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## Hot answers tagged lienard-wiechert

5

I won't try to defend Feynman's derivation, which seems strangely non-relativistic. (A similar argument is used by Schwartz in his "Principles of Electro-Dynamics".) However, I will defend the result (the Liénard-Wiechert potentials), and specifically claim that they are not in conflict with your discrete charge example, at least for the case of uniform ...

5

What your instructor was summarising is the Lienard-Weichert potentials. These are formulae derived from Maxwell’s equations and special relativity that can be used to calculate the EM field due to a charge in general motion. As you can see, the potentials depend on the charge’s velocity, so the field, which is built from the time and spatial derivatives of ...

4

Yes, the Lienard-Wiechert potential is a solution to the wave equation with a point charge source. That's how it is derived. However, if you want to take derivatives in order to show this explicitly from the differential equation, you must be very careful to not neglect any terms in the derivatives, and also be careful to be able to extract the delta ...

3

As my2cts said the equation is not invertible. Here's a quick proof. First we go to momentum space by Fourier transforming both sides of the equation $$(-k^2 \eta^{\mu\nu}+ k^\mu k^\nu )\tilde A_\nu(k) = \frac{4\pi}{c} \tilde j(k)^\mu$$ Now the solution for $\tilde A(k)$ would be simply given by solving the equation algebrically and then doing an inverse ...

3

Although I can't give an accurate in-depth view of what 19th century physicists thought about the results of Electromagnetic Theory, they most likely interpreted the retarded potentials similar to what we interpret nowadays: that "electromagnetic news" travel at the speed of light. It was already known at Maxwell's time that the speed of light was finite ...

3

Both equations (for the instantaneous field of a charge moving with constant velocity $v$) are correct. (Well, maybe the primes should be swapped in the second equation, so that the unprimed frame is that in which the charge is moving.) The first figure is not an accurate representation of the first equation: as Jan Lalinsky stated, the field lines should ...

3

The general rule (see the "Composition with a function" section of the Wikipedia article on Dirac delta functions) is (for suitably well-defined functions): \begin{equation*} \int_{-\infty}^\infty {\mathrm dx \, f(x)\, \delta(g(x))} = \sum_i {\frac{f(x_i)}{|g'(x_i)|} } \end{equation*} where $x_i$ are the roots of $g(x)$, so your "extraction" is justified. ...

3

Let the following symbols for a general curvilinear motion of the charge $\:q$ , see Figure-01. \begin{align} \mathbf{r} & \equiv [\text{position 3-vector of field point}\: \mathrm A] =\left(x,y,z\right) \tag{01a}\\ \mathbf{x}\left(t\right) & \equiv [\text{equation of motion of charge}\: q] \tag{01b}\\ \boldsymbol{\upsilon}\left(t\right) & \...

2

If you look at Feynman Volume II Section 21-6, he walks through this calculation. Your idea and initial assertion look good; the trick is to manage the algebra to get to the final form you want.

2

Everything you've said is correct. Consider if we somehow made the charge suddenly appear, then for us at some $\vec{r}$, we still need to wait a finite amount of time to feel its influence, or we would have superluminal communication. Essentially, in the static Coulomb case, we're within the light-cone of the charge having become stationary, and then we ...

2

The potentials have simple expression (10.19 in Griffiths) only because particular gauge was chosen, where the equations for potentials are simple inhomogeneous wave equations, and also because this is one of two simplest particular solutions of inhomogeneous wave equation in 3 spatial dimensions. In other words, the mathematically simplest case is ...

2

Liénard-Wiechert: $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|}$$ where $n =\overline{r-r_s}$ $\beta = v/c$ This is the first term of $E(r,t)$ from Wikipedia. That thing in the denominator $\frac{1}{1-n \centerdot\beta}$ is not symmetric in the direction of motion. the dot product is positive ...

2

Feynman et. al. are in Ref. 1 deriving a relativistic generalization of Newton's 3rd law in a closed system of point charges. However OP's momentary contact force to particle 1 constitutes an external force unless it is explained via say a 3rd point charge of the system, so OP's 2-particle scenario is at best an incomplete description. Below we sketch a ...

2

I will take you back to Maxwell's equations in Lorenz Gauge. $$\vec \nabla^2 \varphi-\frac{1}{c^2} \frac{\partial^2\varphi}{\partial t^2}=-\frac{\rho}{\varepsilon_0}$$ $$\vec \nabla^2 \bf{A}-\frac{1}{c^2} \frac{\partial^2\bf A}{\partial t^2}=\mu_0\bf J$$ also $$\bf E=-\vec \nabla \varphi-\frac{\partial \bf A}{\partial t}$$ $$\bf B=\vec \nabla \times \bf A$$ ...

2

Advanced solution is just a particular solution for time-dependent EM field, there is no reason to call it "propagating backwards in time". The definition $$F_{rad} = F_{ret} - F_{adv} \tag{*}$$ is the simplest combination of two distinct particular solutions (retarded and advanced solution of Maxwells's equations with sources) that results in a ...

1

If your question is, as the title suggests, whether the second equation can be inverted, the answer is no. Inversion requires 'gauge fixing'. An operator can be inverted only if it establishes a one to one relation between solution and source. The operator in your second equation assigns many potentials to a single source and hence cannot be inverted.

1

Your calculations are correct, total Larmor's power radiated by body 1 is much lower than power of impressed radiation force due to 1 acquired by 2. That it, is it higher not right from the start of interaction (because impressed power is 0 when body 2 has 0 speed) but after some time. We can derive this time from your formulae to be $$t^* = \frac{\frac{2}{... 1 In answer to the last question -- many things, IMHO Be more careful about forces and accelerations, at high enough speeds constant force will not lead to further constant gains in speed. Also I would suggest not working with some fictitious forces. You are accelerating charges, so let your external force will be supplied by electric field. Less chance to ... 1 They are derived from Maxwell's equations, so they satisfy Maxwell's equations, but taking vector derivatives is very complicated with retardation. 1 Retarded solutions are a general feature of the wave equation. The response of the electromagnetic field to a current can be given by a retarded solution, but that's also true of the response of water waves to a splash, or sound waves to a clap. There's nothing inherently relativistic about having such solutions. Light is different because if you boost to ... 1 You get the magnitude of the charges and the separation distance. Consider one charge the source and the other the target. Practically we're talking about the source at the prior time and the target at the later time. Usually they talk like it's source at later time and target at later time, and then they have to go back and figure out where the source was ... 1 The EM theory allows for energy and momentum conservation, quite irrespective of which particular solutions, retarded, advanced or other, we choose to work with. In the picture where only retarded fields are present, which is the most natural one, the answer to your question is as follows. In your reasoning you use the particular EM field \mathbf E_q, \... 1 There is no obvious inconsistency, whether we use retarded, advanced, or any other field. If we use only retarded fields, things go as follows. At the time t=0, we begin to exert force \mathbf{F} on the charge q. It will move with acceleration \mathbf{F}/m for the time interval R/c, where R is the radius of the sphere. At the time t = R/c, ... 1 Firstly, I think you've confused several points in your analysis of the situation. Working with the first picture, by standard electrostatic analysis, we see that there must be a standard Coulomb force between the plates for all times t\ge t_2, not just at t=t_2. This point is not too important, but I thought I'd mention it. Continuing to work with the ... 1 You are working in Coulomb gauge, which is non-standard in radiation problems, so it's a little difficult for me to give you an in-depth analysis of your claims, but I can point out some preliminary mistakes straight away. Firstly, your assumption that the grad and curl terms fall off faster than 1/r^2 is not a valid assumption in the time-dependent case. ... 1 Radar uses the principle of retarded time to calculate distances Since x=ct, dx =c dt! Define dx=x_1-x_2. If x_1 - radar location and x_2 -target location, dt=dx/c=(x1-x2)/dt where dt is the time required to travel to target! So round trip time =2 dt which is recorded by electronic clocks. This is an example of retarded time not special ... 1 An Electric Field is only conservative if it is static. The propagation of E with a L-W field contradicts this, so it is not conservative. 1 Your thought experiment stumbles upon an important idea in electrodynamics which is quite counter-intuitive.The EM field produced as radiation due to the charge in fact produces a reaction force on the charge itself. This is known as the Abraham–Lorentz force which is proportional to rate of change of acceleration of the charge. In SI units it is given by, ... 1 To say that we assume that the mass M of the top particle is so large that its acceleration due to force F is negligible. means to take the limit M \to \infty (since a = \frac{F}{M} shall produce a = 0), since only infinitely heavy things do not accelerate when a force is applied. Under this assumption,$$ \lim_{M\to\infty} E_\text{top} = 0 ...

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