Does constant DC current in curved wire produce EM wave?

[cianfa72]

Note that if $\mathbf{J}$ has no time dependence (like for a continuous DC current loop), then neither does $\mathbf{A}$ so the fields will be DC. That is, it will not radiate.

Why if $\mathbf A$ has not time dependence then the DC current loop does not radiate ?

 

[renormalize]

Why if $\mathbf A$ has not time dependence then the DC current loop does not radiate ?

Because the magnetic field $\vec{B}\left(\vec{x},t\right)=\vec{\nabla}\times\vec{A}\left(\vec{x},t\right)$. If the vector potential has no time dependence then neither does the magnetic field. But radiation fields must oscillate with time, so: no time-dependence $\Rightarrow$ no radiation.

 

[cianfa72]

But radiation fields must oscillate with time, so: no time-dependence $\Rightarrow$ no radiation.

Ok, therefore there is a flow of EM energy in the space surrounding the wires (from the source to the load) as described by the Poynting vector field, however the DC circuit does not radiate.

 

[Baluncore]

A trapezoidal current pulse, with a rise-time, a flat top, and a fall-time, can be described by a Fourier transform. All the harmonics that make up the pulse are present throughout the pulse.

Energy flows continuously to the load during the flat period we call DC. How does that energy reach the load if it is not through a non-zero Poynting vector?

For what duration does the flat top need to be maintained stable, before the harmonics cease to be part of the continuous radiation of energy to the load?

 

[cianfa72]

Energy flows continuously to the load during the flat period we call DC. How does that energy reach the load if it is not through a non-zero Poynting vector?

I'm not sure whether Poynting vector field and Poynting theorem actually makes sense only in case $E$ and $B$ are solutions of the EM wave equation.

Does Poynting vector field actually represents a "real" flux of EM energy only in the above case (for example flowing through a given surface) ?

 

[renormalize]

I'm not sure whether Poynting vector field and Poynting theorem actually makes sense only in case $E$ and $B$ are solutions of the EM wave equation.

As pointed out by Feynman, even static EM fields give rise to Poynting vector fields that represent flows of EM energy that circulate continuously (but without radiating). See the Wikipedia entry Static Fields and its reference to the Feynman lectures.

 

[cianfa72]

See the Wikipedia entry Static Fields and its reference to the Feynman lectures.

From Wikipedia

While the circulating energy flow may seem unphysical, its existence is necessary to maintain conservation of angular momentum. The momentum of an electromagnetic wave in free space is equal to its power divided by c, the speed of light. Therefore, the circular flow of electromagnetic energy implies an angular momentum

Here it talks of momentum of electromagnetic wave in free space (bold is mine). Does it mean that $E$ and $B$ are actually solution of EM wave equation ?

 

[renormalize]

From Wikipedia

Here it talks of momentum of electromagnetic wave in free space (bold is mine). Does it mean that $E$ and $B$ are actually solution of EM wave equation ?

Yes, but since the fields are static, the EM "wave" satisfies a "wave-equation" with vanishing time derivatives, i.e., Laplace's equation: $\nabla^{2}\vec{E}=\nabla^{2}\vec{B}=0$.

 

[cianfa72]

Yes, but since the fields are static, the EM "wave" satisfies a "wave-equation" with vanishing time derivatives, i.e., Laplace's equation: $\nabla^{2}\vec{E}=\nabla^{2}\vec{B}=0$.

There is therefore a flow of EM energy, however it is "confined" in the space surrounding the DC circuit: there is not a flow of EM energy that radiates/propagates out from there.