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Back of the envelope derivation of Larmor's equation

  1. Feb 12, 2009 #1
    "back of the envelope" derivation of Larmor's equation

    Hi all,

    I stumbled on a derivation of the Larmor's equation for the power radiated by an
    accelerating charge that makes use of geometric arguments.
    http://www.cv.nrao.edu/course/astr534/PDFnewfiles/LarmorRad.pdf" [Broken] credits it to EM
    Purcell.

    The derivation seems nice, perhaps for recalling how things work without cumbersome
    calculations. However, there is something I do not completely get.

    The geometric argument shows that the electric field has a component perpendicular
    to the radial direction. This is convincing enough. What I do not completely understand
    is how they calculate the ratio of the radial and orthogonal component of the field.
    The ratio they use applies to the "kink" in the field lines (this is clearer in the second
    derivation). But how are the radial and azimuthal components
    of the "kink" (i.e. a portion of field line) connected to the same components of the field
    itself? Isn't the field amplitude related to the density of the field lines?

    I sort of see that the field lines are denser when [tex]\theta = \pi/2[/tex], but doesn't that
    affect both components equally (radial and azimuthal)?

    Does any of you see a simple argument to relate the field amplitude in a certain direction
    to the projection of the "kink" in that direction?

    Thanks a lot for any insight

    F
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 12, 2009 #2
    Re: "back of the envelope" derivation of Larmor's equation

    All that was used in the derivation was that the fact that the electric field is directed along the 'kink' that was drawn. Field lines describe the direction of the electric field, and the derivation of the formula you mention only needed that direction. Nothing about the strength (magnitude ) of the field went into it.
     
  4. Feb 12, 2009 #3
    Re: "back of the envelope" derivation of Larmor's equation

    Love the concluding remark:
    For a second year E&M course, I like how a few experiments with pith balls, wires and magnets suffice to imply so much (e.g., special relativity and optics). But I think the concluding remark is correct with reference to first year physics (before students acquire the math prerequisite for Maxwell's equations anyway); much simpler to start from the principles of relativity and of electric charges, use lower math, and focus more on the everyday world around us (as opposed to abstractly focussing on fields around different charge/material geometries).

    FranzDiCoccio, note that the qualitative section of the web page is really not referring to a "kink", but to the smooth gradients of a continuously varying field. (Back to the kink, the field lines must join up since the definition of field lines is that they only terminate at the location of a charge - and the number of them crossing a surface is a measure of the amount of charge enclosed. Yes, their spacing - not length - expresses field strength.)
     
  5. Feb 13, 2009 #4
    Re: "back of the envelope" derivation of Larmor's equation

    Hey xboy,

    thanks for your reply. I do not understand it, though. Why are you saying that the field strength is not involved in the derivation?
    The very first equation in the pdf document gives the ratio of the field amplitudes along
    the radial and azimuthal directions, doesnt'it?
    Eq. 2D2, gives the amplitude of the azimuthal component, and it is emphasized that it
    is larger than the radial one in the far-field limit.
    You need the amplitude to evaluate the Poyinting vector that eventually results in the
    Larmor Equation.

    Hi cesiumfrog,

    thanks for your comment. I do realize that the field lines vary smoothly.
    This is seen in the nice pictures in the html document.

    However in this simple construction the field lines are piecewise straight lines, right?
    I was referring to the small piece of each field line contained in the spherical shell as "the kink". That's the bit whose radial and azimuthal components are [tex]c \Delta t[/tex]
    and [tex]\Delta v t \sin \theta[/tex], respectively.

    I'd say that the ratio in the first equation should be evaluated based on the density
    of the azimuthal component of the lines to the density of their radial component.
    What I do not see is why this ratio is simply given by the ratio of the components
    of "the kink".

    I am afraid that this "simple" derivation might be less "self-contained" than it suggests.
    I mean, I like it, but if my question above has no simple and convincing answer, I think
    this derivation is more confusing than useful. You can use to convince yourself that
    there is an azimuthal component, but the proof that this "wins" over the radial one in the
    far field is a bit weak.

    Thanks again
     
  6. Feb 13, 2009 #5
    Re: "back of the envelope" derivation of Larmor's equation

    Franz, I did not mean that the derivation does not require the field strength. What I meant was the formula for the ratio of the two components.

    Suppose we have an arbitrary vector v along a particular direction ( making theta with x-axis, say) Then the ratio : [tex] \ v_y / \v_ x = \ tan \theta [/tex] , isn't it? We know that even if we don't know the magnitudes of the components. So if you know the direction, you know the ratio between the components, even if you do not know the magnitudes.

    The magnitude of the field is, as you rightly say, dependent on the density of field lines. But for the ratio between the two components of the field line, we only need the direction.
     
  7. Feb 13, 2009 #6
    Re: "back of the envelope" derivation of Larmor's equation

    xboy,

    yes, I think now I see what you mean, thanks.
    I guess I was confused about the fact that one can use the value of [tex]E_r[/tex] "just before" the kink. But on second thought, that makes sense.

    Plus, one can use an infinitely thin "pillbox" at the "bending point" to convince oneself that
    the radial component is actually that one. This is discussed in this more detailed "[URL [Broken] version
    [/URL]of the arguments illustrated in http://physics.weber.edu/schroeder/mrr/MRRtalk.html" [Broken] (fig. 4.6).

    Yes, I think now I'm happy of my understanding of this derivation of the Larmor's equation!
    Thanks again

    F

    PS The derivation is credited to JJ Thomson in the "[URL [Broken] notes[/URL].

     
    Last edited by a moderator: May 4, 2017
  8. Jun 30, 2011 #7
    Re: "back of the envelope" derivation of Larmor's equation

    Very interesting.
    If you wish to understand this Larmor em radiation, I recommend reading "Elementary Classical Physics, Vol 2, Weidner and Sells, Acclerating charges and electromagnetic waves. In my edition, it's chapter 41-4, page 1024.
    I used this textbook to teach PHY-112 long, long ago.
    Good luck.
     
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