Cap'n Refsmmat Posted April 9, 2010 Posted April 9, 2010 Right, point me at exactly the correct Mathematica code you'd like me to use, and we can try it on my textbook's example problems. Sound good?
ambros Posted April 9, 2010 Author Posted April 9, 2010 Yes, and that's why your above calculation is wrong for a wire. You say it's for two parallel wires, but it's in fact for two point-like packets of charge. Hence you don't get the same answer. No, I got exactly the same answer. -- Can you not write your integral to be used in Mathematica online? Why not? What result do you get with two perpendicular wires or loop and a straight wire?
mooeypoo Posted April 9, 2010 Posted April 9, 2010 My little Mathematica code for Lorentz force can solve any example in your text book that has to do with any magnetic forces, it can solve it CORRECTLY, so there is no way you can call such equation "wrong". I doubt that very much, but that' sbesides the point. You weren't asked to solve *my* question, you were asked to solve your own. So first you started by throwing blame onto others to hide your own lack of knowledge and understanding, and now you try and shift the subject. Do it or don't, just stop beating around the bush. Really, we all have better things to do than to argue with someone who's not interested in learning. Your attempts to convince us that bad math is true are really quite futile. I think you should go over multi dimentional calculus again, my friend. To call it wrong you should first provide what you believe is correct, or at least you should be able to find some example for which that equation will give the wrong answer. If it always gives correct answer, then it is not wrong, right? Of course, it seems that you also don't read what people write to you. I didn't say it was wrong answer. In fact, i said it was the right answer to the wrong question. But you seem more interested in how to compare sizes rather than actually do math and physics. As a woman, my size competition is different than yours. You have nothing to threaten your position. Can we let go of the size chart now and do some real science? yes? good. We're still eagerly waiting for you to do Capn's exercise. ~moo
Cap'n Refsmmat Posted April 9, 2010 Posted April 9, 2010 No, I got exactly the same answer. -- Can you not write your integral to be used in Mathematica online? Why not? What result do you get with two perpendicular wires or loop and a straight wire? I gave you the Mathematica Player link. I'd rather not have to learn how to use Wolfram Alpha to do all this junk, since it doesn't handle it very well.
mooeypoo Posted April 9, 2010 Posted April 9, 2010 No, I got exactly the same answer. -- Can you not write your integral to be used in Mathematica online? Why not? What result do you get with two perpendicular wires or loop and a straight wire? When Mathematica computes, it produces the code along with the result. Copy/paste it in here, so we see exactly the stages in which you got your answer and we can see either where you were wrong or where we were wrong. It seems you keep getting answers no one else is getting in their calculation. Usually, that's a sign something's wrong with your calculation, but the only way to verify who's right and who's wrong is to go over it step by step. Go ahead and post it.
Cap'n Refsmmat Posted April 9, 2010 Posted April 9, 2010 Oh, yes. Click on the images in Wolfram Alpha and it'll give you the Mathematica input that I can use.
ambros Posted April 9, 2010 Author Posted April 9, 2010 I am waiting any one of you to write down this exact equation in Mathematica code so we can test it on parallel and perpendicular wires: [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] ================================== Here is mine: VARIABLE: c1 c2 r I1 I2 r_hat integrate[0,1] integrate[0,1] 10^-7Newtons/1^2 {1,0,0} x {1,0,0} x {0,-1,0} ...and I repeat - IT WORKS - it is as REAL AS IT GETS and it can be used in real world and applied to any situation, ANY SITUATION, I tell ya! What more do you want?
Cap'n Refsmmat Posted April 9, 2010 Posted April 9, 2010 Here's mine. http://www.scienceforums.net/forum/showpost.php?p=557101&postcount=89
ambros Posted April 9, 2010 Author Posted April 9, 2010 (edited) When Mathematica computes, it produces the code along with the result. Copy/paste it in here, so we see exactly the stages in which you got your answer and we can see either where you were wrong or where we were wrong. It seems you keep getting answers no one else is getting in their calculation. Usually, that's a sign something's wrong with your calculation, but the only way to verify who's right and who's wrong is to go over it step by step. Go ahead and post it. http://www.wolframalpha.com/input/ VARIABLE: c1 c2 r I1 I2 r_hat integrate[0,1] integrate[0,1] 10^-7Newtons/1^2 {1,0,0} x {1,0,0} x {0,-1,0} Now let me see your calculation, ok? It seems you keep getting answers no one else is getting in their calculation. No one else even managed to get any results. No one else, including you, does not have any calculation to show, and especially the one that will compute in Mathematica. Let me see ONE, JUST ONE of those calculations, show me only one of those that I can input in Mathematica and get correct answers for parallel and perpendicular wires, or any result for that matter. I guarantee you will not be able to provide any of those calculations, can you prove me wrong? What say you? Merged post follows: Consecutive posts mergedHere's mine. http://www.scienceforums.net/forum/showpost.php?p=557101&postcount=89 I'm actually downloading that sheeshbab, and curse you if I'm wasting my time again. Edited April 9, 2010 by ambros Consecutive posts merged.
mooeypoo Posted April 9, 2010 Posted April 9, 2010 (edited) http://www.wolframalpha.com/input/ VARIABLE: c1 c2 r I1 I2 r_hat integrate[0,1] integrate[0,1] 10^-7Newtons/1^2 {1,0,0} x {1,0,0} x {0,-1,0} Now let me see your calculation, ok? That's the integral Mr Skeptic explained to you (along with others) is NOT the proper way of dealing with multi dimentional calculus. So, you didn't really try Capn's suggestion. Are you afraid it will actually make sense, or do you not understand it? I'm actually downloading that sheeshbab, and curse you if I'm wasting my time again. It's only fair seeing as we appear to be wasting ours. Edited April 9, 2010 by mooeypoo Consecutive posts merged.
darkenlighten Posted April 9, 2010 Posted April 9, 2010 ambros, you are still not getting the correct results. The force on one wire is not 1 x 10^-7, it is 2 x 10^-7. Taken from http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law "The best-known and simplest example of Ampère's force law, which underlies the definition of the ampere, the SI unit of current, is as follows: For two thin, straight, infinitely long, stationary, parallel wires, the force per unit length one wire exerts upon the other in the vacuum of free space is: [math] F_m = 2k_A\frac{I_1 I_2}{r} [/math]"
Bignose Posted April 9, 2010 Posted April 9, 2010 Gibberish, that's how it is called if worded in English. In mathematics and physics it is called "ERROR - DOES NOT COMPUTE". It took me 10 minutes to figure out what in the world were you talking about and then I double checked to make sure I do not misinterpret you. After 15 minutes I realized you were wasting everyone's time by saying the simplest thing in the most incomprehensible way, and even managed to make a wrong conclusion. In mathematics and physics, correctness is paramount. Oh come on. Really? Exactly what conclusion did I write that was wrong? Please quote it directly. And you really don't need to write so insultingly. I didn't insult you. And if "correctness is paramount", why aren't you paying attention to everyone else that is telling you how to correctly perform integrations? Follow your own words here, ambros, and everyone will be a lot happier here.
Mr Skeptic Posted April 9, 2010 Posted April 9, 2010 What we have here is a serious case of "I'm right and everyone else in the whole world is wrong". These never end well.
ambros Posted April 10, 2010 Author Posted April 10, 2010 Here's mine. http://www.scienceforums.net/forum/showpost.php?p=557101&postcount=89 That is the same code you gave at the beginning which we discussed and I explained many things that are wrong with it, the most important one is that it does not compute in Mathematica, that is simply not correct syntax to produce working integral functions with variable parameters. You and I know that does not work and why or who you are trying to fool is beyond me. [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] I converted your code with syntax used in Mathematica online: *** YOUR CODE: integral[0,1] integral[-infinity, +infinity] dC2[s2] cross dC1[s1] cross (C1[s1]-C2[s2])/Norm[C1[s1]-C2[s2]] /(Norm[C1[s1]-C2[s2]]^2) (does not compute, incorrect syntax & undefined variables) *** MY CODE: integral[0,1] integral[0,1] 10^-7Newtons/1^2 {1,0,0} cross {1,0,0} cross {0,-1,0} (computes, can solve any situation involving magnetic forces) Your code can not be computed as that is not how variables are defined and used in integrals. You need to have very certain input precisely defined with the function you are trying to rewrite, nothing less nothing more. [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] INPUT IS DEFINE BY THE FUNCTION: I1, I2, r, int(dl1), int(dl2) 1.) You can not have C1, dC1 and s1 all at once. You specify all three of these in the beginning with "integral[start,end]", and you must let Mathematica handle the integration and it will inegrate those values for you, if you let it, and instruct it properly that is. 2.) You must have a way to input different magnitudes of I1 and I2, but your vectors are defined outside of the equation, so you first must rewrite the whole thing in vector form and that will help you get rid of duplicate variables. 3.) Make it COMPUTE if you want me to look at it any further, so far the result predicted by your code for any scenario is - ERROR.
Cap'n Refsmmat Posted April 10, 2010 Posted April 10, 2010 (edited) That is the same code you gave at the beginning which we discussed and I explained many things that are wrong with it, the most important one is that it does not compute in Mathematica, that is simply not correct syntax to produce working integral functions with variable parameters. You and I know that does not work and why or who you are trying to fool is beyond me. Have you tried the Mathematica player? I provided you a screenshot of the computed code. Are you suggesting I faked it? Hint: Wolfram Alpha can't do everything Mathematica can. Edited April 10, 2010 by Cap'n Refsmmat
ambros Posted April 10, 2010 Author Posted April 10, 2010 (edited) That's the integral Mr Skeptic explained to you (along with others) is NOT the proper way of dealing with multi dimentional calculus. So, you didn't really try Capn's suggestion. Are you afraid it will actually make sense, or do you not understand it? It's only fair seeing as we appear to be wasting ours. It can not be tried because it does not compute. A.) YOUR CODE DOES NOT EVEN COMPUTE - can you refute this? B.) MY CODE CAN SOLVE ANY MAGNETIC PROBLEM - can you refute this? Yes, I'm afraid, your understanding of this subject is horrifying. You are even so lazy to realize that my code can solve any example involving any magnetic forces. Why? Because it is correct implementation of this equation given by the BIPM that defines ampere unit and all the other equations in that have anything to do with el. currents: [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] Once you are ready to stop waving hands, then you can either find some example that I can not compute with my code, or you can finally make "your" code actually produce something else than ERROR. -- Do you accept the challenge, what say you? Merged post follows: Consecutive posts mergedHave you tried the Mathematica player? I provided you a screenshot of the computed code. Are you suggesting I faked it? Hint: Wolfram Alpha can't do everything Mathematica can. Wolfram Alpha can surely CALCULATE ANY INTEGRAL, finite and indefinite, just as full version can, as long as you can write them correctly. -- You have to have NUMERICAL VALUES and let Mathematica integrate the result, you can not have 15 undefined variables when only 5 are given by the equation, you can not produce numerical result with symbols, so we must be able to enter numerical input which is strictly defined by the BIPM equation and your code must be mathematically identical to that equation, only rewritten in different syntax. I have tried player and it crashed, I can not run you code, but I see the same algorithm in the source code as you gave previously. I simply then converted your CrossProduct functions to mathematical symbols and denoted integral limits at the beginning so it is more readable, but that is that, from there you need to convert it to vector format and get rid of all the variable but 5: I1, I2, dl1, dl2 and r. Edited April 10, 2010 by ambros Consecutive posts merged.
Mr Skeptic Posted April 10, 2010 Posted April 10, 2010 Well, have fun denying reality. I'll be over here, wishing I had those hours of my life back.
ambros Posted April 10, 2010 Author Posted April 10, 2010 Well, have fun denying reality. I'll be over here, wishing I had those hours of my life back. Reality is that you can not express "your theory" so that Mathematica can integrate it properly. There is nothing to deny even, I'm simply being a skeptic, so just as if someone wanted you to start believing in ghosts you would surely need to see one first, or you would not even consider the idea seriously, isn't that right, Mr Skeptic? This Mathematica syntax correctly represents that equation given by the BIPM: integral[0,1] integral[0,1] 10^-7Newtons/1^2 {1,0,0} cross {1,0,0} cross {0,-1,0} ...and anyone can see that for themselves by simply applying it in whatever case scenario involving magnetic forces they can come up with, just by plugging the given values and execute the code here: http://www.wolframalpha.com/input/ [math]F_{12} = \frac {\mu_0*I_1* \int dl_1 \times (I_2* \int dl_2 \times \hat{r}_{21} )}{4 \pi*|r|^2} [/math] VARIABLE: c1 c2 r I1 I2 r_hat integrate[0,1] integrate[0,1] 10^-7Newtons/1^2 {1,0,0} x {1,0,0} x {0,-1,0} Now, you have a chance to test this for yourself easily in Mathematica online, SEE the "ghost" with your own eyes and find out whether it indeed applies to reality, or not. However, if you refuse to even test that code against the reality, than it is you who is in denial. Skepticism is one thing, but when you refuse to "look through my telescope" it's called dogmatism.
mooeypoo Posted April 10, 2010 Posted April 10, 2010 It can not be tried because it does not compute. A.) YOUR CODE DOES NOT EVEN COMPUTE - can you refute this? B.) MY CODE CAN SOLVE ANY MAGNETIC PROBLEM - can you refute this? A - No, you don't know how to compute it. There's a difference. B - Yeah, it's been refuted throughout the thread. If you think I'm going to pretend I need to do it from scratch, you're mistaken.
Mr Skeptic Posted April 10, 2010 Posted April 10, 2010 Reality is that you can not express "your theory" so that Mathematica can integrate it properly. There is nothing to deny even, I'm simply being a skeptic, so just as if someone wanted you to start believing in ghosts you would surely need to see one first, or you would not even consider the idea seriously, isn't that right, Mr Skeptic? What we have here is more like a blind person insisting that colors don't exist because he can't see them -- nevermind that everyone else is telling him they can see them fine and he's just blind. I have no problem believing in many many things I can't see. Most of the things I believe in, I can't see. Atoms, electrons, quarks, the heavier fundamental particles, none of those can I see, yet I believe in all of them. I've not done the experiments myself; I'm just taking other people's word for it. You can't be a skeptic if you are not skeptical of yourself. You are not god. You make mistakes. A saying goes, the intelligent man learns from his mistakes, the wise man learns from the mistakes of others. I'm not sure where it fits in people who don't learn from anyone's mistakes. This Mathematica syntax correctly represents that equation given by the BIPM:integral[0,1] integral[0,1] 10^-7Newtons/1^2 {1,0,0} cross {1,0,0} cross {0,-1,0} No, it doesn't -- you are integrating a constant. You might as well not be doing an integration if all you are going to integrate is a constant. The constant you are integrating is simply a direction, and has nothing to do with any paths. If you don't understand calculus, that's fine, but don't think you can fool us by pretending you know. ...and anyone can see that for themselves by simply applying it in whatever case scenario involving magnetic forces they can come up with, just by plugging the given values and execute the code here: http://www.wolframalpha.com/input/ [math]F_{12} = \frac {\mu_0*I_1* \int dl_1 \times (I_2* \int dl_2 \times \hat{r}_{21} )}{4 \pi*|r|^2} [/math] VARIABLE: c1 c2 r I1 I2 r_hat integrate[0,1] integrate[0,1] 10^-7Newtons/1^2 {1,0,0} x {1,0,0} x {0,-1,0} Yes, I can see it gives the wrong answer, and the integral it gives is wrong. Now, you have a chance to test this for yourself easily in Mathematica online, SEE the "ghost" with your own eyes and find out whether it indeed applies to reality, or not. However, if you refuse to even test that code against the reality, than it is you who is in denial. Skepticism is one thing, but when you refuse to "look through my telescope" it's called dogmatism. I tested it, I looked, and it gave the wrong answer. Your ghost looks more like a dog chasing it's tail. Also, matematica is not reality. Also, it's rather amusing that you are accusing us of dogmatism, when you don't even know what you are talking about. --- Now if you want to get a correct answer that the free online mathematica can interpret, I'm pretty sure you have to do the cross products yourself. Remember to do the cross products for distant points as well as just the points that are closest to each other. Tell me, when you do gravity, do you plug in as the distance the distance between the surfaces of the planets? Cause they're the closest, those are the only ones that count, yes? Because that is what your formula is doing with the wires -- and why you have no trigonometric function in there. You are pretending that the attraction between the farthest points of the wires is the same as between the closest points, because you don't understand calculus, vectors, etc. And then you think that makes you a genius because your answer is unique.
Cap'n Refsmmat Posted April 10, 2010 Posted April 10, 2010 A.) YOUR CODE DOES NOT EVEN COMPUTE - can you refute this? [attach]2456[/attach] Do you believe I am lying?
swansont Posted April 10, 2010 Posted April 10, 2010 Funny, whenever I try to integrate over infinity - Mathematica tells me that I get infinity. So, please, is Mathematica wrong or is your solution wrong, because Mathematica says - IT DOES NOT COMPUTE. Maybe I do not know how to properly do it, especially since I never saw anything even close to what you be pretending to have as a "solution. -- So, can you write down that solution of yours in Mathematica code, please? What result do you get for the loop wire perpendicularly placed around the straight wire? Funny, the answer is actually infinite. What do you expect when you multiply by infinity? That's why the force is given per unit length. The field is finite, though. That solution should compute. I don't have Mathematica, so no, I can't provide you with code.
darkenlighten Posted April 10, 2010 Posted April 10, 2010 I like how he completely ignored my post: postcount 111
swansont Posted April 10, 2010 Posted April 10, 2010 B.) MY CODE CAN SOLVE ANY MAGNETIC PROBLEM - can you refute this? We want it done correctly. Getting an answer and getting the right answer are two very different things. Your solution method has been shown to be incorrect many times.
ambros Posted April 10, 2010 Author Posted April 10, 2010 (edited) Yes' date=' I can see it gives the wrong answer, and the integral it gives is wrong. I tested it, I looked, and it gave the wrong answer. Your ghost looks more like a dog chasing it's tail. [/quote'] Why not say it? That would end this argument to everyone's delight, so why so mysterious once you finally have something? Just tell us what problem do you think I can not solve with that equation and print it out here so I can give it a try. Wouldn't that be fair? Now if you want to get a correct answer that the free online mathematica can interpret, I'm pretty sure you have to do the cross products yourself. Remember to do the cross products for distant points as well as just the points that are closest to each other. Please, print out the complete equation in Mthematica code so everyone can see it and so I can test your one just like you tested mine, wouldn't that be fair? Merged post follows: Consecutive posts mergedWe want it done correctly. Getting an answer and getting the right answer are two very different things. Your solution method has been shown to be incorrect many times. Really? Can you name one or two problems, or practical scenarios that you believe can not be solved by that Mathematica code I gave? What examples did you use to test it? Edited April 10, 2010 by ambros Consecutive posts merged.
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