Download E, B

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.
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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 ' to the 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
with 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 (in IRF).
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E 'L  EL

E 'T  ET  ( v  B)
c
dP' dP'

 qE'
d
dt '
}
}
LT of the fields
Equations relating fields and
force (dP/dt)
dP'L dP'L

 qE'L
d
dt '
dP' dP
dP

 
LT of dP/dt
d
d
dt
dPL
dP'L
dPL / 
dPL



d
d
dt / 
dt
[ since : (PL  P'L  mc)  dPL    dP'L ]
}
Putting everything together:
dPL dP'L

 qEL
dt
dt
dP 1 dP'
v

 q (E T   B )
dt
 dt
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 “strict” electrostatics plus special relativity plus charge
conservation.
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The Maxwell Equations
In an IRF in which charges are at rest,
  E  4 
4 0
J
c
or , in covariant form,  μ F μ 0 
If we apply an LT to change to a new IRF, this equation becomes
 μ F μν 
4 0
J
c
4 ν
J
c
The “time term of this equation is the first Maxwell law. We then
write down the three “space terms” of this 4-vector equation:
Bz By 1 Ex
4




Jx
y
z c t
c
Bz Bx 1 Ey
4



Jy
x
z c t
c
By Bx 1 Ez
4




Jz
x
y c t
c
These three equations can be written in the form of an equation
between 3-vectors:
B 
1 E 4

J
c t
c
which is the 4th Maxwell law.
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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:
E   
1 A
c t
from this we obtain :
  E   
{  
1 A
}    {  A}   B
c t
t
t
the 2nd Maxwell law
And from the:
B   A
we get the 3rd Maxwell law :
  B    (  A)  0
Another way to obtain these last two Maxwell laws is to use the
completely asymmetric tensor of rank 4
ε αβγδ {
=
εμvλρ defined as
0 if at least two indices are equal.
+1 if α β γ δ = 1 2 3 4 or any even permutation.
-1 for any odd permutation of α β γ δ = 1 2 3 4
From this tensor and the field tensor we obtain the so-called
dual tensor F

of the field tensor Fμv
F


1 

F
2
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It turns out that F
Its components are:
F 
μν
0
E
x

Ey

 Ez

 Ex
 Ey
0
Bz
 By
 Bz
0
Bx
is also an asymmetric tensor of rank 4.
 Ez 
By 
 Bx 

0 
0
B

[F ]   x
By

 Bz
 Bx
 By
0
 Ez
Ey
Ez
0
 Ex
 Bz 
 E y 
Ex 

0 
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 the B terms – and vice versa.
And now it is the
μ F
μν
0
that gives us the 2nd and 3rd Maxwell l laws, so that the 4 Maxwell
equations are reduced to the compact form
{
 μ F μν 
Maxwell laws in covariant form:
μ F
μν
4 ν
J
c
0
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.
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EM field invariants
With the two field tensors we can define two SCALARS, which
automatically become Invariants:
1 
F F  B 2  E 2
2
and
1 
F F   E  B
2
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
to find an IRF in which
B>E in all IRFs. It will not be possible
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.
•If B·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.
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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
P  mU
And, instead of deriving wrt t, we derive wrt the proper time τ:
we the write:
F
Where
dP
d
F  { f 0 , f1, , f 2 , f 3}
We want to find the expressions for the fi as a function of the F
coordinates Fi of the Newtonian force F.
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F
d P d P dt
dP
d (mc ) d (mv)



 {
,
}
d
dt d
dt
dt
dt
The space components of
motion, to the γ·d(mγvα
F , fα
are equal, in the equations of
)/dt. Now we impose the condition that, for
β0 , fα = γ·d(mγv)/dt tends to the Newtonian Fα =d(Pα)/dt
Therefore:
d (mv )
f    F  
dt
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:
F  {?, Fx , Fy , Fz }
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:
2
dU 0
dU 3
1 dU
dU1
dU 2
 U0
 U1
U2
U3
0
2 d
d
d
d
d
But…
dU 
f
 
d
m
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U 0  f 0 / m  U1  f1 / m  U 2  f 2 / m  U 3  f / m3  0
c  f 0 / m  vx  f1 / m  v y  f / m2  vz  f 3 / m  0
From this last equation we obtain that :
f0 

c
Fv
So that:

F  { F  v, F}
c
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
3-dimensional Newtonian force.
β0, to the
•It satisfies the equation
F
dP
d
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We have understood the role and meaning of the space components;
and their limit for β0. What about the time_component?
Now, the
F  dP
tells us that its time-component
d
satisfies the condition:
dP 0 
d (mc )
d (mc )
F 
 Fv 

d
c
d
dt
0
And, as a consequence,
d (mc2 )
Fv 
dt
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.
E0 
mc2
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