Mixing
Solution of complicated transcendental equation
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I am trying to reproduce a result from https://arxiv.org/pdf/0811.2230.pdf Particularly, I am trying to compute the total inelasticity and make the same plot as in fig.1. However, I am unable to explicitly express $K_theta$ from eq.21, and therefore I cannot make the integral as stated in eq.22.
The transcendental equation for $K_theta$ is
$(1-K_theta)sqrt{s} = F + betasqrt{(F^2-s_p)}cos(theta) $ $qquad qquad qquad qquad $ (1)
where
$beta = sqrt{1-frac{s}{E^2}}$
$F = frac{1}{2sqrt{s}} (s + s_p - s_x) $
and
$ s=2sqrt{s} , epsilon + s_p $
$ s_p = 2delta_pE_p^2 + m_p^2 $
$ s_x = 2delta_xE_x^2 + m_x^2 $
$E_p = (1-K_theta)E$
$E_x = K_theta E$
When I plugged everything to (1) I obtained
full expresion (in this picture K means $K_theta$, this is what I want to express)
$K_theta$ is a function of $epsilon,E$ and $m_p, m_x, delta_p, delta_x$ are known constants.
I tried to solve this with Maple and Matlab by using the function solve(), but I obtained solution on several pages and further I was unable to make the integral
$int_{0}^{pi} K_theta dtheta $
Thank you very much for any help.
transcendental-equations
asked Jan 27 at 13:07
TomášTomáš
63
$endgroup$
add a comment |
$begingroup$
I am trying to reproduce a result from https://arxiv.org/pdf/0811.2230.pdf Particularly, I am trying to compute the total inelasticity and make the same plot as in fig.1. However, I am unable to explicitly express $K_theta$ from eq.21, and therefore I cannot make the integral as stated in eq.22.
The transcendental equation for $K_theta$ is
$(1-K_theta)sqrt{s} = F + betasqrt{(F^2-s_p)}cos(theta) $ $qquad qquad qquad qquad $ (1)
where
$beta = sqrt{1-frac{s}{E^2}}$
$F = frac{1}{2sqrt{s}} (s + s_p - s_x) $
and
$ s=2sqrt{s} , epsilon + s_p $
$ s_p = 2delta_pE_p^2 + m_p^2 $
$ s_x = 2delta_xE_x^2 + m_x^2 $
$E_p = (1-K_theta)E$
$E_x = K_theta E$
When I plugged everything to (1) I obtained
full expresion (in this picture K means $K_theta$, this is what I want to express)
$K_theta$ is a function of $epsilon,E$ and $m_p, m_x, delta_p, delta_x$ are known constants.
I tried to solve this with Maple and Matlab by using the function solve(), but I obtained solution on several pages and further I was unable to make the integral
$int_{0}^{pi} K_theta dtheta $
Thank you very much for any help.
transcendental-equations
asked Jan 27 at 13:07
TomášTomáš
63
$endgroup$
$begingroup$
For which variable want you to solve the equation?
$endgroup$
– Dr. Sonnhard Graubner
Jan 27 at 14:14
$begingroup$
For the $K_theta$, so I could subsequently integrate it over $theta$
$endgroup$
– Tomáš
Jan 27 at 15:02
add a comment |
$begingroup$
I am trying to reproduce a result from https://arxiv.org/pdf/0811.2230.pdf Particularly, I am trying to compute the total inelasticity and make the same plot as in fig.1. However, I am unable to explicitly express $K_theta$ from eq.21, and therefore I cannot make the integral as stated in eq.22.
The transcendental equation for $K_theta$ is
$(1-K_theta)sqrt{s} = F + betasqrt{(F^2-s_p)}cos(theta) $ $qquad qquad qquad qquad $ (1)
where
$beta = sqrt{1-frac{s}{E^2}}$
$F = frac{1}{2sqrt{s}} (s + s_p - s_x) $
and
$ s=2sqrt{s} , epsilon + s_p $
$ s_p = 2delta_pE_p^2 + m_p^2 $
$ s_x = 2delta_xE_x^2 + m_x^2 $
$E_p = (1-K_theta)E$
$E_x = K_theta E$
When I plugged everything to (1) I obtained
full expresion (in this picture K means $K_theta$, this is what I want to express)
$K_theta$ is a function of $epsilon,E$ and $m_p, m_x, delta_p, delta_x$ are known constants.
I tried to solve this with Maple and Matlab by using the function solve(), but I obtained solution on several pages and further I was unable to make the integral
$int_{0}^{pi} K_theta dtheta $
Thank you very much for any help.
transcendental-equations
asked Jan 27 at 13:07
TomášTomáš
63
$endgroup$
I am trying to reproduce a result from https://arxiv.org/pdf/0811.2230.pdf Particularly, I am trying to compute the total inelasticity and make the same plot as in fig.1. However, I am unable to explicitly express $K_theta$ from eq.21, and therefore I cannot make the integral as stated in eq.22.
The transcendental equation for $K_theta$ is
$(1-K_theta)sqrt{s} = F + betasqrt{(F^2-s_p)}cos(theta) $ $qquad qquad qquad qquad $ (1)
where
$beta = sqrt{1-frac{s}{E^2}}$
$F = frac{1}{2sqrt{s}} (s + s_p - s_x) $
and
$ s=2sqrt{s} , epsilon + s_p $
$ s_p = 2delta_pE_p^2 + m_p^2 $
$ s_x = 2delta_xE_x^2 + m_x^2 $
$E_p = (1-K_theta)E$
$E_x = K_theta E$
When I plugged everything to (1) I obtained
full expresion (in this picture K means $K_theta$, this is what I want to express)
$K_theta$ is a function of $epsilon,E$ and $m_p, m_x, delta_p, delta_x$ are known constants.
I tried to solve this with Maple and Matlab by using the function solve(), but I obtained solution on several pages and further I was unable to make the integral
$int_{0}^{pi} K_theta dtheta $
Thank you very much for any help.
transcendental-equations
transcendental-equations
asked Jan 27 at 13:07
TomášTomáš
63
asked Jan 27 at 13:07
TomášTomáš
63
edited Jan 27 at 15:06
Tomáš
asked Jan 27 at 13:07
TomášTomáš
63
asked Jan 27 at 13:07
TomášTomáš
63
asked Jan 27 at 13:07
TomášTomáš
63
63
$begingroup$
For which variable want you to solve the equation?
$endgroup$
– Dr. Sonnhard Graubner
Jan 27 at 14:14
$begingroup$
For the $K_theta$, so I could subsequently integrate it over $theta$
$endgroup$
– Tomáš
Jan 27 at 15:02
add a comment |
$begingroup$
For which variable want you to solve the equation?
$endgroup$
– Dr. Sonnhard Graubner
Jan 27 at 14:14
$begingroup$
For the $K_theta$, so I could subsequently integrate it over $theta$
$endgroup$
– Tomáš
Jan 27 at 15:02
$begingroup$
For which variable want you to solve the equation?
$endgroup$
– Dr. Sonnhard Graubner
Jan 27 at 14:14
$begingroup$
For which variable want you to solve the equation?
$endgroup$
– Dr. Sonnhard Graubner
Jan 27 at 14:14
$begingroup$
For the $K_theta$, so I could subsequently integrate it over $theta$
$endgroup$
– Tomáš
Jan 27 at 15:02
$begingroup$
For the $K_theta$, so I could subsequently integrate it over $theta$
$endgroup$
– Tomáš
Jan 27 at 15:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Fist of all, Welcome to the site !.
In my humble opinion, do not waste your time. All the work is purely numerical.
- Generate a detailed table of $K$ as a function of $theta$
- Build an interpolating function
- Numerical integration
For the first point, starting with $theta_0=frac pi 2$ makes the problem simple and you get $K_0$. Now $theta_i=theta_{i-1}+Delta$; solve the equation starting with $K_{i-1}$ as a guess to get $K_i$ and continue. Repeat the process with new values $theta_j=theta_{j-1}-Delta$.
answered Jan 27 at 15:29
Claude LeiboviciClaude Leibovici
125k1158135
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add a comment |
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$begingroup$
Fist of all, Welcome to the site !.
In my humble opinion, do not waste your time. All the work is purely numerical.
- Generate a detailed table of $K$ as a function of $theta$
- Build an interpolating function
- Numerical integration
For the first point, starting with $theta_0=frac pi 2$ makes the problem simple and you get $K_0$. Now $theta_i=theta_{i-1}+Delta$; solve the equation starting with $K_{i-1}$ as a guess to get $K_i$ and continue. Repeat the process with new values $theta_j=theta_{j-1}-Delta$.
answered Jan 27 at 15:29
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
Fist of all, Welcome to the site !.
In my humble opinion, do not waste your time. All the work is purely numerical.
- Generate a detailed table of $K$ as a function of $theta$
- Build an interpolating function
- Numerical integration
For the first point, starting with $theta_0=frac pi 2$ makes the problem simple and you get $K_0$. Now $theta_i=theta_{i-1}+Delta$; solve the equation starting with $K_{i-1}$ as a guess to get $K_i$ and continue. Repeat the process with new values $theta_j=theta_{j-1}-Delta$.
answered Jan 27 at 15:29
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
Fist of all, Welcome to the site !.
In my humble opinion, do not waste your time. All the work is purely numerical.
- Generate a detailed table of $K$ as a function of $theta$
- Build an interpolating function
- Numerical integration
For the first point, starting with $theta_0=frac pi 2$ makes the problem simple and you get $K_0$. Now $theta_i=theta_{i-1}+Delta$; solve the equation starting with $K_{i-1}$ as a guess to get $K_i$ and continue. Repeat the process with new values $theta_j=theta_{j-1}-Delta$.
answered Jan 27 at 15:29
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
Fist of all, Welcome to the site !.
In my humble opinion, do not waste your time. All the work is purely numerical.
- Generate a detailed table of $K$ as a function of $theta$
- Build an interpolating function
- Numerical integration
For the first point, starting with $theta_0=frac pi 2$ makes the problem simple and you get $K_0$. Now $theta_i=theta_{i-1}+Delta$; solve the equation starting with $K_{i-1}$ as a guess to get $K_i$ and continue. Repeat the process with new values $theta_j=theta_{j-1}-Delta$.
answered Jan 27 at 15:29
Claude LeiboviciClaude Leibovici
125k1158135
answered Jan 27 at 15:29
Claude LeiboviciClaude Leibovici
125k1158135
answered Jan 27 at 15:29
Claude LeiboviciClaude Leibovici
125k1158135
answered Jan 27 at 15:29
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
add a comment |
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$begingroup$
For which variable want you to solve the equation?
$endgroup$
– Dr. Sonnhard Graubner
Jan 27 at 14:14
$begingroup$
For the $K_theta$, so I could subsequently integrate it over $theta$
$endgroup$
– Tomáš
Jan 27 at 15:02