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EM Radiation versus Transverse Waves

  1. Aug 4, 2005 #1
    Electromagnetic wave are said to travel with velocity c, in vacuum and are also said to be a transverse wave. But consider this:

    Let a positive charge Q be at point P of a three dimensional space.

    Let this charge, at time t=0, begin an uniform circular motion with angular velocity OMEGA = 2 Pi rad/s and with radius R = 1 m. It moves just one turn to make things simpler.

    Supose we (the observers) are very distant from P but in the plane of the particle's circular path. Let this distance be 10 light-seconds for instance.

    At our position we will experience after t = 10 s, modifications on the electric field generated by the charged particle. It just so happens that in some part of the circular path the particle has come nearer to us, and in the other, the particle has gone farther, so that the field has presented not only transverse modifications but logitudinal modifications as well. Aren't these logitudinal modifications considered part of the radiation emitted by the particle ? Does these longitudinal dynamics of the field propagates with c also ?

    Best Regards,

  2. jcsd
  3. Aug 4, 2005 #2

    But the logitudinal components (of the radiation) fall off with distance as [tex] 1/r^3 [/tex] so that
    at any appreciable distance relative to the wavlength you only have the transverse
    components which fall off as [tex] 1/r^2[/tex].
  4. Aug 4, 2005 #3
    Ok, but if it is just a matter of one component falling with distance more rapidly than the other, don't you think that the term "nearly transverse" should be used more often ? At least to shed some light on this aspect of the question ?
  5. Aug 4, 2005 #4
    Well, like so many other cases in Physics the terminology caters
    to the well-versed not to the newcomers.
  6. Aug 4, 2005 #5
    It is interesting to add here that, in the case of dipole emission, along the line conecting the two opposite charges, the field is considered zero. I understand it for there is a subtraction of field strenghs due to the positively and negatively charged particles' contributions to the field in this subspace, but it isn't precisely zero along this line.

    Thank you,

    Best Regards

  7. Aug 4, 2005 #6
    It is not zero anywhere along the line.

    The field of the positive charge points away from it.
    The field of the negative charge points toward it.

    When you add these, they are not zero anywhere in space.
  8. Aug 4, 2005 #7

    Hans de Vries

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    It's actually [tex] 1/r[/tex] for the radiative (transversal) component and
    [tex] 1/r^2[/tex] for the generalized Coulomb (longitudinal) component. :smile:

    The latter can have a small transversal component as well since it
    points to the location were the emitting charge would have been if
    it had continued to move in the same direction at the same speed.

    (Liénard Wiechert potentials)

    Regards, Hans
    Last edited: Aug 4, 2005
  9. Aug 4, 2005 #8
    I stand corrected. I was thinking of radiated power instead of the fields when I wrote that.
  10. Aug 4, 2005 #9
    I know it is not zero along the line, but as I remember, the polar diagram of this emission (supose the charges are located along a horizontal line) reminds the shape of the number eight. (8), so that, horizontally, this diagram suggests no field strengh.

    Could you please confirm if you agree with this statement ?
  11. Aug 4, 2005 #10
    Correct me if I am wrong in the following:

    We can always apply Coulomb's Law to write the field at a given position.
    If the charge is initially at the origin and we are look at the field at the end of the vector R. Then eu could write for static situation:
    E = 9. 10^9 \frac{Q}{R^2}

    in the direction of te vector R (as Q > 0).

    If the charge moves very very slowly, we may adopt the approximative stand point according to which the field has a static configuration at any given instant. (it seems important to this argument to work with an universe with finite volume V).

    But rigorously speaking if the charge was always in the position R0 and, at time t=0, it moves to the position R0 + dR, then we have to take into consideration the limitting velocity c, and the field's spatial configuration at a given instant can be roughly divided in three regions: (1) the already corrected field, the region where the field resembles that of an static charge at Ro + dR, (2) an old fashioned region, where the field resembles that on an static charge still located at R0, and (3) a "being corrected region, where we may face intermediary states between these two configurations ((1) and (2)). It is precisely the third region which represents the radiation.

    I am really not sure about the correctness of this reasoning, although it seems quite reasonable to me.

    Best regards,

  12. Aug 5, 2005 #11

    Hans de Vries

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    You're reasoning is right here.

    In the 1900 Emil Wiechert proves that the Lienard-Wiechert retarded
    potentials give the correct results in cases where the effects due to the
    finite speed of light can't be neglected.

    Calculations must be made by assuming that the potentials, both the electric
    and magnetic vector potential, propagate from the source with the speed of
    light.The fields E and H can be recovered by differentiating in the right way.

    The fields calculated for charges moving at speeds close to c give the same
    results as the Special Theory of relativity (note that this was five years
    before einsteins famous 1905 year )

    Regards, Hans
  13. Aug 5, 2005 #12
    Ok, but why is there so much emphasys in the acceleration of the charge as an important condition for it to emit radiation ?
    According to the arguments I gave before, I see no reason to treat differently one moving electron (with constant velocity) and another one with non zero acceleration.

    Does an electron moving with constant velocity have straight field lines ? I once have seen it but it seemed quite strange to me.
  14. Aug 5, 2005 #13
    You are thinking of the radiation pattern. There will indeed be no
    radiation component (of the transverse type) along the axis of the
    dipole. But in general, there will be a non-zero but tiny radial component
    even along this axis. For an osciallating dipole there will be brief
    moments where the two charges are located at the same place. In this
    moment there will be zero fields and this zero may propogate outwards.

    The acceleration topic is not trivial. But for the uniformly moving charge,
    the field lines are strait in the reference frame of the charge. In any
    other inertial frame they are bent and are accompanied by a magnetic
    field which is only not seen in the reference frame of the charge.

    Edit: Clarification: After Hans de Vries, yes the lines are strait but they
    are tilted with respect to one another.
    Last edited: Aug 5, 2005
  15. Aug 5, 2005 #14

    Hans de Vries

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    The radiative part (which decreases with 1/r) is proportional to the
    acceleration. No acceleration = no radiation. Below a copy from a
    document I wrote:


    The Lienard Wiechert potentials.

    Alfred Lienard (in 1898) and Emil Wiechert (in 1900) determined
    the potentials of an arbitrarily moving and accelerating point
    charge. It is proved that the potentials spread from the source with
    the speed of light. The equations for the electric potential V and
    the magnetic vector potential A are close to the classical ones:

    [tex]V(\vec{x},t) = \frac{q}{4 \pi \epsilon_0 r_{ret}} \left[
    \frac{1}{(1-\beta \cos\phi)} \right] ,
    \ \ \ \
    \vec{A}(\vec{x},t) = \frac{q}{4 \pi \epsilon_0 r_{ret}} \left[
    \frac{\vec{\beta}}{(1-\beta \cos\phi)} \right]

    Where: [tex] \left\{ \begin{array}{lcl}
    \vec{\beta} & = & \mbox{speed vector of the charge: } \vec{v}/c \\
    r_{ret} & = & \mbox{distance from the retarded charge.} \\
    1/(1-\beta \cos\phi) & = & \mbox{compression or 'shockwave' factor }
    \end{array} \right. [/tex]

    Note that we work with the retarded charge: The location where the
    charge was at the moment when the potentials we measure left the
    point charge. The only thing which needs some explanation is the
    compression or 'shockwave' factor. It can become infinite in front
    of the moving charge (psi = 0) when the speed goes to c. This effect
    is equal to the shockwave building up in front of a plane which
    approaches the speed of sound. We see that all the information that
    is needed is: (1) the charge, (2) its location and (3) its speed.

    The Electro Magnetic Fields

    It is interesting that not the electric and magnetic fields E and B
    are propagating away with the speed of light but rather the
    potentials V and A. If we would assemble similar formula's for the
    fields then we get incorrect results. We do get the EM fields back
    however by carefully differentiating the potentials in space and
    time. We get:

    \vec{E}(\vec{x},t) \ \ = \ \ \frac{q}{4 \pi \epsilon_0 r^2_{ret}}
    \left[ \frac{(1-\beta^2) \ \vec{r}_{ph} }{(1-\beta \cos\phi)^3}
    \ \ \ + \ \ \ \ \frac{q}{4 \pi \epsilon_0 r_{ret}} \left[
    \frac{\hat{r}_{ret} \times (\vec{r}_{ph} \times \vec{a})}
    {c^2(1-\beta \cos\phi)^3} \right] ,

    \vec{B}(\vec{x},t) \ \ = \ \ \hat{r}_{ret} \times \vec{E}

    where: [tex] \left\{ \begin{array}{lcl}
    \vec{a} & = & \mbox{accelaration vector of the charge.} \\
    \hat{r}_{ret} & = & \mbox{vector of length 1 from retarded charge towards } (\vec{x},t) \\
    \vec{r}_{ph} & = & \mbox{the vector } (\hat{r}_{ret} -\vec{\beta})
    \mbox{ from the phantom location to } (\vec{x},t)
    \end{array} \quad

    with [itex]v \ll c [/itex] this simplifies to:

    \vec{E}(\vec{x},t) \ \ = \ \ \frac{q}{4 \pi \epsilon_0
    r^2_{ret}} \ [ \hat{r}_{ret} ]
    \ \ \ - \ \ \ \ \frac{q}{4 \pi \epsilon_0 r_{ret}} \ [ \vec{a} / c^2 ]

    We see that the first term is the standard Coulomb field. The
    second term is the radiation term which is proportional to the
    acceleration of the charge. It has a direction opposite to the field
    which would cause the charge to accelerate. The charge opposes a
    change in speed. It is this effect in general which causes

    The 'Phantom' location of the charge.

    Another interesting phenomena is the appearance of the 'phantom'
    location in the general formula. This is the location were the
    charge would be if it would have continued at the same speed after
    the potentials left the charge at the retarded position. The force
    from the generalized Coulomb field (the first term) points to the
    phantom location. At the other hand, the radiation field has its E
    and B components orthogonal to the vector coming from the retarded
    position. We see this effect for instance if we look at the Sun. The
    light comes from the retarded position while the gravitation seems
    to come from the phantom position. Yes, in (Newtonian) gravitation
    it is also the potential which spreads with the speed of light.


    Regards, Hans
    Last edited: Aug 5, 2005
  16. Aug 5, 2005 #15

    Hans de Vries

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    Yes, it has.

    They are straight in any non-accelerating reference frame however not spherical
    symmetric anymore but compressed with a factor [itex]1/(1-v^2/c^2)[/itex] in the direction
    of the motion.

    Regards, Hans
    Last edited: Aug 5, 2005
  17. Aug 5, 2005 #16
    Hans, aren't these the classical potentials?

    Could you elaborate on that? This sounds interesting.
  18. Aug 5, 2005 #17

    Hans de Vries

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    The terms between the square brackets are what makes them different from
    the classical ones.

    It's shown that only if you let V and A propagate away with c from the source
    and then differentiate to obtain E and B you get the right results. This confirms
    what we also see in QM: That the potentials and not the fields are the more
    fundamental quantities.

    Regards, Hans
  19. Aug 5, 2005 #18
    I'm comfortable with that. What I''m not comfortable with is that
    you can't get a correct derivation resorting to the fields alone. This
    seems like a formulational matter in setting up the proper equations.

    That is, if I properly manipulate the field equations they should always
    end up providing the same solutions as if I solve for potentials first
    and differentiate.
  20. Aug 5, 2005 #19

    Hans de Vries

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    The differentiation is not the same in the two cases. A Δx connects potentials
    which are emitted at different times. If you take E and B as propagating away
    with c then you use quantities which are already pre-differentiated presuming a
    source charge at rest. I do presume that, one can not recover the solutions
    in the latter case, taken into account that the differentiation of E and B means
    the loss of the "absolute values" of the potentials.

    Regards, Hans
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