Cancellation of terms, Kittel (first edition) equation 6.4

[Irishdoug]

Homework Statement

I am self studying and cannot understand a cancellation of terms that occurs in the text.

Relevant Equations

In thermal equilibrium at temperature T, with N particles all of mass N, the internal energy is equal to the kinetic energy i.e. U = E

$$E = \frac{1}{2M} \sum_{i=1}^{N} (p_{xi}^{2} + p_{yi}^{2} + p_{zi}^{2})$$

$$E_{average} = U = \frac{\frac{1}{2M} \sum_{\alpha = 1}^{3N} p_{\alpha}^{2} \int_{-\infty}^{\infty} \Pi_{\beta = 1}^{3N} e^{\frac{-p_{\beta}^{2}}{2M k_{b}T} }dp}{\int_{-\infty}^{\infty} \Pi_{\beta = 1}^{3N} e^{\frac{-p_{\beta}^{2}}{2M k_{b}T} }dp}$$

as we know from classical physics that the average value of a quantity A in thermal equilibrium is

$$A_{average} = \frac{A(p,q) \int_{-\infty}^{\infty} e^{\frac{-E}{k_{b}T} }dpdq} {\int_{-\infty}^{\infty} e^{\frac{-E}{k_{b}T} }dpdq}$$

I'm happy I understand why the dq terms cancel out. So,

$$E_{average} = U = \frac{\frac{1}{2M} \sum_{\alpha = 1}^{3N} p_{\alpha}^{2} \int_{-\infty}^{\infty} \Pi_{\beta = 1}^{3N} e^{\frac{-p_{\beta}^{2}}{2M k_{b}T} }dp}{\int_{-\infty}^{\infty} \Pi_{\beta = 1}^{3N} e^{\frac{-p_{\beta}^{2}}{2M k_{b}T} }dp}$$

This term reduces to

$$E_{average} = U = \frac{\frac{1}{2M} \int_{-\infty}^{\infty} p_{\alpha}^{2} e^{\frac{-p_{\alpha}^{2}}{2M k_{b}T}} dp_{\alpha}} {\int_{-\infty}^{\infty} e^{\frac{-p_{\alpha}^{2}}{2M k_{b}T}} dp_{\alpha}}$$

I do not see how this occurs. I have attempted to play around with the exponentials e.g. using $e^{a-b} = \frac{e^{a}}{e^{b}}$ but here I am left with $e^{0}$. I have also explicitedly written out the summation and multiplication but it does not seem to help.

Any help or tips appreciated.

 

[BvU]

Homework Statement:: I am self studying and cannot understand a cancellation of terms that occurs in the text.
Relevant Equations:: In thermal equilibrium at temperature T, with N particles all of mass N, the internal energy is equal to the kinetic energy i.e. U = E

No help so far, so I'll give it a first shot (don't have the full answer - can't find your source; 1st ed ? Thermal physics, or solid state physics?)

You really want to be as accurate as possible when doing things like this, so check what you wrote. You mean
temperature $T$, with $N$ particles all of mass $\bf M$.

$$E = \frac{1}{2M} \sum_{i=1}^{N} (p_{xi}^{2} + p_{yi}^{2} + p_{zi}^{2}) \tag 1 $$

So this is the total energy for $N$ particles in three dimensions. Is that what you meant ? Otherwise a 'small' correction is needed ($\frac 1 N$) .

$$E_{average} = U = \frac{\frac{1}{2M} \sum_{\alpha = 1}^{3N} p_{\alpha}^{2} \int_{-\infty}^{\infty} \Pi_{\beta = 1}^{3N} e^{\frac{-p_{\beta}^{2}}{2M k_{b}T} }dp}{\int_{-\infty}^{\infty} \Pi_{\beta = 1}^{3N} e^{\frac{-p_{\beta}^{2}}{2M k_{b}T} }dp}$$

No, that can't be the case, because

as we know from classical physics that the average value of a quantity A in thermal equilibrium is
$$A_{average} =
\frac{A(p,q) \int_{-\infty}^{\infty} e^{\frac{-E}{k_{b}T} }dpdq}
{\int_{-\infty}^{\infty} e^{\frac{-E}{k_{b}T} }dpdq}
$$

is incorrect. Do you mean$$
A_{average} =
\frac{\int_{-\infty}^{\infty} A(p,q) \ e^{-\frac{E}{kT} }dp\,dq}
{\int_{-\infty}^{\infty} e^{-\frac{E}{kT} }dp\,dq}\quad ?$$

I'm happy I understand why the dq terms cancel out.

Good. We note that $dp\,dq$ is shorthand for $\ d^{3N}p\, d^{3N}q\$ and the $(p, q)$ in $A(p, q)$ is shorthand for $(p_1, \ldots, p_{3N},\,q_1, \ldots ,q_{3N})\$ and, if $A(p,q) = A(p)$ , the integrations over $dq_i$ can be taken out and integrated (namely to what value ?), and then the factors in numerator and denominator cancel.

So, you want to look at $$
E_{average} =
\frac{{\displaystyle \int_{-\infty}^{\infty}} E \ e^{-\frac{E}{kT} }dp\,dq}
{{\displaystyle \int_{-\infty}^{\infty}} e^{-\frac{E}{kT} }dp\,dq}$$
with $E$ as in $(1)$, independent of $q$. We've seen we can drop the $q$.

With an eye on the next equation it looks as if the expectation value of the sum is the sum of the expectation values. Is that a reasonable phrasing of your problem ? i.e. swapping the $\sum$ and the $\int$ ?

$\$

 

[Irishdoug]

Hi BvU. Thankyou for the response. I'm using Kittel's solid state book. Sorry I thought he only had that one. And yes I have the first edition (and the 7th!). I feel from a cursory glance he goes through the maths in a more understandable way in the first, at least in some sections. I am busy today so will get back to going through your response in more detail later, where I may or may not have more questions.

'No, that can't be the case, because'

Yes that was an error I meant to write what you wrote!

Thanks again.

 

[Irishdoug]

No help so far, so I'll give it a first shot (don't have the full answer - can't find your source; 1st ed ? Thermal physics, or solid state physics?)

You really want to be as accurate as possible when doing things like this, so check what you wrote. You mean
temperature $T$, with $N$ particles all of mass $\bf M$.

So this is the total energy for $N$ particles in three dimensions. Is that what you meant ? Otherwise a 'small' correction is needed ($\frac 1 N$) .

No, that can't be the case, because

is incorrect. Do you mean$$
A_{average} =
\frac{\int_{-\infty}^{\infty} A(p,q) \ e^{-\frac{E}{kT} }dp\,dq}
{\int_{-\infty}^{\infty} e^{-\frac{E}{kT} }dp\,dq}\quad ?$$Good. We note that $dp\,dq$ is shorthand for $\ d^{3N}p\, d^{3N}q\$ and the $(p, q)$ in $A(p, q)$ is shorthand for $(p_1, \ldots, p_{3N},\,q_1, \ldots ,q_{3N})\$ and, if $A(p,q) = A(p)$ , the integrations over $dq_i$ can be taken out and integrated (namely to what value ?), and then the factors in numerator and denominator cancel.

So, you want to look at $$
E_{average} =
\frac{{\displaystyle \int_{-\infty}^{\infty}} E \ e^{-\frac{E}{kT} }dp\,dq}
{{\displaystyle \int_{-\infty}^{\infty}} e^{-\frac{E}{kT} }dp\,dq}$$
with $E$ as in $(1)$, independent of $q$. We've seen we can drop the $q$.

With an eye on the next equation it looks as if the expectation value of the sum is the sum of the expectation values. Is that a reasonable phrasing of your problem ? i.e. swapping the $\sum$ and the $\int$ ?

$\$

$$A_{average} =
\frac{\int_{-\infty}^{\infty} A(p,q) \ e^{-\frac{E}{kT} }dp\,dq}
{\int_{-\infty}^{\infty} e^{-\frac{E}{kT} }dp\,dq}\quad ?$$

Yes I meant this indeed.

So I'm viewing it as follows:

p = $$(p_{x1}, p_{y1}, p_{z1} ; p_{x2}, p_{y2}, p_{z2}... p_{xN}, p_{yN}, p_{zN})$$
so
dp = $$(dp_{x1}, dp_{y1}, dp_{z1} ; dp_{x2}, dp_{y2}, dp_{z2}... dp_{xN}, dp_{yN}, dp_dp_{zN})$$

Let's assume N = 1 and for $alpha = 1$ we have

$$ U = \frac{\frac{1}{2M} \int_{-\infty}^{+\infty} (p_{x1}^2 + p_{y1}^2 + p_{z1}^2) (e^{c p_{x1}^2}.e^{c p_{y1}^2}.e^{c p_{z1}^2}.e^{c p_{x2}^2}.e^{c p_{y2}^2}.e^{c p_{z2}^2}.e^{c p_{x3}^2}.e^{c p_{y3}^2}.e^{c p_{z3}^2}) dp_{x1}dp_{y1}dp_{z1}} {\int_{-\infty}^{+\infty} (e^{c p_{x1}^2}.e^{c p_{y1}^2}.e^{c p_{z1}^2}.e^{c p_{x2}^2}.e^{c p_{y2}^2}.e^{c p_{z2}^2}.e^{c p_{x3}^2}.e^{c p_{y3}^2}.e^{c p_{z3}^2} dp_{x1}dp_{y1}dp_{z1} }$$

where $$c = \frac{1}{2Mk_{B}T}$$. We can now split this into 3 different integrals e.g

I can't get this to format, but all the constant terms i.e those not dependent on $dp_{x1}$ come outside in the first integral, those that don't depend on $dp_{x2}$ come outside etc.The terms can be seen to cancel leaving
$$\frac{\int_{-\infty}^{\infty} p_{x1}^2 e^{c p_{x1}^2} dp_{x1}}{\int_{-\infty}^{\infty}e^{c p_{x1}^2} dp_{x1}} etc.$$

Does this sound reasonable?

the integrations over $dq_i$ can be taken out and integrated (namely to what value ?)

I'm thinking 0 but maybe this would cause the whole eqaution to be 0. I will have to think about it more!

 

[BvU]

Does this sound reasonable?

I think that's the idea, yes.

I'm thinking 0

Nope. $\int_{q_{min}}^{q_{max}} {\rm dq}\$ is simply ${q_{max}}-{q_{min}}$ so the lot of them multiplied with each other give you a volume. Which cancels against idem in the denominator.Disclaimer: I'm a little out of my depth here - based on memories from student-times long ago. Perhaps @vanhees71 can shed some more didactically responsible light on this ... ?

$\$

 

[Irishdoug]

I think that's the idea, yes.Nope. $\int_{q_{min}}^{q_{max}} {\rm dq}\$ is simply ${q_{max}}-{q_{min}}$ so the lot of them multiplied with each other give you a volume. Which cancels against idem in the denominator.Disclaimer: I'm a little out of my depth here - based on memories from student-times long ago. Perhaps @vanhees71 can shed some more didactically responsible light on this ... ?

$\$

Well what you have said sounds more reasonable. The integral is given as $\int_{\-infty}^{\infty}$ which is why I said what I did.