Mandelstam Representation

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Mandelstam Representation

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

[math]\Longrightarrow \left({\mathbf P_1^*}- {\mathbf P_1^{'*}}\right)^2=\left({\mathbf P_2^*}- {\mathbf P_2^{'*}}\right)^2\equiv t[/math]

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

s Channel

The s quantity is known as the square of the center of mass energy (invariant mass)

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

[math]s \equiv \left({\mathbf P_1^*}+ {\mathbf P_2^{*}}\right)^2[/math]

[math]s \equiv \mathbf P_1^{*2}+2 \mathbf P_1^* \mathbf P_2^*+ \mathbf P_2^{*2}[/math]

As shown earlier, the square of a 4-momentum is

[math]\mathbf P^{2} \equiv m^2[/math]

[math]s \equiv m^{2}+2 \mathbf P_1^* \mathbf P_2^*+ m_2^{2}[/math]

This gives

[math]s \equiv 2m^{2}+2 \mathbf P_1^* \mathbf P_2^*[/math]

Similarly, the scalar product of two 4-momentums

[math]s \equiv 2m^2+2(E_1^*E_2^*-\vec p_1^* \vec p_2^*)[/math]

In the center of mass frame of reference,

[math]E_1^*=E_2^* \quad and \quad \vec p_1^*=-\vec p_2^*[/math]

[math]s \equiv 2m^2+2(E_1^{*2}+\vec p_1^{*2} )[/math]

Using the relativistic energy equation

[math]E^2 \equiv p^2+m^2[/math]

[math]s \equiv 2m^2+2((m^2+p_1^{*2})+p_1^{*2})[/math]

[math]s=4(m_{CM}^2+p_{CM}^2)[/math]

t Channel

The t quantity is known as the square of the 4-momentum transfer

[math]t \equiv \left({\mathbf P_1^*}- {\mathbf P_1^{'*}}\right)^2=\left({\mathbf P_2^{*}}+ {\mathbf P_2^{'*}}\right)^2[/math]

[math]t \equiv \left({\mathbf P_1^*}- {\mathbf P_1^{'*}}\right)^2[/math]

[math]t \equiv P_1^{*2}-2P_1^*P_1^{'*}+P_1^{'*2}[/math]

[math]t \equiv 2m_1^2-2E_1^*E_1^{'*}+2p_1^*p_1^{'*}[/math]

[math]t \equiv 2m_1^*-2E_1^{*2}+2p_1^{*2}cos\ \theta[/math]

[math]t \equiv -2p_1^{*2}(1-cos\ \theta)[/math]

u Channel

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