How to randomly pick a solution to a system of polynomial equations that has infinite solutions?
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I am a programmer trying to simulate the motion of particles involved in $beta^-$ decay (in 2D). While trying to conserve energy and momentum throughout the process, I end up with a system of equations for the trajectories of the electron ($_e$) and neutrino ($_n$) after $beta^-$ decay,
$$ |v_n|^2 = mathbfv_xn^2 + mathbfv_yn^2 $$
$$ |v_e|^2 = mathbfv_xe^2 + mathbfv_ye^2 $$
$$ p_x = m_e mathbfv_xe + m_n mathbfv_xn $$
$$ p_y = m_e mathbfv_ye + m_n mathbfv_yn $$
Everything not bolded is a known constant and everything bolded are unknowns (4 total). Assuming the constants are given such that this system has infinite solutions (i.e. there is enough energy for $beta^-$ decay to occur in the first place), how can I choose a solution for the unknowns randomly?
EDIT: I realized the equations for momentum included incorrect squares, so I removed them. Now I'm not sure if this system even has infinite solutions, but intuitively I think it should.
polynomials systems-of-equations physics
asked Sep 5 at 8:56
Ken
1185
add a comment |Â
up vote
0
down vote
favorite
I am a programmer trying to simulate the motion of particles involved in $beta^-$ decay (in 2D). While trying to conserve energy and momentum throughout the process, I end up with a system of equations for the trajectories of the electron ($_e$) and neutrino ($_n$) after $beta^-$ decay,
$$ |v_n|^2 = mathbfv_xn^2 + mathbfv_yn^2 $$
$$ |v_e|^2 = mathbfv_xe^2 + mathbfv_ye^2 $$
$$ p_x = m_e mathbfv_xe + m_n mathbfv_xn $$
$$ p_y = m_e mathbfv_ye + m_n mathbfv_yn $$
Everything not bolded is a known constant and everything bolded are unknowns (4 total). Assuming the constants are given such that this system has infinite solutions (i.e. there is enough energy for $beta^-$ decay to occur in the first place), how can I choose a solution for the unknowns randomly?
EDIT: I realized the equations for momentum included incorrect squares, so I removed them. Now I'm not sure if this system even has infinite solutions, but intuitively I think it should.
polynomials systems-of-equations physics
asked Sep 5 at 8:56
Ken
1185
1) Your equation is not linear in the $x_i$ and $y_i;$ it is linear in their squares though. 2) In most cases, this system has a finite number of solutions (four equations, four unknowns). The cases which have infinitely many solutions are very special; the first one I can think of is $e_1=e_2=c=d=0$. Are you really sure that this is what you want? If yes, can you specify your situation? (Please edit your question to clarify.)
– Wrzlprmft
Sep 5 at 9:40
@Wrzlprmft edited, thanks
– Ken
Sep 5 at 19:58
This does not really change the problem. You still have four equations and four unknowns which means a finite number of solutions for most parameter choices. Are you sure you know the neutrino mass? (Sidenote: The symbol for neutrinos is usually ν.)
– Wrzlprmft
Sep 5 at 22:57
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am a programmer trying to simulate the motion of particles involved in $beta^-$ decay (in 2D). While trying to conserve energy and momentum throughout the process, I end up with a system of equations for the trajectories of the electron ($_e$) and neutrino ($_n$) after $beta^-$ decay,
$$ |v_n|^2 = mathbfv_xn^2 + mathbfv_yn^2 $$
$$ |v_e|^2 = mathbfv_xe^2 + mathbfv_ye^2 $$
$$ p_x = m_e mathbfv_xe + m_n mathbfv_xn $$
$$ p_y = m_e mathbfv_ye + m_n mathbfv_yn $$
Everything not bolded is a known constant and everything bolded are unknowns (4 total). Assuming the constants are given such that this system has infinite solutions (i.e. there is enough energy for $beta^-$ decay to occur in the first place), how can I choose a solution for the unknowns randomly?
EDIT: I realized the equations for momentum included incorrect squares, so I removed them. Now I'm not sure if this system even has infinite solutions, but intuitively I think it should.
polynomials systems-of-equations physics
asked Sep 5 at 8:56
Ken
1185
I am a programmer trying to simulate the motion of particles involved in $beta^-$ decay (in 2D). While trying to conserve energy and momentum throughout the process, I end up with a system of equations for the trajectories of the electron ($_e$) and neutrino ($_n$) after $beta^-$ decay,
$$ |v_n|^2 = mathbfv_xn^2 + mathbfv_yn^2 $$
$$ |v_e|^2 = mathbfv_xe^2 + mathbfv_ye^2 $$
$$ p_x = m_e mathbfv_xe + m_n mathbfv_xn $$
$$ p_y = m_e mathbfv_ye + m_n mathbfv_yn $$
Everything not bolded is a known constant and everything bolded are unknowns (4 total). Assuming the constants are given such that this system has infinite solutions (i.e. there is enough energy for $beta^-$ decay to occur in the first place), how can I choose a solution for the unknowns randomly?
EDIT: I realized the equations for momentum included incorrect squares, so I removed them. Now I'm not sure if this system even has infinite solutions, but intuitively I think it should.
polynomials systems-of-equations physics
polynomials systems-of-equations physics
asked Sep 5 at 8:56
Ken
1185
asked Sep 5 at 8:56
Ken
1185
edited Sep 5 at 22:40
asked Sep 5 at 8:56
Ken
1185
asked Sep 5 at 8:56
Ken
1185
asked Sep 5 at 8:56
Ken
1185
1185
1) Your equation is not linear in the $x_i$ and $y_i;$ it is linear in their squares though. 2) In most cases, this system has a finite number of solutions (four equations, four unknowns). The cases which have infinitely many solutions are very special; the first one I can think of is $e_1=e_2=c=d=0$. Are you really sure that this is what you want? If yes, can you specify your situation? (Please edit your question to clarify.)
– Wrzlprmft
Sep 5 at 9:40
@Wrzlprmft edited, thanks
– Ken
Sep 5 at 19:58
This does not really change the problem. You still have four equations and four unknowns which means a finite number of solutions for most parameter choices. Are you sure you know the neutrino mass? (Sidenote: The symbol for neutrinos is usually ν.)
– Wrzlprmft
Sep 5 at 22:57
add a comment |Â
1) Your equation is not linear in the $x_i$ and $y_i;$ it is linear in their squares though. 2) In most cases, this system has a finite number of solutions (four equations, four unknowns). The cases which have infinitely many solutions are very special; the first one I can think of is $e_1=e_2=c=d=0$. Are you really sure that this is what you want? If yes, can you specify your situation? (Please edit your question to clarify.)
– Wrzlprmft
Sep 5 at 9:40
@Wrzlprmft edited, thanks
– Ken
Sep 5 at 19:58
This does not really change the problem. You still have four equations and four unknowns which means a finite number of solutions for most parameter choices. Are you sure you know the neutrino mass? (Sidenote: The symbol for neutrinos is usually ν.)
– Wrzlprmft
Sep 5 at 22:57
1) Your equation is not linear in the $x_i$ and $y_i;$ it is linear in their squares though. 2) In most cases, this system has a finite number of solutions (four equations, four unknowns). The cases which have infinitely many solutions are very special; the first one I can think of is $e_1=e_2=c=d=0$. Are you really sure that this is what you want? If yes, can you specify your situation? (Please edit your question to clarify.)
– Wrzlprmft
Sep 5 at 9:40
1) Your equation is not linear in the $x_i$ and $y_i;$ it is linear in their squares though. 2) In most cases, this system has a finite number of solutions (four equations, four unknowns). The cases which have infinitely many solutions are very special; the first one I can think of is $e_1=e_2=c=d=0$. Are you really sure that this is what you want? If yes, can you specify your situation? (Please edit your question to clarify.)
– Wrzlprmft
Sep 5 at 9:40
@Wrzlprmft edited, thanks
– Ken
Sep 5 at 19:58
@Wrzlprmft edited, thanks
– Ken
Sep 5 at 19:58
This does not really change the problem. You still have four equations and four unknowns which means a finite number of solutions for most parameter choices. Are you sure you know the neutrino mass? (Sidenote: The symbol for neutrinos is usually ν.)
– Wrzlprmft
Sep 5 at 22:57
This does not really change the problem. You still have four equations and four unknowns which means a finite number of solutions for most parameter choices. Are you sure you know the neutrino mass? (Sidenote: The symbol for neutrinos is usually ν.)
– Wrzlprmft
Sep 5 at 22:57
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
With reference to the diagram below, Equations 1 and 2 tell you that for the electron the vector $ (m_e mathbfv_xe,m_e mathbfv_ye) $ has length $ r_e = v_e $, and similarly for the neutrino, $ r_n = v_n $. So in momentum space the vector momenta of the electron and neutrino lie somewhere on two circles about the origin (see sketch on the left). Equations 3 and 4 tell you that the vector sum of the momenta of the electron and neutrino lies at a point $ (p_x, p_y) $, and as you can see from the sketch on the right, there will generally only be two solutions for the momenta of the electron and neutrino. So two solutions, not an infinite number.
answered Sep 5 at 23:22
Penguino
75449
Yes, thanks! I worked this much out on my own just now. Still trying to find those two solutions though. I'm trying polar coordinates, which seems to be simplifying things somewhat...
– Ken
Sep 5 at 23:47
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
With reference to the diagram below, Equations 1 and 2 tell you that for the electron the vector $ (m_e mathbfv_xe,m_e mathbfv_ye) $ has length $ r_e = v_e $, and similarly for the neutrino, $ r_n = v_n $. So in momentum space the vector momenta of the electron and neutrino lie somewhere on two circles about the origin (see sketch on the left). Equations 3 and 4 tell you that the vector sum of the momenta of the electron and neutrino lies at a point $ (p_x, p_y) $, and as you can see from the sketch on the right, there will generally only be two solutions for the momenta of the electron and neutrino. So two solutions, not an infinite number.
answered Sep 5 at 23:22
Penguino
75449
Yes, thanks! I worked this much out on my own just now. Still trying to find those two solutions though. I'm trying polar coordinates, which seems to be simplifying things somewhat...
– Ken
Sep 5 at 23:47
add a comment |Â
up vote
1
down vote
With reference to the diagram below, Equations 1 and 2 tell you that for the electron the vector $ (m_e mathbfv_xe,m_e mathbfv_ye) $ has length $ r_e = v_e $, and similarly for the neutrino, $ r_n = v_n $. So in momentum space the vector momenta of the electron and neutrino lie somewhere on two circles about the origin (see sketch on the left). Equations 3 and 4 tell you that the vector sum of the momenta of the electron and neutrino lies at a point $ (p_x, p_y) $, and as you can see from the sketch on the right, there will generally only be two solutions for the momenta of the electron and neutrino. So two solutions, not an infinite number.
answered Sep 5 at 23:22
Penguino
75449
Yes, thanks! I worked this much out on my own just now. Still trying to find those two solutions though. I'm trying polar coordinates, which seems to be simplifying things somewhat...
– Ken
Sep 5 at 23:47
add a comment |Â
up vote
1
down vote
up vote
1
down vote
With reference to the diagram below, Equations 1 and 2 tell you that for the electron the vector $ (m_e mathbfv_xe,m_e mathbfv_ye) $ has length $ r_e = v_e $, and similarly for the neutrino, $ r_n = v_n $. So in momentum space the vector momenta of the electron and neutrino lie somewhere on two circles about the origin (see sketch on the left). Equations 3 and 4 tell you that the vector sum of the momenta of the electron and neutrino lies at a point $ (p_x, p_y) $, and as you can see from the sketch on the right, there will generally only be two solutions for the momenta of the electron and neutrino. So two solutions, not an infinite number.
answered Sep 5 at 23:22
Penguino
75449
With reference to the diagram below, Equations 1 and 2 tell you that for the electron the vector $ (m_e mathbfv_xe,m_e mathbfv_ye) $ has length $ r_e = v_e $, and similarly for the neutrino, $ r_n = v_n $. So in momentum space the vector momenta of the electron and neutrino lie somewhere on two circles about the origin (see sketch on the left). Equations 3 and 4 tell you that the vector sum of the momenta of the electron and neutrino lies at a point $ (p_x, p_y) $, and as you can see from the sketch on the right, there will generally only be two solutions for the momenta of the electron and neutrino. So two solutions, not an infinite number.
answered Sep 5 at 23:22
Penguino
75449
answered Sep 5 at 23:22
Penguino
75449
answered Sep 5 at 23:22
Penguino
75449
answered Sep 5 at 23:22
Penguino
75449
75449
Yes, thanks! I worked this much out on my own just now. Still trying to find those two solutions though. I'm trying polar coordinates, which seems to be simplifying things somewhat...
– Ken
Sep 5 at 23:47
add a comment |Â
Yes, thanks! I worked this much out on my own just now. Still trying to find those two solutions though. I'm trying polar coordinates, which seems to be simplifying things somewhat...
– Ken
Sep 5 at 23:47
Yes, thanks! I worked this much out on my own just now. Still trying to find those two solutions though. I'm trying polar coordinates, which seems to be simplifying things somewhat...
– Ken
Sep 5 at 23:47
Yes, thanks! I worked this much out on my own just now. Still trying to find those two solutions though. I'm trying polar coordinates, which seems to be simplifying things somewhat...
– Ken
Sep 5 at 23:47
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2906040%2fhow-to-randomly-pick-a-solution-to-a-system-of-polynomial-equations-that-has-inf%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1) Your equation is not linear in the $x_i$ and $y_i;$ it is linear in their squares though. 2) In most cases, this system has a finite number of solutions (four equations, four unknowns). The cases which have infinitely many solutions are very special; the first one I can think of is $e_1=e_2=c=d=0$. Are you really sure that this is what you want? If yes, can you specify your situation? (Please edit your question to clarify.)
– Wrzlprmft
Sep 5 at 9:40
@Wrzlprmft edited, thanks
– Ken
Sep 5 at 19:58
This does not really change the problem. You still have four equations and four unknowns which means a finite number of solutions for most parameter choices. Are you sure you know the neutrino mass? (Sidenote: The symbol for neutrinos is usually ν.)
– Wrzlprmft
Sep 5 at 22:57