10.5: Appendix - Jacobians

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Appendix: Jacobians

A.1. 3D to 1D

A Jacobian is a mathematical entity used to switch partials inside an integral. For the case of velocity in three dimensions being converted into spherical coordinates, where \(v\) is the net velocity and akin to a sphere’s radius:

\[\partial v_x\cdot \partial v_y\cdot \partial v_z=\left|\frac{\partial (v_x,v_y,v_z)}{\partial (v,\theta ,\phi )}\right|\cdot \partial v\cdot \partial \phi \cdot \partial \theta =v^2\cdot

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sin\ (phi )\cdot \partial v\cdot \partial \phi \cdot \partial \theta \nonumber \]

To evaluate the above, we should have some idea how to convert v\({}_{x}\) v\({}_{y}\) v\({}_{z}\) into spherical coordinates to begin with. It’s just a lesson in geometry:

\[v_x=v\cdot

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sin\ (phi )\cdot

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cos\ (theta )v_y=v\cdot

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sin\ (phi )\cdot

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sin\ (theta )v_z=v\cdot

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cos\ (phi ) \nonumber \]

where \(v=\sqrt{v^2_x+v^2_y+v^2_z}\). The Jacobian \(\left|\frac{\partial (v_x,v_y,v_z)}{\partial (v,\theta ,\phi )}\right|\) is the absolute values of the determinant of the following matrix:

\[\left|

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detleft[\frac{∂v_x}{\partial v}frac{∂v_x}{\partial \theta }frac{∂v_x}{\partial \varphi }frac{∂v_y}{\partial v}frac{∂v_y}{\partial \theta }frac{∂v_y}{\partial \varphi }frac{∂v_z}{\partial v}frac{∂v_z}{\partial \theta }frac{∂v_z}{\partial \varphi }right]right| \nonumber \]

In our coordinate system the absolute value of the determinant of the matrix is:

\[\left|

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detleft[

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sin\ (phi )

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cos\ (theta )\ -v

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sin\ (phi )

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sin\ (theta )\ v

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cos\ (phi )

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cos\ (theta )\

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sin\ (phi )

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sin\ (theta )\ v

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sin\ (phi )

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cos\ (theta )\ v

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cos\ (phi )

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sin\ (theta )\

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cos\ (phi )\ 0\ -v

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sin\ (phi )right]right| \nonumber \]

\[=\left|-v^2(phi )(theta )-v^2

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sin\ (phi )(phi )(theta )+0-v^2(phi )(theta )-0-v^2

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sin\ (phi )(phi )(theta )\right|=\left|-v^2(phi )((theta )+(theta ))-v^2

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sin\ (phi )(phi )((theta )+(theta ))\right|=\left|-v^2(phi )-v^2

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sin\ (phi )(phi )\right|=\left|-v^2

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sin\ (phi )((phi )+(phi ))\right|=\left|-v^2

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sin\ (phi )\right| \nonumber \]

Since we take the absolute value of this result (the negative sign goes away), the final answer is: \(\partial v_x\cdot \partial v_y\cdot \partial v_z=v^2\cdot

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sin\ (phi )\cdot \partial v\cdot \partial \phi \cdot \partial \theta\)

When we apply the Jacobian to the Maxwell-Boltzmann formula we see that (bold emphasis added):

\[{\left(\frac{m}{2\cdot \pi \cdot k_B\cdot T}\right)}^{\frac{3}{2}}\cdot e^{\frac{-m\cdot v^2}{2\cdot k_B\cdot T}}\cdot \partial v_x\cdot \partial v_y\cdot \partial v_z={\left(\frac{m}{2\cdot \pi \cdot k_B\cdot T}\right)}^{\frac{3}{2}}\cdot e^{\frac{-m\cdot v^2}{2\cdot k_B\cdot T}}\cdot v^2\cdot

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sin\ (phi )\cdot \partial v\cdot \partial \phi \cdot \partial \theta \nonumber \]

Now we can integrate out the angles:

\[{\left(\frac{m}{2\cdot \pi \cdot k_B\cdot T}\right)}^{\frac{3}{2}}\cdot e^{\frac{-m\cdot v^2}{2\cdot k_B\cdot T}}\cdot v^2\cdot

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sin\ (phi )\cdot \partial v\cdot \int^{2\pi }_0{}

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sin\ (phi )\cdot \partial \phi \cdot \int^{\pi }_0{}\partial \theta \nonumber \]

which leaves us with just a constant:

\[\int^{\pi }_0{}

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sin\ (phi )\cdot \partial \phi \cdot \int^{2\pi }_0{}\partial \theta ={-

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cos\ (phi )]}^{\pi }_0\cdot 2\pi =4\pi \nonumber \]

leaving us with:

\[4\pi \cdot v^2\cdot {\left(\frac{m}{2\cdot \pi \cdot k_B\cdot T}\right)}^{\frac{3}{2}}\cdot e^{\frac{-m\cdot v^2}{2\cdot k_B\cdot T}}\cdot \partial v \nonumber \]

This is the velocity form of the Maxwell-Boltzmann equation.

A.2.

\(\partial v_a\cdot \partial v_b\to \partial G\cdot \partial V_{rel}\): The following Jacobian was used in the calculation of relative velocity:

\[\partial v_a\cdot \partial v_b=\left|\frac{\partial (v_a,v_b)}{\partial (G,V_{rel})}\right|\cdot \partial G\cdot \partial V_{rel} \nonumber \]

The determinant is:

\[\left|\frac{\partial (v_a,v_b)}{\partial (G,V_{rel})}\right|=\left|det\left[\frac{\partial v_a}{\partial V_{rel}}frac{\partial v_a}{\partial G}frac{\partial v_b}{\partial V_{rel}}frac{\partial v_b}{\partial G}right]\right| \nonumber \]

Given that \(v_a=G+\frac{m_b}{m_a+m_b}V_{rel}\) and \(v_b=G-\frac{m_a}{m_a+m_b}V_{rel}\), we can fill out the matrix:

\[\left|det\left[\frac{m_b}{m_a+m_b}\ 1\ -\frac{m_a}{m_a+m_b}\ 1right]\right|=\frac{m_b}{m_a+m_b}+\frac{m_a}{m_a+m_b}=\frac{m_a+m_b}{m_a+m_b}=1 \nonumber \]

Done!

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