Change from vector coordinates in the integral

[n4t3]

I have an issue with a transformation of the variable of integration. The variable of the original problem is the vector distance and it is transformed in order to express the integral in terms of the unit vector and the scalar distance r12. However, I didn't understand the transformation made by the authors. A new integral popped up along with r12 squared and two new variables of integration. Can you guys help me?

 

[mario99]

If you post a picture of the original problem, I may be able to help you. My help is not guaranteed, but I have worked a lot in integrals involving vectors.

 

[n4t3]

If you post a picture of the original problem, I may be able to help you. My help is not guaranteed, but I have worked a lot in integrals involving vectors.

Hi, Mario

The equation comes from a paper (Equation 13 of https://web.archive.org/web/20180722081616id_/https://hal.archives-ouvertes.fr/hal-00513176/document). There are no pictures to illustrate the problem. Someone told me this is a problem of change in the coordinates from cartesian to spherical which should look like:

integration in dvec{r} [cartesian] = integration in r^2*sin(θ)drdθdφ [polar]

where sin(θ)drdθdφ is the diferential of the "solid angle", Ω, and then dΩ = sin(θ)drdθdφ = dversor{r}.

But I'm not sure if this makes sense.

 

[blamocur]

The notation looks very strange to me: if you integrate vectors ([imath]d\overrightarrow{r_{12}}[/imath])shouldn't your result be a vector too? Without having a clue about the subject of the article I'll venture a guess that by vector differential ([imath]d\overrightarrow{r_{12}}[/imath]) they mean volume differential [imath]dV[/imath], and their differential of unit vector [imath]d\hat{r_{12}}[/imath] is actually area differential [imath]dS[/imath] on a unit sphere; or we can probably call it angular differential meaning spacial angles. In such scenario [imath]dV = r^2 dS[/imath] makes sense to me, but take this with a grain of salt.

 

[mario99]

I came to a similar idea of professor blamocur, but a little different.

[imath]V_{\text{exc}}[/imath] is the excluded volume between two molecules. I will include a picture to illustrate that volume.



Vector [imath]\displaystyle \bold{r_{12}}[/imath] goes from the center [imath]A[/imath] of the first molecule to the center [imath]B[/imath] of the second molecule (or reverse). The volume differential element [imath]d\bold{r_{12}}[/imath] can be treated like [imath]d\bold{V}[/imath]. It is complicated to illustrate this, but [imath]d\bold{r_{12}}[/imath] must be a very small volume of that excluded volume, its direction is going from the center [imath]B[/imath] toward the center [imath]A[/imath] (or from the circumeference of [imath]A[/imath] molecule toward the center [imath]B[/imath] horizontally).

If we think of the volume differential element [imath]d\bold{r_{12}}[/imath] like [imath]d\bold{V}[/imath], then a very small volume of a sphere is equal to [imath]4\pi r^2 \ dr[/imath], but we know the radius of our sphere is [imath]|\bold{r_{12}}| = r_{12}[/imath], then we have:

[imath]d\bold{r_{12}} = 4\pi r^2_{12} \ dr_{12}[/imath]

But [imath]d\bold{r_{12}}[/imath] is a vector, so we must give a direction to it, then we have:

[imath]d\bold{r_{12}} = 4\pi r^2_{12} \ dr_{12} \ d\hat{r}_{12}[/imath]

The unit vector [imath]d\hat{r}_{12}[/imath] means that the direction of this differential element is along the centers [imath]A[/imath] and [imath]B[/imath] of the two molecules.

And since I took an arbitrary differential element [imath]d\bold{V}[/imath], I included the [imath]4\pi[/imath]. The book must has worked in a specific coordinate system, which not very obvious to me, so they should get the [imath]4\pi[/imath] during integration.

My final guess is that, the volume differential element is:

[imath]d\bold{r_{12}} = r^2_{12} \ dr_{12} \ d\hat{r}_{12}[/imath], and the [imath] 4\pi[/imath] will eventually appear from calculating the extra stuff inside [imath]V_{\text{exc}}[/imath].

 

[n4t3]

The notation looks very strange to me: if you integrate vectors ([imath]d\overrightarrow{r_{12}}[/imath])shouldn't your result be a vector too? Without having a clue about the subject of the article I'll venture a guess that by vector differential ([imath]d\overrightarrow{r_{12}}[/imath]) they mean volume differential [imath]dV[/imath], and their differential of unit vector [imath]d\hat{r_{12}}[/imath] is actually area differential [imath]dS[/imath] on a unit sphere; or we can probably call it angular differential meaning spacial angles. In such scenario [imath]dV = r^2 dS[/imath] makes sense to me, but take this with a grain of salt.

Thank you for your answer! The subject of the article is about the calculation of the excluded volume of a pair of nonconvex hard particles with different orientations. This calculation involves integrating over the whole angular phase space. For linear particles, this integration can be done in spherical coordinates instead of using Euler angles, for instance, which would be the case for nonlinear particles. However, the contact distance between two nonconvex hard particles, which is necessary to calculate the excluded volume, is not written in terms of the spherical coordinates but in terms of the orientation of the vector distance (versor_{r12} joining the centers of mass of both particles. And this is the problem I can't figure out: does the differential of the versor r12 have the same meaning as sin(theta) dtheta dphi (at least for linear particles)? Anyways, your answer makes sense to me: dV = r^2*dS, which means dS = sin(theta) dtheta dphi = dversor{r_12}.

 

[n4t3]

I came to a similar idea of professor blamocur, but a little different.

[imath]V_{\text{exc}}[/imath] is the excluded volume between two molecules. I will include a picture to illustrate that volume.

View attachment 38110

Vector [imath]\displaystyle \bold{r_{12}}[/imath] goes from the center [imath]A[/imath] of the first molecule to the center [imath]B[/imath] of the second molecule (or reverse). The volume differential element [imath]d\bold{r_{12}}[/imath] can be treated like [imath]d\bold{V}[/imath]. It is complicated to illustrate this, but [imath]d\bold{r_{12}}[/imath] must be a very small volume of that excluded volume, its direction is going from the center [imath]B[/imath] toward the center [imath]A[/imath] (or from the circumeference of [imath]A[/imath] molecule toward the center [imath]B[/imath] horizontally).

If we think of the volume differential element [imath]d\bold{r_{12}}[/imath] like [imath]d\bold{V}[/imath], then a very small volume of a sphere is equal to [imath]4\pi r^2 \ dr[/imath], but we know the radius of our sphere is [imath]|\bold{r_{12}}| = r_{12}[/imath], then we have:

[imath]d\bold{r_{12}} = 4\pi r^2_{12} \ dr_{12}[/imath]

But [imath]d\bold{r_{12}}[/imath] is a vector, so we must give a direction to it, then we have:

[imath]d\bold{r_{12}} = 4\pi r^2_{12} \ dr_{12} \ d\hat{r}_{12}[/imath]

The unit vector [imath]d\hat{r}_{12}[/imath] means that the direction of this differential element is along the centers [imath]A[/imath] and [imath]B[/imath] of the two molecules.

And since I took an arbitrary differential element [imath]d\bold{V}[/imath], I included the [imath]4\pi[/imath]. The book must has worked in a specific coordinate system, which not very obvious to me, so they should get the [imath]4\pi[/imath] during integration.

My final guess is that, the volume differential element is:

[imath]d\bold{r_{12}} = r^2_{12} \ dr_{12} \ d\hat{r}_{12}[/imath], and the [imath] 4\pi[/imath] will eventually appear from calculating the extra stuff inside [imath]V_{\text{exc}}[/imath].

I guess this "4*pi" comes from the integration of the solid angle in spherical coordinates: d\Omega = sin(theta) dtheta dphi. You are integrating theta from 0 to pi, and phi from 0 to 2*pi, which will give you 4*pi. Now, this will always hold true if you integrate dversor{r_12} for spheres, since the contact distance \sigma (last line of equation 13) will not depend on the orientation of the vector distance joining the centers of mass of two spheres. For other nonconvex hard particles, this integral will involve algebraic relations with the surface area, mean radius of curvature and volume, all given by the Isihara-Hadwiger theorem, which is not really my field of expertise =(