Talk:Larmor formula

message: Just got to get the equations correct. Wont be too long. Please bear with me!--Light current 01:14, 23 October 2005 (UTC)
thread_title: Untitled

message: The derivation given here was first published by J. J. Thomson (discoverer of the electron) in 1907. It is derived for the special case where the final velocity of the particle is zero but the Larmor formula is true for any sort of accelerated motion provided that the speed of the particle is always much less than the speed of light.
The energy per unit volume stored in an electric field is
Energy/vol = 1/2 \epsilon_0\|E|^2 
Neglecting the radial component of the field:
:Energy/vol = {q^2 a^2 \sin^2 \theta} \over {32 \pi^2 \epsilon_0 c^4 R^2} 
If the direction in which the energy goes is not important, we can average the energy over all directions. Using a mathematical device, introduce a coordinate system with the origin at the center of the sphere and the x axis along the particle’s original direction of motion. Then for any point (x, y, z) on the spherical shell, cos θ = x/R.
Using angle brackets to denote an average over all points on the shell,  
: \sin^2 \theta = {1 - cos^2 \theta} = 1 - {x^2 \over R^2} \ .
Now since the origin is at the center of the sphere, the average value of x^2 is the same as the average value of y^2 or z^2:
: x^2 = y^2 = z^2 \ 
But this implies that
: x^2 = \frac{1}{3}(x^2 + y^2 + z^2) = \frac{1}{3} R^2 = \frac{R^2}{3} 
since,
: x^2 + y^2 + z^2 = R^2 \ 
and R is constant over the whole shell. Combining equations  gives
: \sin^2 \theta = {1 - R^2 \over 3R^2} = \frac{2}{3} 
So the average energy per unit volume stored in the transverse electric field is
: {{q^2 a^2} \over {48 \pi^2 \epsilon_0 c^4 R^2}} 
To obtain the total energy stored in the transverse electric field, we must multiply equation by the volume of the spherical shell. The surface area of the shell is 4πR2 and its thickness is ct0, so its volume is the product of these factors. Therefore the total energy is
Total energy in electric field =  E = \frac{e^2 a^2}{6 \pi \epsilon_0 c^3}
The total energy is independent of R; that is, the shell carries away a fixed amount of energy that is not diminished as it expands.
There is also a magnetic field, which carries away an equal amount of energy. Many details about magnetic fields have been omitted. A factor of 2 needs inserting. Thus the total energy carried away by the pulse of radiation is twice that of the previous equation  or
Total energy in pulse =  E = \frac{e^2 a^2}{6 \pi \epsilon_0 c^3}
Divide both sides of this equation by t0, the duration of the particle’s acceleration. The left-hand side then becomes the energy radiated by the particle per unit time, or the power given off during the acceleration: Power radiated    
: P = \frac{e^2 a^2}{6 \pi \epsilon_0 c^3}
An example is the electric field around an oscillating charge. A map of the electric field lines around a positively charged particle oscillating sinusoidally, up and down, between the two gray regions near the center. Points A and B are one wavelength apart. If you follow a straight line out from the charge at the center of the figure, you will find that the field oscillates back and forth in direction. The distance over which the direction of the field repeats is called the wavelength. For instance, points A and B in the figure are exactly one wavelength apart. The time that it takes the pattern to repeat once is called the period of the wave, and is equal to the time that the source charge takes to repeat one cycle of its motion. The period is also equal to the time that the wave takes to travel a distance of one wavelength. Since it moves at the speed of light, we can infer that the wavelength and the period are related by
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:Interesting. If this is can be finished and cleaned up, it should moved to the article proper. Especially if this is how Larmor did it. linas 14:38, 4 January 2006 (UTC)
thread_title:  Derivation of formula 

message: What is your source for this? --EMS | Talk 22:13, 31 July 2006 (UTC)
replies: text: Added references Complexica 20:29, 8 August 2006 (UTC)
thread_title:  Equivalence principle 

message: As of Sept 5 2008, this article used \epsilon rather than \varepsilon for the permittivity of vacuum.  I've always see \varepsilon used for this constant.  Indeed, this is what is used in http://en.wikipedia.org/wiki/Permittivity
I changed \epsilon to \varepsilon in this article.  If this is incorrect, please describe why \epsilon should be used in this equation.  —Preceding unsigned comment added by DanHickstein (talk • contribs) 18:59, 5 September 2008 (UTC)
thread_title:  Epsilon or varepsilon for permittivity of free space 

message: The article lacks in the area of superconducting currents. Cited material please.Kmarinas86 (6sin8karma) 20:37, 31 July 2010 (UTC)
thread_title:  Applicability to superconducting currents? 

message: Usually I can stomach arrow notation for vectors, but symbols such as \vec{\dot{\beta}} are truly repulsive. If no one objects, I'm going to change the article's notation to use boldface for vectors. Zueignung (talk) 23:51, 30 December 2012 (UTC)
thread_title:  Yikes, that's some ugly vector notation 

message: there is an erroneous equation in the derivation 1 section 
:\frac{dP}{d\Omega} = \frac{q^2}{4\pi c}\frac{\sin^2(\theta)\, a^2\, \hat{\mathbf{n}}}{c^2 R^2}.
I believe the correct formulas are either
:\mathbf{S} = \frac{q^2}{4\pi c}\frac{\sin^2(\theta)\, a^2\, \hat{\mathbf{n}}}{c^2 R^2}. 
or
:\frac{dP}{d\Omega} = \frac{q^2}{4\pi c}\frac{\sin^2(\theta)\, a^2}{c^2}.
as taken from Jackson 3rd edition pg 665
I'll change it to the latter to flow with the text before the equation
Jhmadden (talk) 00:17, 15 December 2014 (UTC)
replies: text: You are correct: the formula equating dP/dΩ to the expression involving (qa/R)2/c3 must be wrong, based on dimensional analysis alone. dP/dΩ has units of erg s−1 = g cm2 s−3, and (qa/R)2/c3 has units of g s−3. Zueignung (talk) 17:31, 16 December 2014 (UTC)
thread_title:  Derivation 1 typo? 

message: I added a new source, where the formula is actually mentioned.
I couldn't find it in "On a dynamical theory of the electric and luminiferous medium", so this hint shows a paper from the right year and the right author, but is a bit misleading.
141.53.32.79 (talk) 12:01, 7 October 2015 (UTC)
thread_title:  Original source by Larmor 

message: The last sentence in "Atomic Physics" appears to be a serious misunderstanding of the paper of Haus (found in the link to the wikipedia page "Nonradiation condition"), which says that non-accelerating charges do not radiate.  — Preceding unsigned comment added by 47.23.28.187 (talk) 00:27, 30 May 2018 (UTC)
replies: text: I think that last sentence should be removed because it wouldn't apply to electrns. Alfa137 (talk) 15:46, 19 April 2023 (UTC)
thread_title:  non-radiation condition 

message: If electron acceleration is the source of antenna radiation, then at higher frequencies antenna efficiency would increase, which is clearly not the case. One can safely assume that the main thesis of this Wikipedia article is wrong.80.121.121.188 (talk) 09:08, 11 August 2019 (UTC)
replies: text: The electrons in an antenna generally do not move at relativistic speeds, so the gamma factors do not enter. Alfa137 (talk) 16:06, 19 April 2023 (UTC)
thread_title:  Acceleration and Radiation