremely well the exact properties of the system.

##\Delta E_{rms}## above is just ##\sigma _{E}##.
 

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  • #2
Are you aware of the central-limit theorem? It says that the sum of a large number of equally distributed random numbers to a very good approximation follows a Gaussian distribution with mean ##U=\langle E \rangle =N \langle E_1 \rangle## and standard deviation ##\Delta E=\Delta E_1 \sqrt{N}## (corrected in view of #3). The standard deviation is defined by ##\Delta E^2 = \langle (E-U)^2 \rangle=\langle E^2 \rangle - U^2 ##.
 
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  • #3
I think two things are being confused here.
If E is the total energy, i.e. E = ΣEi, then <E> = N<Ei> and ΔE = ΔEi*√N.
If E is the average energy per molecule, i.e. E = ΣEi/N, then <E> = <Ei> and ΔE = ΔEi/√N.
 
  • #4
Sorry, of course you are right: ##\Delta E=\sqrt{N} \Delta E_1##, but ##\Delta E/E=\Delta E_1/(\sqrt{N} \langle E_1 \rangle)##, where ##E## is the total energy.
 
  • #5
The question was not really about why ##\frac{\Delta E_{RMS}}{\langle E\rangle}## can be approximated by ##\frac{1}{\sqrt{N}}##. But, rather, about why the author says the width of the distribution is, approximately, the ratio of the standard deviation to the mean ##\frac{\Delta E_{RMS}}{\langle E\rangle}##, rather than just the standard deviation itself ##\sigma_{E}=\Delta E_{RMS}##, which we all know measures how spread (how wide) a given distribution is.
 
  • #6
I'd say the argument is to consider the (order of magnitude) of the width compared to the (order of magnitude) of the mean of the quantity. In other words you consider the mean an "accurate description" the smaller the standard deviation of the quantity is relative to the magnitude of the quantity itself.
 
  • #7
@vanhees71
you consider the mean an "accurate description" the smaller the standard deviation of the quantity is relative to the magnitude of the quantity itself.

Correct me if I'm wrong, but do you mean by this that the standard deviation can be approximated by the mean when we have a very small standard deviation?
 
  • #8
Aaron121 said:
In these lecture notes about statistical mechanics, page ##10##, we can see the graph below.

View attachment 255318

The author claims that the width of the distribution above is approximately ##\frac{\sigma _{E}}{\langle E\rangle}=\frac{\langle (E-\textrm {U})^{2} \rangle ^{1/2}}{\langle E\rangle}##.

We have to understand what is being graphed on the x-axis. Those notes show calculation for the the relative root-mean-square fluctuation of energy, so I think the x-axis is not total energy, but rather total energy rescaled by dividing it by the mean energy.
 
  • #9
@Stephen Tashi
the x-axis is not total energy, but rather total energy rescaled by dividing it by the mean energy.
No where in the notes is it indicated that the graph is the plot of ##P(E)## against##E/U##. Had it been the case, the x-axis would not have been labeled ##E##. Besides, it is clearly indicated under the graph that it is the plot of##P(E)## against the energy ##E=\sum_{i}^{N}(\frac{mv_{i}^2}{2})##.
 
  • #10
Aaron121 said:
Besides, it is clearly indicated under the graph that it is the plot of##P(E)## against the energy ##E=\sum_{i}^{N}(\frac{mv_{i}^2}{2})##.

Then the notation that the width of the peak is proportional to ##\frac{1}{\sqrt{N}}## is incorrect.

The notes focus on "the thermodynamic limit". Trusting Wikipedia:
The thermodynamic limit, or macroscopic limit,[1] of a system in statistical mechanics is the limit for a large number N of particles (e.g., atoms or molecules) where the volume is taken to grow in proportion with the number of particles.[2] The thermodynamic limit is defined as the limit of a system with a large volume, with the particle density held fixed.[3]

As we increase the number of particles, the mean of the total energy ##U## will approach infinity and the standard deviation of total energy will also approach infinity.

If we want a graph where the the width of the peak about the mean decreases with the number of particles, we must look at quantities like the mean energy per particle or the distribution of some quantity that approaches a finite limit as ##N## approaches infinity. The notes derive the proportionality of ##\sigma_E/ U## to ##\frac{1}{\sqrt {N}}##. The notes don't say that ##\sigma_E## is proportional to ##\frac{1}{\sqrt{N}}##.
 
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  • #11
Of course, in the thermodynamic limit only quantities like the energy and particle density make sense. Of course in real-world systems the volume, the number of particles, the total energy etc. are all finite, the limit "##V \rightarrow \infty##" is a mathematical tool to formalize the simplifying description (and it's far from trivial, as already the treatment of an ideal gas in quantum many-body theory shows, which is only the most simple exactly treatable example).
 

Related to On the width of the kinetic energy distribution of a gas

1. What is the kinetic energy distribution of a gas?

The kinetic energy distribution of a gas refers to the range of kinetic energies that the particles of the gas possess. This distribution is affected by factors such as temperature, pressure, and the mass of the gas particles.

2. How is the width of the kinetic energy distribution of a gas determined?

The width of the kinetic energy distribution of a gas is determined by the temperature of the gas. As the temperature increases, the distribution becomes wider, meaning that the particles have a wider range of kinetic energies.

3. What is the significance of the width of the kinetic energy distribution of a gas?

The width of the kinetic energy distribution of a gas is important because it provides information about the average kinetic energy of the gas particles. It also helps to determine the behavior of the gas, such as its ability to diffuse or undergo phase changes.

4. How does the width of the kinetic energy distribution of a gas relate to the Maxwell-Boltzmann distribution?

The Maxwell-Boltzmann distribution is a mathematical model that describes the distribution of kinetic energies in a gas. The width of the kinetic energy distribution of a gas is directly related to the shape of the Maxwell-Boltzmann curve, with a wider distribution resulting in a flatter curve.

5. Can the width of the kinetic energy distribution of a gas change?

Yes, the width of the kinetic energy distribution of a gas can change depending on the conditions of the gas. For example, increasing the temperature or pressure of the gas will result in a wider distribution, while decreasing the temperature or pressure will result in a narrower distribution.

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