I An interesting question from Veritasium on YouTube

[Baluncore]

Is there a simple formula for that?

Two wire, open line; Zo = ( 120 / Er ) · LOGe( 2 · D / d )
Where; D is separation of centres; d is diameter of the wires; Er = 1, relative permittivity for air.

https://en.wikipedia.org/wiki/Transmission_line#Twin-lead
https://en.wikipedia.org/wiki/Twin-lead#Characteristic_impedance

 

[cmb]

You do not understand transmission lines. They are broad-band up to a certain cutoff frequency in the ...

A two-wire line with 1 m between the wires will have an impedance between about 100 ohm and 1000 ohm, depending on the diameter of the wires. My model used 100 ohms for the lines and for the bulb.

You are assuming that this is a transmission line and I cannot comprehend that.

If you are arguing that two diverging cables running 180 degrees in opposite direction away from an electrical source is a transmission line, then for sure I don't understand that.

It is two lines diverging away from a dipole feed, it is a dipole antenna.

There is only one singular point on each leg of the connections which is equidistant from the source and 1m from each other. The rest are 'something else'.

The electrical source from the battery would NOT propagate down the lines on which the bulb is situated until a propagating EM wave has bridged the gap. The electric potentials on the other side would then be shorted together across the bulb, again wholly dependent on the impedance of the bulb. If the bulb was a dead short (let's arbitrarily use a 1megaAmp bulb) then clearly the leg of the circuit running in parallel 1m away would be a metal rod for all intent and purpose.

IF the battery were located at one narrow end of the circuit and the bulb on the other narrow end, with two parallel lines running between them, then I'd go with a description of a transmission line.

But it is specifically not that.

You call the circuit a transmission line and I say it is clearly not and I am really struggling to understand why you call it that. One feed wire goes one way, the other 180 degrees in the opposite direction. Transmission lines are characterised by conductors running in parallel away from the source.

There is no consistent distance between the one side of the battery to the other. The cable starts out 'very close' and diverges to 1 light year away. I'm unable to see a transmission line here. I say it will behave as a dipole antenna with a dipole receiver 1 m away, and I think it is self-evident it is 'that'. But I am not going to argue with you, I am just going to say again that we disagree and would prefer to leave it at that, for others to form an opinion/solution.

A transmission line consists of a source, a line of constant impedance and a load at the far end of it. This is clearly not 'that'.

Transmission lines work by the energy propagating between and parallel to the conductors. The only conductors that the energy propagates 'between' and parallel to (in the first few months) are two 1 metre legs of wire separated by one light year.

 

[Baluncore]

It is two lines diverging away from a point, it is a dipole antenna.

You call the circuit a transmission line and I say it is clearly not and I am really struggling to understand why you call it that. One feed wire goes one way, the other 180 degrees in the opposite direction. Transmission lines are characterised by conductors running in parallel away from the source.

Two long wires, 1 metre apart, make one two-wire transmission line.
One two-wire transmission line goes one way.
The other transmission line goes in the opposite direction.

 

[cmb]

Two long wires, 1 metre apart, make one two-wire transmission line.
One two-wire transmission line goes one way.
The other transmission line goes in the opposite direction.

What is the load these transmission lines go to?

What is the source placed across these two-wire transmission lines?

 

[Baluncore]

What is the load these transmission lines go to?

The short circuit at the far end of each.

 

[Dale]

Oh, that is what makes it reflect back down the line?

 

[ergospherical]

Is there a simple formula for that?

As a crude analysis from the transmission line equations, one finds solutions $V = V_0 \mathrm{exp}[-i(kz-\omega t)]$ and $I = I_0 \mathrm{exp}[-i(kz-\omega t)]$ with $kV = \omega LI$, and the characteristic impedance of a transmission line is defined $Z \equiv V/I = \sqrt{L/C}$. It remains to determine the capacitance and inductance (per unit length) of two parallel wires of radius $R$ separated by a distance $d \gg R$. For the inductance: the magnetic field of the first wire wire is $B = \dfrac{\mu_0 I}{2\pi r}$ therefore the flux per unit current through the relevant rectangular plane is\begin{align*}
\dfrac{\Phi}{I} = \dfrac{\mu_0 }{2\pi} \int_R^{d-R} \dfrac{dr}{r} = \dfrac{\mu_0 }{2\pi} \mathrm{log}\left( \dfrac{d}{R} \right)
\end{align*}The self-inductance is double this value (each wire contributes a flux), i.e. $L = \dfrac{\mu_0 }{\pi} \mathrm{ln}\left( \dfrac{d}{R} \right)$. For the capacitance: consider two line charges of density $\pm \lambda$, then the field between them is\begin{align*}
E = \dfrac{\lambda}{2\pi \epsilon_0 r} + \dfrac{\lambda}{2\pi \epsilon_0 (d-r)}

\implies V = \int_R^{d-R} E dr &= \dfrac{\lambda}{2\pi \epsilon_0} \int_R^{d-R} \dfrac{1}{r} + \dfrac{1}{d-r} \\
&= \dfrac{\lambda}{\pi \epsilon_0} \mathrm{log}\left( \dfrac{d-R}{R} \right)
\end{align*}which means that $C = \dfrac{\lambda}{V} = \dfrac{\pi \epsilon_0}{\mathrm{log}((d-R)/R)}$. Finally you have for the impedance $Z \sim \dfrac{Z_0 \mathrm{log}(d/R)}{\pi}$ so long as $d \gg R$ and where $Z_0 \sim 377 \ \Omega$ is the impedance of free space.

 

[cmb]

The short circuit at the far end of each.

What's the significance of the bulb then? What part does that play, how will its impedance affect the impedance the source experiences?

 

[Baluncore]

For the first year the impedance of the input end of the transmission lines will appear to be Zo. So the impedance presented to the battery when the switch closes is; Z = Zo + Rbulb + Zo.
Ohms law gives us the current through the bulb. I = V / Z.

After one year the lines suddenly look like short circuits. I = V / Rbulb.

 

[Baluncore]

Oh, that is what makes it reflect back down the line?

I don't need to explain why conservation of energy is important. It is all here.
https://en.wikipedia.org/wiki/Time-domain_reflectometer#Explanation

 

[nsaspook]

What's the big deal? 25:30 on the video
The results are the same using CA Transmission theory and Field theory.

As someone here said once.

Circuit theory describes how much energy is transferred, but it never makes any claim about where energy is located within a circuit element nor where energy crosses a lumped element's boundary. There is no conflict here because circuit theory makes no claim on the question.

 

[Dale]

Normal circuit theory isn’t applicable. The circuit violates the small circuit approximation.

 

[nsaspook]

Normal circuit theory isn’t applicable. The circuit violates the small circuit approximation.

Sure, that's why a engineering transmission line theory (telegrapher's equations ) lumped model is used to solve the problem. The physical interpretation (how energy is transferred) of these solutions seems to be the issue here for some reason.

 

[Interested_observer]

The physical interpretation (how energy is transferred) of these solutions seems to be the issue here for some reason.

Isn't that the WHOLE reason for science?

 

[Baluncore]

The physical interpretation (how energy is transferred) of these solutions seems to be the issue here for some reason.

That is because the problem is deliberately presented to trap the unwary and inexperienced.
1. The transmission line components are drawn as normal wires on the diagram.
2. Some people do not yet understand transmission lines.
3. Two-terminal transmission-line inputs are twice used as series components in the bulb circuit.
4. Instead of “bulb will glow” it should ask when “a circuit current will flow in the load”.
5. The circuit has no relative values, so every analysis can be different. What is the;
a. Wire diameter, or Zo of the transmission line?
b. Battery voltage? Globe voltage?
c. Globe power, or cold/hot resistance relative to Zo?

 

[nsaspook]

Isn't that the WHOLE reason for science?

Sure, but how energy is transferred has been an answered question in engineering and science for more than a century.

 

[Swamp Thing]

With engineers and physicists, it becomes second nature to pick a model that suits the regime that one is currently looking at, knowing all the while that any contradictions between those models are not something to worry about.

When addressing non-specialists, it's a kind of ethical responsibility not to sow confusion at first, only to then seek credit for rescuing the audience from that confusion. Maybe Veritasium has sailed a bit close to the wind in this video.

 

[nsaspook]

With engineers and physicists, it becomes second nature to pick a model that suits the regime that one is currently looking at, knowing all the while that any contradictions between those models are not something to worry about.

When addressing non-specialists, it's a kind of ethical responsibility not to sow confusion at first, only to then seek credit for rescuing the audience from that confusion. Maybe Veritasium has sailed a bit close to the wind in this video.

The Veritasium video IMO is fine for the media audience it's directed to. The electrical misconceptions in the general public from grade-school level simplifications are not Veritasium's fault.

https://www.abc.net.au/science/articles/2014/02/05/3937083.htm

 

[OCR]

Lol. . . Derek Muller's Veritasium video about electric energy not traveling inside wires raised some questions. . .

.

 

[cmb]

You do not understand transmission lines.

I have been thinking more about this and I now completely agree with you about your transmission line model, and your previous circuit. Apologies if it sounded like I doubted you personally, was not the intent and I now see the light.

I still feel that there are different relevant models that could be considered (like it being a dipole or transformer) but it's irrelevant as all the answers should come out the same.

In regards suggesting I don't understand transmission lines, you're partially right because the one thing I did not understand in the first instance was a relationship regarding transmission line capacitance that I could see this would imply, that initially I had been very reluctant to believe, but now accept it.

The implication is this;
As you say, prior to the conductors being charged (forming an electric field that can carry the power 'directly'), the battery will see a load of Z+ B + Z, where Z is the transmission line impedance and B the bulb impedance.

The problem I struggled with was the capacitance of the transmission line. For simplicity I will alter the question and talk here only about what happens when a battery is first placed across both lines of a transmission line (this gets rid of the bulb and splitting the voltage of the battery).

Taking a transmission line, let's now say the propagation velocity of transmission is W (m/s) and battery voltage V (Volts), and capacitance per unit length C' being units of F/m.

The power required to charge up the transmission line is therefore (1/2)*C'*W*V^2, while the power out of the battery is I^2*Z.

So (1/2).C'.W.V^2 = I^2.Z
therefore (1/2)*C'.W.Z = 1, or W = 2/(C'.Z)

This is the thing I could see coming but struggled to believe in the first instance. I felt sure the voltage should come into it, capacitance being a squared term of voltage and power being linear, I couldn't see in my head how those would square up if I wrote out equations.

I now accept that the propagation velocity of the electrical energy (charging up the transmission line), the transmission line's capacitance and its impedance, are in a strict relationship.

Now that I think more about that it seems sort of inevitable in hindsight. I just didn't buy it the first time around. So in that respect, yes, there was indeed something more I needed to understand about transmission lines.

Sorry for taking my time getting there. I don't know if that is a 'known' relationship, I expect it is, but I was not aware of it.

It means the maximum capacitance of a 50 ohm coax is 133pF/m, I would not have suspected there to be such limits on such coax. Make it smaller and the capacitance keeps going up, is what I assumed? I guess not?

 

[OCR]

I now see the light.

Lol. . . .
.

 

[idea2000]

1. If the wires were insulated or shielded, what would happen? How would the energy propagate to the bulb?
2. If someone broke the wire at the point closest to the Sun, how long would the signal take to get to the light bulb?
3. If there were two light bulbs, one at the point closest to the Sun, and one at the spot which is one meter away from the battery, which one would light up first?
4. If the wires were shielded, would the energy take longer to get to the bulb? So, would the bulb take longer to light up?
5. How long does the bulb take to ramp up to full brightness, after 1/c?

 

[Baluncore]

1. If the wires were insulated or shielded, what would happen? How would the energy propagate to the bulb?

Insulated or shielded how? So long as there were still two transmission lines it would still work. The lines could even be a twisted pair, inside a conductive pipe. A solid insulation will cost more than space. The lights in your house still work with insulated wire, so long as the wire was striped where there are connections.

2. If someone broke the wire at the point closest to the Sun, how long would the signal take to get to the light bulb?

1/c. The battery and bulb would not 'know' the line was cut until a reflection got back from the cut. That reflected counter current would cancel the initial current, then turn off the bulb suddenly after maybe 15 minutes.

3. If there were two light bulbs, one at the point closest to the Sun, and one at the spot which is one meter away from the battery, which one would light up first?

The one that is one metre from the battery.

4. If the wires were shielded, would the energy take longer to get to the bulb? So, would the bulb take longer to light up?

No. A metal shield would lower the line impedance, but it would not slow the signal. On the other hand, insulation would slow down the signal because the dielectric constant is greater than space.

5. How long does the bulb take to ramp up to full brightness, after 1/c?

The voltage and current through the bulb take 1/c to stabilise. Then the thermal mass of the filament gradually heats up to the point where it radiates light energy. Maybe 0.1 second.
You can experiment with that. Turn off a filament bulb. Face the bulb. Then shut your eyes just as you switch it on. If you get the timing right, you will see the filament start to glow. You can reverse the process to see the filament dim. Use a low power filament bulb and take care.

 

[cmb]

1. If the wires were insulated or shielded, what would happen? How would the energy propagate to the bulb?

Insulation - All else being equal; the lines would 'charge up' more slowly so the whole circuit would 'close' later (when the circuit finally 'completes' and passes the whole voltage to the bulb). The addition of a dielectric will increase the capacitance of the transmission line, thus slow the propagation.

The impedance of the line might change, depends on geometry and material of insulation, and if so that would affect the bulb brightness when it turns on when the switch is closed.

Given the geometry, almost imperceptibly as a thin dielectric

By way of example; the propagation velocity in a solid PE coax is around 0.6c, whereas it is 0.8c for a foamed dielectric. In a coax that was 'just vacuum' between inner and outer, it'd be at 'c'.

 

[A.T.]

1. If the wires were insulated or shielded, what would happen? How would the energy propagate to the bulb?

Insulated or shielded how? So long as there were still two transmission lines it would still work. The lines could even be a twisted pair, inside a conductive pipe.

What about a coaxial cable? The Wikipedia article seems to suggest that all the power flows between the outer & inner conductor:

https://en.wikipedia.org/wiki/Poynting_vector#Example:_Power_flow_in_a_coaxial_cable

...the power given by integrating the Poynting vector over a cross section of the coax is exactly equal to the product of voltage and current as one would have computed for the power delivered using basic laws of electricity.

 

[Baluncore]

What about a coaxial cable? The Wikipedia article seems to suggest that all the power flows between the outer & inner conductor:

That is true, what about it? It does not matter if the wave is guided by two wires through space or by two surfaces within a closed space. The inside and the outside of the braid are separate, and not connected through the holes. The external voltage wave on the outside of the coax braid has no fixed impedance, and will be quickly limited by inductance. Only the internal wave will continue because it is following a balanced ladder network.

 

[A.T.]

What about a coaxial cable? The Wikipedia article seems to suggest that all the power flows between the outer & inner conductor:
https://en.wikipedia.org/wiki/Poynting_vector#Example:_Power_flow_in_a_coaxial_cable

That is true, what about it?

Would the bulb still light up after 1m/c if instead of two wires, a single coaxial cable was used (laid out like one of the two wires in the Veritasium experiment)?

 

[Baluncore]

Would the bulb still light up after 1m/c if instead of two wires, a single coaxial cable was used (laid out like one of the two wires in the Veritasium experiment)?

The coaxial cable has two terminals at the near end. The two-wire transmission line, composed of two wires 1 m apart, also had two terminals, one on each wire.
If both terminals are connected, it will still work with coax.

 

[A.T.]

If both terminals are connected, it will still work with coax.

So it will still take only 1m/c for the bulb to light up, even when connected via one 1 light-second long coaxial cable?

Is the Wikipedia article wrong in suggesting that the Poynting vector outside of a coaxial cable is zero?
https://en.wikipedia.org/wiki/Poynting_vector#Example:_Power_flow_in_a_coaxial_cable

Outside the entire coaxial cable the magnetic field is identically zero since paths in this region enclose a net current of zero (+I in the center conductor and −I in the outer conductor), and again the electric field is zero there anyway.

Or does "Poynting vector is zero outside of a coaxial cable" apply only in steady state, and not when the current builds up?

 

[cjl]

I feel like there's poor communication going on here. At least the way I read it, Baluncore is assuming that the coax cable is hooked up using the ring connector and the center connector to give the two connections, one hooked to the lightbulb and one hooked to the switch. In this case, yes, it will act as a transmission line with its characteristic impedance.

It seems to me that A.T.'s question though is about using coax in a more conventional manner, with the ring grounded and using the center wire to connect to both the bulb and the switch (which, necessarily, means you'd see two parallel coax cables running off into the distance if you were sitting at the location of the switch. In this case, no, no current would flow through the bulb until the signal has propagated all the way around the circuit.

On a separate note, this is wrong:

The voltage and current through the bulb take 1/c to stabilise. Then the thermal mass of the filament gradually heats up to the point where it radiates light energy. Maybe 0.1 second.

The bulb won't achieve full brightness until at least the time it takes light to go out to the end of the circuit and come back, possibly longer if there's an impedance mismatch that causes several reflections before steady state. Steady state, the battery doesn't see the transmission line impedance, it only sees the DC resistance of the bulb and the wire.

When you first turn it on, the battery sees the transmission line impedance and bulb resistance in series, but you get a substantial ramp up in current after the field propagates fully around the circuit and reaches the bulb, at which point the battery no longer sees the wire as a transmission line. This is very obvious in the simulated behavior seen in this video for example (simulation results start just after 6 minutes in, but the setup is interesting too).

How much brightness you'll see originally vs after the field has fully settled to steady state depends of course on several factors, but you'll always see an increase in current after the field has propagated fully around the wire.