An electric-field amplitude falls across a waveguide region. Has electromagnetic power necessarily been converted into heat there? No. A field’s spatial decay and the transport or dissipation of time-averaged power are different quantities.
The short answer is to examine the electric and magnetic fields together, including their relative phase. A single decaying evanescent component can carry zero average power along its decay direction. A finite lossless region containing two compatible evanescent components can carry nonzero constant power even while its electric-field magnitude decreases.
This guide continues Waveguides under the Gold Standard of Education and How X Works. All amplitudes and lengths below are original fictional teaching cases in a declared single-mode model. They are not measurements, equipment specifications or radio-frequency operating instructions.
eduKateSG · Waveguides
Find your next learning step
Keep the electric–magnetic field pair and its phase before interpreting a decaying amplitude.
ROUTE 1 · CHAPTERS 1–7
Define fields and average power
Establish conventions and test the single decaying component.
ROUTE 2 · CHAPTERS 8–14
Superpose and match boundaries
Recover constant flux from two compatible evanescent components.
ROUTE 3 · CHAPTERS 15–21
Compare port and local readings
Check transmission, inverse length and changed terminations.
ROUTE 4 · CHAPTERS 22–28
Test limits and phase information
Preserve finite boundaries, reference planes and power direction.
ROUTE 5 · CHAPTERS 29–35
Transfer and verify
Distinguish timing, phase measurements and independent field checks.
Chapter contents
Define fields and average power · Chapters 1–7
1. Ask which quantity the apparent attenuation describes
2. State a lossless single-mode boundary problem
3. Use a time convention before assigning complex phases
4. Define normalised electric and magnetic modal partners
5. Write the below-cutoff field equations
6. Calculate a single decaying component's average power
7. Explain why electric magnitude alone cannot supply the answer
Superpose and match boundaries · Chapters 8–14
8. Keep both evanescent components in a finite region
9. Derive the constant-power expression independently
10. Work an original decaying-field, constant-power example
11. Follow the changing phase in that example
12. Interpret the increasing exponential without inventing a source
Compare port and local readings · Chapters 15–21
15. Solve the complete matched-scale finite section
16. Check reflected and transmitted power separately
17. Compare transmission with a local field-magnitude ratio
18. Work an exactly solvable changed length
19. Identify length information only within the declared model
20. Examine why the input is not purely reactive
21. Change the load and check what field presence establishes
Test limits and phase information · Chapters 22–28
22. Inspect the infinite-length limit without changing the boundary silently
23. Check the short-section limit
24. Use a long-section approximation only in its own range
25. Translate the coordinate origin and preserve power
26. Show how identical electric intensities can hide opposite power
Transfer and verify · Chapters 29–35
29. Keep evanescent amplitude decay separate from message speed
30. Solve an independently changed field and load problem
31. Decide what additional phase observation would resolve
32. Audit the field, boundary and power accounts independently
33. Choose a learner route that preserves the field pair
CHAPTER 1 OF 35 · Define fields and average power
1. Ask which quantity the apparent attenuation describes
A plot can show electric-field amplitude, its squared magnitude, a detector response or transmitted power relative to an incident reference. These curves need not represent the same quantity.
In a travelling propagating mode with a known real modal admittance, electric-field squared magnitude can be proportional to forward power. That familiar relationship cannot simply be transferred to every field configuration.
An evanescent field involves a different phase relationship between its electric and magnetic partners. Reflected fields can also make a local total electric amplitude differ from the incident amplitude used to define transmission.
Begin by labelling the plotted quantity and measurement boundary. If the record supplies only electric-field magnitude, ask what magnetic or phase information would be required before interpreting it as power flow or heat generation.
CHAPTER 2 OF 35 · Define fields and average power
2. State a lossless single-mode boundary problem
Consider an ideal guide with perfectly conducting sidewalls and one separable transverse-electric mode. At the chosen frequency, the two outer uniform regions support propagation, while a finite middle region is below that mode’s cutoff.
The cross-section is unchanged. A uniform change of lossless filling can change the longitudinal propagation condition while preserving the chosen transverse pattern. We assume no coupling to other modes in this stipulated idealisation.
Permeability is common across the regions. For the worked matched-scale case, the outer longitudinal propagation constant has the same positive magnitude as the middle region’s evanescent decay constant κ.
These assumptions make the scalar modal boundary calculation possible. A real step in geometry, material loss, mode conversion or imperfect wall requires an appropriate extended analysis. The examples do not imply that every practical discontinuity satisfies this one-mode description.
CHAPTER 3 OF 35 · Define fields and average power
3. Use a time convention before assigning complex phases
Write physical sinusoidal fields as the real parts of complex amplitudes multiplied by exp(−iωt), where i² = −1. A complex amplitude records both magnitude and phase.
For example, amplitudes 1 and i describe equal magnitudes separated by a quarter cycle under this convention. They are not two different kinds of imaginary physical field; the measured time-dependent fields remain real.
Changing the time convention changes several signs in the formulas. It does not change the physical average power when all related definitions are transformed consistently.
For a learner new to phasors, begin with two ordinary sine waves. Their product averaged over a cycle depends on their phase difference. Complex notation compresses that calculation, but it does not remove the need to track which fields are being multiplied.
CHAPTER 4 OF 35 · Define fields and average power
4. Define normalised electric and magnetic modal partners
Let x = κz be dimensionless distance along the middle region. Let e(x) be its normalised transverse electric amplitude and h(x) the magnetic partner whose product with e contributes to axial power.
The electric and magnetic reference scales are chosen consistently with the positive factor κ/(ωμ). A positive transverse-mode integral supplies the remaining physical power scale.
In these normalised quantities, define p = Re[e h*]/2, where the star denotes complex conjugation. Physical axial power is a positive constant times p, so its sign and constancy can be tested using this reduced expression.
This normalisation does not equate electric and magnetic fields in their original units. It places their compatible modal amplitudes on convenient scales. Preserve the scales when translating the result back to dimensional fields or comparing it with another mode.
CHAPTER 5 OF 35 · Define fields and average power
5. Write the below-cutoff field equations
Inside the uniform middle region, the electric modal amplitude satisfies e” = e, where primes denote differentiation with respect to x. Its general solution is e = Aexp(−x) + Bexp(x).
For the stated time and field conventions, h = −ie’ = i[Aexp(−x) − Bexp(x)]. The two amplitudes are therefore linked; they cannot be assigned independent phases after choosing e.
The increasing exponential is part of the general finite-interval solution. Whether it is present depends on the boundaries. A condition at a far load can require both terms even though neither is a freely propagating travelling mode in this region.
Before interpreting either coefficient, check the differential equation and the magnetic relation. A guessed electric decay curve alone omits the partner that determines average axial power.
CHAPTER 6 OF 35 · Define fields and average power
6. Calculate a single decaying component's average power
Take B = 0. Then e = Aexp(−x) and h = iAexp(−x). The magnetic partner is a quarter cycle out of phase with the electric amplitude.
Their product e h* = −i|A|²exp(−2x) is purely imaginary. Therefore p = 0 everywhere along the decay direction, despite a nonzero electric field.
The field can store and exchange energy over a cycle without carrying nonzero time-averaged power in that direction. This is different from a propagating wave whose forward power gradually dissipates into heat.
The result does not mean an evanescent field has no physical effects, or that it cannot interact with a nearby load. It describes one isolated decaying component under the stated lossless model and the specified axial power measurement.
CHAPTER 7 OF 35 · Define fields and average power
7. Explain why electric magnitude alone cannot supply the answer
The single component’s electric squared magnitude is |A|²exp(−2x). It decreases exponentially, while its average axial power remains zero rather than decreasing from some positive input value.
Interpreting that squared-magnitude curve as a dissipating-power curve would assign a heat loss that the stipulated lossless model does not contain. The missing step is the electric–magnetic phase relation.
A comparison with an ordinary propagating mode helps. There, a real positive modal admittance can make forward power proportional to electric squared magnitude. Below cutoff the corresponding relationship is reactive, so that proportionality no longer has the same meaning.
A good rule is to retain the field pair until the relevant power expression has been formed. The optical appearance of a falling curve is not a substitute for identifying what physical quantity that curve represents.
CHAPTER 8 OF 35 · Superpose and match boundaries
8. Keep both evanescent components in a finite region
For a finite interval, write a = Aexp(−x) and b = Bexp(x) temporarily. Then e = a + b and h = i(a − b).
Expanding e h* gives self terms involving |a|² and |b|², plus cross terms between a and b. The self terms are purely imaginary. The real average power comes from their interference.
The result is p = Im(A*B). The exponential factors cancel in this cross product, so p is constant in x for the lossless uniform region.
This is the important extension beyond the single-component result. Neither exponential on its own carries real axial power, yet their compatible superposition can. Saying that each separate component has zero real power does not permit discarding the cross terms of their combined physical fields.
CHAPTER 9 OF 35 · Superpose and match boundaries
9. Derive the constant-power expression independently
Since h = −ie’, the normalised power can also be written p = Im(e*e’)/2. Differentiate it with respect to x.
The derivative is Im[e’*e’ + e*e”]/2. The first term is |e’|² and is real. The field equation gives e*e” = |e|², also real. Hence dp/dx = 0.
This derivation checks conservation directly from the differential equation rather than only expanding the exponential solution. It applies to every solution of the stipulated lossless equation, not only the numerical example we choose next.
It also reveals what the amplitude plot omits. The conserved quantity involves both the amplitude and its complex spatial derivative. A declining |e|² does not by itself specify whether this particular bilinear combination vanishes, stays positive or has another sign.
CHAPTER 10 OF 35 · Superpose and match boundaries
10. Work an original decaying-field, constant-power example
Take a finite region from x = 0 to x = 1. Set A = 1 and B = i exp(−2). Then p = exp(−2), about 0.1353, throughout the interval.
| Position x | Electric squared magnitude | Normalised axial power p |
|---|---|---|
| 0 | 1 + exp(−4) ≈ 1.0183 | exp(−2) ≈ 0.1353 |
| 1/2 | exp(−1) + exp(−3) ≈ 0.4177 | exp(−2) ≈ 0.1353 |
| 1 | 2exp(−2) ≈ 0.2707 | exp(−2) ≈ 0.1353 |
The electric squared magnitude falls across these positions. The axial power does not. This supplies the promised counterexample to equating local electric decay with dissipated power.
The numbers use a normalised modal scale. They are not watts or a detector calibration until the appropriate physical scale and measurement model are supplied.
CHAPTER 11 OF 35 · Superpose and match boundaries
11. Follow the changing phase in that example
For A = 1 and B = i exp(−2), the electric and magnetic squared magnitudes are equal at each position, but their relative phase changes.
At x = 1, e = h = exp(−1)(1 + i), so their phase difference is zero. At x = 0, the cosine of their phase difference is 2exp(−2)/[1 + exp(−4)], about 0.2658.
As the field magnitudes decrease toward the right boundary, their phase relationship changes in a way that preserves Re[e h*]/2. Looking at magnitude alone would miss that compensation.
This is a useful concrete interpretation of the cross term. The average power is not created by the smaller field at the output. It is the same conserved flux described by a different mixture and phase relationship inside the region.
CHAPTER 12 OF 35 · Superpose and match boundaries
12. Interpret the increasing exponential without inventing a source
The term Bexp(x) grows as x increases. In a finite boundary-value problem, that mathematical behaviour does not imply active gain or energy generation inside the middle region.
Its coefficient is fixed together with A by the left and right boundary conditions. The increasing term can be viewed as the component required to account for the far boundary, not as a separate independently powered object inside the material.
The numerical example remains finite because x is restricted to a finite interval. Extending that same coefficient to arbitrarily large x would change the problem and violate the intended bounded condition at infinity.
Keep the domain alongside the solution. A component that is admissible between two boundaries may be inadmissible on an infinite half-line. The boundary conditions, not the word growing, decide whether a field representation is physically consistent.
CHAPTER 13 OF 35 · Superpose and match boundaries
13. Apply a matched propagating load at the right boundary
Let the middle interval have dimensionless length L. Under the matched-scale outer-region assumption, a purely outgoing propagating field at its right boundary has h(L) = e(L).
Substituting the two exponentials gives B = iAexp(−2L). Our earlier A = 1, L = 1 example therefore satisfies this particular outgoing-load condition.
Its constant power is |A|²exp(−2L), which is positive when A is nonzero. The load has required two evanescent components inside the finite middle region, even though the field beyond it is a propagating outgoing wave.
This result depends on the specified modal scales and load. A different external propagation constant or termination changes the boundary relation. It would be wrong to treat B = iAexp(−2L) as a universal property of all below-cutoff structures.
CHAPTER 14 OF 35 · Superpose and match boundaries
14. Separate incident and reflected amplitudes at the input
In the left propagating region, use the same matched-scale normalisation. Let f be the incident electric amplitude and g the reflected amplitude at the input boundary.
Then e(0) = f + g and h(0) = f − g. The minus sign reflects the reversed power direction of the reflected propagating wave.
The incident power is |f|²/2, the reflected power magnitude is |g|²/2, and net input power is their difference. The total local electric squared magnitude |f + g|² is another quantity.
This distinction matters before computing transmission. A coefficient defined relative to incident power should not use total boundary electric amplitude as its denominator. Near a reflecting interface, interference can make those two amplitude references differ substantially even in a completely lossless problem.
CHAPTER 15 OF 35 · Compare port and local readings
15. Solve the complete matched-scale finite section
Set the incident amplitude f = 1. The boundary conditions give transmitted amplitude t = sech L = 1/cosh L and reflected amplitude g = −i tanh L.
One way to check the result is to propagate the middle-region state backwards from its outgoing right boundary. The relations are e(0) = (cosh L − i sinh L)t and h(0) = (cosh L + i sinh L)t.
Adding these equations gives 2 = 2cosh L·t, so t = sech L. Subtracting the incident amplitude from e(0) then gives g = −i tanh L.
These amplitudes follow from the stipulated single-mode matching problem. They are not a universal insertion-loss formula for arbitrary guide steps, where other modes, unequal permeability or a different modal-admittance ratio may matter.
CHAPTER 16 OF 35 · Compare port and local readings
16. Check reflected and transmitted power separately
The power transmission coefficient is T = |t|² = sech²L. The reflected fraction is R = |g|² = tanh²L.
Their sum is one because cosh²L − sinh²L = 1. The difference between incident and transmitted power has been reflected, not dissipated in the lossless middle region.
At L = 1, T ≈ 0.4200 and R ≈ 0.5800. With unit incident amplitude, incident normalised power is 1/2; transmitted power is about 0.2100 and reflected power magnitude about 0.2900.
This power balance is different from asserting that electric magnitude does not change. Fields can vary strongly within a lossless boundary problem. The conservation law concerns the appropriately formed flux and the complete input–output account.
CHAPTER 17 OF 35 · Compare port and local readings
17. Compare transmission with a local field-magnitude ratio
In the earlier A = 1, L = 1 representation, the electric squared-magnitude ratio between the right and left boundaries is 2exp(−2)/[1 + exp(−4)], about 0.2658.
That is not the power transmission coefficient 0.4200. The left boundary electric amplitude includes both incident and reflected fields, while the transmission coefficient uses incident power alone.
The mismatch is not an inconsistency in the solution. It identifies two distinct denominators. Multiplying the entire solution by another complex amplitude changes its scale but leaves both ratios and their difference intact.
A useful learner check writes the denominators in full words: local total electric squared magnitude at the input, or incoming propagating power. A familiar-looking output-over-input expression can mean different things when the input quantity has not been named precisely.
CHAPTER 18 OF 35 · Compare port and local readings
18. Work an exactly solvable changed length
Take L = ln2. Then cosh L = 5/4 and sinh L = 3/4. The transmitted amplitude is 4/5 and the reflected amplitude is −3i/5.
Thus T = 16/25 = 0.64 and R = 9/25 = 0.36. The boundary fields are e(0) = 1 − 0.60i and h(0) = 1 + 0.60i, while e(L) = h(L) = 0.80.
Net power at both boundaries is 0.32. Yet electric squared magnitude falls from 1.36 at the input to 0.64 at the output, giving a local ratio 8/17 rather than 0.64.
The fractions make the comparison independently reproducible. A learner can multiply e by h* directly at each boundary, then compare that result with the electric squared-magnitude calculation and explain why only one is the axial-power account.
CHAPTER 19 OF 35 · Compare port and local readings
19. Identify length information only within the declared model
If a measurement supplies T = 0.64 and every matched-scale assumption is independently justified, the model inverse gives L = arcosh(1/√T) = ln2.
But L = κd_phys is a product of physical section length and evanescent decay constant. Transmission alone does not separate those two quantities. A known physical length or a supported dispersion model is needed for that next inference.
The same observed T could also arise under a different interface model with other parameters. A correct inverse within one specified family does not prove that family describes an unknown device.
The nearby cutoff-uncertainty guide treats a distinct above-cutoff timing calculation and explicitly leaves finite below-cutoff fields outside that freely propagating branch.
CHAPTER 20 OF 35 · Compare port and local readings
20. Examine why the input is not purely reactive
For A = 1 and B = i exp(−2L), write b₀ = exp(−2L). The input modal admittance is h(0)/e(0) = (b₀ + i)/(1 + ib₀).
Its real part is 2b₀/(1 + b₀²), which is positive for finite L. Its imaginary part is (1 − b₀²)/(1 + b₀²).
The real part permits positive net average power into the finite loaded section. Treating the middle region as if it contained only the isolated decaying component would instead give a purely imaginary local relationship and miss the load-mediated power transfer.
This does not require dissipation inside the section. The power enters from the left and leaves through the propagating load at the right. The complete boundary problem determines the input relationship, even though the local differential equation remains below cutoff throughout the middle.
CHAPTER 21 OF 35 · Compare port and local readings
21. Change the load and check what field presence establishes
Replace the outgoing matched load with a boundary h(L) = 0. The exponential coefficients must then satisfy B = Aexp(−2L), so A*B is real and p = 0.
The electric field at the right boundary is generally nonzero: e(L) = 2Aexp(−L). A finite field at that boundary therefore does not by itself establish delivered average power.
A boundary e(L) = 0 instead gives B = −Aexp(−2L), again producing zero real power. These idealised terminations differ in field shape while both carrying no net average power into a dissipative load.
The examples are boundary conditions in a paper model, not instructions for configuring equipment. Their role is to show why load information matters. Field presence, transmission to a propagating output and absorption are three different claims requiring the corresponding evidence.
CHAPTER 22 OF 35 · Test limits and phase information
22. Inspect the infinite-length limit without changing the boundary silently
For the matched finite section, T = sech²L tends to zero as L increases, while R tends to one. The coefficient ratio B/A = i exp(−2L) also tends to zero.
The input then approaches the single-decaying-component behaviour of an infinite below-cutoff half-line: finite field near the entrance, reactive input and no transmitted average power at infinity.
This limiting result is consistent with the earlier zero-power calculation. The finite-section result did not contradict it; it used a second boundary at a finite distance, which required the second exponential.
A careful explanation states which boundary is moving in the limit. Holding a finite nonzero B while extending exp(x) to infinity would not represent the same matched-load family and would fail the bounded-field condition of the half-line problem.
CHAPTER 23 OF 35 · Test limits and phase information
23. Check the short-section limit
As L approaches zero in this stipulated matched-scale family, sech²L approaches one and tanh²L approaches zero. The middle section becomes vanishingly short, and its two interfaces coincide in the limit.
This does not say a finite below-cutoff guide supports an ordinary freely propagating middle mode. Its local mode equation remains evanescent; the boundary-value solution controls the finite-section transmission.
The limit also provides a check on approximations. A formula that predicts a transmitted fraction greater than one near L = 0 cannot be an exact passive power coefficient for this family.
For a learner, keep local propagation classification separate from the complete device’s transmission. A mode can be evanescent within one finite region while the full lossless system transfers power between propagating outer regions through the compatible field superposition.
CHAPTER 24 OF 35 · Test limits and phase information
24. Use a long-section approximation only in its own range
For large positive L, sech L is approximately 2exp(−L), so T is approximately 4exp(−2L). The exponential captures the leading length dependence, while the prefactor comes from the boundary matching.
Using only exp(−2L) misses that factor in this particular matched-scale family. Using 4exp(−2L) at very small L can give a value above one, revealing that the large-L approximation has been used outside its range.
An exponential-looking measurement therefore does not by itself identify a bulk dissipative loss coefficient. A below-cutoff finite section can produce exponential length dependence in transmission while the missing power is reflected.
The existing coherent-interface guide analyses a different propagating-section interference model. Both examples make interface assumptions visible, but their field equations and length dependencies should not be interchanged.
CHAPTER 25 OF 35 · Test limits and phase information
25. Translate the coordinate origin and preserve power
Move the origin to x = x₀ and use a new coordinate ξ = x − x₀. The same electric field has coefficients A’ = Aexp(−x₀) and B’ = Bexp(x₀).
Their product A’*B’ equals A*B because the two real scale factors cancel. The conserved power Im(A*B) is therefore unchanged by the coordinate translation.
The numerical values assigned to the separate decreasing and increasing coefficients depend on the reference plane. They are not independently invariant measures of physical power.
This provides a simple reference check for calculations that compare fields reported at different planes. Transform the complete pair before interpreting its coefficients. A change in one coefficient after moving the origin need not describe a change in the physical field or in the power carried by that field.
CHAPTER 26 OF 35 · Test limits and phase information
26. Show how identical electric intensities can hide opposite power
Take A = 1 and B = b₀exp(iφ), with b₀ positive. The electric squared magnitude is exp(−2x) + b₀²exp(2x) + 2b₀cosφ.
The normalised axial power is b₀sinφ. Replacing φ by −φ preserves the entire electric squared-magnitude profile because cosine is unchanged, while reversing the power direction because sine changes sign.
For φ = π/2, the two profiles with B = ±ib₀ are identical in electric intensity and carry opposite flux. The negative-flux case requires compatible excitation or boundary conditions supplying power from the other direction.
This is an information counterexample, not a claim that a passive left-driven matched load can arbitrarily choose either sign. Once that load and input are known, they constrain the phase. Electric intensity alone omits information that the boundary model may supply separately.
CHAPTER 27 OF 35 · Test limits and phase information
27. Distinguish real power from reactive exchange
The complex product e h*/2 contains both a real and an imaginary part. The real part is the cycle-averaged axial power in our normalisation. The imaginary part records the quadrature relationship relevant to reactive field behaviour.
It should not be read as an additional stream of dissipated watts. Nor does a nonzero imaginary part imply that the real part must vanish in a finite loaded section.
The earlier single-component example had purely imaginary product. The two-component loaded example can have both parts, with constant real flux despite position-dependent fields.
Keeping the complex product until the final interpretation avoids two opposite mistakes: calling every decaying field an absorbing field, or calling every evanescent region incapable of participating in real power transfer when a second boundary and compatible superposition are present.
CHAPTER 28 OF 35 · Test limits and phase information
28. Recognise what actual dissipation would require
In the stated lossless middle region, average axial power is constant. If a new model includes electrical conductivity, dielectric loss or imperfect conducting walls, electromagnetic energy can be converted into heat.
The appropriate power balance then includes the dissipative terms and any other power paths. It cannot be recovered simply by assigning a heat-loss interpretation to the old electric squared-magnitude curve.
A lower transmitted fraction also needs a full port account. Reflection, radiation into another mode or loss can each reduce one measured output, and their contributions require suitable evidence.
The unequal-interface inverse guide constructs a different ambiguity between interface reflectivities and propagation loss. Its lesson complements this one: a restricted power observation may leave the internal explanation unresolved.
CHAPTER 29 OF 35 · Transfer and verify
29. Keep evanescent amplitude decay separate from message speed
The real exponential exp(−κz) describes spatial amplitude variation in the stated monochromatic below-cutoff region. It does not provide the phase progression of an ordinary travelling mode in that region.
A group-delay expression derived for an above-cutoff freely propagating branch should not be applied by replacing an imaginary quantity with its absolute value. A finite pulse through a loaded structure needs the complete frequency-dependent transfer problem.
Nothing in the present field and power calculation establishes faster-than-light signalling or a universal transit time. Its subject is the relation between local fields, boundaries, reflection and cycle-averaged power.
This boundary keeps the extension focused. The Waveguides root introduces dispersion, and its dedicated cutoff article tests timing-domain conditions. The current examples address an omitted power-accounting question without reusing a timing formula outside its supported branch.
CHAPTER 30 OF 35 · Transfer and verify
30. Solve an independently changed field and load problem
Let L = ln2 and A = 1. For the outgoing matched-scale load, B = i/4 and p = 1/4. At the right boundary e = h = (1 + i)/2, whose product gives the same p.
Change only the right boundary to h(L) = 0. Then B = 1/4, p = 0 and e(L) = 1. The electric field is larger at that boundary than in the previous case, yet delivered average power has fallen to zero.
The change is not a mysterious failure of energy conservation. It changed the termination and therefore the relative phase and coefficients of the finite-interval solution.
For transfer, ask which earlier quantities stay fixed: κ, length and A remain supplied, but incident and reflected amplitudes on the left generally change. Holding A fixed is not the same as holding the incident propagating amplitude fixed after changing the load.
CHAPTER 31 OF 35 · Transfer and verify
31. Decide what additional phase observation would resolve
If the complete complex electric amplitude is known as a function of position and the modal equation is justified, its spatial derivative supplies h = −ie’. The missing information is therefore not an unmeasurable extra quantity; it is information absent from electric intensity alone.
Write e(x) = M(x)exp[iφ(x)] where M is positive. Substituting into the conserved expression gives p = M²φ’/2. Both magnitude and spatial phase gradient contribute to the flux.
In the original A = 1, B = i exp(−2) example, φ = arctan[exp(−2)exp(2x)]. Its derivative is 2exp(−2)/[exp(−2x) + exp(−4)exp(2x)]. As electric squared magnitude decreases, this combined-field phase gradient increases, preserving p.
At x = 0 that derivative is about 0.2658, and at x = 1 it is one. Multiplying each by its corresponding M²/2 reproduces exp(−2).
This spatial phase belongs to the superposed finite-region field. It should not be relabelled as the real propagation constant of an isolated freely travelling middle mode. The local field representation and the classification of its separate components remain different concepts.
For an evidence question, the distinction suggests a purposeful next measurement: supported phase-sensitive field information can distinguish alternatives left identical by more intensity samples. The required instrument model, spatial resolution and phase reference would still need justification before applying that reasoning to an actual device.
CHAPTER 32 OF 35 · Transfer and verify
32. Audit the field, boundary and power accounts independently
A useful calculation record names the time convention, mode, coordinate scale, loss assumptions, modal reference scales and input and output boundaries. It distinguishes electric intensity from incident, reflected, transmitted and net axial power.
Check e” = e and h = −ie’. Expand the power product, verify its derivative vanishes, and substitute the boundary coefficients into both the input and output conditions.
Independently solve the matched-boundary linear equations and compare T + R with one. For the changed termination, verify zero real power without assuming the electric field must vanish everywhere.
These tests establish consistency of the stipulated model. They do not validate a particular waveguide or detector. The modal-aperture guide develops another distinct question about phase and incomplete spatial observations using its own explicitly limited scalar reading model.
CHAPTER 33 OF 35 · Transfer and verify
33. Choose a learner route that preserves the field pair
For repair, begin with the cycle average of two sinusoids and the zero real power of the single decaying component. Explain what electric magnitude leaves unknown.
For stabilisation, reconstruct the finite two-component example and compare its falling electric squared magnitude with its constant p. Then derive the matched-load transmission and reflection fractions.
For extension, use the coordinate translation, opposite-flux intensity pair, inverse length and changed termination. Identify which extra boundary information removes each ambiguity.
Independent practice should change the measurement or the load. Supply electric intensity without magnetic phase, or replace the outgoing condition with a reactive boundary. A stable explanation begins by rebuilding the compatible field pair rather than assigning power to a familiar-looking exponential by appearance.
Does electric amplitude decay prove dissipated power?
No. A single lossless evanescent component can decay while carrying zero average power along its decay direction.
Can a finite evanescent region carry real axial power?
Yes, under compatible boundaries its two-component superposition can have nonzero constant real flux.
Is a low transmitted fraction necessarily absorption?
No. In the matched lossless example the remainder is reflected, with T + R = 1.
Does a growing exponential imply active gain?
No. It can be a bounded component required by the far boundary of a finite interval.
Can electric intensity alone identify power direction?
Not generally. The opposite-phase example preserves its whole intensity profile while reversing flux, subject to compatible boundary excitation.
CHAPTER 35 OF 35 · Transfer and verify
35. Continue through the verified waveguide routes
Return to Waveguides for modes, cutoff and propagation context. The four linked specialists address modal aperture readings, above-cutoff delay uncertainty, coherent propagating-section interfaces and the unequal-interface inverse problem. The present work adds the explicit two-evanescent-component power account.
Primary field context is MIT OpenCourseWare’s Electromagnetics and Applications, Chapter 9, checked 26 October 2026. Its guided-wave section explains below-cutoff fields and the zero real axial power of an isolated evanescent component. The finite loaded-section examples, normalised amplitudes, interference calculations and inverse checks here are independently constructed educational derivations. No source figures or measured device data are reproduced.
Continue through the Education parent and How X Works when another subject asks whether a local magnitude records the conserved flow or the mechanism that removes it.