pengkuan Posted March 8, 2012 Share Posted March 8, 2012 Internal Lorentz force paradox A current in a magnetic field feels the Lorentz force. For isolated coil, the Lorentz force on a current segment is due to the magnetic field of the coil itself. This force is internal to the coil and its resultant is 0. Let us see the internal force of the coil shown in the Figure 1. This coil is composed of a curved wire part and a straight wire part. According to the Lorentz force law, the magnetic force is always perpendicular to the current, as indicated by the forces Fcurv and Fstr in the Figure 1. The resultant force on the curved part is Rcurv and that on the straight part is Rstr. Rcurv and Rstr are forces internal to the coil, thus their sum must be 0. But, is the sum predicted by the Lorentz force law 0? I have done a calculation for this coil in dimensionless form. The length of the straight part is 2, the height of the curved part is 1 and the inclination 45°. The numeric values of dimensionless forces are as follow: The resultant force on the curved part: Rx= -6.4618, Ry =9.5457 The resultant force on the straight part: Rx= 0 Ry =-16.8930 The resultant force on the curved and straight parts are calculated as follow: These forces are drawn in the Figure 1. As we see, Rcurv and Rstr are not parallel and give rise to a resultant force Rres. If the real internal force conforms to this prediction, one could make the following process: Make a coil with inclined curve and straight wires, put the current on, and then let the coil move in the direction of the resultant force Rres. This force would do a work in the movement. Since the magnetic flux passing through the coil is constant, the current will not do any work. Thus a quantity of energy is created by Rres. This is impossible because energy cannot be created, hence an inconsistency. Why the result forces on the 2 parts of the coil are not parallel? Because Lorentz force is always perpendicular to current. On the straight part, the Lorentz force is vertical. But on the curved part, the Lorentz forces on every segment are inclined due to the direction of the current. So, their resultant cannot be vertical. In fact, because of the direction of the segments of currents, for any coil of non symmetric shape, the resultant Lorentz force internal to the coil cannot be 0. Thus, the Lorentz force law does not predict correct internal magnetic force. I call this inconsistency the "Internal Lorentz force paradox". For reading the solution and more information, please read this link or the pdf document. My link Aharonov_net.pdf Link to comment Share on other sites More sharing options...
swansont Posted March 8, 2012 Share Posted March 8, 2012 Rcurv and Rstr are forces internal to the coil, thus their sum must be 0. But, is the sum predicted by the Lorentz force law 0? I'm not sure what you mean by this. But: the projection of the curved part onto the straight part has a length of 2, and by symmetry the x components of the curved part have to give the same magnitude answer, so I can't reconcile your numbers with these observations. By inspection, I think your math has to be flawed. Link to comment Share on other sites More sharing options...
pengkuan Posted March 9, 2012 Author Share Posted March 9, 2012 I'm not sure what you mean by this. But: the projection of the curved part onto the straight part has a length of 2, and by symmetry the x components of the curved part have to give the same magnitude answer, so I can't reconcile your numbers with these observations. By inspection, I think your math has to be flawed. Thank you for your comment. It makes me to think for those who do not do numerical calculation about the resultant force immediately; i.e almost everyone. Here is a simple graphical explanation. I will do a mathematical demonstration if necessary. Take a rectangle coil. The Lorentz forces from the point A on a segment of the vertical sides are F left and F right. When the rectangle is sheared an angle to the right, the distance rA will shorten and r'A lengthen, and F left will increase and F right decrease. For the symmetric point B, there is the same variation of the forces. The x components will vary in the same sense approximately. So, there will be a resultant force on the upper sides with a x component. Link to comment Share on other sites More sharing options...
pengkuan Posted March 9, 2012 Author Share Posted March 9, 2012 I'm not sure what you mean by this. But: the projection of the curved part onto the straight part has a length of 2, and by symmetry the x components of the curved part have to give the same magnitude answer, so I can't reconcile your numbers with these observations. By inspection, I think your math has to be flawed. Here is a mathematical proof that the x force is not 0. See the drawing and the pdf. parallelepipede.pdf The full article with the parallelepiped proof integrated.internal.pdf Link to comment Share on other sites More sharing options...
swansont Posted March 9, 2012 Share Posted March 9, 2012 Why are you using such a convoluted geometry? I think that's what is getting you in trouble; your use of sin(phi) and sin(beta) is not clear at all. Why not just use right triangles? Link to comment Share on other sites More sharing options...
pengkuan Posted March 9, 2012 Author Share Posted March 9, 2012 Why are you using such a convoluted geometry? I think that's what is getting you in trouble; your use of sin(phi) and sin(beta) is not clear at all. Why not just use right triangles? Because the sides and the forces F right and F left are parallel . I can sum them and find Fx. I have used a triangle in my previous numerical calculation. It was not better. Link to comment Share on other sites More sharing options...
swansont Posted March 9, 2012 Share Posted March 9, 2012 Because the sides and the forces F right and F left are parallel . I can sum them and find Fx. I have used a triangle in my previous numerical calculation. It was not better. Yes, the forces are. So why not use proper right triangles that have identical angles and sides instead of the triangles you used? The drawing has symmetry. The force components will reflect this. Theta and (90-delta) are equal, for starters. Link to comment Share on other sites More sharing options...
pengkuan Posted March 10, 2012 Author Share Posted March 10, 2012 Yes, the forces are. So why not use proper right triangles that have identical angles and sides instead of the triangles you used? The drawing has symmetry. The force components will reflect this. Theta and (90-delta) are equal, for starters. Seeing your drawing, I have understood what is the "trouble" you are talking about. You are considering the resultant force on each side from all the other sides. That is, in the figure here, the back arrows, and their sum is 0. This is not my view. I am considering the lateral sides plus the top side as a whole part, the red part, and the bottom side another part, the bleu part. The tree upper sides feel a resultant force from the bottom, the red arrow, and the bottom from the 3 upper sides, the bleu arrow. The bleu and red arrows do not cancel out. The Newton's law is a law, no matter how you cut your coil, the resultant forces must cancel out. However, when you cut it into red and bleu parts, the resultant force remains non null. This is the problem of the Lorentz force law. The raison I have used curved and straight coil is that you cut intuitively the coil into curved and straight parts and this leads to trouble. Our discussion is good in that it makes clear the cutting of the coil. Link to comment Share on other sites More sharing options...
swansont Posted March 10, 2012 Share Posted March 10, 2012 Your issue is with vectors, i.e. math, not physics. And you have the math wrong. Link to comment Share on other sites More sharing options...
pengkuan Posted March 10, 2012 Author Share Posted March 10, 2012 Your issue is with vectors, i.e. math, not physics. And you have the math wrong. I respect your point of view. But if you do not point out what is the error, I cannot explain why. Link to comment Share on other sites More sharing options...
swansont Posted March 10, 2012 Share Posted March 10, 2012 I respect your point of view. But if you do not point out what is the error, I cannot explain why. You have two pairs of vectors that are each equal in magnitude and opposite in direction. They cancel. It doesn't matter how you do the details of the math. If you get an answer where they don't cancel, you've made a math error. You don't show enough detail in how/why you set the problem up this way for me to find the exact error. Link to comment Share on other sites More sharing options...
pengkuan Posted March 11, 2012 Author Share Posted March 11, 2012 (edited) You have two pairs of vectors that are each equal in magnitude and opposite in direction. They cancel. It doesn't matter how you do the details of the math. If you get an answer where they don't cancel, you've made a math error. You don't show enough detail in how/why you set the problem up this way for me to find the exact error. Ok. Let us take the problem only physically. No math at all. Take 2 rods, W1 and W2. Their mutual force are f1 and f2. f1+f2=0 because Newton 3 and the rods do not move. Now add a third rod, W3, on which the force from 1 and 2 is F3. The Force on the other rods have changed. Let denote the mutual forces on each rod as F1,F2,F3, and their sum Fr=F1+F2+F3. Because Newton 3 and the rods do not move, Fr=0. This must be true for any kind of force, electrostatic, gravitation, magnetic….. Question: If F1, F2 and F3 were Lorentz force, do we have Fr =F1+F2+F3=0 or Fr ≠0? What is your answer? We must suppose Fr =0. If not, Newton 3 is violated already. Now close the coil with rod 4. The connections (dashed lines) are soft and do not transmit force. Even with current on, the internal force (Lorentz force) within the ensemble 3 rods W1+W2+W3 must be 0, that is Fr=0. So, we can study the Lorentz force that the rode W4 exerts on W1+W2+W3 without drawing F1, F2, F3 Fr. The Lorentz force is originated from the Biot-Savart law, that is, it is inversely proportional to the distance between the interacted currents. We know that the average distance between W4-W3 (r left) is shorter than that of W4-W1 (r right). So, the resultant Lorentz force F left is greater than F right. Thus, F left- F right>0 The resultant Lorentz force from W4 on the ensemble W1+W2+W3 has a x component to the left, denoted Fx. The Lorentz force on the W4 is vertical, thus the x component on the ensemble W1+W2+W3+W4 is Fx. This is not allowed by Newton 3. Version 2 https://docs.google....0bGlOMVVLRVhIZw My Email adress:titang78@gmail.fr Edited March 11, 2012 by pengkuan Link to comment Share on other sites More sharing options...
swansont Posted March 11, 2012 Share Posted March 11, 2012 Ok. Let us take the problem only physically. No math at all. Take 2 rods, W1 and W2. Their mutual force are f1 and f2. f1+f2=0 because Newton 3 and the rods do not move. Now add a third rod, W3, on which the force from 1 and 2 is F3. The Force on the other rods have changed. Let denote the mutual forces on each rod as F1,F2,F3, and their sum Fr=F1+F2+F3. Because Newton 3 and the rods do not move, Fr=0. This must be true for any kind of force, electrostatic, gravitation, magnetic….. You're mixing up concepts here. You do not sum force-pairs you get from Newton's third law in the context of "no net force" because the forces act on different objects. If you have two parallel wires with current passing through them, there is most definitely a force they exert on each other and it is not zero — you can construct a current balance and see this and is in fact how the Ampere is defined. And yet [math]F_{1-2} = - F_{2-1}[/math] Question: If F1, F2 and F3 were Lorentz force, do we have Fr =F1+F2+F3=0 or Fr ≠0? What is your answer? We must suppose Fr =0. If not, Newton 3 is violated already. [attachment not included] Now close the coil with rod 4. The connections (dashed lines) are soft and do not transmit force. Even with current on, the internal force (Lorentz force) within the ensemble 3 rods W1+W2+W3 must be 0, that is Fr=0. So, we can study the Lorentz force that the rode W4 exerts on W1+W2+W3 without drawing F1, F2, F3 Fr. The Lorentz force is originated from the Biot-Savart law, that is, it is inversely proportional to the distance between the interacted currents. We know that the average distance between W4-W3 (r left) is shorter than that of W4-W1 (r right). So, the resultant Lorentz force F left is greater than F right. Thus, F left- F right>0 The resultant Lorentz force from W4 on the ensemble W1+W2+W3 has a x component to the left, denoted Fx. The Lorentz force on the W4 is vertical, thus the x component on the ensemble W1+W2+W3+W4 is Fx. This is not allowed by Newton 3. Saying that the force from the left wire is different that that from the right is NOT the same as saying that the Newton's third law force-pairs are not the same, since you are looking at different pairs. Link to comment Share on other sites More sharing options...
pengkuan Posted March 11, 2012 Author Share Posted March 11, 2012 You're mixing up concepts here. Saying that the force from the left wire is different that that from the right is NOT the same as saying that the Newton's third law force-pairs are not the same, since you are looking at different pairs. Well, you can critic me for not saying everything. I have using a rigid coil. That is, the W1, W2 and W3 are connected rigidly so that all force W1 receives are transferred to W2 and W3. But, to make this idea obvious, I bound W1, W2 and W3 on a wooden plate, the yellow surface in the figure. This way, we can sum all forces that the wires receive and their resultant force R res is on a point within the plate. R res does not balance with F w4, breaking Newton 3. Link to comment Share on other sites More sharing options...
swansont Posted March 12, 2012 Share Posted March 12, 2012 R res does not balance with F w4, breaking Newton 3. How can it not? The only difference in the math (if done properly) is a minus sign. Link to comment Share on other sites More sharing options...
pengkuan Posted March 12, 2012 Author Share Posted March 12, 2012 (edited) How can it not? The only difference in the math (if done properly) is a minus sign. If you affirm only the minus sign exists, I suppose it is not an affirmation based on approximate feeling and you have done the math properly. If not, here is my proper math. See the figure. F1-2 is the Lorentz force on the rod W2 due to the magnetic field from W1. F2-1 is the Lorentz force on the rod W1 due to the magnetic field from W2. Globally, there are 3 Lorentz forces on each rod. The resultant force on the W4 is: R4=F1-4 +F2-4+ F3-4 The resultant force on the W1+W2+W3 is: Rres=F2-1+F3-1+F4-1+F1-2 +F3-2 F4-2+F1-3+F2-3+F4-3 Physically, the couple W1-W2 is the same as W3-W4, and W1-W4 is the same as W3-W2. So, F2-1=-F4-3 and F4-1=-F2-3, so they cancel out. Symmetrically, F3-1 cancels F1-3. So, Rres=F1-2 +F3-2 +F4-2 By symmetry, we have: F1-4 =- F3-2 , F2-4=- F4-2, F3-4=- F1-2 So, Rres=-( F1-4 +F2-4+ F3-4) In conclusion, we have: Rres=-R4 Is-it this math right? Version 3 of the article https://docs.google.com/open?id=0B5-G5ctz2vmbUl8tNGQ2Nk9UQlM2eG80UlIzVFlXZw Edited March 12, 2012 by pengkuan Link to comment Share on other sites More sharing options...
swansont Posted March 12, 2012 Share Posted March 12, 2012 If you affirm only the minus sign exists, I suppose it is not an affirmation based on approximate feeling and you have done the math properly. It is true by definition. Rres=F1-2 +F3-2 +F4-2 By symmetry, we have: F1-4 =- F3-2 , F2-4=- F4-2, F3-4=- F1-2 So, Rres=-( F1-4 +F2-4+ F3-4) In conclusion, we have: Rres=-R4 Is-it this math right? It looks like this says that the force on W4 is opposite of that exerted by W4, exactly as one would expect. Link to comment Share on other sites More sharing options...
pengkuan Posted March 12, 2012 Author Share Posted March 12, 2012 It is true by definition. It looks like this says that the force on W4 is opposite of that exerted by W4, exactly as one would expect. I have to say sorry because this math is wrong. Let us take the resultant force on the W1+W2+W3: Rres=F2-1+F3-1+F4-1+F1-2 +F3-2 F4-2+F1-3+F2-3+F4-3 We divide Rres into two groups: Rres=(F1-2 +F1-3+F2-1+F2-3+F3-1+F3-2 )+(F4-1+F4-2+F4-3) We denote: R123= F1-2 +F1-3+F2-1+F2-3+F3-1+F3-2 R4-123= F4-1+F4-2+F4-3 R4-123 is the force from W4 to W1+W2+W3. R123 is the sum of the forces from W1 to W1+W2+W3 from W2 to W1+W2+W3 from W3 to W1+W2+W3. That is, R123 is internal to the ensemble 3 rods W1+W2+W3 bounded to the wooden plate (yellow parallelepiped). So, Newton 3 requires R123=0. But, with the above consideration, Symmetrically, F3-1 cancels F1-3. we have: R123= F1-2 +F3-2+ F2-1+F2-3 As the average distance between W2-W1 is shorter than that between W3-W2, We have F2-1//F2-3 and the x component (F2-1+F2-3)x>0. So, the resultant force R123 is non null: R123≠0. The force R123 internal to W1+W2+W3 violates Newton 3. So, if we had Rres=-R4 in the previous math, it was because R123 violated Newton 3. We have tried to save the Lorentz force law which violated Newton 3 for a 4 sides parallelepiped by using a internal force that violated Newton 3 for 3 sides. It is not good physics. Link to comment Share on other sites More sharing options...
swansont Posted March 12, 2012 Share Posted March 12, 2012 But, with the above consideration, Symmetrically, F3-1 cancels F1-3. we have: R123= F1-2 +F3-2+ F2-1+F2-3 As the average distance between W2-W1 is shorter than that between W3-W2, We have F2-1//F2-3 and the x component (F2-1+F2-3)x>0. Not sure what F2-1//F2-3 means. Is it supposed to be F2-1≠F2-3? If so, who cares? You also have F1-2≠F3-2. The differences are equal & opposite. They cancel. F1-2+F2-1 = 0 and F3-2+F2-3 = 0 Link to comment Share on other sites More sharing options...
pengkuan Posted March 12, 2012 Author Share Posted March 12, 2012 (edited) Not sure what F2-1//F2-3 means. Is it supposed to be F2-1≠F2-3? If so, who cares? You also have F2-1≠F2-3 The differences are equal & opposite. They cancel. F1-2+F2-1 = 0 and F3-2+F2-3 = 0 Sorry. F2-1//F2-3 means they are parallel and can be added together. So (the component x of F2-1 + F2-3 ) >0 F1-2 and F3-2 are vertical. They do not have x component. I understand that you do not have enough time to reflect carefully, and you have mixed mutual Lorentz forces and direct action-reaction forces. You said F1-2+F2-1 = 0 and F3-2+F2-3 = 0 This is true if they were direct action-reaction forces. But F1-2 and F2-1 are mutual Lorentz forces and are at an angle (please use right hand rule to find the direction of Lorentz force) so F1-2+F2-1 = R12≠0 and F3-2+F2-3 = R23≠0 And they do not cancel out. Edited March 12, 2012 by pengkuan Link to comment Share on other sites More sharing options...
swansont Posted March 13, 2012 Share Posted March 13, 2012 I understand that you do not have enough time to reflect carefully, and you have mixed mutual Lorentz forces and direct action-reaction forces. You said F1-2+F2-1 = 0 and F3-2+F2-3 = 0 Yes, my mistake. But there is no reason that R123 must be zero. Try this: rotate your coordinate system to regain the symmetry of the problem. 1 Link to comment Share on other sites More sharing options...
pengkuan Posted March 13, 2012 Author Share Posted March 13, 2012 Yes, my mistake. But there is no reason that R123 must be zero. Try this: rotate your coordinate system to regain the symmetry of the problem. In physics, there are quantities that are invariant. The Lorentz force is invariant. That is, it does not depend on reference frame. The Newton 3 law is also invariant, that is, it gives the same result what ever the coordinate system. If you charge the W1+W2+W3 with electric charge, the resultant force is 0, what ever the coordinate system you use. If there is current in W1+W2+W3, the Newton 3 applies too and R123 must be 0. Why do you say But there is no reason that R123 must be zero. can you give an argument? Link to comment Share on other sites More sharing options...
swansont Posted March 13, 2012 Share Posted March 13, 2012 In physics, there are quantities that are invariant. The Lorentz force is invariant. That is, it does not depend on reference frame. The Newton 3 law is also invariant, that is, it gives the same result what ever the coordinate system. Yes indeed. So if you can look in a coordinate system that is easy to analyze and see that there is no net force on the system, it must be true in the other coordinate systems. If you charge the W1+W2+W3 with electric charge, the resultant force is 0, what ever the coordinate system you use. If there is current in W1+W2+W3, the Newton 3 applies too and R123 must be 0. Why do you say can you give an argument? 123 does not form a closed loop. There can be no current in 123 without the fourth section of the wire; to assume there is a closed loop is unphysical. You are ignoring another force that's exerted on the wires (from W4) and still assuming that the resultant will be zero. Link to comment Share on other sites More sharing options...
pengkuan Posted March 13, 2012 Author Share Posted March 13, 2012 You are ignoring another force that's exerted on the wires (from W4)… I am not ignoring the other force, please read my previous post. We divide Rres into two groups: Rres=(F1-2 +F1-3+F2-1+F2-3+F3-1+F3-2 )+(F4-1+F4-2+F4-3) We denote: R123= F1-2 +F1-3+F2-1+F2-3+F3-1+F3-2 R4-123= F4-1+F4-2+F4-3 R4-123 is the force from W4 to W1+W2+W3. R123 is the sum of the forces from W1 to W1+W2+W3 from W2 to W1+W2+W3 from W3 to W1+W2+W3. But if you say … and still assuming that the resultant will be zero. Would you mean that I should not assume R123=0 if I take R4-123 into account? That is, if I add the force R4-123 on W1+W2+W3, the resultant force R123 would change its value? This reasoning is like saying that: The force between 2 electrons is (C*qq/r²). But if I put a proton near them, the force (C*qq/r²) will change. Why? Is the distance r changed? Or the charge of electrons? Or the coefficient C? You said: 123 does not form a closed loop. There can be no current in 123 without the fourth section of the wire; to assume there is a closed loop is unphysical. Right. But there is a physical way to prevent that the wire W4 exerts a force on W1+W2+W3. Look at the figure. I put the wire W4 in a magnetic shield; Thus, the magnetic field of W4 does not influence W1+W2+W3, neither W1+W2+W3 influence W4. The shield does not carry current. Then I hang the coil plus wooden plate through a string. In this case, the resultant force on the coil is only R123 which is non null. Will the coil move or turn? If yes, there is creation of energy. If not, R123 is 0. That is, the value of R123 predicted by Lorentz force law is wrong. You said in a previous post: But there is no reason that R123 must be zero. You mean, R123 can have a non null value for you. Do you still think so? Link to comment Share on other sites More sharing options...
swansont Posted March 13, 2012 Share Posted March 13, 2012 Would you mean that I should not assume R123=0 if I take R4-123 into account? That is, if I add the force R4-123 on W1+W2+W3, the resultant force R123 would change its value? You already established that mutual Lorentz forces for non-parallel wire segments do not cancel. Why do you assume that they cancel here? 1 Link to comment Share on other sites More sharing options...
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