Talk:Larmor formula

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

Derivation of formula
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


 * 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)

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

Epsilon or varepsilon for permittivity of free space
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)

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

total amount of energy radiated over a full cycle
what is the total amount of energy radiated over a full cycle from a charge oscillating back and forth with a simple sine wave motion? Just granpa (talk) 20:02, 28 November 2010 (UTC)


 * http://maxwellsociety.net/PhysicsCorner/Transformations/LienardLarmor/Covariance%20of%20the%20Larmor.html
 * http://www.phys.lsu.edu/~jarrell/COURSES/ELECTRODYNAMICS/Chap14/hmwk14.pdf
 * Just granpa (talk) 20:40, 28 November 2010 (UTC)


 * a = (d/dt)(d/dt)(d*sin(f*2*pi*t))
 * a = -4 pi^2 d f^2 sin(2 pi f t)


 * integral from t=0 to t=1 of a^2
 * integral from t=0 to t=1 of (4 pi^2 d f^2 sin(2 pi f t) )^2


 * = 2 pi^3 d^2 f^3 (4 pi f-sin(4 pi f))


 * sin(2*pi) = 0 so it reduces to


 * 2 pi^3 d^2 f^3 (4 pi f) = 8 * pi^4 * d^2 * f^4


 * However, thats for one unit of time not one cycle


 * Just granpa (talk) 06:59, 23 November 2015 (UTC)

Inconsistent with Bremsstrahlung??
Bremsstrahlung discusses the power radiated from an accelerated charge, but the formulas are different from this article. I don't see how to reconcile them. At the very least I expect the two articles to cross-reference each other and explain the different assumptions that lead to the different formulas. It's also possible that one of the two articles is flat-out wrong. Someone knowledgeable can please help?? Thanks in advance!! --Steve (talk) 14:19, 12 December 2012 (UTC)


 * The non-relativistic formula in the lede of Larmor formula is:
 * $$ P = \frac{e^2 a^2}{6 \pi \varepsilon_0 c^3} \mbox{ (SI units)} $$
 * The formula in Bremsstrahlung is
 * $$P = \frac{q^2 \gamma^4}{6 \pi \varepsilon_0 c}

\left( \dot{\beta}^2 + \frac{(\vec{\beta} \cdot \dot{\vec{\beta}})^2}{1 - \beta^2}\right)$$
 * In the non-relativistic limit, the second term in parentheses can be neglected, and the first term reduces to a^2/c^2. Set q = e and gamma = 1 and the formulas become identical. Where's the problem? Art Carlson (talk) 11:07, 13 December 2012 (UTC)


 * Bremsstrahlung says "The general expression for the total radiated power is
 * $$P = \frac{q^2 \gamma^4}{6 \pi \varepsilon_0 c} \left( \dot{\beta}^2 + \frac{(\vec{\beta} \cdot \dot{\vec{\beta}})^2}{1 - \beta^2}\right)$$
 * Larmor says
 * $$P = \frac{2q^2\gamma^6}{3c}\left((\vec{\dot{\beta}})^2 - (\vec{\beta}\times\vec{\dot{\beta}})^2\right)$$
 * I'll put Larmor in SI units for a better comparison
 * $$P_{\text{Larmor article}} = \frac{q^2\gamma^6}{6\pi \varepsilon_0 c}\left((\vec{\dot{\beta}})^2 - (\vec{\beta}\times\vec{\dot{\beta}})^2\right)$$
 * I'll pull out gamma^2 in Bremsstrahlung's for a better comparison
 * $$P_{\text{Brem article}} = \frac{q^2 \gamma^6}{6 \pi \varepsilon_0 c} \left( (1 - \beta^2)\dot{\beta}^2 + (\vec{\beta} \cdot \dot{\vec{\beta}})^2 \right)$$
 * Let $$\theta$$ be the angle between velocity and acceleration...
 * $$P_{\text{Larmor article}} = \frac{q^2\gamma^6}{6\pi \varepsilon_0 c}\left((\dot{\beta})^2 - \beta^2 \dot{\beta}^2 \sin^2 \theta\right)$$
 * $$P_{\text{Brem article}} = \frac{q^2 \gamma^6}{6 \pi \varepsilon_0 c} \left( (\dot{\beta})^2 - (\dot{\beta}^2)(\beta^2) + (\dot{\beta}^2)(\beta^2) \cos^2 \theta \right)$$
 * Ah, OK, I see, sorry about that. --Steve (talk) 14:31, 13 December 2012 (UTC)

Yikes, that's some ugly vector notation
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)

Derivation 1 typo?
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)


 * 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&minus;1 = g cm2 s&minus;3, and (qa/R)2/c3 has units of g s&minus;3. Zueignung (talk) 17:31, 16 December 2014 (UTC)

Original source by Larmor
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)

non-radiation condition
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)


 * I think that last sentence should be removed because it wouldn't apply to electrns. Alfa137 (talk) 15:46, 19 April 2023 (UTC)

Acceleration and Radiation
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)wabi


 * 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)

Units
The meaning of the formula $$\mu_0\frac{q^2a^2}{6\pi c}$$ is independent of a specific system of units, so what is the idea behind the parenthesis $$\text{(SI units)}$$ and how does it not apply just as well when working in the cgs system? --Lambiam 18:23, 23 January 2023 (UTC)


 * I think this last term is wrong and should be removed, as this µ0, stands for something like 1 over ϵ_0 c^3, which I think is meaningless and certainly not the expression of a dipole moment. This is already obvious by checking the (SI) units. Snoopy Urania (talk) 15:02, 2 April 2023 (UTC)
 * I don't understand this. The vacuum permittivity is related to the vacuum permeability by $$\varepsilon_0 =(\mu_0c^2)^{{-}1}.$$ This implies that the last two expressions in the chain of equalities are equivalent – if the last one is wrong, so is the one but last. So that one is then wrong too and should also be removed. But it is clearly equivalent to the one before it, so ...
 * None of this is relevant to the suggested dependence on the system of units. In the literature some texts have a factor $$4\pi\varepsilon_0$$ and others don't. The formulas on the first line have dimension $$\mathsf{T}^{{-}3}\mathsf{L}^2\mathsf{M},$$ the same as power, Those of the second line have dimension $$\mathsf{T}\mathsf{L}^{{-}1}\mathsf{I}^2$$ if I'm not mistaken, in any case another dimension than power.  --Lambiam 18:59, 2 April 2023 (UTC)
 * ALL of the equations in that section depend on the system of units used.
 * The $\mu_0$ there is not meant to be the expression of a dipole moment, and shouldn't be. Alfa137 (talk) 15:56, 19 April 2023 (UTC)
 * If all equations depend on the system of units, the sentence about how the power radiated by a single electron is described in either unit system by the same formula is wrong.
 * Can you explain how the formula for $$P$$ in cgs units has the dimension of power? What are the dimensions of $$q,~a$$ and $c$?
 * (I think $$\mu_0$$ stands here for the vacuum permeability, a physical constant. The electric dipole moment is a completely incommensurate physical quantity and not a constant. It is not referenced in the article and I have no idea why it was mentioned here.) --Lambiam 19:54, 19 April 2023 (UTC)
 * In cgs units, a has the dimension cm/sec^2, c has dimension cm/sec. Then, q^2a^2/c^3 has the dimension q^2/(cm-sec).

q^2/cm has the dimension of energy in ergs, So q^2a^2/c^3 has the dimension of power in ergs/sec. q has the dimension of 'statcoulombs', but that doesn't have to be mentioned if q^2/cm is recognized as energy in ergs. — Preceding unsigned comment added by Alfa137 (talk • contribs) 01:38, 20 April 2023 (UTC)