5 ms·
His presentation is not very clear & imo the way he answers the question is not actually helpful in actually understanding what’s going on, but he is correct th
by pja 5y ago
His presentation is not very clear & imo the way he answers the question is not actually helpful in actually understanding what’s going on, but he is correct that current will flow in the far side almost immediately. As is made clearer in the documents he shared which contain his discussions with actual physicists, that current will be much smaller than the final current that flows all the way round the circuit, but it is there nevertheless. (In fact that current will initially flow whether or not the two halves of the flat ring are connected - if you think about it things have to work like this, because the middle of the system cannot “know” whether the far ends are connected until light has had time to get there & back. Relativity is absolute.)
Take a look at the video made by YouTuber AlphaPhoenix where he constructs the entire thing on a smaller scale and demonstrates the effect with an oscilloscope: https://www.youtube.com/watch?v=2Vrhk5OjBP8 https://www.youtube.com/watch?v=2Vrhk5OjBP8
One way of viewing what’s going on here is to consider the two sides as a pair of large dipole antennas. When the switch (or in AlphaPheonix’s case, a transistor) is turned on, a pulse of electrons is pushed down one antenna, setting up a transient wave in the electromagnetic field around it which induces a current in the opposite direction in the “receiving” antenna on the other side. This happens regardless of whether the wired are connected in a loop or not!
Physics is great.
- deleted 5y ago[deleted]
- marcan_42 5y agoI found the AlphaPhoenix video a bit disappointing because he didn't clearly explain that the capacitive interaction between both ends is also light speed limited, and very dependent on distance, i.e. the geometry of the test. Had the test been done in a big round loop, current wouldn't start flowing at the far end immediately. Otherwise you'd have made faster than light communications. I'm also not even sure if his testing methodology is solid; he was getting a ton of common mode junk in the target measurement probes, and subtracting it out in the scope. That works in steady state, but when trying to measure fast transients like this it depends on things like probe skew and more. It's just not clean enough for me to be able to trust the results are caused by the explanation he gave. Measuring subtle phenomena like this with long transmission lines involved with a single ground referenced scope and no true differential probes... ick.
- elisbce 5y agoHe has an upcoming part2 of the video which he listed big round loop as one of the follow-up experiments. We should just wait for that.
- pja 5y ago>Had the test been done in a big round loop, current wouldn't start flowing at the far end immediately. Otherwise you'd have made faster than light communications. Yes, this is obviously the case. But it's also not the setup that was being tested. (Apparently he's editing a part2 which contains many weird & wonderful topological arrangements.) Honestly though, I'm not sure what your problem is here. His results match very closely what you'd expect theoretically given the experimental setup. Not only that, if you model the setup in software using a suitable electromagnetic modelling code (there are videos on YouTube of people doing this) you get exactly the same result. Common mode noise on both sides is an irrelevance - you can just subtract it away without loss of generality.
- marcan_42 5y agoYou can only subtract common mode noise out if it really is common mode and perfectly equal on both sides of the system. With two discrete probes, that's going to give you a lot worse common mode rejection than with a proper differential probe. He can't even have proper ground references the way he did it (most of his probe ground clips are unconnected). It's just not a good rig. Sure, it might work, and his result might be good, I just don't feel confident about it given the experimental set-up. If you look at his scope plot, the common mode voltage is kind of a mess and much larger than the actual differential voltage. That doesn't give me a lot of confidence in the final result. I'd like to see a more controlled version of this experiment with better equipment. Source: I've done the same thing he did, using two probes to measure a differential voltage, and I can tell you it kind of works, but not well. Especially not at higher frequencies.
- formerly_proven 5y agoThe common mode problem can be avoided by making the experiment not differential (which coincidentally saves half the wiring) but even if you do that, you still get a large influence on the result due to the coupling between the twin-lead line and ground. It's actually a pretty good demonstration why using baluns for twin-lead is necessary; only the differential mode of the line works well, because it's not actually a two-conductor line, as it's an open line it is also coupled to ground. In my experiment (with ~6 cm wire spacing, which makes relatively little difference as the impedance of the line is proportional to the log of the spacing) I found that suspending it around a meter above ground gave almost identical differential and common mode impedances. That of course rather significantly reduces the amplitude you're going to see in the experiment.