Difference between revisions of "Scattering Amplitude"

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<center><math>-i \mathfrak{M}_{1}=ie^{2}\left (\frac{(\mathbf p_{1}+\mathbf p_{1}^{'})_{\mu} (\mathbf p_{2}+\mathbf p_{2}^{'})^{\mu}}{(\mathbf p_{2}^{'}-\mathbf p_{2})^{2}} \right ) \qquad \qquad -i \mathfrak{M}_{2}=ie^{2}\left (\frac{(\mathbf p_{1}+\mathbf p_{2}^')_{\mu} (\mathbf p_{2}+\mathbf p_{1}^{'})^{\mu}}{(\mathbf p_{1}^{'}-\mathbf p_{2})^{2}} \right ) </math></center>
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<center><math>-i \mathfrak{M}_{1}=ie^{2}\left (\frac{\left(\mathbf p_{1}+\mathbf p_{1}^{'}\right)_{\mu} \left(\mathbf p_{2}+\mathbf p_{2}^{'}\right)^{\mu}}{\left(\mathbf p_{2}^{'}-\mathbf p_{2}\right)^{2}} \right ) \qquad \qquad -i \mathfrak{M}_{2}=ie^{2}\left (\frac{\left(\mathbf p_{1}+\mathbf p_{2}^{'}\right)_{\mu} \left(\mathbf p_{2}+\mathbf p_{1}^{'}\right)^{\mu}}{\left(\mathbf p_{1}^{'}-\mathbf p_{2}\right)^{2}} \right ) </math></center>
  
  

Revision as of 21:09, 29 December 2018

[math]\underline{\textbf{Navigation}}[/math]

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Scattering Amplitude

In the Møller scattering [math](\mathbf P_{1} + \mathbf P_{2} \rightarrow \mathbf P_{1}^{'} + \mathbf P_{2}^{'})[/math] we have identical particles in the initial and final states. This that the amplitude to be symmetric under interchange of particles [math](\mathbf P_{1}^{'} \leftrightarrow \mathbf P_{2}^{'} [/math] or [math] \mathbf P_{1} \leftrightarrow \mathbf P_{2})[/math]. Due to this symmetry we can determine two 1st level Feynman diagrams to describe this scattering.

Feynman1stLevel.png

The amplitudes of the individual Feynman diagrams add linearly to form the total amplitude

[math]\mathfrak{M}=\mathfrak{M}_{1}+\mathfrak{M}_{2}[/math]


Using the Feynman rules, each vertex contribute a factor

[math]ie(\mathbf p_{initial} + \mathbf p_{final})^{\mu}[/math]

and the propagator gives

[math] \frac{−ig_{\mu \nu}}{q^2}[/math]

where q is the momentum of the photon

[math]q \equiv \mathbf p_{final}-\mathbf p_{initial}[/math]

and [math]g_{\mu \nu}[/math] is the Mandelstam metric which allows the transformation from the contravariant to covariant form needed for tensor multiplication. Examining both Feynman diagrams seperately, we find for their individual amplitudes


[math]-i \mathfrak{M}_{1}=ie(\mathbf p_{1}+\mathbf p_{1}^{'})^{\mu} \left (\frac{-ig_{\mu \nu}}{q^2} \right ) ie ( \mathbf p_{2}+\mathbf p_2^')^{\nu} \qquad \qquad -i \mathfrak{M}_{2}=ie(\mathbf p_{1}+\mathbf p_{2}^')^{\mu} \left (\frac{-ig_{\mu \nu}}{q^2} \right ) ie ( \mathbf p_{2}+\mathbf p_1^')^{\nu}[/math]


[math]-i \mathfrak{M}_1=ie(\mathbf p_{1}+\mathbf p_{1}^{'})^{\mu} \left (\frac{-ig_{\mu \nu}}{(\mathbf p_{2}^{'}-\mathbf p_{2})^{2}} \right ) ie( \mathbf p_{2}+\mathbf p_{2}^{'})^{\nu} \qquad \qquad -i \mathfrak{M}_{2}=ie(\mathbf p_{1}+\mathbf p_{2}^{'})^{\mu} \left (\frac{-ig_{\mu \nu}}{(\mathbf p_{1}^{'}-\mathbf p_{2})^{2}} \right ) ie( \mathbf p_{2}+\mathbf p_{1}^{'})^{\nu}[/math]


[math]-i \mathfrak{M}_{1}=ie^{2}\left (\frac{\left(\mathbf p_{1}+\mathbf p_{1}^{'}\right)_{\mu} \left(\mathbf p_{2}+\mathbf p_{2}^{'}\right)^{\mu}}{\left(\mathbf p_{2}^{'}-\mathbf p_{2}\right)^{2}} \right ) \qquad \qquad -i \mathfrak{M}_{2}=ie^{2}\left (\frac{\left(\mathbf p_{1}+\mathbf p_{2}^{'}\right)_{\mu} \left(\mathbf p_{2}+\mathbf p_{1}^{'}\right)^{\mu}}{\left(\mathbf p_{1}^{'}-\mathbf p_{2}\right)^{2}} \right ) [/math]


Without loss of generality, we can extend this to the center of mass frame


[math]-i \mathfrak{M}_{e^-e^-}=-ie^2 \left ( \frac{(\mathbf p_1^*+\mathbf p_{1}^{'*})_{\mu}(\mathbf p_2^*+\mathbf p_{2}^{'*})^{\mu}}{(\mathbf p_{2}^{'*}-\mathbf p_2^*)^2}- \frac{(\mathbf p_1^*+\mathbf p_{2}^{'*})_{\mu}(\mathbf p_2^*+\mathbf p_{1}^{'*})^{\mu}}{(\mathbf p_{1}^{'*}-\mathbf p_2^*)^2} \right )[/math]


[math] \mathfrak{M}_{e^-e^-}= e^2\left ( \frac{\mathbf P_1^* \mathbf P_2^*+\mathbf P_{1}^{'*} \mathbf P_{2}^{'*}+\mathbf P_{1}^{'*} \mathbf P_2^*+\mathbf P_1^* \mathbf P_{2}^{'*}}{(\mathbf P_{2}^{'*}-\mathbf P_2^*)^2}- \frac{\mathbf P_1^* \mathbf P_2^*+\mathbf P_{2}^{'*} \mathbf P_{1}^{'*}+\mathbf P_{2}^{'*} \mathbf P_2^*+\mathbf P_1^* \mathbf P_{1}^*}{(\mathbf P_{1}^{'*}-\mathbf P_2^*)^2} \right )[/math]



Using the fact that [math]\mathbf P_1^{'*} \mathbf P_2^{'*}=\mathbf P_1^* \mathbf P_2^* \quad \quad \mathbf P_1^{'*} \mathbf P_1^{*}=\mathbf P_2^{'*} \mathbf P_2^* \quad \quad \mathbf P_1^{*} \mathbf P_2^{'*}=\mathbf P_2^* \mathbf P_1^{'*}[/math]


[math] \mathfrak{M}_{e^-e^-}= e^2\left ( \frac{2\mathbf P_1^* \mathbf P_2^*+2\mathbf P_{1}^{'*} \mathbf P_2^*}{(\mathbf P_{2}^{'*2}-2\mathbf P_{2}^{'*}\mathbf P_2^{*}+\mathbf P_2^{*2})}- \frac{2\mathbf P_1^* \mathbf P_2^*+2\mathbf P_1^* \mathbf P_{1}^{'*}}{(\mathbf P_{1}^{'*2}-2\mathbf P_{1}^{'*}\mathbf P_2^{'*}+\mathbf P_2^{'*2})} \right )[/math]


[math] \mathfrak{M}_{e^-e^-}= e^2\left ( \frac{2\mathbf P_1^* \mathbf P_2^*+2\mathbf P_{1}^{'*} \mathbf P_2^*}{(\mathbf P_2^{*2}-2\mathbf P_2^{*}\mathbf P_{2}^{'*}+\mathbf P_{2}^{'*2})}- \frac{2\mathbf P_1^* \mathbf P_2^*+2\mathbf P_1^* \mathbf P_{1}^{'*}}{(\mathbf P_2^{'*2}-2\mathbf P_2^{'*}\mathbf P_{1}^{'*}+\mathbf P_{1}^{'*2})} \right )[/math]


[math] \mathfrak{M}_{e^-e^-}= e^2\left ( \frac{2\mathbf P_1^* \mathbf P_2^*+2\mathbf P_{1}^{'*} \mathbf P_2^*}{(\mathbf P_2^*-\mathbf P_{2}^{'*})^2}- \frac{2\mathbf P_1^* \mathbf P_2^*+2\mathbf P_1^* \mathbf P_{1}^{'*}}{(\mathbf P_2^{'*}-\mathbf P_{1}^{'*})^2} \right )[/math]



[math] \mathfrak{M}_{e^-e^-}= e^2\left ( \frac{ (\mathbf P_1^{*2}-2 \mathbf P_1^* \mathbf P_2^{'*}+ \mathbf P_2^{'*2})-(\mathbf P_1^{*2}+2 \mathbf P_1^* \mathbf P_2^*+ \mathbf P_2^{*2})}{t}- \frac{(\mathbf P_1^{*2}-2 \mathbf P_1^* \mathbf P_1^{'*}+ \mathbf P_1^{'*2})-( \mathbf P_1^{*2}+2 \mathbf P_1^* \mathbf P_2^*+ \mathbf P_2^{*2})}{u} \right )[/math]


[math]\mathfrak{M}_{e^-e^-}=e^2 \left (\frac{u-s}{t}+\frac{t-s}{u} \right )[/math]



[math]\underline{\textbf{Navigation}}[/math]

[math]\vartriangleleft [/math] [math]\triangle [/math] [math]\vartriangleright [/math]