RETARDED POTENTIALS

RETARDED POTENTIALS RETARDED POTENTIALS ARE SO CALLED BECAUSE THE COTRIBUTION OF EACH ELEMENT OF CHARGE AND CURRENT IS DELAYED BY THE TIME TAKEN FOR LIGHT TO TRAVEL TO THE POINT WHERE THE POTENTIAL IS BEING CALCULATED. INFORMATION ABOUT THE PRESENT VALUE OF CHARGE AND CURRENT PROPAGATESOUTWARDS AT A SPEED C. IF THE CHARGE DISTRIBUTION NEAR Q CHANGES WITH TIME, THE INFORMATION THAT IT HAS CHANGED CAN ONLY BE APPRECIATED AT THE POINT P A TIME |r-r`|/C AFTER THE CHANGE. THIS IS THE TIME IT TAKES ELECTROMAGNETIC RADIATION TO TRAVEL THE DISTANCE |r-r`| PROCEDURE FOR SOLVING PROBLEMS OF RETARDED POTENTIALS WE START BY USING THE LORENTZ TRANSFORMATION TO FIND THE POTENTIALS GENERATED BY A SMALL CHARGE WHEN IT IS VIEWED FROM A MOVING FRAME OF REFERENCE. THEN THE POTENTIALS CAUSED BY ANY DISTRIBUTION OF CHARGE AND CURRENT CAN BE FOUND BY SUPERPOSITION. IF WE WISH TO RELATE THE FIELDS TO THE CHANGING CHARGE AND CURRENT DISTRIBUTION WHICH PRODUCE THEM, IT IS EASIER TO WORK WITH THE POTENTIALS; THAN WITH THE FIELDS DIRECTLY. WE WILL OUTLINE THE STEPS TAKEN TO OBTAIN THE POTENTIALS IN TERMS OF GIVEN CHARGE AND CURRENT DISTRIBUTIONS. ONCE THE POTENTIALS HAVE BEEN DETERMINED THE FIELDS CAN BE OBTAINED FROM THE RELATIONSHIPS BELOW B = CURL A (*) B≡MAGNETIC FIELD B A≡MAGNETIC VECTOR POTENTIAL H≡MAGNETIC FIELD INTENSITY(STRENGTH) E =-(𝜕A/𝜕t) – gradΦ (**) Φ ≡ ELECTROSTATIC POTENTIAL (IF THE FIELDS ARE INDEPENDENT OF TIME) OTHERWISE IF VARYING WITH TIME JUST ELECTRIC POTENTIAL E ≡ ELECTRIC FIELD D ≡ ELECTRIC DISPLACEMENT FIRST WE DERIVE DIFFERENTIAL EQUATIONS OBEYED BY THE POTENTIALS A AND Φ SEPARATELY. WE ONLY CONSIDER CHANGING CHARGE AND CURRENT DISTRIBUTION IN FREE SPACE (FREE SPACE OR AIR.) MAXWELL’S EQUATIONS BECOME div E = ρ/ε0 (***) divB = 0 Curl E = -𝜕B/𝜕t (1/𝜇0)Curl B = j + ε0𝜕E/𝜕t (@) where: D = εε0E H=B/𝜇𝜇0 SUBSTITUTE FOR E IN (***) USING (**) TO GET -∇2Φ – (𝜕/𝜕t)(divA) = ρ/ε0 (****) NB. IF f(x,y,z) IS A TWICE DIFFERENTIABLE SCALAR FUNCTION, THEN grad f =(𝜕f/𝜕x)i+(𝜕f/𝜕y)j+(𝜕f/𝜕z)k THE div V= (𝜕V1/𝜕x)+(𝜕V2/𝜕y)+(𝜕V3/𝜕z) = ∇.V = (𝜕/𝜕xi+𝜕/𝜕yj+𝜕/𝜕zk).(V1i+V2j+V3k) = div(grad f) = (𝜕2f/𝜕x2)+(𝜕2f/𝜕y2)+(𝜕2f/𝜕z2) div(grad f ) =∇2.f THIS IS THE LAPLACIAN OF f THE POTENTIALS A AND Φ ARE NOT UNIQUELY SPECIFIED BY EQUATIONS (*) AND (**) WE CAN ADD TO THE VECTOR A ANY FUNCTION WHOSE CURL IS ZERO AND THE SAME MAGNETIC FIELD RESULTS. WE CAN ADD TO THE POTENTIAL Φ ANY FUNCTION WHOSE GRADIENT IS ZERO AND THE SAME ELECTRIC FIELD RESULTS. WE ARE AT LIBERTY TO IMPOSE ANY CONDITION ON THE FUNCTION A AND Φ THAT DOES NOT CONTRADICT MAXWELL’S EQUATIONS. IN THIS DISCUSSION OF STEADY MAGNETIC FIELD WE ADDED THE CONDITION div A = 0 WITH TIME – DEPENDENT FIELDS, IT IS CONVINIENT TO GENERALIZE THIS TO div A= - ε0𝝻0𝜕Φ/𝜕t (&) THE RESTRICTION ON THE POTENTIALS IMPOSED BY EQUATION & ABOVE IS KNOWN AS THE LORENTZ CONDITION. IT INTRODUCES A SYMMETRY INTO THE EQUATIONS FOR THE POTENTIAL. IT ALSO ENABLES ELECTROMAGNETISM TO BE BROUGHT INTO LINE WITH THE SPECIAL THEORY OF RELATIVITY. SUBS (&) INTO (****) TO GET -∇2Φ + ε0𝝻0𝜕2Φ/𝜕t2 =ρ/ ε0 (&&) NOTICE THAT THIS EQUATION INVOLVES Φ ALONE. Ie THE POTENTIAL Φ IN THE AREAS OF SPACE WHERE ρ IE CHARGE DENSITY IS ZERO, THE POT5ENTIAL Φ OBEYS THE SAME WAVE EQUATION AS THE FIELDS E AND B. THE EQUATION FOR THE VECTOR POTENTIAL A CAN BE OBTAINED BY SUBSTITUTING (*), AND (**)IN (@). THEN USING LORENTZ CONDITION LEADS TO AN EQUATION SIMILAR TO EQUATION (&&) ABOVE. -∇2A + ε0𝝻0𝜕2A/𝜕t2 =𝝻0j (&&&) THESE ARE THE DIFFERENTIAL EQUATIONS FOR THE POTENTIALS. A SOLUTION OF THESE EQUATIONS GIVE RISE TO THE POTENTIALS A AND Φ IN TERMS OF THE CHARGE AND CURRENT DISTRIBUTIONS WE WILL SOLVE THIS PROBLEM BY INFERENCE IE NOT DIRECTLY BUT USE THE ANOLOGY WITH THE SOLUTION OF TIME INDEPENDENT PROBLMS DISCUSSED IN ELECTROSTATICS AND MAGNETOSTATCS. IN ELECROSTATICS, WE SOLVE THE PROBLEM -∇2Φ =ρ/ ε0 NB: THE FIELDS ARE STATIC HENCE THE DIFFERENTIAL OF THE FIELD IS ZERO. IE. THAT COMPONENT REMOVED FROM THE TIME VARYING POTENTIAL DIFFERENTIAL EQUATION. (ε0𝝻0𝜕2Φ/𝜕t2 ≡ FOR TIME VARYING FIELDS) THE SOLUTION IS Φ(r) = (1/4πε0)∫v((ρ(r’)/(|r-r’|))dτ’) Where V: Volume occupied by the Charge distribution ρ This is the solution of the equation: -∇2Φ = ρ/ ε0