# A question about 3D integrals.

1. Oct 7, 2011

### Beer-monster

1. The problem statement, all variables and given/known data

If we have volume integral of a Gaussian function, in phase space for example.

$$F= \int^{\infty}_{-\infty} e^{-aq} d^{3}q$$

Now, I think the the answer would be the standard answer for a Gaussian integral cubed wouldn't it?

$$F=\left(\frac{\pi}{a}\right) ^{3/2}$$

I was wondering if this answer could apply in general to all 3d volumes regardless of their coordinate system. For example a system with cylindrical symmetry

$$C = \int^{\infty}_{-\infty} e^{-aq} rdr d \phi dz$$

Would this have a similar answer? If not can it be found if I don't explicitly know the radial, angular or vertical components of the displacements q?

2. Oct 8, 2011

### vela

Staff Emeritus
q should be squared in the exponential.

Yes. You should be able to see it by inspection by writing $q^2 = q_x^2+q_y^2+q_z^2$.
Your question isn't very clear. Are you asking if you integrate the same function over the same volume, if you will get the same answer regardless of which coordinate system you use to evaluate the integral? If so, then yes, of course you would. The answer can't depend on which coordinate system you choose.

If instead you're saying $q = f(\vec{r})$, then no. Why would you expect the same answer when you're integrating a different function?

If you're saying $\vec{q}$ is the vector specified by cylindrical coordinates (r, θ, z) or spherical coordinates (r, θ, Φ), you should find it straightforward to express q2 in terms of the relevant variables.

3. Oct 9, 2011

### Beer-monster

Sorry about the late reply. Lets see if I can be more specific.

We were taught in Stat Mech that if we want to calculate the partition function of a gas of N particles with an arbitrary potential (U(q)) we can use a phase space integration. The process shown in most texbooks gives the following result as a general solution:

$$Z= \frac{1}{N!h^{3N}}\int \Pi_{i} d^{3}pd^{3}q e^{-\beta \Sigma_{i} \frac{p_{i}^{2}}{2m} -\beta \Sigma_{i} U(q_{i})} = \frac{1}{N!\lambda^{3}}Z_{c}$$

Where $\lambda$ is the Debroglie wavelength and Zc is the configurational integral (over the spacial degree of freedom).

So, I guess I'm wondering if this result applies in all cases. If we used a cylindrical coordinate system and assumed cylindrical symmetry then, I think, the momentum integral would become.

$$2 \pi L \int_{0}^{\infty} \Pi_{i} dp \: p e^{-\beta \Sigma_{i} \frac{p_{i}^{2}}{2m}}$$

Which has a very different result (no sqrt factors) from the first method, although, as far as I can tell, we only switched coordinate systems.

Incidently, I have a question about the standard result I quoted at the start of this post. You can see it separately http://www.pma.caltech.edu/~mcc/Ph127/b/Lecture1.pdf" [Broken] in equation (3).

Basically, I can't follow what happened to the sum in the exponent. I believe the powers of N in the final result arsis from the product term (big pi) which would leave N multiples of the gaussian integral. However, I can't see here how the sum was dealt with, though I believe it should have been treated as a product itself (sum in exponential = geometric series of exponential functions).

Last edited by a moderator: May 5, 2017
4. Oct 9, 2011

### vela

Staff Emeritus
If we look at the momentum for just one particle, you get
\begin{align*}
\iiint dp_x\,dp_y\,dp_z \exp\left(-\beta\frac{p_x^2+p_y^2+p_z^2}{2m}\right)
&= \iiint dp_x\,dp_y\,dp_z \exp\left(-\beta \frac{p_x^2}{2m}\right) \exp\left(-\beta \frac{ p_y^2}{2m}\right) \exp\left(-\beta \frac{ p_z^2}{2m}\right) \\
&= \int dp_x\,\exp\left(-\beta \frac{ p_x^2}{2m}\right) \int dp_y\,\exp\left(-\beta \frac{ p_y^2}{2m}\right) \int dp_z\,\exp\left(-\beta \frac{ p_z^2}{2m}\right) \\
&= \left[\int dp\,\exp\left(-\beta \frac{p^2}{2m}\right)\right]^3
\end{align*}
That's what's happening to the summation.

In cylindrical coordinates, you need to use the fact that $p_x^2+p_y^2+p_z^2 = p^2+p_z^2$, so the integral becomes
\begin{align*}
\iiint dp_x\,dp_y\,dp_z \exp\left(-\beta\frac{p_x^2+p_y^2+p_z^2}{2m}\right)
&= \iiint dp\,d\theta\,dp_z\, p\exp\left(-\beta \frac{p^2}{2m}\right) \exp\left(-\beta \frac{ p_z^2}{2m}\right)
\end{align*}
In spherical coordinates, you have $p_x^2+p_y^2+p_z^2 = p^2$, and you get
\begin{align*}
\iiint dp_x\,dp_y\,dp_z \exp\left(-\beta\frac{p_x^2+p_y^2+p_z^2}{2m}\right)
&= \iiint dp\,d\phi\,d\theta\, p^2\sin\theta\,\exp\left(-\beta \frac{p^2}{2m}\right)
\end{align*}
I'll leave it to you to convince yourself you get the same answer regardless of which coordinate system you use.

5. Oct 9, 2011

### Beer-monster

I think I understand the second part, but Im still not sure about the summation.

Firstly, doesn't this only apply if $p_{x}=p_{y}=p_{z}=p$

Also, if we consider N particles shouldn't we have an extra power of N.

$$\exp\left(-\beta \frac{p_{1}^{2}...p_{2}^{2}...p_{n}^{2}}{2m}\right) =\exp\left(-\beta \frac{p^{2}}{2m}\right)^{N}=\exp\left(-\beta N \frac{p^{2}}{2m}\right)$$

Making the integral:

$$\left[\int dp\,\exp\left(-\beta N \frac{p^2}{2m}\right)\right]^3$$

However, this extra factor of N doesn't appear in the textbooks so I wonder where it has gone?

Last edited: Oct 9, 2011
6. Oct 9, 2011

### vela

Staff Emeritus
Yes, you typically assume the momentum is distributed isotropically, so each component can be treated identically.
You can't simplify the exponential the way you did because of the integrations. For example, in the one particle case, you get the integral cubed, not the integral of the cube. You have the same thing in the N particle case. You get the integral to the power of 3N, not what you wrote.

7. Oct 9, 2011

### Beer-monster

Thanks again. But could I go over that last part again, just so I'm clear.

I understand that the cube of integral and the cube of the exponential are not the same thing. However, by rewriting the sum in the exponent as a series of products like so:

$$\exp\left(-\beta \Sigma^{N}_{i} \frac{p_{i}^{2}}{2m}\right ) = \exp\left(-\beta \frac{p_{1}^{2}+p_{2}^{2}+...p_{N}^{2}}{2m}\right ) = \exp\left(-\beta \frac{p_{1}^{2}}{2m}\right )\exp\left(-\beta \frac{p_{2}^{2}}{2m}\right )\exp\left(-\beta \frac{p_{3}^{2}}{2m}\right )....\exp\left(-\beta \frac{p_{N}^{2}}{2m}\right )$$

And using the isotropic distribution of momentum we already covered:

$$p_{x}=p_{y}=p_{z}=p$$

Shouldn't we get:

$$\exp\left(-\beta \frac{p^{2}}{2m}\right)^{N}$$

Which we would then integrate and the gaussian integral, including a factor of N, to the power of 3?

I guess I'm saying, that by doing this we're not changing the function to be integrated (are we?) but just simplifying it, by realising the sum in the exponential can be expressed as a product of N exponential functions.

I'm sure I'm wrong I'm just not seeing how yet and would like to know what I'm missing :D.

Gievn that I am wrong. Does this mean that other expressions should be treated by the integration before simplication. Or am I just not seeing how what I'm doing above is beyond simplifying the term.

For example the expression of the form:

$$\int \exp\left(-\beta \Sigma_{r}^{N} A ln(\frac{x_{r}}{B})\right) dp$$

Is it okay (if x is isotropic) to resolve use the exponential to cancel the logarithm before integrating i.e. rewriting as

$$\int -\beta \frac{x}{B}^{-\beta NA} dx$$

Last edited: Oct 9, 2011
8. Oct 9, 2011

### vela

Staff Emeritus
You can't set p=px=py=pz before the integration. That would mean the components are correlated, which they are not. That would pick out (1,1,1) as a preferred direction. It's not consistent with isotropy.

But the integral over px is equal to the integral over, say, py because of isotropy. There's no difference between integrating over the x component and the y component because everything looks the same regardless of which direction you look.

9. Oct 9, 2011

### Beer-monster

So even though the exponent can be represented as a multiple of N exponential functions, they are not all equivalent (as px1 does not equal px2 does not equal py etc). This means I can't roll everything into one exponential. However, the integral of p can be considered isotropic and then I can roll them together?

Does that make sense?

In that case what roll does the product sequence have in this as I originally thought the power of N comes from that?

Sorry, to keep badgering you about this but it seems like an important bit of maths and I want to make sure I understand it fully,

10. Oct 10, 2011

### vela

Staff Emeritus
I shouldn't have agreed with you earlier when you asked about p=px=py=pz. The components aren't equal. On the other hand, the components "look" the same. This is reflected in the fact that they appear in the integral in the same way, which is why you can do what I do below.

I assume you understand this part:
$$\iiint dp_x\,dp_y\,dp_z \exp\left(-\beta \frac{p_x^2}{2m}\right) \exp\left(-\beta \frac{ p_y^2}{2m}\right) \exp\left(-\beta \frac{ p_z^2}{2m}\right) = \int dp_x\,\exp\left(-\beta \frac{ p_x^2}{2m}\right) \int dp_y\,\exp\left(-\beta \frac{ p_y^2}{2m}\right) \int dp_z\,\exp\left(-\beta \frac{ p_z^2}{2m}\right)$$
So now px is simply a dummy variable in the first integral, right? So I can change the name:
$$\int dp_x\,\exp\left(-\beta\frac{p_x^2}{2m}\right) = \int dp\,\exp\left(-\beta\frac{p^2}{2m}\right)$$
Similarly, I can do the same thing with py and pz:
\begin{align*}
\int dp_y\,\exp\left(-\beta\frac{p_y^2}{2m}\right) &= \int dp\,\exp\left(-\beta\frac{p^2}{2m}\right) \\
\int dp_z\,\exp\left(-\beta\frac{p_z^2}{2m}\right) &= \int dp\,\exp\left(-\beta\frac{p^2}{2m}\right)
\end{align*}So if we let
$$I=\int dp\,\exp\left(-\beta \frac{p^2}{2m}\right)$$we have
$$\iiint dp_x\,dp_y\,dp_z \exp\left(-\beta \frac{p_x^2}{2m}\right) \exp\left(-\beta \frac{ p_y^2}{2m}\right) \exp\left(-\beta \frac{ p_z^2}{2m}\right) = I \times I \times I = I^3$$
With N particles, you have 3N identical integrals, so the result is I3N.

I assume by "product sequence", you're referring to the $\Pi_i$. That's just there to say you're integrating over $d^3p_1\,d^3p_2\dots d^3p_n$.

11. Oct 10, 2011

### Beer-monster

I think I get it now. Even though the components are not equal the form of their role in the functions are the same, so we can solve the integral for the general form rather than the specific components. Then we even the books, by noting there are three of these functions in a product , leading to the integral raised to the 3rd power.

With regards to the $\Pi_{i}$, I'm reading your reply as meaning that it is there as a notational convenience and it does not actually add anything (such as an additional power of N) to the result?

Now that we cannot say the components are equal/isotropic can I check how that would affect my hypothetical logarithmic function from post 7.

$$\int \exp\left(-\beta \Sigma_{r}^{N} A ln(\frac{x_{r}}{B})\right) dp$$

The sum of all of the logarithms would be expressed as a product inside the logarithm.

$$\Sigma_{r}^{N} A \ln(\frac{x_{r}}{B})=A \ln(\frac{x_{1} x_{2}...x_{N}}{B}$$

If I use the exponential to cancel out the logarithm I'd get.

$$-\beta \frac{x_{1}x_{2}...x_{N}}{B}^{-\beta NA}$$

If I then integrate with a dummy x I'd get:

$$\left( \int \frac{x}{B}^{-\beta A} dx \right) ^{N}$$

I tried to approach that based on my understanding of what you've tried to explain. Does is seem okay or am I missing something?