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Need help understanding electromagnetics and relativity

  1. Oct 4, 2015 #1
    Hello
    I'm having a little trouble understanding how two observers in two different inertial frames of reference would view the same simple electromagnetic event.

    Let's call the first frame K, and we can use cartesian coordinates x, y, z, and t for time in K. There is an electric field E in K that is constant in time, and is everywhere directed in the y-direction with the same magnitude. At t=0, a small positively charged free-moving particle, q, is fired with a gun from the origin in the x-direction with an initial velocity v.

    I would think that in K, we'd see the motion of this particle describe a parabola starting at the origin, and arcing upward in the x-y plane. That's because the only force on it is q[E] which gives it constant acceleration in the y-direction as it continues to move in the x-direction with velocity v.

    The second frame is K'. This frame moves with velocity v in the x-direction with respect to K -- the same velocity as the x-component of the charged particle -- and at t=0 and t'=0, the coordinate axes of K and K' all coincide. I'm very confused in trying to understand how an observer in K' would view the motion of this particle. On the one hand , may seem quite obvious: since he's moving along the x-axis of K at the same speed as the particle, all he sees is its acceleration in the y'-direction, and therefore he observes it moving in a straight line in the y'-direction with constant acceleration.

    But this would seem inconsistent with the electromagnetic field transformation from K to K'. That's because in K', there is a field B' in the negative-z' direction. And so in K', as E' is accelerating the free particle in the y'-direction, I would think B' would be causing it to spiral.

    And so either I'm wrong about it describing a parabola in K, or I'm wrong about it spiralling in K'. Maybe both. Thanks!
     
    Last edited: Oct 4, 2015
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  3. Oct 4, 2015 #2

    Orodruin

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    The point is that you are trying to take one relativistic effect (the emergence of the B field in K') into account while neglecting others (by stating that the motion is a parabola in K). Increasing the momentum in the y direction while keeping the momentum in the x direction fixed will actually decrease the velocity in the x direction.
     
  4. Oct 4, 2015 #3

    vanhees71

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    Indeed, you have to solve the relativistic equation of motion for the charged particle (here neglecting radiation reaction). Then you can boost the solution without problems to the other frame ##K'##, where you have both electric and magnetic field components. Then you'll see that your transformed solution of the equation of motion satisfies the equation of motion in ##K'## with the full Lorentz force.

    The manifestly covariant equation of motion reads
    $$m c \frac{\mathrm{d} u^{\mu}}{\mathrm{d \tau}}=q F^{\mu \nu} u_{\nu}, \quad u^{\mu}=\frac{1}{c} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}.$$
    For details, see
    http://fias.uni-frankfurt.de/~hees/pf-faq/srt.pdf
     
    Last edited: Oct 5, 2015
  5. Oct 4, 2015 #4

    Let's consider the frame K'. At first the Lorentz force points to the negative-x direction, so charged particle accelerates to that direction. If we stay at the same x-coordinate with the charged particle measuring the magnetic field, we will notice that the magnetic field decreases and becomes zero at some time. After that point there will be no magnetic field and no spiraling.

    Our buddy that we told to stay still at our original xy-coordinate, still measures a magnetic field, and sees a charged particle moving in that magnetic field without experiencing a Lorentz force. Strange.
     
  6. Oct 4, 2015 #5

    vanhees71

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    I thought in K is only a static homogeneous electric field. It is easy to solve the equation of motion there. Then you can transform this equation of motion with a Lorentz boost to describe it from the point of view of K'. You'll find in addition to an electric also a magnetic Lorentz force. This is so, because physics is Poincare invaraint, but of course only when you use the correct relativistic equations. Particularly the spatial trajectory is not a parabola as in the non-relativistic approximation.
     
  7. Oct 4, 2015 #6

    bcrowell

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    I don't think you can conclude that it would spiral in K'. A particle in a pure magnetic field spirals. The field in K' is not pure magnetic. Note that ##E^2-B^2## is invariant, so the electric field will dominate for all values of ##v##.

    In K', the fields are ##E'=\gamma E## and ##B'=\gamma v E##. Let u' be the velocity of the charge in K'. Early in the motion, u' is nonrelativistic. Approximating the solution to the equations of motion to lowest order in K', I get:

    ## u'_y \approx \frac{q\gamma E}{m} t' ##

    ## u'_x \approx \frac{q^2\gamma^2E^2v}{2m^2}t'^2 ##

    If you like, you could transform this velocity back into K and see if you get the expected results to some order in ##t##. Or you could take the motion in K, approximated for nonrelativistic ##u##, and transform into K' to see if it matches the above expressions to the given order in ##t'##.
     
    Last edited: Oct 4, 2015
  8. Oct 4, 2015 #7

    Okay. So in K, its velocity in the x-direction is not constant, as my non-relativistic model led me to incorrectly conclude, but actually decreases. And I'm guessing, this is having something to do with the mass increase of the particle as its acceleration in the y-direction increases its kinetic energy?

    One thing that is a little confusing to me: Assume the velocity in the x-direction of the particle is decreasing in K. Is this not equivalent to saying that it has an acceleration in the negative x-direction, and if so, must there not be a force to cause this? In other words, is there any force acting on the particle in K other than q[E] ?

    I apologize for perhaps being in a little over my head, and everybody's patience is sure appreciated. Thanks!
     
  9. Oct 4, 2015 #8

    Orodruin

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    Yes.

    No. This is again a non-relativistic assumption. Also in classical mechanics, force is usually not defined as ##F = ma##, but rather as ##F = dp/dt##, where ##p## is the momentum. Now, for an object of constant mass, ##p = mv## results in ##F = ma## (classically).

    Relativistic momentum is not given by ##\vec p = m\vec v##, but rather by ##\vec p = m\gamma \vec v## and again ##F## will not be equal to ##ma##.

    Relativistic mass is out of fashion and not in use by most actual physicists. I suggest you read this.
     
  10. Oct 4, 2015 #9

    bcrowell

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    Another way of putting what Orodruin said in #8 is that although we can write down a version of Newton's second law that works in special relativity, it looks like ##F=m\gamma a_\perp+m\gamma^3 a_\parallel##, where ##a_\perp## and ##a_\parallel## are the components of the acceleration perpendicular to and parallel to the velocity. Because the two terms have different powers of ##\gamma##, the force and acceleration vectors are not in general parallel to one another. The object acts as though it has less inertia in the direction perpendicular to its motion. In your frame K, this causes the particle's trajectory to curve more strongly than in the parabolic approximation. This is observed as an acceleration in the negative x direction.

    I spent some time looking around on the web for whether there is a closed-form solution for the motion. This type of problem breaks down into various cases depending on the invariants ##E^2-B^2## and ##E\cdot B##. For the case where the particle starts out at rest in a pure E field, and radiation reaction is neglected, the solution is that the velocity four-vector is given by

    ## v^a = (\cosh k\tau,\sinh k\tau,0,0),##

    where ##k=qE/m##. This is motion with constant proper acceleration (also known as hyperbolic motion). It's not obvious to me whether there is any way to get a closed-form solution in the case where the particle is never at rest. There's probably a treatment buried somewhere in Jackson.
     
    Last edited: Oct 4, 2015
  11. Oct 4, 2015 #10

    vanhees71

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    I hope, it's allowed to give the complete solution, although it's a B-question, but often it's simpler to do the math than to guess around (at least that's my experience when learning a new subject).

    So let's do it. For our purposes, since we want to work in two inertial frames and we want to understand the transformation properties of the various quantities involved, it's most convenient to work in the manifestly covariant way.

    In K the solution is quite simple, because there are only electric components, namely ##\vec{E}=E \vec{e}_2=\text{const}##. Let's first translate this into the Minkowski-space language. The electromagnetic field is represented by the antisymmetric field-strength tensor (Faraday tensor). In our case it's given by
    $$(F^{\mu \nu})=\begin{pmatrix}
    0 & 0 & E & 0\\
    0 & 0 & 0 & 0\\
    -E & 0 & 0 &0\\
    0 & 0 & 0 &0
    \end{pmatrix}$$
    The equation of motion is written in terms of the proper time of the particle, which is an invariant (scalar) quantity under Lorentz transformations. In this form the covariant equation of motion for the particle reads
    $$mc \mathrm{d}_{\tau} u^{\mu}=q F^{\mu \nu} u_{\nu}, \quad u^{\mu}=\frac{1}{c} \mathrm{d}_{\tau} x^{\mu}.$$
    We have the constraint
    $$u_{\mu} u^{\mu}=0,$$
    which follows from the definition of proper time.

    In our case the equations of motion for the four-velocity (a four-vector) thus reads
    $$\mathrm{d}_{\tau} u^0 = \omega u^2, \\
    \mathrm{d}_{\tau} u^1 =0, \\
    \mathrm{d}_{\tau} u^2 =\omega u^0,\\
    \mathrm{d}_{\tau} u^3 =0.
    $$
    Let's assume the particle initially moves with some velocity in ##1##-direction. Then we have ##u^{\mu}(\tau=0)=(\sqrt{1+(u_0^1)^2},u_0^1,0,0)##. The only non-trivial task involves ##u^0## and ##u^2##. Deriving the equation for ##u^2## by ##\tau## and using the equation for ##u^0## gives
    $$\mathrm{d}_{\tau}^2 u^2=\omega^2 u^2,$$
    which has the solution
    $$u^2(\tau)=A \sinh(\omega \tau)$$
    with some integration constant ##A##. The initial condition ##u^2(\tau=0)=0## is already used here.

    To determine ##\tau## we use the equation for ##u^2## again:
    $$u^0=\frac{1}{\omega} \mathrm{d}_{\tau} u^2=A \cosh(\omega \tau).$$
    From the initial condition we get ##A=\sqrt{1+(u_0^1)^2}##.

    For the time-position four-vector we find, assuming the initial condition ##x^{\mu}(\tau=0)=(0,0,0,0)##
    $$x^{\mu}(\tau)=\int_0^{\tau} \mathrm{d} \tau' c u^{\mu}(\tau')=(A c/\omega \sinh(\omega \tau),u_0^1 \tau,A c/\omega [\cosh(\omega t)-1],0).$$
    Now you can boost this solution to K', with the Lorentz-boost matrix
    $${\Lambda^{\mu}}_{\nu}=\begin{pmatrix}
    \gamma & -\beta \gamma & 0 & 0 \\
    -\beta \gamma & \gamma & 0 & 0 \\
    0 & 0 & 1 & 0\\
    0& 0 &0 & 1
    \end{pmatrix}.$$
    Then you get
    $$\tilde{F}^{\mu \nu}={\Lambda^{\mu}}_{\alpha} {\Lambda^{\nu}}_{\beta} F^{\alpha \beta}=\begin{pmatrix}
    0 & 0 &-\gamma E & 0\\
    0 & 0 & \beta \gamma E & 0 \\
    \gamma E & -\beta \gamma E & 0 \\
    0 & 0 & 0 & 0
    \end{pmatrix}.$$
    This means that in K' we have electric and magnetic field components
    $$\vec{E}'=\gamma \vec{E}=\gamma E \vec{e}_y', \quad \vec{B}'=-\gamma \vec{\beta} \times \vec{E}=-\gamma \beta E \vec{e}_3.$$
    It's also easy to show by direct calculation of the proper-time derivatives that the transformed solution
    $$\tilde{x}^{\mu}(\tau)={\Lambda^{\mu}}_{\alpha} x^{\alpha}(\tau)$$
    solves the equation of motion in K'.
     
  12. Oct 4, 2015 #11
    Thanks so much, everybody!

    I'm wondering if there are any tutorials out there on understanding some of the mathematical language that has been used throughout this thread involving matrices, tensors and suchlike to describe vectors, because I'm sure I'd find it helpful. I know I'm missing out by not being conversant in it.
     
  13. Oct 4, 2015 #12

    bcrowell

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    Could you tell us your background in math and physics?
     
  14. Oct 4, 2015 #13

    bcrowell

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    I find the simplicity of vanhees71's nice derivation counterintuitive. I guess its simplicity arises because the equations of motion can be written as a homogeneous first-order differential equation in terms of the four-velocity.

    Since that derivation shows that the x component of the four-velocity ##u^1## is constant, it follows that the x component of the three-velocity is just ##u^1/\gamma##, which verifies that the x motion decelerates.
     
  15. Oct 4, 2015 #14

    vanhees71

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    Sure, this only works, because the field is homogeneous. Of course, the non-covariant Hamilton equations of motion in a fixed inertial frame are usually the most convenient ones. You can also solve the relativistic Kepler problem, which famously lead Sommerfeld to the fine structure of the hydrogen spectrum, however by pure chance ;-)).
     
  16. Oct 4, 2015 #15

    vanhees71

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    A good introduction is found in the Feynman lectures and also Landau Lifshitz vol. 2.
     
  17. Oct 4, 2015 #16
    Thanks Ben

    No real academic or professional background in either. But what I do know today, I've learned mostly on my own, having become fascinated with electromagnetic theory.

    Some years back I read a textbook on electromagnetics by an author/physicist named Nannapaneni Narayana Rao called "Elements of Engineering Electromagnetics", and to understand what he was saying, I needed to teach myself a lot of calculus, namely partial derivatives and integrals.
     
  18. Oct 4, 2015 #17
    Thank you, vanhees
     
  19. Oct 4, 2015 #18

    Orodruin

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    I would say this is exactly as expected from the fact that there is no force in the x direction. This leads to the x component of the four momentum being constant and therefore the x component of the four velocity as they are proportional.
     
  20. Oct 4, 2015 #19

    bcrowell

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    Neither of those is going to work for the OP, since s/he only knows a little differential calculus.

    @mairzydoats: For the kind of math we've been using in this thread, I would suggest something like the following. First, finish learning some basic integral calculus. Then you need to learn some linear algebra and differential equations. For linear algebra, I like the free online book by Hefferon.
     
  21. Oct 4, 2015 #20

    bcrowell

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    I was referring to the fact that the differential equation is homogeneous, not to the fact that the field is uniform. That is, the differential equation has the form ##v'=Av## rather than ##v'=Av+b##. It's first-order, linear, and homogeneous.

    In general, we would expect the equations of motion for a particle acted on by a force to be second order, nonlinear, and inhomogeneous.
     
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