|Feb12-09, 01:51 PM||#1|
"back of the envelope" derivation of Larmor's equation
I stumbled on a derivation of the Larmor's equation for the power radiated by an
accelerating charge that makes use of geometric arguments.
This document credits the derivation to JJ Thompson, while this site credits it to EM
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
|Feb12-09, 02:31 PM||#2|
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.
|Feb12-09, 04:17 PM||#3|
Love the concluding remark:
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.)
|Feb13-09, 03:22 AM||#4|
"back of the envelope" derivation of Larmor's equation
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
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.
|Feb13-09, 06:25 AM||#5|
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.
|Feb13-09, 06:47 AM||#6|
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 pdf version
of the arguments illustrated in this webpage (fig. 4.6).
Yes, I think now I'm happy of my understanding of this derivation of the Larmor's equation!
PS The derivation is credited to JJ Thomson in the pdf notes.
|Jun30-11, 12:52 PM||#7|
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.
|Similar Threads for: "back of the envelope" derivation of Larmor's equation|
|Rigid Pendulum "g" derivation equation||Advanced Physics Homework||1|
|"Back of envelope calculations" for thurst, specific impulse, etc.||Engineering Systems & Design||3|
|A "Dirac-Like" Derivation of the Dirac-Kemmer Equation||General Physics||0|
|Two Reviews: "God's Equation" and "Universe In A Nutshell"||Science Textbook Discussion||9|