[Linear Algebra]- Simultaneous equation. Deriving the general solution.

In summary: Actually, in this case, where the two particles are equal in mass, the result is very simple. You either have v3 = v1, v4 = v2, or v3 = v2, v4 = v1. You've written the conservation of momentum equation. Let's call momentum p. In relativity, the energy of a particle is related to the momentum by e^2-p^2c^2=m^2c^4, so e=\sqrt{m^2c^4+p^2c^2}. Your energy equation is e1 + e2 = e3 + e4, which works out to\sqrt{\frac{c^6 m^2}{c^2
  • #1
randombill
81
0
Hello,

I need help with solving the following simultaneous equation. I tried using matrix solution using Gaussian elimination but it became too difficult early on.

Known constants are a,A,b, and B.
Known variables are x and y.
Unknown variables are m and n.

(a[itex]\times[/itex]x)[itex]/[/itex](A - x) + (b[itex]\times[/itex]y)[itex]/[/itex](B - y) =
(a[itex]\times[/itex]m)[itex]/[/itex](A - m) + (b[itex]\times[/itex]n)[itex]/[/itex](B - n)

I tried the matrix solution below. I set the known constants to 1 for the second row.

[(a[itex]\times[/itex]x)[itex]/[/itex](A - x) (b[itex]\times[/itex]y)[itex]/[/itex](B - y) = c1 ]
[(x)[itex]/[/itex](1 - x) (y)[itex]/[/itex](1 - y) = c2]
 
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  • #2
randombill said:
Hello,

I need help with solving the following simultaneous equation. I tried using matrix solution using Gaussian elimination but it became too difficult early on.

Known constants are a,A,b, and B.
Known variables are x and y.
Unknown variables are m and n.

(a[itex]\times[/itex]x)[itex]/[/itex](A - x) + (b[itex]\times[/itex]y)[itex]/[/itex](B - y) =
(a[itex]\times[/itex]m)[itex]/[/itex](A - m) + (b[itex]\times[/itex]n)[itex]/[/itex](B - n)

I tried the matrix solution below. I set the known constants to 1 for the second row.

[(a[itex]\times[/itex]x)[itex]/[/itex](A - x) (b[itex]\times[/itex]y)[itex]/[/itex](B - y) = c1 ]
[(x)[itex]/[/itex](1 - x) (y)[itex]/[/itex](1 - y) = c2]

Hey randombill and welcome to the forums.

I'm sorry but I can't figure out what linear system this is representing.

Also I notice you have a few division signs in your post. Linear systems have only variables with powers of 1, so if those any of the denominators involve variables, you won't be able to use linear algebra to find a solution (if one exists).
 
  • #3
Ok, thanks for your help. I guess that solves the problem.

Linear systems have only variables with powers of 1, so if those any of the denominators involve variables, you won't be able to use linear algebra to find a solution (if one exists).

So I guess numerically it may be possible.
 
  • #4
randombill said:
Hello,

I need help with solving the following simultaneous equation. I tried using matrix solution using Gaussian elimination but it became too difficult early on.

Known constants are a,A,b, and B.
Known variables are x and y.
Unknown variables are m and n.

(a[itex]\times[/itex]x)[itex]/[/itex](A - x) + (b[itex]\times[/itex]y)[itex]/[/itex](B - y) =
(a[itex]\times[/itex]m)[itex]/[/itex](A - m) + (b[itex]\times[/itex]n)[itex]/[/itex](B - n)
This NOT a "system" of equations, it is a single equation in two unknowns and so has no unique solution.

I tried the matrix solution below. I set the known constants to 1 for the second row.

[(a[itex]\times[/itex]x)[itex]/[/itex](A - x) (b[itex]\times[/itex]y)[itex]/[/itex](B - y) = c1 ]
[(x)[itex]/[/itex](1 - x) (y)[itex]/[/itex](1 - y) = c2]

This is NOT a matrix (matrices do not have "=" inside them) nor does it correspond to the equation you had before. Assuming the "= c1" and "= c2" were supposed to be outside the matrix, this corresponds to the equations
[tex]\frac{ax}{A- x}+ \frac{by}{B-y}= c1[/tex]
[tex]\frac{x}{1- x}+ \frac{y}{1-y}= c2[/tex]

Not the single equation you had before. I have no idea what problem you really meant!
 
  • #5
This NOT a "system" of equations, it is a single equation in two unknowns and so has no unique solution.


So I suppose this equation would have the same situation and the square roots make it even more difficult to work with.


[itex] \frac{mv1}{\sqrt{1-v1/c^{2}}}[/itex] + [itex] \frac{mv2}{\sqrt{1-v2/c^{2}}}[/itex] = [itex] \frac{mv3}{\sqrt{1-v3/c^{2}}}[/itex] + [itex] \frac{mv4}{\sqrt{1-v4/c^{2}}}[/itex]

where v1 and v2 are the initial velocities and v3 and v4 are the final velocities and v1,v2 are known.
 
  • #6
randombill said:
So I suppose this equation would have the same situation and the square roots make it even more difficult to work with.


[itex] \frac{mv1}{\sqrt{1-v1/c^{2}}}[/itex] + [itex] \frac{mv2}{\sqrt{1-v2/c^{2}}}[/itex] = [itex] \frac{mv3}{\sqrt{1-v3/c^{2}}}[/itex] + [itex] \frac{mv4}{\sqrt{1-v4/c^{2}}}[/itex]

where v1 and v2 are the initial velocities and v3 and v4 are the final velocities and v1,v2 are known.
Ah! Finally it becomes clear what you're working on! You're trying to figure out what are the possible outcomes of a collision between two particles, right?

In one dimension, if you know v_1 and v_2, then the outcome (v_3 and v_4) is almost completely determined (assuming an elastic collision). But you need another equation: conservation of energy. One way to find the solution is to boost into the center of mass frame. In that frame, where v_1 = -v_2, there are only two possible results: (1) v'_3 = v'_1, v'_4 = v'_2, or (2) v'_3 = -v'_1, v'_4 = -v'_2. (v' = velocity in the CM frame.) So then the only problem is to find the boost that gets you in the CM frame.

Hope this helps.
 
  • #7
pmsrw3 said:
Ah! Finally it becomes clear what you're working on! You're trying to figure out what are the possible outcomes of a collision between two particles, right?

Yup!
In one dimension, if you know v_1 and v_2, then the outcome (v_3 and v_4) is almost completely determined (assuming an elastic collision). But you need another equation: conservation of energy. One way to find the solution is to boost into the center of mass frame. In that frame, where v_1 = -v_2, there are only two possible results: (1) v'_3 = v'_1, v'_4 = v'_2, or (2) v'_3 = -v'_1, v'_4 = -v'_2. (v' = velocity in the CM frame.) So then the only problem is to find the boost that gets you in the CM frame.

Hope this helps.

Can you give an example, please. I'm not sure what equations to use.
 
  • #8
randombill said:
Can you give an example, please. I'm not sure what equations to use.
Actually, in this case, where the two particles are equal in mass, the result is very simple. You either have v3 = v1, v4 = v2, or v3 = v2, v4 = v1.

You've written the conservation of momentum equation. Let's call momentum p. In relativity, the energy of a particle is related to the momentum by [itex]e^2-p^2c^2=m^2c^4[/itex], so [itex]e=\sqrt{m^2c^4+p^2c^2}[/itex]. Your energy equation is e1 + e2 = e3 + e4, which works out to
[tex]
\sqrt{\frac{c^6 m^2}{c^2-\text{v1}^2}}+\sqrt{\frac{c^6 m^2}{c^2-\text{v2}^2}}=\sqrt{\frac{c^6 m^2}{c^2-\text{v3}^2}}+\sqrt{\frac{c^6 m^2}{c^2-\text{v4}^2}}
[/tex]

It should be obvious, looking at that equation and the conservation of momentum equation that you've already posted, that the two answers I gave are solutions. It takes a little more work to show that they're the only solutions, but actually, no messy solving of equations is necessary. Just notice that momentum always has the same sign as velocity, and always increases as velocity increases. This tells you that, if you have a solution of the momentum equation, any other solution will require compensating changes in v3 and v4 -- if one goes up, the other has to come down. Energy, on the other hand, is always positive, and increases as the absolute value of velocity. If you think hard about this, you'll see that it means there are only two ways to solve both equations.
 
  • #9
It should be obvious, looking at that equation and the conservation of momentum equation that you've already posted, that the two answers I gave are solutions. It takes a little more work to show that they're the only solutions, but actually, no messy solving of equations is necessary. Just notice that momentum always has the same sign as velocity, and always increases as velocity increases. This tells you that, if you have a solution of the momentum equation, any other solution will require compensating changes in v3 and v4 -- if one goes up, the other has to come down. Energy, on the other hand, is always positive, and increases as the absolute value of velocity. If you think hard about this, you'll see that it means there are only two ways to solve both equations.

It is obvious when the masses are the same but does the above equation work for unequal masses and if not, how would I solve for that.

thanks
 
  • #10
randombill said:
It is obvious when the masses are the same but does the above equation work for unequal masses and if not, how would I solve for that.
The equations work for unequal masses, except of course that, instead of having just "m" everywhere, you'll have m_1 and m_2 in the parts of the expressions referring to the momenta and energies of particles 1 and 2. Solving the equations is a lot messier when the masses are unequal (except for some simple special cases). I would have to think a bit about how to do that. Maybe someone else who is familiar with this problem will jump in.
 
  • #11
pmsrw3 said:
The equations work for unequal masses, except of course that, instead of having just "m" everywhere, you'll have m_1 and m_2 in the parts of the expressions referring to the momenta and energies of particles 1 and 2. Solving the equations is a lot messier when the masses are unequal (except for some simple special cases). I would have to think a bit about how to do that. Maybe someone else who is familiar with this problem will jump in.

I know its not easy, I was expecting you to surprise me but thanks anyways. Hopefully someone will chime in.
 

1. What is a simultaneous equation?

A simultaneous equation is a set of two or more equations that contain two or more unknown variables. The goal is to find the values of the unknown variables that satisfy all of the equations in the set.

2. How do you solve simultaneous equations?

There are various methods for solving simultaneous equations, such as substitution, elimination, and graphing. However, in linear algebra, the most common method is using matrix operations to find the general solution.

3. What is the general solution of simultaneous equations?

The general solution of simultaneous equations is the set of all possible solutions that satisfy the equations. It is usually expressed in terms of the unknown variables, and it can be written in a vector form using the variables as components.

4. How do you derive the general solution of simultaneous equations?

The general solution can be derived using matrix operations, specifically row operations, on the augmented matrix of the equations. By reducing the augmented matrix to reduced row echelon form, the general solution can be obtained by expressing the leading variables in terms of the free variables.

5. Can simultaneous equations have infinite solutions?

Yes, simultaneous equations can have infinite solutions if the equations are dependent or if there are more unknown variables than equations. This results in one or more free variables, which can take on any value, leading to an infinite number of solutions.

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