Cap'n Refsmmat

Please, let's not start fighting on page 2 of a thread. We can stay civil for longer than this.

Hmm. This thread would be a whole lot easier if we could establish one or two big details. So, ambros, I have a challenge for you, that I'd like to use to better understand your position. Perhaps I can figure out how you're doing this integration this way. So, here's the situation. I have two 1m lengths of wire. One of these is curled into a loop of 1m circumference. They are arranged something like this: Ignore the multiple loops -- I just wanted a diagram that gives a general idea. We have one wire going vertically and another looping around that wire. The loop is exactly in the middle of the vertical wire, 0.5m from the bottom and 0.5m from the top. There is one amp of current going from the top to the bottom of the center wire. There is one amp of current going counterclockwise (as viewed from above) in the loop. What is the net force on the loop due to the magnetic field? Use 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] Then I can understand exactly how you're doing this.

Whoops. Sorry, that's the equation for the magnetic field around an infinite wire. There's no length because it's for an infinitely long wire. If you want to calculate the magnetic field for a finite wire, use Biot-Savart.

Please learn to read. You were talking about this equation: [math] B = \frac{\mu_0 I}{2\pi r} [/math] and that is the equation my comment is relevant to. The above equation has units of Newtons per meter. (Well, the proper form of it does. The numerator should have the currents of both wires in it, not just one wire, but I believe it was set up for a situation where the wires have equal currents.) Incidentally, it is Ampere's force law. Take a look on Wikipedia or on this site. Incidentally, that link plugs the numbers in and solves for the force on one wire. You might be interested.

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. 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 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. Merged post follows: Consecutive posts merged 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.

Why even allocate neurons to ponder if you should not allocate neurons to ponder the first question?

[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? It does mean it has to be a definite integral. No. Go buy Mathematica if you want. They are the variables being integrated, like x in: [math]\int_0^4 x \, dx[/math] 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]. I can -- in the screenshots. It's how Mathematica works.

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. 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. 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. 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.} Merged post follows: Consecutive posts merged 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.

No. The derivative of C2 is {1, 0, 0}. The 1 in C2 is a constant, so it does not appear in the derivative. Doesn't matter what I name it. I just chose t because that's what I always used in path integrals. http://en.wikipedia.org/wiki/Ampere%27s_force_law So I'm making one wire infinitely long, by making the path infinitely long. [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. 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. http://en.wikipedia.org/wiki/Ampere%27s_force_law Not Ampere's force law? My hammer's plenty big enough, if you know what I mean. Merged post follows: Consecutive posts merged << 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.}

I think this should settle things nicely... (click to see large version) Here's what I did in Mathematica: I defined C1 and C2, the paths representing the wires, as in the original 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] You can see that they are 1 meter apart, since C2 is one meter higher in the y dimension. Next, I specified their derivatives, which gives us [math]d \mathbf{s_1}[/math] and [math]d \mathbf{s_2}[/math]. Next, I specified [math]\mu_0[/math]. Then I wrote out the Ampere force law as an integral. You'll note the limits on the integral. Ampere's force law gives us the force on a piece of wire -- its units are Newtons, not Newtons per meter. So one wire is infinite, with limits from negative infinity to positive infinity. For the other wire, we will consider a one-meter segment, so we establish the force on one meter of one wire. Regarding the Norm[]s: Norm[] computes the magnitude of a vector. C1 - C2 gives us the vector pointing from one to the other, as you know from basic vector analysis. The fraction (C1 - C2 divided by its norm) gives us a vector pointing from one to the other, of unit length -- [math]\hat{\mathbf{r}}_{12}[/math]. I evaluated it. You can see that the result is [math]2 \times 10^{-7} \mbox{ N}[/math] in the y direction. Not [math]1 \times 10^{-7} \mbox{ N}[/math]. The one meter segment of wire experiences a force of [math]2 \times 10^{-7} \mbox{ N}[/math] from its interaction with the infinite wire. If I extended the one meter wire to infinity, the force it experiences would become infinitely large, of course, so we have to calculate the force per meter. That's why I limit the integral to one meter.

Sigh. I did dimensional analysis. I did not compute the value of the function. I have shown how to compute its value before.

I suppose. But there's no sense doing something the easy way when you can make it more complicated. We've been over this before, haven't we?

It doesn't matter what symbols you use, as long as you know what they mean. I believe Bignose is implying that the units can be correct but the math can still be wrong. But I won't put words in his mouth.

Sense is made! First, let's move that junk out of the integral. [math]F_{12} = \frac {I_1 I_2 \mu_0}{4 \pi} \int \int \frac {dl_1 \times (dl_2 \times \hat{r}_{21} )}{|r|^2} = \frac {I_1 I_2 \mu_0}{4 \pi} \int dl_1 \times \int \frac {dl_2 \times \hat{r}_{21} }{|r|^2}[/math] Now, let's look at that last integral specifically. Here's how its units work out: [math]\int \frac {dl_2 \times \hat{r}_{21} }{|r|^2}= \int \frac{\mbox{m} \times \mbox{1}}{\mbox{m}^2} = \frac{1}{\mbox{m}}[/math] (since r hat is a unit direction vector, it doesn't have units) So. [math]F_{12} = \frac {I_1 I_2 \mu_0}{4 \pi} \int dl_1 \times \frac{1}{\mbox{m}} = \frac {\mbox{A A N}}{\mbox{A}^2} \mbox{m} \frac{1}{\mbox{m}} = \mbox{N}[/math] I think that's right. So the equation at least has the right units.

Hmm. I suppose if you multiply by the unit of the infinitesimal, you'd be fine. Hrm. Would cross products yield multiplied units? The magnitude of the cross product of two vectors is the same as the area of the parallelogram spanned by them, so I'd assume that cross products are like multiplications for units.

You'll note that if you, say, integrate velocity with respect to time, the result will be a distance, not a velocity. Integration matters. Damn. Not even the fuzzy pink unicorns from last night? I swear they farted rainbows. I'm sure swansont's flattered to be lumped with the teenagers. Getting a PhD in physics while a teenager is a pretty good feat. Sixty-seven, I think. It's a far more pleasant number.

http://news.sciencemag.org/sciencenow/2010/04/scienceshot-animals-that-live-wi.html?rss=1 So, life can exist without oxygen. Does this mean there are far, far more possible places (and planets) that can support life?

For very large values of "several."

NBC would be effective against an agent that causes immediate effects, like mustard gas or a short-term biological agent, but any agent with a month-long incubation period would have already infected everyone with NBC training before they realize there's something to defend against. In any case, the US is prohibited from developing and stockpiling biological weapons under the Biological Weapons Convention.

Do you say this because you enjoy releasing classified information or because you think it just has to be true?

Can you actually perform this calculation for us, as well as the one you say gives contradictory results?

I've sent off ten questions, and we may be able to ask a few more as follow-ups. But for the moment we just have to wait on responses.

You forgot that these are definite integrals along paths in the original formulas. You need to perform definite integration. Now, let's not go around in the same circles again, shall we? Or this thread may have an extremely short life.

I usually don't -- I just compare the mixed partial derivatives to be sure they're equal, as given in the last section of my link.

http://www.scienceforums.net/forum/showpost.php?p=555940&postcount=131 Definite integration. I integrate along [math]\theta[/math] from [math]\theta_1[/math] to [math]\theta_2[/math], just like when I do: [math]\int_0^4 1 \, dx = x |_0^4 = (4) - (0) = 4[/math] Fundamental theorem of calculus. If you do not understand this, you clearly do not understand vector calculus at all. I could apply both to real world situations by adjusting the paths I integrate along accordingly. I'd have to work out the geometry and integrate this. It would be a rather complicated integration. Theta is not the angle of the magnetic field. It is the angle from a point on one wire to a point on another wire. Check my previous post for links. Merged post follows: Consecutive posts merged Then show me exactly where I made a mistake, rather than demonstrating your lack of understanding of basic definite integration, and of the concept of "radians" (fundamental to any vector calculus course).