Calculate the distribution of energies from momenta

In summary: No, he doesn't.In summary, the author is trying to derive the probability distribution for the two energies given the particle momenta, but is having difficulty because the equations are not well-defined. He uses a heuristic to arrive at the result, and states that the density cannot be written as a product of the densities in the subsystems. He then states that the joint momentum distribution cannot be written as a product of the joint distributions in the subsystems, and that p(E_j, E_k)=p(E_j-E_k).
  • #1
SchroedingersLion
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TL;DR Summary
Question to a derivation in the book "An Introduction to Statistical Mechanics and Thermodynamics" by R. H. Swendsen (second edition).
Greetings!

I started working through a book of statistical mechanics, and I struggle with properly deriving a formula that makes intuitive sense. I will describe the situation and then my own solution attempts.

Suppose we have two systems ##j## and ##k## of volumes ##V_j## and ##V_k## and particle numbers ##N_j## and ##N_k##. The particles do not interact. The systems are isolated from each other. Suppose we are given the total energy ##E## and we are interested in the probability distribution ##P(E_j, E_k)## where ##E_j + E_k = E = const##. We make the assumption that all particle momenta are equally likely, but they satisfy the constraint ## E=\sum_i^N \frac{p_i^2}{2m} ##, where ##p_i## is the momentum of the ##i-th## particle, ##m## the particle mass, and ##N=N_j + N_k##. The same energy-momentum relation holds in each subsystem.

Now consider this section that the author allegedly makes use of:

book1.png

I already found a heuristic proof for this, so let's accept this as a given fact.

Now he wants to derive the distribution of the two energies from the momenta making use of (5.33). He writes:

book2.png


---
I don't understand how he actually ended up at (6.3). Given (5.33), what we can write is
$$\begin{align}
p_{E_j}(E_j) & = \int_{-\infty}^{\infty} p_{p_j}(p_{j,1},p_{j,2},...) \delta \left( E_j - \sum_{i}^{N_j} \frac{p_{j,i}^2}{2m} \right) \, dp_j , \\
p_{E_k}(E_k) &= \int_{-\infty}^{\infty} p_{p_k}(p_{k,1},p_{k,2},...) \delta \left( E_k - \sum_{i}^{N_k} \frac{p_{k,i}^2}{2m} \right) \, dp_k,
\end{align}$$
where ##p_{p_x}## and ##p_{E_x}## are the distributions of momenta and energies in system ##x##.

The joint distribution of all momenta should be writable as
$$
p(p_{1}, p_{2},...) = \frac{ \delta \left( E- \sum_i^N \frac{p_i^2}{2m}\right) } {\int_{-\infty}^{\infty} \delta \left( E- \sum_i^N \frac{p_i^2}{2m}\right) \, dp}.
$$

Now, all the delta functions are there, one just has to combine them properly...
It is clear that ##P(E_j, E_k) \neq p_{E_j}(E_j) p_{E_k}(E_k)## (since the energies in the subsystems are linearly dependent, i.e. fully correlated). In the same way, the joint momentum distribution cannot be written as a product of the joint distributions in the subsystems...
It is also clear that ##p_{E_k}(E_k) = p_{E_j}(E-E_k) ##.
If one interprets the delta-"functions" as living under energy integrals, one can also write
$$
\delta \left( E- \sum_i^N \frac{p_i^2}{2m}\right) = \delta \left( E_j - \sum_{i}^{N_j} \frac{p_{j,i}^2}{2m} \right)\delta \left( E_k - \sum_{i}^{N_k} \frac{p_{k,i}^2}{2m} \right). $$

Any idea from here?
I am always frustrated by the lack of derivations and / or mathematical rigor in physics books. I spent more time thinking about derivations such as these than actually learning the physics...
 
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  • #2
Your last formula with the relation
[tex]E=E_j+E_k[/tex]
says
LHS: spherical shell in 3N dimension p-space
RHS: spherical shell in 3N_j dimension * spherical shell in N_k dimension, p-spaces
RHS is cut of sherical shell in two sub spaces.
so RHS should be integrated to be right one wrt LHS as
[tex]\delta(E-...)=\int \int \delta(E_j-...) \delta(E_k-...) \delta(E-E_j-E_k)dE_jdE_k[/tex]
 
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  • #3
I agree to that, but how does this relation help me in deriving (6.3) from above?
 
  • #4
As for (6.3)
Numerator : (volume of spherical shell in ##3N_i## dimension p-space for radius^2##=E_i## )*(volume of spherical shell in ##3N_j## dimension p-space for radius^2=##E_j##)
Denominator: volume of spherical shell in ##3N=3(N_i+N_j)## dimension p-space for radius^2##=E##

Numerator consists of cut two parts of Denominator by plane ##E_j=E-E_i## for specific ##E_i## and ##E_j##. So by integrating cut parts
[tex] \frac{\int \int \delta(E-E_i-E_j) Numerator \ dE_idE_j}{Denominator}=1[/tex]

It means
[tex]\frac{Numerator}{Denominator}=p(E_i,E_j)[/tex]
 
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  • #5
anuttarasammyak said:
As for (6.3)
Numerator : (volume of spherical shell in ##3N_i## dimension p-space for radius^2##=E_i## )*(volume of spherical shell in ##3N_j## dimension p-space for radius^2=##E_j##)
Denominator: volume of spherical shell in ##3N=3(N_i+N_j)## dimension p-space for radius^2##=E##
I understand this part.
anuttarasammyak said:
Numerator consists of cut two parts of Denominator by plane ##E_j=E-E_i## for specific ##E_i## and ##E_j##. So by integrating cut parts
[tex] \frac{\int \int \delta(E-E_i-E_j) Numerator \ dE_idE_j}{Denominator}=1[/tex]
I don't quite get the "cut two parts" stuff. I see that the equation is correct, but just because the integral is 1 does not mean that the following is the correct density:

anuttarasammyak said:
It means
[tex]\frac{Numerator}{Denominator}=p(E_i,E_j)[/tex]

Does he even make use of eq. (5.33)?
 
  • #6
SchroedingersLion said:
I don't quite get the "cut two parts" stuff.
For an example 3D sphere shell
[tex]x^2+y^2+z^2=c^2[/tex]
is cut with cross section of 2D sphere shell (circle) and 1D sphere shell (##\pm b##) ,i.e.
[tex]x^2+y^2=a^2[/tex]
[tex]z^2=b^2[/tex]
where ##a^2+b^2=c^2##.

example #2
4D sphere shell
[tex]x^2+y^2+z^2+s^2=c^2[/tex]
is cut with cross section of two 2D sphere shells (circles) ,i.e.
[tex]x^2+y^2=a^2[/tex]
[tex]z^2+s^2=b^2[/tex]
where ##a^2+b^2=c^2##.

Integration of the cross sections with all the possible b and c restores the original sphere shell.
Area of cross sections (=Numerator) depends on the way of cutting and since we postulate that number density of states is homogeneous on the sphere shell of the defined volume ( =Denominator) , it seems reasonable that Numerator / Denominator equals probability that such a cut, i.e. division of energy to ##E_i## and ##E_j## to each group, takes place, i.e. ##p(E_i,E_j)##.

In case of example #2 the largest cross section, i.e. most probable case is
[tex]x^2+y^2=c^2/2[/tex]
[tex]z^2+s^2=c^2/2[/tex]
equal distribution of energy to the two groups.
 
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  • #7
Thanks for the examples, @anuttarasammyak , I am beginning to understand this geometric approach.

But shouldn't the density then be given by
$$
p(E_j, E_k) = \frac{\delta\left(E-E_j - E_k\right) numerator }{denominator}?
$$
Since integrating this over ##E_j## and ##E_k## gives 1. And integrating this numerator over small energy intervals would give fractions of the area of the 3N-sphere. Dividing this fraction by the total area given in the denominator gives the probability to find the energies in the intervals.

That would imply that (6.3) is incomplete.
 
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  • #8
By author's definition
[tex]\int \int p(E_j,E_k)\delta(E-E_j-E_k)dE_jdE_k = \int p(E_j,E-E_j)dE_j=1[/tex]
By your defnition
[tex]\int \int p(E_j,E_k)dE_jdE_k = \int \frac{Numerator(E_j,E-E_j)}{Denominator}dE_j=1[/tex]
So it depends on how the normalization is defined.
 
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  • #9
Sorry guys, coming back to this, I still don't see a rigorous way to understand this.
Is there no way to derive this only with properties of the dirac delta and with (5.33)?

My new approach:
I only know the total energy E and how many particles are in each subsystem. If all momenta are equally likely given E, then the joint probability density for the momenta are given by
$$
P(p_{1}, p_{2},...) = \frac{ \delta \left( E- \sum_i^N \frac{p_i^2}{2m}\right) } {\int_{-\infty}^{\infty} \delta \left( E- \sum_i^N \frac{p_i^2}{2m}\right) \, dp} =: \frac {1}{Z} \delta \left( E- \sum_i^N \frac{p_i^2}{2m}\right).
$$

Now, according to (5.33), the density of ##E_j## is given by
$$\begin{align}
P(E_j) &=\int_{-\infty}^{\infty}P(p_{1}, p_{2},...) \delta \left( E_j - \sum_{i}^{N_j} \frac{p_{j,i}^2}{2m} \right) \, dp \\
&= \frac {1}{Z}\int_{-\infty}^{\infty} \delta \left( E- \sum_i^N \frac{p_i^2}{2m}\right) \delta \left( E_j - \sum_{i}^{N_j} \frac{p_{j,i}^2}{2m} \right)\, dp
\end{align}$$
But I am actually interested in ##P(E_j, E_k) = P(E_j) P(E_k | E_j)## with conditional density ##P(E_k | E_j)##. Since ##E_k = E - E_j##, the latter can be written as ##P(E_k | E_j) = \delta \left( E_k - (E-E_j)\right)##.
We thus arrive at
$$
P(E_j, E_k) = \frac {1}{Z}\int_{-\infty}^{\infty}\delta \left( E_k - (E-E_j)\right) \delta \left( E- \sum_i^N \frac{p_i^2}{2m}\right) \delta \left( E_j - \sum_{i}^{N_j} \frac{p_{j,i}^2}{2m} \right)\, dp
$$

Now, the integral somehow must be reasoned to be equal to the author's numerator in (6.3)...
 
  • #10
Hello.

RHS of (4) has two energy parameters ##E## and ##E_j## so LHS should be written as
[tex]P(E_j\cap E)[/tex]
where I would say E_j has vale E_j and E has value E. Your ##P(E_j,E_k)=P(E_j \cap E_k) ## is same with it ?
 
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  • #11
The trick is that you just have to calculate the microcanonical integral
$$P(E)=\int \mathrm{d}^{3N} p \delta[E-\sum_{j=1}^N \vec{p}_j^2/(2m)].$$
Then you get the probability for a energy partition into ##E_A## and ##E_B## as
$$P_{\text{part}}(E_A)=\frac{P(E_A) P(E_B)}{P(E)}=\frac{P(E_A) P(E-E_A)}{P(E)}.$$
 
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  • #12
vanhees71 said:
The trick is that you just have to calculate the microcanonical integral
$$P(E)=\int \mathrm{d}^{3N} p \delta[E-\sum_{j=1}^N \vec{p}_j^2/(2m)].$$
Then you get the probability for a energy partition into ##E_A## and ##E_B## as
$$P_{\text{part}}(E_A)=\frac{P(E_A) P(E_B)}{P(E)}=\frac{P(E_A) P(E-E_A)}{P(E)}.$$
Thank you Vanhees, I think that's what I need. Even though I still don't understand why your second formula is obvious. Rewritten:
$$
P_{part}(E_A)P(E) = P(E_A)P(E_B)
$$

##E_A## is the event of having ##N_A## particles at energy ##E_A##. ##E_B## is the event of having ##N-N_A## particles at ##E_B##.
Are the events ##E_A## and ##E_B## independent, i.e. do we have at this point that ##P(E_A\cap E_B)=P(E_A)P(E_B)##?

Now ##E## is the event of having ##N## particles at energy ##E##. ##part(E_A)## then is the event of having energy ##E## distributed over the subsystems as ##E=E_A + E_B## ? Are those independent, i.e. do I have ##P(E \cap part(E_A)) = P(E)P(part(E_A))## ?

Now in case that both the independencies hold, how can I be sure that I have ##P(E_A\cap E_B)=P(E \cap part(E_A)) ## ?

Sorry if I am a bit slow...
 
  • #13
I believe it is just a statement of Bayes theorem actually. Part of stuff worth knowing.
 
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  • #14
hutchphd said:
I believe it is just a statement of Bayes theorem actually. Part of stuff worth knowing.
Bayes rule for two events ##A,B##:
$$
P(A|B) = \frac{P(B|A)P(A)}{P(B)}
$$

So written for my events in my previous post:
$$
P(E_A | E) = \frac{P(E | E_A)P(E_A)}{P(E)}
$$

In order to obtain the desired shape given by @vanhees71 we need to have that ##P(E | E_A)=P(E_B)##. Does this make sense?
 
  • #15
It does to me. If it doesn't to you, tell us why.
 
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  • #16
So if I know a priori that there are formally three systems ##(E,N), (E_A, N_A), (E_B,N_B)## with ##E=E_A+E_B## and ##N=N_A+N_B##, and now I am given ##E_A##... I would say that ##P(E|E_A)=P(E_B|E_A)##.
If I now assume that ##E_B## and ##E_A## are independent, then yes, we have ##P(E|E_A)=P(E_B)##.

But couldn't I also assume that ##E## and ##E_A## are independent? Then I had ##P(E)=P(E_B|E_A)##...
How do I know which independency to assume?
 
  • #17
You choose them to suit your needs. But the constraint holds regardless.
 
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  • #18
Right... I understand that both the quations ##P(E|E_A)=P(E_B)## and ##P(E)=P(E_B|E_A)## make sense under the assumption that the constraint ##E=E_A+E_B## holds. I am just not used to being able to "choose" between two independencies. But it makes sense that if I have three systems and only a single constraint, I can always choose two of them to be independent of each other.
 

FAQ: Calculate the distribution of energies from momenta

1. What is the relationship between energy and momentum?

The relationship between energy and momentum is described by the equation E = pc, where E is energy, p is momentum, and c is the speed of light. This equation is known as the relativistic energy-momentum relation and is a fundamental principle in physics.

2. How do you calculate the distribution of energies from momenta?

The distribution of energies from momenta can be calculated using the relativistic energy-momentum relation, E = pc. This equation allows us to determine the energy associated with a given momentum value, and by considering a range of momentum values, we can create a distribution of energies.

3. What factors affect the distribution of energies from momenta?

The distribution of energies from momenta is affected by several factors, including the mass of the particles, their velocity, and the angle at which they are moving relative to the observer. These factors can impact the overall shape and spread of the energy distribution.

4. How is the distribution of energies from momenta used in scientific research?

The distribution of energies from momenta is used in a variety of scientific research, particularly in the fields of particle physics and astrophysics. By understanding the distribution of energies, scientists can gain insights into the properties and behavior of particles and objects at the subatomic and cosmic levels.

5. Can the distribution of energies from momenta be applied to everyday situations?

While the concept of energy and momentum may seem abstract, the distribution of energies from momenta can be observed and applied in everyday situations. For example, the distribution of energies can be seen in the behavior of light and other electromagnetic waves, as well as in the motion of objects in our daily lives.

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