Luminosity in Accelerator Physics (Derivations)
Derivations from Luminosity Page
1a. Luminosity of Round, Gaussian Beams
Starting from this definition of Luminosity:
\[\mathcal{L} = KfN_1N_2N_b\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \rho_1(x, y, s, -s_0) \rho_2(x, y, s, s_0) dx dy ds ds_0\]For a gaussian distribution for one beam:
\(\rho_{iz}(z) = \frac{1}{\sqrt{2\pi}\sigma_{iz}}\exp(-\frac{z^2}{\sigma_{iz}}); i = 1,2; z = x, y\)
\(\rho_{is}(s, s_0) = \frac{1}{\sqrt{2\pi}\sigma_{is}}\exp(-\frac{(s + s_0)^2}{\sigma_{is}}); i = 1,2\)
For two beams heading toward each other (\(v_1 = - v_2\)), then \(K = 2\)
And assumming a round beam:
\(\sigma_{x1} = \sigma_{x2} = \sigma_{x}\) = constant
\(\sigma_{y1} = \sigma_{y2} = \sigma_{y}\) = constant
\(\sigma_{s1} = \sigma_{s2} = \sigma_{s}\) = constant
Then:
\(\Rightarrow \rho_1\rho_2 = \frac{1}{(2\pi)^3(\sigma_x\sigma_y\sigma_z)^2}\exp(-(\frac{x}{\sigma_x})^2 + (\frac{y}{\sigma_y})^2 + \frac{(s + s_0)^2 + (s - s_0)^2}{2\sigma_z^2})\)
\(= \frac{1}{(2\pi)^3(\sigma_x\sigma_y\sigma_z)^2}\exp(-(\frac{x}{\sigma_x})^2 + (\frac{y}{\sigma_y})^2 + \frac{s^2 + 2ss_0 + s_0^2 + s^2 - 2ss_0 + s_0^2}{2\sigma_z^2})\)
\(= \frac{1}{(2\pi)^3(\sigma_x\sigma_y\sigma_z)^2}\exp(-(\frac{x}{\sigma_x})^2 + (\frac{y}{\sigma_y})^2 + (\frac{s}{\sigma_z})^2 + (\frac{s_0}{\sigma_z})^2)\)
\(\Rightarrow \mathcal{L} = \frac{2fN_1N_2N_b}{(2\pi)^3(\sigma_x\sigma_y\sigma_z)^2} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp(-(\frac{x}{\sigma_x})^2 + (\frac{y}{\sigma_y})^2 + (\frac{s}{\sigma_z})^2 + (\frac{s_0}{\sigma_z})^2) dx dy ds ds_0\)
Using:
\(\int_{-\infty}^{\infty}\exp(-ax^2)dx = \sqrt{\frac{\pi}{a}}\)
We arrive at:
\(\Rightarrow \mathcal{L} = \frac{fN_1N_2N_b}{4\pi\sigma_x\sigma_y}\)
1b. Luminosity of two different short bunches:
\(\sigma_{x1} \neq \sigma_{x2}\)
\(\sigma_{y1} \neq \sigma_{y2}\)
\(\sigma_{z1} = \sigma_{z2} = \sigma_{z}\)
RMS = \(\sqrt{\frac{1}{n}\sum_ix_i^2}\)
\(\Rightarrow \sigma_i = \sqrt{\frac{1}{2}(\sigma_{i1}^2 + \sigma_{i2}^2)}\)
Then from 1:
\(\Rightarrow \mathcal{L} = \frac{fN_1N_2N_b}{4\pi\sqrt{\frac{1}{2}(\sigma_{x1}^2 + \sigma_{x2}^2)}\sqrt{\frac{1}{2}(\sigma_{y1}^2 + \sigma_{y2}^2)}}\) \(= \frac{fN_1N_2N_b}{2\pi\sqrt{\sigma_{x_1}^2 + \sigma_{x_2}^2}\sqrt{\sigma_{y_1}^2 + \sigma_{y_2}^2}}\)
2. Crossing Angle For Round Gaussian Beams
The rotation matrices explicitly give the following transformations:
\(x_1 = x \cos \frac{\phi}{2} - s \sin \frac{\phi}{2}\)
\(x_2 = x \cos \frac{\phi}{2} + s \sin \frac{\phi}{2}\)
\(s_1 = s \cos \frac{\phi}{2} + x \sin \frac{\phi}{2}\)
\(s_2 = s \cos \frac{\phi}{2} - x \sin \frac{\phi}{2}\)
For crossing angle case: \(K = 2\cos^2(\frac{\phi}{2})\)
\(\Rightarrow \mathcal{L} = \frac{2\cos^2(\frac{\phi}{2})fN_1N_2N_b}{(2\pi)^3(\sigma_x\sigma_y\sigma_s)^2}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp(-\frac{(x_1^2 + x_2^2)}{2\sigma_x^2}) \exp(-\frac{y^2}{\sigma_y^2}) \exp(-\frac{(s_1 - s_0)^2 + (s_2 + s_0)^2}{2\sigma_s^2}) dx dy ds ds_0\)
Integrate wrt \(y\):
\(\Rightarrow \mathcal{L} = \frac{\cos^2(\frac{\phi}{2})fN_1N_2N_b}{4\pi^{5/2}(\sigma_x\sigma_s)^2\sigma_y}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp(-\frac{(x_1^2 + x_2^2)}{2\sigma_x^2}) \exp(-\frac{(s_1 - s_0)^2 + (s_2 + s_0)^2}{2\sigma_s^2}) dx ds ds_0\)
\(\Rightarrow \mathcal{L} = \frac{\cos^2(\frac{\phi}{2})fN_1N_2N_b}{4\pi^{5/2}(\sigma_x\sigma_s)^2\sigma_y}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp(-\frac{(x_1^2 + x_2^2)}{2\sigma_x^2}) \exp(-\frac{s_1^2 + s_2^2 + 2s_0^2 + 2s_0(s_2 - s_1)}{2\sigma_s^2}) dx ds ds_0\)
Using:
\(\int_{-\infty}^{\infty}\exp(-(ax^2 + bx + c))dx = \sqrt{\frac{\pi}{a}}\exp(\frac{b^2}{4a} - c)\)
To integrate over \(s_0\):
\(\Rightarrow \mathcal{L} = \frac{\cos^2(\frac{\phi}{2})fN_1N_2N_b}{4\pi^2\sigma_x^2\sigma_y\sigma_s}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp(-\frac{x_1^2 + x_2^2}{2\sigma_x^2}) \exp(\frac{(s_2 - s_1)^2}{4\sigma_s^2} - \frac{s_1^2 + s_2^2}{2\sigma_s^2}) dx ds\)
Substitute change of coordinates and simplifying:
\(\Rightarrow \mathcal{L} = \frac{\cos^2(\frac{\phi}{2})fN_1N_2N_b}{4\pi^2\sigma_x^2\sigma_y\sigma_s}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp(-\frac{x^2\cos(\frac{\phi}{2})}{\sigma_x^2}) \exp(-s^2(\frac{\sin(\frac{\phi}{2})}{\sigma_x^2} + \frac{\cos(\frac{\phi}{2})}{\sigma_s^2})) dx ds\)
Integrate wrt \(x\):
\(\Rightarrow \mathcal{L} = \frac{\cos(\frac{\phi}{2})fN_1N_2N_b}{4\pi^{3/2}\sigma_x\sigma_y\sigma_s}\int_{-\infty}^{\infty} \exp(-s^2(\frac{\sin(\frac{\phi}{2})}{\sigma_x^2} + \frac{\cos(\frac{\phi}{2})}{\sigma_s^2})) ds\)
If \(\sigma_x, \sigma_y\) depend on \(s\) due to hourglass effect, the expression can no longer be simplified, and \(\sigma_x, \sigma_y\) go in the \(s\) integrand.
Integrate wrt \(s\) for constant \(\sigma_x, \sigma_y\):
\(\Rightarrow \mathcal{L} = \frac{\cos(\frac{\phi}{2})fN_1N_2N_b}{4\pi\sigma_x\sigma_y\sigma_s}(\frac{\sin(\frac{\phi}{2})}{\sigma_x^2} + \frac{\cos(\frac{\phi}{2})}{\sigma_s^2})^{-\frac12}\)
Simplifiying further gives:
\(\Rightarrow \mathcal{L} = \frac{fN_1N_2N_b}{4\pi\sigma_x\sigma_y}\frac{1}{\sqrt{1 + (\frac{\sigma_s}{\sigma_x})^2\tan^2(\frac{\phi}{2})}}\)
Finally:
\(\mathcal{L_{CA}} = L_0 S(\phi)\)