[Split]Technical split from: Pear Cable CEO Calls James Randi's $1 Million Offer a Ho

[sol invictus]

As for the force being attractive, I've already agreed with you on that so I guess we can put this one to bed.

OK.

I guess I would quibble with that a bit. I mean, would a living coelacanth only be "evidence" of coelacanths living today?

That would be rather strong evidence . That's possible because you picked an absolute and very general statement (there are NO coelacanths alive today). Those are very easy to falsify.

But here we're not talking about an absolute claim - "cables never make an audible difference" - because such a claim is obviously false. We're talking about something more nuanced, like: "reasonable lengths of properly designed speaker cable are not audibly distinguishable". Being adults, we're able to handle the necessary (but small) degree of ambiguity present in that statement without freaking out.

You mean there's no more research being done with regard to human aural perception? Didn't know the book had been closed on that.

I'm not an expert, but my understanding is that that is indeed the case, at least with regard to gross questions such as what the thresholds of audibility are. The research being done has long since moved on to more detailed questions, such as precisely how the inner ear works, what the neural interface is with the brain, etc.

I certainly agree that there are many other effects which are much larger, though I don't think that fact has any particular relevance.

Of course it's relevant. If I design a bridge and take into account the effect a piece of chewing gum dropped in the center of the span would have, but not the effect of trucks driving across, I haven't done my job very well, have I? And if I then issue a report into the stability of the structure and mention only the chewing gum, it's a false and misleading report.

I've been as critical of the claims made by cable manufacturers (to the point it's got me banned from certain forums), but despite that, I don't believe most are insincere in their beliefs.

Being one yourself you should know, so I'll take your word for it.

 

[Paulhoff]

It's not my claim to prove. It's Pear Cable's claim to prove, just as your claim is yours to prove.

se

Again, I am not the one claiming new BS, they are. I don't have to prove anything, they do, or if you what you can. I've had electronics, I am also a Ham, I have no problem understand that they are selling BS, do you.

Paul



Why do you think that the test has gone anywhere, daaaaaaaaaaaa.

 

[Paulhoff]

Here, write to JJ, James Johnston.

http://home.comcast.net/~retired_old_jj/

Paul

 

[Dan O.]

http://theory.uwinnipeg.ca/physics/mag/node10.html

It is a good exercise to show that if the wires were carrying currents in the opposite directions that the resulting forces will have the same magnitude as in Eq.(1.11) but are such as to cause a repulsion between the wires.

http://en.wikipedia.org/wiki/Railgun

There are also forces acting on the rails attempting to push them apart, but since the rails are firmly mounted they cannot move.

Now, exactly what is the magnitude of the forces pushing these wires apart and how much acoustic energy could we expect to be generated by a pair of 10 ft. cables?

 

[Paulhoff]

If this BS that they sell was true, Radio would never work.

Paul

 

[Paulhoff]

This is a silly test. There are differences in cables, it's easy to spot with good headphones, but its a totally different matter are these differences true to the signal. Any cable can color the sound one way or another. Impedance mismatch will make obvious changes in bass quantity even though the cable will measure the same way etc.


Snake oil. It's funny that snake oil salesmen may have been onto something. There's an article at sciam.com, but I can't post the link because I have under 15 posts.

You're funny, yes you are.

Paul

 

[sol invictus]

http://theory.uwinnipeg.ca/physics/mag/node10.html

As I have already said, that is correct only if there is no current or voltage source connected to the wires. In that case the current will not remain constant as the wires move, due to Lenz's law. In the case of audio cables connected to an amp the force is attractive. Again as I said before, this is a common error.

The magnitude of that force will be extremely small.

 

[Dan O.]

As I have already said, that is correct only if there is no current or voltage source connected to the wires. In that case the current will not remain constant as the wires move, due to Lenz's law. In the case of audio cables connected to an amp the force is attractive. Again as I said before, this is a common error.

So you are saying that the wires somehow (magically?) know when the current passing through them came from an amplifier that they should move together but when the current is in a closed superconducting loop they will move apart!?

I can do this experiment. I can hang 2 wires so the loops are free to swing together or apart and apply a current source such as a battery at one end with the far end shorted. Will the wires swing together or apart when the connection is made? Will it make a difference if the source is an audio amp instead of a battery and the far end is connected to speakers?

The magnitude of that force will be extremely small.

For a 100W amp driving 4 ohm speakers through conductors spaced .25cm apart you get a force of about .002Nm per meter of cable. For #12 aluminum wire this would create an acceleration of about 0.22m/s/s.

 

[Paulhoff]

The current is opposite in two wires, always opposite, therefore the little force that is there is always an attractive one.

Paul

 

[Dan O.]

The current is opposite in two wires, always opposite, therefore the little force that is there is always an attractive one.

Here I suspended 2 parallel wires so they could swing.



This is a close-up with the current off


And this is a close-up with the current on.


(Sorry about the blurr. I could probably get a better shot and wouldn't burn my fingers if I limited the current )

 

[robinson]

Wow. This is getting interesting.

 

[sol invictus]

So you are saying that the wires somehow (magically?) know when the current passing through them came from an amplifier that they should move together but when the current is in a closed superconducting loop they will move apart!?

Not magic, conservation of energy and electric forces. But it's more subtle than what I said above, which I think now was incorrect.

Normally, force is minus the derivative of the energy. In the case of superconducting wires with no source, the energy (if the currents are opposite) will decrease as they move apart. That's true because the magnetic flux through the loop will stay constant. The flux is f=BA if B is the field and A is the area of the loop, but the energy is B^2A = f^2/A, so as A gets bigger the energy decreases. Therefore the force is repulsive. Of course you can also get the same answer from the Lorentz force law.

If the wires are connected to a current source, then the energy in the field increases as the wires move apart (this is clear because if the wires coincide the field is zero). If the force were minus the derivative of that energy, it would mean the wires attract, as I was saying.

However it's more subtle than that - the total energy includes a component due to the work the current source does to maintain the current. I think what happens is that even though both the energy in the field is increasing AND the kinetic energy in the wires is increasing as the wires move apart (which normally would violate conservation of energy), the current source is providing the work necessary to make that happen, but without exerting any force on the wires.

Confusing, but I guess that's how it works. Sorry for the mistake, and nice little experiment.

 

[sol invictus]

Incidentally, there is a very similar situation with a parallel plate capacitor.

Suppose you have two plates a distance d apart. Charge them up to charge +Q and -Q by connecting a battery for a while. Now disconnect the battery.

Obviously there will be an attractive force, since +Q attracts -Q. We can also see this from the energy, which is (1/2)Q^2/C, where C goes as 1/d. So the energy is proportional to Q^2 d, and therefore decreases as the plates come together and so the force is attractive.

However now suppose you leave the battery attached. In that case Q isn't constant, but V is. The energy is CV^2~V^2/d, which increases as the plates come together. Does that mean the force is repulsive?

Evidently not, and it must be for the same reason - that the battery does work to allow the plates to come together and increase the energy in the field, and again the force is not given by the derivative of the energy in the field.

 

[pmurray]

This is incorrect. Two wires with opposing currents, connected to a constant current source such as an amplifier, attract, not repel.

I am fairly certain that you are incorrect on this one. The design of the Z-machine particle acellerator, for instance, relies on tungsten wires carrying a current to be attracted to one another.

That the correct answer must be an attractive force is obvious from the fact that when the wires are very close the magnetic field is zero, which means that configuration has lower energy than when they are far apart.

The magnetic field between the two conductors is not zero, no matter how close you put 'em.

 

[pmurray]

Normally, force is minus the derivative of the energy.

Normally, energy is force times distance. No derivatives involved.

That two currents in the same direction are attracted to one another explains the magnetic pinch effect, and the existence of "streams" of ions.

 

[MRC_Hans]

MMm, where did this quote come from?

• Triboelectric effect
• Movement relative to an electric field
• Movement relative to a magnetic field

It is important to realize that there are two separate energy sources for this vibration.


The first energy source comes from the loudspeakers themselves. As music is played, the vibrations in the air (sound) vibrate “hairs” in your ears (so you hear the sound), as well as everything else in the room. This vibration includes your audio cables. This mechanical vibration, induced by the sound itself, is a major enemy of cable producers. Unfortunately, this problem is rarely talked about, and frequently overlooked.


The second energy source comes from the current running through the audio cables. As mentioned previously, the current will produce a magnetic field. Since all of the wires in our audio systems have current running in opposing directions (+ and – lead, this applies to AC as well) opposing magnetic fields are set up in the conductors. These opposing magnetic fields mechanically push the conductors apart, thus causing mechanical vibration. This is a major design consideration for loudspeaker cables. However, in audio interconnects, the current is so small, that this effect can truly be neglected, so we will not elaborate on this second mechanism for now.

Anyhow, it is nonsense. The first part: Of course, there will be some mechanical vibration in the cables, provided you play loud enough, but so what? Unless they are literally jumping around (and probably not even then), any signals induced into them will be unmeasurable.

Second part: Yes, wires running together and being in the same circuit, like speaker wires, will attract. And again, so what? Again unless there is a violent movement, and that will require currents way beyond even the crazy audiophile's monster amplifier, it will have no influence on the signal.

Hans

 

[MRC_Hans]

The magnetic field between the two conductors is not zero, no matter how close you put 'em.

Not between them, but the field reaching from them approaches zero as they get closer. What does also approach zero is the inductivity of the wires, because the fields cancel out. That is the effect exploited in transmission lines.

Hans

 

[MRC_Hans]

Incidentally, there is a very similar situation with a parallel plate capacitor.

Is there any other kind?

Suppose you have two plates a distance d apart. Charge them up to charge +Q and -Q by connecting a battery for a while. Now disconnect the battery.

Obviously there will be an attractive force, since +Q attracts -Q. We can also see this from the energy, which is (1/2)Q^2/C, where C goes as 1/d. So the energy is proportional to Q^2 d, and therefore decreases as the plates come together and so the force is attractive.

Wrong on several points. First of all, the energy, or charge, does not change (you know, conservation of energy and all that). Now since the charge is constant, but the capacitance increases as you move the plates closer, the voltage will fall. The falling voltage will reduce the attraction, but since the plates move closer, this will make the attraction bligger, resulting in a net zero; the attraction stayes the same.

However now suppose you leave the battery attached. In that case Q isn't constant, but V is. The energy is CV^2~V^2/d, which increases as the plates come together. Does that mean the force is repulsive?

Now, as the capacitance increases, and the voltage is kept constant, the charge will increase. Since the voltage is constant and the distance falls, the attraction will increase.

Evidently not, and it must be for the same reason - that the battery does work to allow the plates to come together and increase the energy in the field, and again the force is not given by the derivative of the energy in the field.

Ehrm, not directly. The force is basically a product of the voltage, but it is also a product of the area of the plates, and since a larger area absorbs a larger charge (energy) to obtain the same voltage, the energy is actually perfectly correlated to the attractive force.

Hans

 

[maarek99]

I don't get it. I mean, isn't it obvious that there are differences in cables? A steel cable will definitely sound worse than a copper cable. Ofcourse after you get over a certain threshold it doesn't matter anymore and studio quality cables should be enough for everybody.

I have here one of the many replacement cables (snake oil if you will) for the hd650 phones. Now there's definitely a difference between it and the stock one. The difference is simple, it has less bass and more treble. That's it really.

Is it ********? Am I hearing differences? No. I actually prefer the stock cable so I'm not even biased to prefer the "upgraded" cable. But the difference is there anyway and should be pretty easy to spot blindfolded.

You can change the sound frequency of a phone by changing the output impedance somewhat and I think that's what most of these cables are simply doing.

Now what about amplifiers, do they all sound the same? Not by a long shot. There's another candidate for the 1 million dollar challenge

 

[sol invictus]

Normally, energy is force times distance. No derivatives involved.

The formula you're thinking is for the work performed when a force acts on something over a distance . It won't help us here, where we're asking about the force acting on a wire held in place. The one we need is F = -dU/dx, where U is the potential energy.

The magnetic field between the two conductors is not zero, no matter how close you put 'em.

The total energy in the magnetic field is the integral of (1/2)B^2. B is a vector, and in some places the fields of the two wires will (partially) cancel. In the case of two long straight wires with opposite currents, B adds in between them, and mostly cancels everywhere else. As you move them close together the cancellation gets better and better, thus decreasing the energy in the fields. If they're right on top of each other there's no net current and therefore no field at all.

Is there any other kind?

Yes. Any two disconnected conductors have a capacitance. An easy example is two concentric spherical shells.

Wrong on several points. First of all, the energy, or charge, does not change (you know, conservation of energy and all that). Now since the charge is constant, but the capacitance increases as you move the plates closer, the voltage will fall. The falling voltage will reduce the attraction, but since the plates move closer, this will make the attraction bligger, resulting in a net zero; the attraction stayes the same.

The energy does change unless you keep track of all parts of it. If you hold the plates, but allow them to slowly be attracted together, you will be exerting a force to keep them from getting out of control, and doing work on the system. Therefore the energy in the field must be changing. Similarly if you let them go - in that case the kinetic energy of the plates would increase, and that energy had to come from somewhere. In the case the plates are disconnected it could only come from the field.

Ehrm, not directly. The force is basically a product of the voltage, but it is also a product of the area of the plates, and since a larger area absorbs a larger charge (energy) to obtain the same voltage, the energy is actually perfectly correlated to the attractive force.

First of all, the area is totally irrelevant - we're talking about forces per unit area, energy per unit area, etc. If you insist on keeping it, multiply everything I wrote by A.

At constant voltage, which is appropriate when the plates are attached to a battery, the energy in the electric field increases as the plates approach each other. Here's another way to see that: the electric field is simply E = V/d in this case, and the energy in the field is proportional to E^2 d (times A if you want), so the energy is V^2/d.

So you simply can't analyze it that way, and in retrospect it's clear why. The battery is changing Q as the plates move so as to maintain the voltage, and so clearly it's doing some work. If you ignore that work you get a wrong answer.