Reflections on a Transmission Line


Many years ago there was a series of articles in Wireless World about classical electromagnetic theory, and much discussion in the letters pages. I wrote a letter to the editor which was never published, but some years later in 1996 I rewrote and expanded it to become this article. I mentioned on my 'Latest News' page that I considered the article somewhat flawed. The version of EM theory I was originally criticising was, I believed, an incorrect interpretation of the classical theory, and I wanted to present a correct version. One problem is that classical theory is anyway incorrect, and the correct theory should include quantum mechanical effects. I should therefore try to explain what I mean by a 'correct classical theory'.
What I really mean is a theory which is 'sufficiently correct for the intended purpose.' For example if we were using a transmission line to connect a radio transmitter to an aerial it could be useful to understand scattering parameters or Smith Charts, but almost entirely unhelpful to understand Quantum Electrodynamics (QED). The point is not that QED is incapable of more accurately predicting the behaviour of transmission lines, but that to do so would be difficult, and in practice completely pointless. The conventional classical treatment is good enough for this and many other purposes. However, there are different levels of incorrectness, and the simple description of a transmission line in terms of inductance, resistance and capacitance per unit length, although sufficient for many purposes, is not the best we can do within the classical framework, it is just an approximation, and if we choose to ask a few awkward questions such as how energy is reflected from the open end of a line then the simplified theory soon gets into trouble. The 'less incorrect' version I present here also gets into trouble towards the end, leading me reluctantly to mention virtual photons, an entirely non-classical concept.

Reflections on a Transmission Line

A interesting problem relating to transmission lines is the question of what happens to the electromagnetic field energy travelling along an open ended line when it reaches the end. It is well known of course that the energy is reflected back, but how can this happen? Does it stop, turn around and accelerate back to the speed of light? Clearly not. The energy has mass and momentum, and Newtons laws of motion suggest that it will therefore continue in the same direction in a straight line unless acted on by a force. The only force known to act on e.m. energy is gravity, which is not significant in this situation. So what is the standard theory explanation of the reflection?

Elementary treatments of transmission line theory divide the line into elements of length dx, each with inductance dL and capacitance dC. For this 'capacitance' however, the separation between the conductors means that anything happening in one 'plate' of capacitance dC will have no immediate effect on the other. There is a time delay because of the finite velocity of e.m. energy, which has some effect and prevents dC being correctly described as a simple capacitor obeying the equation I = C dv/dt. The prediction that e.m. energy is reflected from the open end of the line is produced magically by the standard treatment with no explanation of how this occurs. If the line is described purely in terms of incremental capacitors and inductors with no reference to the space between the conductors, then it is inevitable that the energy stays in the line, there is nowhere else for it to go in this analysis. But, if we are interested in the underlying physical mechanism involved, then we must abandon the standard analysis and look more closely at what is happening.

To start from Maxwell's equations and try to analyse what is happening is difficult. Fortunately there is a different but equivalent formulation of e.m. theory which helps us in this instance. For the step function we know fairly well what the charges in the conductors are doing at various positions along the line. So, all we need to know is the e.m. field of each charge in the conductors and then add them all together. The field of a charge depends on the position, velocity, and acceleration of the charge:

Consider a moving charge q, and some point p a distance r away. u is a unit vector in the direction from q to p. The motion of the charge now determines the field at p a time r/c later. If we call this field E' to indicate that it is measured at a different time to the other variables in the equation, then:

This equation was apparently first derived by Richard Feynman for a series of lectures around 1950. The first term in the bracket in Eq.l is just the normal 'Coulomb field' of a static charge, and is inversely proportional to the square of the distance from the charge. The second term is the rate of change of the Coulomb field multiplied by r/c, the time delay. For a charge moving with constant velocity the effect of this term is to almost exactly compensate for the time delay, so that the electric field is almost what we would expect if the coulomb field changed instantaneously to follow the charge without time delay at large distances. This is what we would expect from the theory of relativity. An observer moving along with the charge would see a static charge, and therefore just the usual inverse-square law, radially symmetric electric field. This 'instantaneous coulomb field' is not exact. Transforming from the moving reference frame back to our own frame does have some effect, but at velocities far below the speed of light the effect is negligible and the field is almost perfectly radially symmetric, and we can forget about time delays. The third term in Eq. 1 is related to the acceleration of the charge, and at large distances is inversely proportional to distance. This is the radiated field, which in the case of a radio aerial is responsible for the transmission of the radio signal.

Note there is still some dispute about that third term, which seems to suggest a charge undergoing constant acceleration will radiate. This is a problem because the principle of equivalence then suggests a charge held stationary in a gravitational field will also radiate, which appears not to happen. What I suggest is relevant is the choice of reference frame of the observer of the field, in the first case we assume a 'stationary' non-accelerating observer, so in the second case with the gravitational field we need an observer in freefall to achieve equivalence, so we need to consider the field seen by an observer accelerating past a stationary charge in the reference frame of the charge. Anyway, we are not here concerned with constant acceleration so that question can be ignored.

Considering a step function propagating along the transmission line, we can now understand the origin of the field between the wires. Ahead of the step the wires have no net charge or current. Behind the wavefront the wires are charged and a current flows along them. A magnetic field is associated with the current, and an electric field with the charge. The excess electrons in one of the wires gives it a net negative charge and a deficiency of electrons in the other wire gives it a net positive charge. The conduction electrons in both wires drift slowly along the wires. For a current of 1 amp flowing along a copper wire of 1mm diameter the average drift velocity of the electrons is just a fraction of a millimetre per second, so we are certainly justified in treating the electric field as just the inverse square law static field of the charges in the wires. The only acceleration of charges related to the step function occurs during the rise-time, and some energy will be lost as radiation as a result of this, but most of the e.m. energy in the field travels parallel to the conductors and is the local, inverse square law field of the charges in the wires. It is the first two terms in Eq. 1 which describe this field, and the behaviour of the field on reaching the open end of the transmission line is a consequence of this. For comparison consider a purely radiated field, described by the third term in Eq. 1, such as a beam of light from a torch directed parallel to the wires. On reaching the end of the line it will continue on its way out into the space beyond the line. Although originally generated by accelerating electrons in the filament of the torch bulb the radiation is now independent of these electrons. The local field of the electrons is not independent however, and cannot leave the electrons when they reach the end of the line. The electrons themselves will generally not have enough energy to leave the wires. If one electron left the wire there would be a net positive charge at the location where it left, and this would produce a force pulling the escaped electron back to the conductor. It might be thought that because there are an excess of electrons in the negatively charged conductor, there would actually be a repulsive force pushing the escaped electron away, but in, for example, a 300 ohm transmission line carrying 1 amp. the negative conductor has an excess of conduction electrons only of the order of 1 part in 1012. The net positive charge produced at the location where the electron tries to leave will far outweigh the effect of these excess electrons.

So, the field travelling along the line cannot leave at the end of the line because it is the local field of the charges, which are themselves unable to escape. The conduction electrons arriving at the end of the conductor rapidly accumulate there and build up the potential difference between the conductors. When the potential difference has doubled this will act as a signal source to generate a step travelling in the reverse direction just sufficient to reduce the current behind this new step to zero and prevent any more charge reaching the ends of the wires.

But, there are still a few problems with this description. For one thing, the static fields of the electrons travel along with them, typically at less than a millimetre per second, but we know that e.m. energy travels at the speed of light in a transmission line. (The mm/sec electron velocity mentioned is just the average drift velocity. Individual electrons have much higher thermal velocities, but these are in random directions so that the effects almost entirely cancel. The uncancelled part can be observed as the thermal noise voltage and current of a conductor.)

The flow of energy in an e.m. field is given by the Poynting vector, S = k E x B where k is a constant and B the magnetic field, which is perpendicular to the electric field E. This gives the flow of energy per unit area per second in the direction of S. Applying this to our twin wire transmission line we find that this accounts for virtually all of the energy travelling along the line. The electrons do carry a small amount of kinetic energy because of their drift velocity, but this is negligible in comparison to the energy in the field, and in any case the velocity is in opposite directions in the two conductors. But, when we work out the mass of an electron in the classical theory, this includes the energy of the static field, and so the kinetic energy of the electrons already appears to include the flow of energy in the field.

To see where the high energy originates start with two electrons initially far apart and bring them closer together. Work must be done to overcome the repulsive force between the two negative charges, so what happens to the energy put in to bring them closer? The electric field is the vector sum of the two individual fields, so if the electrons are close together the distant field will be close to double the field of one electron. But energy density is proportional to the square of the field, and so this is increased by a factor of 4. Adding more electrons close together gives further increase in field energy, and for n electrons the distant energy density is n2 times that of a single electron, and therefore n times larger than the sum of the individual energy densities of n isolated electrons. It is the close proximity of large numbers of electrons which gives the high field energy transmitted along a transmission line, not the individual kinetic energies of the electrons.

The static magnetic fields of the electrons will, generally, be in random directions and will not add in the same way as the electric fields, so these can be ignored, but if a large number of electrons are flowing in the same direction along a line then there will be a significant magnetic field associated with this current. If there are n electrons close together flowing in the same direction then the distant magnetic field energy will be n2 times greater than for a single electron, so again it is the close proximity of electrons which gives a high energy.

If one electron needs a certain energy to accelerate it up to the drift velocity, then surely the same energy would accelerate another electron in the same way? Then we have up to 4 times the energy in the magnetic field, but have only put in twice as much, so it appears that we have gained energy from nowhere! Unfortunately this is not so. If we start with two electrons at rest, one ahead of the other, and accelerate them both to the same velocity, keeping the separation constant, then the following electron feels a repulsive force from the field of the leading charge. At rest this is equal and opposite to the force on the leading electron from the field of the following charge, so no additional energy would be needed to overcome these forces if they remained equal and opposite during acceleration. There is, however, the time delay to take into account. The field experienced by the following electron is determined not by the current position of the leading charge, but by its position at an earlier time, because of the finite velocity, c, at which changes in the field travel. For constant velocity the two forces are still equal and opposite (because of the second term in the field equation mentioned earlier). But, for acceleration the following electron experiences a field produced by the leading electron at an earlier time when it was further back, so the field is greater, and in opposition to the acceleration. The leading electron experiences a reduced force from the other electron which was further away at the earlier time. The two forces no longer balance and there is a net force in opposition to the acceleration, and extra energy must be put in to overcome this, to account for the extra magnetic field energy.

Incidentally, this mechanism was proposed to account for the inertial mass of the electron in the classical theory. The idea is that in an electron of non-zero size each part of the electron would exert a force on every other part, and the resulting imbalance in opposition to acceleration would account for the inertial mass. There are problems with the classical theory of the electron, but the principle just described is valid for groups of individual electrons as in our transmission line problem.

The above account is applicable to electrons in line along the direction of acceleration, but if they are side by side then the acceleration term in the formula for the electric field of a moving charge given earlier shows that there is an electric field propagated along a direction perpendicular to the direction of acceleration. This is the 'radiated field' and for two accelerated charges side by side the direction of this electric field component from one charge is in the right direction to oppose the acceleration of the other charge. Each charge therefore has to overcome an additional force, and so again more energy is required to accelerate the two charges than the sum of the energies for two separate charges.

Having accounted for the energy in the line we now turn to the problem of its velocity. The energy is apparently still contained in the static 'coulomb field' of the charges, which travel at very low speed, so how can the energy travel at the speed of light? The important point to realise is that the extra energy previously shown to exist as a result of increased electron density can travel at a different velocity to the individual electrons. To see how this can happen imagine an electromagnetic wave in the form of a short pulse travelling along adjacent to a wire at velocity c. Suppose it is not travelling exactly parallel to the wire, and so there is a small component of the electric field along the wire so that as it passes each conduction electron it gives it a short impulse along the wire sufficient to accelerate it up to 1mm per sec. Consider a series of these electrons originally spaced 1mm apart, and assume that the impulses are short in duration in comparison to the time for the wave to travel from one electron to the next.

The field first accelerates e1 up to 1mm per sec. The field travels at velocity 3 x 108 m per sec, so it arrives at the second electron e2 a time 3.3 x 10-12 sec later, and then accelerates that electron up to 1mm per sec. By this time, however, e1 will have travelled a short distance along the line, so even though e1 and e2 are now moving at the same velocity and remain a constant distance apart, the distance is less than the original 1mm. The reduction is 3.3 x 10-12 mm. Similarly, by the time the field reaches e3 , e2 has travelled a distance 3.3 x 10-12 mm. along the line, so the distance between these two charges also is reduced by this amount. This happens to each pair of electrons along the wire, and so the electron density increases, and this increase propagates along the line at the speed of the field producing the acceleration, i.e. at velocity c, or to be more precise slightly less than c because the field was not travelling exactly parallel to the wire. The increase in electron density, and also the increase in electron velocity, lead to increased energy in the electric and magnetic fields which therefore also travels almost at velocity c. So, although the charges producing the fields travel very slow, the additional energy associated with increased charge density and the increase in velocity of the many conduction electrons travels at velocity close to c.

The origin of the accelerating field must be explained rather than simply referring to an external field which just happens to be passing by as in the above analysis. A simple explanation might be to say that as each electron is accelerated its field begins to change to correspond to the new motion, and the change travels at velocity c, and the changes in all other electron fields also travel at velocity c, so once we have started a current flow along the line at the input end there will be a field travelling along the line which can accelerate electrons it encounters and produce a current further along the line. Another explanation we could try is to consider the repulsive force between two electrons. If we move one electron along the line then it will exert an additional force on another electron further along the line and cause it to move in the same direction. The field causing the force will not arrive instantaneously, but travels at velocity c.

There are several possible objections to this. One objection is that there should be a delay due to the time taken for each electron to accelerate up to its drift velocity before it can pass on its field to the next electron in the series. This is roughly what happens in the case of the transfer of mechanical momentum along an iron bar. When we apply an impulse to one end (e.g. hit it with a hammer) the transfer of energy takes place from one atom to the adjacent atom via the electromagnetic field between the atoms, and travels along the bar far below the speed of light, i.e. at the speed of sound through iron. It is easy to show that the delay is negligible in the case of the transmission line by calculating how much of the energy is carried by the field, and how much by the kinetic energy of the electrons. The kinetic energy component is a minute proportion of the total, so we would expect the average velocity of the total energy to be close to the speed of the field, i.e. the speed of light, c. In the case of the iron bar there is only a relatively small component of electromagnetic energy in the space between adjacent atoms when the outer electrons are pushed closer together. The atoms are electrically neutral overall so there is no addition of external fields to give the high field energy possible in a transmission line.

There is good reason to believe that in the rising edge of a voltage step travelling along the line there is excess charge throughout the conductor. To see why this is so consider the constant current flowing after the rising edge has passed by. This is the flow of conduction electrons, and although the excess charges responsible for giving the conductors a nett charge may then reside at the surface the conduction electrons exist throughout the wire, and the current is more or less evenly distributed throughout the wire also (According to Feynman the distribution is the one which gives the minimum increase in entropy, but for the present purpose a uniform distribution is accurate enough). So how are electrons within the wire accelerated up to their drift velocity to give this current? There must have been a component of the electric field within the conductor during the rising edge, and this would generate an excess charge density just the same as at the surface. These excess charges will repel each other and be accelerated out towards the surface, which is where they will end up in the steady state, but until then, during the rising edge there will be a distribution of excess charge inside the wire. This does depend on the rise time, and if this is fast enough the e-m energy will be restricted to the 'skin depth'. This effect is a result of the refractive index of a metal having a real and an imaginary component of the same magnitude (at low frequency!) with the result that an e-m wave falls exponentially in amplitude as it travels into the metal, so it is effectively confined to a surface layer. At high frequencies metals can become transparent to e-m radiation. In this context 'high' is about 1012 Hz for copper. At 'low' frequencies the skin depth increases as we go down in frequency, so for the step function we would expect the e-m field to extend further into the conductor as time passes after an initial fast rise time. There is an additional problem that as the external field progresses into the conductor the resulting current creates a magnetic field inside the conductor, with associated field energy, and so there must be an additional flow of energy from the external field to supply this magnetic field energy, as well as the kinetic energy of the conduction electrons. In a real conductor there are resistive losses which require a flow of energy from the field whenever a current is flowing to replace the energy lost as heat.
The following diagram ignores resistive loss but shows the required curvature of the field required to give a flow of energy into the conductor during the rising edge:

What this shows is the electric field of the excess charge on the wire. After the edge of the step the charge is shown on the surface, although in reality there will be some delay before all the excess charge reaches the surface. During the step a charge distribution inside the wire is shown together with an internal field. This field includes the component of the field which has been shown to exist inside and along the wire, and the component between the wires, and adding the two together we can see that there is a curvature of the electric field between the conductors, so the direction of the Poynting vector perpendicular to the electric field lines shows that there is a component of energy flow into the wire. The diagram is very approximate and the curvature of the field is highly exaggerated. In reality it will make an angle with the surface of the conductor differing from 90 degrees by a small fraction of a degree.

One problem we have not solved is what happens to the momentum in the field when it is reflected. The three x, y and z components of momentum can each have three components of flow, so the momentum flow is not simply a vector, and it is commonly combined with energy density and flow to give the 'electromagnetic stress tensor', which can be found in many standard text-books. Although we can employ this to show the path of the momentum from the field to the wire when reflection takes place, this is only a description, not an explanation. If there is an explanation it is possibly tied in with another question, of why the static field of a charged particle is 'attached' to the charge, and is unable to leave it. Classical theory appears to have nothing to say about this, and we must resort to quantum theory, where there are two types of photon, 'real' and 'virtual'. Real photons are free to travel through space independent of their source, but virtual photons do not have the necessary energy and momentum to be 'real' and only exist by virtue of the uncertainty principle, so this provides an explanation of why there are two types of field.

Having tried to explain the reflection from the end of the line, it should be mentioned that there is never 100% reflection. The ends of the wires act as a small dipole aerial, and if a sinewave signal is used with wavelength comparable to the seperation between the wires, then there may be significant transmission of energy out from the end of the line. There will also be radiation out perpendicular to the line as explained in the following footnote.

A few years ago two diagrams appeared in two different electronics magazines published around the same time. Both show a parallel pair of conductors connected to a signal source, and both have arrows showing the direction of travel of electro-magnetic energy.

So why does one diagram show the energy radiating out perpendicular to the conductors, while the other has it confined between the conductors and travelling parallel to them? The first diagram appeared in an article about radio transmitter aerials, and the seperation between the conductors is chosen to be half a wavelength. The second diagram is from an article about transmission lines and shows a step function with a vertical wavefront travelling along the line.

The simplest solution to this problem is to say that with any signal source (other than a constant voltage) connected to any parallel pair of conductors, both effects occur, but it is possible to choose the relationship between the signal and the length and separation of the conductors to minimise or maximise either effect. Take the transmitting aerial first. A single wire will radiate equally in all perpendicular directions. The radiated electric field produced by each element of the conductor is proportional to the acceleration of the electric charge in the element, which is proportional to the rate of change of current. This will in general vary along the length of the conductor and the total radiated field is the vector sum of the contributions from different parts. Time delays must of course be taken into account and the field at a distance x from an element of conductor depends on what was happening in that element x/c sec. earlier, where c is the speed of light. The total length of the conductor determines the transmission efficiency, and with a sine wave signal a length of a quarter wavelength is commonly used for transmitting aerials. If a second conductor is placed nearby the resulting field is simply the vector sum of the two contributions. Fig. 1 shows the effect for a spacing of one wavelength with conductors fed in opposite phase with a sine wave signal. For clarity only the e.m. wave travelling from left to right is shown. The signals are out of phase and subtract, while in Fig.2 they are in phase and add because the separation is half a wavelength.

Of course the amplitudes reduce with distance, so the field originating from the left hand conductor will be smaller than that from the right, because it has travelled further, so there will not be exact cancellation in Fig. 1. Also, there will be some directions where the signals do add in phase and there will be e.m. energy transmitted in these directions. Different radiation patterns can be produced by changing the spacing between the conductors. If the spacing is made small compared to the wavelength then in any perpendicular direction the effects of opposite phase currents in the conductors will tend to cancel, and the closer together the conductors the more accurately the radiated field will approach zero. For any non-zero separation, however, it is possible to choose a frequency sufficiently high for the separation to be half a wavelength, and then there will be significant radiation of e.m. energy in the perpendicular direction.

Turning now to a step function travelling along a twin-wire transmission line: A perfect step function with zero rise-time is impossible to achieve in practice, and for the sort of rise-times commonly encountered in electronic circuits the conductors used can be kept sufficiently close to keep radiated energy loss to acceptably low levels. The usual design objectives of avoiding large current loops and keeping path lengths as short as possible help prevent such problems. Before and after the step the current is constant and there is no radiation.