Issues concerning the 3-particle phase space in d dimensions
$begingroup$
I am not sure whether this problem fits better in the physics section or not.
Recently I came across a problem concerning the 3-particle phase space. I am trying to show, that the 3-particle phase space for massless particles with momenta $p_1$, $p_2$ and $k$ is given by
$$
dPhi_3=frac{s^{1-2epsilon}}{16(2pi)^3}frac{(4pi)^{2epsilon}}{Gamma(2-2epsilon)}frac{dx_1dx_2}{[(1-x_1)(1-x_2)(1-x_3)]^{epsilon}},
$$
where I start with
$$
dPhi_3=frac{1}{8(2pi)^{2d-3}}(p_1^0)^{d-3}(p_2^0)^{d-3}frac{1}{k^0}dp_1^0dp_2^0dOmega_1^{d-3}dOmega_2^{d-2}(1-cos^2(theta_{12}))^{frac{d-4}{2}}dcos(theta_{12})delta(p_1^0+p_2^0+k^0-sqrt{s}).
$$
To clarify notation: $d=4-2epsilon$, $theta_{12}$ is the angle of $vec{p_2}$ with respect to $vec{p_1}$, $x_i=frac{2p_i^0}{sqrt{s}}$ and $dOmega^{d-1}$ is an element of solid angle for the surface of the d dimensional unit ball.
First of all: Is this formula correct?
For the $cos(theta_{12})$ part, I stick to the instructions provided here: https://www.ippp.dur.ac.uk/~krauss/Lectures/QuarksLeptons/Basics/PS_2.html
Therefore I get after integration
$$
(1-cos^2(theta_{12}))^{frac{d-4}{2}}dcos(theta_{12})delta(p_1^0+p_2^0+k^0-sqrt{s})frac{1}{k^0}=(1-u_0^2)^{-epsilon}frac{4}{x_1x_2s}
$$
with $u_0=frac{2}{x_1x_2}+x_1+x_2-1-frac{1}{2}x_1x_2$.
What I am finally left with is
$$
dPhi_3=frac{s^{1-2epsilon}}{16(2pi)^3}frac{(4pi)^{2epsilon}}{Gamma(2-2epsilon)}left(frac{2}{x_1x_2}right)^{epsilon}dx_1dx_2[1-u_0^2]^{-epsilon}.
$$
Looks rather similar to the desired result, but I cannot manage the final steps in my calculation. Any suggestions how to deal with this problem?
integration physics
asked Dec 5 '18 at 23:42
SchnarcoSchnarco
1518
$endgroup$
add a comment |
$begingroup$
I am not sure whether this problem fits better in the physics section or not.
Recently I came across a problem concerning the 3-particle phase space. I am trying to show, that the 3-particle phase space for massless particles with momenta $p_1$, $p_2$ and $k$ is given by
$$
dPhi_3=frac{s^{1-2epsilon}}{16(2pi)^3}frac{(4pi)^{2epsilon}}{Gamma(2-2epsilon)}frac{dx_1dx_2}{[(1-x_1)(1-x_2)(1-x_3)]^{epsilon}},
$$
where I start with
$$
dPhi_3=frac{1}{8(2pi)^{2d-3}}(p_1^0)^{d-3}(p_2^0)^{d-3}frac{1}{k^0}dp_1^0dp_2^0dOmega_1^{d-3}dOmega_2^{d-2}(1-cos^2(theta_{12}))^{frac{d-4}{2}}dcos(theta_{12})delta(p_1^0+p_2^0+k^0-sqrt{s}).
$$
To clarify notation: $d=4-2epsilon$, $theta_{12}$ is the angle of $vec{p_2}$ with respect to $vec{p_1}$, $x_i=frac{2p_i^0}{sqrt{s}}$ and $dOmega^{d-1}$ is an element of solid angle for the surface of the d dimensional unit ball.
First of all: Is this formula correct?
For the $cos(theta_{12})$ part, I stick to the instructions provided here: https://www.ippp.dur.ac.uk/~krauss/Lectures/QuarksLeptons/Basics/PS_2.html
Therefore I get after integration
$$
(1-cos^2(theta_{12}))^{frac{d-4}{2}}dcos(theta_{12})delta(p_1^0+p_2^0+k^0-sqrt{s})frac{1}{k^0}=(1-u_0^2)^{-epsilon}frac{4}{x_1x_2s}
$$
with $u_0=frac{2}{x_1x_2}+x_1+x_2-1-frac{1}{2}x_1x_2$.
What I am finally left with is
$$
dPhi_3=frac{s^{1-2epsilon}}{16(2pi)^3}frac{(4pi)^{2epsilon}}{Gamma(2-2epsilon)}left(frac{2}{x_1x_2}right)^{epsilon}dx_1dx_2[1-u_0^2]^{-epsilon}.
$$
Looks rather similar to the desired result, but I cannot manage the final steps in my calculation. Any suggestions how to deal with this problem?
integration physics
asked Dec 5 '18 at 23:42
SchnarcoSchnarco
1518
$endgroup$
add a comment |
$begingroup$
I am not sure whether this problem fits better in the physics section or not.
Recently I came across a problem concerning the 3-particle phase space. I am trying to show, that the 3-particle phase space for massless particles with momenta $p_1$, $p_2$ and $k$ is given by
$$
dPhi_3=frac{s^{1-2epsilon}}{16(2pi)^3}frac{(4pi)^{2epsilon}}{Gamma(2-2epsilon)}frac{dx_1dx_2}{[(1-x_1)(1-x_2)(1-x_3)]^{epsilon}},
$$
where I start with
$$
dPhi_3=frac{1}{8(2pi)^{2d-3}}(p_1^0)^{d-3}(p_2^0)^{d-3}frac{1}{k^0}dp_1^0dp_2^0dOmega_1^{d-3}dOmega_2^{d-2}(1-cos^2(theta_{12}))^{frac{d-4}{2}}dcos(theta_{12})delta(p_1^0+p_2^0+k^0-sqrt{s}).
$$
To clarify notation: $d=4-2epsilon$, $theta_{12}$ is the angle of $vec{p_2}$ with respect to $vec{p_1}$, $x_i=frac{2p_i^0}{sqrt{s}}$ and $dOmega^{d-1}$ is an element of solid angle for the surface of the d dimensional unit ball.
First of all: Is this formula correct?
For the $cos(theta_{12})$ part, I stick to the instructions provided here: https://www.ippp.dur.ac.uk/~krauss/Lectures/QuarksLeptons/Basics/PS_2.html
Therefore I get after integration
$$
(1-cos^2(theta_{12}))^{frac{d-4}{2}}dcos(theta_{12})delta(p_1^0+p_2^0+k^0-sqrt{s})frac{1}{k^0}=(1-u_0^2)^{-epsilon}frac{4}{x_1x_2s}
$$
with $u_0=frac{2}{x_1x_2}+x_1+x_2-1-frac{1}{2}x_1x_2$.
What I am finally left with is
$$
dPhi_3=frac{s^{1-2epsilon}}{16(2pi)^3}frac{(4pi)^{2epsilon}}{Gamma(2-2epsilon)}left(frac{2}{x_1x_2}right)^{epsilon}dx_1dx_2[1-u_0^2]^{-epsilon}.
$$
Looks rather similar to the desired result, but I cannot manage the final steps in my calculation. Any suggestions how to deal with this problem?
integration physics
asked Dec 5 '18 at 23:42
SchnarcoSchnarco
1518
$endgroup$
I am not sure whether this problem fits better in the physics section or not.
Recently I came across a problem concerning the 3-particle phase space. I am trying to show, that the 3-particle phase space for massless particles with momenta $p_1$, $p_2$ and $k$ is given by
$$
dPhi_3=frac{s^{1-2epsilon}}{16(2pi)^3}frac{(4pi)^{2epsilon}}{Gamma(2-2epsilon)}frac{dx_1dx_2}{[(1-x_1)(1-x_2)(1-x_3)]^{epsilon}},
$$
where I start with
$$
dPhi_3=frac{1}{8(2pi)^{2d-3}}(p_1^0)^{d-3}(p_2^0)^{d-3}frac{1}{k^0}dp_1^0dp_2^0dOmega_1^{d-3}dOmega_2^{d-2}(1-cos^2(theta_{12}))^{frac{d-4}{2}}dcos(theta_{12})delta(p_1^0+p_2^0+k^0-sqrt{s}).
$$
To clarify notation: $d=4-2epsilon$, $theta_{12}$ is the angle of $vec{p_2}$ with respect to $vec{p_1}$, $x_i=frac{2p_i^0}{sqrt{s}}$ and $dOmega^{d-1}$ is an element of solid angle for the surface of the d dimensional unit ball.
First of all: Is this formula correct?
For the $cos(theta_{12})$ part, I stick to the instructions provided here: https://www.ippp.dur.ac.uk/~krauss/Lectures/QuarksLeptons/Basics/PS_2.html
Therefore I get after integration
$$
(1-cos^2(theta_{12}))^{frac{d-4}{2}}dcos(theta_{12})delta(p_1^0+p_2^0+k^0-sqrt{s})frac{1}{k^0}=(1-u_0^2)^{-epsilon}frac{4}{x_1x_2s}
$$
with $u_0=frac{2}{x_1x_2}+x_1+x_2-1-frac{1}{2}x_1x_2$.
What I am finally left with is
$$
dPhi_3=frac{s^{1-2epsilon}}{16(2pi)^3}frac{(4pi)^{2epsilon}}{Gamma(2-2epsilon)}left(frac{2}{x_1x_2}right)^{epsilon}dx_1dx_2[1-u_0^2]^{-epsilon}.
$$
Looks rather similar to the desired result, but I cannot manage the final steps in my calculation. Any suggestions how to deal with this problem?
integration physics
integration physics
asked Dec 5 '18 at 23:42
SchnarcoSchnarco
1518
asked Dec 5 '18 at 23:42
SchnarcoSchnarco
1518
asked Dec 5 '18 at 23:42
SchnarcoSchnarco
1518
asked Dec 5 '18 at 23:42
SchnarcoSchnarco
1518
asked Dec 5 '18 at 23:42
SchnarcoSchnarco
1518
1518
add a comment |
add a comment |
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