Lecture 3
8.251 Spring 2007
Lecture 3 - Topics
• Relativistic electrodynamics.
• Gauss’ law
• Gravitation and Planck’s length
Reading: Zwiebach, Sections: 3.1 - 3.6
Electromagnetism and Relativity
Maxwell’s Equations
Source-Free Equations:
� =−
�×E
�
1 ∂B
c ∂t
(1)
� =0
�·B
(2)
With Sources (Charge, Current):
� =ρ
�·E
(3)
�
� − 1 ∂E = 1 J�
�×B
c
c ∂t
(4)
Notes:
1. E and B have same units.
3. ρ is charge density [charge/volume]. Here no
get messy in higher dimensions.
4. J� is current density [current/area]
0
or 4π - those constants would
�, B
� are dynamical variables.
E
�
�
d�
p
1
�
�
= q E + �v × B
dt
c
1
Lecture 3
8.251 Spring 2007
Solve the source free equations
� = 0 solved by B
� =�×A
� . (Used to have � × E
� = 0, E = −�Φ)
�·B
True equation:
� ��
1 ∂
1
∂A
�
�
�
�×E+
(� × A) = � × E + � ×
c ∂t
c
∂t
�
�
�
� + 1 ∂A = 0
=�× E
c ∂t
So:
�
� + 1 ∂A = −�Φ
E
c ∂t
(Φ scalar)
Thus:
�
� = −�Φ − 1 ∂A
E
c ∂t
� B
�
� ) encoded as (Φ, A)
(E,
Φ, A are the fundamental quantities we’ll use
Gauge Transformations
�→A
�� = A
�+�
A
� � = � × A� = � × (A + �) = B
�
B
function of �x,t. � function = vector.
1 ∂
Φ → Φ� = Φ −
c ∂t
�
�
� � = −�(Φ� ) = −� Φ − 1 ∂ − 1 ∂ (A + �) = E
�
E
c ∂t
c ∂t
� and B
� fields unchanged!
So under gauge transformations, E
↔
� g.t. (Φ� , A
��)
(Φ, A)
(Physically equivalent)
� ’s and B
� ’s. Not guaranteed to be
Suppose 2 sets of potentials give the same E
gauge-related.
� ) then Aµ = (−Φ, A
�)
Suppose we have 4-vector Aµ = (Φ, A
2
Lecture 3
8.251 Spring 2007
Take ∂x∂µ . Have indices from ∂x∂µ and from Aµ so will get a 4x4 matrix. Have
two important quantities (E and B) with 3 components each ⇒ 6 important
quantities. Hint that we should get a symmetric matrix.
Fµν =
∂Aν
∂Aµ
−
= ∂µ Aν − ∂ν Aµ
∂xµ
∂xν
Fµν = −Fνµ
1 ∂Ai
∂
−
(−Φ) = −Ei
c ∂t
∂xi
F12 = ∂x Ay − ∂y Ax = Bz
Foi =
⎛
Fµν
0
⎜ Ex
⎜
=⎝
Ey
Ez
−Ex
0
−Bz
By
−Ey
Bz
0
−Bx
⎞
−Ez
−By ⎟
⎟
Bx ⎠
0
What happens under gauge transformation?
Aµ → A�µ = Aµ + ∂µ
Then get:
�
Fµν
= ∂µ Aν� − ∂ν A�µ
= ∂µ (Aν + ∂ν ) − ∂ν (Aµ + ∂µ )
= Fµν + ∂µ ∂ν − ∂ν ∂µ
= Fµν
Define:
Tλµν = ∂x Fµν + ∂µ Fνλ + ∂ν Fλµ
Note indices are cyclic.
Some interesting symmetries:
Tλµν = −Tµλν
Tλµν = −Tλνµ
So Tλµν is totally antisymmetric. A totally symmetric object in 4D has only 4
nontrivial components so Tλµν = 0 gives you 4 equations.
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Lecture 3
8.251 Spring 2007
Tλµν = 0 = ∂λ (−∂ν Aµ ) + ∂µ (∂ν Aλ ) + ∂ν (∂λ Aµ − ∂µ Aλ )
Charge Q is a Lorentz invar. Not everything that is conserved is a Lorentz
invar. eg. energy. Since Q is both conserved and a Lorentz invar, (cρ, J�) form
a 4-vector J µ
Now let’s do what a typical theoretical physicist does for a living: guess the
equation!
F µν ≈ J µ
No, derivatives not right.
∂F µν /∂xν ≈ J µ
F µν /∂xµ =
1 µ
J
c
No, constants not right.
Correct, amazingly! (even sign)
µ = 0:
∂F 0ν /∂xν = ρ
∂F 0i /∂xi = ρ
F0i = −Ei
F 0i = Ei
� = ρ verified!
So � · E
Electromagnetism in a nutshell:
Fµν = ∂µ Aν − ∂ν Aµ
∂F µν
J µ
=
ν
∂x
c
Consider electromagnetism in 2D xy plane. Get rid of Ez component:
⎛
⎞
0 −Ex −Ey
0
Bz ⎠
Fµν = ⎝ Ex
Ey −Bz
0
But what about Bz ? Doesn’t push particle out of the plane (v × Bz with v in
the xy plane remains in xy plane) but rename Bz as B, a scalar.
4
Lecture 3
8.251 Spring 2007
How about in 4D spatial dimensions?
⎛
−Ex −Ey
⎜
0
∗
⎜
F =⎜
0
⎜
⎝
−Ez
∗
∗
0
−EN
∗
∗
∗
0
⎞
⎟
⎟
⎟
⎟
⎠
So get tensor B!
It’s a coincidence that in our 3D spacial world E and B are both vectors.
� = ρ in all dimensions.
Let’s look at � · E
Notation: Circle S � is a 1D manifold, the boundary of a ball B 2
Sphere S 2 (R) : x21 + x22 + x23 = R2
Ball B 3 (R) : x21 + x22 + x23 ≤ R2
5
Lecture 3
8.251 Spring 2007
When talking about S 2 (R), call it S 2 (R = 1 implied)
Vol(S 1 ) = 2π
Vol(S 2 ) = 4π
Vol(S 3 ) = 2π 2
d
Vol(S d−1 ) =
2π 2
Γ( d2 )
All you need to know about the Gamma function:
√
Γ(1/2) =
π
Γ(1) = 1
Γ(x + 1) = xΓ(x)
Γ(n) = (n − 1)! for n ∈ Z
� ∞
Γ(x) =
dte−t tx−1 for x > 0
0
� in d = 3 and general d dimensions.
Calculating � · E
d = 3:
�
�
� · Ed(vol)
=
B 3 (r)
�
ρd(vol) = q
B 3 (r)
� through S 2 (r)
This represents the flux of E
E(r) · vol(S 2 (r)) = q
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Lecture 3
8.251 Spring 2007
E(r) · 4πr2 = q
E(r) =
1 q
4π r2
This falls off much faster at large r and increases much faster as small r.
General d:
�
�
� · Ed(vol)
=
B d (r)
�
ρd(vol) = q
B d (r)
� through S d−1 (r)
This represents the flux of E
E(r) · vol(S d−1 (r)) = q
E(r) =
Γ(d/2) q
2π d/2 rd−1
Electric field of a point charge in d dimensions.
If there are extra dimensions, then would see larger E at very small distances.
7