Newest questions tagged lienard-wiechert - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-06-19T03:54:05Z https://physics.stackexchange.com/feeds/tag?tagnames=lienard-wiechert&sort=newest http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/469555 0 About density charged in the Liénard - Wiechert Potential to Point Charge? PCat27 https://physics.stackexchange.com/users/117996 2019-03-30T15:15:14Z 2019-03-30T16:00:41Z <p>I'm reading Griffiths Ch. 10. In the 10.3.1 section, there's a proof of this integral</p> <p><span class="math-container">$$\int \rho(r^\prime, t_r) \mathrm{d} \tau^\prime$$</span> which is not equal to the charge of the particle, but really I don't understand what is the reason, because if the system is isolated, all density charge will be the same in the time retarded, no? so, it will be equal to the total charge.</p> <p>I have another question,</p> <p>Griffiths shows that</p> <p><span class="math-container">$$\int \rho(r^\prime, t_r) \mathrm{d} \tau^\prime = \frac{q}{1 -\frac{ |\vec{r}-\vec{r'}| \cdot \vec{v}}{c}}$$</span></p> <p>And deriving it, he used this example:</p> <p><a href="https://i.stack.imgur.com/El1Au.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/El1Au.png" alt="enter image description here"></a></p> <p>he writes <span class="math-container">$$\frac{L'}{c} =\frac{L'-L}{v}$$</span></p> <p>But I think, there are many assumptions that aren't necessarily always right. For example:</p> <ul> <li>Is the final time the same to the light as to the train? if the Light travels faster, how it can be possible? </li> </ul> <p>I don't know, I'm very confused about it, I will appreciate any help. </p> https://physics.stackexchange.com/q/463347 0 Feynman-Heaviside formula and Mach's principle John Eastmond https://physics.stackexchange.com/users/22307 2019-02-27T17:47:36Z 2019-02-27T17:54:17Z <p>I was wondering if the <a href="http://www.feynmanlectures.caltech.edu/II_21.html" rel="nofollow noreferrer">Feynman-Heaviside formula</a> for the electric field of a moving charge could be used to write down the force/reaction force between charges <span class="math-container">$q_1$</span> and <span class="math-container">$q_2$</span> in a Machian purely relational way.</p> <p>The retarded electric force <span class="math-container">$\vec{F_{12}}$</span>, on a charge <span class="math-container">$q_2$</span> that is at rest at its current position <span class="math-container">$\vec{r_2}(t)$</span>, due to a moving charge <span class="math-container">$q_1$</span> at its earlier position <span class="math-container">$\vec{r_1}(t-R/c)$</span> is <span class="math-container">$$\vec{F_{12}} = \frac{q_1 q_2}{4 \pi \epsilon_0} \left[\frac{\vec{n}}{R^2} + \frac{R}{c}\frac{d}{dt} \left(\frac{\vec{n}}{R^2}\right) + \frac{1}{c^2} \frac{d^2\vec{n}}{dt^2}\right]\tag{1}$$</span> Where <span class="math-container">$$\begin{eqnarray} \vec{R} &amp;=&amp; \vec{r_2}(t)-\vec{r_1}(t-R/c)\tag{2}\\ R &amp;=&amp; |\vec{R}|\\ \vec{n} &amp;=&amp; \frac{\vec{R}}{R} \end{eqnarray}$$</span></p> <p>The expression for the force <span class="math-container">$\vec{F_{12}}$</span> in Eqn. <span class="math-container">$(1)$</span> is written entirely in terms of the magnitude and direction of the relative vector between the current position of charge <span class="math-container">$q_2$</span> and the earlier position of charge <span class="math-container">$q_1$</span>. No variables defined in terms of an absolute reference frame are used.</p> <p>The first two terms on the righthand side of Eqn. <span class="math-container">$(1)$</span> are the near-field Coulomb term and its correction that fall off like <span class="math-container">$1/R^2$</span> whereas the last is the far-field radiative term that falls of like <span class="math-container">$1/R$</span>. </p> <p>The advanced reaction force <span class="math-container">$\vec{F_{21}}$</span> back on charge <span class="math-container">$q_1$</span> at its earlier position <span class="math-container">$\vec{r_1}(t-R/c)$</span> due to charge <span class="math-container">$q_2$</span> at its current position <span class="math-container">$\vec{r_2}(t)$</span> is then just</p> <p><span class="math-container">$$\vec{F_{21}} = - \vec{F_{12}}\tag{3}$$</span></p> <p>Thus Newton's third law of action and reaction is obeyed through an influence that travels at the speed of light forward in time from <span class="math-container">$q_1$</span> to <span class="math-container">$q_2$</span> and then backward in time from <span class="math-container">$q_2$</span> to <span class="math-container">$q_1$</span>. </p> <p>This reaction force provides an electromagnetic inertial force back on charge <span class="math-container">$q_1$</span> at the earlier time <span class="math-container">$t-R/c$</span> due to the presence of the charge <span class="math-container">$q_2$</span> at the current time <span class="math-container">$t$</span>.</p> <p>One could test for this electromagnetic inertia by accelerating an electron of charge <span class="math-container">$-e$</span> inside an insulating charged sphere of radius <span class="math-container">$R$</span> and charge <span class="math-container">$Q$</span>. The electron’s inertia should be increased by an amount ~ <span class="math-container">$eQ/(4 \pi \epsilon_0 c^2 R)$</span>.</p> <p>Finally this Feynman-Heaviside force could be generalized to the <a href="https://en.wikipedia.org/wiki/Gravitoelectromagnetism" rel="nofollow noreferrer">weak-field</a> gravitational case simply by substituting masses for charges and Newton’s constant <span class="math-container">$G$</span> for <span class="math-container">$-1/(4 \pi \epsilon_0)$</span>. Thus standard inertia could be explained as the result of the gravitational advanced Feynman-Heaviside reaction force acting back on an accelerated test mass from all the other masses in the Universe.</p> <p>Does this make sense?</p> https://physics.stackexchange.com/q/456087 0 Electric field produced by a moving charged particle above a planar dielectric interface Mathemagician https://physics.stackexchange.com/users/173300 2019-01-23T10:20:41Z 2019-01-23T10:46:15Z <p>The electrostatic field of a single charged particle above a planar dielectric interface is a standard example given in many books (see example 4.4 in Griffiths or <a href="https://en.wikipedia.org/wiki/Method_of_image_charges#Reflection_in_a_dielectric_planar_interface" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Method_of_image_charges#Reflection_in_a_dielectric_planar_interface</a>). I was wondering what happens in the case where the charge is moving?</p> <p>More precisely, suppose that we split <span class="math-container">$\mathbb{R}^3$</span> into two halfspaces: for <span class="math-container">$z \geq 0$</span> we have vacuum and for <span class="math-container">$z \leq 0$</span> we have a linear medium with electric and magnetic constants <span class="math-container">$\epsilon_1, \mu_1$</span>. Suppose that we have a single charged particle moving in the upper region <span class="math-container">$z \geq 0$</span>. </p> <p>Q1: What are the electric and magnetic fields of such a particle?</p> <p>Q2: More specifically, if at <span class="math-container">$t = 0$</span> the particle was stationary and the electric field was the aforementioned electrostatic field (magnetic field was <span class="math-container">$0$</span>), then given the trajectory for <span class="math-container">$t \geq 0$</span> is the solution uniquely determined (from the interface conditions and the respective Maxwell equations in each medium) ? i.e., is this initial-boundary value problem well posed.</p> <p>Q3: Can the solution be described in terms of Liénard–Wiechert potentials of certain moving image charges?</p> <p>Q4: Does this explain reflection and refraction?</p> https://physics.stackexchange.com/q/415367 0 Lienard-Wiechert fields in D-dimentions user199512 https://physics.stackexchange.com/users/0 2018-07-06T05:51:55Z 2018-07-06T05:51:55Z <p>I tried to calculate Lienard-Wiechert fields in D-dimentions and got some expression with sums of derivatives, very ugly. Does anybody have "nice" expression for it?</p> https://physics.stackexchange.com/q/376821 7 Has the Helmholtz decomposition of the $\mathbf{E}$ field from the Liénard–Wiechert potentials been worked out? Sean E. Lake https://physics.stackexchange.com/users/47360 2017-12-28T22:54:44Z 2017-12-29T04:04:59Z <p>If you look at Maxwell's equations for $\mathbf{E}(\mathbf{x},t)$ they split neatly into two categories. They are: \begin{align} \nabla\cdot\mathbf{E}(\mathbf{x},t)&amp;=\frac{\rho(\mathbf{x},t)}{\epsilon_0},\ \mathrm{and} &amp; \nabla\times\mathbf{E}(\mathbf{x},t) &amp; = -\frac{\partial \mathbf{B}(\mathbf{x},t)}{\partial t}. \end{align} Examined in light of <A HREF="https://en.wikipedia.org/wiki/Helmholtz_decomposition" rel="nofollow noreferrer">Helmholtz decomposition</A>, these equations could be read as: electric charge produces irrotational electric fields, and changing magnetic fields produce solenoidal electric fields.</p> <p>The electric field from the <A HREF="https://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential" rel="nofollow noreferrer">Liénard–Wiechert potential</A> is given by equation 14.14 from <A HREF="https://www.wiley.com/en-us/Classical+Electrodynamics%2C+3rd+Edition-p-9780471309321" rel="nofollow noreferrer">Jackson's "Classical Electrodynamics" (3rd Ed)</A> as $$\mathbf{E}_{LW}(\mathbf{x},t) = \frac{q}{4\pi\epsilon_0}\left[\frac{\mathbf{n}-\boldsymbol{\beta}}{\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R^2} \right]_{\mathrm{ret}} + \frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\times\left[(\mathbf{n}-\boldsymbol{\beta})\times \dot{\boldsymbol{\beta}}\right]}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}},$$ where Jackson's emphasis is on the separation of radiation fields (second term) from the near field (first term).</p> <p>Has someone already done the Helmholtz decomposition of $\mathbf{E}_{\mathrm{LW}}(\mathbf{x},t)$, and if so, what is it?</p> <p>Equally useful for my purposes would be the Helmholtz decomposition of <A HREF="https://en.wikipedia.org/wiki/Jefimenko%27s_equations" rel="nofollow noreferrer">Jefimenko's equation</A> for the electric field: \begin{align} \mathbf{E}_J(\mathbf{x},t) &amp;= \frac{1}{4\pi\epsilon_0} \int \left[\left(\frac{\rho(\mathbf{x}',t_r)}{\left|\mathbf{x}-\mathbf{x}'\right|^3} + \frac{1}{c\left|\mathbf{x}-\mathbf{x}'\right|^2} \frac{\partial \rho(\mathbf{x}',t_r)}{\partial t}\right)(\mathbf{x}-\mathbf{x}') - \frac{1}{|\mathbf{x}-\mathbf{x}'|c^2} \frac{\partial\mathbf{J}(\mathbf{x}',t_r)}{\partial t}\right]\operatorname{d}^3x' \\ t_r&amp;\equiv t-\frac{|\mathbf{x}-\mathbf{x}'|}{c}. \end{align}</p> <p>A big part of the reason for being interested in the HD of $\mathbf{E}$ using either of the above formulae for it is because the separation is trivial in the <A HREF="https://en.wikipedia.org/wiki/Gauge_fixing#Coulomb_gauge" rel="nofollow noreferrer">Coulomb gauge</A> because when $\mathbf{A}$ is solenoidal it doesn't contribute to the irrotational part of $\mathbf{E}$. Thus \begin{align} \mathbf{E}_{\mathrm{irrot}}(\mathbf{x},t) &amp; = \frac{1}{4\pi\epsilon_0} \int \frac{\mathbf{x} - \mathbf{x}'}{|\mathbf{x} - \mathbf{x}'|^3} \rho(\mathbf{x}',t)\operatorname{d}^3x', \ \mathrm{and} \\ \mathbf{E}_{\mathrm{sol}}(\mathbf{x},t) &amp; = \frac{1}{4\pi\epsilon_0} \nabla \times \int \left[\int_0^{|\mathbf{x}-\mathbf{x}'|/c} t' \frac{\mathbf{x}-\mathbf{x}'}{\left|\mathbf{x}-\mathbf{x}'\right|^3} \times \dot{\mathbf{J}}(\mathbf{x}',t-t') \operatorname{d}t'\right] \operatorname{d}^3x', \end{align} and I cannot see what property of $\mathbf{E}_{\mathrm{sol}}(\mathbf{x},t)$ allows it to cancel the apparent instantaneous nature of $\mathbf{E}_{\mathrm{irrot}}(\mathbf{x},t)$.</p> https://physics.stackexchange.com/q/353202 1 Liénard-Wiechert Fields for a static particle Alienbash https://physics.stackexchange.com/users/164955 2017-08-22T17:17:43Z 2017-08-22T18:17:27Z <p>Note: There was already a similar question to mine, but it did not actually answer my question: <a href="https://physics.stackexchange.com/questions/61693/retarded-time-lienard-wiechert-potential">Retarded time Lienard Wiechert potential</a></p> <p>When considering the Liénard-Wiechert fields, which are the electromagnetic fields of a moving charge carrying particle and taking the special case, that the particle is static, i.e. v=0, then we end up with the following expression for the electric field: $$\vec{E}(\vec{r},t) = \frac{e}{4\pi\varepsilon_0}\frac{\vec{r}}{r^2}\vert_{ret},$$ which is almost the expression we would get from the Coulomb potential, except for the fact that it is evaulated at $t=t_{ret}$. </p> <p>My question is: What is the physical explanation of this? Is it just the fact, that when placing an electron somewhere, even the static Coulomb Field does not instantaneously "fill space", but takes finite time for that? Evaluating $t_{ret} =t-\frac{\vert\vec{r}-\vec{r'}\vert}{c}$ would mean that $\vec{r'}$ is just the position of the particle itself, correct?</p> https://physics.stackexchange.com/q/303642 0 Confusion about Lorenz Gauge assumption in derivation of Liénard Wiechert Potentials/Fields Donkey Kong https://physics.stackexchange.com/users/141402 2017-01-08T02:40:13Z 2017-01-09T08:16:18Z <p>I have been going through Griffith's 'Introduction To Electrodynamics" 3rd Edition chapter 10 on potentials and fields and I am a little confused about the derivation of the Liénard Wiechert potentials, equations 10.39 and 10.40:</p> <p>$V(\textbf{r},t)=\frac{1}{4\pi\epsilon_0}\frac{qc}{(rc-\textbf{r}\cdot\textbf{v})}$</p> <p>$\textbf{A}(\textbf{r},t)=\frac{\textbf{v}}{c^2}V(\textbf{r},t)$</p> <p>If my understanding proves correct, these equations are derived as such:</p> <ol> <li><p>Assume static, time-independent fields which leads to our familiar Poisson equation for the two potentials, $V(\textbf{r},t)$ and $\textbf{A}(\textbf{r},t)$, $\textbf{without}$ any gauge assumptions on the potentials; no choice of gauge was made.</p></li> <li><p>Extract the integral, static form of $V(\textbf{r},t)$ and $\textbf{A}(\textbf{r},t)$ (equation 10.17).</p></li> <li><p>Argue that charge and current densities are to be evaluated at the retarded time due to the finite speed of light and show, in doing so, the Lorenz gauge (10.12) with be satisfied along with its subsequent d'Alembertian-form inhomogeneous wave equations (equations 10.16), despite making no decision in gauge.</p></li> <li><p>Argue the geometrical doppler like effect for the charge and current densities and evaluate the integral forms of $V(\textbf{r},t)$ and $\textbf{A}(\textbf{r},t)$ (equation 10.17) to reach the final results of equations 10.39 and 10.40.</p></li> </ol> <p>My confusion is that the Lorenz gauge did not seem to play a role in the above arguments, except maybe for point 3. But, the fact that point 3 was satisfied seemed like sheer coincidence and more so like the result of the solid argument that the information needs time to travel and be received; hence evaluation of the potentials at the retarded time. </p> <p>So, are we forced to remain in the Lorenz gauge if we wish to use equations 10.39 and 10.40 for $V(\textbf{r},t)$ and $\textbf{A}(\textbf{r},t)$?</p> <p>Was the fact that we must evaluate the potentials at the retarded time somehow automatically convoluted/correlated/ingrained with the Lorenz gauge?</p> https://physics.stackexchange.com/q/281115 0 Lienard -Wiechert Potential Equation in Landau Vol 2 HYW https://physics.stackexchange.com/users/130611 2016-09-19T14:34:30Z 2016-09-19T17:04:53Z <p>I am reading Laudau-Lifshitz The Classical Theory of Field (4th edition). In (63.2), it says</p> <blockquote> <p>In the system of reference in which the particle is at rest at time $t^\prime,$ the potential at the point of observation at time t is just the Coulomb potential:</p> <p>$$\varphi = \frac e{R(t^\prime)} \tag{63.2}$$</p> </blockquote> <p>This $R(t^\prime)$ is the distance between the moving point charge and the point of observation in the observation frame. I thought if the frame is changed to that in which the particle is at rest, there should be Lorentz Transformation (contraction) of $R(t^\prime).$ Why does it use the same $R(t^\prime)$ in that equation? </p> https://physics.stackexchange.com/q/220591 1 Retarded potentials and fields Athis Startis https://physics.stackexchange.com/users/59960 2015-11-26T03:12:13Z 2015-12-02T00:27:34Z <p>Why can't we use retarded times to make an expression for retarded fields instead of potentials? As far as I know it doesn't work, since the solutions produced ("retarded fields") don't satisfy Maxwell's equations, but I would like a more physical explanation, if that is possible. As a reference, one can look up Griffiths "Introduction to electrodynamics" p.423. What is it that makes potentials "special", since we know that the wave equations that potentials have to obey are derived from Maxwell's equations (that fields have to obey).</p> <p>Thanks!</p> https://physics.stackexchange.com/q/206249 2 Deriving the Lienard-Wiechert Potentials Si Chen https://physics.stackexchange.com/users/38053 2015-09-10T15:19:59Z 2016-06-06T11:29:31Z <p>Let $\mathbf{w}(t)$ be the trajectory of a moving charge. Let the observation event be $(\mathbf{r},t)$.</p> <p>The scalar potential is:</p> <p>$$\varphi = \frac{q}{4\pi\epsilon_0}\int \frac{\delta\left(\mathbf{r'} - \mathbf{w}\left(t - \frac{|\mathbf{r} - \mathbf{r'}|}{c}\right)\right)}{|\mathbf{r'}-\mathbf{r}|} \mathrm d^3\mathbf{r'}$$</p> <p>It can be shown that at most only ONE event on the trajectory of the charge produces the potential at the observation event. This is the event $(\mathbf{w}(t_r),t_r)$, where $t_r$ is such that $|\mathbf{r}-\mathbf{w}(t_r)| = c(t-t_r)$. </p> <p>Because the delta function is 0 apart from at one point, it seems to make sense that $\mathbf{w}(t_r)$ must be the point that it picks out. Is it then legitimate to write the scalar potential as:</p> <p>$$\varphi = \frac{q}{4\pi\epsilon_0|\mathbf{r} - \mathbf{w}(t_r)|}\int \delta\left(\mathbf{r'} - \mathbf{w}\left(t - \frac{|\mathbf{r} - \mathbf{r'}|}{c}\right)\right) \mathrm d^3\mathbf{r'}\;?$$</p> <p>If not, why not? And what is the best way to calculate the remaining delta function integral?</p> https://physics.stackexchange.com/q/176451 0 Poynting vector from 1st term in Lienard-Wiechert field newt https://physics.stackexchange.com/users/77937 2015-04-16T10:18:39Z 2016-03-20T15:08:03Z <p>I start with 1st (non-radiative) term from Lienard-Wiechert fields:</p> <p>$$\vec{E} = q (1-v^2) \frac{\vec{R_{t'}} - \vec{v}R_{t'}}{(R_{t'} - \vec{v}\vec{R_{t'}})^3}$$</p> <p>$$\vec{H} = - q (1-v^2) \frac{\vec{R_{t'}} \times \vec{v}}{(R_{t'} - \vec{v}\vec{R_{t'}})^3}$$</p> <p>for particle with charge $q$ and constant velocity $v$, where $R_{t'}$ is radius-vector from position of a particle to observer at retarded moment $t'$.</p> <p>The particle moves along $z$-axis with $x=0$, $y=0$. </p> <p>These vectors can be written using present time $t$:</p> <p>$$\vec{E} = q (1-v^2) \frac{\vec{R}}{(R^*)^3}$$</p> <p>$$\vec{H} = - q (1-v^2) \frac{\vec{R} \times \vec{v}}{(R^*)^3}$$</p> <p>where </p> <p>$$R^* = \sqrt{(1-v^2)(x^2 + y^2) + (z-vt)^2}$$ and everything is taken at $t$.</p> <p>So, we have cylindrical symmetry and we can parametrize:</p> <p>$$x = r \cos{\phi}$$ $$y = r \sin{\phi}$$</p> <p>then</p> <p>$$\vec{E} = q (1-v^2) \frac{1}{(R^*)^3} (r\cos{\phi} \vec{e_x} + r\sin{\phi} \vec{e_y} + (z-vt) \vec{e_z})$$</p> <p>$$\vec{H} = - q (1-v^2) \frac{1}{(R^*)^3} (vr\cos{\phi} \vec{e_x} - vr\sin{\phi} \vec{e_y} + 0 \vec{e_z})$$</p> <p>Poynting vector $\vec{S} = [\vec{E} \times \vec{H}]$ is (in cartesian coordinates):</p> <p>$$\vec{S} = \frac{q^2(1-v^2)^2}{(R^*)^6} (-vr\cos{\phi} (z-vt) \vec{e_x} - vr\sin{\phi}(z-vt) \vec{e_y} + vr^2 \vec{e_z})$$</p> <p>Finally, in cylindrical coordinates we have:</p> <p>$$\vec{S} = \frac{q^2(1-v^2)^2}{(R^*)^6} (-v(z-vt) r \vec{e_r} + vr^2 \vec{e_z})$$</p> <p>The fact that $r$-component of $\vec{S}$ has different signs in different space-points is strange for me, because for some space-regions $\vec{S}$ directs towards the particle. Of course, this term is non-radiative, but I have some doubt in this expression for $\vec{S}$.</p> <p><strong>Question:</strong> If my expression is right, how to explain it?</p> https://physics.stackexchange.com/q/176271 10 From Liénard-Wiechert to Feynman potential expression Jodocus https://physics.stackexchange.com/users/77836 2015-04-15T13:34:24Z 2017-12-19T12:52:34Z <p>When studying the potential of an uniformly moving charge in vacuum, Feynman proposes to apply a Lorentz transformation on the Coulomb potential, which reads in the rest frame</p> <p>$\phi'(\mathbf r',t') = \frac{q}{4\pi\epsilon_0} \frac{1}{r'}$,</p> <p>where $|\mathbf r'| = r'$. In a frame with constant velocity $\mathbf v$ along the x-axis, he obtains the following expression: $$\phi(\mathbf r, t) = \frac{\gamma q}{4\pi\epsilon_0} \dfrac{1}{\sqrt{(\gamma(x-vt))^2+y^2+z^2}} \tag 1$$</p> <p>by transforming $\phi = \gamma\left(\phi'+\dfrac{A'_xv}{c^2}\right)$, where $\gamma = \dfrac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ and the vector potential $\mathbf A'$ vanishes within the rest frame. Another Lorentz transformation of the time and space coordinates $(\mathbf r', t') \rightarrow (\mathbf r,t)$ yields (1). I suspect that (1) describes the potential at a given point for the <i>instantaneous</i> time t. What I am wondering is how this formula is connected to the expression of Liénard and Wiechert, namely $$\phi(\mathbf r, t)=\dfrac{q}{4\pi\epsilon_0}\dfrac{1}{|\mathbf r - \mathbf x(t_{ret})| - \frac{1}{c}\mathbf v(t_{ret})\cdot(\mathbf r - \mathbf x(t_{ret}))} \tag 2,$$</p> <p>where $\mathbf x(t_{ret})$ describes the position of the charge and $\mathbf v(t_{ret}) = \frac{d}{dt}\mathbf x(t)\bigg|_{t=t_{ret}}$ its velocity at the retarded time $t_{ret}(\mathbf r,t) = t-\frac{|\mathbf r - \mathbf x(t_{ret})|}{c}$, respectively.</p> <p>In the case of uniform motion, we have $\mathbf x(t) = (vt,0,0)^\intercal$. </p> <p><b>How do I get now from (2) to (1)?</b></p> <p>My idea is to actually calculate an explicit expression for the retarded time and plug it into (2), which should yield (1) if I understand it correctly. By asserting that $c^2(t-t_{ret})^2 = (x-vt_{ret})^2+y^2+z^2$, $t_{ret}$ can be found in terms of solving the quadratic equation, leading to the solutions</p> <p>$t_{ret}^\pm = \gamma\left(\gamma(t-\frac{vx}{c^2})\pm\sqrt{\gamma^2(t-\frac{vx}{c^2})^2-t^2+\frac{r^2}{c^2}}\right) = \gamma\left(\gamma t'\pm\sqrt{\gamma^2t'^2-\tau^2}\right)$ where $t'$ is the Lorentz transformation of $t$ and $\tau = \frac{t}{\gamma}$ looks like some proper time. Plugging this into (2) looks nothing like (1), what am I missing?</p> https://physics.stackexchange.com/q/162420 1 Is the Liénard-Wiechert electric field conservative? Sofia https://physics.stackexchange.com/users/63535 2015-01-30T15:39:36Z 2015-03-06T11:51:50Z <p>I know that an accelerated charge should emit an e.m. field and loose energy. Therefore, the Liénard-Wiechert (L.W.) electric field of an accelerated charge should be non-conservative. </p> <p>But I checked first what happens when the charge is not accelerated, i.e. moves with a constant velocity. I expected to find a conservative field as in the case when the charge is at rest. A charge moving with constant velocity doesn't radiate. But it seems that this is not what happened.</p> <p>Given the scalar potential $\phi$ and vector potential $\vec A$, the electric field is</p> <p>$$\ (1) \ \vec E = - \nabla \phi - \frac {∂ \vec A}{∂t},$$</p> <p>where</p> <p>$$(2) \ \phi (r, t) = \frac {1}{4 \pi \epsilon _0} \left( \frac {q}{(1 - \vec n \vec \beta _s)|\vec r - \vec r_s|} \right)_{t_r},$$</p> <p>$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3) \ \vec A = \frac {\mu _0 c}{4 \pi} \left( \frac {q \ \vec {\beta} _s}{(1 - \vec n \vec \beta _s)|\vec r - \vec r_s|} \right)_{t_r} = \frac {\vec \beta _s (t_r)}{c} \phi (r, t).$$</p> <p>see <a href="http://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential#Definition_of_Li.C3.A9nard.E2.80.93Wiechert_potentials" rel="nofollow">the article</a>.</p> <p>I assume that for constant velocity of the charge, $t_r = t$. A field that obeys </p> <p>$$\ (4) \ \vec F(\vec r) = \nabla V(r)$$</p> <p>is conservative, i.e. </p> <p>$$\ (5) \ \int_{\vec {r_1}}^{\vec {r_2}} \vec F \ d \vec {\ell} = V(\vec {r_2}) - V(\vec {r_1}).$$</p> <p>So, I expected that for the constant velocity the formula (1) will turn into (4), i.e. that I would get that $\vec A$ does not depend on time. But this doesn't happen. Why? A charge in movement with constant velocity shouldn't radiate, its electric field should be conservative.</p> <p>Do I make a confusion, do I make a mistake?</p> https://physics.stackexchange.com/q/162378 2 Falling charged objects: energy conservation paradox? John Eastmond https://physics.stackexchange.com/users/22307 2015-01-30T11:51:24Z 2015-01-30T14:12:18Z <p>Imagine that we start with two oppositely charged objects on the ground, separated by a distance $d$, with charges $+q$, $-q$ and masses $m$. </p> <p>We raise them both up to a height $h$. </p> <p>In doing so we only move them at constant velocity so that neither charge produces a radiative electromagnetic field in the vicinity of the other.</p> <p>Also assume that the objects move simultaneously on frictionless poles so that their Coulomb attraction is always horizontal and thus plays no part in the experiment.</p> <p>The energy that we have put into the system is simply:</p> <p>$$E_{grav} = 2mgh$$</p> <p>Now we simultaneously release the two objects and they fall back to the ground.</p> <p>The force of gravity or weight, $mg$, acts on each object over the distance $h$ so that we end up with the final kinetic energy of the objects, <em>assuming the effect of gravity alone</em>, given by:</p> <p>$$KE_{grav} = 2mgh$$</p> <p>Therefore we seem to have an energy balance as expected.</p> <p>But this is not the end of the story. As the charged objects accelerate to the ground they each produce a <a href="http://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential#Corresponding_values_of_electric_and_magnetic_fields" rel="nofollow">radiative electric field</a> at a distance $d$ given by:</p> <p>$$\mathbf{E_{rad}} = -\frac{\pm q}{4 \pi \epsilon_0c^2d}\mathbf{g}$$</p> <p>Therefore there is an extra downward force on each object given by:</p> <p>$$\mathbf{F_{em}} =\frac{q^2}{4 \pi \epsilon_0c^2d}\mathbf{g}$$</p> <p>As this force acts on each object over a distance $h$ then the <em>extra</em> kinetic energy of the objects when they reach the ground due to the mutual effect of their radiative electric fields is:</p> <p>$$KE_{em} = \frac{2 q^2 g h}{4 \pi \epsilon_0c^2d}$$</p> <p>We seem to be getting more energy out than we have put into the system.</p> <p>What's wrong with this calculation?</p> <p><strong>Postscript</strong></p> <p>Mark Mitchison has pointed out that I need to find the acceleration for the case where both the electromagnetic force and the gravitational force are acting. I can't assume that it is just the gravitational acceleration $\mathbf{g}$.</p> <p>Let me try to work out the equation of motion for one of the objects, where both objects have a downward acceleration $a$.</p> <p>We have:</p> <p>$$F = m a$$</p> <p>The total force $F$ on one of the objects is the sum of the gravitational and electromagnetic components:</p> <p>$$F = m g + \frac{q^2a}{4 \pi \epsilon_0 c^2 d}$$</p> <p>If we define:</p> <p>$$m_{e} = \frac{q^2}{4 \pi \epsilon_0 c^2 d}$$</p> <p>then the total force on an object is given by:</p> <p>$$F = m g + m_{e} a$$</p> <p>The equation of motion is then:</p> <p>$$m a = m g + m_{e} a$$</p> <p>Thus the acceleration $a$ is given by:</p> <p>$$a = \frac{mg}{m - m_{e}}$$</p> <p>The total energy gained by an object under the influence of both the gravitational force and the electromagnetic force as it falls a distance $h$ is then:</p> <p>$$E = F h$$</p> <p>$$E = \left(mg + m_{e} \cdot \frac{mg}{m - m_{e}}\right)h$$</p> <p>$$E = mgh \cdot \frac{m}{m-m_{e}}$$</p> <p>This is <em>more</em> energy than the $mgh$ that we put into the object in the first place.</p> <p>So the paradox still stands.</p> <p>P.S. If $m_e$ is small then we have:</p> <p>$$E \approx mgh + m_{e}gh$$</p> <p>which is consistent with my original calculation above where I took the acceleration to be simply $g$.</p> https://physics.stackexchange.com/q/159766 1 Energy conservation in electrodynamic system John Eastmond https://physics.stackexchange.com/users/22307 2015-01-16T16:04:55Z 2015-11-21T23:48:08Z <p>Consider two charged particles initially at rest in the configuration below.</p> <p><img src="https://i.stack.imgur.com/OaY50.jpg" alt="diag"></p> <p>Let us assume the following:</p> <ol> <li>Starting at time $t=0$, we apply a constant force $f$ to the the bottom particle so that it has a constant acceleration $a=f/m$.</li> <li>The top particle has a large mass $M$.</li> <li>The distance $r$ is large enough so that the Coulomb repulsion between the particles, which is inversely proportional to $r^2$, is negligible.</li> </ol> <p>Under these conditions the <a href="http://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential" rel="nofollow noreferrer">Lienard-Wiechert</a> retarded radiative electric field due to the bottom particle produces a force $F$ on the top particle given by:</p> <p>$$F(t)=\frac{qQa(t-r/c)}{4\pi\epsilon_0c^2r}.$$</p> <p>For simplicity we assume that the mass $M$ of the top particle is so large that its acceleration due to force $F$ is negligible. Thus it does not produce a significant radiative electric field back at the position of the bottom particle.</p> <p>Now let us calculate the energy $E_{in}$ that we supply to the system.</p> <p>Let us assume that we apply a force $f$ to the bottom particle for a time interval $\delta t$.</p> <p>During time interval $\delta t$ the bottom particle travels a distance $d$ so that the energy supplied $E_{in}$ is given by:</p> <p>$$E_{in} = f \times d$$</p> <p>The bottom particle has a constant acceleration $a$ so the distance it travels in time interval $\delta t$ is given by:</p> <p>$$d = \frac{1}{2}a \delta t^2$$</p> <p>Using the expression for the acceleration of the bottom particle, $a=f/m$, we find from the above two relations that the energy supplied to the system during time interval $\delta t$ is given by:</p> <p>$$E_{in} = \frac{f^2\delta t^2}{2m}$$</p> <p>Where has this energy gone?</p> <p>The kinetic energy, $KE$, of the bottom particle after the time interval $\delta t$ is:</p> <p>$$KE = \frac{1}{2} m v^2$$</p> <p>The velocity of the bottom particle after a time interval $\delta t$ is given by:</p> <p>$$v = a \delta t$$</p> <p>Since the acceleration $a=f/m$ the above two equations imply that the kinetic energy of the bottom particle is given by:</p> <p>$$KE = \frac{f^2 \delta t^2}{2 m^2}$$</p> <p>Therefore, as expected, all the energy $E_{in}$ that we supplied during time $\delta t$ has gone into the kinetic energy $KE$ of the bottom particle.</p> <p>But, as stated above, since the bottom particle is accelerating, after a slight time delay $t=r/c$, there is a force $F$ acting on the top particle. During the time interval $\delta t$ an energy $E_{top}$ is supplied to the top particle given by:</p> <p>$$E_{top} = \frac{F^2\delta t^2}{2M}$$</p> <p>My question is where has this energy come from given that all the energy we supplied, $E_{in}$, is fully accounted for in the kinetic energy of the bottom particle alone?</p> https://physics.stackexchange.com/q/158606 0 Lienard-Wiechert fields for low velocity source John Eastmond https://physics.stackexchange.com/users/22307 2015-01-10T13:16:14Z 2015-01-10T14:02:35Z <p>I would like to use the <a href="http://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential" rel="nofollow">Lienard-Wiechert</a> <strong>E</strong> and <strong>B</strong> field expressions for a slowly moving charge where $\beta = v/c &lt;&lt; 1$.</p> <p>Is there an accepted approximate form to use?</p> <p>Can one just set $\gamma = 1$ and the retardation factors $1 - \mathbf{n} . \mathbf{\beta}=1$ in the standard expressions for <strong>E</strong> and <strong>B</strong>?</p> https://physics.stackexchange.com/q/138583 6 Feynman's proof for Liénard-Wiechert's potential of a moving charge guillefix https://physics.stackexchange.com/users/29621 2014-10-03T19:41:00Z 2019-04-06T16:55:17Z <p>Feynman's <a href="http://www.feynmanlectures.caltech.edu/II_21.html#Ch21-S5" rel="nofollow noreferrer">proof</a> utilizes a geometrical and fundamental integration argument. I like it, except this bit:</p> <p><img src="https://i.stack.imgur.com/dTzVB.png" alt="FLP"></p> <p>What makes me unconfortable somehow is that in (c) we are counting in some of the charge we counted at (b). It seems to me that it is this extra counting which makes the potential to be larger than expected, and I am uncomfortable with it. To see why, consider the following situation with discrete charges:</p> <p><img src="https://i.stack.imgur.com/nlFrF.png" alt="enter image description here"></p> <p>Here, the yellow line represents the light cone (the observer being, of course at its apex), and the blue dots, the places where the observer "sees" each of the constituent charges. However, it is clear that if the charge cloud was small enough, or if we were far enough, the potential would be just the potential for a point charge of charge equal to the total charge of the cloud, as no charge is "overcounted" (something which is also due to the cloud's speed being less than c).</p> <p>This is in disagreement with what Feynman derives and with the Liénard-Wiechert's potential of a moving charge. So why is my argument wrong? Is the continuity of the cloud, somehow crucial for the proof? If so why?</p> https://physics.stackexchange.com/q/127391 3 How can I calculate the divergence of the lienard wiechert eletric field? user55782 https://physics.stackexchange.com/users/55782 2014-07-20T17:36:43Z 2014-07-20T17:36:43Z <p>I was reading Introduction to Eletrodynamics by Griffiths and I see that´s nothing there about to prove the gauss law for charges in arbitrary motion and non constant velocity . So I try to calculate the divergence of the lienard wiechert fields to show that the divergence of the lienard wiechert eletric field is equal to the charge density but I am getting stuck in the calculations ! Somebody know how to do these calculations ???</p> https://physics.stackexchange.com/q/93390 1 Field of moving charge / Lorentz; Liénard-Wiechert simon.reiger https://physics.stackexchange.com/users/37261 2014-01-12T19:39:11Z 2017-11-10T08:38:57Z <p>First question here. I'm really confused at the moment. An electron moves at constant velocity, no acceleration</p> <p>Wikipedia says here <a href="http://en.wikipedia.org/wiki/Biot%E2%80%93Savart_law#Point_charge_at_constant_velocity" rel="nofollow noreferrer">Lorentz</a>: $$\mathbf E=\frac{q}{4\pi\epsilon_0}\frac{1-v^2/c^2}{1-v^2\sin^\theta/c^2}\frac{\hat{\mathbf r}}{r^2},$$ which yields something like this: </p> <p><img src="https://i.stack.imgur.com/5Ut38.gif" alt="enter image description here"></p> <hr> <p>Whereas here, Wikipedia says <a href="http://en.wikipedia.org/wiki/Relativistic_electromagnetism#The_field_of_a_moving_point_charge" rel="nofollow noreferrer">this</a> and <a href="http://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential#Corresponding_values_of_electric_and_magnetic_fields" rel="nofollow noreferrer">this</a>, $$\frac{E'_y}{E'_x} = \frac{E_y}{E_x\sqrt{1-v^2/c^2}} = \frac{y'}{x'},$$ which yields something like this:</p> <p><img src="https://i.stack.imgur.com/yi9Zl.gif" alt="enter image description here"></p> <p>Which one is correct? If you could explain me exactly the reason why one of them is correct, I give you a big imaginary hug.</p> <p>Last question: In none of those fields is there any radiated energy, since there is no acceleration, correct?</p> https://physics.stackexchange.com/q/81361 2 Surely force on shell can't be balanced by field momentum? John Eastmond https://physics.stackexchange.com/users/22307 2013-10-19T18:49:01Z 2018-07-13T08:10:52Z <p>Imagine a particle with charge $q$ at rest at the origin.</p> <p>It is surrounded by a concentric spherical insulating shell, also at rest, with charge $Q$ and radius $R$.</p> <p>At time $t=0$ I apply a constant horizontal acceleration $\mathbf{a}$ to the particle.</p> <p>The electromagnetic field spreads out radially in all directions from the charge at the speed of light (Appendix 1).</p> <p>The momentum density in the field is directed radially and also spreads out from the charge at the speed of light (Appendix 2).</p> <p>As the momentum density at opposite field points cancel the total momentum in the field is always zero (Appendix 2).</p> <p>At time $t=R/c$ the field reaches the charged shell and imparts a horizontal force on it (Appendix 3).</p> <p><em>What balances this force if the rate of change of horizontal momentum in the field is always zero?</em></p> <p>I think it has to be balanced by an opposite force on the particle that develops due to an advanced electromagnetic interaction from the spherical shell at time $t$ backwards in time to the particle at time $t_r$.</p> <p><strong>Appendix 1: EM field of an accelerated charge with velocity zero</strong> </p> <p>Specializing the <a href="http://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential" rel="nofollow">Lienard-Wiechert fields</a> for an accelerating charge with zero velocity we find that the electric field is given by:</p> <p>$$\mathbf{E}(\mathbf{r},t)=\frac{q}{4\pi\epsilon_0}\left(\frac{\mathbf{\hat r}}{r^2}+\frac{\mathbf{\hat r}\times(\mathbf{\hat r}\times \mathbf{a})}{c^2r}\right)_{t_r}$$</p> <p>and the magnetic field is given by:</p> <p>$$\mathbf{B}(\mathbf{r},t)=\frac{\mathbf{\hat r}(t_r)}{c}\times\mathbf{E}(\mathbf{r},t)$$</p> <p>where $\mathbf{\hat r}$ is the unit vector in the direction of the field point and the retarded time $t_r$ at the charge $q$ is given by:</p> <p>$$t_r=t-\frac{r(t_r)}{c}.$$</p> <p><strong>Appendix 2: Zero total momentum in EM field from an accelerated charge</strong></p> <p>The momentum density $\mathbf{g}(\mathbf{r},t)$ in the electromagnetic field is given by:</p> <p>$$\mathbf{g}(\mathbf{r},t)=\epsilon_0\mathbf{E}(\mathbf{r},t)\times\mathbf{B}(\mathbf{r},t).$$</p> <p>The momentum density is directed radially so that:</p> <p>$$\mathbf{g}(\mathbf{r},t)=\frac{\epsilon_0E^2}{c}\mathbf{\hat r}$$</p> <p>where the magnitude of the electric field $E$ is given by</p> <p>$$E = -\frac{q a \sin \theta}{4\pi\epsilon_0c^2r}$$</p> <p>and $\theta$ is the angle between acceleration $\mathbf{a}$ and radial direction $\mathbf{\hat r}$.</p> <p>As $\sin(\pi-\theta)=\sin(\theta)$ then we have:</p> <p>$$\mathbf{g}(-\mathbf{r},t)=-\mathbf{g}(\mathbf{r},t)$$</p> <p>so that the momentum density at opposite field points cancel.</p> <p>Thus the total momentum in the fields is always zero.</p> <p><strong>Appendix 3: Total force on charged spherical shell from accelerated charge</strong></p> <p>The electric field at the spherical shell consists of a static radial part and an acceleration part. The total force from the static radial part will cancel out in opposite pairs so that we only have to worry about the acceleration part.</p> <p>The total horizontal force on the spherical shell with charge $Q$ and radius $R$ is given by:</p> <p>$$F = \int_{sphere} E\sin \theta\ \sigma\ dA$$</p> <p>where</p> <p>$$E = -\frac{q a \sin \theta}{4\pi\epsilon_0c^2R}$$</p> <p>$$\sigma=\frac{Q}{4\pi R^2}$$</p> <p>$$dA = 2 \pi R^2 \sin \theta\ d\theta$$</p> <p>Performing this integration over the sphere using:</p> <p>$$\int_0^\pi \sin^3 \theta\ d\theta=\frac{4}{3}$$</p> <p>We find that the total horizontal force is given by:</p> <p>$$F = -\frac{2}{3}\frac{qQa}{4\pi\epsilon_0c^2R}.$$</p> https://physics.stackexchange.com/q/80596 1 Accelerated charge inside sphere (again!) John Eastmond https://physics.stackexchange.com/users/22307 2013-10-13T13:23:35Z 2013-10-30T20:18:08Z <p>Sorry to go on about this scenario again but I think something is going on here.</p> <p>Imagine a stationary charge $q$, with mass $m$, at the center of a stationary hollow spherical dielectric shell with radius $R$, mass $M$ and total charge $-Q$.</p> <p>I apply a force $\mathbf{F}$ to charge $q$ so that it accelerates:</p> <p>$$\mathbf{F} = m \mathbf{a}$$</p> <p>The accelerating charge $q$ produces a retarded (forwards in time) radiation electric field at the sphere. When integrated over the sphere this field leads to a total force $\mathbf{f}$ on the sphere given by:</p> <p>$$\mathbf{f} = \frac{2}{3} \frac{qQ}{4\pi\epsilon_0c^2R}\mathbf{a}.$$</p> <p>So I apply an external force $\mathbf{F}$ to the system (charge + sphere) but a total force $\mathbf{F}+\mathbf{f}$ operates on the system.</p> <p>Isn't there an inconsistency here?</p> <p>As the acceleration of charge $q$ is constant there is no radiation reaction force reacting back on it from its electromagnetic field - so that's not the answer.</p> <p>Instead maybe there is a reaction force back from the charged shell, $-\mathbf{f}$, to the charge $q$ so that the equation of motion for the charge is given by:</p> <p>$$\mathbf{F} - \mathbf{f} = m\mathbf{a}\ \ \ \ \ \ \ \ \ \ \ (1)$$</p> <p>This reaction force might be mediated by an advanced electromagnetic interaction going backwards in time from the shell to the charge so that it acts at the moment the charge is accelerated.</p> <p>Now the total force acting on the system is the same as the force supplied:</p> <p>$$\mathbf{F} - \mathbf{f} + \mathbf{f} = \mathbf{F}.$$</p> <p>If one rearranges Equation (1) one gets:</p> <p>$$\mathbf{F} = (m + \frac{2}{3} \frac{qQ}{4\pi\epsilon_0c^2R}) \mathbf{a}$$</p> <p>Thus the effective mass of the charge $q$ has increased.</p> https://physics.stackexchange.com/q/78097 6 Reaction-at-a-distance: Do charged plates immediately repel each other? John Eastmond https://physics.stackexchange.com/users/22307 2013-09-21T16:37:59Z 2015-09-25T01:33:49Z <p><img src="https://i.stack.imgur.com/YMVBx.jpg" alt="Retarded"></p> <p>Imagine that we have a pair of parallel plates, $A$ and $B$, separated by some distance as in Fig. $1$ above.</p> <p>At time $t_1$ we simultaneously charge both the plates. This could be done by previously sending a light signal to a charging apparatus at each plate from a source located at the mid-point between them.</p> <p>According to standard electromagnetic theory a retarded electric influence travels at the speed of light from $A$ to $B$ and vice-versa.</p> <p>At time $t_2$ the electric influence from $A$ produces a force at $B$ and vice-versa. </p> <p>There are two points that I would like to raise about this description:</p> <ol> <li>There are no reaction forces. It is as if a pair of boxers each punched the other on the nose simultaneously but neither felt a reaction back on their boxing glove.</li> <li>Once the electric influences have left the charged plates at time $t_1$, and before they have produced forces on the opposite plates at time $t_2$, they must exist "somewhere". That somewhere is the electromagnetic field. </li> </ol> <p>Now consider the picture described in Fig $2$ below which includes both retarded and advanced interactions.</p> <p><img src="https://i.stack.imgur.com/zBS6w.jpg" alt="Retarded and Advanced"></p> <p>Again at time $t_1$ we simultaneously charge both the plates.</p> <p>Now as well as a retarded electric influence that travels at the speed of light from $A$ to $B$ we also have an advanced electric influence which travels backwards in time from $B$ to $A$. Thus the force at plate $B$ at time $t_2$ is balanced by an equal and opposite force on plate $A$ at time $t_1$ (and vice-versa). </p> <p>Now as soon as we charge the plates up we measure electric forces on them.</p> <p>At first glance it seems that we have "action at a distance" but in fact we only have "reaction at a distance". In terms of spacetime, each plate at time $t_1$ is linked with the opposite plate at time $t_2$ in a manner that is consistent with the principle of locality provided we include advanced interactions.</p> <p>As there is no delay between charging the plates and measuring forces then there is no time interval during which the influences could be said to be in transit in the form of an electromagnetic field.</p> <p>Thus in this picture we have:</p> <ol> <li>Reaction forces</li> <li>No electromagnetic field</li> </ol> <p>Could one perform such an experiment to see if charged plates immediately repel each other?</p> https://physics.stackexchange.com/q/64371 2 Mathematical equivalence between Liénard-Wiechert potential and 4-potential in Rindler coordinates Ana S. H. https://physics.stackexchange.com/users/15672 2013-05-13T00:32:22Z 2013-05-13T00:58:33Z <p>I'm studying the problem of the radiation of an uniformly accelerated point charge:</p> <p>$$x^{\mu}(\lambda)\to(g^{-1}\sinh g\lambda,0,0,g^{-1}\cosh g\lambda)$$</p> <p>I found that when a point charge is moving along the $z$ axis with a constant acceleration $g$, the components of the 4-potential can be found using Rindler coordinates:</p> <p>$$z=Z\cosh g\tau, \qquad t=Z\sinh g\tau \qquad \mathrm{I}$$ $$z=Z\sinh g\tau, \qquad t=Z\cosh g\tau \qquad \mathrm{II}$$ $$z=-Z\cosh g\tau, \qquad t=-Z\sinh g\tau \qquad \mathrm{III}$$ $$z=-Z\sinh g\tau, \qquad t=-Z\cosh g\tau \qquad \mathrm{IV}$$</p> <p>with the metric</p> <p>$$ds^{2}=\epsilon(-g^{2}Z{}^{2}\, d\tau^{2}+dZ^{2})+dx^{2}+dy^{2}$$</p> <p>where $\epsilon=+1$ in regions I and III, and $\epsilon=-1$ in regions II and IV. And the numbers indicate the space-time region: <img src="https://i.stack.imgur.com/LBqaG.png" alt="enter image description here"></p> <p>Now, I understand that the potentials obtained with the Rindler coordinates must be equivalent to the Liénard-Wiechert potential, because the change of coordinates (Minkowski->Rindler) is equivalent to change from an inertial frame of reference to an accelerated one. The problem is, the article </p> <blockquote> <p>Radiation from a uniformly accelerated charge. D G Boulware. <a href="http://dx.doi.org/10.1016/0003-4916(80)90360-7" rel="nofollow noreferrer"><em>Ann. Phys.</em> <strong>124</strong> no.1 (1980), pp. 169–188</a>. <a href="http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.205.5420" rel="nofollow noreferrer">Available on CiteSeerX</a></p> </blockquote> <p>finds that the components of the 4-potential (Rindler coordinates) are:</p> <p>$$A_{\tau}=-\frac{eg}{4\pi}\frac{\epsilon Z^{2}+\rho^{2}+g^{-2}}{\left[\left(\epsilon Z^{2}+\rho^{2}+g^{-2}\right)^{2}-4 \epsilon Z^{2}g^{-2}\right]^{1/2}}$$</p> <p>where $\rho^{2}:=x^{2}+y^{2}$. $A_x=A_y=0$, and </p> <p>$$A_{Z}=-\frac{e}{4\pi Z}$$</p> <p>In regions I and II.</p> <p>My <strong>question</strong> is: How can I conclude that this components are actually equivalent to the components of the Liénard-Wiechert potential?</p> https://physics.stackexchange.com/q/61693 2 Retarded time Lienard Wiechert potential shilov https://physics.stackexchange.com/users/21956 2013-04-20T11:48:32Z 2013-07-06T21:49:18Z <p>In a potential which needs to be evaluated at the retarded time, is this the time which represents the actual time the "physics" occurred? So $t_{\text{ret}}=t-\frac{r}{c}$, not just because it may be that you are receiving a signal at light speed but because "causality" spreads out at the maximum speed, $c$, is this correct?</p> <p>The Lienard-Wiechert 4-potential for some point charge ($q$): $A^\mu=\frac{q u^\mu}{4\pi \epsilon_0 u^\nu r^\nu}$ where $r^\nu$ represents the 4-vector for the distance from the observer. In the rest frame of the charge $A^i$ for $i=1,2,3$ is clearly zero but from what has been said about the retarded time we have that $A^0=\frac{q}{4\pi\epsilon_0c(t-r/c)}$.</p> <p>Obviously I would like to get $A^0=-\frac{q}{4\pi\epsilon_0 r}$, so where is the misunderstanding of retarded time and instantaneous time? Unless we would like the time since the signal was emitted which is $r/c$? Or if $t$ itself is already $t'-r/c$ and we need to return to the instantaneous time $t$, when the signal was emitted.</p> https://physics.stackexchange.com/q/59730 3 Electromagnetic inertia due to advanced radiation? John Eastmond https://physics.stackexchange.com/users/22307 2013-04-01T20:46:57Z 2016-01-13T21:08:04Z <p>The scalar potential $\phi$ and vector potential $A$ at a distance $r$ from a charge $q$ are given approximately by</p> <p>$$\phi = \frac{q}{r}$$</p> <p>$$\mathbf{A} = \frac{q\mathbf v}{r}$$</p> <p>where the constants have been suppressed.</p> <p>The corresponding electric and magnetic fields are given by</p> <p>$$\mathbf{E} = -\nabla \phi - \frac{\partial \mathbf A}{\partial t}$$</p> <p>$$\mathbf{B} = \nabla \times \mathbf A$$</p> <p>Now the gradient and curl terms fall off as $1/r^2$ so that at large distances from the charge $q$ we only have an electric field $E$ given approximately by</p> <p>$$\mathbf{E} = - \frac{\partial \mathbf A}{\partial t}\ \ \ \ \ \ \ \ \ (1)$$</p> <p>where</p> <p>$$\mathbf{A} = \frac{q\mathbf v}{r}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$$</p> <p>Thus at large distances an accelerating charge produces an electric field that is proportional to its acceleration. This is the standard (retarded) radiation field of an accelerated charge.</p> <p>But perhaps this is not the only solution to equation (1). Imagine that the charge is fixed with respect to an inertial frame and that the observer is moving with velocity $-v$.</p> <p>In the rest frame of the observer the charge is moving with velocity $v$. This implies that in the observer's rest frame there is an $A$-field.</p> <p>Now if the observer changes his velocity then in his comoving frame he experiences a changing $A$-field.</p> <p>My proposal is that Equation (1) still applies in this situation so that this "apparently" changing $A$-field induces an electric field in the observer's accelerating frame. This field causes any charges that are co-moving with the accelerating observer to feel a kind of electromagnetic inertial force.</p> <p>I realise that this is non-standard as Maxwell's equations are usually only applied in an inertial frame. My hypothesis is that the retarded solutions apply to an inertial frame whereas the previously unused advanced solutions apply to accelerated frames.</p> https://physics.stackexchange.com/q/57797 1 Advanced Heaviside-Feynman formula implies electromagnetic inertia? John Eastmond https://physics.stackexchange.com/users/22307 2013-03-23T15:59:09Z 2013-04-01T11:00:49Z <p>The Heaviside-Feynman formula (see Feynman Lectures vol I Ch.28, vol II Ch. 21) gives the electric and magnetic fields measured at an observation point $P$ due to an arbitrarily moving charge $q$</p> <p>$$\mathbf{E} = -\frac{q}{4 \pi \epsilon_0} \left\{ \left[ \frac{\mathbf{\hat r}}{r^2} \right]_{ret} + \frac{[r]_{ret}}{c} \frac{\partial}{\partial t} \left[ \frac{\mathbf{\hat r}}{r^2} \right]_{ret} + \frac{1}{c^2} \frac{\partial^2}{\partial t^2} [\mathbf{\hat r}]_{ret} \right\}$$ $$\mathbf{B} = -\frac{1}{c} [\mathbf{\hat r}]_{ret} \times \mathbf{E}$$</p> <p>where $[\mathbf{\hat r}]_{ret}$ and $[r]_{ret}$ is the unit vector and distance <em>from</em> the observation point $P$ at time $t$ <em>to</em> the retarded position of charge $q$ at time $t - [r]_{ret}/c$ (hence the minus signs). </p> <p>This formula is remarkable in that it is completely relational. It does not refer to any external reference frame. The fields at point $P$ only depend on the vector $\mathbf{r}$ from point $P$ to the retarded position of charge $q$ and its first and second order rates of change with respect to local time $t$.</p> <p>Now one can imagine two ways in which the vector $\mathbf{r}$ from point $P$ to charge $q$ can change. One could move charge $q$ and keep point $P$ fixed or one could move point $P$ and keep charge $q$ fixed. The above formula is valid for the former situation giving the fields at a fixed point $P$ due to a moving charge $q$. This is the conventional retarded solution of Maxwell's equations.</p> <p>But what about the latter situation in which the observation point $P$ moves and the charge $q$ is fixed. The relational nature of the formula implies to me that it should still apply in this situation. Perhaps this is the situation in which the <em>advanced</em> Heaviside-Feynman formula is valid given by</p> <p>$$\mathbf{E} = -\frac{q}{4 \pi \epsilon_0} \left\{ \left[ \frac{\mathbf{\hat r}}{r^2} \right]_{adv} - \frac{[r]_{adv}}{c} \frac{\partial}{\partial t} \left[ \frac{\mathbf{\hat r}}{r^2} \right]_{adv} + \frac{1}{c^2} \frac{\partial^2}{\partial t^2} [\mathbf{\hat r}]_{adv} \right\}$$ $$\mathbf{B} = -\frac{1}{c} [\mathbf{\hat r}]_{adv} \times \mathbf{E}$$</p> <p>where $[\mathbf{\hat r}]_{adv}$ and $[r]_{adv}$ is the unit vector and distance from the observation point $P$ at time $t$ to the advanced position of charge $q$ at time $t + [r]_{adv}/c$. The advanced Heaviside-Feynman formula is the time-reverse of the conventional retarded formula.</p> <p>This interpretation of the advanced formula, if valid, implies that an accelerated observer inside a fixed charged insulating spherical shell would measure an electric field whose strength is proportional to the acceleration. This implies that an accelerated charge feels a kind of electromagnetic inertial force and thus has an electromagnetic inertia due to the presence of the charged spherical shell. </p> <p>For example imagine an electron with charge $-e$ inside such a fixed charged insulating spherical shell at potential $+V$ volts. Using the above advanced Heaviside-Feynman formula one can calculate that this electromagnetic inertia $m_{em}$ is given by $$m_{em} = \frac{2}{3} \frac{eV}{c^2}$$ For a shell charged to a high voltage $V=1000000$ volts this electromagnetic inertia would be of similar order to the electron's native mass and should therefore be easily observable. It would probably be important that the spherical shell is a charged insulator rather than a conductor because it is assumed that the charges inside the shell remain fixed.</p> <p>Finally, there is a close analogy between Maxwell's equations and Einstein's field equations in the limit of weak gravitational fields. There is a clear gravitational analogue of the advanced Heaviside-Feynman formula. Thus one would expect that a mass accelerated inside a fixed spherical shell of mass should experience a gravitational inertial force in a manner analogous to the above electrical example (just substitute mass $m$, gravitational potential $\phi$ for charge $e$, electrical potential $V$ in above expression). </p> <p>Perhaps this is the origin of inertia as hypothesised in Mach's Principle?</p> https://physics.stackexchange.com/q/44971 3 What is the physical meaning of retarded time? Ana S. H. https://physics.stackexchange.com/users/15672 2012-11-24T03:04:53Z 2016-12-04T09:43:01Z <p>Consider this figure <img src="https://i.stack.imgur.com/epLm6.jpg" alt="A charge $e$ moves with certain velocity $\mathbf v_e$"></p> <p>Now, when I measure a field produced by the charge $e$ at the point $\mathbf r$, at the time $t=t_1$, it means that the charge sent the signal field at the time $t=t_r$, where $t_1$ and $t_r$ are related by $$t_{r}=t_1-\frac{||\mathbf{r}-\mathbf{r}_{e}(t)|{}_{t=t_{r}}||}{c}$$ Now, my question is, how is it possible that we can take $t_r$ like a time variable? I mean, when we want to measure the velocity of the charge $\mathbf v_e$, I must derive $\mathbf r_e$ respect to $t_r$: $$\mathbf{v}_{e}=\frac{d\mathbf{r}_{e}}{dt_{r}}$$ but why? I mean, why not to derive respect to $t$? So, what is the physical meaning of $t_r$? Or, in other words, how can we interpret the time $t_r$? Are there actually two time-axes?</p> https://physics.stackexchange.com/q/33225 3 why is advanced radiation absent? lurscher https://physics.stackexchange.com/users/955 2012-07-31T17:45:20Z 2015-01-26T12:00:56Z <p>the Lienard-Wiechert green functions have future and past null cones of radiation. Maxwell equations allow for a continuous range of mixtures between the retarded and advanced components, but we have observed so far only the retarded emission components</p> <p>or so it goes the story, but is that really accurate? It looks to me the advanced component is not radiating at all but actually absorbing; if a reverting wavefront is arranged to converge where a electron is going to be, then it will be left afterward with <em>more</em> energy, not less, and the mixture will be temporarily reversed by this artificial arrangement of incoming radiation, with an advanced absorbing component and a retarded radiative component which will be zero or very small</p> <p>Does it make sense an advanced component that is radiating, i.e: the electron is left will less energy? by symmetry under time reflection, the existence of radiating advanced wavefronts would imply the existence of <em>absorbing</em> retarded wavefronts (i.e: retarded wavefronts of negative electromagnetic energy) which we don't see either</p>