9 ms·

Have you read the slides cited in the description on YouTube, and found the error? There's both a theoretical argument (which I haven't followed), and lab resul

Have you read the slides cited in the description on YouTube, and found the error? There's both a theoretical argument (which I haven't followed), and lab results using a 15m (?) transmission line. Both seem to confirm the claim in the video. If they're wrong, I'd really like to understand where the error is. E&M can be a little tricky, even for experienced circuit designers.

sebzim4500 5y ago
The slides look correct to me but the video massively misrepresents what is actually going on. For example, you could cut the wire at both ends and the answer would not change, so the fact that there is a 'circuit' is irrelevant.
tsimionescu 5y ago
The video gives a wrong impression that most current is flowing along the Poynting vector. In fact, only a tiny amount of current is flowing directly, and the vast majority is flowing along the wires, as expected. The tiny amount flowing directly through the air is much better understood as an antenna effect, unrelated to regular current flow, and has nothing really to do with the Poynting vector.
- tom-thistime 5y ago
  I agree that current doesn't follow the Poynting vector. Power follows the Poynting vector. That's true for antennas and for transmission lines.
  - tsimionescu 5y ago
    Yes, but most power is in fact following the Poynting vector along the wires, not the direct Poynting vector between the battery and the bulb, as the video misleadingly suggests (but doesn't outright claim).
    - mannykannot 5y ago
      > Yes, but most power is in fact following the Poynting vector along the wires. Only in the steady state does the Poynting vector follow the wires all around the loop (a point made in the original video.) In the period of interest - well before the EM waves launched by closing the switch have reached the far ends of the transmission lines - it does not, yet during this period, the load is passing a non-trivial current (and therefore dissipating power) that is determined by the voltage of the battery, the resistance of the load, and the characteristic impedances of the two transmission lines. For example, for 24 awg superconducting wire (~0.5mm in diameter) the characteristic impedance for each of the transmission lines would be a little less than 1 kΩ, so with a 12 volt battery, a 2kΩ load would be passing about 3mA and dissipating about 18mW (that is fully 50% of the current it will pass in the steady state.) During this period, there is no energy-transferring chain of physical cause-and-effect that goes all around the wire loop, and it would, of course, upend physics if there were! In fact, during this period, it makes no difference whether the far ends of the transmission lines are open- or closed-circuit.
      - tsimionescu 5y ago
        > During this period, there is no energy-transferring chain of physical cause-and-effect that goes all around the wire loop, and it would, of course, upend physics if there were! Even before the steady state is achieved, there is an energy-transferring chain of physical cause-and-effect that is moving (at limited speed, ~c) along the wire (but not all around the wire loop yet) - this is of course the moving electric and magnetic field caused by the "chain" of electrons being pushed or pulled. Of course, that will only reach the lightbulb after ~1 second in the original experiment. At the same, time, there is a different, smaller electric field "leaking" through the air/void in a straight line from the battery to everywhere else, which will reach the lightbulb much sooner in the original configuration, and of course this is also carrying some energy. The situation is analogous to what would happen with mechanical waves (at least qualitatively if not quantitatively) - if you were to have the same wires but a source of sound instead of the battery, and a sound detector (or a seismograph) instead of the lightbulb, you would get a pretty similar effect - the "main" sound wave would travel along the wires at the speed of sound in that medium, while some other waves would also travel through the air and ground through the much shorter straight-line distance, and register much more quickly on the seismograph. Of course, there are some significant quantitative differences - there is no sound super-conductor, so some amount of mechanical energy will always be lost along the wires; and sound travels much more slowly through air than through metal wires, so the speed difference will be less noticeable.

sebzim4500 5y ago

The slides look correct to me but the video massively misrepresents what is actually going on. For example, you could cut the wire at both ends and the answer would not change, so the fact that there is a 'circuit' is irrelevant.

tsimionescu 5y ago

The video gives a wrong impression that most current is flowing along the Poynting vector. In fact, only a tiny amount of current is flowing directly, and the vast majority is flowing along the wires, as expected. The tiny amount flowing directly through the air is much better understood as an antenna effect, unrelated to regular current flow, and has nothing really to do with the Poynting vector.

tom-thistime 5y ago
I agree that current doesn't follow the Poynting vector. Power follows the Poynting vector. That's true for antennas and for transmission lines.
- tsimionescu 5y ago
  Yes, but most power is in fact following the Poynting vector along the wires, not the direct Poynting vector between the battery and the bulb, as the video misleadingly suggests (but doesn't outright claim).
  - mannykannot 5y ago
    > Yes, but most power is in fact following the Poynting vector along the wires. Only in the steady state does the Poynting vector follow the wires all around the loop (a point made in the original video.) In the period of interest - well before the EM waves launched by closing the switch have reached the far ends of the transmission lines - it does not, yet during this period, the load is passing a non-trivial current (and therefore dissipating power) that is determined by the voltage of the battery, the resistance of the load, and the characteristic impedances of the two transmission lines. For example, for 24 awg superconducting wire (~0.5mm in diameter) the characteristic impedance for each of the transmission lines would be a little less than 1 kΩ, so with a 12 volt battery, a 2kΩ load would be passing about 3mA and dissipating about 18mW (that is fully 50% of the current it will pass in the steady state.) During this period, there is no energy-transferring chain of physical cause-and-effect that goes all around the wire loop, and it would, of course, upend physics if there were! In fact, during this period, it makes no difference whether the far ends of the transmission lines are open- or closed-circuit.
    - tsimionescu 5y ago
      > During this period, there is no energy-transferring chain of physical cause-and-effect that goes all around the wire loop, and it would, of course, upend physics if there were! Even before the steady state is achieved, there is an energy-transferring chain of physical cause-and-effect that is moving (at limited speed, ~c) along the wire (but not all around the wire loop yet) - this is of course the moving electric and magnetic field caused by the "chain" of electrons being pushed or pulled. Of course, that will only reach the lightbulb after ~1 second in the original experiment. At the same, time, there is a different, smaller electric field "leaking" through the air/void in a straight line from the battery to everywhere else, which will reach the lightbulb much sooner in the original configuration, and of course this is also carrying some energy. The situation is analogous to what would happen with mechanical waves (at least qualitatively if not quantitatively) - if you were to have the same wires but a source of sound instead of the battery, and a sound detector (or a seismograph) instead of the lightbulb, you would get a pretty similar effect - the "main" sound wave would travel along the wires at the speed of sound in that medium, while some other waves would also travel through the air and ground through the much shorter straight-line distance, and register much more quickly on the seismograph. Of course, there are some significant quantitative differences - there is no sound super-conductor, so some amount of mechanical energy will always be lost along the wires; and sound travels much more slowly through air than through metal wires, so the speed difference will be less noticeable.