Inertia from the Poynting vector

Let us study extra inertia in the case of electric charges.

A tube of a small negative charge sliding on a rod of a positive charge

                          E'               E

                   -   -   -   -

        ++++++++++++++++++  rod

                   -   -   -   -    R

                    1 meter

                  tube  v ---->

Let us calculate the Poynting vector. We assume that the length of the tube is one meter, its radius is R, and the electric field of the tube E' is much weaker than the electric field E of the rod.

The magnetic field of the moving tube is

       B = E' v / c².

The Poynting vector is defined as

       S = E × B / μ.

The energy flow per a unit area around the tube is

       1 / (μ c²)  E E' v.

The energy flows to the opposite direction from v.

Let us contract R a little, so that the normal area outside the tube grows by A. The combined electric field loses the energy density

       D = ε E E'

           = 1 / (μ c²)  E E'

there. Let us interpret this that along with the tube, the negative energy density D flows at the speed v to the right. Now we see that the Poynting vector correctly calculates the positive energy flow to the left.

We have claimed in this blog that the tube acquires extra inertia since field energy must move around. In this case, the Poynting vector captures our idea.

The extra inertia in this case is as if the tube would be carrying the potential energy that we harvested as we let R contract from infinite to its present value.

We should measure the extra inertia in practice, to be certain that the analysis is correct.

In some other cases, the extra inertia probably is not as nicely linked to the harvested potential energy.

The energy of a magnetic field is twice the kinetic energy of the corresponding electric field

The energy density of an electric field is

       1/2 ε E²

and of a magnetic field

       1/2 * 1 / μ * B².

If an electric field E moves at a velocity v << c normal to the field lines, it generates a magnetic field

       B = E v / c²,

where c is the speed of light.

The kinetic energy (1/2 m v²) density of a moving electric field is

       1/4 ε E² / c² * v²

       = 1/4 1 / (μ c²) E² v² / c²

       = 1/4 * 1 / μ * B².

We see that the kinetic energy density of the electric field is 1/2 of the energy density of the corresponding magnetic field.

A circular loop of positive charge surrounded by a tube with the equivalent negative charge

We let the tube slide around the loop at a velocity v. We adjust the charges in such a way that the electric field is zero outside the tube.

In this case, the Poynting vector does not show any energy flow at all because the electric field is zero outside the tube.

The magnetic field B, on the other hand, differs from zero outside the tube. The energy of the magnetic field is twice of what would be the kinetic energy of the electric field of the tube if there were no positive charge inside the tube.

What is the extra inertia in this case? Let I be the inertia of the tube in a rotating motion without the loop inside. Let us conjecture that the inertia without the electric field outside the tube would be I - m. The magnetic field B adds 2 m to the inertia. The inertia is then

        I + m.

The extra inertia m is as if the tube would be carrying the potential energy which we harvested as we let the radius R of the tube to contract from infinite to its present value.

Let us contract the tube to such R that we have recovered the entire mass-energy M of the tube while lowering it. The mass-energy M is the energy of the electric field of the rod from R outwards. Let us assume that the inertia of the tube, I, in this (strange) case would be zero without the magnetic field. The inertia from the magnetic field is then 2 M. We see that, with these assumptions, the inertia is I + m, using the notation above.

We should measure the extra inertia in practice, to be sure.

Conclusions

In the case of a tube with a very small charge, the Poynting vector might calculate the extra inertia which is associated with the interaction with the charged rod.

But in the case of the tube and the rod with equal charges, the Poynting vector is zero. In this case, the energy of the associated magnetic field may indicate the amount of extra inertia. Even though the electric field is zero outside the tube, it may still possess a lot of momentum, in the form of a magnetic field.

What did we learn? Even though an electric field is seemingly static, or even zero in the second case, it may still possess a lot of momentum. We are not sure how to determine the extra inertia. We used very different methods in these two cases.