Difference between revisions of "Limits based on Mandelstam Variables"

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<center><math>\underline{\textbf{Navigation}}</math>
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[[VanWasshenova_Thesis#Moller_Scattering|<math>\triangle </math>]]
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=Limits based on Mandelstam Variables=
 
=Limits based on Mandelstam Variables=
  
<center><math>\Longrightarrow \left({\mathbf P_1^*}+ {\mathbf P_2^{*}}\right)^2=\left({\mathbf P_1^{'*}}+ {\mathbf P_2^{'*}}\right)^2\equiv s</math></center>
 
  
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Since the Mandelstam variables are the scalar product of 4-momenta, which are invariants, they are invariants as well.  The sum of these invariant variables must also be invariant as well.  Find the sum of the 3 Mandelstam variables when the two particles have equal mass in the center of mass frame gives:
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<center><math>s+t+u=(4(m^2+ p \ ^{*2}))+(-2 p \ ^{*2}(1-cos\ \theta))+(-2 p \ ^{*2}(1+cos\ \theta))</math></center>
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<center><math>s+t+u \equiv 4m^2</math></center>
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Since
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<center><math>s \equiv 4(m^2+\vec p \ ^{*2})</math></center>
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This implies
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<center><math>s \ge 4m^2</math></center>
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In turn, this implies
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<center><math> t \le 0  \qquad u \le 0</math></center>
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At the condition both t and u are equal to zero, we find
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<center><math> t = 0  \qquad u = 0</math></center>
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<center><math>-2 p \ ^{*2}(1-cos\ \theta) = 0  \qquad -2 p \ ^{*2}(1+cos\ \theta) = 0</math></center>
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<center><math>(-2 p \ ^{*2}+2 p \ ^{*2}cos\ \theta) = 0  \qquad (-2 p \ ^{*2}-2 p \ ^{*2}cos\ \theta) = 0</math></center>
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<center><math>2 p \ ^{*2}cos\ \theta = 2 p \ ^{*2}  \qquad -2 p \ ^{*2}cos\ \theta = 2 p \ ^{*2}</math></center>
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<center><math>\cos\ \theta = 1  \qquad \cos\ \theta = -1</math></center>
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<center><math>\Rightarrow  \theta_{t=0} = \arccos \ 1=0^{\circ}  \qquad \theta_{u=0} = \arccos \ -1=180^{\circ}</math></center>
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Holding u constant at zero we can find the minimum of t
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<center><math>s+t_{max} \equiv 4m^2</math></center>
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<center><math>\Rightarrow t_{max}=4m^2-s</math></center>
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<center><math>t_{max}=4m^2-4m^2- 4p \ ^{*2}</math></center>
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The maximum transfer of momentum would be
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<center><math>t_{max}=-4p \ ^{*2}</math></center>
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<center><math>-2 p \ ^{*2}(1-cos\ \theta_{t=max})=-4p \ ^{*2}</math></center>
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<center><math>(1-cos\ \theta_{t=max})=2</math></center>
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<center><math>-cos\ \theta_{t=max}=1</math></center>
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<center><math> \theta_{t=max} \equiv \arccos -1</math></center>
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The domain of the arccos function is from −1 to +1 inclusive and the range is from 0 to π radians inclusive (or from 0° to 180°).  We find as expected for u=0 at <math>\theta=180^{\circ}</math>
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<center><math>\theta_{t=max}=180^{\circ}</math></center>
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However, from the definition of s being invariant between frames of reference
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<center><math>s \equiv \overbrace{\left({\mathbf P_1^*}+ {\mathbf P_2^{*}}\right)^2=\left({\mathbf P_1^{'*}}+ {\mathbf P_2^{'*}}\right)^2}^{CM\ FRAME}=\overbrace{\left({\mathbf P_1}+ {\mathbf P_2}\right)^2 = \left({\mathbf P_1^{'}}+ {\mathbf P_2^{'}}\right)^2}^{LAB\ FRAME}</math></center>
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In the center of mass frame of reference,
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<center><math>E_1^*=E_2^*=E^* \quad and \quad \vec p \ _1^*=-\vec p \ _2^*= \vec p \ ^*</math></center>
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<center><math>s \equiv 2m^2+2(E_1^{*2}+\vec p \ ^{*2} )</math></center>
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Using the relativistic energy equation
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<center><math>E^2 \equiv \vec p \ ^2+m^2</math></center>
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<center><math>s \equiv 2m^2+2((m^2+\vec p \ ^{*2})+\vec p \ ^{*2})</math></center>
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<center><math>s=4m^2+4 \vec p \ ^{*2})</math></center>
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<center><math>\frac{s-4m^2}{4}= \vec p \ ^{*2}</math></center>
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<center><math>t=-2 p \ ^{*2}(1-cos\ \theta)=\frac{-2(s-4m^2)}{4}(1-cos\ \theta)</math></center>
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<center><math>\frac{-2t}{s-4m^2}=(1-cos\ \theta)</math></center>
  
In the center of mass frame, the momentum of the particles interacting are equal and opposite, i.e. <math>p_1=-p_2</math>.  However, the 4-momentum still retains an energy component, which as a scalar quantity, can not be countered by another particle's direction of motion.
 
  
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<center><math>cos\ \theta=1-\frac{-2t}{s-4m^2}</math></center>
  
<center><math>{\mathbf P_1^*}\equiv \left(\begin{matrix} E_1\\ p_{x_1} \\ p_{y_1} \\ p_{z_1} \end{matrix} \right) \ \ \ \ {\mathbf P_2^*}\equiv \left(\begin{matrix} E_2\\ p_{x_2} \\ p_{y_2} \\ p_{z_2} \end{matrix} \right)</math></center>
 
  
  
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----
  
<center><math> \left({\mathbf P_1^*}+ {\mathbf P_2^{*}}\right)^2=\left( \left(\begin{matrix} E_1\\ p_{x_1} \\ p_{y_1} \\ p_{z_1} \end{matrix} \right)+\left(\begin{matrix} E_2\\ p_{x_2} \\ p_{y_2} \\ p_{z_2} \end{matrix} \right) \right)</math></center>
 
  
  
<center><math> \left({\mathbf P_1^*}+ {\mathbf P_2^{*}}\right)^2=\left( \left(\begin{matrix} E_1\\ p_{x_1} \\ p_{y_1} \\ p_{z_1} \end{matrix} \right)+\left(\begin{matrix} E_2\\ -p_{x_1} \\ -p_{y_1} \\ -p_{z_1} \end{matrix} \right) \right)</math></center>
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<center><math> \left({\mathbf P_1^*}+ {\mathbf P_2^{*}}\right)^2= \left(\begin{matrix} E_1+E_2\\0 \\ 0 \\ 0 \end{matrix} \right)</math></center>
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</center>

Latest revision as of 19:07, 1 January 2019

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

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Limits based on Mandelstam Variables

Since the Mandelstam variables are the scalar product of 4-momenta, which are invariants, they are invariants as well. The sum of these invariant variables must also be invariant as well. Find the sum of the 3 Mandelstam variables when the two particles have equal mass in the center of mass frame gives:


[math]s+t+u=(4(m^2+ p \ ^{*2}))+(-2 p \ ^{*2}(1-cos\ \theta))+(-2 p \ ^{*2}(1+cos\ \theta))[/math]


[math]s+t+u \equiv 4m^2[/math]


Since

[math]s \equiv 4(m^2+\vec p \ ^{*2})[/math]


This implies

[math]s \ge 4m^2[/math]


In turn, this implies


[math] t \le 0 \qquad u \le 0[/math]


At the condition both t and u are equal to zero, we find


[math] t = 0 \qquad u = 0[/math]


[math]-2 p \ ^{*2}(1-cos\ \theta) = 0 \qquad -2 p \ ^{*2}(1+cos\ \theta) = 0[/math]


[math](-2 p \ ^{*2}+2 p \ ^{*2}cos\ \theta) = 0 \qquad (-2 p \ ^{*2}-2 p \ ^{*2}cos\ \theta) = 0[/math]


[math]2 p \ ^{*2}cos\ \theta = 2 p \ ^{*2} \qquad -2 p \ ^{*2}cos\ \theta = 2 p \ ^{*2}[/math]


[math]\cos\ \theta = 1 \qquad \cos\ \theta = -1[/math]


[math]\Rightarrow \theta_{t=0} = \arccos \ 1=0^{\circ} \qquad \theta_{u=0} = \arccos \ -1=180^{\circ}[/math]

Holding u constant at zero we can find the minimum of t


[math]s+t_{max} \equiv 4m^2[/math]


[math]\Rightarrow t_{max}=4m^2-s[/math]


[math]t_{max}=4m^2-4m^2- 4p \ ^{*2}[/math]



The maximum transfer of momentum would be


[math]t_{max}=-4p \ ^{*2}[/math]



[math]-2 p \ ^{*2}(1-cos\ \theta_{t=max})=-4p \ ^{*2}[/math]


[math](1-cos\ \theta_{t=max})=2[/math]


[math]-cos\ \theta_{t=max}=1[/math]


[math] \theta_{t=max} \equiv \arccos -1[/math]


The domain of the arccos function is from −1 to +1 inclusive and the range is from 0 to π radians inclusive (or from 0° to 180°). We find as expected for u=0 at [math]\theta=180^{\circ}[/math]

[math]\theta_{t=max}=180^{\circ}[/math]


However, from the definition of s being invariant between frames of reference

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


In the center of mass frame of reference,

[math]E_1^*=E_2^*=E^* \quad and \quad \vec p \ _1^*=-\vec p \ _2^*= \vec p \ ^*[/math]


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


Using the relativistic energy equation

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


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


[math]s=4m^2+4 \vec p \ ^{*2})[/math]


[math]\frac{s-4m^2}{4}= \vec p \ ^{*2}[/math]



[math]t=-2 p \ ^{*2}(1-cos\ \theta)=\frac{-2(s-4m^2)}{4}(1-cos\ \theta)[/math]


[math]\frac{-2t}{s-4m^2}=(1-cos\ \theta)[/math]


[math]cos\ \theta=1-\frac{-2t}{s-4m^2}[/math]




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

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