Electromagnetism and special relativity We have nearly finished our program on special relativity… And then we can go on about how it affects electromagnetism, and how we have to change our familiar quantities –charges, currents, potentials, fields, energy (and momentum!) density and flow, etc. Lots of vectors in electromagnetism….! We still have to discuss only the concept of force in connection with special relativity, and its aspects at high velocity. We shall do that one of the next lessons. For now, what we need to anticipate of the force and related subjects is that, at low energy, we can write, in three dimensions, the Newton law under the form F dP dt Which can easily be rewritten in 4 dimensions since we only need to substitute τ for t, while P… we already know how to use its 4-vector form. Let us summarize the cornerstones of special relativity that we shall make use of, as we did when sorting out 4-velocity etc. PRINCIPLES •Equivalence of Inertial Reference Systems (IRFs) •The fundamental physics laws have the same form in all IRFs. This means that the mathematical expression is the same independently of IRF, that the two terms on the sides of the “equal” sign have the same transformation properties for a change of IRF. AND… that the basic physics constants (including the speed of light) have the same value in all IRFs. Advanced EM Master in Physics 2011-2012 1 RESULTS of Special Relativity •Physics phenomena take place in a pseudo-euclidean 4-dimensional space, called Space-Time. •Basic physics laws have the same form in all IRFs. •The coordinate transformation between two IRFs (for a change between two systems with the same direction of axes but relative uniform motion) is the Lorentz transformation LT. We shall use the representation of 4-vectors, 4-tensors etc. based on the use of contravariant and covariant vectors. The convention is that the 4 components are expressed with indices 0, 1, 2, 3 where index 0 indicates the “time” component. The LT – for a relative motion along axis x1, i.e. “x”, is x0 1 x x2 0 3 x 0 0 0 x 0 ' 0 0 x1 ' 2 0 0 1 0 0 1 x ' 3 x ' where we have used for time and space the same units (c=1). The “norm” of a 4-vector A is: A 2 A A A0 A0 A1 A1 A2 A2 A3 A3 Advanced EM Master in Physics 2011-2012 2 Covariance of the physical laws Physical laws are written in the form of Equalities between two terms which: • Have well-defined transformation properties for the LT. • Such properties are the same for the two members of the equality (scalars, 4-vectors etc.) • A LT transforms them into equations that have the same form: p.ex.: pμ pμ m2 p'μ p'μ m2 If we are talking fundamental laws of physics, not applications to special systems with a preferred direction, then if we find that a particular component of a vector is equal to the same component of another vector then the two vectors are equal! Electrostatics and special relativity Starting assumptions: 1. The force on a charged particle at rest is due only to the electric field at the particle’s position. 2. IF – in the particle’s rest frame, charges and currents which generate fields are static, the electric field is: E(r ) r r ' r r' 3 dV ' 3. The electric charge is an invariant for Lorentz Transformation. Advanced EM Master in Physics 2011-2012 3 What has been called “Starting assumptions” in the previous slide are in fact experimentally established facts. The Coulomb law and the principle of superposition are proved experimentally. And so is the invariance of the electric charge for Lorentz transformations . This is demonstrated by the neutrality of matter independent of temperature. As we heat some matter, electrons’ velocity increases much faster than protons’, but the total charge does not change! Charge density ρ, Current density J, and Special Relativity In a given IRF, J = ρ·v where v is the velocity of the charges. Now, the question: How do ρ and J transform under a LT? Now, ρ is defined as charges’ rest system is and J will be zero. dQ dQ dV dx dy dz 0 which, in the dQ dx dy dz If I look at that system from an IRF’ in motion (β, γ) the charge will not change. What will change is dx, which will become and therefore dx' dx / ' 0 Advanced EM Master in Physics 2011-2012 4 Moreover, the observer in IRF’ will record a charge density ρ’ moving at speed – β. Let us use now a system in which c is not equal to one. A current distribution J’ = -ρ’·v will be seen in IRF’. And it will be precisely J' 'v 0 v So, our charge density at rest in its own system becomes a different charge density when seen in IRF’, and on top a current density is generated. The whole thing with coefficients that seem to be borrowed by the LT. Let us then check if an object made with 4 components: { , J x / c, J y / c, J z / c} {, J / c} does transform like a 4-vector, and then is a 4-vector? Well, under a LT such object in the charge’s rest frame is {0, 0} and {0, 0} { 0, 0, ,0,0} where Λ is the matrix of the LT. The new values the LT gives for the new charge and current distributions are exactly what is seen! Then…. The charge density and current distribution are components of a 4-vector. This makes a big change wrt electromagnetism in 3-dimensional space. We used to consider charge density as a scalar, but now it turns out to be the zero-component of a 4-vector! This fact will have important consequences. Advanced EM Master in Physics 2011-2012 5 First, how is modified the equation of the charge conservation? J 0 t The derivatives are the components of the 4-dimensional nabla, while J and ρ are the components of the 4-current. And, in Minkowski space language, the equation of charge conservation becomes: μ μ J 0 Well then, now that we have found the 4-current, we can go on with adjusting the EM formulas to spacetime. The charge and current densities appear in many EM equations, let us start with the simplest: 2 4 Let us examine with Mr. Lorentz’s eye how this equation behaves: pretty badly. In the first term, we have a laplacian which in 4space is not a physical quantity – but could easily become a D’Alambertian, since (remember? we are in electrostatics) the time-derivatives are null. And… the D’Alambertian is a scalar. The potential Φ, well, in electrostatics it is a scalar but…… What settles it is the second term. The equation is obviously a scalar equations, i.e. both members are just one number. But they are obviously not scalars from the point of view of LT!!! Because the second term is, apart from a multiplying constant, the zero_component of a 4-vector. We have here found a case of an equality between the 2 corresponding components of two 4-vectors: the equality must therefore hold for the 4-vectors. Advanced EM Master in Physics 2011-2012 6 “Two” 4-vectors? That in the second term there must a 4-vector, can be agreed upon. But the first term?? Φ in 3-D electrostatics is a scalar. Well, the second term is the zero-component of a 4vector, so there must be three other equations for the other three components of the 4-current; and on the left side there will be the three space-components of a “vector potential” that has still to be defined (which is the least problem), and especially understood in terms of physics: what does it represent? Well, the Poisson equation of electrostatics is now written: 2 4J 0 The relativity then entails the existence of this vector-potential A in 3 dimensions, and that it satisfies the three equations 2 A 4J / c Remark how with this formula we have also found that the components of vector A are determined as functions of J same way that Φ was determined by ρ. In the time-independent case, it is: (r ) r r ' r r' dV ' Then So far for the potential… and now: The Electric Field. Well, we now know what to do: examine the electrostatics equation, recognize the terms whose relativistic properties we know, i.e.: are they scalars, 4vectors or what? We start from the field equation in electrostatics: Advanced EM Master in Physics 2011-2012 7 E The equation above can now be re-written in Minkowski space terms: Ex 1 A0 Ey 2 A0 Ez 3 A0 These equations tell us that the electric field is not a 4-vector, since it has 2 indices: it is a part of a two-indices object, i.e. a 4-tensor. And since those three terms in the right side transform as elements of a tensor, so must do the components of the electric field. Ex is really a F10 Ey is really a F20 Ez is really a 0 F3 } components of the first line of a Tensor A tensor of rank 2 has 16 components – of which we already know as many as three! So, we have only 13 components to find out. Now, rank 2 tensors come in three types: symmetrical ( Tij = Tji ); antisymmetric ( Tij = -Tji ); and generic, i.e. neither the one nor the other. Any such tensor can anyway be written as the sum of a symmetric plus an antisymmetric tensor. It is easy to find that a symmetric tensor has 10 independent components, while an antisymmetric one has only six. If the field tensor were an antisymmetric one we would need only 3 new components. Remark, btw, that the LT transforms symmetric tensors into symmetric ones, and the same for antisymmetric ones. Of course, it would simpler if our tensor were antisymmetric. Advanced EM Master in Physics 2011-2012 8 That the field tensor is an antisymmetric one can be demonstrated. To start with, let us use the Newton law F=ma. In relativistic matters the quantity “force” loses meaning, so we will replace it by dP/dτ. The Newton’s law is now written as Fi dPi 0 qFi d There must then exist other components that satisfy the equations: p qF 0 Now in the rest system pμ equation becomes =[m0c, 0, 0, 0]= pμ,and the previous p q p F m0 c This equation is now written as an equality between 4-vectors and, as such, it is valid in all IRFs. We can then make the scalar product with the covariant momentum pμ. p 2q 2 p p p F m0c 2 2q [p ] p p F 0 m0c The derivative wrt τ is the derivative of a constant and therefore always null. 2q Then also m c p p F is null, which can only happen if 0 μν = ν F F μ Advanced EM Master in Physics 2011-2012 9 Since that equality holds for any value of it is the field which is antisymmetric. p p then it must be that Since, on the other hand, the field is some sort of A , then a term that satisfies both this requirement and the requirement of antisymmetry is : F A A where A is what we have called “the vector potential”, without having the slightest idea of what it might be. But we know that this formulas, when applied to the first row of the field tensor reads: E 1 A c t Of course adding in the elements that second term A , which for the electric field becomes (1/c)dAi/dt, the complete field tensor now reads: 0 1 Ax x c t 1 Ay y c t 1 Az z c t 1 Ax x c t 0 Ax Ay y x Az Ax x z 1 Ay y c t Ax Ay y x 0 Ay z Az y 1 Az z c t Ax Az z x Ay Az z y 0 We find again in row 1 and again in column 1 the well known formula for the electric field E, valid now also in the case of charges and currents varying in time. But we have also 3 new terms, for which we need an explanation. We remark that they have the structure of the curl of the 3-vector A. Let us call them “B“. Advanced EM Master in 10 Physics 2011-2012 We have just defined a new vector, and called it B. B A And we can re-write the field tensor: F μν 0 E x Ey E z Ex Ey 0 Bz Bz 0 By Bx Ez B y Bx 0 What does the “B“ vector is still to be understood. What is known is that under a LT it transform as the components of a tensor. F ' F μν μν 1 If the LT represented by Λ has β aligned along the X axis then F ' μν 0 Ex ( E y Bz ) ( E z B y ) E 0 ( B E ) ( B E ) x z y y z ( E y Bz ) ( Bz E y ) 0 Bx ( E B ) ( B E ) B 0 z y y z x Which can be re-written as { B'x Bx E 'x Ex E 'y ( E y Bz ) E y ( v B) y c E 'z ( E z By ) E z ( v B) z c Advanced EM Master in Physics 2011-2012 11 Which can be written in the new, more explicit, form: E ' L EL E 'T ET where the indices “L “ and “T “ stand c | vB | for longitudinal, i.e. parallel to v and transverse, i.e. perpendicular to it. Note that the vector product is already perpendicular to v. The Lorentz force We have now seen the fields – we must use the plural now, since we have two fields to work with. And…. two fields which are the same thing, since they mix when a LT is applied as a consequence of change of IRF. Note that we not only have two fields, but also have the formulas to compute them as functions of the charge and current distributions. Now what we still miss is what one of these fields, the one we called B, is doing to the charges. In order to find what this field does in a generic situation, i.e. with moving charges, we can do (actually, will do) the following: •Study the interaction with one charge only. It will be only too easy to extend this case to the presence of many charges. •The general case studied is that of a charge moving in an environment in which there are fields, E and B. •We want to find the force acting on that charge in that IRF Advanced EM Master in Physics 2011-2012 12 How we shall proceed is the following: We have the velocity of the charge: we shall make a LT to the IRF’ in which the charge is at rest: in that system, for one of the preliminary assumptions, the force acting on the particle is only that due to E’. We shall use as indicator of the force the instantaneous momentum change dP ' / dt . Then we will transform again the dP ' / dt original IRF frame in which the particle is in motion, and see what corresponds to the values of dP / dt . Since the LT has different values for longitudinal and transverse coordinates, we shall keep the two separate. In the charge’s rest frame: { dP ' dP ' qE ' d dt' dP 'T wit h especially qE ' T d F' 1. What we want to calculate is dP/dt as a function of the fields the charged particle sees in the IRF (the lab). 2. We express the force in IRF’ (Coulomb law): we will have to separate the longitudinal and transverse components of force and fields since they behave differently under a LT. 3. We establish the relation between dP/dt and dP’/dt’ (LT). Again, keep transverse and longitudinal components separate. 4. We write the LT of longitudinal and transverse electric fields E’ as a function of E, B Advanced EM Master in Physics 2011-2012 13 E 'L EL E 'T ET c ( v B) dP' dP' qE' d dt ' Fields in Rest Frame (IRF’) dP' L dP'L qE' L d dt ' dP' dP dP d d dt dPL dP'L dP / dP L L d d dt / dt NB : (dPL dP'L ) Equations of motion in IRF’ Lorentz Transformation of motion from IRF’ (rest) to IRF. Putting everything together: dPL dP'L qE L dt d dP dP' v q (E T B ) dt d c This is the (3-dimensional) force applied by the fields on a moving charge. Observe the force generated by the “vector” B. The Lorentz force which had been found experimentally is deduced from electrostatics plus special relativity plus charge conservation. Advanced EM Master in Physics 2011-2012 14

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