Consider the four differential equations in the table given on wikipedia  and assume there is no charge distribution at any point in time, and thus also no current. If there is no charge, then the four equations reduce to the following: $\nabla\cdot E = 0$ $\nabla\cdot B = 0$ $\frac{\partial B}{\partial t} = -\nabla\times E$ $\frac{\partial E}{\partial t} = c^2\nabla\times B$ The last two equations tell us how both the magnetic and electric fields change over time respectively, thus given some initial magnetic and electric fields, one should be able to determine any future state of both field. This makes the first two equations seem redundant to me and thus the system seems over determined. However they are clearly necessary, so I must be missing something. Are the first two equations simply initial conditions?
98190
specify two vector and two scalar (differential) equations. That implies 8 components in the equations. But between vector fields $\vec{E}=(E_x,E_y,E_z)$ and $\vec{B}=(B_x,B_y,B_z)$, there are only 6 unknowns. So we have 8 equations for 6 unknowns. Why isn't this a problem?  As far as I know, the answer is basically because the equations aren't actually independent but I've never found a clear explanation. Perhaps the right direction is in  on arXiv.  Apologies if this is a repost. I found some discussions on PhysicsForums but no similar question here.
120563
I am trying to formally learn electrodynamics on my own (I only took an introductory course). I have come across the differential form of Gauss's Law.   $$ \nabla \cdot \mathbf E = \frac {\rho}{\epsilon_0}.$$  That's fine and all, but I run into what I believe to be a conceptual misunderstanding when evaluating this for a point charge.   I know the math looks better in spherical coordinates, but I will be using Cartesian.   So when I calculate the divergence I obtain:  $$ \nabla \cdot \mathbf E = \nabla \cdot kQ\langle\frac{x}{(x^2+y^2+z^2)^{\frac{3}{2}}},\frac{y}{(x^2+y^2+z^2)^{\frac{3}{2}}},\frac{z}{(x^2+y^2+z^2)^{\frac{3}{2}}}\rangle = \frac{-3(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{\frac{5}{2}}}+\frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}}.$$  This can further be simplified:  $$\frac{-3(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{\frac{5}{2}}}+\frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}} = \frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}}-\frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}} = \frac{3-3}{(x^2+y^2+z^2)^{\frac{3}{2}}}.$$   Now instinctively I would say that 3-3 is zero and then the while thing is zero everywhere. I am confused as to why (purely mathematically) this expression is not equal to zero at the origin. I completely understand why it physically has to be that way. And I also understand that it is modeled with the delta dirac function. But what (again, mathematically) is stopping me from saying that equation is just zero even at the origin?
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Consider a system of the two identical positive point charges situated in free space (isolated from the influence of any other external fields) as shown in the attached diagram. Particle $1$ is at $(a,a,0)$ and particle $2$ is at $(0,a,0)$. Their velocities, at the considered instant of time, are as shown in the diagram ($\mathbf{v}_1$ along the $+x$ axis, $\mathbf{v}_2$ along the $+z$ axis).    Now, by applying the Biot-Savart law, we find that the magnetic field due to particle $2$ at the position of particle $1$ is along $+y$ axis, which means that the force acting on particle $1$ is along $+z$ axis according to Fleming’s left-hand rule. A similar analysis shows that there is no magnetic force on particle $2$ as the magnetic field of particle $1$ should vanish at positions (relative to particle $1$) located along its velocity vector.  Now, if we observe the net torque on the system of two particles about the $y$ axis then it is non-zero and is directed along the $-y$ axis. Here, there are no external forces or torques acting on the isolated two-particle system, and yet the net torque, as well as the net force on the particles, are nonzero. Why? Also, Newton's third law of motion seems to be broken in this scenario. Why?    Edit $1$  I have come to know from the responses that the electromagnetic field itself takes away some momentum and angular momentum about the considered axis. However, I think that if I consider only two charged particles as my system then the Abraham-Lorentz force can be assumed to be acting upon the system, and that is sufficient to make sure that we have considered the momentum being carried away by the electromagnetic field itself.   Even after considering the action of the Abraham-Lorentz force for the two-particle system, the scenario breaks both Newton's third law and the conservation of linear and angular momentum. This is because the Abraham Lorentz forces do not exactly counterbalance the force and the torques, on the two-particle system under consideration, due to the magnetic field.    Edit $2$  The previous edit was a result of confusion and misunderstanding on my part. The force associated with the momentum carried away by the electromagnetic field as a result of the electromagnetic interaction described in the question is the Lorentz force itself which simply doesn't obey the third law of motion. The Abraham-Lorentz force is a different story. It is associated with the momentum carried away by the radiation emitted by the accelerated charged particles. This is an additional force apart from the Lorentz force and corresponds to an additional carriage of momentum by the electromagnetic field. The momentum carried away by the electromagnetic field in correspondence with the Lorentz force doesn't correspond to radiation.
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I'm having trouble understanding the diagrams of elctromagnetic waves.  I have no problem with any concept in classical mechanics, and I think this can be answered without any relativity (which I don't know yet). I hope for an answer which is as intuitive as possible but any answer will be good.  So in the common diagram of an electromagnetic wave:     What exactly do the vectors represent? The wave is propagating though an eletromagnetic field. Are the lines just the field created by a particle?  We learned in physics that a charged particle creates a field in all of space, Is this a picture of changes in the field when it moves? And if so, how does it move? does it move up and down with the amplitude of this wave?  Lastly, It's not clear to me why the E-field part is not in the direction of the motion of the wave, because usually an electric field is depicted as so:    i.e it is in a direction "outwards" from the source of the charge.  This is really running through my mind for some time, so thanx for reading this!
71848
I was trying to derive Lagrangian for the electromagnetic field from Lorentz's force formula $$\mathbf{F}= q(\mathbf{E} + \mathbf{v}\times \mathbf{B})$$ I have to find the potential by using the line integral of the force. I have written the above equation in terms of scalar and vector potential which is given by. $$F=q(-\nabla\phi-\frac{1}{c}\frac{d\mathbf{A}}{dt}+\frac{1}{c}\nabla(\mathbf{v}\cdot \mathbf{A}))$$ now I have to find potential by using line integral $$V=-\int \mathbf{F}\cdot d\mathbf{r}$$ now I have to show that the integral of the second term is zero so that I can get exact potential which is given by$$V=q(\phi-\frac{1}{c}\mathbf{v}\cdot \mathbf{A})$$ So my question is how to show that the integration of second term is zero?
22771
As I have read that "time varying electric field is source of changing magnetic field and time varying magnetic field is source of changing electric field "  Hence, I have the following doubt : If time varying nature of one field originates other changing field. Then, why are they in phase?  One is changing and producing the second thing hence second must depend upon first's derivative, I. e. If one is sine then second will be cosine Hence a phase difference of pi/2 and not in phase. Where am I wrong?
47885
For a point charge that moves with the trajectory $ \vec r(t)$, we know that it has the singular charge and current densities  $$ \rho (\vec x, t) = q \delta^3(\vec x - \vec r(t)) $$ $$ \vec J(\vec x, t) = q \frac{d\vec r}{dt} \delta ^3 (\vec x - \vec r (t))$$  How do you prove that these densities satisfy the continuity equation?  $$ \frac{\partial \rho}{\partial t} + \nabla \cdot \vec J = 0 $$  I considered using the property of the dirac delta  $$ \delta (f(x)) = \sum_i^N  \frac{\delta (x - x_i)}{|\frac{df}{dx}|_{x = x_i}}$$  where $x_i$s are the roots of $f(x)$. In the problem at hand, we see that the dummy variable x is t, and $f(x) = \vec x - \vec r(t)$ so that the only root is $t_0$ which makes $\vec r(t_0) = \vec x$.  However I just don't know how to apply the derivative to the dirac delta. I know that   $$ \delta '[f] = f'(0)$$  Any help please? Thank you very much!
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I've started reading Peskin and Schroeder on my own time, and I'm a bit confused about how to obtain Maxwell's equations from the (source-free) lagrangian density $L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ (where $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ is the field tensor).  Substituting in for the definition of the field tensor yields $L = -\frac{1}{2}[(\partial_\mu A_\nu)(\partial^\mu A^\nu) - (\partial_\mu A_\nu)(\partial^\nu A^\mu)]$.  I know I should be using $A^\mu$ as the dynamical variable in the Euler-Lagrange equations, which become $\frac{\partial L}{\partial A_\mu} - \partial_\mu\frac{\partial L}{\partial(\partial_\mu A_\nu)} = - \partial_\mu\frac{\partial L}{\partial(\partial_\mu A_\nu)}$, but I'm confused about how to proceed from here.  I know I should end up with $\partial_\mu F^{\mu\nu} = 0$, but I don't quite see why.  Since $\mu$ and $\nu$ are dummy indices, I should be able to change them: how do the indices in the lagrangian relate to the indices in the derivatives in the Euler-Lagrange equations?
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I was wondering why Faraday's law of induction and Maxwell-Ampere's law (without sources) are not totally symmetric in the sense that Maxwell-Ampere's law has a $\epsilon_0 \mu_0$ term on the right (in SI units) while Faraday's law doesn't, as symmetry is an important feature in most physical laws.   \begin{align} \nabla\times\mathbf E&amp;=-\frac{\partial\mathbf B}{\partial t} \\ \nabla\times\mathbf B&amp;=\color{blue}{\mu_0\varepsilon_0}\frac{\partial\mathbf E}{\partial t} \end{align}  A popular reference book states the reason being "that we use SI units". Can anyone tell me how using a particular unit can affect the symmetry of physical laws written in their mathematical form?
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I have just finished learning the basics of magnetism, and it should be noted that I am not very familiar with Maxwell's equations.  Note:    In the question, when I say , I am referring to the equation without Maxwell's correction. Also, when I say , I am referring to the equation: $\mathrm dB= (\mu_0/4\pi)(I)(\mathrm dL~ X~\hat r)/r^2$     Consider an infinitely long straight wire, carrying a time varying current I(t) such that dI(t)/dt is non-zero. Also consider a point P which is at a distance r from the wire. Using Biot Savart Law, we find out that the magnetic field is $\mu_0\cdot I(t)/2\pi \cdot r$, at any instant t.   Now, I have read that Ampere's Circuital law is NOT valid for cases in which the currents are time varying. However, if we consider an Amperian loop along a circle of radius r and centre at the perpendicular from P to the wire, using symmetry arguments, we obtain the same value of field: $\mu_0 I(t)/2\pi\cdot r$. Since Ampere's law is invalid for such a current, the expression mentioned for the magnetic field must be incorrect.  So, can Biot Savart Law also NOT be used for time varying currents? Also, just out of curiosity, what would be the actual value of the magnetic field at time t?  My book (Halliday and Resnick) derives the equation for the magnetic field created due to a moving point charge. However, after the derivation, it states that the result obtained is not really valid, since "a point charge cannot be assumed as a steady current by any stretch of imagination". This makes me believe that even Biot-Savart Law is only true for non time varying currents. Am I right or wrong?
24767
In electromagnetism, we can re-write the electric field in terms of the electric scalar potential, and the magnetic vector potential.  That is:  $E = -\nabla\phi - \frac{\partial A}{\partial t}$, where $A$ is such that $B = \nabla \times A$.  I have an intuitive understanding of $\phi$ as the electric potential, as I am familiar with the formula $F = -\nabla V$, where $V$ is the potential energy.  Therefore since $E = F/q$, it is easy to see how $\phi$ can be interpreted as the electric potential, in the electrostatic case.  I also know that $F = \frac{dp}{dt}$, where $p$ is momentum, and thus this leads me to believe that $A$ should be somehow connected to momentum, maybe like a "potential momentum".  Is there such an intuitive way to understand what $A$ is physically?
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I am not sure which form of  is fundamental, integral form or differential form. Imagine an ideal infinitely long solenoid. When a current is changing in time, can we detect classical effects outside a solenoid, for example generating a circular current around solenoid by Faraday's law. If the differential form is fundamental, we won't get any current, but the integral form is fundamental we will get a current.
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We can think that the electric field and the gravitational field operate similarly in the sense that the forms of their governing laws (namely, Coulomb's law and Newton's law respectively) are strikingly similar. The only difference one can point out is that while the electric charges come in two flavors, the gravitational masses come in just one.  Now, I have read that when a charged particle moves, the electric field lines associated with are distorted in a fashion because of the finite time required for the information about the change in the position of the charge to get propagated. And, I have led to the understanding that this is the cause of the existence of the magnetic field (and that if calculus is used it can be proven mathematically).   So (if this is true then) why doesn't the same thing happen to the gravitational field? Why is there nothing like a gravitational magnetic field? Or, is there?  Note  I have changed the language and the tone of the question massively. Although the question was fairly well received, I believe it was really ill-posed. As pointed out by ACuriousMind in the comment, the ''reason''described here behind the existence of the magnetic field is something that cannot be found a good support for. But still, due to the similarity between the equations describing their static behavior of electric and gravitational fields, one can still ask as to whether a boost would create some sort of a gravitational magnetic field if the original frame only had a static gravitational field. As the accepted answer points out, the answer is, roughly, a yes but the Maxwell-type equations for gravity aren't as well-behaved as are the original Maxwell equations of electromagnetism--one should note. In particular, the equations for gravity take the Maxwell-like form in appropriate weak limits only in some appropriately chosen gauges--and aren't Lorentz covariant.
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As we know, the  are applicable for conservative electric fields. Now it is applicable for circuits where inductors are present but the field there is not .  So how are the  Kirchhoff's laws  applicable there?
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I read in  the statement that the subset of Maxwell equations comming from the Bianchi identity:  $$ \nabla \cdot {\bf B} = 0, \quad \nabla \times {\bf E} + \frac{1}{c}\frac{\partial {\bf B} }{\partial t} = 0 $$ is actually invariant under Galilean transformations. I came up with this reasoning, is it correct, or am I missing something?  First $c$ has to be assumed a constant, otherwise it will not work, I think the author assumes this, though he does not say it.  A Galilean transformation, say on the $x$ axis, transforms $x' = x -ut$, $z'=z$, $y'=y$, $t' =t$. It does not mix electric and magnetic fields, so they only mix components amongst themselves. The transformation matrix for the electric field say, will be   $$ E'_i (x') = \frac{\partial x^k}{\partial x'^i} E_k(x) = E_{i}(x) $$ and something identical for the magnetic field. Also derivatives will be trivially transformed $\nabla =\nabla', \quad \frac{\partial {\bf } }{\partial t} = \frac{\partial {\bf } }{\partial t'}$ so invariance follows readily.  Then the author says the whole set of Maxwell equations is not invariant due to the displacement current term. I do not understand this statement. Presumably he reffers to the existence of electromagnetic waves propagating with speed $c$ and this will not be constant under Galilean transformations, other than that the equations would retain their same form due to the argument I gave above.  But if we started assuming $c$ is not a constant, then we could not even prove the invariance of the first two equations (the point is perhaps that in such case we cannot interpret $c$ as the speed of anything?).
89160
The  is a well-known and very useful tool for solving problems in electrostatics.  Unfortunately, when I was taught this method, it was presented simply as an algorithm. No real physical justification was given for its usage, and there was a complete absence of rigorous mathematics. The method has always been a little hazy to me since.  If someone could give a rigorous first-principles derivation of this method from Maxwell's equations (perhaps simply Coulomb's law?), it would be most appreciated.
49719
Greiner in his book "Field Quantization" page 173, eq.(7.11) did this calculation:     ${\mathcal L}^\prime=-\frac{1}{2}\partial_\mu A_\nu\partial^\mu A^\nu+\frac{1}{2}\partial_\mu A_\nu\partial^\nu A^\mu-\frac{1}{2}\partial_\mu A^\mu\partial_\nu A^\nu $   $\space\space\space\space=-\frac{1}{2}\partial_\mu A_\nu\partial^\mu A^\nu+\frac{1}{2}\partial_\mu[A_\nu(\partial^\nu A^\mu)-(\partial_\nu A^\nu) A^\mu]$        The last term is a four-divergence which has no influence on the field equations. Thus the dynamics of the electromagnetic field (in the Lorentz gauge) can be described by the simple Lagrangian        ${\mathcal L}^{\prime\prime}=-\frac{1}{2}\partial_\mu A_\nu\partial^\mu A^\nu$     Yes, if it is a four-divergence of a vector whose 0-component doesn't contain time derivatives of the field, indeed according to the variational principle this four-divergence will not influence the field equation.  And actually I calculated the time-derivative dependence of 0-component of $[A_\nu(\partial^\nu A^\mu)-(\partial_\nu A^\nu) A^\mu]$, in which only $[A_0(\partial^0 A^0)-(\partial_0 A^0) A^0]$ could possibly contain time-derivative, which vanishes fortunately, so whatever the general case it doesn't matter in this present case.  But how can he seem to claim that it holds for a general four-divergence term,The last term is a four-divergence which has no influence on the field equations?  EDIT: I only assumed the boundary condition to be $A^\mu=0$ at spatial infinity, not at time infinity. And the variation of the action $S = \int_{t_1}^{t_2}L \, dt$ is due to the variation of fields which vanish at time, $\delta A^\mu(\mathbf x,t_1)=\delta A^\mu(\mathbf x,t_2)=0$, not having the knowledge of $\delta \dot A^\mu(\mathbf x,t_1)$ and $\delta \dot A^\mu(\mathbf x,t_2)$, which don't vanish generally, so the four-divergence term will in general contribute to the action, $$\delta S_j=\delta \int_{t_1}^{t_2}dt\int d^3\mathbf x \partial_\mu j(A(x),\nabla A(x),\dot A(x))^\mu =\delta \int_{t_1}^{t_2}dt\int d^3\mathbf x \dot j^0 =\int d^3\mathbf x [\delta j(\mathbf x, t_2)^0-\delta j(\mathbf x, t_1)^0]$$ which does not vanish in general!