You mean their garden.There are also fairies living at the bottom of your garden!
Uh..., where have you been for last 25 years? It's been their daily diet for at least that long.
Couldn't have said it better
You took the words out of my mouth.Look at this as an opportunity to build up a useful ignore list. For some reason it takes a lot of mouse clicks, perhaps as a reminder that good things should not be too easy
just for fun those of you guys who have made your own cables would you mind posting a couple of photos ive never tried and would find it quite interesting to see the various creations of the inventive people here if its too much trouble don't worry but it would be a cool thing for everyone to see
if its too much trouble don't worry but it would be a cool thing for everyone to see 
well anybody used this cables? i bet if you use this cables youll hear something that you have never heard with normal cables! the ultimate fidility cables!
An externally hosted image should be here but it was not working when we last tested it.
Where are the lifters, you might have to super size them.
Ya go to all the trouble to make a cable with a characteristic impedance of 1 or 2 ohms by paralleling 50 or 75 twisted pairs, you get the lumped cable inductance into the 2 to 5 nanohenry per foot range with capacitance in the 600 to 1200 pf per foot.....
Then you split the pair of conductors at each end giving 1 to 4 microhenries at each end, a factor of 20 to 50 higher than the cable...and a terminal resistance many times that of the cable?
What were they thinking?
Jn
This is what people often say. I believe it comes from a misunderstanding of Poynting's theorem - which is sometimes known as Poynting's conjecture.Galu said:
For an isolated cable, far from any other fields, yes. There is no a priori reason why in any given situation the two fields have to be at right angles; it all depends in the detailed geometry and things like frequency.Galu said:
No. The definition of the Poynting vector as a cross-product is what makes it point at right angles to the electric and magnetic field.Galu said:
You cannot determine signal flow or the Poynting vector just from considering current. Voltage matters too.
Of course.
It has not helped me, but perhaps my reply has helped you understand it a little better?
Glad you agree with me on one point!
However, nothing you have said actually disproves the validity of the other points!
I was disagreeing, not disproving. Disproving would take too long and might involve vector calculus which I forgot years ago.Galu said:
If you look at derivations of Poynting's theorem you will find that they almost always have to include a volume V surrounded by a surface S (so the theorem applies to volumes, not points) and they almost always assume that the fields are varying in time (so the theorem may say less about static and quasistatic situations such as DC and audio). Hence the Poynting vector is a useful tool for calculating where the energy goes but it doesn't necessarily tell you how it got there. As I said, in a simple DC circuit the energy is transported by potential energy of the electrons, not the fields. An audio interconnect is almost DC because the physical size of the apparatus is much smaller than the wavelength of the excitation. People trying to explain how audio interconnects sound different by invoking the Poynting vector are merely showing their ignorance of EM - just like those who claimed to invent a new radio antenna which uses 'Poynting vector synthesis'.
Warning, nitpicking ahead
.The physical interpretation of the Poynting vector in static fields is a non-trivial question, it may easily be considered as an open question today's EM theory. Mr. Bybee may have something to say
, since Feynman addressed this very question is his lectures The Feynman Lectures on Physics Vol. II Ch. 27: Field Energy and Field Momentum but did not expand too much in his argumentation.To summarize the dilemma, we need to consider what is called the continuity equation, valid (one way or another) in all branches of physics
divS=d(Ws)/dt (sorry, no symbol for partial derivative available)
essentially saying that "something" is conserved. Ws is the density of "something" while S is the flux density of "something".
In the particular case of the EM field, Ws=W=(E^2+B^2)/2 then from Maxwell it follows that
divP=dW/dt
where P is now the Poynting vector ExB. For the static fields, if follows that divP=0 however the problem is, divP=0 does NOT necessary mean (as a solution) that P=0. Considering a closed surface for integrating divP=0, there are two possibilities, either P that it enters the volume surrounded by S is the same that the one that it leaves it (but how something static may "enter" or "leave"?), or simply P=0. To support the first assertion, some are arguing that since |P| varies as x^-4 (so it decreases quickly with distance), if we integrate divP=0 over a surface which is large enough we indeed may consider the P flux as zero. so there's no conflict. But since we know the static P=ExB is NOT zero though, and considering the integration result depending on the integration surface is (at least) a stretch, it follows that the static Poynting vector has nothing to do, from a physical interpretation for the case of static fields, with the general energy transport interpretation.
The most common interpretation (I also happen to agree with) is that if E and H are static, nothing is happening (mathematically, nothing 'flows') until an object is placed in the field. Then a force will be exerted on the object and, as a result, the physical interpretation of the Poynting vector changes. To understand why, a solution of divP=0 is any vector P=P'+F where div x F=0. The presence of F implies an angular momentum, which also needs to conserve. Therefore, we can think of a static EM field Poynting vector simply creating an angular momentum in the circuit, which of course is too small to be measured by simple methods. The conservation laws are therefore saved once again, and the conflict between "divP=0 does not imply P=0" is also solved.
This is not to say that this "mechanical" interpretation is without flaws. Calculations on certain geometries, like a spherical capacitor, shows a non uniform strange distribution of the angular momentum, across the capacitor electrodes. Mathematically, this is because changing from rectangular coordinates (for the current density) to the spherical coordinates for the capacitor... Such a non uniformity still has to be interpreted.
So we don't need the Poyinting vector to explain the energy transport in a DC circuit, that remains for AC (time dependent EM field) only. The static Poynting vector only creates a angular momentum over the circuit (that is, it tries to bend your PCB
).
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Ummm... yesterday was a long time ago!
I was trying to get at some aspect of believing they had a great car (cable) but someone else could see it was no better than any other car (cable) other than it had high-end branding on it. The buyer is very pleased with their "supercar" as they drive it slowly around their private grounds, never to find out it doesn't perform any better than the re-badged Ford that it is..
They're response against any criticism of the McLaren-badged Ford is "Do you have one? Well then, you have no clue what you're talking about then.."
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