I Do Electric field lines propagate by themselves away from a charge?

[gionole]

Energy travels in the EM field wherever $\vec E \times \vec B$ is nonzero. So if you understand a rope wave by focusing on energy traveling, then you should do the same for EM.

Again, you understand by looking at energy.

What has that got to do with anything? The EM fields still transfer energy just like the rope and the sound. The problem is that you abandoned a sensible approach for understanding waves, looking at the energy, and instead focused on the E field lines. That is a useless approach here.

And now that you have found it is not easier, why do you persist in going down this unfruitful path?

Yes. This is an EM wave.

Field lines do not have a speed.

Field lines don’t have an age. There are no old or new field lines.

I don’t understand that drawing. But I don’t need you to fix it. I prefer to abandon this whole approach.

The E field lines for an EM wave are indeed not straight.

And once you succeed in that, what have you gained? Do you think this produces an understanding of EM waves? Maybe it does. I haven’t seen anyone try this approach, so I cannot say that it will definitely fail. It just seems like there has been no progress so far using this approach.

Why not take a break from it for a while. Either pursue the energy approach or the potentials approach for a bit. See if a different approach helps. If it does, great, if not then you can always come back to this approach.

Absolutely agree with it. Though a couple of statements from my side.

1. When you say field lines dont have a speed, I agree but it is still good to imagine they do. Otherwise when charge moves with constant speed, dont you have a doubt why field lines even trillion meters away from the charge automatically follow charge direction ? Somehow the field lines even very far away already know with what constant velocity charge is moving, so in a kind of analogy, you could say field lines have same speed as charge’s speed in charge’s direction and once acceleration happens, thats where they dont update instantly as this change needs to propagate.

2. Lets follow the wave description then. As it turns out, if I focus on waves as energy propagation, then I can tell what could be interesting here. So if charge got accelerated, E and B will start increasing but this will happen one point at a time. So this change will be gradually in space. When E changes at 10m away, after some small time, E will change 10.01m away so it is gradually. This is clear to me that E and B at each point increase/decrease in a sinusoidal shape. Though, here comes a tricky part. In sound wave, one particle gains kinetic energy and collides with another and transfering its energy. In EM wave, there is no collision, so how does the transfer of energy happen from two neighboring points in space ? If we imagine 2 points in space(neighbouring ones - x1 and x2), then E got increased at X1, and after so small time, E now got increased at x2. So somehow energy must be traveling from charge acceleration point to outwards. How does energy then travel in vacuum ? I hope I expressed the confusion.

True that I can forget about Kinks. The funny thing is I was even calling kink the curved shape where as you were calling kink only the straight line. So I think we were pretty much close to each other’s understanding. Thanks so much and I hope there is not much time until I fully get the logic.

 

[vanhees71]

The uniformly moving charge is a special case. There you have a pure Coulomb self-field in the rest frame of the particle, which only is Lorentz boosted to the frame, where the charge is moving with the given constant velocity. It's immediately clear that there is no radiation field in this case. Of course, also in this case you can also calculate the fields with the retarded propgator though it's much less convenient than just using the Lorentz boost. The corresponding calculation, which is most elegantly done within the manifestly covariant description can be found in Sect. 4.8 of

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

The calculation with the boost can be done most easily by introducing the four-velocity of the charge. In its rest frame you have $(u^{\prime \mu})=(1,0,0,0)$ and the four-potential is given by
$$(A^{ \prime \mu}(x')) = u^{\prime \mu} \frac{q}{4 \pi |\vec{x}'|} = u^{\prime \mu} \frac{q}{4 \pi \sqrt{(u' \cdot x')-x' \cdot x'}}.$$
This is in a manifestly covariant form, i.e., in the frame where the charge is goving with the constant velocity $\vec{v}$ you imply have to set $(u^{\mu})=\gamma (1,\vec{v})$. and write
$$A^{\mu}(x)=\frac{u^{\mu}}{4 \pi \sqrt{u \cdot x-x \cdot x}}.$$
The fields in this frame are then
$$\vec{E}=-\dot{\vec{A}}-\vec{\nabla} A^0, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$

 

[Dale]

When you say field lines dont have a speed, I agree but it is still good to imagine they do

I disagree that it is good. I see no evidence that it has helped you learn this so far.

Furthermore, you will have to unlearn the concept if you want to ever learn relativity. So if you want to continue that path then you will have to explore it without me, I will not assist or support you in what I view is a mistaken approach.

This is clear to me that E and B at each point increase/decrease in a sinusoidal shape. Though, here comes a tricky part. In sound wave, one particle gains kinetic energy and collides with another and transfering its energy. In EM wave, there is no collision, so how does the transfer of energy happen from two neighboring points in space ?

Good. If you understand that there is an E and B at each point then you can easily know the energy transfer.

The energy transfer in the E and B fields is given by $\vec S = \frac{1}{\mu_0} \vec E \times \vec B$. So at all points energy flows in a direction that is perpendicular to both E and B, it is proportional to both E and B, and it is proportional to the sin of the angle between them.

Note that this energy transfer applies for EM waves, but it is more general than that. This energy transfer is also the way that energy is transferred in standard circuits, including DC circuits. Any time you have a battery powering a light bulb, the energy is transferred from the battery to the light bulb using $\vec S$.

 

[gionole]

I don't know where field lines come into this. EM waves are propagating disturbances in the EM field and their measurable effects are an oscillation in the electric and magnetic field vectors of the EM field. That is, if we place a bunch of detectors in a row and cause an EM wave to pass by them the detectors will show the field vectors oscillating back and forth around some mean value. This is analogous to placing a bunch of buoys in water and watching as their height changes upon the passage of a water wave (aka a gravity wave). Or the change in air pressure as a sound wave passes.

A field line is a line made by connecting different points in the field such that the line is tangent to the vector at every point. That's it. This in itself has nothing to do with EM waves.

Well, I think I was even calling a "curved joinining field" between old and new lines as a kink. and you guys have only been calling `kink` the straight joining line` and thats where our misconception arised.

I disagree that it is good. I see no evidence that it has helped you learn this so far.

Furthermore, you will have to unlearn the concept if you want to ever learn relativity. So if you want to continue that path then you will have to explore it without me, I will not assist or support you in what I view is a mistaken approach.Good. If you understand that there is an E and B at each point then you can easily know the energy transfer.

The energy transfer in the E and B fields is given by $\vec S = \frac{1}{\mu_0} \vec E \times \vec B$. So at all points energy flows in a direction that is perpendicular to both E and B, it is proportional to both E and B, and it is proportional to the sin of the angle between them.

Note that this energy transfer applies for EM waves, but it is more general than that. This energy transfer is also the way that energy is transferred in standard circuits, including DC circuits. Any time you have a battery powering a light bulb, the energy is transferred from the battery to the light bulb using $\vec S$.

When you got battery, electrons moving from one point to another and it is easier to understand energy transfer.

Q1. What is interesting is that unless E and B change at each point gradually, we don't have this energy transfer. But note that charge moving with constant speed does not have EM wave. What I mean is that if charge moves constantly, at each point, E and B still change gradually. This is because even if charge moves constantly, it increases or decreases distance between each point on some axis. So you got changing E field. Which for sure is what happens for B as well. So even though E and B still change gradually with constant speed charge, we don't call it EM wave. So just saying changing E and B is what EM wave is seems wrong to me. This is why I was trying to understand my bad approach pathway. This all makes me believe that it is not E and B changing that produces energy, otherwise if it did, we would have EM wave for constant speed charge as well. Thoughts ?

Q2. Energy got produced at the point of charge acceleration. Since it travels, does E and B that change at each point help it move forward ? How do they help it ?

 

[Dale]

When you got battery, electrons moving from one point to another and it is easier to understand energy transfer.

Except that the energy transfer does not happen through the electrons. The energy goes from the battery to the light bulb in the fields, not the electrons.

The electrons are involved in changing energy from chemical to electromagnetic at the battery and in changing the energy from electromagnetic to thermal at the light bulb. They are not involved in moving the energy from the battery to the light bulb, except for the fact that the current produces the B field.

If you do not understand $\vec S = \frac{1}{\mu_0} \vec E \times \vec B$ then you do not understand the energy flow in a DC circuit. Once you do understand the energy flow in a DC circuit then you also understand it for an EM wave. They are the same.

What is interesting is that unless E and B change at each point gradually, we don't have this energy transfer

This is not true. Static E and B fields, such as those in a simple DC circuit as I mentioned above, have this same transfer. Anywhere that S is nonzero, energy is transferred, regardless of whether the field is static or not.

So even though E and B still change gradually with constant speed charge, we don't call it EM wave.

That is correct. The difference is where the energy goes. In a constant speed charge the energy goes forward and stays close to the charge. In an EM wave the energy goes out away from the charge. Similarly with a DC circuit. There is no wave because the energy stays close to the wires at all times.

 

[gionole]

@Dale

I think thats where my confusion is Dale and why I have been focusing this much time on "kinks".

Basically, I will mention couple of confusions from my side.

First: I don't know what you call "static E and B". If charge moves with constant speed, E still changes at every point in space due to distance. So If E changes, B changes at each point as well due to distance. So In what sense do you call something "static" if it still changes ?

Second: I think what I don't get is what you mean by: "In a constant speed charge the energy goes forward and stays close to the charge. In an EM wave the energy goes out away from the charge". If something moves constantly, that means no force acts on it since we got no acceleration. So what energy do you mean here that moves forward(in charge direction ?) ? and if it's energy, then we could say charge still emits radiation, but in one direction only, but where would this energy come from ?

 

[renormalize]

If charge moves with constant speed, E still changes at every point in space due to distance. So If E changes, B changes at each point as well due to distance. So In what sense do you call something "static" if it still changes ?

"Static" means something is independent of time. That something can still depend on position in space, like the $1/r^{2}$ dependence of a charge's E-field.

If something moves constantly, that means no force acts on it since we got no acceleration. So what energy do you mean here that moves forward(in charge direction ?)

Of course a charge in uniform motion carries energy that moves forward with it: it's called kinetic energy! For uniform motion, KE is conserved, so none of it is converted to radiated energy that moves away from the charge.

 

[gionole]

"Static" means something is independent of time. That something can still depend on position in space, like the $1/r^{2}$ dependence of a charge's E-field.

Of course a charge in uniform motion carries energy that moves forward with it: it's called kinetic energy! For uniform motion, KE is conserved, so none of it is converted to radiated energy that moves away from the charge.

1. And what's that something that's independent of time ? if charge moves with constant speed, E will change at observable point for sure, but why not dependent on time ? at `t=2`, E can be 10 and at `t=3`, E can be 12. So as time passes, charge moves more which changes E at each point. I might be asking the obvious, but with this logic, it's unclear.

2. ah, true. In that case, charge doesn't even emit energy in its own direction, it's just a carrier of energy. Seems like Newton's first law holds very true here. Correct ?

3. If 2 charges kind of collide, there'll always be at least 2 EM waves produced(one from each charge, since they both experienced acceleration). There could be more waves depending on whether the charges stay accelerated or deaccelerated. right ?

4. If charge was moving constant speed and then acceleration became 2m/s, but stayed 2, then this will continue producing EM waves as long as acceleration is non zero. right ? it's just won't be sinusoidal form, meaning E and B will only go in one direction(either increasing or decreasing) at each point. Correct ?

5. If electron had KE as `E1`. Then it got accelerated, and energy quickly became `E2` (E2>E1). As far as I understand some portion from E2 will be radiated in a way that => E2 - radiated energy > E1. Is this correct ?

6. What actually defines how much energy charge radiates ? I know you will say the more acceleration it got, the more energy it will radiate. True, but if you focus on a different view such as If charge got accelerated to E2, and we know by the formula, that it radiated the energy = `X`. Why `X` and not `X-2`. Whats the logical explanation that it radiated this very specific energy ?

 

[Dale]

I don't know what you call "static E and B".

A static E and B is one that does not change over time. For example, a DC circuit, like a battery, a light bulb, and two connecting wires.

If charge moves with constant speed, E still changes at every point in space due to distance.

Yes, an inertially moving charge is not a static E or B.

If something moves constantly, that means no force acts on it since we got no acceleration. So what energy do you mean here that moves forward(in charge direction ?) ?

The E field and the B field each contain potential energy. That potential energy moves forward with the charge.

then we could say charge still emits radiation

No. The energy stays localized around the charge. So it is not radiation. Not all energy flow is radiation.

where would this energy come from ?

That is what the Poynting theorem tells you. It tells you where the EM field energy comes from, where it goes, and how it gets there.

 

[Dale]

"Static" means something is independent of time. That something can still depend on position in space, like the $1/r^{2}$ dependence of a charge's E-field.

Of course a charge in uniform motion carries energy that moves forward with it: it's called kinetic energy! For uniform motion, KE is conserved, so none of it is converted to radiated energy that moves away from the charge.

While the charge does have KE, that is not the energy I am talking about. I am talking about the EM field energy density: $u=(E^2+B^2)/2$ with whatever constants are needed to make the units work

 

[renormalize]

While the charge does have KE, that is not the energy I am talking about. I am talking about the EM field energy density: $u=(E^2+B^2)/2$ with whatever constants are needed to make the units work

Yes, but doesn't the EM field around a moving charge contribute to its total kinetic energy due to the so-called "electromagnetic mass": https://www.feynmanlectures.caltech.edu/II_28.html ?

 

[Dale]

Yes, but doesn't the EM field around a moving charge contribute to its total kinetic energy due to the so-called "electromagnetic mass": https://www.feynmanlectures.caltech.edu/II_28.html ?

Yes, definitely. But although the mass from the field energy contributes to the kinetic energy, it is much larger than the kinetic energy. So again, you are not wrong that there is KE, but I was specifically referring to the EM field energy.

I am not correcting you, I am clarifying my own statement.

 

[vanhees71]

Except that the energy transfer does not happen through the electrons. The energy goes from the battery to the light bulb in the fields, not the electrons.

The electrons are involved in changing energy from chemical to electromagnetic at the battery and in changing the energy from electromagnetic to thermal at the light bulb. They are not involved in moving the energy from the battery to the light bulb, except for the fact that the current produces the B field.

If you do not understand $\vec S = \frac{1}{\mu_0} \vec E \times \vec B$ then you do not understand the energy flow in a DC circuit. Once you do understand the energy flow in a DC circuit then you also understand it for an EM wave. They are the same.

This is not true. Static E and B fields, such as those in a simple DC circuit as I mentioned above, have this same transfer. Anywhere that S is nonzero, energy is transferred, regardless of whether the field is static or not.

That is correct. The difference is where the energy goes. In a constant speed charge the energy goes forward and stays close to the charge. In an EM wave the energy goes out away from the charge. Similarly with a DC circuit. There is no wave because the energy stays close to the wires at all times.

That's very important and easy to observe when switchin on the light in your room. There are some meters distance between the switch and the light source, and the light turns on very quickly. As it turns out when you analyze what happens using Maxwell's equations is that the signal velocity is the speed of light. You can calculate this for simple geometries (e.g., a coax cable) even using the simplifying telegrapher's equation.

If it were the electrons transporting the energy from the source to the light bulb it would take minutes until you get light after switching it on, because the velocity of the electrons in the cable is a few mm/s (millimeters per second!).

 

[gionole]

Thanks so much everyone. I will study couple of things to get ready to better understand all this and will reply as soon as possible.