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I'm trying to construct a variational calculus approach to solving the angle at which a neutrino beam has to be fired through the earth, to hit a coin on the other side of the planet. For example, hitting an Australian dollar in Perth, with a beam fired from CERN.

What I would like to produce is a function for the direction I would need to fire at, based on the position of the start and end location, along with the velocity of the neutrino beam.

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My thoughts on the portions of the problem so far:

  1. There is an effect due to gravity. Light travelling next to the earth will bend by a couple of milli-arc seconds. I have done a quick calculation to test this, but depending on the speed of the beam, this effect seems to be negligible. However, I would like to characterise it somewhat to be sure that it actually is negligible.
  2. Rotation of the earth. If the earth is rotating about its rotational poles, then there would be a required adjustment in the initial direction that the beam would have to be sent in. Some quick maths shows that if travelling across the centre of the earth, starting at the equator, the positional error will be about 20m. This is substantial.

Do I even need variational calculus to solve this?

How would you approach this problem?

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$\def \b {\mathbf}$ $\def \theta {\lambda}$

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the Neutrino is going from point 1 to point 2 both are on a sphere surface which describe with this equation

$$\b R_i=r\,\left[ \begin {array}{c} \cos \left( \phi_i \right) \sin \left( \lambda_i \right) \\ \sin \left( \phi_i \right) \sin \left( \lambda \right) \\ \cos \left( \lambda_i \right) \end {array} \right]\quad,i=1,2 $$

thus the neutrino goes on the straight line $~\b g=\b R_2-\b R_1~$ and the distance is $~d=|\b g|~$

we put a local coordinate system at $~\phi_1~,\lambda_1~$ with the axis $~\hat{\b{e}}_{\phi_1}~,\hat{\b{e}}_{\lambda_1}~,\hat{\b{r}}~$ the coordinates of the vector $~\b g~$ in the local system are $$ g_L=\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}=[~\hat{\b{e}}_{\phi_1}~,\hat{\b{e}}_{\lambda_1}~,\hat{\b{r}}~]^T\,\b g$$

from here the angle $~\psi=\arctan(y/z)~$

Kinematic

The position vector to the neutrino mass is

$$\b R_N=s\,\hat{\b{g}}_L\quad, \dot{\b{R}}_N=\dot s\,\hat{\b{g}}_L$$

where $~s~$ is the generalized coordinate, $~0\le s\le d~$ and $\dot s=v$

If the velocity v is constant, then $~s=v\,t~$ with $~0\le t\le d/v~$


\begin{align*} &\vec{R}=\left[ \begin {array}{c} r\cos \left( \phi \right) \sin \left( \theta \right) \\ r\sin \left( \phi \right) \sin \left( \theta \right) \\ r\cos \left( \theta \right) \end {array} \right]\quad, \vec e_r=\frac{\partial\vec R}{\partial r}=\hat{e}_r\quad, \frac{1}{r}\,\vec{e}_\theta=\frac{\frac{\partial\vec R}{\partial \theta}}{|\frac{\partial\vec R}{\partial \theta}|}=\hat{\b{e}}_\theta\quad, \frac{1}{r\,\sin(\theta)}\, \vec{e}_\phi=\hat{\b{e}}_\phi\\\\ &[~\hat{\b{e}}_{\phi_1}~,\hat{\b{e}}_{\lambda_1}~,\hat{\b{r}}~]= \left[ \begin {array}{ccc} -\sin \left( \phi \right) &\cos \left( \phi \right) \cos \left( \lambda \right) &\cos \left( \phi \right) \sin \left( \lambda \right) \\ \cos \left( \phi \right) &\sin \left( \phi \right) \cos \left( \lambda \right) &\sin \left( \phi \right) \sin \left( \lambda \right) \\ 0 &-\sin \left( \lambda \right) &\cos \left( \lambda \right) \end {array} \right] \end{align*}

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