Suppose one has an integral of the following form, $$ \int \text{d} \Omega_{1} \text{d} \Omega_{2} f(\gamma). $$ Where gamma is the relative angle between $(\theta_1, \phi_1)$ and $(\theta_2, \phi_2)$, $$ \cos \gamma = \cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2 \cos(\phi_1-\phi_2) $$ Based on symmetry arguments one could argue that as the integrand is only dependent the relative angle, we can fix $\Omega_1$, integrate over $\Omega_2$, and multiply the result with $4 \pi$ to compensate for fixing $\Omega_1$. Additionally we can argue that if we fix $\Omega_1$ along the $z$ axis the integration over $\phi_2$ just gives a factor $2 \pi$. So we have, $$ \int \text{d} \Omega_{1} \text{d} \Omega_{2} f(\gamma) = 8 \pi^{2} \int \sin \gamma \, \text{d}\gamma f(\gamma). $$ However I fail to make an explicit mathematical derivation of the above reasoning. If I try to write the solid angle differentials in terms of $\gamma$ i get quite lengthy and ugly looking expressions.
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I think you should explain what you consider "an explicit mathematical derivation". To me, your argument constitutes just that. – joriki May 18 '13 at 15:38
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I was just wondering how would one do this derivation without being "clever" and attempt a transformation of variables from $\theta_1 \phi_1, \theta_2 \phi_2$ to $\gamma + $ other variables that will be integrated out to give $8 \pi^2$. Because I do not see how you would do that. So without using symmetry arguments but using only basic transformations of variables. – camelthemammel May 18 '13 at 16:40
1 Answers
Ok, I found the answer. In case someone else has/will have the same problem I'm posting it.
Suppose we have an integral which looks like, \begin{align*} \int \textrm{d} \Omega_1 \int \textrm{d} \Omega_2 \, f( \gamma ), \end{align*} where $\gamma$ is the relative angle between $ \Omega_1$ and $ \Omega_2$ given by $\cos \gamma = \cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2 \cos (\phi_2 - \phi_1)$. Now we can make the following reasoning based on symmetry. As the integrand is only depend on the relative angle $y$ we can fix $(\theta_1, \phi_1)$ at the $z$-axis. The integration over $\phi_2$ simply gives $2 \pi$ while $\theta_2 \equiv \gamma$. Now to compensate for fixing $(\theta_1, \phi_1)$ we have to multiply the result with $4 \pi$. So we get, \begin{align*} \int \textrm{d} \Omega_1 \int \textrm{d} \Omega_2 \, f( \gamma ) = \int \textrm{d}\theta_1 \sin \theta_1 \int \textrm{d}\phi_1 \int \textrm{d} \sin \theta_2 \int \textrm{d} \phi_2 \, f( \gamma ) = 8 \pi^2 \int \textrm{d} \gamma \sin(\gamma) f(\gamma). \end{align*} But what about making this a hard proof using only transformations of variables? Let's begin. We introduce the following transformation of variables, \begin{align*} \{ \theta_1, \phi_1, \theta_2, \phi_2 \} \rightarrow \{ \theta_1, \phi_1, \gamma, \omega \} \end{align*} With $\gamma$ the relative angle and $\omega$ a rotation angle around the $ \theta_1, \phi_1$-direction (which obviously preserves $\gamma$, or $\gamma$ and $\omega$ are independent). First we construct the expressions for the new variables in terms of the old variables \begin{align*} \theta_1 &= \theta_1 \\ \phi_1 &= \phi_1 \\ \cos \gamma &= \cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2 \cos (\phi_2 - \phi_1) \\ \sin \omega &= \frac{ \sin \theta_2 \sin (\phi_2 - \phi_1)}{\sin \gamma} \end{align*} The Jacobian is given by, \begin{align*} \left( \begin{array}{c c c c} \frac{\partial \theta_1}{\partial \theta_1} & \frac{\partial \theta_1}{\partial \phi_1} & \frac{\partial \theta_1}{\partial \gamma} & \frac{\partial \theta_1}{\partial \omega} \\ \frac{\partial \phi_1}{\partial \theta_1} & \frac{\partial \phi_1}{\partial \phi_1} & \frac{\partial \phi_1}{\partial \gamma} & \frac{\partial \phi_1}{\partial \omega} \\ \frac{\partial \theta_2}{\partial \theta_1} & \frac{\partial \theta_2}{\partial \phi_1} & \frac{\partial \theta_2}{\partial \gamma} & \frac{\partial \theta_2}{\partial \omega} \\ \frac{\partial \phi_2}{\partial \theta_1} & \frac{\partial \phi_2}{\partial \phi_2} & \frac{\partial \phi_2}{\partial \gamma} & \frac{\partial \phi_2}{\partial \omega} \\ \end{array} \right) = \left( \begin{array}{c c c c} 1 & 0 & \frac{\partial \theta_1}{\partial \gamma} & \frac{\partial \theta_1}{\partial \omega} \\ 0 & 1 & \frac{\partial \phi_1}{\partial \gamma} & \frac{\partial \phi_1}{\partial \omega} \\ 0 & 0 & \frac{\partial \theta_2}{\partial \gamma} & \frac{\partial \theta_2}{\partial \omega} \\ 0 & 0 & \frac{\partial \phi_2}{\partial \gamma} & \frac{\partial \phi_2}{\partial \omega} \\ \end{array} \right) = \frac{\partial \theta_2}{\partial \gamma}\frac{\partial \phi_2}{\partial \omega} - \frac{\partial \theta_2}{\partial \omega}\frac{\partial \phi_2}{\partial \gamma} \end{align*} The expression for $\theta_2$ and $\phi_2$ in function of $\gamma$ and $\omega$ do not appear to be easily computable. Let's take instead the inverse transformation $ \{ \theta_1, \phi_1, \gamma, \omega \} \rightarrow \{ \theta_1, \phi_1, \theta_2, \phi_2 \}$ and invert the jacobian afterwards. The Jacobian of the inverse transformation is given by, \begin{align*} \left( \begin{array}{c c c c} \frac{\partial \theta_1}{\partial \theta_1} & \frac{\partial \theta_1}{\partial \phi_1} & \frac{\partial \theta_1}{\partial \theta_2} & \frac{\partial \theta_1}{\partial \phi_2} \\ \frac{\partial \phi_1}{\partial \theta_1} & \frac{\partial \phi_1}{\partial \phi_1} & \frac{\partial \phi_1}{\partial \theta_2} & \frac{\partial \phi_1}{\partial \phi_2} \\ \frac{\partial \gamma}{\partial \theta_1} & \frac{\partial \gamma}{\partial \phi_1} & \frac{\partial \gamma}{\partial \theta_2} & \frac{\partial \gamma}{\partial \phi_2} \\ \frac{\partial \omega}{\partial \theta_1} & \frac{\partial \omega}{\partial \phi_1} & \frac{\partial \omega}{\partial \theta_2} & \frac{\partial \omega}{\partial \phi_2} \\ \end{array} \right) = \left( \begin{array}{c c c c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \frac{\partial \gamma}{\partial \theta_1} & \frac{\partial \gamma}{\partial \phi_1} & \frac{\partial \gamma}{\partial \theta_2} & \frac{\partial \gamma}{\partial \phi_2} \\ \frac{\partial \omega}{\partial \theta_1} & \frac{\partial \omega}{\partial \phi_1} & \frac{\partial \omega}{\partial \theta_2} & \frac{\partial \omega}{\partial \phi_2} \\ \end{array} \right) = \frac{\partial \gamma}{\partial \theta_2}\frac{\partial \omega}{\partial \phi_2} - \frac{\partial \gamma}{\partial \phi_2}\frac{\partial \omega}{\partial \theta_2} \end{align*} Let's calculate the partial derivatives one by one, \begin{align*} \frac{\partial \gamma}{\partial \theta_2} = \frac{\partial \gamma}{\partial \cos \gamma} \frac{\partial \cos \gamma}{\partial \theta_2} = - \frac{1}{\sin \gamma} \frac{\partial \cos \gamma}{\partial \theta_2} \end{align*} \begin{align*} \frac{\partial \gamma}{\partial \phi_2} = \frac{\partial \gamma}{\partial \cos \gamma} \frac{\partial \cos \gamma}{\partial \phi_2} = - \frac{1}{\sin \gamma} \frac{\partial \cos \gamma}{\partial \phi_2} \end{align*} \begin{align*} \frac{\partial \omega}{\partial \theta_2} = \frac{\partial \omega}{\partial \sin \omega} \frac{\partial \sin \omega}{\partial \theta_2} &= \frac{1}{\cos \omega} \frac{\partial \sin \omega}{\partial \theta_2} \\ &= \frac{\sin(\phi_2 - \phi_1)}{\cos \omega} \frac{\partial}{\partial \theta_2} \frac{\sin \theta_2}{\sin \gamma} \\ &= \frac{\sin(\phi_2 - \phi_1)}{\cos \omega \sin^{2} \gamma} \left( \cos \theta_2 \sin \gamma - \sin \theta_2 \frac{\partial \sin \gamma }{ \partial \theta_2} \right) \\ &= \frac{\sin(\phi_2 - \phi_1)}{\cos \omega \sin^{2} \gamma} \left( \cos \theta_2 \sin \gamma - \sin \theta_2 \frac{\partial \sin \gamma }{ \partial \cos \gamma} \frac{ \partial \cos \gamma}{\partial \theta_2} \right) \\ &= \frac{\sin(\phi_2 - \phi_1)}{\cos \omega \sin^{2} \gamma} \left( \cos \theta_2 \sin \gamma + \sin \theta_2 \cot \gamma \frac{ \partial \cos \gamma}{\partial \theta_2} \right) \end{align*} \begin{align*} \frac{\partial \omega}{\partial \phi_2} = \frac{\partial \omega}{\partial \sin \omega} \frac{\partial \sin \omega}{\partial \phi_2} &= \frac{1}{\cos \omega} \frac{\partial \sin \omega}{\partial \phi_2} \\ &= \frac{\sin \theta_2}{\cos \omega} \frac{\partial}{\partial \phi_2} \frac{\sin (\phi_2 - \phi_1)}{\sin \gamma}\\ &= \frac{\sin \theta_2}{\cos \omega \sin^2 \gamma} \left( \cos (\phi_2 - \phi_1) \sin \gamma - \sin (\phi_2 - \phi_1) \frac{\partial \sin \gamma}{\partial \phi_2}\right) \\ &= \frac{\sin \theta_2}{\cos \omega \sin^2 \gamma} \left( \cos (\phi_2 - \phi_1) \sin \gamma + \sin (\phi_2 - \phi_1) \cot \gamma \frac{\partial \cos \gamma}{\partial \phi_2}\right) \end{align*} So the total expression becomes, \begin{multline*} \frac{\partial \gamma}{\partial \theta_2}\frac{\partial \omega}{\partial \phi_2} - \frac{\partial \gamma}{\partial \phi_2}\frac{\partial \omega}{\partial \theta_2} = - \frac{\partial \cos \gamma}{\partial \theta_2}\frac{\sin \theta_2}{\cos \omega \sin^3 \gamma} \left( \cos (\phi_2 - \phi_1) \sin \gamma + \sin (\phi_2 - \phi_1) \cot \gamma \frac{\partial \cos \gamma}{\partial \phi_2}\right) \\ + \frac{\partial \cos \gamma}{\partial \phi_2} \frac{\sin(\phi_2 - \phi_1)}{\cos \omega \sin^{3} \gamma} \left( \cos \theta_2 \sin \gamma + \sin \theta_2 \cot \gamma \frac{ \partial \cos \gamma}{\partial \theta_2} \right) \\ = \frac{1}{\ \cos \omega \sin^{3} \gamma} \left( - \sin \theta_2 \cos (\phi_2 - \phi_1) \sin \gamma \frac{\partial \cos \gamma}{\partial \theta_2} - \sin \theta_2 \sin (\phi_2 - \phi_1) \cot \gamma \frac{\partial \cos \gamma}{\partial \phi_2}\frac{\partial \cos \gamma}{\partial \theta_2} \right.\\ + \left. \cos \theta_2 \sin(\phi_2 - \phi_1) \sin \gamma \frac{\partial \cos \gamma}{\partial \phi_2} + \sin \theta_2 \sin(\phi_2 - \phi_1) \cot \gamma \frac{ \partial \cos \gamma}{\partial \theta_2} \frac{\partial \cos \gamma}{\partial \phi_2} \right) \\ = \frac{1}{\ \cos \omega \sin^{2} \gamma} \left( \cos \theta_2 \sin(\phi_2 - \phi_1) \frac{\partial \cos \gamma}{\partial \phi_2} - \sin \theta_2 \cos (\phi_2 - \phi_1) \frac{\partial \cos \gamma}{\partial \theta_2} \right). \end{multline*} With, \begin{align*} \frac{\partial \cos \gamma}{\partial \theta_2} &= - \cos \theta_1 \sin \theta_2 + \sin \theta_1 \cos \theta_2 \cos (\phi_2 - \phi_1 ), \\ \frac{\partial \cos \gamma}{\partial \phi_2} &= - \sin \theta_1 \sin \theta_2 \sin (\phi_2 - \phi_1), \end{align*} we get, \begin{multline} \frac{\partial \gamma}{\partial \theta_2}\frac{\partial \omega}{\partial \phi_2} - \frac{\partial \gamma}{\partial \phi_2}\frac{\partial \omega}{\partial \theta_2} = \frac{1}{\ \cos \omega \sin^{2} \gamma} \Big( - \cos \theta_2 \sin \theta_1 \sin \theta_2 \sin^{2} (\phi_2 - \phi_1) \\ + \sin^{2} \theta_2 \cos \theta_1 \cos (\phi_2 - \phi_1) - \sin \theta_2 \sin \theta_1 \cos \theta_2 \cos^{2} (\phi_2 - \phi_1 ) \Big) \\ = \frac{\sin \theta_2}{\ \cos \omega \sin^{2} \gamma} \left( \sin \theta_2 \cos \theta_1 \cos (\phi_2 - \phi_1) - \sin \theta_1 \cos \theta_2 \right) \end{multline} Using the standard cosine rule we find, \begin{align*} \cos \theta_2 = \cos \theta_1 \cos \gamma + \sin \theta_1 \sin \gamma \cos \omega \Rightarrow \cos \omega = \frac{\cos \theta_2 - \cos \theta_1 \cos \gamma}{\sin \theta_1 \sin \gamma} \end{align*} Hence we get, \begin{align*} \frac{\partial \gamma}{\partial \theta_2}\frac{\partial \omega}{\partial \phi_2} - \frac{\partial \gamma}{\partial \phi_2}\frac{\partial \omega}{\partial \theta_2} &= \frac{\sin \theta_1 \sin \theta_2}{\ \sin \gamma} \left( \frac{ \sin \theta_2 \cos \theta_1 \cos (\phi_2 - \phi_1) - \sin \theta_1 \cos \theta_2 }{ \cos \theta_2 - \cos \theta_1 \cos \gamma} \right) \\ &=\frac{\sin \theta_1 \sin \theta_2}{\ \sin \gamma} \left( \frac{ \sin \theta_2 \cos \theta_1 \cos (\phi_2 - \phi_1) - \sin \theta_1 \cos \theta_2 }{ \cos \theta_2 - \cos^{2} \theta_1 \cos \theta_2 - \sin \theta_1 \cos \theta_1 \sin \theta_2 \cos ( \phi_2 - \phi_1)} \right) \\ &=\frac{\sin \theta_1 \sin \theta_2}{\ \sin \gamma} \left( \frac{ \sin \theta_2 \cos \theta_1 \cos (\phi_2 - \phi_1) - \sin \theta_1 \cos \theta_2 }{ \cos \theta_2 \sin^{2} \theta_1 - \sin \theta_1 \cos \theta_1 \sin \theta_2 \cos ( \phi_2 - \phi_1)} \right) \\ &= - \frac{\sin \theta_1 \sin \theta_2}{\ \sin \gamma \sin \theta_1 } \left( \frac{ \sin \theta_2 \cos \theta_1 \cos (\phi_2 - \phi_1) - \sin \theta_1 \cos \theta_2 }{ \sin \theta_2 \cos \theta_1 \cos ( \phi_2 - \phi_1) - \sin \theta_1 \cos \theta_2 } \right) \\ &= - \frac{\sin \theta_2}{\ \sin \gamma } \end{align*} Now we have derived the Jacobian of the inverse transformation. The jacobian of the original transformation is given by \begin{align*} |J| = \frac{ \sin \gamma }{\sin \theta_2 } \end{align*} Hence we get, \begin{align*} \int \textrm{d}\theta_1 \sin \theta_1 \int \textrm{d}\phi_1 \int \textrm{d} \sin \theta_2 \int \textrm{d} \phi_2 \, f( \gamma ) &\rightarrow \int_{0}^{\pi} \textrm{d}\theta_1 \sin \theta_1 \int_{0}^{2 \pi} \textrm{d}\phi_1 \int_{0}^{2 \pi} \textrm{d} \omega \int_{0}^{\pi} \textrm{d} \gamma \sin \theta_2 \, \frac{ \sin \gamma }{\sin \theta_2 } f( \gamma ) \\ &= 8 \pi^{2} \int_{0}^{\pi} \textrm{d} \gamma \, \sin (\gamma) \, f( \gamma ) \,\,\, \square. \end{align*} Which is exactly the result we expected.
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