c348 mathematics for general relativity chapters 3 and 4 (ucl)

7/30/2019 C348 Mathematics for General Relativity Chapters 3 and 4 (UCL)

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Chapter 3

Special Relativity

Really this is what is meant by the Fourth Dimension, though some people who talk about the FourthDimension do not know they mean it. It is only another way of looking at Time. There is no differencebetween time and any of the three dimensions of space except that our consciousness moves along it. Butsome foolish people have got hold of the wrong side of that idea. You have all heard what they have tosay about this Fourth Dimension?

3.1 Minkowski Space-time

In modern terms, special relativity is the study of physics in a universe governed by the Minkowski metric,

equation (2.97). The Minkowski metric has coordinates

(X0,X1,X2,X3) = (ct,x,y,z) (3.1)

where t is time and c is the speed of light. Note that the time component, by convention, is distinguishedby being given the index 0. Also, the metric is diagonal,

ab = Diagonal(1,1,1,1), (3.2)

with the time component of opposite sign to the spatial components.We will examine the geometry of the Minkowski metric in detail. First recall that the spatial part of

ab is Cartesian, apart from an overall minus sign. We could, of course, transform to another coordinatesystem such as spherical polars. However, this would introduce an apparent dependence of the metric

elements on position, which would obscure the simplicity and symmetry of the metric. Thus for thischapter we will exclusively use Cartesian spatial coordinates.We first note that the Minkowski metric is independent of position and time. This fact gives it the

property ofDefinition 3.1 Homogeneity An object or physical law is homogeneous if it has the same form at

all places, i.e. (its form is invariant to translations).

A rotation of the spatial (x, y, z) axes leaves the Minkowski metric unchanged. Exercise 3.1

Show that the Minkowski metric is unchanged by a rotation by an angle about the z axis, where

t t; (3.3)x x cos y sin ; (3.4)y

y cos + x sin ; (3.5)

z z. (3.6)

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3.2 Units 35

Thus it has the property of isotropy.

Definition 3.2 Isotropy An object or physical law is isotropic if it has the same in all directions,i.e. its from is invariant to rotations, about any central point and axis.

Objects can be homogeneous but not isotropic.Examples:

A uniform magnetic field. The field looks the same at all points in space, but points in a particulardirection.

A regular crystal. The crystal structure may appear the same at different places, but the molecularbonds are oriented in particular directions

Most fabrics are woven with a warp and a weft, with the result that their ability to stretch dependson direction.

It is not possible to be isotropic but non-homogeneous. For example, consider a dis-tribution of stars. If the distribution is non-isotropic, more stars are seen in some direc-tions than others. But then regions seen indifferent directions (A and B) must be differ-ent and therefore non-homogeneous.

Isotropic Homogeneous . (3.7)Also note that spherical symmetry about some

central point does not imply isotropy about allpoints.

3.2 Units

The zeroth coordinate in Minkowski space-time is X0 = ct. The presence of the factor of c makes manyof the equations more complicated. But we need this factor because traditional units for time and spaceare different. In order to understand space and time in a unified way, we need to employ a system ofunits which treats space and time more equally.

Exercise 3.2 Suppose there were a move to convert the measure of distance on British roads tokilometers. However, this move was fiercely resisted by half of the population. In a political compromise,

it was decided to measure East-West distances with kilometers, and North-South distance with miles.Imagine coping with this mixed system. What would be the distance from London to Manchester?

What would speedometers and odometers look like?

Relativistic UnitsIn conventional units the speed of light in a vacuum is c = 2.997 . . . 108 m s1.In a relativistic system of units c = 1. There are two ways of constructing such systems.

a. Use a basic unit of time; the length unit will be the distance travelled by light in that time.

A) choose the time unit to be the second (s), and define the unit of length to be the light-second ( s)

1 s = 3 108 m. (3.8)In these units, c = 1 s/ s. Usually, we do not bother writing the s/ s, and so c = 1.


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36 Special Relativity

B) Time unit: year (y); length unit: light-year ( y).

1 y 3 108 ms 3.4 107 s (3.9)

1016 m (3.10)

and c = 1 y/ y. Again we will ignore the y/ y and just say c = 1.

b. Use a basic unit of length; the time unit will be the interval of time needed for light to travel thatdistance.

Choose the length unit to be the metre ( m), and the time unit to be the light-metre ( m)

1 m 3 109 s (3.11)

and c = 1 m/ m = 1.

Example 3.1 Express Watts in relativistic units with basic units second, kilogram.

Solution

1W = 1 kg m2 s3 (3.12)

= 1 kg s1m2

s2

l s

3 108 m2

(3.13)

= 19 1016 kg s

1 l ss

2(3.14)

1.1 1017 kg s1 (3.15)

N.B. We could have reached this by multiplying by c or c1 until only the units kg and s were left(cancel out as many factors of c as necessary to get the units right).

In reverse: what is 1kg s1 in Watts?

Solution Multiply by c2 to obtain the right units

1 k g s1 = 9 1016 kg m2 s3 (3.16)= 9 1016W. (3.17)

Exercise 3.3 The gravitational constant is G = 6.67108cm3g1s2. Express G in relativisticunits, with the basic units being grams and centimetres (time measured in light-cms). Next in relativisticunits calculate the escape velocity V from the surface of the earth. Also calculate 1. (Recall that thegravitational potential energy of an object of mass m at the surface of the Earth isGMm/R, whereM is the mass of the Earth and R is its radius.)Earth mass: M = 610

27 g.Earth radius: R = 6.4 103 km.

Exercise 3.4 The acceleration due to gravity at the Earths surface is 1g = 9.8ms2

. Express thisin relativistic units with basic unit being the year (i.e. lengths are measured in light years). (1 year

3.2 107

s.)


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3.3 Einsteins Axioms of Special Relativity 37

3.3 Einsteins Axioms of Special Relativity

The Minkowski metric, as we have seen, is invariant to translations and spatial rotations. However, ina four dimensional space-time manifold, we can also consider rotations involving both space and timecoordinates. Such mixing of space and time coordinates may seem mysterious, but actually the effect issimple: the spatial origin x = y = z = 0 in the rotated system moves at a constant velocity with respectto the original system. This is called boost.

Definition 3.3 Boost A boost is a transformation to a coordinate system moving at a constantrelative velocity with respect to the original system.

Einsteins famous 1905 paper demonstrated that we needed a new conception of space-time if we wereto have a theory of electromagnetism which looked the same in all coordinate systems, especially onesreached via a boost transformation. He started out with the idea of a coordinate system, or referenceframe in which there are no inertial forces such as centrifugal or Coriolis forces.

Definition 3.4 Inertial Reference Frame An inertial reference frame (IRF) is a co-ordinatesystem for space-time with Cartesian spatial co-ordinates, and where there exist no inertial (fictitious)forces.

The invariance of Maxwells equations led Einstein to believe that the speed of light does not changewhen boosting to a moving reference frame. He wrote down two axioms for physics in general, andelectromagnetic radiation in particular:

a. The laws of physics are invariant to translations, rotations and boosts.

b. The speed of light is the same in all IRFs.

3.4 Space-Time Diagrams

Space-time diagrams place time on the vertical axis, with one space dimension on the horizontal axis (ortwo in a horizontal plane).

Definition 3.5 Worldline The worldline of an object is the path it traces in space-time.

If we just look at one space dimension (x), the velocity of the object is dx/ dt. Since the verticalaxis is time, however, the slope of the world line is dt/ dx.

For photons c = dx/ dt = 1. In special relativity light moves at an angle of arctan(1) = /4 on aspace-time diagram.

Choose a point on an objects world-line, P, to be = 0. Then let be the arclength away from P,

=

QP

gab dXa dXb. (3.18)

Definition 3.6 Proper Time The arclength along a worldline is called the proper time.


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x

t

y

Figure 3.1: The object with the worldline to the left is at rest. The object to the right ismoving in a circular orbit.

Q)(

(P)

t

x

P

Q

Definition 3.7 Event A space-time event Pis a point in space-time.Suppose at event Pa camera flash goes off, sending an expanding sphere of light into space (and

time). We can most easily picture this expanding sphere by suppressing one dimension. For example,suppose the z co-ordinate ofPis 0. We can then consider how the light travels in the z = 0 plane. Theexpanding sphere of light intersects this plane as an expanding circle. In a space-time diagram showingthe x, y, and t directions, the expanding circle traces out a cone.

The future light cone at an event Pshows how a pulse of light emitted at Ptravels through space-


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3.4 Space-Time Diagrams 39

time. A light-cone splits space-time into space-like and time-like parts. Objects moving slower than light

(i.e. all massive objects!) can only reach the time-like parts. One would need to move faster than lightto reach the space-like parts.

Figure 3.2: The interval PQ is light-like ( = 0), as Q is on Ps past light cone. The intervalPR is space-like ( < 0), while PSis time-like ( > 0).

A particle with 3-velocityV = ( dx/ dt, dy/ dt, dz/ dt) satisfiesV2 dt2 = dx2 + dy2 + dz2. (3.19)

For a photon,V2 = c = 1, and so dt2 = dx2+ dy2+ dz2 between any two events along a photons

world-line. By Einsteins second axiom, this is true for a photon as seen in any IRF.Suppose we define a small interval between two points d by

d2 dt2 dx2 dy2 dz2. (3.20)

Then

d2

= 0 (3.21)

along a photon path in all IRFs. This suggests that d behaves like a metric line element. In fact, it isprecisely the line element

d2 = gab dXa dXb (3.22)

resulting from the Minkowski metric

gab = ab =

1 0 0 00 1 0 00 0 1 00 0 0 1

. (3.23)

Space-time equipped with ab is called Minkowski space or M4. A world-line is a curve in M4.

Theorem


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Proper time equals clock time in the rest frame of the object.

Proof: In the rest frame R,

dR1 = dR2 = dR3 = 0, (3.24)

so d2 = dR02

= dt2R

. We generally chose to have coordinate time increase in the samedirection as proper time, so we can take the positive square root:

d = dtR

. (3.25)

The tangent vector to the world line is the 4-vector

Ua = dXa

d=

dt/ d

dx/ ddy/ ddz/ d

(3.26)What is the corresponding form Ua?

Ua = abUb (3.27)

=

1 0 0 00 1 0 00 0 1 00 0 0 1

dt/ ddx/ ddy/ d

dz/ d

(3.28)

=

dt

d, dx

d, dy

d, dz

d

. (3.29)

|U|2 = UaUa (3.30)

=

dt

d

2

dx

d

2

dy

d

2

dz

d

2(3.31)

=dt2 dx2 dy2 dz2

d2(3.32)

= ds2

d2= d

d

2= 1 (3.33)

UaUa = 1 in all reference frames (3.34)This is easiest to see in the rest frame, where Ua = (1, 0, 0, 0).

Exercise 3.5 Suppose a spaceship moves at speedV in the Earth frame. What is

dtE/ d where tE is the Earth time and is proper time inside the spaceship? Next sup-pose the spaceship is investigating some scalar function of position f(tE, xE, yE, zE) (e.g.temperature of the interplanetary medium). The ship measures and records f() as it trav-els through space. Find df / d in terms of f

tEand the spatial gradient ( f

xE, fyE

, fzE

).

Can you express your result in 4-vector notation?


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3.5 The Poincare and Lorentz Groups 41

3.5 The Poincare and Lorentz Groups

These are the sets of transforms from one IRF to another i.e. that preserve gab = ab.Thus going from coordinates XA to XB with transform LAB gives

gAab =

1 0 0 00 1 0 00 0 1 00 0 0 1

; gBab =

1 0 0 00 1 0 00 0 1 00 0 0 1

(3.35)

Example 3.2 Rotation about Axes.

xA

xB

yAyB

LAB has the rule

tB = tA (3.36)

xB = cos xA + sin yA (3.37)

yB = sin xA + cos yA (3.38)zB = zA (3.39)

LAB = Ba

Ab=

1 0 0 00 cos sin 00 sin cos 00 0 0 1

(3.40)

Example 3.3 Translations.

Consider the transformation PAB

tB = tA + 3 yB = yA 5xB = xA 2 zB = zA 4 (3.41)

The orientation of the axes does not change: BA

= I4. However, the origin moves theorigin of the B system is at (ta, xa) = (5, 4).

The set ofall ab preserving transformations is called the Poincare group, while thosewhich leave the origin fixed (no translation, just rotation) are the Lorentz group. TheLorentz group is a subgroup of the Poincare group.


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3.5.1 Group Axioms

a. Closure:

XAPAB XB

PBC XC (3.42)

= XAPAC XC. (3.43)

Let PAB and PBC be any elements of the Poincare group. Then: PAC =PBCPAB (composition of the two transformations) preserves ab if both PAB andPBC do.

b. Identity:If XA = XB, then

PAB =

1 0 0 00 1 0 00 0 1 00 0 0 1

. (3.44)

c. Inverse:

P1AB

= PBA

(3.45)

d. Associative:

(PABPBC) PCD = PAB (PBCPCD) (3.46)

Theorem:For any Lorentz transform LAB

|det(LAB)| = 1 (3.47)Proof:

Both A and B are inertial frames, so gAab = ab and gBab = ab. Thus

ab =Bc

AaBd

Ab cd (3.48)

= LcaLdb cd. (3.49)

where L = LAB. Now, the determinant of the product of two matrices is the product ofthe determinants, so

det() = det(L)2 det() (3.50)

= det(L)2 = 1. (3.51)

|det(LAB)

|= 1 . (3.52)

Definition 3.8 Proper and Improper Transforms


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3.6 Lorentz Boosts 43

Proper Lorentz transforms have det(L) = 1

Improper Lorentz transforms have det(L) = 1Example 3.4 Mirror Transform

txyz

B

=

1 0 0 00 1 0 00 0 1 00 0 0 1

txyz

A

. (3.53)

This improper transform reflects objects in the x direction.

3.6 Lorentz Boosts

3.6.1 Deriving the transformation matrix

Suppose a spaceship moves, velocity vi w.r.t. Earth:

Ships rest frame: S (3.54)

Earths rest frame: E (3.55)

Assume the origins coincide Sa = 0 is the same event as Ea = 0. There will be manyLorentz transformations which go from the Earth frame to the Ships frame; these will

differ by rotations in space. We can guess, however, that there will be a simple one wherethe y and z coordinates do not change: yS = yE and zS = zE. Thus we will try transformsof the form:

txyz

S

=

? 0 0 00 ? 0 00 0 1 00 0 0 1

txyz

E

(3.56)

=

tx

S

=

tx

E

(3.57)

1) c = 1 in both frames

(t,-t)

(t,t)

t

x

Pho

ton

Photon

In all frames, a photon moving to theright passes through events with

tx

=

tt

. (3.58)

A photon moving to the left, on theother hand, passes through

tx = tt . (3.59)


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Suppose in the Earth frame a photon passes through the eventtx

E

=

11

E

.

In ships co-ordinates, tx

S

=

tt

S

=

11

E

(3.60)

Thus

tS = + (3.61)= + (3.62)

+ = + . (3.63)

A photon could also pass through

tx

E

=

11E

, which has ships co-ordinates

tx

S

=

11E

(3.64)

= ttS (3.65) tS = (3.66)

= (3.67) = (3.68)

Combining these two results gives

+ = + (3.69)

= (3.70) 2 = 2 (3.71)

= , = (3.72)

We now have tx

S

=

tx

E

. (3.73)

2) Follow spatial origin in ships co-ordinates


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3.6 Lorentz Boosts 45

On the ship, txS = t0S. But, Earthlings see this move at speed V

tx

E

=

t

V t

(3.74)

tx

S

=

t0

S

(3.75)

=

t

V t

E

(3.76)

And so,

tS = tE + V tE (3.77)

= (+ V) tE (3.78)

xS = tE + V tE (3.79)

= (+ V) tE (3.80)

= 0 (3.81)

+ V = 0 (3.82) = V (3.83)

Thus, we have tx

S

= 1 VV 1

tx

E

. (3.84)

3) Apply det(L) = 1

det

V

V

= 1 (3.85)

= 2 2V2 (3.86)= 2

1 V2

(3.87)

2 = 11 V2 (3.88)

=1

1 V2 . (3.89)

4) Inverse TransformationFrom the ships frame to the Earths frame

tx

E

=

V

V

tx

S

(3.90)

This is the inverse transform to the one from Earth to ship, as V V.


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3.7 Simultaneity

A surface of simultaneity is a set of points where t =constant in some reference frame.Lets look at the Ships surface of simultaneity in the Earths co-ordinate frame.

The line tS = 0 contains events occur-ring simultaneously in the ships frame.

tE

xEQ

R

P

line t =0

line t =0

E

S

From the inverse transformation,

tE = V xS (3.91)xE = xS (3.92)

tE = V xE . (3.93)This is the line in the Earths frame of reference corresponding to tS = 0.Events P, Q are simultaneous in the Earths frame, but not in the Ships. Events P, R

are simultaneous in the Ships frame, but not in the Earths.

Exercise 3.6 Suppose a spaceship moves at speedV x with respect to the Earth, whereV = 1/2. Let the coordinates in the rest frame of the ship be

Sa = (S0,S1) = (tS, xS), (3.94)

(ignoring the y and z components). Similarly, letEa = (E0,E1) = (tE, xE) be coordinatesin the rest frame of the Earth. Also let (0, 0)S = (0, 0)E. Draw a space-time diagramwhere the horizontal axis gives xE and the vertical axis gives tE. On this diagram drawthe line xS = 0 (the time axis in the ship rest frame) and the line tS = 0 (the space axisin the ship rest frame). What is the angle between these lines?

3.8 Length Contraction

noindentDefinition Length: The length of an object is the spatial distance between theends, measured simultaneously in some reference frame.


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3.8 Length Contraction 47

tRE

V

P Q

R

L

L

S

E

Consider a metre stick at rest on thespaceship; The space travellers mea-sure the position of the ends of thestick simultaneously at P, R. Earth-lings see P, Q as simultaneous eventscorresponding to the ends of the stickat tE = 0.

Thus,

tE xE tS xSP 0 0 0 0Q 0 LE ? ?R ? ? 0 LS

Apply the Lorentz transform to find the ?s.

Q :

tQxQ

=

V

V

0LE

=

V LELE

(3.95)

R :

tRxR

=

V

V

0

LS

=

V LSLS

(3.96)

The right end of the stick moves from Q R with speed V w.r.t. Earth xRE = xQE + V(tRE tQE) (3.97)

= xQE + V tRE (3.98)

If we combine this with the Lorentz transform, we get

LS = LE + V(V LS) (3.99)

= LE = LS( V2) (3.100)= LS(1 V2) (3.101)= LS

2 (3.102)

= LS1 (3.103)

LE = LS1 (3.104)


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and since > 1,

LE < LS . (3.105)

3.9 Relativistic Dynamics

3.9.1 The 4-momentum

The three-momentum of an object is defined as p = mV. In space-time, we use Uinstead of

V. For an object travelling at speed

V, a Lorentz transformation from the

rest frame of the object gives

Ua =

V

. (3.106)

We can check that |U|2 = 1:

UaUa = (,V)

V

= 2(1 V2) = 1. (3.107)

Next, we extend momentum from the three dimensional p to a four-dimensionalobject. We do this by including the energy E. This makes intuitive sense: classical me-chanics conserves E as well as the three components of momentum. Another rational for

combining energy and momentum follows from the symmetries of space-time. Noetherstheorem (see chapter 7) shows that momentum conservation is a direct consequence of thehomogeneity of space (i.e. invariance to spatial translations). But Noether also showedthat energy conservation follows in the same way from the homogeneity of time (invari-ance to time translations). Thus combining space and time into space-time correspondsto combining energy and momentum into one object as well.

As we will see later, when analyzing orbits, the 4-momentum works most naturallyas a form rather than a vector. In other areas of physics momentum also appears asdual or conjugate to vectors, in particular the position vector. Recall, for example, thatin quantum theory momentum appears paired with the position vector x , e.g. in theFourier transform term exp(i

p

x /). In these casesp combines with

x to form a

scalar, just as a form combines with a vector to form a scalar. This is why we will beginwith its definition as a form. We are free to give names to the components of p. In

fact, we will give the symbols E to p0 and p to the other components. Afterwards, wejustify these names, showing that E acts like energy and p acts like the non-relativisticmomentum.

noindentDefinition 4-momentum The 4-momentum is a form p defined by

pa mgabUb = mUa. (3.108)The components of the 4-momentum will be given names:

pa (E,p ). (3.109)
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3.9 Relativistic Dynamics 49

The raised form of the 4-momentum pa

gabpb is simply

pa = mUa. (3.110)

In Special Relativity where gab = ab the raised form of the 4-momentum is

pa = abpb =

Ep

. (3.111)

This implies

pa =

m

mV

(3.112)

so that p = mV. (3.113)The right hand side gives the non-relativistic 3-momentum, apart from the factor of (which is very nearly 1 in non-relativistic situations). Thus we are justified in our choiceof the symbol p for the spatial components of the 4-momentum.

The 0th component of p resembles the non-relativistic energy, plus just a bit extra:

E = p0 = m (3.114)

= m

1 V2

1/2

(3.115)

= m1 + 12V2 + O(V4) (3.116) m + 1

2mV2 + O(V4). (3.117)

We interpret this as the rest mass (m) + kinetic energy (mV2/2) + relativistic correction(O(V4)).

In the rest frame = 1 and we have Einsteins famous equation

E = m . (3.118)

(or in non-relativistic units E = mc2).

Exercise 3.7

a. Show that if the particle has three-velocityV, p = EV.

b. Show that E2 = |p |2 + m2.

3.9.2 Forces

Newton:Classically: F = maIn Special Relativity, this becomes

fa = maa (3.119)


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where

aa dUad

= d2Xad2

. (3.120)

Theorem:aaUa = 0 (3.121)

i.e. the 4-acceleration is perpendicular to the 4-velocity.Proof:

aU = dUd

U (3.122)

=

1

2

d U Ud (3.123)=

1

2

d

d(1) (3.124)

= 0. (3.125)

Corollary Since the force f = ma, with m scalar, we also have

f U = 0 . (3.126)

3.9.3 Energy-Momentum Conservation

Consider 2 particles colliding:

p1

p

2

p3

p4

Total 4-momentum

before pa1 + pa2 (3.127)

after pa3 + pa4 (3.128)

Conservation of Energy and Momen-tum:

= pa1 + pa2 = pa3 + pa4 (3.129)

3.9.4 Photons

The four-velocity becomes ill-defined when V 1:

Ua

= V V . (3.130)


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3.9 Relativistic Dynamics 51

Fortunately, the 4-momentum still makes sense. Let m

0 while V

1, keeping

E = m constant:

pa = (m,mV) = (E,EV). (3.131)Note that |p|2 = m2.

The 3-vectorV becomes a unit vector as its modulus V 1. Let k = limV1V.

Then pa = (E,Ek). Here k tells us the direction of travel of the photon.In quantum theory, E = for a photon, where is the angular frequency of the

light. This implies

pa = (,k ) (3.132)

where k = k. We will write ka = (,k ) for the wave-number= p = k . (3.133)

N.B. The world-line of a photon cannot be parameterized by (proper time), sinceproper time does not exist for a photon (d = 0 along the path of a photon). We can stilluse other parameters, for example the coordinate time t in some reference frame.


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Chapter 4

Maxwells Equations in Tensor Form

Well, I do not mind telling you I have been at work upon this geometry of Four Dimensions for some

time. Some of my results are curious. For instance, here is a portrait of a man at eight years old, another

at fifteen, another at seventeen, another at twenty-three, and so on. All these are evidently sections, as it

were, Three-Dimensional representations of his Four-Dimensioned being, which is a fixed and unalterable

thing.

4.1 Maxwells Equations Review

We will use units where 0 = 0 = c = 1 (Vacuum Equations)

4.1.1 Internal Structure Equations

The internal structure equations involve the fields only; matter terms involving chargesand currents do not appear.

B = 0, (4.1)E + tB = 0. (4.2)

Equation (4.1) implies there are no magnetic monopoles - lines of magnetic flux haveno endpoints. The meaning of Equation (4.2) can be seen by integrating over a surface Sbounded by a curve C:

52
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4.1 Maxwells Equations Review 53

n

C

S

S

E

n d2x =

S

tB n d2x (4.3)

= ddt

S

B n d2x (4.4)

= ddt

[magnetic flux through S] (4.5)

but, by Stokes theoremS

E

n d2x =

C

E dl (4.6)

= [electric power round C] (4.7)

Thus, changes in the magnetic flux produce electric power (and vice-versa).

4.1.2 Source Equations

E = c, (4.8)

B tE = J . (4.9)Equation (4.8) implies that electric field lines start and stop at electric charges.

For non-relativistic applications, tE is

small, and equation (4.9) gives

B J (4.10)i.e. Magnetic field lines circle currents

J

B

Maxwells equations give us 4 vector equations, but 8 component equations.
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54 Maxwells Equations in Tensor Form

4.1.3 Lorentz Force

The source equations tell us how matter generates fields. We need a supplemental equationto see how fields affect matter - the Lorentz Force equation. For a particle of charge q

F = q

E +

V B

. (4.11)

4.1.4 Charge Conservation

Charge conservation is expressed by the equation

tc + J = 0. (4.12)If we integrate this over a volume V, bounded by the surface S, containing charge Q;

V

tc d3x =

V

J d3x (4.13)

= ddt

V

c d3x =

S

J n d2x (By the Divergence Theorem) (4.14)

=

d

dt

(Q) =

[flow of charge out of V] (4.15)

J

S

Q

V

A physical system of fields and matter can be represented as follows:


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4.2 The Faraday Tensor 55

4.2 The Faraday Tensor

We define the Faraday Tensor as the antisymmetric tensor

Fab =

b

0 Ex Ey EzEx 0 Bz ByEy Bz 0 BxEz By Bx 0

a. (4.16)

Definition 4.1 two-forms A two-form is an antisymmetric second rank tensor withtwo lower indices.

Thus the Faraday tensor is a two-form. In general, we will define fields as forms (likegradients), and particles (i.e. 4-velocities, currents) as vectors. However, at times we canraise or lower using the metric. e.g. Ua = gabU

b.

Exercise 4.1

a. Find the raised version of Fab, i.e. findFcd = cedfFef. Be careful if you use matrix

multiplication!!

b. Next find the dual Faraday tensor

Fab

1/2 abcdFcd. (4.17)

Answer:

Fab =

0 +Bx +By +BzBx 0 Ez EyBy Ez 0 ExBz Ey Ex 0

(4.18)

4.3 Internal Structure Equations

aFbc + bFca + cFab = 0 (4.19)True for any a,b,c = 0, 1, 2, 3.

Example 4.1 a = 1, b = 2, c = 3

1F23 + 2F31 + 3F12 = 0

= x

Bx +

yBy +

zBz = 0

B = 0which is the first Maxwell equation.

Note: There are 64 combinations of a,b,c, but most are useless!


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Example 4.2 a = 1, b = 2, c = 2

1F22 + 2F21 + 2F12 = 0

= x

(0) +

y(Bz) +

y(Bz) = 0

= 0 = 0Which is true automatically, and tells us nothing.

Only 4 choices of a,b,c are useful those where all three are different.

Exercise 4.2 Consider the equation

bFab

= 0. (4.20)

Find the four equations for E and B generated by letting a = 0, a = 1, a = 2, and a = 3.Show that these are just the Internal Maxwell equations

B = 0, E + Bt

= 0. (4.21)

4.4 Source Equations

bFab = ja . (4.22)

where: j0 = c, (j1, j2, j3) =

J , and

Fab = acbdFcd, (Special Relativity); (4.23)

Fab = gacgbdFcd, (General Relativity). (4.24)

Special Relativity:

Fab = 0 Ex Ey Ez

Ex 0 Bz ByEy Bz 0 BxEz By Bx 0

. (4.25)There are 4 equations, for a = 0, 1, 2, 3E.g. a = 0:

bF0b = j0

0F00 + 1F

01 + 2F02 + 3F

03 = j0

t(0) + xEx + yEy + zEz = c

E = c


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4.5 Charge Conservation 57

4.5 Charge Conservation

ct

+ J = 0 (4.26) 0j0 +

1j

1 + 2j2 + 3j

3

= 0 (4.27)

oraj

a = 0 . (4.28)

This equation follows immediately from the Source equation:

aja = abF

ab = 0 (4.29)

as ab is antisymmetric while Fab

is symmetric.The 4-divergenceIn general, when the 4-divergence, aV

a, of a vector field vanishes, we say that Va isconserved.

The 4-current is ja =

cJ

. The time component j0 = c gives the amount of charge

moving in the time direction per unit (space) volume. The components ofJ give the

amount of charge moving in each space direction per unit time.

Exercise 4.3 Let inertial frame B move at speed Vx with respect to inertial frameA. Suppose in frame A the magnetic field components vanish. Using the Faraday Tensor,

find the magnetic field components and electric field components in frame B.

4.6 Lorentz Force

fa = qUbFba . (4.30)

Example 4.3 What is the a = 1 component of the force? Solution

f1 = qUbFb1

= f1 = q

F01 VxF11 VyF21 VzF31

(Ua = (,V))

= q(Ex

Vx(0)

Vy(

Bz)

VzBy)

= q(Ex + (VyBz VzBy))= q

Ex + (

V B )x

and so

f1, f2, f3

= q

E + B

Checking that fU = 0

fU = qUF

U= qFUU


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but F is anti-symmetric, while UU is symmetric; the double contraction therefore

gives 0.Exercise 4.4 Express the Lorentz scalars

FabFab,FabFab (4.31)

in terms of E and B. Suppose that E vanishes in some inertial frame. Show that E mustbe perpendicular to B in all frames. Is it possible for B = 0, E = 0 in one frame andE = 0, B = 0 in another?

4.7 Potential Form

Definition 4.2 Electromagnetic potential The electromagnetic potential is given by theform

= (e,A) (4.32)where e is the static electric potential and

A is the magnetic vector potential.

The Faraday tensor involves antisymmetric derivatives of :

Fab = ba ab . (4.33)This definition is consistent with the assignments

E = e tA, (4.34)B = A. (4.35)

4.7.1 Advantage Internal Structure Equations

These are automatically satisfied:

aFbc + bFca + cFab = ?

Using the electromagnetic potential,

= a (cb bc) + b (ac ca) + c (ba ab) = 0just by cancellations.

4.7.2 Advantage Source Equations

The equationbF

ab = ja (4.36)

becomesb

ba ab = ja (4.37)

where a = gabb.


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4.8 Gauge Transformations 59

We can write this as

2a a bb = jawhere:

2 dAlembertian

= bb

=2

t2

2

x2

2

y2

2

z2

=2

t22

Thus, Maxwells equations reduce to a single source equation, with the internal equa-tions being automatic and no longer needed.

4.8 Gauge Transformations

Recall thatB = A. Suppose we apply the gauge transformationA = A + for

some function . Then

B = A (4.38)

= A + (4.39)= A + 0 (4.40)=

B . (4.41)

Similarly, if = + , then

Fab = b (a + a) a (b + b)= ba ab + (ba ab)= Fab.

The potentials are therefore not unique, and we are free to choose the most convenientpotential, a, to solve the problem

4.9 Lorentz Gauge

Suppose we try a potential, , where

aa = h

for some function h. Then we may apply the gauge transformation

= +

where satisfiesaa = h.


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This equation can be shown to always have a solution, and so we are left with a new

potential which satisfies

aa = 0 . Lorentz Gauge (4.42)

In Lorentz gauge, the source equation becomes

2a = ja . (4.43)

4.10 Light Waves

In the vacuum, ja = 0

= 2a = 0 (4.44)= bba = 0 (4.45)

= a

t2

a

x2

a

y2

a

z2= 0 (4.46)

which is the wave equation, with solution of the form

a = Caeikbxb

(4.47)

= Caei(tk x ). (4.48)

Here k = (,k ) is the wave vector, and Ca is the amplitude.Check:

2a

t2= 2a (4.49)

2a

x2= k2xa (4.50)

etc. . . (4.51)

(4.52)

Substituting these into the wave equation gives

2 +k 2 = 0 (4.53)or

= k . (4.54)

Exercise 4.5 Suppose that magnetic monopoles exist in nature. Then, in addition tothe electric charge-current 4-vector je, there is a magnetic charge-current 4-vector jm =(m, jmx, jmy, jmz) where m is the magnetic charge density and jmx is the current ofmagnetic charge in the x direction. The Maxwell equations become

bFab = jae

bFab = jam.


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4.10 Light Waves 61

a. Consider the second equation bFab = jam. Find the four equations for

E and

B

generated by letting a = 0, a = 1, a = 2, and a = 3.

b. Show that magnetic charge is conserved; i.e. show that

ajam = 0.

c. The Lorentz force on a magnetic monopole of charge qm and 4-velocity Ua is

fa =dpa

d= qmUb

Fab.

Find the four equations generated by letting a = 0, a = 1, a = 2, anda = 3. Express

these in terms of the three-velocityV = dx /dt and = (1 V2

)1/2

.d. Show that the Lorentz force in the previous item is perpendicular to U in the sense

thatf U = 0.

e. Suppose that the Faraday tensor Fab can be written in the form

Fab = ab bafor some four-potential . Show that the magnetic current 4-vector must vanish, i.e.

jm = 0.

c348 mathematics for general relativity chapters 3 and 4 (ucl)

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