k
,
(9.35)
v w
and for the frequency perceived by the receiver, Eq. (34) yields
v v
w
r
'
.
(9.36)
v w
On the other hand, if the receiver and the medium are at rest in the reference frame 0 ’, while the wave source is bound to the frame 0 (so that v = – v s), Eq. (35) should be replaced with
'
k k'
,
(9.37)
v w
and Eq. (34) yields a different result:
v
w
'
,
(9.38)
v v
w
s
Finally, if both the source and detector are moving, it is straightforward to combine these two results to get the general relation
v v
w
r
'
.
(9.39)
v v
w
s
19 Strictly speaking, Eq. (32) is valid to an additive constant, but for notation simplicity, it may be always made equal to zero by selecting (as has already been done in all relations of Sec. 1) the reference frame origins and/or clock turn-on times so that at t = 0 and x = 0, t’ = 0 and x’ = 0 as well.
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At low speeds of both the source and the receiver, this result simplifies,
v v
' 1
r
s
,
with
,
(9.40)
v w
but at speeds comparable to v w we have to use the more general Eq. (39). Thus, the usual Doppler effect is generally affected not only by the relative speed ( v r – v s) of the wave’s source and detector but also by their speeds relative to the medium in which the waves propagate.
Somewhat counter-intuitively, for the electromagnetic waves the calculations are simpler because for them the propagation medium (aether) does not exist, the wave velocity equals c in any reference frame, and there are no two separate cases: we can always take k = / c and k’ = ’/ c.
Plugging these relations, together with the Lorentz transform (19a), into the phase-invariance condition (32), we get
ct' x'
'
( x' ct' )
x' 't'
.
(9.41)
c
c
c
This relation has to hold for any x’ and t’, so we may require that the net coefficients before these variables vanish. These two requirements yield the same equality:
' 1
( ) .
(9.42)
This result is already quite simple, but may be transformed further to be even more illuminating: 1
1 1 1/2
'
.
(9.43)
2
1 1/2
1 1
At any sign before , one pair of parentheses cancel, so that20
1/ 2
Longitudinal
1
Doppler
'
effect
.
(9.44)
1
Thus the Doppler effect for electromagnetic waves depends only on the relative velocity v = c between the wave source and detector – as it should be, given the aether’s absence. At velocities much lower than c, Eq. (44) may be approximated as
1 / 2
'
1 ,
(9.45)
1 / 2
i.e. in the first approximation in v/ c it tends to the corresponding limit (40) of the usual Doppler effect.
If the wave vector k is tilted by angle to the vector v (as measured in frame 0), then we have to repeat the calculations, with k replaced by kx, and components ky and kz left intact at the Lorentz transform. As a result, Eq. (42) is generalized as
20 It may look like the reciprocal expression of via ’ is different, violating the relativity principle. However, in this case, we have to change the sign of , because the relative velocity of the system is opposite, so that we return to Eq. (44) again.
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' 1 cos .
(9.46)
For the case cos = 1, Eq. (46) reduces to our previous result (42). However, at = /2 (i.e. cos = 0), the relation is rather different:
Transverse
'
.
(9.47) Doppler
2
1 1/2
effect
This is the transverse Doppler effect – which is absent in non-relativistic physics. Its first experimental evidence was obtained using electron beams (as had been suggested in 1906 by J. Stark), by H. Ives and G. Stilwell in 1938 and 1941. Later, similar experiments were repeated several times, but the first unambiguous measurements were performed only in 1979 by D. Hasselkamp et al. who confirmed Eq. (47) with a relative accuracy of about 10%. This precision may not look too spectacular, but besides the special tests discussed above, the Lorentz transform formulas have been also confirmed, less directly, by a huge body of other experimental data, especially in high energy physics, agreeing with calculations incorporating this transform as their part. This is why, with due respect to the spirit of challenging authority, I should warn the reader: if you decide to challenge the relativity theory (called
“theory” by tradition only), you would also need to explain all these data. Best luck with that! 21
9.3. 4-vectors, momentum, mass, and energy
Before proceeding to the relativistic dynamics, let us discuss the mathematical formalism that makes all calculations more compact – and more beautiful. We have already seen that the three spatial coordinates { x, y, z} and the product ct are Lorentz-transformed similarly – see Eqs. (18)-(19) again. So it is natural to consider them as components of a single four-component vector (or, for short, 4-vector),
{ x , x , x , x } ct
,
(9.48)
0
1
2
3
, r
with components
Space
x ct, x x, x y, x z .
(9.49) -time
0
1
2
3
4-vector
According to Eqs. (19), its components are Lorentz-transformed as
3
Lorentz
x
L x' ,
(9.50) transform:
j
jj' j'
j'0
4-form
where Ljj’ are the elements of the following 44 Lorentz transform matrix
0 0
0 0
Lorentz
.
(9.51) transform
0
0
1 0
matrix
0
0
0 1
Since such 4-vectors are a new notion for this course and will be used for many more purposes than just the space-time transform, we need to discuss the general mathematical rules they obey. Indeed, 21 The same fact, ignored by crackpots, is also valid for other favorite directions of their attacks, including the Universe expansion, quantum measurement uncertainty, and entropy growth in physics, and the evolution theory in biology.
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as was already mentioned in Sec. 8.9, the usual (three-component) vector is not just any ordered set ( string) of three scalars { Ax, Ay, Az}; if we want it to represent a reference-frame-independent physical reality, the vector’s components have to obey certain rules at the transfer from one reference frame to another. In particular, in the non-relativistic limit the vector’s norm (its magnitude squared), 2
2
2
2
A A A A ,
(9.52)
x
y
z
should be invariant with respect to the transfer between different reference frames. However, a naïve extension of this approach to 4-vectors would not work, because, according to the calculations of Sec. 1, the Lorentz transform keeps intact the combinations of the type (7), with one sign negative, rather than the sum of all components squared. Hence for the 4-vectors, all the rules of the game have to be reviewed and adjusted – or rather redefined from the very beginning, for example as follows.22
An arbitrary 4-vector is a string of 4 scalars,23
General
4-vector
A , A , A , A ,
(9.53)
0
1
2
3
whose components Aj, as measured in the reference frames 0 and 0 ’ shown in Fig. 1, obey the Lorentz transform relations similar to Eq. (50):
Lorentz
3
transform:
A
L A' .
(9.54)
j
general
jj' j'
j'0
4-vector
As we have already seen in the example of the space-time 4-vector (48), this means in particular that 3
3
Lorentz
2
A
A
A'
A'
.
(9.55)
0
invariance
2
j
02 j2
j1
j1
This is the so-called Lorentz invariance condition for the 4-vector’s norm. (The difference between this relation and Eq. (52), pertaining to the Euclidian geometry, is the reason why the Minkowski space is called pseudo-Euclidian.) It is also straightforward to use Eqs. (51) and (54) to check that the evident generalization of the norm, the scalar product of two arbitrary 4-vectors, 3
Scalar
4-product
A B
A B ,
(9.56)
0
0
j j
j1
is also Lorentz-invariant.
Now consider the 4-vector corresponding to a small interval between two close world events: dx
{
, dx , dx , dx }
;
(9.57)
0
1
2
3
cdt, r
d
its norm,
3
Interval
2
2
2
2
2
2
( ds) dx dx c ( dt) ( dr) ,
(9.58)
0
j
j 1
22 The most prominent alternative, which has both advantages and drawbacks, is to use 4-vectors with one imaginary component – for example, the imaginary time ict instead of the real product ct in Eq. (48).
23 Such vectors are said to reside in so-called 4D Minkowski spaces – called after Hermann Minkowski who was the first one to recast (in 1907) the special relativity relations in a form in which the spatial coordinates and time (or rather ct) are treated on an equal footing.
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is of course also Lorentz-invariant. Since the speed of any particle (or signal) cannot be larger than c, for any pair of world events that are in a causal relation with each other, ( dr)2 cannot be larger than ( cdt)2, i.e. such time-like interval ( ds)2 cannot be negative. The 4D surface separating such intervals from space-like intervals ( ds)2 < 0 is called the light cone (Fig. 9).
time-like interval ds 2 0
t
(causal relation possible)
space-like interval ds 2 < 0
x 2
(causal relation impossible)
0
r ct
x
Fig. 9.9. A 2+1 dimensional image of
1
the light cone – which is actually 3+1
dimensional.
Now let us consider two close world events that happen with the same point moving with velocity u. Then in the frame moving with the point (v = u), the last term on the right-hand side of Eq.
(58) equals zero, while the involved time is the proper one, so that
ds
cd ,
(9.59)
where d is the proper time interval. But according to Eq. (21), this means that we can write
dt
d
,
(9.60)
where dt is the time interval in an arbitrary (besides being inertial) reference frame, while u
1
1
β
and
(9.61)
c
2
1 1/2
2 2
1 u / c 1/2
are the parameters (17) corresponding to the point’s velocity (u) in that frame, so that ds = cdt/.24
Let us use Eq. (60) to explore whether a 4-vector may be formed using the spatial Cartesian components of the point’s velocity
dx dy dz
u ,
,
.
(9.62)
dt dt dt
Here we have some problem: as Eqs. (22) show, these components do not obey the Lorentz transform.
However, let us use d dt/, the proper time interval of the point, to form the following string:
dx dx dx dx
dx dy dz
0
,
1 ,
2 ,
3 c
,
,
,
c,
u .
(9.63) 4-velocity
d d d d
dt dt dt
24 I have opted against using special indices (e.g., u and u) to distinguish Eqs. (17) and (61) here and below, in a hope that the suitable velocity (of either a reference frame or a particle) will be always clear from the context.
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As it follows from the comparison of the middle form of this expression with Eq. (48), since the time-space vector obeys the Lorentz transform, and is Lorentz-invariant, the string (63) is a legitimate 4-vector; it is called the 4-velocity of a point – or of a point particle.
Now we are well equipped to proceed to relativistic dynamics. Let us start with such basic notions as the momentum p and the energy E – so far, for a free particle.25 Perhaps the most elegant way to “derive” (or rather guess26) the expressions for p and E as functions of the particle’s velocity u, is based on analytical mechanics. Due to the conservation of v, the trajectory of a free particle in the 4D
Minkowski space { ct, r} is always a straight line. Hence, from the Hamilton principle,27 we may expect its action S, between points 1 and 2, to be a linear function of the space-time interval (59): 2
2
t 2
Free
dt
particle:
S
action
ds c d c ,
(9.64)
1
1
t
1
where is some constant. On the other hand, in analytical mechanics, the action is defined as t 2
S L dt ,
(9.65)
t 1
where L is the particle’s Lagrangian function.28 Comparing these two expressions, we get 1/ 2
2
c
u
L
c 1
.
(9.66)
2
c
In the non-relativistic limit ( u << c), this function tends to
u 2
u 2
L c 1
c
.
(9.67)
2 c 2
2 c
In order to correspond to the Newtonian mechanics,29 the last (velocity-dependent) term should equal mu 2/2. From here we find = – mc, so that, finally,
Free
1/ 2
2
2
particle:
u
mc
2
Lagrangian
L mc 1
.
(9.68)
2
function
c
Now we can find the Cartesian components pj of the particle’s momentum as the generalized momenta corresponding to the corresponding components rj ( j = 1, 2, 3) of the 3D radius-vector r:30
25 I am sorry for using, just as in Sec. 6.3, the same traditional notation (p) for the particle’s momentum as had been used earlier for the electric dipole moment. However, since the latter notion will be virtually unused in the balance of this course, this may hardly lead to confusion.
26 Indeed, such a derivation uses additional assumptions, however natural (such as the Lorentz-invariance of S), i.e. it can hardly be considered as a real proof of the final results, so that they require experimental confirmation.
Fortunately, such confirmations have been numerous – see below.
27 See, e.g., CM Sec. 10.3.
28 See, e.g., CM Sec. 2.1.
29 See, e.g., CM Eq. (2.19b).
30 See, e.g., CM Sec. 2.3, in particular Eq. (2.31).
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L
L
2
1/ 2
u 2 u 2 u 2
mu
1
2
3
j
p
mc
1
m u .
(9.69)
j
r
u
u
c 2
1/ 2
2
2
j
j
j
1 u / c
j
Thus for the 3D vector of momentum, we can write the result in the same form as in non-relativistic mechanics,
p m u u
M ,
(9.70) Relativistic
momentum
using the reference-frame-dependent scalar M (called the relativistic mass) defined as m
M m
,
(9.71) Relativistic
1 u 2
2
/
m
c 1/ 2
mass
m being the non-relativistic mass of the particle. (More often, m is called the rest mass, because in the reference frame in which the particle rests, Eq. (71) yields M = m.)
Next, let us return to analytical mechanics to calculate the particle’s energy E (which for a free particle coincides with its Hamiltonian function H):31
1/ 2
3
2
2
2
mu
u
mc
2
E H p u L p u L
mc 1
. (9.72)
j
j
j 1
2 2
1 u / c 1/2
2
c
2 2
1 u / c 1/2
Thus, we have arrived at the most famous of Einstein’s formulas – and probably of physics as a whole: 2
2
E m c Mc ,
(9.73) E = Mc 2
which expresses the relation between the free particle’s mass and its energy.32 In the non-relativistic limit, it reduces to
2
2
2
mc
u
mu
2
2
E
mc
mc
(9.74)
1
2
u / 2
c
1
,
1/ 2
2 2
c
2
the first term mc 2 being called the rest energy of a particle.
Now let us consider the following string of 4 scalars:
E
E
4-vector of
, p , p , p
.
(9.75)
energy-
1
2
3
, p
c
c
momentum
Using Eqs. (70) and (73) to represent this expression as
E
, p m c,
u ,
(9.76)
c
31 See, e.g., CM Eq. (2.32).
32 Let me hope that the reader understands that all the layman talk about the “mass to energy conversion” is only valid in a very limited sense of the word. While the Einstein relation (73) does allow the conversion of “massive”
particles (with m 0) into particles with m = 0, such as photons, each of the latter particles also has a non-zero relativistic mass M, and simultaneously the energy E related to this M by Eq. (73).
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and comparing the result with Eq. (63), we immediately see that, since m is a Lorentz-invariant constant, this string is a legitimate 4-vector of energy-momentum. As a result, its norm,
2
E
2
p ,
(9.77a)
c
is Lorentz-invariant, and in particular, has to be equal to the norm in the particle-bound frame. But in that frame, p = 0, and according to Eq. (73), E = mc 2, and the norm is just 2
2
2
E
mc
mc2 ,
(9.77b)
c
c
so that in an arbitrary frame
2
E
2
2
p ( mc) .
(9.78a)
c
This very important relation33 between the relativistic energy and momentum (valid for free particles only!) is usually represented in the form34
Free
particle:
2
E mc 2
2
pc2 .
(9.78b)
energy
According to Eq. (70), in the so-called ultra-relativistic limit u c, p tends to infinity, while mc 2 stays constant so that pc/ mc 2 . As follows from Eq. (78), in this limit E pc. Though the above discussion was for particles with finite m, the 4-vector formalism allows us to consider compact objects with zero rest mass as ultra-relativistic particles for which the above energy-to-moment relation, E pc,
for m 0 ,
(9.79)
is exact. Quantum electrodynamics35 tells us that under certain conditions, the electromagnetic field quanta (photons) may be also considered as such massless particles with momentum p = k. Plugging (the modulus of) the last relation into Eq. (78), for the photon’s energy we get E = pc = kc = . Please note again that according to Eq. (73), the relativistic mass of a photon is not equal to zero: M = E/ c 2 =
/c 2 , so that the term “massless particle” has a limited meaning: m = 0. For example, the relativistic mass of an optical phonon is of the order of 10-36 kg. On the human scale, this is not too much, but still a noticeable (approximately one-millionth) part of the rest mass m e of an electron.
The fundamental relations (70) and (73) have been repeatedly verified in numerous particle collision experiments, in which the total energy and momentum of a system of particles are conserved –
at the same conditions as in non-relativistic dynamics. (For the momentum, this is the absence of external forces, and for the energy, the elasticity of particle interactions – in other words, the absence of alternative channels of energy escape.) Of course, generally only the total energy of the system is conserved, including the potential energy of particle interactions. However, at typical high-energy 33 Please note one more simple and useful relation following from Eqs. (70) and (73): p = (E/ c 2)u.
34 It may be tempting to interpret this relation as the perpendicular-vector-like addition of the rest energy mc 2 and the “kinetic energy” pc, but from the point of view of the total energy conservation (see below), a better definition of the kinetic energy is T( u) E (u) – E (0).
35 It is briefly reviewed in QM Chapter 9.
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particle collisions, the potential energy vanishes so rapidly with the distance between them that we can use the momentum and energy conservation laws using Eq. (73).
As an example, let us calculate the minimum energy Emin of a proton (p a), necessary for the well-known high-energy reaction that generates a new proton-antiproton pair, p a + p b p + p + p + p , provided that before the collision, proton p b had been at rest in the lab frame. This minimum corresponds to the vanishing relative velocity of the reaction products, i.e. their motion with virtually the same velocity (ufin), as seen from the lab frame – see Fig. 10.
frame
lab
frame
c.o.m.
min
E
u
u
Fig. 9.10. A high-energy proton
p
min
p
a
b
fin
reaction at E Emin – schematically.
Due to the momentum conservation, this velocity should have the same direction as the initial velocity (umin) of proton pa. This is why two scalar equations: for energy conservation, 2
2
mc
4 mc
2
mc
,
(9.80a)
2
2
1 u
/ c
1 u / c
min
1/2
2 2
fin
1/2
and for momentum conservation,
mu
4 mu
0
,
(9.80b)
2
2
1 u
/ c
1 u / c
min
1/
fin
2
2
2
fin
1/2
are sufficient to find both u min and u fin. After a rather tedious solution of this system of two nonlinear equations, we get
4 3
3
u
c 990
.
0
c,
u
c 866
.
0
c .
(9.81)
min
7
fin
2
Finally, we can use Eq. (72) to calculate the required energy; the result is Emin = 7 mc 2. (Note that at this threshold, only a minor 2 mc 2 part of the kinetic energy T min = Emin – mc 2 = 6 mc 2 of the initially moving particle, goes into the “useful” proton-antiproton pair production.) The proton’s rest mass, m p 1.6710-27 kg, corresponds to m p c 2 1.50210-10 J 0.938 GeV, so that Emin 6.57 GeV.
The second, more intelligent way to solve the same problem is to use the center-of-mass ( c.o.m.) reference frame that, in relativity, is defined as the frame in which the total momentum of the system vanishes.36 In this frame, at E = Emin, the velocity and momenta of all reaction products are vanishing, while the velocities of the protons p a and p b before the collision are equal and opposite, with an initially unknown magnitude u’. Hence the energy conservation law becomes
2
2 mc
2
4 mc ,
(9.82)
2
2
1 u' / c 1/2
36 Note that according to this definition, the c.o.m.’s radius-vector is R = kMkr k/ kMk k kmkr k/ k kmk, i.e. is generally different from the well-known non-relativistic expression R = kmkr k/ kmk .
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readily giving u’/ c = 3/2. (This is of course the same result as Eq. (81) gives for u fin.) Now we can use the fact that the velocity of the proton p a in the c.o.m. frame is (– u’), to find its lab-frame speed, using the velocity transform (25):
2 u'
u
.
(9.83)
min
2
2
1 u' / c
With the above result for u’, this relation gives the same result as the first method, u min/ c = 43/7, but in a simpler way.
9.4. More on 4-vectors and 4-tensors
This is a good moment to introduce a formalism that will allow us, in particular, to solve the same proton collision problem in one more (and arguably, the most elegant) way. Much more importantly, this formalism will be virtually necessary for the description of the Lorentz transform of the electromagnetic field, and its interaction with relativistic particles – otherwise the formulas would be too cumbersome.
Let us call the 4-vectors we have used before,
Contravariant
A A ,
,
(9.84)
0
A
4-vectors
contravariant, and denote them with top indices, and introduce also covariant vectors, Covariant
A
4-vectors
A , ,
(9.85)
0
A
marked by bottom indices. Now if we form a scalar product of these two vectors using the standard (3D-like) rule, just as a sum of the products of the corresponding components, we immediately get
2
2
A A A A A A
.
(9.86)
0
Here and below the sign of the sum of four components of the product has been dropped.37 The scalar product (86) is just the norm of the 4-vector in our former definition, and as we already know, is Lorentz-invariant. Moreover, the scalar product of two different vectors (also a Lorentz invariant), may be rewritten in any of two similar forms:38
Scalar
product's
A B A B
;
(9.87)
0
0
A B
A
B
forms
again, the only caveat is to take one vector in the covariant, and the other one in the contravariant form.
Now let us return to our sample problem (Fig. 10). Since all components (E/ c and p) of the total 4-momentum of our system are conserved at the collision, its norm is conserved as well:
p p
.
(9.88)
a
b p
p
a
b
(4 p) (4 p)
37 This compact notation may take some time to be accustomed to, but is very convenient (compact) and can hardly lead to any confusion, due to the following rule: the summation is implied when (and only when) the same index is repeated twice, once on the top and another at the bottom. In this course, this shorthand notation will be used only for 4-vectors, but not for the usual (3D spatial) vectors.
38 Note also that, by definition, for any two 4-vectors, A B = B A.
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Since now the vector product is the usual math construct, we know that the parentheses on the left-hand side of this equation may be multiplied as usual. We may also swap the operands and move constant factors through products as convenient. As a result, we get
p
2
16
.
(9.89)
a pa
pb pb pa pb
p p
Thanks to the Lorentz invariance of each of the terms, we may calculate it in the reference frame we like. For the first two terms on the left-hand side, as well as for the right-hand side term, it is beneficial to use the frames in which that particular proton is at rest; as a result, according to Eq. (77b), each of the two left-hand-side terms equals ( mc)2, while the right-hand side equals 16( mc)2. On the contrary, the last term on the left-hand side is more easily evaluated in the lab frame, because in it, the three spatial components of the 4-momentum pb vanish, and the scalar product is just the product of the scalars E/ c for protons a and b. For the latter proton, being at rest, this ratio is just mc so that we get a simple equation,
2
2
min
E
2
( mc) ( mc) 2
mc
(
16 mc) ,
(9.90)
c
immediately giving the final result Emin = 7 mc 2, already obtained earlier in two more complex ways.
Let me hope that this example was a convincing demonstration of the convenience of representing 4-vectors in the contravariant (84) and covariant (85) forms,39 with Lorentz-invariant norms (86). To be useful for more complex tasks, this formalism should be developed a little bit further.
In particular, it is crucial to know how the 4-vectors change under the Lorentz transform. For contravariant vectors, we already know the answer (54); let us rewrite it in our new notation: Lorentz
transform:
A L A' .
(9.91) contravariant
vectors
where
L is the matrix (51), generally called the mixed Lorentz tensor:40
0 0
0 0
Mixed
L
,
(9.92) Lorentz
0
0
1 0
tensor
0
0
0 1
Note that though the position of the indices and in the Lorentz tensor notation is not crucial, because this tensor is symmetric, it is convenient to place them using the general index balance rule: the difference of the numbers of the upper and lower indices should be the same in both parts of any 4-vector/tensor equality. (You may check that all the formulas above do satisfy this rule.) 39 These forms are 4-vector extensions of the notions of contravariance and covariance, introduced in the 1850s by J. Sylvester (who also introduced the term “matrix” in its mathematical sense) for the description of the change of the usual 3-component spatial vectors at the transfer between different reference frames – e.g., resulting from the frame rotation. In this case, the contravariance or covariance of a vector is uniquely determined by its nature: if the Cartesian coordinates of a vector (such as the non-relativistic velocity v = dr/ dt) are transformed similarly to the radius-vector r, it is called contravariant, while the vectors (such as f ) that require the reciprocal transform, are called covariant. In the 4D Minkowski space, both forms may be used for any 4-vector.
40 Just as the 4-vectors, 4-tensors with two top indices are called contravariant, and those with two bottom indices, covariant. The tensors with one top and one bottom index are called mixed.
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In order to rewrite Eq. (91) in a more general form that would not depend on the particular orientation of the coordinate axes (Fig. 1), let us use the contravariant and covariant forms of the 4-vector of the time-space interval (57),
dx cdt, d
r ,
dx
cdt, d
r ;
(9.93)
then its norm (58) may be represented as41
2
2
2
( ds) ( cdt) ( dr) dx
dx
dx dx .
(9.94)
Applying Eq. (91) to the first, contravariant form of the 4-vector (93), we get
dx L dx' .
(9.95)
But with our new shorthand notation, we can also write the usual rule of differentiation of each component x, considering it a function (in our case, linear) of four arguments x’, as follows:42
x
dx
dx' .
(9.96)
x'
Comparing Eqs. (95) and (96), we can rewrite the general Lorentz transform rule (92) in a new form, Lorentz
x
transform:
A
A' .
(9.97a)
general form
x'
which does not depend on the coordinate axes’ orientation.
It is straightforward to verify that the reciprocal transform may be represented as
Reciprocal
Lorentz
x'
transform
A'
A .
(9.97b)
x
However, the reciprocal transform has to differ from the direct one only by the sign of the relative velocity of the frames, so that for the coordinate choice shown in Fig. 1, its matrix is 41 Another way to write this relation is ( ds)2 = g dx dx = g dx dx, where double summation over indices
and is implied, and g is the so-called metric tensor,
1 0
0
0
0 1 0
0
g
g
,
0
0
1 0
0 0
0
1
which may be used, in particular, to transfer a covariant vector into the corresponding contravariant one and back: A = g A, A = g A. The metric tensor plays a key role in general relativity, in which it is affected by gravity – “curved” by particles’ masses.
42 Note that in the index balance rule, the top index in the denominator of a fraction is counted as a bottom index in the numerator, and vice versa.
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0 0
x'
0 0
,
(9.98)
x
0
0
1 0
0
0
0 1
Since according to Eqs. (84)-(85), covariant 4-vectors differ from the contravariant ones by the sign of their spatial components, their direct transform is given by matrix (98). Hence their direct and reciprocal transforms may be represented, respectively, as
x'
x
Lorentz
,
transform:
A
A'
A'
A ,
(9.99)
x
x'
covariant
vectors
evidently satisfying the index balance rule. (Note that primed quantities are now multiplied, rather than divided as in the contravariant case.) As a sanity check, let us apply this formalism to the scalar product A A. As Eq. (96) shows, the implicit-sum notation allows us to multiply and divide any equality by the same partial differential of a coordinate, so that we can write:
x'
x
x'
A A
A' A'
A' A'
A' A'
A' A' ,
(9.100)
x
x'
x'
i.e. the scalar product A A (as well as A A) is Lorentz-invariant, as it should be.
Now, let us consider the 4-vectors of derivatives. Here we should be very careful. Consider, for example, the following 4-vector operator
, ,
(9.101)
x
( ct)
As was discussed above, the operator is not changed by its multiplication and division by another differential, e.g., x’ (with the corresponding implied summation over all four values of ), so that
x'
.
(9.102)
x
x
x'
But, according to the first of Eqs. (99), this is exactly how the covariant vectors are Lorentz-transformed! Hence, we have to consider the derivative over a contravariant space-time interval as a covariant 4-vector, and vice versa.43 (This result might be also expected from the index balance rule.) In particular, this means that the scalar product
A
A
0
A
(9.103)
x
( ct)
should be Lorentz-invariant for any legitimate 4-vector. A convenient shorthand for the covariant derivative, which complies with the index balance rule, is
,
(9.104)
x
43 As was mentioned above, this is also a property of the reference-frame transform of the “usual” 3D vectors.
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so that the invariant scalar product may be written just as A. A similar definition of the contravariant derivative,
,,
(9.105)
x ( ct)
allows us to write the Lorentz-invariant scalar product (103) in any of the following two forms: A
0
A
A
A .
(9.106)
( ct)
Finally, let us see how the general Lorentz transform changes 4-tensors. A second-rank 44
matrix is a legitimate 4-tensor if the 4-vectors it relates obey the Lorentz transform. For example, if two legitimate 4-vectors are related as
A T B ,
(9.107)
we should require that
A' T' B' ,
(9.108)
where A and A’ are related by Eqs. (97), while B and B’, by Eqs. (99). This requirement immediately yields
Lorentz
x
x
x'
x'
T
T' ,
T'
T ,
(9.109)
transform
x'
x'
x
x
of 4-tensors
with the implied summation over two indices, and . The rules for the covariant and mixed tensors are similar.44
9.5. Maxwell equations in the 4-form
This 4-vector formalism background is sufficient to analyze the Lorentz transform of the electromagnetic field. Just to warm up, let us consider the continuity equation (4.5),
j 0 ,
(9.110)
t
which expresses the electric charge conservation, and as we already know, is compatible with the Maxwell equations. If we now define the contravariant and covariant 4-vectors of electric current as 4-vector
j
of electric
j c, ,
j
c, j,
(9.111)
current
then Eq. (110) may be represented in the form
Continuity
equation:
j
j 0
,
(9.112)
4-form
showing that the continuity equation is form-invariant 45 with respect to the Lorentz transform.
44 It is straightforward to check that transfer between the contravariant and covariant forms of the same tensor may be readily achieved using the metric tensor g: T = g T g, T = g T g.
45 In some texts, the formulas preserving their form at a transform are called “covariant”, creating a possibility for confusion with the covariant vectors and tensors. On the other hand, calling such formulas “invariant” would not distinguish them properly from invariant quantities, such as the scalar products of 4-vectors.
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Of course, such a form-invariance of a relation does not mean that all component values of the 4-vectors participating in it are the same in both frames. For example, let us have some static charge density in frame 0; then Eq. (97b), applied to the contravariant form of the 4-vector (111), reads
x'
j'
j ,
with j c, ,
0 ,
0
0 .
(9.113)
x
Using the particular form (98) of the reciprocal Lorentz matrix for the coordinate choice shown in Fig.
1, we see that this relation yields
Lorentz
' ,
j' c v
,
j' j' 0 .
(9.114) transforms
x
y
z
of and j
Since the charge velocity, as observed from frame 0 ’, is (–v), the non-relativistic results would be ’ =
, j ’ = –v. The additional factor in the relativistic results is caused by the length contraction: dx’ =
dx/, so that to keep the total charge dQ = d 3 r = dxdydz inside the elementary volume d 3 r = dxdydz intact, (and hence jx) should increase proportionally.
Next, at the end of Chapter 6 we have seen that Maxwell equations for the electromagnetic potentials and A may be represented in similar forms (6.118), under the Lorenz (again, not “Lorentz”, please!) gauge condition (6.117). For free space, this condition takes the form
1
A
0 .
(9.115)
2
c t
This expression gives us a hint of how to form the 4-vector of electromagnetic potentials:46
A , A ,
A
4-vector
,A ;
(9.116)
c
c
of potentials
indeed, this vector satisfies Eq. (115) in its 4-form:
Lorenz
A
A 0
.
(9.117) gauge:
4-form
Since this scalar product is Lorentz-invariant, and the derivatives (104)-(105) are legitimate 4-vectors, this implies that the 4-vector (116) is also legitimate, i.e. obeys the Lorentz transform formulas (97), (99). Even more convincing evidence of this fact may be obtained from the Maxwell equations (6.118) for the potentials. In free space, they may be rewritten as
2
2
c
2
c ,
.
(9.118)
2
2
0
2 A
j
2
0
( ct)
c
c
( ct)
0
Using the definition (116), these equations may be merged to one:47
Maxwell
equation
A j ,
(9.119)
0
for
4-potential
where is the d’Alembert operator,48 which may be represented as either of two scalar products, 46 In the Gaussian units, the scalar potential should not be divided by c in this relation.
47 In the Gaussian units, the coefficient 0 in Eq. (119) should be replaced, as usual, with 4/ c.
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D’Alembert
2
2
.
(9.120)
operator
2
( ct)
and hence is Lorentz-invariant. Because of that, and the fact that the Lorentz transform changes both 4-vectors A and j in a similar way, Eq. (119) does not depend on the reference frame choice. Thus we have arrived at a key point of this chapter: we see that the Maxwell equations are indeed form-invariant with respect to the Lorentz transform. As a by-product, the 4-vector form (119) of these equations (for potentials) is extremely simple – and beautiful!
However, as we have seen in Chapter 7, for many applications the Maxwell equations for the field vectors are more convenient; so let us represent them in the 4-form as well. For that, we may express all Cartesian components of the usual (3D) field vector vectors (6.7),
A
E
,
B A,
(9.121)
t
via those of the potential 4-vector A. For example,
A
A
E
x
c
x
c
,
(9.122)
x
0 1 1 0
A
A
x
t
x
c ( ct)
A
A
B
z
y
.
(9.123)
x
2 3 3 2
A
A
y
z
Completing similar calculations for other field components (or just generating them by appropriate index shifts), we find that the following antisymmetric, contravariant field-strength tensor,
F
A A ,
(9.124)
may be expressed via the field components as follows:49
0
E / c E / c E / c
x
y
z
Field-
strength
E / c
0
B
B
x
z
y
tensors
F
,
(9.125a)
E / c
B
0
B
y
z
x
E / c
B
B
0
z
y
x
so that the covariant form of the tensor is
0
E / c E / c E / c
x
y
z
E / c
0
B
B
x
z
y
F g F g
.
(9.125b)
E / c
B
0
B
y
z
x
E / c B
B
0
z
y
x
48 Named after Jean-Baptiste le Rond d’Alembert (1717-1783), who has made several pioneering contributions to the general theory of waves – see, e.g., CM Chapter 6. (Some older textbooks use notation 2 for this operator.) 49 In Gaussian units, this formula, as well as Eq. (131) for G, do not have the factor c in all the denominators.
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If Eq. (124) looks a bit too bulky, please note that as a reward, the pair of inhomogeneous Maxwell equations, i.e. two equations of the system (6.99), which in free space (D = 0E, B = 0H) may be rewritten as
E
E
c ,
B
j,
(9.126)
0
0
c
( ct) c
may now be expressed in a very simple (and manifestly form-invariant) way,
Maxwell
F
j ,
(9.127)
0
equation
for tensor F
which is comparable with Eq. (119) in its simplicity – and beauty. Somewhat counter-intuitively, the pair of homogeneous Maxwell equations of the system (6.99),
B
E
,
0
B ,
0
(9.128)
t
look, in the 4-vector notation, a bit more complicated:50
F F F 0
.
(9.129)
Note, however, that Eqs. (128) may be also represented in a much simpler 4-form,
0
G
,
(9.130)
using the so-called dual tensor
0
B
B
B
x
y
z
B
0
E / c
E / c
x
z
y
G
,
(9.131)
B
E / c
0
E / c
y
z
x
B
E / c
E / c
0
z
y
x
which may be obtained from F, given by Eq. (125a), by the following replacements: E
E
B, B
.
(9.132)
c
c
Besides the proof of the form-invariance of the Maxwell equations with respect to the Lorentz transform, the 4-vector formalism allows us to achieve our initial goal: to find out how the electric and magnetic field components change at the transfer between two (inertial!) reference frames. For that, let us apply to the tensor F the reciprocal Lorentz transform described by the second of Eqs. (109).
Generally, it gives, for each field component, a sum of 16 terms, but since (for our choice of coordinates, shown in Fig. 1) there are many zeros in the Lorentz transform matrix, and the diagonal components of F equal zero as well, the calculations are rather doable. Let us calculate, for example, E’x – cF’ 01. The only non-zero terms on the right-hand side are
x'
0
01
x'
1
x'
0
10
x'
1
E
E' cF' c
F
F 01 c 2
2
x
1
E .
(9.133)
x
1
0
0
1
x
x
x
x
x
c
50 To be fair, note that just as Eq. (127), Eq. (129) this is also a set of four scalar equations – in the latter case with the indices , , and taking any three different values of the set {0, 1, 2, 3}.
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Repeating the calculation for the other five components of the fields, we get very important relations E' E ,
B' B ,
x
x
x
x
E' E vB
B'
(9.134)
y
B vE / 2 c
y
