This page is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
I believe Nigel is confused about radio propagation and Poynting's vector and that he is only considering the so called 'induction field' (near field) antennas whose signal strength decreases as the square of the distance from the antenna as opposed the the radiation fields that I am talking about (these field strengths are inversely prop to distance). Induction fields can, in my opinion, pass no energy as they consist of magnetic fields only or electric fields only. Both fields combined together in an em wave are needed to transfer energy from one place to another. If you look at proper high frequency antennas, you find that they are shaped like horns. This is done in order to match the impedance of the transmitter to the impedance of free space (377 ohms). Nigels antenna model is merely an induction antenna transmitting short range electric of magnetic fields only. Thier efficiency at radiating proper em waves is very low.-- Light current 01:07, 23 January 2006 (UTC)
Nigel have a look at this. See if you agree with it!. [1]-- Light current 03:42, 23 January 2006 (UTC)
Yes I tend to agree with the last part of the above paragraph. But are you not changing your story here?. I thought you said before that you can have radiation from one isolated conductor. I said you couldnt! However, your interpretation is I feel slightly wrong. Radiation does not travel from one conductor to the other, iit travels between the conductors as per well knoewn TL theory. Poyntings vector etc. Im ignoring the small near (induction) field effects in open wire lines. And, of course, in coax there are no leakage fields.-- Light current 04:17, 24 January 2006 (UTC)
Well Nigel, it looks like you are now 'stealing my clothes'. However, one place we do differ is in the mechanism of high frequency currents being caused to flow in the conductors. You say there is some sort of mutual induction. Well, sort of! I say any currents that can be detected in the conductors will be induced by the em energy flow between them. Think about waveguides! The walls only act to reflect the waves. The only currents in the walls are induced currents. Regarding Maxwell, I agree that in a capacitor, the so called displacement current is indentically equal to the conduction current in the wires. That is trivially obvious. However that does not mean that displacement current actually flows across the plates from one to the other. THe current gets to one side from the other by means if the electromagnetic energy that is liniking both conductors (or plates). My previous posts have a fuller explained this view.-- Light current 04:32, 24 January 2006 (UTC)
No, Kevin, it involved measurements of radiated field strength with an oscillscope: the radio emission is proportional to di/dt fed into the aerial. You are the one who appears to not have a grasp of this fact. Sine waves are indeed an ideal case because it maximises the overall emission (a square wave input doesn't provide any radio emission at all during the flat parts, just a series of spaced pulses during the rises and falls of the wave. Thanks, Nigel 172.203.223.44 11:50, 24 January 2006 (UTC)
As far as I am aware, no one has the authority on WP to delete posts on talk pages. If this has happened, it is what we term as vandalism (or an error). The perpetrator will be warned not to repeat this action. -- Light current 00:36, 24 January 2006 (UTC)
The external fields around the open wire TL are, in my opinion, near field effects. If any energy were absorbed from these fields, it must affect the transmission of energy along the conductors. However, I wish to restrict the argument to coaxial cable as this simplifies matters greatly in that there are no leakage fields to worry about.-- Light current 04:40, 24 January 2006 (UTC)
Yes, well, the only problem I have with Bill Beaty's description is that he seems to be restricting the cause of fields to charge and charge movements. We all know that charges cannot move at light speed, so the description does not answer all the questions. However, the em energy (energy current) induction theory does appear to explain displacement current (ie that it doesnt exist) and how 'conduction' current can appear to get from one side of a capacitor to the other without actually passing between the plates thro the (vacuum) dielectric. (Thats a subtle one!)-- Light current 04:50, 24 January 2006 (UTC)
Kevin, the orthodox view on capacitor charging has been completely blown away by read Catts article on @diplacement current (and how to get rid of it). Even Nigel agrees with that!! So I dont see what point you are making in this post except that you realise that Catt is proposing reciprocating energy currents. Yes! I agree that reciprocating energy currents are one answer to DC situations as I have been saying for about 3 months now!-- Light current 04:58, 24 January 2006 (UTC)
OK Nigel I think everyone should reply outside the body of others' posts. I'll start now. It would help though if the posts were kept shorter and deal with few points at one time. Rgarding your last post, yuo say
an isolated conductor connected to a charge (battery terminal) radiates energy as it charges up. However, this plainly cannot be the case as there is no return path for the current- unless of course the other end of the capaciance is connected to the other battery terminal. In this case you have a two conductor system and the capacitance must be taken into account in the analysis. BTW, charging a capacitor does not radiate energy, unless youre very unlucky! You then say For a pair of conductors connected to the two terminals of a battery (Catt's anomaly situation) the current in each conductor is the opposite of the other, so the radio emission cancels out beyond the system,
I agree with this mostly. In a coax of coarse there is no radiation at all-- Im not sure how much radiation ther is from a pair of open wires.
BTW you keep talking about a step as only having effect at that instant. Do a Fourier analyisis of the step to see what sinusodal components it contains! -- Light current 14:27, 24 January 2006 (UTC)
Kevin???(dont you mean LC), coax or a waveguide (for microwave frequencies) is used in many practical cases, but let's keep to the simple physics of a DC TEM wave step propagated by two straight aerial-like conductors, as the Catt anomaly uses. We get electromagnetic radiation (radio emission) from a net time-varying current in a conductor. If you have two such conductors, with each having an inverted form of the signal in the other, they exchange energy which induces the current in the other. But there is no long distance propagation of this energy due to exact interference, so the coupling is perfect. This is how the front of the logic step propagates: each conductor causes the current in the opposite conductor by simple electromagnetic radiation due to the time-varying current as it rises. Catt simply missed out this mechanism, and is now too prejudiced against the mainstream to back these facts. Best wishes, Nigel 172.203.152.63 23:18, 24 January 2006 (UTC)
Light current: you say "there must be a return path for the current". Radio is the return path. Radio is not just a sine wave phenomena. You cannot stop radio (self-propagating emission of energy) if the current changes with time. The current in a single wire falls rapidly, the energy being propagated away as radio. Since the self-inductance is infinite, the current continues falling rapidly, although the whole wire still "charges" up to the battery terminal voltage at light speed. Thanks, Nigel 172.212.87.209 15:40, 27 January 2006 (UTC)
Nigel, you appear not to have heard of Mr Fourier and his famous Fourier analysis which says that any repetitive waveform can be represented by a sum of sine and /or cosine waveforms of appropriate amplitude. Now in your experiment, the square wave that you used would have a fundamental plus all the odd harmonics decreasing in amplitude proportional to the harmonic. So when you say 'the flat part doesn't do anything', strictly speaking, you are incorrect or your statement is careless.-- Light current 05:36, 25 January 2006 (UTC)
Nigel: If only a single step, you may be correct. Is that what you are referring to in your arguments? However, to say that no energy flows to the load after the initial step I think is not correct.-- Light current 15:44, 27 January 2006 (UTC)
I found this blog that has a lot of information about Ivor Catt, I am kind of new to editing, would this be a worthwile source to quote:
If so let me know and I will go ahead and do the work
This page is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
I believe Nigel is confused about radio propagation and Poynting's vector and that he is only considering the so called 'induction field' (near field) antennas whose signal strength decreases as the square of the distance from the antenna as opposed the the radiation fields that I am talking about (these field strengths are inversely prop to distance). Induction fields can, in my opinion, pass no energy as they consist of magnetic fields only or electric fields only. Both fields combined together in an em wave are needed to transfer energy from one place to another. If you look at proper high frequency antennas, you find that they are shaped like horns. This is done in order to match the impedance of the transmitter to the impedance of free space (377 ohms). Nigels antenna model is merely an induction antenna transmitting short range electric of magnetic fields only. Thier efficiency at radiating proper em waves is very low.-- Light current 01:07, 23 January 2006 (UTC)
Nigel have a look at this. See if you agree with it!. [1]-- Light current 03:42, 23 January 2006 (UTC)
Yes I tend to agree with the last part of the above paragraph. But are you not changing your story here?. I thought you said before that you can have radiation from one isolated conductor. I said you couldnt! However, your interpretation is I feel slightly wrong. Radiation does not travel from one conductor to the other, iit travels between the conductors as per well knoewn TL theory. Poyntings vector etc. Im ignoring the small near (induction) field effects in open wire lines. And, of course, in coax there are no leakage fields.-- Light current 04:17, 24 January 2006 (UTC)
Well Nigel, it looks like you are now 'stealing my clothes'. However, one place we do differ is in the mechanism of high frequency currents being caused to flow in the conductors. You say there is some sort of mutual induction. Well, sort of! I say any currents that can be detected in the conductors will be induced by the em energy flow between them. Think about waveguides! The walls only act to reflect the waves. The only currents in the walls are induced currents. Regarding Maxwell, I agree that in a capacitor, the so called displacement current is indentically equal to the conduction current in the wires. That is trivially obvious. However that does not mean that displacement current actually flows across the plates from one to the other. THe current gets to one side from the other by means if the electromagnetic energy that is liniking both conductors (or plates). My previous posts have a fuller explained this view.-- Light current 04:32, 24 January 2006 (UTC)
No, Kevin, it involved measurements of radiated field strength with an oscillscope: the radio emission is proportional to di/dt fed into the aerial. You are the one who appears to not have a grasp of this fact. Sine waves are indeed an ideal case because it maximises the overall emission (a square wave input doesn't provide any radio emission at all during the flat parts, just a series of spaced pulses during the rises and falls of the wave. Thanks, Nigel 172.203.223.44 11:50, 24 January 2006 (UTC)
As far as I am aware, no one has the authority on WP to delete posts on talk pages. If this has happened, it is what we term as vandalism (or an error). The perpetrator will be warned not to repeat this action. -- Light current 00:36, 24 January 2006 (UTC)
The external fields around the open wire TL are, in my opinion, near field effects. If any energy were absorbed from these fields, it must affect the transmission of energy along the conductors. However, I wish to restrict the argument to coaxial cable as this simplifies matters greatly in that there are no leakage fields to worry about.-- Light current 04:40, 24 January 2006 (UTC)
Yes, well, the only problem I have with Bill Beaty's description is that he seems to be restricting the cause of fields to charge and charge movements. We all know that charges cannot move at light speed, so the description does not answer all the questions. However, the em energy (energy current) induction theory does appear to explain displacement current (ie that it doesnt exist) and how 'conduction' current can appear to get from one side of a capacitor to the other without actually passing between the plates thro the (vacuum) dielectric. (Thats a subtle one!)-- Light current 04:50, 24 January 2006 (UTC)
Kevin, the orthodox view on capacitor charging has been completely blown away by read Catts article on @diplacement current (and how to get rid of it). Even Nigel agrees with that!! So I dont see what point you are making in this post except that you realise that Catt is proposing reciprocating energy currents. Yes! I agree that reciprocating energy currents are one answer to DC situations as I have been saying for about 3 months now!-- Light current 04:58, 24 January 2006 (UTC)
OK Nigel I think everyone should reply outside the body of others' posts. I'll start now. It would help though if the posts were kept shorter and deal with few points at one time. Rgarding your last post, yuo say
an isolated conductor connected to a charge (battery terminal) radiates energy as it charges up. However, this plainly cannot be the case as there is no return path for the current- unless of course the other end of the capaciance is connected to the other battery terminal. In this case you have a two conductor system and the capacitance must be taken into account in the analysis. BTW, charging a capacitor does not radiate energy, unless youre very unlucky! You then say For a pair of conductors connected to the two terminals of a battery (Catt's anomaly situation) the current in each conductor is the opposite of the other, so the radio emission cancels out beyond the system,
I agree with this mostly. In a coax of coarse there is no radiation at all-- Im not sure how much radiation ther is from a pair of open wires.
BTW you keep talking about a step as only having effect at that instant. Do a Fourier analyisis of the step to see what sinusodal components it contains! -- Light current 14:27, 24 January 2006 (UTC)
Kevin???(dont you mean LC), coax or a waveguide (for microwave frequencies) is used in many practical cases, but let's keep to the simple physics of a DC TEM wave step propagated by two straight aerial-like conductors, as the Catt anomaly uses. We get electromagnetic radiation (radio emission) from a net time-varying current in a conductor. If you have two such conductors, with each having an inverted form of the signal in the other, they exchange energy which induces the current in the other. But there is no long distance propagation of this energy due to exact interference, so the coupling is perfect. This is how the front of the logic step propagates: each conductor causes the current in the opposite conductor by simple electromagnetic radiation due to the time-varying current as it rises. Catt simply missed out this mechanism, and is now too prejudiced against the mainstream to back these facts. Best wishes, Nigel 172.203.152.63 23:18, 24 January 2006 (UTC)
Light current: you say "there must be a return path for the current". Radio is the return path. Radio is not just a sine wave phenomena. You cannot stop radio (self-propagating emission of energy) if the current changes with time. The current in a single wire falls rapidly, the energy being propagated away as radio. Since the self-inductance is infinite, the current continues falling rapidly, although the whole wire still "charges" up to the battery terminal voltage at light speed. Thanks, Nigel 172.212.87.209 15:40, 27 January 2006 (UTC)
Nigel, you appear not to have heard of Mr Fourier and his famous Fourier analysis which says that any repetitive waveform can be represented by a sum of sine and /or cosine waveforms of appropriate amplitude. Now in your experiment, the square wave that you used would have a fundamental plus all the odd harmonics decreasing in amplitude proportional to the harmonic. So when you say 'the flat part doesn't do anything', strictly speaking, you are incorrect or your statement is careless.-- Light current 05:36, 25 January 2006 (UTC)
Nigel: If only a single step, you may be correct. Is that what you are referring to in your arguments? However, to say that no energy flows to the load after the initial step I think is not correct.-- Light current 15:44, 27 January 2006 (UTC)
I found this blog that has a lot of information about Ivor Catt, I am kind of new to editing, would this be a worthwile source to quote:
If so let me know and I will go ahead and do the work