拉莫爾方程式

八木天線，無線電波可由天線中的加速電子產生。此為相干性過程，因此輻射總功率和電子加速度的平方成正比。

在電動力學的領域中，拉莫方程式（Larmor formula，請勿和古典核磁共振的拉莫爾進動混淆）是用來計算非相對論性點電荷在有加速度的狀態下釋放電磁波的總功率。本公式是由約瑟夫·拉莫爾於1897年提出的光的波動理論一部分。

當任何點電荷（例如電子）有正或負的加速度時，會以電磁輻射的形式釋放能量。對於遠小於光速的狀態下，總幅射能量可用如下方程式表示：

$P={\frac {e^{2}a^{2}}{6\pi \varepsilon _{0}c^{3}}}{\mbox{ }}$（SI單位制）

$P={2 \over 3}{\frac {e^{2}a^{2}}{c^{3}}}{\mbox{ }}$（CGS單位制）

公式中 $a$ 是加速度，$e$ 是電荷，$c$ 是光速。相對論狀況下則由黎納-維謝勢給定。

推導

推導1：數學逼近

首先必須要找到電場和磁場的形式，可使用以下形式表示（更完整推導請參見黎納-維謝勢）：

${\vec {E}}({\vec {x}},t)=q\left({\frac {{\vec {n}}-{\vec {\beta }}}{\gamma ^{2}(1-{\vec {\beta }}\cdot {\vec {n}})^{3}R^{2}}}\right)_{\rm {ret}}+{\frac {q}{c}}\left({\frac {{\vec {n}}\times [({\vec {n}}-{\vec {\beta }})\times {\vec {\dot {\beta }}}]}{(1-{\vec {\beta }}\cdot {\vec {n}})^{3}R}}\right)_{\rm {ret}}$

以及

${\vec {B}}={\vec {n}}\times {\vec {E}}$

此處

$\mathbf {\beta }$ 是電荷速度和光速 c 的商

$\mathbf {\dot {\beta }}$ 是電荷加速度和光速 c 的商

$\mathbf {n}$ 是 $\mathbf {r} -\mathbf {r} _{0}$ 方向上的單位向量

$R$ 是 $\mathbf {r} -\mathbf {r} _{0}$ 方向的純量

而以下公式右方則是計算推遲時間

$t'=t-{R \over c}$

場方程式本身會分離為速度場和加速度場。速度場只與 β 相關，而加速度場則和 $\beta \,$ 與 ${\dot {\beta }}$ 之間的夾角相關。因為速度場和 $R^{-2}$ 成正比，會使它隨距離增加快速下降。另一方面，加速度場和 $R^{-1}$ 成正比，代表它隨距離增加的下降率遠低於速度場。正因為如此，加速度場代表了輻射場，並且從電荷攜帶了大多數的能量。

而輻射場的能量通量密度可使用坡印廷向量得知：

${\vec {S}}={\frac {c}{4\pi }}{\vec {E}}_{a}\times {\vec {B}}_{a}$

上式中下標 'a' 則強調是在加速度場的分量。以磁場和電場關係進行取代，同時假設粒子在特定時間 $t_{r}$ 是靜止的，可得到：

${\vec {S}}={\frac {q^{2}}{4\pi c}}\left|{\frac {{\vec {n}}\times ({\vec {n}}\times {\vec {\dot {\beta }}})}{R}}\right|^{2}.$

當 $\beta \left(t_{r}\right)\neq 0$ 時證明會較困難（參見 Griffiths）

如果讓加速度和觀測者方向的向量夾角相等於 $\theta$，就可用以下公式表示：

${\vec {S}}={\frac {q^{2}}{4\pi c^{3}R^{2}}}\sin ^{2}{\theta }|{\vec {\dot {V}}}|^{2}{\hat {n}}.$

其實這是電荷輻射在每單位立體角下的單位功率。可使用積分計算完整立體角下的功率，即：

$P={\frac {2}{3}}{\frac {q^{2}|{\vec {\dot {V}}}|^{2}}{c^{3}}}.$

以上公式就是非相對論性下有加速度電荷所釋放輻射的功率，功率大小與加速度相關。這公式明確表示加速度較高電荷會產生較強輻射，我們可以期望輻射場和加速度相關。

推導2：爱德华·珀塞尔逼近

The full derivation can be found here.

Here is an explanation which can help understanding the above page.

This approach is based on the finite speed of light. A charge moving with constant velocity has a radial electric field $E_{r}$ (at distance $R$ from the charge), always emerging from the future position of the charge, and there is no tangential component of the electric field $(E_{t}=0)$. This future position is completely deterministic as long as the velocity is constant. When the velocity of the charge changes, (say it bounces back during a short time) the future position "jumps", so from this moment and on, the radial electric field $E_{r}$ emerges from a new position. Given the fact that the electric field must be continuous, a non-zero tangential component of the electric field $E_{t}$ appears, which decreases like $1/R$ (unlike the radial component which decreases like $1/R^{2}$).

Hence, at large distances from the charge, the radial component is negligible relative to the tangential component, and in addition to that, fields which behave like $1/R^{2}$ cannot radiate, because the Poynting vector associated with them will behave like $1/R^{4}$.

The tangential component comes out (SI units):

$E_{t}={{ea\sin(\theta )} \over {4\pi \varepsilon _{0}c^{2}R}}.$

And to obtain the Larmour formula, one has to integrate over all angles, at large distance $R$ from the charge, the Poynting vector associated with $E_{t}$, which is:

$\mathbf {S} ={E_{t}^{2} \over \mu _{0}c}\mathbf {\hat {r}} ={{e^{2}a^{2}\sin ^{2}(\theta )} \over {16\pi ^{2}\varepsilon _{0}c^{3}R^{2}}}\mathbf {\hat {r}}$

giving (SI units)

$P={{e^{2}a^{2}} \over {6\pi \varepsilon _{0}c^{3}}}.$

This is mathematically equivalent to:

$P={{\mu _{0}e^{2}a^{2}} \over {6\pi c}}.$

相對論性狀況下

Covariant Form

We can do this by rewriting the Larmor formula in terms of momentum and then using the four vector generalisation of momentum (see four momentum), $P^{\mu }$. We know that the power is a Lorentz invariant, so all we have to show is that our generalisation is also invariant and that it reduces to the Larmor formula in the low velocity limit. So;

$P={\frac {2}{3}}{\frac {q^{2}}{c^{3}m^{2}}}\left({\frac {d{\vec {p}}}{dt}}\cdot {\frac {d{\vec {p}}}{dt}}\right).$

Assume the generalisation;

$P=-{\frac {2}{3}}{\frac {q^{2}}{m^{2}c^{3}}}{\frac {dP^{\mu }}{d\tau }}{\frac {dP_{\mu }}{d\tau }}.$

When we expand and rearrange the energy-momentum four vector product we get;

${\frac {dP^{\mu }}{d\tau }}{\frac {dP_{\mu }}{d\tau }}={\frac {v^{2}}{c^{2}}}\left({\frac {dP}{d\tau }}\right)^{2}-\left({\frac {d{\vec {p}}}{d\tau }}\right)^{2}$

where I have used the fact that ${\frac {dE}{d\tau }}={\frac {Pc^{2}}{E}}{\frac {dP}{d\tau }}=v{\frac {dP}{d\tau }}$. When you let $\beta$ tend to zero, $\gamma$ tends to one, so that $d\tau$ tends to dt. Thus we recover the non relativistic case.

This is an interesting equation. It says that the power radiated by the particle into space depends upon its rate of change of momentum with respect to its time. It also says that the power radiated is proportional to the charge squared and inversely proportional to the mass squared. Thus for a highly charged, extremely small particle the radiation will be much greater than that for a large particle with a small charge.

Non-Covariant Form

To obtain the non-covariant form of the generalisation we first substitute $p^{\mu }=(\gamma mc^{2},\gamma m{\vec {v}})$ in to the above and then performing the differentiation as follows (for brevity I have omitted the constants from the calculation below);

${\frac {dp^{\mu }}{d\tau }}{\frac {dp_{\mu }}{d\tau }}=-\left({\frac {d{\vec {p}}}{d\tau }}\right)^{2}+{\frac {1}{c^{2}}}\left({\frac {dE}{d\tau }}\right)^{2}$

${\frac {dp^{\mu }}{d\tau }}{\frac {dp_{\mu }}{d\tau }}=-\gamma ^{2}\left({\frac {d\gamma m{\vec {v}}}{dt}}\right)^{2}+{\frac {\gamma ^{2}}{c^{2}}}\left({\frac {d\gamma mc^{2}}{dt}}\right)^{2}$

${\frac {dp^{\mu }}{d\tau }}{\frac {dp_{\mu }}{d\tau }}=-\gamma ^{2}[-(\gamma m{\vec {\dot {v}}}+\gamma ^{3}m{\vec {v}}({\vec {\beta }}\cdot {\vec {\dot {\beta }}}))^{2}+{\frac {1}{c^{2}}}(\gamma ^{3}{\vec {\beta }}\cdot {\vec {\dot {\beta }}}mc^{2})^{2}]$

${\frac {dp^{\mu }}{d\tau }}{\frac {dp_{\mu }}{d\tau }}=\gamma ^{8}m^{2}c^{2}[({\vec {\beta }}\cdot {\vec {\dot {\beta }}})^{2}-({\vec {\beta }}({\vec {\beta }}\cdot {\vec {\dot {\beta }}})+{\frac {\vec {\dot {\beta }}}{\gamma ^{2}}})^{2}]$

$\Rightarrow {\frac {dp^{\mu }}{d\tau }}{\frac {dp_{\mu }}{d\tau }}=\gamma ^{8}m^{2}c^{2}\left(-{\frac {1}{\gamma ^{2}}}({\vec {\beta }}\cdot {\vec {\dot {\beta }}})^{2}-{\frac {{\vec {\dot {\beta }}}^{2}}{\gamma ^{4}}}\right).$

Although the above is correct as it stands, it is not immediately obvious what sort of relationship the radiated power has to the velocity and the acceleration of the particle. If we make this relationship more explicit then it will be clear how the radiation depends on the particle's motion, and what happens in different cases. We can obtain this relation by adding and subtracting ${\frac {{\vec {\beta }}^{2}\cdot {\vec {\dot {\beta }}}^{2}}{\gamma ^{2}}}$ to the above giving;

$\gamma ^{6}m^{2}c^{2}[({\vec {\beta }}^{2}{\vec {\dot {\beta }}}^{2}-({\vec {\beta }}\cdot {\vec {\dot {\beta }}})^{2})-{\vec {\dot {\beta }}}^{2}].$

If we apply the vector identity;

$({\vec {\beta }}\times {\vec {\dot {\beta }}})\cdot ({\vec {\beta }}\times {\vec {\dot {\beta }}})=({\vec {\beta }}^{2}{\vec {\dot {\beta }}}^{2}-({\vec {\beta }}\cdot {\vec {\dot {\beta }}})^{2}).$

Then we obtain;

$P={\frac {2q^{2}\gamma ^{6}}{3c}}\left(({\vec {\dot {\beta }}})^{2}-({\vec {\beta }}\times {\vec {\dot {\beta }}})^{2}\right)$

where I have replaced all the constants and the negative sign dropped earlier.

This is the Lienard result, which was first obtained in 1898. The $\gamma ^{6}$ means that when $\gamma$ is very close to one (i.e. $\beta <<1$) the radiation emitted by the particle is likely to be negligible. However when $\gamma$ is greater than one (i.e. $\beta \rightarrow 1$) the radiation explodes as the particle tries to lose its energy in the form of EM waves. It's also interesting that when the acceleration and velocity are orthogonal the power is reduced by a factor of ${\vec {\beta }}\cdot {\vec {\dot {\beta }}}$. The faster the motion becomes the greater this reduction gets. In fact, it seems to imply that as $\beta$ tends to one the power radiated tends to zero (for orthogonal motion). This would suggest that a charge moving at the speed of light, in instantaneously circular motion, emits no radiation. However, it would be impossible to accelerate a charge to this speed because the $\gamma ^{6}$ would explode to $\infty ^{6}$, meaning that the particle would radiate a gigantic amount of energy which would require you to put more and more energy in to keep accelerating it. This would imply that there is a cosmic speed limit, namely c. Such a connection was not made until 1905 when Einstein published his paper on Special Relativity.

We can use Lienard's result to predict what sort of radiation losses to expect in different kinds of motion.

相關議題與影響

輻射反作用力

來自帶電粒子的輻射會攜帶有能量和動量。為了滿足能量和動量守恆，帶電粒子會在釋放電磁輻射時出現反衝現象，因此在帶電粒子上必須要有一個額外的力。在非相對論性下這個力量就是阿布拉罕-勞侖茲力，而在相對論性狀態下則是阿布拉罕-勞侖茲-狄拉克力。

原子物理

在古典物理中，環繞原子核的電子會被加速且釋放出電磁輻射，會使電子失去能量並以螺旋路徑墜入原子核。在古典力學之下原子應是不穩定的，因此穩定的電子軌道違反了古典力學的預測。這個原子物理學的問題由量子力學或隨機電動力學解決。

參見

• 原子理論
• 回旋辐射
• 電磁波方程式
• 弯曲时空中的麦克斯韦方程组
• 波动方程
• 阿布拉罕-勞侖茲力
• 惠勒－費曼吸收體理論

延伸閱讀

• J. Larmor, "On a dynamical theory of the electric and luminiferous medium", Philosophical Transactions of the Royal Society 190, (1897) pp. 205–300 (Third and last in a series of papers with the same name).

• Griffiths, David J. Introduction to Electrodynamics (3rd ed.). Addison-Wesley. 1999. ISBN 9780138053260. 

• Jackson, John D. Classical Electrodynamics (3rd ed.). Wiley. 1998. ISBN 0-471-30932-X. 

• Misner, Charles; Thorne, Kip S. & Wheeler, John Archibald. Gravitation. San Francisco: W. H. Freeman. 1973. ISBN 0-7167-0344-0. 

• R. P. Feynman, F. B. Moringo, and W. G. Wagner. Feynman Lectures on Gravitation. Addison-Wesley. 1995. ISBN 0-201-62734-5.