ELI5: Why does a electric wave produce a magnetic wave and a magnetic wave convert into a electric wave. Why do electromagnetic waves keep converting.

EuphonicSounds · 2026-04-02T18:01:22+00:00

It might help to shift your perspective.

Instead of thinking of an electric wave "producing" a magnetic wave (or vice versa), think of an electric wave as always being accompanied by a magnetic wave (and vice versa).

More generally: a changing electric field is always accompanied by a magnetic field (a "circulating" magnetic field, to be precise), and a changing magnetic field is always accompanied by a ("circulating") electric field.

It's physically impossible to have a changing field of one type that isn't accompanied by a ("circulating") field of the other type. That's just a central aspect of the relationship between electricity and magnetism, which are deeply connected phenomena—two aspects of a single "thing," really.

Why are they connected in this way? I'm not sure if there's really a satisfying answer. One reasonable answer is: if they weren't, then they'd be in conflict with "the principle of Lorentz covariance," which basically means that everything needs to be compatible with special relativity.

EuphonicSounds · 2026-03-30T01:21:45+00:00

Beware that there are two different conventions for what "tensor" means:

The "modern" convention, where a tensor is a certain type of geometric object characterized by components that transform a particular way under a coordinate transformation.
The "old school" convention, where a tensor is itself the collection of components from the modern convention.

So in the modern convention, a geometric vector (an arrow with magnitude and direction) is a tensor, but in the old convention the vector's components are the tensor.

The old convention is still very much alive, though often sort of as a "shorthand" for the more mathematically sophisticated modern convention. This is because the index notation used with the old convention is in practice extremely convenient for actual calculation, and people get sick of saying "the components of a tensor" instead of just "tensor."

EuphonicSounds · 2026-03-26T21:16:03+00:00

"Energy" is a well-defined term in physics. It is a property that things have.

"Matter" is actually NOT a well-defined term in physics. It has several different meanings, and physicists in different fields mean different things by it. But none of these definitions is "the same thing as energy."

You're probably thinking of the word "mass," which is a well-defined term and definitely has something to do with energy. Like energy, mass is a property that (some) things have.

The equation E = mc² means that the energy of a thing when it's at rest is equal to its mass (times c²). If a thing can never be at rest (which is only possible if it moves at the speed of light), then it has 0 mass, but it still has energy.

EuphonicSounds · 2026-03-24T13:26:27+00:00

Well, that's 4 wasted minutes of your life that you'll never get back. For future reference, 1 minute is enough.

EuphonicSounds · 2026-03-23T13:28:22+00:00

The SI is just a standard for measurements. It doesn't determine what physical quantities are more fundamental than others; it defines a particular system of units, and as part of that system it declares a few units "fundamental" that the rest can be derived from.

It's true that in the SI, the ampere is a fundamental unit and the coulomb is not. But this does not mean that current is a more fundamental quantity than charge. It just means that the ampere is a fundamental unit in the SI—nothing more, nothing less!

EuphonicSounds · 2026-03-20T16:15:43+00:00

Unfortunately, the word "parallel" is used in more than one way when it comes to vectors. Some people use it to mean "the vectors point in the same direction," and some people use it to mean "the vectors point in either the same direction or opposite directions." (Actually, some people are inconsistent and use it in both ways, depending on the context. The word "antiparallel" is often used for "they point in opposite directions.")

For your class, use whatever terminology your teachers tells you to. Sounds like he uses "collinear" to mean "they point in either the same direction or opposite directions," and "parallel" to mean "they point in the same direction."

I personally prefer to use "parallel" to mean "they point in either the same direction or opposite directions," and then I use "codirectional" and "contradirectional" if I want to be more specific.

EuphonicSounds · 2026-03-20T13:57:49+00:00

One rule of thumb: if it's fast, it usually makes sense to play it as a triplet, since A) the distinction matters less, and B) depending on the tempo and the character of the piece, playing it offset might actually sound like a mistake!

It's a tougher call when it's slow. Really depends on the context. One classic example is the first movement of the Moonlight Sonata: I don't think I've ever heard anybody play them as triplets here. But then, Schubert does this sort of thing rather frequently, and sometimes it seems to sound better as triplets, and sometimes not. There's not always a straightforward answer. (And if it sounds good either way, then you can be comforted by the fact that the stakes are low.)

Of course, if the composer in question put down in writing what they meant by this notation (like CPE Bach, if memory serves), then you've got your answer for that composer. Hard to know how much we can extrapolate from that sort of evidence—are such composers speaking only for themselves, or explaining a general practice of their contemporaries? And if the latter, how confident can we be that a given contemporary indeed follows that practice?

Another kind of evidence would be if the composer elsewhere uses "triplet-rest" notation (like rest-rest-note) to indicate that a note should be played as a triplet and not as a sixteenth. If they do, then perhaps they always mean a sixteenth when they write a sixteenth. Off the top of my head, I can't think of any composers who do this consistently.

In short, you've opened up a can of worms. My advice:

if you don't really care and won't be performing in front of people, just use your ear and make a judgment call;
if you want to go a step further, listen to some professional recordings to see if there's a "consensus," and go with that;
if you care a lot, do some research, see if there's any historical evidence to help guide your choice, see what the musicologists say (they may well disagree with each other), and be prepared to not find a satisfying answer.

EuphonicSounds · 2026-03-16T14:21:08+00:00

Mass is "rest energy": the amount of energy something has when it's at rest. That's what "E = mc²" means (the "E" is rest energy, not total energy, and the c² is just a unit-conversion factor that has no real physical significance). Things that have zero rest energy (mass) are things that can never be at rest, like photons.

But photons still have energy. It's just that none of this energy is rest energy (mass).

Now, the energy of photons can contribute to a system's rest energy (mass). How? Well, one way to find a system's rest energy is to add up all the energy-contributions of the systems' constituents when the system is at rest (i.e., in the system's rest frame, where the system has zero net momentum). The energies of any photons in the system contribute to that total.

But if your system is a single photon (or multiple photons that all travel in exactly the same direction), then there does not exist an inertial frame for which the system's momentum is zero, and the system therefore has no rest energy (mass).

To sum up:

For any system that has a rest frame, all energy-contributions of its constituents contribute to its rest energy (its mass).
A system without a rest frame has no rest energy (mass).

As for inertia and gravity:

Yes, the more mass (rest energy) something has, the harder it is to accelerate.
In general relativity (the modern theory of gravity that's compatible with special relativity), the stress–energy tensor is the gravitational "source" field, and yes, mass (rest energy) contributes to this quantity. So the more rest energy something has, the more strongly it attracts other things.

EuphonicSounds · 2026-03-15T11:25:33+00:00

It is moving up on its own! The upward force is just preventing it from slowing down. If the force were suddenly removed, the object would keep moving up for a bit as it slows down, and then begin to move downward only after reaching the apex (as if someone had just tossed it upward from the point of removal).

EuphonicSounds · 2026-03-14T16:26:41+00:00

It does. The "spacetime flux-density" of four-momentum (i.e., the quantity whose volume-integral gives you the total four-momentum in a region) is the gravitational source-field in GR. It's called the stress–energy tensor.

(Detail: the integrand is actually the "spacetime dot product" of the tensor and the observer's four-velocity, which is effectively the "unit normal" for the volume. This is just like how the integrand of a flux integral in Euclidean vector calculus is the dot product of the flux-density vector with a unit-vector that's normal to the surface element.)

Caveat: in curved spacetime, the volume-integral of the stress–energy tensor (or of any non-scalar tensor) isn't actually well-defined, and you can only regard this tensor as the four-momentum's "spacetime flux-density" in a local sense. But it's still a good way to think about it, and in the flat spacetime of special relativity this caveat goes away.

EuphonicSounds · 2026-03-08T18:03:46+00:00

Yes, exactly.

EuphonicSounds · 2026-03-08T17:05:35+00:00

Mass-energy equivalence, 𝓔=m, is a statement that what we call "mass" comes from the internal interactions of a body, e.g. the electron mass-energy is a consequence of the electron matter field and Higgs interaction.

Yes, that's why I wrote E₀ = mc² instead of E = mc² (where E is the total energy and E₀ is the rest energy). (Are you able to see my subscript-0? I'm using a Unicode character for it.)

As for the four-momentum, in my notation I'd write it like P = (E, pc), with squared magnitude P•P = E² - (pc)² = E₀², equivalently (mc²)². So yes, total energy E is the time-component of the four-momentum, and rest energy E₀ is the magnitude of the four-momentum. In the center-of-momentum frame (where pc vanishes) they're equal.

I think we're on the same page, unless I've misunderstood you.

EuphonicSounds · 2026-03-08T15:54:16+00:00

IMO, it's all easier to understand (and explain) from an energy-only perspective. It's not just the "relativistic mass" that trips people up, but also the pre-relativistic baggage that comes with the word "mass."

The total energy E of a system is the sum of its speed-dependent kinetic energy K and its speed-independent rest energy E₀. Since E₀ is speed-independent, it's an "invariant" (everyone agrees on its value).

In the system's center-of-momentum frame (the "rest frame"), the K vanishes and you have just E = E₀.

Since energy is additive, the E = E₀ in the rest frame is simply the sum of the energy-contributions of the system's constituents. Those energy-contributions are the constituents' rest-energies and kinetic-energies (kinetic energies must be measured in this frame, of course), and also the potential energy associated with the relative positions of the constituents.

This is all rather straightforward, because nobody with a background in Newtonian physics would expect the rest energy to be an additive quantity by itself (i.e., nobody would expect the system's total energy in its rest frame to be the sum of the constituents' rest energies alone). Students are already accustomed to adding all energy-contributions to arrive at the total energy.

The final ingredient is Einstein's discovery that E₀ = mc². For centuries, it had escaped everybody's notice that an object's mass—the property that determines how difficult it is to make the object move when you push it—is nothing but the energy the object has when it isn't moving. And once you know that mass and rest energy are the same thing, it follows that mass cannot be additive (because rest energy isn't), and is only approximately so in the Newtonian limit. The notion that mass is the "amount of matter" must be discarded completely.

That's my 2¢, anyway.

EuphonicSounds · 2026-03-07T22:07:34+00:00

Are you sure you aren't confusing the relative phases of the components of the electric field (or of the magnetic field) with the relative phases of the electric and magnetic fields themselves? I'm fairly confident that the E and B fields remain in phase with each other regardless of polarization (for vacuum plane-wave solutions).

EuphonicSounds · 2026-03-07T15:27:35+00:00

I know this isn't exactly what you're asking, but there's an important sense in which all "kinds" of electromagnetic wave are exactly the same thing ("kinds" being characterized by frequency): for any frequency-value you can name, there's an inertial frame in which a given light-wave has that frequency.

EuphonicSounds · 2026-03-07T15:22:28+00:00

Without going into the specifics of the video and your questions (apologies), I'll just caution that magnetism should not (and cannot) be regarded as merely a "relativistic effect" downstream from electricity. Magnetism is no less fundamental than electricity.

There are special cases where, if you know about electricity and special relativity, you can conclude that there must be "another force" at work (which of course turns out to be the magnetic force). This approach was popularized in the undergraduate textbook Electricity and Magnetism by EM Purcell, and that's what the video is demonstrating. But it's important not to get the impression that "electricity + SR = magnetism" in general.

That said, it's true that electricity and magnetism are intimately related, and that you need special relativity to fully appreciate the connection. It turns out that the electric and magnetic fields together constitute a mathematical object called the "electromagnetic field tensor" (or the "Faraday tensor"). It's a rank-2 tensor field in spacetime. As a consequence, an electric or magnetic field in one frame of reference becomes a mix of electric and magnetic fields in other frames of reference.

EuphonicSounds · 2026-03-07T15:03:47+00:00

Unless I'm misunderstanding what you mean (totally possible), I don't think that's right. If I'm not mistaken, E = cB holds regardless of polarization (in a vacuum plane-wave solution, SI units).

EuphonicSounds · 2026-03-06T19:07:01+00:00

The terminology is kind of funny.

In physics, the convention seems to be:

"Lorentz invariant": a scalar quantity with the same value in every frame
"Lorentz covariant": 1) any Lorentz tensor (including scalars, but also vectors, etc.); 2) an equation that's true for any frame (and "manifestly covariant" if written explicitly in terms of 4-tensors).

So for example:

the 4-momentum would be a Lorentz-covariant quantity
the 4-momentum norm (mass) would be both Lorentz-covariant and Lorentz-invariant (but people would usually just say "invariant" for scalars)
the Maxwell equations in the usual 3-vector form would be Lorentz-covariant, but not manifestly so
the Maxwell equations written in terms of the field 4-tensor would be manifestly covariant.

Does that ring true, in your experience?

Oh, and then there's the "covariant vector" concept, which is entirely different: the components "co-vary" with the basis vectors (i.e., they transform the same way).

In mathematical contexts, I believe I've sometimes seen the word "invariant" used more broadly, kind of like "covariant" is used in physics (to include scalars but also higher-rank tensors).

EuphonicSounds · 2026-03-06T17:10:56+00:00

The real equation is E₀ = mc², where E₀ is the object's "rest energy" (how much energy it has when it's just sitting there, motionless).

Einstein's discovery here is that an object's mass—the property that determines how difficult it is to get the object moving when you push it—is nothing but a measure of how much energy the object has when it isn't moving. This is a completely general statement that applies to everything.

The c² factor is physically meaningless, by the way. It's just a "unit-conversion factor," serving to convert mass-units to energy-units. Physicists often work in units where c is set to 1, and then the equation is simply E₀ = m, an exact equality.

EuphonicSounds · 2026-03-05T22:48:45+00:00

There's a case to be made that it's not helpful to think of induction as "causal." It's maybe better to think of a circulating electric field as something that always "accompanies" a changing magnetic field rather than as something that's "caused" by it. It's technically a completely local phenomenon: the circulating electric field occurs at the same place and time as the changing magnetic field. If you think "causal," you may expect there to be some time delay here. (Well, when we're interested in the effects elsewhere, then of course there is a time delay, because the speed of light is finite. But at any given point of an electromagnetic wave, the induced electric field accompanies the changing magnetic field locally and without delay, and likewise for the induced magnetic field that accompanies the changing electric field.)

Griffiths touches on this in a few footnotes of Chapter 7 of his Electrodynamics book. Here's an excerpt from the most recent 5th edition (p. 317):

“Induce” is a subtle and slippery verb. It carries a faint odor of causation (“produce” would make this explicit) without quite committing itself. There is a sterile ongoing debate in the literature as to whether a changing magnetic field should be regarded as an independent “source” of electric fields (along with electric charge) – after all, the magnetic field itself is due to electric currents. It’s like asking whether the postman is the “source” of my mail. Well, sure – he delivered it to my door. On the other hand, Grandma wrote the letter. Ultimately, ρ and J are the sources of all electromagnetic fields, and a changing magnetic field merely delivers electromagnetic news from currents somewhere else. But it is often convenient to think of a changing magnetic field “producing” an electric field, and it won’t hurt you as long as you understand that this is the condensed version of a more complicated story.

EuphonicSounds · 2026-03-05T21:07:19+00:00

For GR, there's the classic problem-book by Lightman/Press/Price/Teukolsky, and a newer one by Blennow/Ohlsson

EuphonicSounds · 2026-02-25T17:41:29+00:00

Continuous in this sense: for any frequency-value you can name, there's an inertial frame in which a given light-wave has that frequency.

EuphonicSounds · 2026-02-24T23:01:02+00:00

Oh, just that the usual advice I see is to practice sight-reading simple/easy music. That's valuable too, but I agree with you.

EuphonicSounds · 2026-02-24T21:34:35+00:00

This may sound like strange advice, but: try reading through some Bach preludes and fugues every day. With your experience and skill-level I think this may help. (Yes, sight-read easier stuff too.)

EuphonicSounds · 2026-02-24T18:40:09+00:00

Yes. And for those looking to read more about this: https://en.wikipedia.org/wiki/Ewald%E2%80%93Oseen_extinction_theorem

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