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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°). This implies for arccos -1, the range will include imaginary numbers. Knowing that the range of the cosine function is -1 to +1 inclusive and the domain to be any angle | | 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°). This implies |
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| | | <center><math>\theta_{max}=180^{\circ}</math></center> |
| <center><math>z = \arccos{-1}</math></center> | |
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| <center><math>\cos{z} = \cos{\arccos{-1}}</math></center>
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| <center><math>\cos{z} = 3</math></center>
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| From Euler's formula
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| <center><math>\cos{x} = \frac{e^{i z} + e^{-i z}}{2}</math></center>
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| <center><math> \frac{e^{i z} + e^{-i z}}{2} = 3</math></center>
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| <center><math> e^{i z} + e^{-i z} = 6</math></center>
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| Multiply with <math>e^{i z}</math>
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| <center><math>e^{2i z} + 1 = 6e^{i z}</math></center>
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| <center>Letting <math>y = e^{i z}</math></center>
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| We get an quadratic equation:
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| <center><math>y^2 - 6y + 1 = 0</math></center>
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| <center><math>y = (6 ± √32)/2</math></center>
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| <center><math>y_1 = 5.828427 = e^{i z}</math></center>
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| <center><math>y_2 = 0.171573 = e^{i z}</math></center>
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| Apply the natural log on both sides gives the solution for arccos 3:
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| <center><math>z_1 = \frac{ln(5.828427) }{ i}=-1.76275i</math></center>
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| <center><math> z_2 = \frac{ln(0.171573) }{ i}=1.76275i</math></center>
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| Converting to polar coordinates:
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| <center><math>r \equiv |z|=sqrt{z^2}=1.76275</math></center>
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| Since
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| <center><math>z \equiv a+bi</math></center>
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| <center><math>\theta \equiv \arctan{\frac{\pm b}{a}}=\pm 90^{\circ}</math></center>
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| Since there is only an imaginary component.
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| <center>[[File:Imaginary_plane.png|thumb|center|500px|alt=Angle Theta with respect to imaginary plane|'''Figure 2.1:''' Angle theta measured with respect to imaginary plane. ]]</center>
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[math]\displaystyle{ \textbf{\underline{Navigation}} }[/math]
[math]\displaystyle{ \vartriangleleft }[/math]
[math]\displaystyle{ \triangle }[/math]
[math]\displaystyle{ \vartriangleright }[/math]
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]\displaystyle{ s+t+u=(4(m^2+ p \ ^{*2}))+(-2 p \ ^{*2}(1-cos\ \theta))+(-2 p \ ^{*2}(1+cos\ \theta)) }[/math]
[math]\displaystyle{ s+t+u \equiv 4m^2 }[/math]
Since
[math]\displaystyle{ s \equiv 4(m^2+\vec p \ ^{*2}) }[/math]
This implies
[math]\displaystyle{ s \ge 4m^2 }[/math]
In turn, this implies
[math]\displaystyle{ t \le 0 \qquad u \le 0 }[/math]
At the condition both t and u are equal to zero, we find
[math]\displaystyle{ t = 0 \qquad u = 0 }[/math]
[math]\displaystyle{ -2 p \ ^{*2}(1-cos\ \theta) = 0 \qquad -2 p \ ^{*2}(1+cos\ \theta) = 0 }[/math]
[math]\displaystyle{ (-2 p \ ^{*2}+2 p \ ^{*2}cos\ \theta) = 0 \qquad (-2 p \ ^{*2}-2 p \ ^{*2}cos\ \theta) = 0 }[/math]
[math]\displaystyle{ 2 p \ ^{*2}cos\ \theta = 2 p \ ^{*2} \qquad -2 p \ ^{*2}cos\ \theta = 2 p \ ^{*2} }[/math]
[math]\displaystyle{ \cos\ \theta = 1 \qquad \cos\ \theta = -1 }[/math]
[math]\displaystyle{ \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 maximum of t
[math]\displaystyle{ s+t_{max} \equiv 4m^2 }[/math]
[math]\displaystyle{ \Rightarrow t_{max}=4m^2-s }[/math]
[math]\displaystyle{ t_{max}=4m^2-4m^2- 4p \ ^{*2} }[/math]
[math]\displaystyle{ t_{max}=-4p \ ^{*2} }[/math]
[math]\displaystyle{ -2 p \ ^{*2}(1-cos\ \theta)=-4p \ ^{*2} }[/math]
[math]\displaystyle{ (1-cos\ \theta)=2 }[/math]
[math]\displaystyle{ -cos\ \theta=1 }[/math]
[math]\displaystyle{ \theta_{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°). This implies
[math]\displaystyle{ \theta_{max}=180^{\circ} }[/math]