Difference between revisions of "Limit of Scattering Angle Theta in Lab Frame"

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This gives
This gives
<center><math>51072727.272728\ MeV\approx (.511000181\ MeV- 13.6\ eV) \cos \theta_{1\ 2^{'}}</math></center>
<center><math>51072727.272728\ MeV\approx (.511000181\ MeV)- (13.6\ eV) \cos \theta_{1\ 2^{'}}</math></center>
<center><math>51072727.272728\ MeV\approx (.510986400181\ MeV \cos \theta_{1\ 2^{'}}</math></center>

Revision as of 20:09, 15 March 2018

The quantity is known as the

[math]u \equiv \left({\mathbf P_1^*}- {\mathbf P_2^{'*}}\right)^2=\left({\mathbf P_2^{*}}- {\mathbf P_1^{'*}}\right)^2[/math]

In the CM Frame

[math]{\mathbf P_1^{*}}=-{\mathbf P_2^{*}}[/math]

[math]{\mathbf P_1^{'*}}=-{\mathbf P_2^{'*}}[/math]


[math]\left | \vec p_1^* \right |=\left | \vec p_1^{'*} \right |=\left | \vec p_2^* \right |=\left | \vec p_2^{'*} \right |[/math]

[math]u =\left({\mathbf P_1^*}- {\mathbf P_2^{'*}}\right)^2=\left({\mathbf P_2^{*}}- {\mathbf P_1^{'*}}\right)^2[/math]

[math]u={\mathbf P_1^{*2}}+ {\mathbf P_2^{'*2}}-2 {\mathbf P_1^*} {\mathbf P_2^{'*}}={\mathbf P_2^{*2}}+ {\mathbf P_1^{'*2}}-2 {\mathbf P_2^*} {\mathbf P_1^{'*}}[/math]

[math]u=2m^2-2E_1^*E_2^{'*}+2 \vec p_1^* \vec p_2^{'*}=2m^2-2E_2^*E_1^{'*}+2 p_2^* p_1^{'*}[/math]

[math]u=2m^2-2E_1^{*2}+2 \left | p_1^{*2}\right | \cos \theta_{1^*\ 2^{'*}}=2m^2-2E_2^{*2}+2 \left | p_2^{*2}\right | \cos \theta_{2^*\ 1^{'*}}[/math]

where [math]\theta_{1^*\ 2^{'*}}[/math] and [math]\theta_{2^*\ 1^{'*}}[/math]is the angle between the before and after momentum in the CM frame

Using the relativistic relation [math]E^2=m^2+p^2[/math] this reduces to

[math]u=-2p_1^{*2}+2 \left | p_1^{*2}\right | \cos \theta_{1^*\ 2^{'*}}=-2p_2^{*2}+2 \left | p_2^{*2}\right | \cos \theta_{2^*\ 1^{'*}}[/math]

[math]u=-2p_1^{*2}(1- \cos \theta_{1^*\ 2^{'*}})=-2p_2^{*2}(1-\cos \theta_{2^*\ 1^{'*}})[/math]

[math]\theta \approx 180^{\circ}[/math]

There is no scattering, or no momentum transfer at 180 degrees since the incident momentum direction is the same as the scattered momentum direction for the CM frame Moller electron. However, at a certain angle enough momentum must be transferred to provide the ionization energy to create a Moller electron.


For [math]\theta_{1^*1^{'*}}=90^{\circ}[/math], by symmetry this implies [math]\theta_{1^*2^{'*}}=270^{\circ}[/math]


This can be rewritten again using the relativistic energy relation [math]E^2=m^2+p^2[/math]


In the Lab Frame

[math]u={\mathbf P_1^{2}}+ {\mathbf P_2^{'2}}-2 {\mathbf P_1} {\mathbf P_2^{'}}={\mathbf P_2^{2}}+ {\mathbf P_1^{'2}}-2 {\mathbf P_2} {\mathbf P_1^{'}}[/math]

[math]u=2m^2-2E_1E_2^{'}+2 \vec p_1 \vec p_2^{'}=2m^2-2E_2E_1^{'}+2 p_2 p_1^{'}[/math]

with [math]p_2=0[/math]

and [math]E_2=m[/math]

[math]u=2m^2-2E_1E_2^{'}+2 \vec p_1 \vec p_2^{'}=2m^2-2mE_1^{'}[/math]

[math]u=2m^2-2E_1E_2^{'}+2 \left | p_1 \right | \left | p_2^{'} \right | \cos \theta_{1\ 2^{'}}=2m^2-2mE_1^{'}[/math]

Minimum Moller Scattering Angle Theta in Lab Frame

Since u is invariant between frames

[math]u=2m^2-2E_1E_2^{'}+2 \left | p_1 \right | \left | p_2^{'} \right | \cos \theta_{1\ 2^{'}}=2(m^2-E_2^{*2})[/math]

with[math] E_2^{*} \approx 53\ MeV[/math] for [math]E_1=11000\ MeV[/math]

As found earlier, the Moller electron has a maximum energy possible of:

[math]E_2^{'}=5500\ MeV[/math]

Using the relativistic energy relation, [math]E^2=m^2+p^2[/math]

[math]p=\sqrt {E^2-m^2} \rightarrow p \approx E[/math] for [math]E \gg m[/math]

[math]\therefore p_2^{'} \approx E_2^{'}[/math]

Rewriting the expression relating the terms

[math]u=2m^2-2E_1E_2^{'}+2 \left | p_1 \right | \left | p_2^{'} \right | \cos \theta_{1\ 2^{'}}=2(m^2-E_2^{*2})[/math]

[math]u=2m^2-2(11000\ MeV)(5500\ MeV) +2(11000\ MeV)(5500\ MeV) \cos \theta_{1\ 2^{'}}=2(m^2-(53\ MeV)^{2})[/math]

Solving for the angle theta

[math]2m^2-2(11000\ MeV)(5500\ MeV) +2(11000\ MeV)(5500\ MeV) \cos \theta_{1\ 2^{'}}=2(m^2-(53\ MeV)^{2})[/math]

[math](11000\ MeV)(5500\ MeV)(1 - \cos \theta_{1\ 2^{'}})=(53\ MeV)^{2}[/math]

[math](1 - \cos \theta_{1\ 2^{'}})=\frac{(53\ MeV)^{*2}}{(11000\ MeV)(5500\ MeV)}[/math]

[math]1 - \cos \theta_{1\ 2^{'}}=4.64297520661\times 10^{-5}[/math]

[math]\cos \theta_{1\ 2^{'}}=1-4.64297520661\times 10^{-5}[/math]

[math]\theta_{1\ 2^{'}}=\arccos .999953570248[/math]

[math]\theta_{1\ 2^{'}}=.009636400914\ radians=.55^{\circ}[/math]

The Moller electron is traveling near, but not on the beamline.

Maximum Moller Scattering Angle Theta in Lab Frame

Since u is invariant between frames

[math]u=2m^2-2E_1^*E_2^{'*}+2 \left | p_1^* \right | \left | p_2^{'*} \right | \cos \theta_{1\ 2^{'}}=2m^2-2E_1E_2^{'}+2 \left | p_1 \right | \left | p_2^{'} \right | \cos \theta_{1\ 2^{'}}[/math]

[math]E_1^*=53\ MeV=p_1^*[/math]

[math]E_1=11000\ MeV=p_1[/math]

[math]-E_1^*E_2^{'*}+ \left | p_1^* \right | \left | p_2^{'*} \right | \cos \theta_{1\ 2^{'}}\approx -E_1E_2^{'}+ \left | p_1 \right | \left | p_2^{'} \right | \cos \theta_{1\ 2^{'}}[/math]

[math]E_1^*E_2^{'*}- \left | p_1^* \right | \left | p_2^{'*} \right | \cos \theta_{1\ 2^{'}}\approx E_1E_2^{'}- \left | p_1 \right | \left | p_2^{'} \right | \cos \theta_{1\ 2^{'}}[/math]

[math]E_1^{*2}(1- \cos \theta_{1\ 2^{'}})\approx E_1(E_2^{'}- \left | p_2^{'} \right | \cos \theta_{1\ 2^{'}})[/math]

[math](53\ MeV)^{*2}(1- \cos \theta_{1\ 2^{'}})\approx (11000\ MeV)(E_2^{'}- \left | p_2^{'} \right | \cos \theta_{1\ 2^{'}})[/math]

[math](2553636.36364\ MeV)(1- \cos \theta_{1\ 2^{'}})\approx (E_2^{'}- \left | p_2^{'} \right | \cos \theta_{1\ 2^{'}})[/math]

At [math]\theta \approx 180^{\circ}[/math]

[math]2(2553636.36364\ MeV)\approx (E_2^{'}- \left | p_2^{'} \right | \cos \theta_{1\ 2^{'}})[/math]

[math]5107272.7272728\ MeV\approx (E_2^{'}- \left | p_2^{'} \right | \cos \theta_{1\ 2^{'}})[/math]

For [math]p \approx 13.6\ eV[/math]

[math]E=\sqrt {(13.6 eV)^2+(.511MeV)^2}=\sqrt {184.96+261121000000}\ eV=\sqrt{261121000185}\ eV=511000.000181\ eV[/math]

This gives

[math]51072727.272728\ MeV\approx (.511000181\ MeV)- (13.6\ eV) \cos \theta_{1\ 2^{'}}[/math]