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A system of differential equations - Lorentz Force

  1. Apr 25, 2008 #1
    I was wonder if anybody might know how to solve this (this is not a homework problem, btw).

    The Lorentz Force is given by:

    [tex]\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})[/tex]

    Now afer working that out, I get three differential equations:

    [tex]\frac{dv_x}{dt}=\frac{q}{m}(E_x+v_yB_z-v_zB_y)[/tex]
    [tex]\frac{dv_y}{dt}=\frac{q}{m}(E_y+v_zB_x-v_xB_z)[/tex]
    [tex]\frac{dv_z}{dt}=\frac{q}{m}(E_z+v_xB_y-v_yB_x)[/tex]

    So far so good. But I can't solve these differential equations because we have [tex]\frac{dv_x}{dt}[/tex] as a function of [tex]v_y[/tex] and [tex]v_z[/tex]

    I would be thankful if anyone would show me how to solve such a system of differential equations.
     
  2. jcsd
  3. Apr 25, 2008 #2
    Could I treat these as a "regular" set of linear equations and solve it that way?

    For example:

    [tex]\left(\begin{array}{ccc}D&-\frac{q}{m}B_z&\frac{q}{m}B_y\\\frac{q}{m}B_z&D&-\frac{q}{m}B_x\\-\frac{q}{m}B_y&\frac{q}{m}B_x&D\end{array}\right)\vec{v}=\frac{q}{m}\vec{E}[/tex]

    Where [tex]D=\frac{d}{dt}[/tex]
     
    Last edited: Apr 25, 2008
  4. Apr 25, 2008 #3
    If you differentiate the equation for the velocity in the x direction then you will have time derivatives of the velocity in the y and z directions which can then be substituted from the other two equations.

    What you have is called a system of differential equations, and you want to decouple the equations by repeated differentiation.

    If the E and B fields are uniform and constant then the solution will be a cycloid.
     
  5. Apr 25, 2008 #4

    Mute

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    If the magnetic field is constant (in space, at least), you might as well choose your coordinate system so that [itex]\mathbf{B} = B_z \hat{\mathbf{z}}[/itex]. That will simplify things a little.
     
  6. Apr 28, 2008 #5
    Not sure that will work in this case.

    This

    [tex]\dot{v_x}=\frac{q}{m}(E_x+v_yB_z-v_zB_y)[/tex]

    leads to this:

    [tex]\ddot{v_x}=\frac{q}{m}(\dot{v_y}B_z-\dot{v_z}B_y)[/tex]

    And I have this from above:

    [tex]\dot{v_y}=\frac{q}{m}(E_y+v_zB_x-v_xB_z)[/tex]
    [tex]\dot{v_z}=\frac{q}{m}(E_z+v_xB_y-v_yB_x)[/tex]


    And if I make the substitution I'll still have [tex]\ddot{v_x}[/tex] as a function of [tex]v_y[/tex] and [tex]v_z[/tex]:

    [tex]\ddot{v_x}=\frac{q}{m}((\frac{q}{m}(E_y+v_zB_x-v_xB_z))B_z-(\frac{q}{m}(E_z+v_xB_y-v_yB_x))B_y)[/tex]

    Simplification:

    [tex]\ddot{v_x}=(\frac{q}{m})^2((E_yB_z+v_zB_xB_z-v_xB_z^2)-(E_zB_y+v_xB_y^2-v_yB_xB_y))[/tex]
     
  7. Apr 28, 2008 #6

    Mute

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    Since your magnetic and electric fields don't depend on time, if you choose your coordinate system such that [itex]B_x=B_y=0[/itex], however, then that equation becomes:

    [tex]\ddot{v_x}=\left(\frac{qB_z}{m}\right)^2\left(\frac{E_y}{B_z}-v_x\right)[/tex]

    And similarly for [itex]v_y[/itex]. The equation for [itex]v_z[/itex] is then just

    [tex]\frac{dv_z}{dt}=\frac{q}{m}E_z[/tex]

    So, all the velocity components are given by decoupled inhomogeneous equations in the coordinate basis where [itex]\mathbf{B} = B_z \mathbf{\hat{z}}[/itex] (the equations for v_x and v_y are 2nd order and v_z is first order). If the electric field were zero, this would just be cyclotron motion. If you then want to put things back in terms of a general magnetic field, just rotate your coordinate system with a rotation matrix.
     
    Last edited: Apr 28, 2008
  8. May 5, 2008 #7
    I'm looking for a general solution that can be used for any [itex]B_x[/itex], [itex]B_y[/itex] and [itex]B_z[/itex].
     
  9. May 5, 2008 #8

    Mute

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    Last thing I said in my post above was:

    For a constant magnetic field the motion is going to be cyclotron motion about some axis. You might as well solve for the motion about that axis, and then just rotate your coordinate system. (technically it won't quite be cyclotron motion since the electric field isn't zero, so the coordinate system in which the mag field is along the z direction will have a different electric field than your general case above).

    If you really want to solve it generally anyways, then I guess that since you have three coupled inhomogeneous equations, writing it as [itex]\dot{\mathbf{v}} = \mathsf{A}\mathbf{v} + \mathbf{v}_0[/itex], you diagonalize the matrix A, giving you A = P^{-1}DP, then mutliply the whole equation by P, and define some new velocity v' = Pv. Solve the resulting uncoupled inhomogeneous differential equation, dv'/dt = Dv' + Pv_0. This is, of course, going to be equivalent to choosing your coordinate system such that B is along the z direction and then rotating everything afterwards to get back to the original coordinate system.
     
    Last edited: May 5, 2008
  10. May 6, 2008 #9
    I played with this a bit too, and am posting this after having done so (perhaps too late if it's considered solved by picking an appropriate axis of symmetry ).

    Here's my approach, which essentially replays the rigid body rotation derivation backwards (such as the one found in

    http://www.damtp.cam.ac.uk/user/tong/dynamics/three.pdf

    ). In that Lorentz Force Law, we essentially have an equation like the rigid body rotation equation [itex]y' = \omega \times y + x_0'[/itex], that resulted from [itex]y = R x + x_0[/itex]. That's a good hint about the antiderivative required for this Lorentz problem, so we have to go backwards from the cross product, and solve for the rotation:

    [tex]
    mv' = q/m ( E + v \times B )
    [/tex]
    [tex]
    v = (q/m) E + \Omega v
    [/tex]

    Here [itex]\Omega[/itex] is:

    [tex]
    -q/m
    \begin{bmatrix}
    0 & B_3 & -B_2 \\
    -B_3 & 0 & B_1 \\
    B_2 & -B_1 & 0 \\
    \end{bmatrix}
    [/tex]

    Following the rigid body treatment (in Tong above) this antisymmetric matrix can be expressed in terms of a rotation matrix (ie: essentially it is a rotation matrix derivative with a rotation factored out of it). So, let

    [tex]
    \Omega = R' R^T
    [/tex]

    , and use one more trick from the rigid body analysis:

    [tex]
    (RR^T)' = R' R^T + R {R'}^T = I' = 0
    [/tex]

    [tex]
    v' - R' R^T v = v' + R {R'}^T v = R (R^T v' + {R'}^T v)
    \implies
    R (R^T v)' = (q/m) E
    [/tex]

    Thus the solution can be written as two equations, one explicit for v, and one matrix differential equation to solve for R:

    [tex]
    v = (q/m) E t + R^T C
    [/tex]

    [tex]
    R' = -(q/m)
    \begin{bmatrix}
    0 & B_3 & -B_2 \\
    -B_3 & 0 & B_1 \\
    B_2 & -B_1 & 0 \\
    \end{bmatrix}
    R
    [/tex]

    Does this look reasonable?
     
    Last edited: May 6, 2008
  11. May 6, 2008 #10

    Mute

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    Whoops, I should have paid more attention to the problem. Given Peeter's post, I guess you can't diagonalize what I labelled as A, and so what I said about solving it by diagonalization isn't correct. However, there should still be a coordinate system representation that does make B along the z direction only - this might be what Peeter's analysis is doing.
     
  12. May 6, 2008 #11
    It appears that I also mixed up the sign of [itex]\Omega[/itex], and thus also the sign where I explicitly wrote out [itex]R' = \Omega R[/itex]. I haven't tried solving that last equation numerically (or analytically), but intuition says that diagonalization would do the trick. ie: with a solution having a term of the form:

    [tex]
    e^{Dt}
    [/tex]

    I should try this with an example to see if it all holds together (at the bare minimum, if I did things right, it should work for B along the [itex]\hat{z}[/itex] axis;)

    I've also found there's a treatment of this problem in the book Geometric Algebra for Physicists (chapter 5 on spacetime algebra). So far reading that book, I'd temporarily skipped that chapter for some easier stuff in chapter 6 (vector calculus chapter). Going back and reading just this fragment, I can't say I fully understand their treatment. It's interesting looking though;) They end up reformulating the equation as:

    [tex]
    m v' = q F \cdot v
    [/tex]

    where F is a combined electrodynamic field:

    [tex]
    F = E + I B
    [/tex]

    Then they introduce a rotor parameterization of the velocity, and an equation to solve for the rotor that's similar to the rotation matrix equation I had (factor of two difference because the rotor is a double sided half angle operator unlike the single sided rotation matrix) :

    [tex]
    R' = q/2m F R
    [/tex]

    There's a lot of similarities to what I hacked up, but it will probably take me a while before I can get to the point to digest and compare the two. Their rotor equation ends up with terms for both electric field and magnetic field whereas mine is magnetic only, ... that makes more sense to me. My rationalization for this is a guess that this is a side effect of the cross product Lorentz force "law" as stated above not being correct relativistically, whereas theirs is.
     
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