Dimensional analysis

[darkenlighten]

CAN YOU RESPOND TO THIS:

 

[math]

I\int dl \ => \ 1 A = 1\tfrac C s \ => \ \mathrm{C} = 1 \mathrm{A} \cdot 1 \mathrm{s}

[/math]

 

What speed is that of 1C/s?

 

[math] I\int d\mathbf{l} \Rightarrow 1 A\cdot meter [/math]

 

dl is not dimensionless, because it is the length of the path.

[ambros]

Wasn't it?

 

When you put it that way, I agree.

 

 

Speed has units of distance over time. This isn't a speed, it's a flow.

 

Yes, but I forgot now what point I was trying to make. On a brighter side I just e-mailed BIPM request to explain the definition, equation and their relation in regards to what is actual force on one wire and what is the actual result predicted by that equation.

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[math] I\int d\mathbf{l} \Rightarrow 1 A\cdot meter [/math]

 

dl is not dimensionless, because it is the length of the path.

 

What is the length of that path in meters?

 

 

 

[math]1 \mathrm{C} = 1 \mathrm{A} \cdot 1 \mathrm{s}[/math]

 

Where did that meter disappear from here? Those two do not equate, or do they?

Edited April 8, 2010 by ambros
Consecutive posts merged.

[Mr Skeptic]

[math] I\int d\mathbf{l} \Rightarrow 1 A\cdot meter [/math]

 

dl is not dimensionless, because it is the length of the path.

 

Good catch!

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Yes, but I forgot now what point I was trying to make. On a brighter side I just e-mailed BIPM request to explain the definition, equation and their relation in regards to what is actual force on one wire and what is the actual result predicted by that equation.

 

If their answer disagrees with yours, will you believe them? I hope they answer.

 

What is the length of that path in meters?

 

Undefined. Without knowing what limits there are on the integral, you don't get a definite answer.

 

[math]1 \mathrm{C} = 1 \mathrm{A} \cdot 1 \mathrm{s}[/math]

 

Where did that meter disappear from here? Those two do not equate, or do they?

 

No, the result of that integral would have units of charge times velocity (ampere*meters if you prefer). What were you trying to calculate? Does this surprise you?

 

Your other equation however is correct for the definitions of coulomb and ampere.

Edited April 8, 2010 by Mr Skeptic
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[Cap'n Refsmmat]

Please copy/paste that here in text format so I can easier refer to it.

 

* int[54] is wrong, move y to 1

No. The derivative of C2 is {1, 0, 0}. The 1 in C2 is a constant, so it does not appear in the derivative.

 

* why is the x coordinate marked as "t", some relation to time?

Doesn't matter what I name it. I just chose t because that's what I always used in path integrals.

 

* where did you get idea that 2nd integral is from - to + infinity?

http://en.wikipedia.org/wiki/Ampere%27s_force_law

The double line integration sums the force upon each element of circuit 2 due to each element of circuit 1,

ds1 and ds2 are infinitesimal vector elements of the paths C1 and C2, respectively, with the same direction as the conventional current (usually measured in metres)

 

So I'm making one wire infinitely long, by making the path infinitely long.

 

* "t" is not initialized and it does not appear in the final equation?

[math]f(x) = 4x[/math]

 

[math]t= 7[/math]

 

[math]f(t^2 - 7) = 189[/math]

 

x is not initialized in the final equation here, but it's not necessary. Same goes for t.

 

Will you please plug in the equation EXACTLY as it is written, which means that both of those two integrals integrate over the same distance - IN PARALLEL... and see what Mathematica will tell you.

Integrals don't work in parallel, as I'm sure you already know. I'm just integrating along the paths.

 

The equation I have in Mathematica is already exactly as it is written. The integrals are path integrals (hence the circle in [imath]\oint_{C_1}[/imath], and I parametrized the paths.

 

Perhaps you can suggest what you'd like me to compute in Mathematica to get a result identical to yours? Choose the limits, C1, and C2 to give the "right" answer.

 

 

 

 

Close, but you are still mixing two very different equations. Magnetic force IS NOT DEFINED with Ampere's force law, it is defined with Biot-Savart law and Lorentz force equations. This is Ampere's force law, it is some OLD equation that specifically relates only to parallel wires and force per unit length:

[math]F_m = F/1_m = 2 k_A \frac {I_1 I_2 } {r}[/math]

 

 

This below is NOT Ampere's force law, and these two equations ARE NOT EQUAL:

[math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_1} \oint_{C_2} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2}

[/math]

 

...this is Biot-Savart and Lorentz force combined, and the difference should be obvious, beside that the units are different, as you finally have realized now. We are getting closer to truth. Very good, I'm proud of you.

 

http://en.wikipedia.org/wiki/Ampere%27s_force_law

A more general formulation of Ampère's force law for arbitrary geometries is based upon line integrals, and is as follows:

[math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_1} \oint_{C_2} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2}

[/math]

 

Not Ampere's force law?

 

Yes, and if that doesn't work just use a bigger hammer.

 

My hammer's plenty big enough, if you know what I mean.

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<< VectorAnalysis`

C1[t_] := {t, 0, 0};

dC1[t_] := {1, 0, 0};

C2[t_] := {t, 1, 0};

dC2[t_] := {1, 0, 0};

mu = 4 \[Pi]*10^-7;

(mu * 1 * 1)/(4 \[Pi]) \!\(
\*SubsuperscriptBox[\(\[integral]\), \(0\), \(1\)]\(
\*SubsuperscriptBox[\(\[integral]\), \(-\[infinity]\), \(\[infinity]\)]
FractionBox[\(CrossProduct[dC2[s2], \ CrossProduct[dC1[s1], 
\*FractionBox[\(C1[s1] - C2[s2]\), \(Norm[C1[s1] - C2[s2]]\)]]]\), 
SuperscriptBox[\(Norm[
       C1[s1] - 
        C2[s2]]\), \(2\)]] \[DifferentialD]s1 \[DifferentialD]s2\)\) \
// N

{0., 2.*10^-7, 0.}

Edited April 8, 2010 by Cap'n Refsmmat
Consecutive posts merged.

[swansont]

[math]1 \mathrm{C} = 1 \mathrm{A} \cdot 1 \mathrm{s}[/math]

 

Where did that meter disappear from here? Those two do not equate, or do they?

 

It didn't "disappear." [math]I\int dl \[/math] isn't an equation that gives you the charge

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Sadly, no one managed to realize what are the units of this equation and point if there are any mistakes in this dimensional analysis:

 

[math]F_{12} = \frac {\mu_0}{4 \pi} \int \int \frac {I_1*dl_1 \times (I_2*dl_2 \times \hat{r}_{21} )}{|r|^2} = \frac {\mu_0*I_1* \int dl_1 \times (I_2* \int dl_2 \times \hat{r}_{21} )}{4 \pi*|r|^2}[/math]

 

I don't think the validity of this equation, or the Biot-Savart law, were ever in question. It's the result one gets from evaluating the integrals (or not) that was in question.

[ambros]

The derivative of C2 is {1' date=' 0, 0}. The 1 in C2 is a constant, so it does not appear in the derivative.

[/quote']

 

It does appear, it is unreadable, but that much everyone can see.

Regardless if it is constant or not you made wire 2 cross over to wire 1.

 

C1= {t, 0, 0}: X=0, Y=0
dC1= {1, 0, 0}: X=1, Y=0

C2= {t, 1, 0}: X=0, Y=1
dC2= {1, 0, 0}: X=1, Y=0

    Y
      ^
      |
      |
      |        C2
 y=1 -C....................D------------------wire 2--->
      |   . 
      |         .    
      |                 .
 y=0 -A....................B------------------wire 1---> 
      |        C1          |                             X
     x=0                  x=1

 

Line C1 goes from A(0,0) to B(1,0)

 

Line C2 goes from C(0,1) to B(1,0) ---> it should be D(1,1)

 

If you want your el. current to go along x axis you need Y coordinate to be CONSTANT.

 

 

 

So I'm making one wire infinitely long, by making the path infinitely long.

 

And when you come back to real life where both wires are ACTUALLY 1 meter long, then this equation will not work correctly? And since one wire is much shorter than it is SUPPOSED to be, that then means the result in reality would be much less than predicted by your result, right?

 

 

Perhaps you can suggest what you'd like me to compute in Mathematica to get a result identical to yours? Choose the limits, C1, and C2 to give the "right" answer.

 

Those are not really definite integrals, it's more of Antiderivative actually. Your choice to make one wire 1m long and the other infinite is completely arbitrary, you should solve that with indefinite integrals or antiderivatives, alternatively, make them both 1m, both infinite, of make them both infinitesimal.

 

(mu * 1 * 1)/(4 \[Pi]) \!\(
\*SubsuperscriptBox[\(\[integral]\), \(0\), \(1\)]\(
\*SubsuperscriptBox[\(\[integral]\), \(-\[infinity]\), \(\[infinity]\)]
FractionBox[\(CrossProduct[dC2[s2], \ CrossProduct[dC1[s1], 
\*FractionBox[\(C1[s1] - C2[s2]\), \(Norm[C1[s1] - C2[s2]]\)]]]\), 
SuperscriptBox[\(Norm[
       C1[s1] - 
        C2[s2]]\), \(2\)]] \[DifferentialD]s1 \[DifferentialD]s2\)\) \
// N

{0., 2.*10^-7, 0.}

 

You are not hiding anything, eh?

Write the equation properly so it is readable.

[swansont]

And when you come back to real life where both wires are ACTUALLY 1 meter long, then this equation will not work correctly? And since one wire is much shorter than it is SUPPOSED to be, that then means the result in reality would be much less than predicted by your result, right?

 

You adjust the limits of the integral appropriate to the length of the wire. Yes, the results will be smaller, as people have been telling you all along.

 

But this is supposed to be a thread about unit analysis.

 

For example, if one were to solve the Biot-Savart law for a long wire

 

[math]B® = \frac{\mu_0 I}{4\pi r}[/math]

 

This gives us units of (Newtons/Ampere^2) * Ampere * 1/meter = Newtons/(meter*Ampere) (correct units)

 

[math]B® = \frac{\mu_0 I}{4\pi r^2}[/math]

 

This gives us units of (Newtons/Ampere^2) * Ampere * 1/meter^2 = Newtons/(meter^2*Ampere) (incorrect units)

[ambros]

If their answer disagrees with yours' date=' will you believe them? I hope they answer.

[/quote']

 

I'm the one who contacted them, it kind of implies I regard their opinion highly, but they will get some pepper in either case. If I'm wrong then they are stupid for saying: "force between two wires", instead of: "force on one wire". This is not supposed to be so vague, something stinks, whatever the case.

 

 

Undefined. Without knowing what limits there are on the integral, you don't get a definite answer.

 

Infinity is undefined, so how can we define the ampere unit with that?

 

 

No, the result of that integral would have units of charge times velocity (ampere*meters if you prefer). What were you trying to calculate? Does this surprise you?

 

Length units get canceled in a way to show that distance does not matter, but I don't think I wish to pursue that argument, it's kind of boring and I really forgot what was my punch-line.

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You adjust the limits of the integral appropriate to the length of the wire. Yes, the results will be smaller, as people have been telling you all along.

 

But this is supposed to be a thread about unit analysis.

 

For example, if one were to solve the Biot-Savart law for a long wire

 

[math]B® = \frac{\mu_0 I}{4\pi r}[/math]

 

This gives us units of (Newtons/Ampere^2) * Ampere * 1/meter = Newtons/(meter*Ampere) (correct units)

 

[math]B® = \frac{\mu_0 I}{4\pi r^2}[/math]

 

This gives us units of (Newtons/Ampere^2) * Ampere * 1/meter^2 = Newtons/(meter^2*Ampere) (incorrect units)

 

You made me repeat so many times. Do you have a memory loss problem?

 

[math]B® = \frac{\mu_0 I * dl}{4\pi r^2} \ N/A*m[/math]

 

 

 

----

These equations are what defines all of the electronics and electrics, that people DO USE in the real world. These equation are EXACT and not approximation, we would not used them to define units if they were not, would we? -- If they were not exact or inapplicable to regular wires in real world, than no one would be able to properly measure anything that has to do with electric currents. -- How do you imagine instruments that measure amperes work and based on what formulas? Are you saying there are some better formulas that one given by BIPM?

Edited April 8, 2010 by ambros
Consecutive posts merged.

[Cap'n Refsmmat]

It does appear, it is unreadable, but that much everyone can see.

Regardless if it is constant or not you made wire 2 cross over to wire 1.

 

C1= {t, 0, 0}: X=0, Y=0
dC1= {1, 0, 0}: X=1, Y=0

C2= {t, 1, 0}: X=0, Y=1
dC2= {1, 0, 0}: X=1, Y=0

    Y
      ^
      |
      |
      |        C2
 y=1 -C....................D------------------wire 2--->
      |   . 
      |         .    
      |                 .
 y=0 -A....................B------------------wire 1---> 
      |        C1          |                             X
     x=0                  x=1

 

Line C1 goes from A(0,0) to B(1,0)

 

Line C2 goes from C(0,1) to B(1,0) ---> it should be D(1,1)

 

If you want your el. current to go along x axis you need Y coordinate to be CONSTANT.

Yup. Fortunately, the y coordinates are constant in all of my equations, as you can see. t is only involved in the x coordinates.

 

Or perhaps I should be more clear: in no case do I go from C1 to dC1 or C2 to dC2, which would screw up directions. No, dC1 and dC2 are not position vectors but direction vectors, so what matters is that they point in the correct direction.

 

Even more clearly, C1 and C2 parametrize the paths, whereas dC1 and dC2 are the tangent vectors to those paths. This is standard vector calculus.

 

And when you come back to real life where both wires are ACTUALLY 1 meter long, then this equation will not work correctly? And since one wire is much shorter than it is SUPPOSED to be, that then means the result in reality would be much less than predicted by your result, right?

If both wires are 1 meter long, the result is [math]8.28427\times 10^{-8}\mbox{ N}[/math]. Fortunately, the BIPM stipulates infinite wires so we can look at the force on one meter of an infinite wire.

 

Those are not really definite integrals, it's more of Antiderivative actually. Your choice to make one wire 1m long and the other infinite is completely arbitrary, you should solve that with indefinite integrals or antiderivatives, alternatively, make them both 1m, both infinite, of make them both infinitesimal.

No. They are line integrals. They are to be integrated along the lines as definite integrals. Look:

 

[math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_1} \oint_{C_2} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2}

[/math]

 

See the C₁ and C₂? Those are paths to be integrated along. You do the definite integrals along those paths. If these were indefinite integrals, there would be no need for the paths.

 

Consider: If the wires had some weird geometry (formed into squares or pyramids), the forces between them would be different, yes? But the equation is a general one, since you just parametrize the paths described by each wire and you'll get the correct answer.

 

You are not hiding anything, eh?

Write the equation properly so it is readable.

http://www.scienceforums.net/forum/attachment.php?attachmentid=2456

 

I just copied and pasted that out of Mathematica, and that's what I got. Can't blame me for Mathematica's weirdness.

 

If it helps, here's a rewritten form of the equation in Mathematica syntax:

In[10]:= (mu * 1 * 1)/(4 \[Pi]) * 
 Integrate[
  Integrate[(CrossProduct[dC2[s2], 
      CrossProduct[
       dC1[s1], (C1[s1] - C2[s2])/Norm[C1[s1] - C2[s2]]]])/(Norm[
      C1[s1] - C2[s2]]^2), {s1, -\[infinity], \[infinity]}], {s2, 0, 
   1}] // N

Out[10]= {0., 2.*10^-7, 0.}

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These equations are what defines all of the electronics and electrics, that people DO USE in the real world. These equation are EXACT and not approximation, we would not used them to define units if they were not, would we? -- If they were not exact or inapplicable to regular wires in real world, than no one would be able to properly measure anything that has to do with electric currents. -- How do you imagine instruments that measure amperes work and based on what formulas? Are you saying there are some better formulas that one given by BIPM?

 

I don't know any current sensors that measure the force between two infinitely long wires, no. But you can calculate the force from shorter wires of complex geometry using the same formula I gave (just change the paths to something more complex), and you can build a sensor to measure that.

Edited April 8, 2010 by Cap'n Refsmmat

[swansont]

I'm the one who contacted them, it kind of implies I regard their opinion highly, but they will get some pepper in either case. If I'm wrong then they are stupid for saying: "force between two wires", instead of: "force on one wire".

 

No, that terminology is pretty standard and would not tend to trip up people who have studied physics. Since Newton's third law mandates that [math]F_{12} = - F_{21}[/math], it is understood that it's the force that either one exerts on the other, since they have the same magnitude. Using the sum makes no physical sense, and it would not tend to occur to a physicist to do that.

 

 

You made me repeat so many times. Do you have a memory loss problem?

 

[math]B® = \frac{\mu_0 I * dl}{4\pi r^2} \ N/A*m[/math]

 

No, I think my memory is working fine in this instance. Clicking on the little arrow icon takes you to the post in question; these are three separate instances where you made this claim.

 

[math]B®= \frac{\mu_0 I d\mathbf{l} \times \mathbf{r}}{4\pi r^2} = \frac{\mu_0 I sin(angle)}{4\pi r^2} = \frac{\mu_0 I sin(90)}{4\pi r^2} = \frac{\mu_0 I * 1}{4\pi r^2} = \frac{\mu_0 I}{4\pi r^2}[/math]

 

[math]\mathbf{B} = \frac{\mu_0 q \mathbf{v}}{4\pi} \times \frac{\mathbf{\hat r}}{r^2} = \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2} =wires=> \frac{\mu_0 I}{4\pi r^2}[/math]

 

 

I SAY THIS IS CORRECT: [imath]B = \frac{\mu_0 I}{4\pi r^2} [/imath]

 

YOU SAY THIS IS CORRECT: [imath]B = \frac{\mu_0 I}{2\pi r} [/imath]

 

———

 

These equations are what defines all of the electronics and electrics, that people DO USE in the real world. These equation are EXACT and not approximation, we would not used the to define units if they were, would we? -- If they were not exact or inapplicable to regular wires in real world, than no one would be able to properly measure anything that has to do with electric currents. -- How do you imagine instruments that measure amperes work and based on what formulas? Are you saying there are some better formulas that one given by BIPM?

 

The irony here is so delicious. (The irony being you lecturing me on metrology practices)

 

Yes, I am telling you that these exact solutions are not the ones used in the real world. The equations are derived are ideal cases (e.g. infinite wires), that do not work to infinite precision in the real world, though the scientists, engineers and technicians try and get as close to the ideal case when they can.

 

In the ampere balance case, using a 1m length of wire vs an infinite wire would introduce only a small error for small separations of the wire that in a college physics class lab you wouldn't bother to correct the equation. (You can see this by adjusting the integration limits) But a standards lab, aiming for many more orders of magnitude of precision and accuracy, would most assuredly take this into account. Or they would take an approach such that a correction was not necessary.

 

I even provided you with a quote from a metrology site that explains this (BTW, I don't think you ever answered if you really thought the BIPM had an ampere balance with infinite wires. Where do they keep it?)

 

Here's one from the BIPM directly

 

The realization to high accuracy of the ampere (a base unit of the SI), the ohm and the volt (derived units of the SI) directly in terms of their definitions is difficult and time consuming. The best such realizations of the ampere are now obtained through combinations of realizations of the watt, the ohm and the volt. The watt realized electrically is compared by balance experiments with the watt realized mechanically. These experiments employ a coil in a magnetic flux and are devised in such a way that it is not necessary to know either the dimensions of the coil or the magnitude of the flux density.

 

http://www.bipm.org/en/si/si_brochure/appendix2/electrical.html

 

Got that? They use a Watt balance, not an Ampere balance, and it's a coil, not straight wires. Why? Because that's the best way of doing it precisely. They solve the equations for their particular circumstance, and choose a geometry that minimizes extra measurements (since any measurement introduces error) — in this case, by choosing a geometry such that the loop size doesn't impact the answer, so they don't have to measure the loop size. That gives a more precise result than a geometry that does depend on some dimension that must be measured.

[ambros]

Yup. Fortunately, the y coordinates are constant in all of my equations, as you can see.

 

No, I can not see and you obviously do not care to be understood. If you wish me to look at it, and also make it easier for everyone else, then you will write it down.

 

 

 

No. They are line integrals. They are to be integrated along the lines as definite integrals. Look:

 

Uh, think as you wish, line or not has nothing to do with whether it should be solved with antiderivative or not. Enough has been said in a form of an opinion, show your work and let us see the math. I can not make any meaningful comment on any of that unless the equation is on the table, sorry.

 

 

It does not work with ether of these two:

 

http://integrals.wolfram.com

http://www.wolframalpha.com/input/

 

Can you make it work online? -- What are the s1 and s2, what numerical value(s) do they have? What value did you say 't' has in your equation? How come online version is able to print out those integrals in normal notation and you can not do that? Or, can you?

Edited April 8, 2010 by ambros

[Mr Skeptic]

I'm the one who contacted them, it kind of implies I regard their opinion highly, but they will get some pepper in either case. If I'm wrong then they are stupid for saying: "force between two wires", instead of: "force on one wire". This is not supposed to be so vague, something stinks, whatever the case.

 

The force of each wire on the other. Both experience the same force.

 

Infinity is undefined, so how can we define the ampere unit with that?

 

Just because infinity is undefined, does not mean that it is not useful. Mathematicians often take the limit of something as one variable tends toward infinity (and the same for division by infinitesimals). In fact, I don't know how you could understand calculus if you don't understand the useages of infinity (an integral is the limit of a sum as the number of summed elements tends toward infinity, for example).

 

You made me repeat so many times. Do you have a memory loss problem?

 

And another thread starts marching toward where threads go to die. (Though that's not my prerogative this time). Let's not go the route of endless repetition again.

 

[math]B® = \frac{\mu_0 I * dl}{4\pi r^2} \ N/A*m[/math]

 

That has the correct units but makes no sense. The infinitesimal is going to make the result zero unless you integrate.

Edited April 8, 2010 by Mr Skeptic

[ambros]

No' date=' that terminology is pretty standard and would not tend to trip up people who have studied physics. Since Newton's third law mandates that [math']F_{12} = - F_{21}[/math], it is understood that it's the force that either one exerts on the other, since they have the same magnitude. Using the sum makes no physical sense, and it would not tend to occur to a physicist to do that.

 

I said already those are TWO DIFFERENT OBJECTS. Do we cancel force vectors when they do NOT act on the same object? No! -- Why don't you try to put your finger BETWEEN TWO permanent magnets and tell me if the force you feel would be combined (magnitude sum), or that of only one magnet?

 

 

No, I think my memory is working fine in this instance. Clicking on the little arrow icon takes you to the post in question; these are three separate instances where you made this claim.

 

Ay, caramba! I was solving the problem numerically and I FORGOT that term as it made no difference then, but we discussed it later and I corrected it, so it was suppose to be concluded - post #106:

 

http://www.scienceforums.net/forum/showpost.php?p=555790&postcount=106

 

[math]\mathbf{B} = \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2} = \frac{\mu_0 I*d\mathbf{l}*sin(alpha)}{4\pi r^2} = \frac{\mu_0 I*d\mathbf{l}*sin(90)}{4\pi r^2} = \frac{\mu_0 I*d\mathbf{l}}{4\pi r^2} \ N/A*m[/math]

 

 

This is standard derivation and what can be converted to point charge with the term:

 

 

I*int(dl) = q*v = 1C*1m/s

 

...while "your" equation has only 2Pi and can not be equated like that. Which is the point I forgot about earlier on. My equation does have speed, but your one is missing LENGTH.

 

YOU: 1C = 1A*1s

 

ME: 1C*1m/s = 1A*1m

 

 

The irony here is so delicious.

 

Hehe, it's a double irony. It is my ultimate point to show how ampere unit is the most ridiculous unit and circularly defined mess of self referencing nonsense, which makes all but three equations rather dubious. -- However, those three equations are accurate and applicable to the real world without infinite wires, the errors are not related to integration at all, as these equation can be very successfully applied to free charges, like electron beams, Z-pinch applications and such.

Edited April 8, 2010 by ambros

[Mr Skeptic]

The irony here is so delicious. (The irony being you lecturing me on metrology practices)

 

What, just because your clock is fancier than mine?

 

Hehe, it's a double irony. It is my ultimate point to show how ampere unit is the most ridiculous unit and circularly defined mess of self referencing nonsense, which makes all but three equations rather dubious. -- However, those three equations are accurate and applicable to the real world without infinite wires, the errors are not related to integration at all, as these equation can be very successfully applied to free charges, like electron beams, Z-pinch applications and such.

 

Just so you know, a lot more people ask him what time it is than ask you about any measure.

Edited April 8, 2010 by Mr Skeptic

[Cap'n Refsmmat]

No, I can not see and you obviously do not care to be understood. If you wish me to look at it, and also make it easier for everyone else, then you will write it down.

[math]\mathbf{C1}(t) = (t, 0, 0)[/math]

 

[math]\frac{\partial}{\partial t} \mathbf{C1}(t) = \mathbf{dC1}(t) = (1,0,0)[/math]

 

[math]\mathbf{C2}(t) = (t, 1, 0)[/math]

 

[math]\frac{\partial}{\partial t} \mathbf{C2}(t) = \mathbf{dC2}(t) = (1,0,0)[/math]

 

[math]\mathbf{F}_{12} = \frac{\mu_0 I_1 I_2}{4 \pi} \int_0^1 \int_{-\infty}^{\infty} \frac{\mathbf{dC2}(s2) \times \left( \mathbf{dC1}(s1) \times \frac{\mathbf{C1}(s1) - \mathbf{C2}(s2)}{||\mathbf{C1}(s1) - \mathbf{C2}(s2)||}\right) }{||\mathbf{C1}(s1) - \mathbf{C2}(s2)||^2}[/math]

 

That clear enough for you?

 

Uh, think as you wish, line or not has nothing to do with whether it should be solved with antiderivative or not.

It does mean it has to be a definite integral.

 

Can you make it work online?

No. Go buy Mathematica if you want.

-- What are the s1 and s2, what numerical value(s) do they have?

They are the variables being integrated, like x in:

 

[math]\int_0^4 x \, dx[/math]

 

What value did you say 't' has in your equation?

None. It's an argument to the path function. C1 and C2 are functions. C1[s1] substitutes s1 for t in the equation, of course. Just like [imath]f(x) = 4x[/imath] and [imath]f(4) = 16[/imath].

 

How come online version is able to print out those integrals in normal notation and you can not do that? Or, can you?

I can -- in the screenshots. It's how Mathematica works.

[Mr Skeptic]

Uh, think as you wish, line or not has nothing to do with whether it should be solved with antiderivative or not. Enough has been said in a form of an opinion, show your work and let us see the math. I can not make any meaningful comment on any of that unless the equation is on the table, sorry.

 

 

It does not work with ether of these two:

 

http://integrals.wolfram.com

http://www.wolframalpha.com/input/

 

Sure it works. You just got to tell it what path it's integrating on. You gotta find what dl is.

 

You don't even need math to prove that this has to be a line integral. Consider current flowing in a closed loop: does it generate a magnetic field, or doesn't it? If you don't use a path integral, you integrate over zero distance and get zero, which doesn't make sense.

[ambros]

[math]\frac{\partial}{\partial t} \mathbf{C1}(t) = \mathbf{dC1}(t) = (1' date='0,0)[/math']

 

[math]\frac{\partial}{\partial t} \mathbf{C2}(t) = \mathbf{dC2}(t) = (1,0,0)[/math]

 

So, your variables do not have initial values and yet you managed to produce that result of 2*10^-7 all by itself, out of symbols? Those are either completely unnecessary or your program does not compute at all. -- Do you realize those two paths share the same coordinates?

 

 

[math]

\mathbf{F}_{12} = \frac{\mu_0 I_1 I_2}{4 \pi} \int_0^1 \int_{-\infty}^{\infty} \frac{\mathbf{dC2}(s2) \times \left( \mathbf{dC1}(s1) \times \frac{\mathbf{C1}(s1) - \mathbf{C2}(s2)}{||\mathbf{C1}(s1) - \mathbf{C2}(s2)||}\right) }{||\mathbf{C1}(s1) - \mathbf{C2}(s2)||^2}

[/math]

 

 

You are not making any angles, this term equals +/- ONE: [math]\frac{\mathbf{C1}(s1) - \mathbf{C2}(s2)}{||\mathbf{C1}(s1) - \mathbf{C2}(s2)||} [/math]

 

 

They are the variables being integrated, like x in:

 

You have to tell me exactly how do you initialize those values to obtain that result of 2*10^-7 N, numbers can not become out of symbols..

 

t = 0? 1? 0 to 1? 0 to infinity? - to + infinity? something else?

s1 = 0? 1? 0 to 1? 0 to infinity? - to + infinity? something else?

s2 = 0? 1? 0 to 1? 0 to infinity? - to + infinity? something else?

Edited April 8, 2010 by ambros

[swansont]

Ay, caramba! I was solving the problem numerically and I FORGOT that term as it made no difference then, but we discussed it later and I corrected it, so it was suppose to be concluded - post #106:

 

http://www.scienceforums.net/forum/showpost.php?p=555790&postcount=106

 

[math]\mathbf{B} = \int\frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2} = \frac{\mu_0 I*d\mathbf{l}*sin(alpha)}{4\pi r^2} = \frac{\mu_0 I*d\mathbf{l}*sin(90)}{4\pi r^2} = \frac{\mu_0 I*d\mathbf{l}}{4\pi r^2} \ N/A*m[/math]

 

As has been pointed out numerous times, this equation is wrong because the angle is a variable, and a function of the variable of integration dl, and dl should not remain in the equation if you've done the integral. Mathematically, this is a disaster; two of the four "=" signs are incorrect.

[ambros]

Sure it works. You just got to tell it what path it's integrating on. You gotta find what dl is.

 

Ok, can you then tell us how to input this equation with different parameters of integrals paths:

 

[math]

F_{12} = \frac {\mu_0}{4 \pi} \int \int \frac {I_1*dl_1 \times (I_2*dl_2 \times \hat{r}_{21} )}{|r|^2} = \frac {\mu_0*I_1* \int dl_1 \times (I_2* \int dl_2 \times \hat{r}_{21} )}{4 \pi*|r|^2}

[/math]

 

 

You don't even need math to prove that this has to be a line integral. Consider current flowing in a closed loop: does it generate a magnetic field, or doesn't it? If you don't use a path integral, you integrate over zero distance and get zero, which doesn't make sense.

 

There is no such thing as zero distance, just like there is no edge to the universe. I explained this in previous thread. -- It is 'line integral', why is that under question? The trick is that if you do "point integral" it still works.

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As has been pointed out numerous times, this equation is wrong because the angle is a variable, and a function of the variable of integration dl, and dl should not remain in the equation if you've done the integral. Mathematically, this is a disaster; two of the four "=" signs are incorrect.

 

I laugh at your hands waving through the air. -- Your equation is wrong because it does not have and reference to LENGTH which you insist is so important. -- If it is "mathematical disaster" than you should be able to demonstrate it by using MATHEMATICS, stop waving hands.

 

Can you show you work?

 

 

 

[math]B = \frac{\mu_0*I}{2\pi r} [/math]

 

Where is the length, and where did 2 Pies go?

 

Where are the freaking variable angles in your equation???

Edited April 8, 2010 by ambros
Consecutive posts merged.

[darkenlighten]

There is no such thing as zero distance, just like there is no edge to the universe. I explained this in previous thread. -- It is 'line integral', why is that under question? The trick is that if you do "point integral" it still works.

 

The only thing I know that is close to a "point integral" is an impulse. And this is clearly not what we have. This is the ENTIRE argument ambros. It has nothing to do with which equation is right or wrong. It's the fact you are calculating it WRONG, end of story.

 

How bout this look: [math] I \int d\mathbf{l} = I x [/math] where x is the length of your straight wire. We will take out the integral since we are working with a straight path. Try it now, you will see the problem for the infinite wire.

[ambros]

...calculating it WRONG' date=' end of story.

[/quote']

 

Q1: Did you ever use this equation to calculated the force of attraction between two parallel wires 1m apart and both having steady current of 1A:

 

[math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_1} \oint_{C_2} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2}

[/math]

 

YES/NO?

 

 

Q2: Are you hallucinating, laying and fulling yourself, or you actually can SHOW YOUR WORK so we can see your LOGIC and MATHEMATICS? -- Until then, you're only amusing me and fooling yourself.

Edited April 8, 2010 by ambros

[Mr Skeptic]

Ok, can you then tell us how to input this equation with different parameters of integrals paths:

 

[math]

F_{12} = \frac {\mu_0}{4 \pi} \int \int \frac {I_1*dl_1 \times (I_2*dl_2 \times \hat{r}_{21} )}{|r|^2} = \frac {\mu_0*I_1* \int dl_1 \times (I_2* \int dl_2 \times \hat{r}_{21} )}{4 \pi*|r|^2}

[/math]

 

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

 

In short, convert f(l) * dl to f(l(t)) * dl/dt * dt and integrate with respect to t. You need to know the function of l. The t need not be time.

 

There is no such thing as zero distance, just like there is no edge to the universe. I explained this in previous thread. -- It is 'line integral', why is that under question? The trick is that if you do "point integral" it still works.

 

Oh? What's the distance between a point and itself? (let's say you measure it in meters).

Edited April 8, 2010 by Mr Skeptic

[Cap'n Refsmmat]

So, your variables do not have initial values and yet you managed to produce that result of 2*10^-7 all by itself, out of symbols? Those are either completely unnecessary or your program does not compute at all. -- Do you realize those two paths share the same coordinates?

These are not paths. They are direction vectors. The paths are C1 and C2.

 

Tell me, if I did this:

 

[math]f(x) = x^2[/math]

 

[math]\int_0^2 f(y) \, dy[/math]

 

... do I need to supply a value for x?

 

No. f(x) is a function, not an expression. C1 and C2 are functions. C1[s1] is {s1, 0, 0}. As s1 varies according to the limits of the integral, the value changes.

 

You are not making any angles, this term equals +/- ONE: [math]\frac{\mathbf{C1}(s1) - \mathbf{C2}(s2)}{||\mathbf{C1}(s1) - \mathbf{C2}(s2)||} [/math]

Nope. The numerator is a vector. The denominator is a scalar. This term gives us a unit vector.

 

(A vector minus another vector is a vector.)

 

t = 0? 1? 0 to 1? 0 to infinity? - to + infinity? something else?

s1 = 0? 1? 0 to 1? 0 to infinity? - to + infinity? something else?

s2 = 0? 1? 0 to 1? 0 to infinity? - to + infinity? something else?

t is unimportant, as I explain above.

 

s1 and s2 vary according to the limits of the integral. [math]-\infty \leq s1 \leq \infty[/math] and [math]0\leq s2 \leq 1[/math], as you can see by the integrals.

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I laugh at your hands waving through the air. -- Your equation is wrong because it does not have and reference to LENGTH which you insist is so important. -- If it is "mathematical disaster" than you should be able to demonstrate it by using MATHEMATICS, stop waving hands.

 

As has been explained (and shown through dimensional analysis), that's because it's an equation for the force per unit length between two infinite wires. Newtons per meter.

[darkenlighten]

Q1: Did you ever used this equation to calculated the attraction of two parallel wires 1m apart with both having steady current of 1A:

 

[math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_1} \oint_{C_2} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2}

[/math]

 

YES/NO?

 

Yes I did.

 

Q2: Are you hallucinating, laying and fulling yourself, or you actually can SHOW YOUR WORK so we can see your LOGIC and MATHEMATICS?

 

Until then, you're only amusing me and fooling yourself.

 

I wish I was.

 

I showed you how this equation was derived. You ignored it. I showed you the steps of integration. You ignored it. I've put in real variables/constants to show you how you get 2 x 10^-7 N per unit length. You ignored that.

 

I'm feeling a strong trend going on here.

[ambros]

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

 

In short, convert f(l) * dl to f(l(t) * dl/dt * dt and integrate with respect to t. You need to know the function of l. The t need not be time.

 

Can you print out the complete code for the complete equation? -- There is no 't', no time, as I tried to explain at the beginning. I'm the only one here who actually got any result that makes sense from that equation, and I did it in three different ways, here are two:

 

 

TWO PARALLEL WIRES, ampere unit setup: I1=I2= 1A; r= 1m

*DO NOT MODIFY EQUATIONS, USE THEM & CALCULATE: No.2

================================================

 

[math]F_{12} = \frac {\mu_0}{4 \pi} \oint_{C_1} \oint_{C_2} \frac {I_1*dl_1 \times (I_2*dl_2 \times \hat{r}_{21} )}{|r|^2}[/math]

 

 

 

[math]=> \frac {\mu_0*Q_1*v_1 \times (Q_2*v_2 \times \hat{r}_{21} )}{4 \pi*|r|^2}[/math]

 

[math]=> \frac {4 \pi*10^-7_{N/A^-2}*1_C*1_{m/s} \times (1_C*1_{m/s} \times 1)}{4 \pi*1_{m^2}}[/math]

 

[math]=> \frac {4 \pi*10^-7_{N/A^-2}*1_A*1_m \times (1_A*1_m)}{4 \pi*1_{m^2}} \ =OR= \ 10^-7_{N/A^-2}*1_{C/s} \times (1_{C/s})[/math]

 

[math]=> 10^-7_{N/A^-2}*1_A \times 1_A \ = \ 10^-7 \ N[/math]

 

 

 

TWO PARALLEL WIRES, ampere unit setup: I1=I2= 1A; r= 1m

*DO NOT MODIFY EQUATIONS, USE THEM & CALCULATE: No.3

================================================

 

[math]B= \frac{\mu_0*Q*v \times \hat{r}}{4\pi*|r|^2}[/math]

 

 

[math]=> \frac{4\pi*10^-7_{N/A^-2}*1_C*1_{m/s} \times 1}{4\pi*1_{m^2}}[/math]

 

[math]=> \frac{10^-7_{N/A^-2}*1_A*1_m }{1_{m^2}} \ =OR= \ \frac{10^-7_{N/A^-2}*1_{C/s}}{1_m}[/math]

 

[math]=> \frac{10^-7_{N/A}}{1_m} \ = \ 10^-7 \ N/A*m \ (1Tesla)[/math]

 

 

[math]F = Q*v \times B[/math]

 

[math]=> 1_C*1_{m/s} \times 10^-7_{N/A*m}[/math]

 

[math]=>1_{C/s}*1_m \times 10^-7_{N/A*m} \ =OR= \ 1_A*1_m \times 10^-7_{N/A*m} \ = \ 10^-7 \ N[/math]

 

 

Oh? What's the distance between a point and itself? (let's say you measure it in meters).

 

That one would be zero, but it does not qualify as distance as long as there are not at least TWO entities, BETWEEN which the distance can be measured.

 

 

I even provided you with a quote from a metrology site that explains this (BTW' date=' I don't think you ever answered if you really thought the BIPM had an ampere balance with infinite wires. Where do they keep it?)

[/quote']

 

Huh?? It is you who is saying those equations do not work in real life and that ONLY work with infinite wires. I can quote you saying that. -- I'm saying they work just the same. It is *Maxwell's equations* that have trouble with the ENDS of the wire (cylinder cuts) because their two-dimensionality, and that is the only reason why we have 'infinite wires'.

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I've put in real variables/constants to show you how you get 2 x 10^-7 N per unit length. You ignored that.

 

You are dreaming. Go ahead, show us that post where we can see how you put REAL VARIABLES of: I1= 1A, I2= 1A and r=1m, in THIS EQUATION:

 

[math]

\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_1} \oint_{C_2} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2}

[/math]

 

What post, what number? What say you?

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As has been explained (and shown through dimensional analysis)' date=' that's because it's an equation for the [b']force per unit length[/b] between two infinite wires. Newtons per meter.

 

This is ridiculous.

 

post #23: http://www.scienceforums.net/forum/showpost.php?p=556655&postcount=23

 

 

[math]

\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_1} \oint_{C_2} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2}

[/math]

 

Cap'n Refsmmat: ...gives us the force on a piece of wire -- its units are Newtons, not Newtons per meter.

 

 

* Ampere's force law IS in N/m. Fm =MEANS= F/1meter

 

* That equation given by BIPM is in Newtons (N) and that is NOT Ampere's force law but Biot-Savart-law and Loerntz force as demonstrated above.

Edited April 8, 2010 by ambros
Consecutive posts merged.