Download VectorCalcTheorems

PHY2061 Enriched Physics 2 Lecture Notes
Gauss’ and Stokes Theorem
Vector Calculus Theorems
Disclaimer: These lecture notes are not meant to replace the course textbook. The
content may be incomplete. Some topics may be unclear. These notes are only meant to
be a study aid and a supplement to your own notes. Please report any inaccuracies to the
professor.
Gauss’ Theorem (Divergence Theorem)
Consider a surface S with volume V. If we divide it in half into two volumes V1 and V2
with surface areas S1 and S2, we can write:


S
E  dA 

S1
E  dA 

S2
E  dA
S1
V1
S2
D
V2
since the electric flux through the boundary D between the two volumes is equal and
opposite (flux out of V1 goes into V2).
Now let’s continue this process of dividing the original volume into a great number of
infinitesimal volumes, each cubic in shape:


S
E  dA   i  E  dA
Si
Consider now one of these small cubic volumes. Consider one corner of this cube at
position  x0 , y0 , z0  . The length of each side is x, y, z , and each face is perpendicular
to one of the coordinate axes.
D. Acosta
Page 1
6/19/2017
PHY2061 Enriched Physics 2 Lecture Notes
Gauss’ and Stokes Theorem
y
y
 x0 , y0 , z0 
z
x
x
E
z
We are interested in computing the flux passing through this small volume. The flux
through the top and bottom faces will only depend on E y since E  dA  0 for the other 4
faces. Since the cube is infinitesimal, we can do a Taylor expansion of the field about
 x0 , y0 , z0  and find the y component of the field at the center of the bottom face
x
z 

, y, z 
x
:
2
2 

 x  E
Eybottom  Ey  x0 , y0 , z0     y
 2  x
x0 , y0 , z0
 z  E
  y
 2  z
x0 , y0 , z0

x
z 

, y  y, z 
Similarly for the top face  x 
 we have:
2
2 

E
 x  E
 z  E
Eytop  Ey  x0 , y0 , z0     y x0 , y0 , z0    y x0 , y0 , z0  y y
y
 2  x
 2  z
x0 , y0 , z0
So the net flux between top and bottom is:
top-bottom  xzEytop  xzEybottom
The negative sign arises because the electric field points into one surface (chosen to be
the bottom) and out of the other (top). Thus,
D. Acosta
Page 2
6/19/2017
PHY2061 Enriched Physics 2 Lecture Notes
 top-bottom
 top-bottom
Gauss’ and Stokes Theorem
E y 

 x  E y  z  E y
 
 y
 E y  x0 , y0 , z0    

y 
 2  x  2  z
 x  z 


 x  E y  z  E y
 
  E y  x0 , y0 , z0    

 2  x  2  z


E
 x y z y
y
We can apply the same procedure to the other two pairs of sides:
Ex
x
Ez
 xyz
z
 front-back  xyz
 left-right
So the total flux passing through this infinitesimal volume is:
 E E y Ez 
E  dA  xyz  x 


Si
y
z 
 x
i  dV   E
i 

Here we have introduced the volume element dV  xyz , and the divergence
operator:
E E E
  E  div E  x  y  z
x
y
z
Where the gradient operator is defined by:
  xˆ



 yˆ
 zˆ
x
y
z
Now we can sum the contributions from all infinitesimal volumes comprising the full
volume V:
   i i   i  E  dA
Si
  i dVi    E  x , y , z   dV   E
i
i
i
  E  dA   dV  E
S
V
This forms Gauss’ Theorem, or the Divergence Theorem. It states that the surface
integral of E  dA can be related to the volume integral of  E .
D. Acosta
Page 3
6/19/2017
PHY2061 Enriched Physics 2 Lecture Notes
Gauss’ and Stokes Theorem
Differential form of Gauss’ Law
So how does Gauss’ Theorem relate to what we have learned in electromagnetism?
Consider Gauss’ Law, which relates the electric flux through a closed surface to the net
enclosed charge:


S
E  dA 
qenc
0
We can relate the enclosed charge to the volume integral of the differential charge density
per unit volume, , in the enclosed volume:
qenc    dV
V
And by Gauss’ Theorem, we can relate the surface integral of E  dA to the volume
integral of  E , giving us:

V
dV   E 
1
0

V
 dV
Now there is nothing special about what volume V we choose, so the above expression
must hold for all volumes. That means that the integrands must be equal:
E 

0
This is the differential form of Gauss’ Law. It holds for every point in space. When
combined with further differential laws of electromagnetism (see next section), we can
derive a differential equation for electromagnetic waves.
For example, consider a constant electric field: E  E0 xˆ . It is easy to see that the
divergence of E will be zero, so the charge density =0 everywhere. Thus, the total
enclosed charge in any volume is zero, and by the integral form of Gauss’ Law the total
flux through the surface of that volume must be zero.
On the other hand, if E  E0 xxˆ  E ' yˆ , then   E 
Ex
 E0     0 E0
x
Note: If one integrates the obtained charge density within an certain enclosed volume,
you will get exactly the same amount of enclosed charge as if you used the integral form
of Gauss’ Law (try it, it really works!)
D. Acosta
Page 4
6/19/2017
PHY2061 Enriched Physics 2 Lecture Notes
Gauss’ and Stokes Theorem
Stokes Theorem
Consider the line integral of a vector function around a closed curve C:
   F  ds
C
A
C
ds
F
This integral is called the “circulation”. We use the right-hand rule to define the direction
of the area vector (perpendicular to the surface) with respect to the integration direction
(counter-clockwise in this case).
Now suppose we subdivide this surface into two regions, and calculate the circulation of
each around closed curves C1 and C2:
A
C1
C2
B
1 

C1
F  ds
2 

C2
F  ds
It should be clear that in both cases, line segment AB is traversed in opposite directions,
so the contribution to 1 is equal and opposite to 2.
Thus:   1   2
We can continue subdividing the surface into N subregions and we’ll get:
  i1 i
N
Let’s let these regions become infinitesimal is size, and calculate the circulation for just
one infinitesimal area aligned in the x-y plane.
D. Acosta
Page 5
6/19/2017
PHY2061 Enriched Physics 2 Lecture Notes
Gauss’ and Stokes Theorem
 x0  x, y0  y 
y
 x0 , y0 
dA
x
F
z
To find the circulation for this infinitesimal contour, let’s Taylor expand F about the
bottom left corner at  x0 , y0  to find the value at the center of each of the 4 line
segments:
x Fx
Fxbottom  Fx  x0 , y0  

2 x x0 , y0
Fxtop  Fx  x0 , y0  
Fyleft  Fy  x0 , y0  
x Fx
2 x
 y
x0 , y0
y Fy
2 y
Fyright  Fy  x0 , y0  
y Fy
2 y
Fx
y

x0 , y0

x0 , y0
 x
x0 , y0
Fy
x

x0 , y0
Now the circulation of this infinitesimal loop is:
  Fbot  s  Fleft  s  Ftop  s  F right  s
  Fxbot x  Fyright y  Fxtop x  Fyleft y
(taking into account the dot products and integration direct
where we take into account the integration direction with the dot products. When we plug
in our Taylor expansion of F, we get:
D. Acosta
Page 6
6/19/2017
PHY2061 Enriched Physics 2 Lecture Notes

x Fx
  x  Fx  x0 , y0  
2 x

Gauss’ and Stokes Theorem


y Fy
  y  Fy  x0 , y0  
2 y


x0 , y0 

F
x Fx
x  Fx  x0 , y0  
 y x
2 x x0 , y0
y

 Fy Fx 
   xy 



x
y 

 x
x0 , y0


y Fy
  y  Fy  x0 , y0  
2 y

x0 , y0 



x x , y 
0 0 
Fy


x0 , y0 

If we define the infinitesimal area vector in the +z direction based on the RH rule:
 F F 
 z  Az  y  x 
y 
 x
Now suppose we did this for a loop in the y-z plane instead. Then xy and yz in the
above formula and we get:
 F F 
 x  Ax  z  y 
z 
 y
and if we did this for the z-x plane we get:
 F F 
 y  Ay  x  z 
x 
 z
So summarizing this with vector notation:
  A     F 
where we have introduced a curl operator:
 F F 
 F F 
 F F 
  F  curl  F    z  y  xˆ   z  x  yˆ   y  x  zˆ
z 
z 
y 
 x
 y
 x
Thus our infinitesimal line integral for the circulation is:
i 

Ci
F  ds  A i     F 
Now, coming back to the total circulation which is the sum of all the individual
contributions:
  i 1 i  i 1 Ai   F 
N
N
which when we take the infinitesimal limit gives us:
D. Acosta
Page 7
6/19/2017
PHY2061 Enriched Physics 2 Lecture Notes


C
Gauss’ and Stokes Theorem
F  ds   dA   F 
S
This is Stoke’s Theorem. It transforms a closed line integral of a vector function into a
surface integral of the curl of that function.
Why is this useful? Well…
Differential form of Faraday’s and Ampere’s Laws
Recall the integral form of Faraday’s Law of Induction:
 

C
dB
dt
E  ds  

B
B  dA   
 dA

S
S
t
t
Now let’s use the newly derived Stoke’s Theorem to transform the left side of the
equation involving the electric field:

C
E  ds      E   dA
S
B
 dA
t
Now both and left sides of Faraday’s Law involves a surface integral over surface S. This
should be true for any arbitrary surface S, so the integrands must be equal:
     E   dA   
S
E  
S
B
t
This is the differential form of Faraday’s Law! It holds for every point in space.
Now recall the integral form of Ampere’s Law:

C
B  ds  0ienc  0  j  dA
S
where we have introduced the current density j, whose integral across the surface S gives
us the total current passing through it
Now Maxwell noted that to complete the symmetry between magnetic and electric fields,
there should be an additional term added to Ampere’s Law equivalent to Faraday’s Law
where a changing electric field induces a magnetic field (rather than vice versa):
D. Acosta
Page 8
6/19/2017
PHY2061 Enriched Physics 2 Lecture Notes
Gauss’ and Stokes Theorem
dE
 0ienc
C
dt
E
C B  ds  0 0 S t  dA  0 S j  dA

B  ds   0  0
Now by Stoke’s Theorem we can re-write this as:

C
B  ds      B   dA
S
E
 dA  0  j  dA
S t
S
     B   dA  0 0 
S
Which must be true for all surfaces:
  B   0 0
E
 0 j
t
This is the differential form of Ampere’s + Maxwell’s Laws.
D. Acosta
Page 9
6/19/2017