The Lorentz force

1. 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, expressed in terms of the fields E and B. Advanced EM - Master in Physics 2010-2011

2. In the charge’s rest frame: 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 . Then we will transform again the to the original IRF frame in which the particle is in motion, and see what corresponds to the values of . Since the LT has different values for longitudinal and transverse coordinates, we shall keep the two separate. • What we want to calculate is dP/dt as a function of the fields the charged particle sees in the IRF (the lab). • 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. • We establish the relation between dP/dt and dP’/dt’ (LT). Again, keep transverse and longitudinal components separate. • We write the LT of longitudinal and transverse electric fields E’ as a function of E, B (in IRF). Advanced EM - Master in Physics 2010-2011

3. LT of the fields Equations relating fields and force (dP/dt) LT of dP/dt Putting everything together: 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 “strict” electrostatics plus special relativity plus charge conservation. Advanced EM - Master in Physics 2010-2011

4. The Maxwell Equations In an IRF in which charges are at rest, If we apply an LT to change to a new IRF, this equation becomes The “time term of this equation is the first Maxwell law. We then write down the three “space terms” of this 4-vector equation: These three equations can be written in the form of an equation between 3-vectors: which is the 4th Maxwell law. Advanced EM - Master in Physics 2010-2011

5. And from the: One 4-vector equation covers two Maxwell equations: one scalar and the other a 3-vector equation. We may expect that the other two Maxwell equations (the second and the third) will correspond to another 4-vector equation. From the first column of the field tensor we have: Another way to obtain these last two Maxwell laws is to use the completely asymmetric tensor of rank 4 εμvλρ defined as ε αβγδ={ 0 if at least two indices are equal. +1 if αβγδ = 1 2 3 4 or any even permutation. -1 for any odd permutation of αβγδ = 1 2 3 4 From this tensor and the field tensor we obtain the so-called dual tensor of the field tensor Fμv Advanced EM - Master in Physics 2010-2011

6. { It turns out that is also an asymmetric tensor of rank 4. Its components are: This dual tensor contains the same information, the same components as the original field tensor; but in a different order. It is easy to notice that, with the exception of a few sign changes the main effect is to replace the E with theB terms – and vice versa. And now it is the that gives us the 2nd and 3rd Maxwell l laws, so that the 4 Maxwell equations are reduced to the compact form Maxwell laws in covariant form: The dual tensor, with its structure symmetrical wrt that of the field tensor, lend itself nicely to be contracted with the field tensor to find field invariants. Advanced EM - Master in Physics 2010-2011

7. EM field invariants With the two field tensors we can define two SCALARS, which automatically become Invariants: • It is easy just looking at the two tensors why one product gives the scalar product of the fields with themselves and the other product gives the scalar product of one field with the other. • If in an IRF B>E, then B>E in all IRFs. It will not be possible to find an IRF in which B=0. And vice versa. • If B=E in one IRF then B=E in all IRFs (radiation!) • If B·E≠0, this property will hold in any IRF. There is no reference system in which either B or E can be null. • IfB·E=0 in an IRF, then either B=E, and the two fields are orthogonal in any IRF (radiation); or an IRF exists in which either is null. Advanced EM - Master in Physics 2010-2011

8. The force as a 4-vector • We want to find a 4-vector that • Satisfies an equation of type F=ma, but of course with 4-vectors. • For β0 it has a “space” sector that tends to 3-dimension force. • Can be expressed in terms of the 3-dimension force F in systems IRF’ in which charges – acted upon by the force F are relativistic. With such 4-vector we can try to cover the same route that brought us (in 3 dimensions) to the Energy conservation. We shall start from F=dP/dt , which obviously are 3-vectors. To transform this equation into one for 4-vectors we can use And, instead of deriving wrt t, we derive wrt the proper time τ: we the write: Where We want to find the expressions for the fi as a function of theF coordinatesFiof the Newtonian force F. Advanced EM - Master in Physics 2010-2011

9. The space components of , fα are equal, in the equations of motion, to the γ·d(mγvα )/dt. Now we impose the condition that, for β0 , fα= γ·d(mγv)/dt tends to the Newtonian Fα =d(Pα)/dt Therefore: The term mγv for β0 tends to mv, the Newton momentum and, therefore, the fα tends to the Newtonian d(Pα)/dt . First result: This is a good result: the 4-vector force” is defined in terms of the components of the 3-vector force. We still miss the f0 coordinate. Now.. We know that the squared module of the 4-velocity is a constant. Then its derivative is null: But… Advanced EM - Master in Physics 2010-2011

10. From this last equation we obtain that : So that: • We obtain the unexpected result that the force exerted on an object depends on the object’s velocity. • This 4-force is called “Minkowski force” • It is a 4-vector • Its space components tend, for β0, to the 3-dimensional Newtonian force. • It satisfies the equation Advanced EM - Master in Physics 2010-2011

11. We have understood the role and meaning of the space components; and their limit for β0. What about the time_component? Now, the tells us that its time-component satisfies the condition: And, as a consequence, Now we remember that F·v is the work done by the force on the object. Its physical interpretation is as a consequence a dE /dt, where E is the object’s energy. Therefore, it tells us that the object’s energy is mγc². This result has been obtained in much the same way as the newtonian result, i.e. integrating the work done by a force. In this case though it tells us what is the rest energy of an object, i.e. Advanced EM - Master in Physics 2010-2011