Some Basic theory

Presuppositions

When a wire is put into the beam, the number of interactions is proportional to the beam density integration along the wire path. We expect therefore to see in the photomultipliers a proportional increase in counting rate. This remains always true in our case, except if the acquisition system saturates. Unfortunately, our PMTs voltage were not always correctly adjusted, and we have had some saturation.

A perfect beam

If we have a perfect beam spot spot with a Gaussian distribution, then we can write the beam density function in a plane perpendicular to the beam axis as :

$$f(x,y) = A e^{-\frac{x^{2}+y^{2}}{\sigma^{2}}}$$

If we assume the wire to be thin compared to the $\sigma$ of the beam distribution, then a scan on the x axis will give :

$$F_{x}(x) = \int_{-\infty}^{+\infty} f(x,y) dy$$

So, using the fact that $\int_{-\infty}^{+\infty}e^{-x^{2}}dx=\sqrt{\pi}$ :

$$F_{x}(x) = A \sigma \sqrt{\pi} e^{-\frac{x^{2}}{\sigma^{2}}}$$

As there is a cylindrical symmetry, this result remains the same for any direction of scanning.

One can then conclude that if the beam is a perfect Gaussian beam, a scan on any direction will give exactly the $\sigma$ of the beam. Yeah, should have been trivial, but better do it anyway.

The ellipsoidal beam

If the beam is ellipsoidal, then we expect a more complex result, with the exception of the trivial case when the beam ellipticity axes are aligned with the harp scanning wires.

Then we simply have :

 

 $$f(x,y) = A e^{-\frac{x^{2}}{\sigma_{1}^{2}}+\frac{y^{2}}{\sigma_{2}^{2}}}$$

And a scan on x and y will give us the exact value of $\sigma_{1}$ and $\sigma_{2}$.

Lets go to the general case when the beam has an ellipticity $\frac{\sigma_{1}^{2}}{\sigma_{2}^{2}}$ in a direction $\theta$.

Starting from the previous equation and applying a rotation $\theta$ we get :

$$f(x,y) = A e^{ -x^{2}(\frac{c^{2}}{\sigma_{1}^{2}}+\frac{s^... ...) -xysc (\frac{1 }{\sigma_{2}^{2}}-\frac{1 }{\sigma_{1}^{2}}) }$$

Where $c = \cos \theta$ and $s = \sin \theta$.

Now, the beam does not appear anymore as Gaussian on the x and y axes of the scan.

Knowing that [16] :

$$\int_{-\infty}^{+\infty} e^{-p^2 x^2 \pm qx} dx = e^{\frac{q^2}{4p^2}} \frac{\sqrt{\pi}}{p}$$

We can now compute the x and y scan :

 

 $$F_{x}(x)=Ae^{-x^2(\alpha - \frac{\gamma^2}{4 \beta}) \sqrt{\frac{\pi}{\beta}}}$$

and

$$F_{y}(y)=Ae^{-y^2(\beta - \frac{\gamma^2}{4 \alpha}) \sqrt{\frac{\pi}{\alpha}}}$$

Where $\alpha = \frac{c^{2}}{\sigma_{1}^{2}}+\frac{s^{2}}{\sigma_{2}^{2}}$, $\beta = \frac{s^{2}}{\sigma_{1}^{2}}+\frac{c^{2}}{\sigma_{2}^{2}}$ $\gamma = sc\frac{1 }{\sigma_{2}^{2}}-\frac{1 }{\sigma_{1}^{2}}$.

A clever fit of the scan of such a beam using this parametrisation will gives us $p_{1} = \alpha - \frac{\gamma^2}{4 \beta }$, and $p_{2} = \beta - \frac{\gamma^2}{4 \alpha}$.

But there is no way to deduce the values of $\alpha$ and $\beta$, because we don't have enough known parameters.

The best parameter we can use is still :

$$\frac{p_{1}}{p_{2}} = \frac{s^{2}\sigma_{1}^{2} + c^{2}\sigma_{2}^{2}} {c^{2}\sigma_{1}^{2} + s^{2}\sigma_{2}^{2}}$$

Even if we know that this term is only smaller that the value we would have seen $\sigma_{2}^{2}/\sigma_{1}^{2}$.

Adding another wire is certainly enough to solve this problem, as we have 3 unknown terms ($\sigma_{1}$, $\sigma_{2}$ and $\theta$), and as each scan will give us at least another independent parameter. When starting the calculation I was hopping to get more than one parameter when fitting the result of the scan, unfortunately, a Gaussian distribution does not give us many places to hang on it.

I tried to extend this works for 3 wires, but didn't succeed analytically, and didn't took time to investigate the question further, as we know that the beam profile is usually in the scan axes, and that our answer will anyway give a good approximation of the real shape of the beam.