Top Document: Invariant Galilean Transformations On All Laws Previous Document: 15. But The Transform Won't Work On Wave Equations? Next Document: 17. First and Second Derivative differential equations. See reader questions & answers on this topic!  Help others by sharing your knowledge Oh? Just what is the magical term in them that prevents (x'x.c')=(xvt  x.c+vt)=(xx.c) from holding true? It turns out not to be magic, but reality, that interferes with the application of the galilean transforms to the gen eralized coordinate form(s) of Maxwell: there are no coordi nates to transform! When True Believer crackpots are shown the simple demonstration that the galilean transform on generalized cartesian coordinates is invariant, their first defense is usually an incredibly stupid "x0'=x0, because the coordinate of a circle center, or point of emission, etc, is a constant and can't be transformed." The last defense is "but Maxwell's equations are not invariant under that coordinate transform." When asked just what magic occurs in Maxwell that would prevent the simple algebra (x'x0')=[ (xvt)(x0vt) ]=(xx0) from working, and when asked them for a demonstration, they will never do so, however many hundreds of times their defense is asserted. The reason may help you understand part of Einstein's 1905 paper in which he gave us his absurd Special Relativity derivation: THERE ARE NO COORDINATES IN THE EQUATIONS TO BE TRANSFORMED. Einstein gave the electric force vector as E=(X,Y,Z) and the magnetic force vector as B=(L,M,N), where the force components in the direction of the x axis are X and L, Y and M are in the y direction, Z and N in the z direction. Those values are not, however, coordinates, but values very much like acceleration values. BTW, the current fad is that E and B are 'fields', having been 'force fields' for a while, after being 'forces'. So, when Einstein says he is applying his coordinate transforms to the Maxwell form he presented, he is either delusive or lying. (a) there are no coordinates in the transform equations he gives us for the Maxwell transforms, where B=beta=1/sqrt(1(v/c)^2): X'=X. L'=L. Y'=B(Y(v/c)N). M'=B(M+(v/c)Z). Z'=B(Z+(v/c)M). N'=B(N(v/c)Y). X is in the same direction as x, but is not a coordinate. Ditto for L. They are not locations, coordinates on the xaxis, but force magnitudes in that direction. Similarly for Y and M and y, Z and N and z. (b) the v of the "coordinate transforms" is in Maxwell before any transform is imposed; Einstein's transform v is the velocity of a coordinate axis, not the velocity of a particle, which is what was in the equation before he touched it. (c) if they were honest Einsteinian transforms, they'd be incompetent. The direction of the particle's movement is x, which means it is X and L that are supposed to be transformed, not Y and M, and Z and N. And when SR does transform more than one axis, each axis has its own velocity term; using the v along the xaxis as the v for a yaxis and zaxis transform is thus trebly absurd: the axes perpendicular to the motion are not changed according to SR, the v used is not their v, and the v is not a transform velocity anyway. (d) as everyone knows, the effect of E and B are on the particle's velocity, which is a speed in a particular direction. Both the speed and direction are changed by E and B, but v  the speed  is a constant in SR. As absurd as are the previously demonstrated Einsteinian blunders, this one transcends error and is an incredible example of True Believer delusion propagating over decades. The components of E and B do differ from point to point, and in the variations that are not coordinate free, they are subject to the usual invariant galilean trans formation when put in the generalized coordinate form.  The SR crackpots don't know what coordinates are. The various things they call coordinates include coordin nates, but also include a variety of other quantities.  1. One may express coordinates in a oneaxisatatime manner [like x^2+y^2=r^2] but it is the use of vector notation that shows us what is going on. In vector notation the triplet x,y,z [or x1,x2,x3, whatever] represents the three spatial coordinates, but there are socalled basis vectors that underlie them. Those may be called i,j,k. Thus, what we normally treat as x,y,z is a set of three numbers TIMES a basis vector each. 2. These e*i, f*j, g*k products can have a lot of meanings. If e, f, j are distances from the origin of i,j,k then e*i, f*j, g*k are coordinates: distances in the directions of i,j,k respectively, from their origin. That makes the triplet a coordinate vector that we describe as being an x,y,z triplet; perhaps X=(x,y,z). The e*i, f*j, g*k products could be directions; take any of the other vectors described above or below and divide the e,f,g numbers by the length of the vector [sqrt(e^2+f^2+g^2)]. That gives us a vector of length=1.0, the e,f,g values of which show us the direction of the original vector. That makes the triplet a direction vector that we describe as being an x,y,z triplet; perhaps D=(x,y,z). The e*i, f*j, g*k products could be velocities; take any of the unit direction vectors described above and multiply by a given speed, perhaps v. That gives a vector of length v in the direction specified. That makes the triplet a velocity vector that we describe as being an x,y,z triplet; perhaps V=(x,y,z). Each of the three values, e,f,g, is the velocity in the direction of i,j,k respectively. The e*i, f*j, g*k products could be accelerations; take any of the unit direction vectors described above and multiply by a given acceleration, perhaps a. That gives a vector of length a in the direction specified. That makes the triplet an acceleration vector that we describe as being an x,y,z triplet; perhaps A=(x,y,z). Each of the three values, e,f,g, is the acceleration in the direction of i,j,k respectively. The e*i, f*j, g*k products could be forces (much like accel erations); take any of the unit direction vectors described above and multiply by a given force, perhaps E or B. That gives a vector of length E or B in the direction specified. That makes the triplet a force vector that we describe as being an x,y,z triplet; perhaps E=(x,y,z) or B=(x,y,z). Each of the three values, e,f,g, is the force in the direction of i,j,k respectively. Einstein's  and Maxwell's  E and B are not coordinate vectors. ============================================================ There is another variety of intellectual befuddlement that misinforms the idea that Maxwell isn't invariant under the galilean transform: confusions about velocities. Velocities With Respect to Coordinate Systems.  Aaron Bergman supplied the background in a post to a sci.physics.* newsgroup: =============================================================== Imagine two wires next to each other with a current I in each. Now, according to simple E&M, each current generates a magnetic field and this causes either a repulsion or attraction between the wires due to the interaction of the magnetic field and the current. Let's just use the case where the currents are parallel. Now, suppose you are running at the speed of the current between the wires. If you simply use a galilean transform, each wire, having an equal number of protons and electrons is neutral. So, in this frame, there is no force between the wires. But this is a contradiction. ================================================================ First of all, the invariance of the galilean transform (x'x.c') =(xx.c), insures that it is an error to imagine there is any difference between the data and law in one frame and in another; the usual, convenient rest frame is the best frame and only frame required for universal analysis. [Well, (x'<>x, x,c'<>x.c, but (x'x.c')=(xx.c).] Second, given that you decide unnecessarily to adapt a law to a moving frame, don't confuse coordinate systems with meaningful physical objects, like the velocity relative to a coordinate system instead of relative to a physical body or field. In other words, what does current velocity with respect to a coordinate system have to do with physics? Nothing. Certainly not anything in the example Bergman gave. What is relevant is not current velocity with respect to a coordinate system, but current velocity with respect to wires and/or a medium. The velocity of an imaginary coordinate sys tem has absolutely nothing to do with meaningful physical vel ocity. You can  if you are insightful enough and don't violate item (e)  identify a coordinate system and a relevant physical object, but where some v term in the pretransformed law is in use, don't confuse it with the velocity of the coordinate transform. Velocities With Respect to ... What?  Albert Einstein opened his 1905 paper on Special Relativity with this ancient incompetency: =============================================================== The equations of the day had a velocity term that was taken as meaning that moving a magnet near a conductor would create a current in the conductor, but moving a conductor near a wire would not. This was belied by fact, of course. The important velocity quantity is the velocity of the magnet and conductor with respect to each other, not to some absolute coordinate frame (as far as we know) and not to an arbitrary coordinate system. One possible cause was the idea: "but the equation says the magnet must be moving wrt the coordinate system" or "... the absolute rest frame". There not being anything in the equation(s) to say either of those, it is amazing that folk will still insist the velocity term has nothing to do with velocity of the two bodies wrt each other.  User Contributions:Comment about this article, ask questions, or add new information about this topic:Top Document: Invariant Galilean Transformations On All Laws Previous Document: 15. But The Transform Won't Work On Wave Equations? Next Document: 17. First and Second Derivative differential equations. Single Page [ Usenet FAQs  Web FAQs  Documents  RFC Index ] Send corrections/additions to the FAQ Maintainer: Thnktank@concentric.net (Eleaticus)
Last Update March 27 2014 @ 02:12 PM
