text stringlengths 16 3.88k | source stringlengths 60 201 |
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4.4. Suppose Ct is a general pencil of smooth genus g plane curves
acquiring an ordinary cusp (a singularity whose local equation is given by y2 = x3).
Describe the stable limit of this family of curves.
Exercise 4.5. Read and do the exercises in Chapter 3 Section C of [HM].
5. Deligne-Mumford Stacks
In this secti... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
ζ), then there is a
unique morphism π : C ≤ � C ≤≤ such that ω ∩ π = ζ and p(π ) = f .
Example 5.2. Recall that a Deligne-Mumford stable curve (or simply a stable
curve) of genus g → 2 over a scheme S is a proper, flat family β : C � S whose
geometric fibers are reduced, connected, one dimensional schemes Cs satisfyi... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
and � is an element of the
set F (X). A morphism between (X, �) and (Y, λ) is a morphism f : X � Y such
that F (f )(λ) = �. In particular, this construction allows us to view schemes as
groupoids. To a scheme X we can associate its functor of points Hom(�, X). Since
this is a contravariant functor from schemes to s... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
� � B, the set of isomorphisms in T (B≤) between f �(X) and
f �(Y ).
In the case of Deligne-Mumford stable curves, given any two stable curves C and
C ≤ , IsomX (C, C ≤) associates to any morphism f : Y � X the set of isomorphisms
between f �(C) and f �(C ≤). Recall that C and C ≤ are both canonically polarized
by... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
/G] defined in Example 5.4 is a stack. Let e, e≤ be
two objects in [X/G](Y ) corresponding to two principal G-bundles E, E ≤ � Y with
G-equivariant maps f, f ≤ to X, respectively. IsomY (e, e≤) is empty unless E = E≤
and f = f ≤ . In the latter case the isomorphisms correspond to the subgroup of
G that stabilizes th... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
each
scheme as a stack. In the litterature stacks that arise this way are usually referred
to as schemes meaning that the stack associated to the scheme. We will also indulge
in this habit.
A morphism of stacks f : T � T ≤ is representable if for any map of a scheme
X � T ≤ the fiber product T ×T X is represented... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
surjective
S-morphism U � F .
15
The F is a Deligne-Mumford stack.
A consequence of this theorem is that if X/S is a Noetherian scheme of finite
type and G/S is a smooth group scheme acting on X with with finite and reduced
stabilizers, then [X/G] is a Deligne-Mumford stack. The conditions on the stabiliz
ers (th... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
of g1 and g2.
Theorem 5.11 (The valuative criterion of properness). If f is separated, then f
is proper if and only if, for any discrete valuation ring R with field of fractions
K and any map Spec R � T which lifts over Spec K to a map to T , there is a
finite extension K ≤ of K such that the lift extends to all of S... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
to
directly take the G.I.T. quotient of the Hilbert scheme parameterizing n-canonically
embedded stable curves. The advantage of the first approach is that it does away
16
�
�
�
with delicate calculations describing the stable and semi-stable loci of this action.
The first approach may also be used to construct mod... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
geometrically reductive if for every
linear action of G on kn and every non-zero invariant point v ⊗ kn, there exists an
invariant homogeneous polynomial that does not vanish on v. The group is called
linearly reductive if the homogeneous polynomial may be taken to have degree
one. Finally a group is called reducti... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
in the
subspace spanned by the Hi. Pick a basis for this subspace h1, . . . , hn. We obtain
a rational representation of G on this subspace, hence a linear action on kn making
the morphism β : X � kn given by β(x) = (h1(x), . . . , hn(x)) into a G-morphism.
Since G is geometrically reductive there is an invariant p... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
ductive group acting rationally
on a finitely generated k-algebra R. Then the ring of invariants RG is finitely
generated.
In view of these theorems A(X)G is finitely generated. Hence we can let Y =
Spec A(X)G . The inclusion of A(X)G � A(X) induces a morphism ζ : X � Y .
�
The claimed properties are easy to check f... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
If not, we can replace it by
a minimal G-invariant subspace, which is finite-dimensional by the argument in
Lemma 6.1.) We thus obtain a linear G action on the subspace spanned by fi by
setting
Let S = k[X1, . . . , Xn]. There is an action of G on S by setting
f g =
i
�
j
�i,j (g)fj .
X g =
i
�
j
�i,j (g)Xj .... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
for
confusion between hg and (h)t . The first denotes the g-translate of h, the second
denotes the t-th power of h. To distinguish between these two, we will put paren
theses around h in the latter case.] Since J is invariant, hg − h is in J for every g.
We conclude that M ≤ J has codimension 1 in M . We can write e... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
finite RG/(J ≤ RG)-module.
Choose a non-zero homogeneous element f of RG of degree at least one. If
f is not a zero-divisor, f R ≤ RG = f RG . Since RG/f RG is finitely generated,
(RG/f RG)+ is finitely generated as an ideal. Hence RG is finitely generated as an
+
ideal in RG . Hence RG is a finitely generated k-algebr... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
matrices Mn by conjugation. The space of matrices is isomorphic to affine
space An . Hence, the coordinate ring is k[ai,j ], 1 ∼ i, j ∼ n. Any conjugacy class
19
2
has a representative in Jordan canonical form which is unique upto a permutation
of the Jordan blocks. Since the set of eigenvalues of a matrix is invar... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
on X.
Definition 6.9. A point x ⊗ X is called semi-stable if there exists an invariant
homogeneous polynomial that does not vanish on x. A point x ⊗ X is called stable
if there exists an invariant polynomial f that does not vanish on x, the action of
G on Xf is closed and the dimension of the orbit of x is equal to ... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
is important to determine the stable and semi-stable
loci for reductive group actions on projective varieties. Unfortunately, this in gen
eral is a very difficult problem. There is one instance where stability and semi-
stability is easy to determine.
Definition 6.12. A one-parameter subgroup is a homomorphism � : Gm �... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
1 given by sending g to gˆ
x is a lift of x is not proper. By the
g ⊗ SL(m, K) such that ¯x ⊗ Rn+1 ,
valuative criterion of properness, there exists ¯
but ¯ ⊗ SL(m, R). We can, however, clear denominators so that T rg¯ ⊗ SL(m, R)
for some r. The ring R is a P.I.D., hence we can decompose ¯ = g¯1dg¯2 where g1 and
g2 a... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
as a basis of K n+1 . Then
g −1dg2ei = T ri ei. In particular,
2
−1 g¯ = g
−1 g¯1dg¯2 = (g −1dg2)g
xi �
g¯1
−1
g
2
Therefore, the i-th component of g −1 g¯1
−1
g
2
in R, we conclude that ri → 0.
−1
2
g¯1
2
¯ x. Consequently, the i-th component of g −1 g2 ˆ
g2 ˆ
2
gˆ
g¯2.
−1 ¯x is T ri times the i-th compon... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
is
stable if and only if it does not have any zeros with multiplicity → d/2. Similarly,
a homogeneous polynomial is semi-stable if and only if it does not have any zeros
with multiplicity > d/2.
d−i
i
j
Example 6.16 (Cubic plane curves). Consider the action of SL(3) on the homo
geneous polynomials of degree 3 in t... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
the stable locus in this case
constructs the j-line.
3
Exercise 6.17. Try to generalize the previous example to the action of SL(3) on
homogeneous polynomials of degree 4, 5, 6, .... In particular, describe what kinds of
singularities are allowed on stable curves of degree 4, 5, 6...
6.2. The construction of M g ... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
(n + 1) on
Pl¨
ucker coordinate { Yj1 , . . . , YjP (m) } where Yji =
is given by
mji ,r
xr
r
�
�
wr mji ,r.
�
i,r
The Hilbert-Mumford criterion for semi-stability then translates to the condition
that for each one parameter subgroup, there should be a non-vanishing Pl¨ucker
coordinate whose weight is non-pos... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
has h1 = 0).
(4) If C ≤ is a complete subcurve of C of arithmetic genus g≤ meeting the rest
of the curve C in k points, then the following estimate holds
degC (OC (1)) −
d
g − 1
k
(gC − 1 + )
2
k
.
2
∼
�
�
�
�
�
�
�
�
Remark 6.20. Observe that if C ≤ is a smooth rational curve meeting the rest of
the cur... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
−g is a
connected curve with semi-stable m-th Hilbert point, then C is potentially stable.
The proof of this theorem is quite lengthy eventhough the strategy is straight
forward. We suppose C has a geometric property that violates potential stability.
Under this assumption we construct a one-parameter subgroup that... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
H ssˆ .
Proof. To show that H ss is closed we need to show that the inclusion H ss � ˆ
H ss
is proper. By the valuative criterion of properness it suffices to check that given
a map from the spectrum of a DVR to ˆH ss whose generic point lies in H ss, the
closed point also lies in H ss . Given such a map consider the ... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
degD (OCR (1)) −
degC (OCR (1)|C )
2
degC
2
(�|C
2
)
2 degD (�|C
2
) ∼
k
.
2
�
Lemma 6.23. Every curve C whose Hilbert point lies in H ss is Deligne-Mumford
stable.
Proof. By the potential stability theorem C is semi-stable. In order to show that
it is stable we need to check that there are no ra... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
scheme ˆH ss by the action of the special linear group is projective,
after a base change we can extend the map to ˆH ss . Since H ss is closed, the image
of the map lies in H ss . Pulling back the universal curve we obtain a semi-stable
reduction of a family of stable curves. By the uniqueness of semi-stable reduct... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
gebrizable and the generic fiber is smooth.
˜
Let [C] ⊗ H ss be a point. Let C be the universal formal deformation of C over
B = Spec k[[t1, . . . , tr]]. Set S be the formal completion of H ss at [C]. By the
universal property of the Hilbert scheme we get a map S � H ss . By the universal
property there exists a un... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
exhibit every smooth curves as a branched cover of P1 . When
the number of branch points is large relative to the degree of the map, using the
combinatorics of the symmetric group one may show that the space of branched
covers of P1 is irreducible. Suppose now that the characteristic of the field k is
positive. Let ... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
contained in a hyperplane.
If the curve is degenerate, then the map H 0(OPn (1)) � H 0(Cred,
OCred (1)) has non-trivial kernel. Use the filtration that assigns weight −1 to sec
tions vanishing on Cred and weight w > 0 to the others so that the average weight is
0. There exists an integer q such that the q-th power o... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
potential stability hold. The final step is to show that the estimate in (4) holds.
This is done by showing that if not the filtration FC is destabilizing.
References
[Ab]
[DM]
[Ed]
[Fan]
[G]
[Gr]
S. S. Abhyankar. Resolution of singularities of arithmetical surfaces. In Arithmetical
Algebraic Geometry (Proc. C... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
emes d’existence en g´
emas de Hilbert. In S´
alg´
221, 249–276. Soc. Math. France, Paris, 1995.
etrie
eminaire Bourbaki, Vol. 6, pages Exp. No.
ebrique. IV. Les sch´
eom´
eor`
[Hab] W. J. Haboush. Reductive groups are geometrically reductive. Ann. of Math. (2)
[HM]
[Ha]
[KM]
[K]
102(1975), 67–83.
J. Harris a... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
algebraic geometry. In Snowbird lectures in
algebraic geometry, volume 388 of Contemp. Math., pages 119–136. Amer. Math. Soc.,
Providence, RI, 2005.
[Li]
[Mum1] D. Mumford. Further pathologies in algebraic geometry. Amer. J. Math. 84(1962), 642–
648.
[Mum2] D. Mumford. Lectures on curves on an algebraic surface. ... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-intersection-theory-on-moduli-spaces-spring-2006/06c4e979c3dadcba7d0f3dd69efdb315_const.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
6.641 Electromagnetic Fields, Forces, and Motion, Spring 2005
Please use the following citation format:
Markus Zahn, 6.641 Electromagnetic Fields, Forces, and Motion, Spring
2005. (Massachusetts Institute of Technology: MIT OpenCourseWare).
http://ocw.mit.edu (accessed MM DD, ... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf |
].
Q
inside V = − (cid:118)∫
q N d
S
i
da
= ∫ ρP dV
V
paired charge or
equivalently
polarization
charge density
Qinside V = − P i da = − ∇ i P dV =
(cid:118)∫
S
∫
V
ρ dV
P
∫
V
(Divergence Theorem)
P = q N d
∇ i P = −ρP
B. Gauss’ Law
∇ i (ε E) = ρ
o
total
= ρ + ρ = ρ − ∇ i P
u
P
u
unpaired charge
densit... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf |
su
sp
S
V
D. Polarization Current Density
∆Q = q N dV
=
q N d da
i
=
P
i da
[Amount of Charge passing through
surface area element da ]
d ip =
∂∆Q
∂t
=
∂P
∂t
i da
= Jp i da
polarization current density
[Current passing through surface
area element da ]
Jp =
∂P
∂t
Ampere’s law:
∇ x H = Ju + Jp + εo... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf |
θ) = ε E (r = R,
o r
θ) = − ε ∂Φ
o ∂r r R
=
=
ε oEo
⎡
⎢1 +
⎣
3
2R
3
r
⎤
⎥ cos
r R ⎦=
θ
= εo o
3 E cos θ
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 5 of 27
IV. Artificial Dielectric
E =
v
d
, σ = ε E =
s
ε v
d
q = σsA =
ε A
d
v
C =
q
v
=
ε A
d
E
d
... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf |
lectric constant)
ε
ε=
r
ε
o
ε =
⎡1 + ψ ⎤ = ε
e ⎦
o ⎣
⎛
⎛
⎜1 + π 4
⎜
o ⎜
⎝
⎝
R
s
3 ⎞
⎞
⎟
⎟ ⎟
⎠ ⎠
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 6 of 27
V. Demonstration: Artificial Dielectric
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
6.... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf |
x 10 −
12
farads/m
ω =
p−
2q n
−
m − ε
−
≈ 5.6 x 10
11
rad/s
fp − =
ωp − ≈ 9 x 10
2π
10 Hz
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 9 of 27
B. Drift-Diffusion Conduction [Neglect inertia]
0
m +
+dv
dt
= q
E
+
− m ν v −
+
+
+
+
∇ (n k T )
n+
⇒ v... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf |
+
D − =
k T
m − ν −
charge mobilities
Molecular
Diffusion
Coefficients
D + =
µ
+
D − =
µ
−
k T = thermal voltage (25 mV @ T ≈ 300o K)
q
Einstein’s Relation
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 10 of 27
C. Drift-Diffusion Conduction Equilibrium (J+ = J−... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf |
∇ Φ −
Φ
2 = 0
L d
1 ∂ ⎛ 2 ∂Φ ⎞
r
r2 ∂r ⎜
∂r ⎟
⎠
⎝
=
1 ∂2
r ∂r
2 (r Φ)
E. Ohmic Conduction
⇒
d2
)
(
dr2 r Φ −
r Φ
L2
d
= 0
0
r Φ = A e
1
−r / Ld
+ A e
2
+r / L d
Φ ( )r =
Q
4 π ε r
−r / L d
e
J = ρ µ E
+
+
+
− D +
∇ρ
+
J = −ρ µ E
−
−
−
− D −
∇ρ
−
If charge density gradients small, then ∇ρ ± ... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf |
− Φ =
n
p
2
q ND xn
2ε
+
2
q NA xp
2ε
=
q ND xn
2ε
(xn + xp ) =
q N D xn
2ε
2 ⎛
⎜1 +
⎝
ND ⎞
⎟
NA ⎠
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 15 of 27
VII. Relationship Between Resistance and Capacitance In Uniform Media Described
by ε and σ .
C =
uq
v
... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf |
R 2
4 π σ
R =
, C =
4 π ε
1
1
−
R 1 R 2
ε
⇒ RC =
σ
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 17 of 27
VIII. Change Relaxation in Uniform Conductors
∇ i Ju +
∂ ρ
∂ t
u = 0
ρ
∇ i E =
u
ε
J =
u
σ E
σ ∇ i E
+
∂ ρ
∂ t
u = 0
⇒
∂ ρ
u
∂ t
+
σ
ε
ρ = 0
u
ρ
u ... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf |
. Markus Zahn
Lecture 7
Page 19 of 27
ρu (t = 0) =
ρ0
r R 1<
0
r > R 1
Q T =
4
3
π R 1
3 ρ0
( )
ρu t =
ρ e t
0
− τ
e
r < R
1
(τ = ε σ)
e
0
r > R 1
− τ
ρ0 r e t e
3 ε
=
Q r e t e
− τ
4 π ε R 3
1
0 <
r
< R 1
r (
E r, t
) =
Q e t e
− τ
4 π ε r 2
Q
4 π ε0 r 2
R <
1
r R
<
2
r > R
2
σ su
... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf |
�� ⎢
⎥
⎥ ⎢
⎥
⎥ ⎢V(cid:108)
⎥
⎥ ⎢
⎥
⎥ ⎢
⎥
⎥
V(cid:108)
n C i
3 ⎥
⎦ ⎢
⎦
⎣
2
= 0
⎡n Ci
⎢
⎢
⎢0
⎢
⎢
⎢
⎣Cs
Cs
n C
i
0
det = 0
From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Le... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf |
, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 23 of 27
XII. Maxwell’s Capacitor
A. General Equations
_
( ) x
E t i
a
0 < x < a
_
( ) x
E t i
b
−
b x < 0
<
E =
a
∫ x
E d = v t = E t b + E t a
a
x
b ( )
( )
( )
−b
⎡
i ⎣ a
n J
− Jb ⎤ +
⎦
dσsu = 0 ⇒ σ E ( ) − σ E... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf |
+ σ +
⎜ a
⎝
σ a ⎞
b E t
⎟ a
b ⎠
( ) =
( )
ε
b v t
σ
b dv
+
b dt
b
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 24 of 27
B. Step Voltage: v t
( )
= V u t
( )
Then
dv
dt
= V δ t (an impulse)
( )
At t=0
⎛
⎜εa +
⎝
εb
εba dE a =
dv
⎞
⎟
b dt
b ⎠ dt
ε... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf |
σ + ω ε +
j
Ea
⎢ a
⎣
a
a (σ + ω
b
j ε )⎤
⎥
⎦
b
b
=
V(cid:108)
b ⎣ b
⎡σ + ω ε ⎤ = 0
j
b ⎦
E(cid:3)
j ω ε + σ
a
b
b
=
E(cid:3)
j ω ε + σ
b
a
a
=
V(cid:108)
⎡b (σ + ω
⎣
j ε ) + a (σ + ω ε )⎤
b ⎦
j
a
a
b
σ(cid:3)
su = ε E(cid:3) − ε E(cid:3)
b
a
b
a
=
(ε σ − ε σ ) V(cid:108)
b (σ + ω ε ) + a
⎡
... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf |
, C = b
b
b
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 27 of 27 | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/06ced4886eba1fc1f0eb037b210e5afb_lecture7.pdf |
18.997 Topics in Combinatorial Optimization
March 4, 2004
Lecture 8
Lecturer: Michel X. Goemans
Scribe: Constantine Caramanis
This lecture covers the proof of the Bessy-Thomass´e Theorem, formerly known as the Gallai
Conjecture. Also, we discuss the cyclic stable set polytope, and show that it is totally dual int... | https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf |
cycle C, denoted iO(C), is
the number of backward arcs in C. Recall from the last lecture that the index is well defined,
since the index is invariant under the equivalence operations defined above.
4. We say that an ordering O is valid if for any arc (u, v) ∈ A, there exists a cycle C containing
that arc, with index... | https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf |
, A), and a valid ordering O, if αO denotes
the size of the largest cardinality cyclic stable set, then
�
αO = min
iO(Ci),
where the cycles {C1, . . . , Cp} cover the vertex set V .
{C1,...,Cp}
The inequality
αO ≤ min
�
iO(Ci),
{C1,...,Cp}
is straightforward (as each vertex of a cycle stable set must be con... | https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf |
the right hand
This follows because we can can compute the quantity min
side of the theorem above, efficiently. We can do this by formulating a network flow problem that
computes the minimization. To do this, fix an enumeration of the ordering. Attach a cost of 0 to
every forward arc in the digraph under the given enum... | https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf |
is a cyclic stable
set.
Suppose, to the contrary, that S has no forward arcs, but S is not a cyclic stable. Let vi
Proof:
be the first element of the enumeration in S. If we rotate the enumeration so that vi becomes v1,
no forward paths are either created or destroyed in S, so we may assume, without loss of general... | https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf |
enumeration with respect to which as many elements of S as possible are
the first elements of the enumeration. Since S is assumed not to be a cyclic stable set, there must be some
element w sandwiched by elements of S.
the enumeration, by S> the remaining elements of S, and by W the elements after the last element
o... | https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf |
, to w. However, there also can be
no arc from w to any vertex before it in the enumeration. This follows because O was assumed to
be a valid ordering. If there were such an arc, say (w, v) for v earlier in the enumeration, because
we assume there are no forward arcs from any vertex coming before w, to w, and that w... | https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf |
enumeration. Therefore W must be empty, and S is indeed a cyclic stable set.
We now move to the proof of the Bessy-Thomass´e Theorem.
Proof:
The Main Idea: We want to show that the size of the maximum cyclic stable set equals the minimum
total index of a family of cycles covering V . Essentially the proof relies on... | https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf |
the given ordering.
(cid:1)
We form an acyclic digraph D(cid:1) = (V (cid:1), A(cid:1)) from D as follows. Let V (cid:1) = {v1, . . . , vn, v1, . . . , vk
(cid:1)
}
(we duplicate the elements of S) so that |V (cid:1)| = n + k. Next, if (v, w) ∈ A is a forward arc, then
(v, w) ∈ A(cid:1). If (w, vi ) ∈ A is a backw... | https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf |
, . . . , vk } and
8-4
(cid:1)
(cid:1)
{v1, . . . , vk
} have no incoming and outgoing arcs, respectively, they are both antichains in T . This
is also evident from Figure 2. We show that they are in fact maximum size antichains. Consider any
antichain I. As the ordering is valid, for any vert... | https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf |
are no forward paths between any two
elements of ID . Therefore, by Lemma 2, ID is a cyclic stable set. Therefore ID , and consequently
I, can have size at most equal to the size of S, that is, αO . We have thus shown that the size of the
largest antichain in T is equal to the cyclic stability number αO of D.
Now c... | https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf |
proof is
complete. If this is not the case, then a cycle in D obtained by joining together the paths Pi that
correspond to a cycle of σ may in fact intersect itself. Suppose that i and j are in the same cycle
of σ and the paths Pi and Pj intersect, in say v. We can then replace the paths Pi and Pj by two
other path... | https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf |
x(C) ≤ iO(C), ∀ directed cycles C
P = x � xv ≥ 0,
∀v ∈ V
�
.
We show in this section that the polytope P is totally dual integral (TDI) (see lecture 5 for more
on TDI system of inequalities).
Given a cyclic stable set S (cyclic stable with respect to the given ordering), let xS denote its
incidence vector, i.e.,... | https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf |
�
≤ min : � C
s.t. :
iO(C)yC
C : v∈C
yC ≥ 0, ∀C
yC ∈ {0, 1}.
yC ≥ 1, ∀v ∈ V
But this last quantity is exactly the minimum total index of a cycle cover of V , and thus by the
Bessy-Thomass´e Theorem, the final quantity equals αO . Therefore equality must hold throughout.
Recall that in order to prove that the d... | https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf |
≤ wu. From our reasoning above, we
know that the linear program associated to the digraph D(cid:1) (now we have wv = 1 for every v ∈ V (cid:1))
produces an integral solution that corresponds to a maximum size cyclic stable set in D(cid:1). Note that
if xv,i is in the stable set S(cid:1) for D(cid:1), then we can als... | https://ocw.mit.edu/courses/18-997-topics-in-combinatorial-optimization-spring-2004/06d9723de3290adf809e3e61e4dadc8a_co_lec8.pdf |
Massachusetts Institute of Technology
Department of Electrical Engineering and Computer Science
6.685 Electric Machines
Class Notes 1: Electromagnetic Forces
c 2003 James L. Kirtley Jr.
(cid:13)
1
Introduction
Bearings
Shaft
End Windings
Stator
Stator
Conductors
Rotor
Air
Gap
Rotor
Conductors
Figure 1: Form of Electric... | https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf |
indicated rotor conductors, but sometimes the rotor has permanent magnets
either fastened to it or inside, and sometimes (as in Variable Reluctance Machines) it is just an
oddly shaped piece of steel. The stator is, in this drawing, on the outside and has windings. With
most of the machines we will be dealing with, the... | https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf |
is just:
T
Vr
= 2 < τ >
Now, determining what can be done in a volume of machine involves two things. First, it is
clear that the volume we have calculated here is not the whole machine volume, since it does not
include the stator. The actual estimate of total machine volume from the rotor volume is actually
quite comp... | https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf |
the sum of electric power inputs to the different phase
terminals:
Mechanical power is torque times speed:
Pe =
viii
i
X
Pm = T Ω
3
Electric Power In
Electro-
Mechanical
Converter
Mechanical Power Out
Losses: Heat, Noise, Windage,...
Figure 3: Energy Conversion Process
And the sum of the losses is the difference:
Pd = P... | https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf |
is:
P m = f e dx
dt
The difference between these two is the rate of change of energy stored in the system:
dWm = P e
dt
−
P m
It is then possible to compute the change in energy required to take the system from one state to
another by:
Wm(a)
Wm(b) =
a
idλ
f edx
−
where the two states of the system are described by a = (... | https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf |
L(x)i = λ too early in the
derivation produces erroneous results: in the case of a linear system it produces a sign error, but
in the case of a nonlinear system it is just wrong.
3.1.1 Example: simple solenoid
Consider the magnetic actuator shown in cartoon form in Figure 5. The actuator consists of
a circular rod of f... | https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf |
dx L(x)
1
d 1
dx L
=
1
2
0πR N
µ
2
f e
(x) =
λ2
0
2 µ
1
0πR N 2
2
−
Given that the system is to be excited by a current, we may at this point substitute for flux:
λ = L(x)i =
µ0πR2N i
x + g
and then total force may be seen to be:
f e
=
0πR2N 2 i2
µ
(x + g)2 2
−
The force is ‘negative’ in the sense that it tends to reduc... | https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf |
that force produced is:
dW
′
m =
λkdik + f dx
e
k
X
fe =
∂W ′
m
∂x
3.2 Example: Synchronous Machine
Stator
Gap
Rotor
C
A’
B
θ
B’
A
C’
F’
µ
F
F
F’
C’
A
B’
B
µ
C
A’
Figure 6: Cartoon of Synchronous Machine
Consider a simple electric machine as pictured in Figure 6 in which there is a single winding
on a rotor (call it th... | https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf |
multiple excitation it
is important to exercise some care in taking the coenergy integral (to ensure that it is taken over
a valid path in the multi-dimensional space). In our case there are actually five dimensions, but
only four are important since we can position the rotor with all currents at zero so there is no
con... | https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf |
ia0 = Ia cos ωt
ib0 = Ia cos ω
t
(cid:18)
ic0 = Ia cos ω
t +
2π
− 3
2π
3
(cid:19)
(cid:19)
if 0 = If
(cid:18)
9
and assume the rotor is turning at synchronous speed:
pθ = ωt + δi
Noting that cos x sin y = 1 sin(x
2
−
y) + 1 sin(x + y), we find the torque expression above to be:
2
Te =
pM IaIf
−
(cid:18)
+
+
1
2
(cid:18... | https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf |
happens in electric machines.
4.1 Field Description of Energy Flow: Poyting’s Theorem
Start with Faraday’s Law:
and Ampere’s Law:
~E =
~∂B
− ∂t
∇ ×
~H = J
~
∇ ×
Multiplying the first of these by H and the second by E and taking the difference:
~
~
~H
~E
~E
−
· ∇ ×
· ∇ ×
~H =
∇ ·
~E
~
H
×
(cid:16)
(cid:17)
=
~H
−
∂B~
dt −... | https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf |
, if there is motion of any material within the system, we
can use the empirical expression for transformation of electric field between observers moving with
respect to each other. Here the ’primed’ frame is moving with respeect to the ’unprimed’ frame
with the velocity ~v
′
~E
~= E + ~v
~B
×
This transformation descri... | https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf |
This is the Lorentz Force Law, which describes the interaction of current with magnetic field
to produce force. It is not, however, the complete story of force production in electromechanical
systems. As we learned earlier, changes in geometry which affect magnetic stored energy can also
produce force. Fortunately, a com... | https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf |
)
·
~H H µ
~
∇
(cid:17)
~H
~H
µ
·
∇
(cid:17)
(cid:17)
~H
~H
×
(cid:17)
∇ ×
(cid:16)
~ =
H
~
H
(cid:16)
· ∇
(cid:17)
~H
1
− 2 ∇
~F = µ
= µ
~H
~H
~H
(cid:16)
~H
(cid:16)
·
∇
· ∇
(cid:17)
(cid:17)
1
2
µ
−
− ∇ (cid:18)
∇
1
2
~H
(cid:16)
µ
·
~H
(cid:16)
~
H
(cid:17)
~H
·
−
(cid:17)(cid:19)
~H
~H
·
(cid:17)
~
H
(cid:16)
~H
·... | https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf |
H
2
n
!
n
X
where we have used the Kroneker delta δik = 1 if i = k, 0 otherwise.
Note that this force density is in the form of the divergence of a tensor:
Fk =
∂
∂xi
Tik
or
∇ ·
In this case, force on some object that can be surrounded by a closed surface can be found by
~F =
T
using the divergence theorem:
~f =
vol
Z
... | https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf |
cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)
(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)
(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid... | https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf |
amplitudes:
The y- component of Faraday’s Law is, assuming the problem is uniform in the z- direction:
H y = z + K s
K
jkg
−
jω B
s
′
y = jkEz
−
or
ω
s µ0H
k
A bit of algebraic manipulation yields expressions for the complex amplitudes of rotor surface
′
z =
−
E
y
current and gap magnetic field:
Ks
=
H y =
σ
j µ0ωs s
k2... | https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf |
shows what we already suspected: the electromagnetic power flow from the stator is the
force density on the shuttle times the wave velocity. The electromagnetic ower flow into the shuttle
is the same force density times the ’slip’ velocity. The difference between these two is the power
converted to mechanical form and it ... | https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf |
can be picked to
be perpedicular to each of the current filaments since the divergence of current is zero. The flux λ
is calculated over a path that coincides with each current filament (such paths exist since current
has zero divergence). Then the flux is:
Z
Now, if we use the vector potential A for which the magnetic flux... | https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf |
~
C
×
∇ ·
(cid:16)
~
D
~
= D
(cid:17)
·
∇ ×
(cid:16)
~
C
·
−
~
C
(cid:17)
′
Wm =
Then, noting that B =
~
∇ ×
−∇ ·
vol
Z
~
A:
~
~
A dM dv +
×
(cid:16)
(cid:17)
vol
Z
∇ ×
(cid:16)
·
(cid:17)
~D
Now,
∇ ×
(cid:16)
~
A
(cid:17)
~
dM dv
W
′
m =
−
(cid:13)
Z
Z
~dM d~s +
~A
×
~dM dv
~B
·
vol
Z
The first of these integrals (clos... | https://ocw.mit.edu/courses/6-685-electric-machines-fall-2013/06deccf046b565a8295e5ada36b252df_MIT6_685F13_chapter1.pdf |
8.022 (E&M) – Lecture 6
Topics:
(cid:132) More on capacitors
(cid:132) Mini-review of electrostatics
(cid:132) (almost) all you need to know for Quiz 1
Last time…
(cid:132) Capacitor:
(cid:132) System of charged conductors
(cid:132) Capacitance:
C
=
Q
V
(cid:132)
It depends only on geometry
+Q
+
+
+
+
+
+
--
-Q
... | https://ocw.mit.edu/courses/8-022-physics-ii-electricity-and-magnetism-fall-2004/07002b8b4ec36c93c70bfc4697be3789_lecture6.pdf |
:132) Parallel plates capacitor:
C
=
Q Q
=
Ed
V
=
A
dπ
4
(cid:132) Add a dielectric between the plates:
(cid:132) Dielectric’s molecules are not spherically symmetric
(cid:132) Electric charges are not free to move
(cid:198) E will pull + and – charges apart and orient them // E
+
+
+
+
+
+
+
+
+
+
E=4πσ
_ +
_ +
_ +
... | https://ocw.mit.edu/courses/8-022-physics-ii-electricity-and-magnetism-fall-2004/07002b8b4ec36c93c70bfc4697be3789_lecture6.pdf |
=
V
=
Q
2
=
1
V V
+
Q
2
=
1
C
1
+
1
C
2
B
-
+
-
+
-
+
-
+
-
+
-
+-
This is 1 conductor that
starts electrically neutral
(cid:198) Q1=Q2
G. Sciolla – MIT
1
−
C
⎞
⎟
⎠
i
8.022 – Lecture 6
⎛
= ⎜
⎝
1
C
i
∑
Capacitors in parallel
(cid:132)
Let’s connect 2 capacitors C1 and C2 in the following way:
Q1
Q2
V1
V2
V
?
V
(cid:1... | https://ocw.mit.edu/courses/8-022-physics-ii-electricity-and-magnetism-fall-2004/07002b8b4ec36c93c70bfc4697be3789_lecture6.pdf |
++++++
++++++
…
-------
++++++
80µF
60W
(cid:132)
Total capacitance: 960 µF
(cid:132)
Discharged on a 60 W light bulb when capacitors are charged at:
(cid:132)
V = 100 V, 200 V, V = 300 V
(cid:132) What happens?
(cid:132)
(cid:132)
Energy stored in capacitor is U= ½ CV2
(cid:198) V = V0: 2xV0 : 3xV0 (cid:198) U = U0 ... | https://ocw.mit.edu/courses/8-022-physics-ii-electricity-and-magnetism-fall-2004/07002b8b4ec36c93c70bfc4697be3789_lecture6.pdf |
Q
r
V
Q
(cid:71)
F
Q
i N
=
= ∑
i
=
1
q Q
i
2
r
|
i
|
ˆ
r
i
(cid:71)
QF =
∫
V
dq Q
2
|r|
ˆ
r
=
∫
V
ρ dV Q
2
|r|
ˆ
r
G. Sciolla – MIT
8.022 – Lecture 6
13
The Importance of Superposition
Extremely important because it allows us to transform complicated
problems into sum of small, simple problems that we know how to
s... | https://ocw.mit.edu/courses/8-022-physics-ii-electricity-and-magnetism-fall-2004/07002b8b4ec36c93c70bfc4697be3789_lecture6.pdf |
W
− > = −
2
1
∫
1
(cid:71)
F
C
(cid:71)
ds
•
w here
F
C oulom b
=
ˆ
Q qr
2
r
y
1
2
P1
P2
3
x
(cid:132) NB: integral independent of path: force conservative!
(cid:132) Assembling a system of charges costs energy. This is the energy
stored in the electric field:
U
=
1
2
∫
Volume
with
charges
ρφ
dV
=
2
E
8
π
dV
∫
E... | https://ocw.mit.edu/courses/8-022-physics-ii-electricity-and-magnetism-fall-2004/07002b8b4ec36c93c70bfc4697be3789_lecture6.pdf |
not always useful:
Symmetry is needed!
(cid:132) Main step: choose the “right” gaussian surface so that
E is constant on the surface of integration
G. Sciolla – MIT
8.022 – Lecture 6
18
9
From ρ (cid:198) φ
(cid:132) General case:
(cid:132) For a point charge:
(cid:132) Superposition principle:
φ =
q
r
φ = ∫
V
N... | https://ocw.mit.edu/courses/8-022-physics-ii-electricity-and-magnetism-fall-2004/07002b8b4ec36c93c70bfc4697be3789_lecture6.pdf |
φ is always continuous
(cid:132) E is not always continuous: it can “jump”
(cid:132) When we have surface charge distributions
(cid:132) Remember problem #1 in Pset 2
(cid:198) When solving problems always check for consistency!
G. Sciolla – MIT
8.022 – Lecture 6
21
Summary
ρ(x,y,z)
2
φ πρ
4
∇ = −
G a uss’s la w... | https://ocw.mit.edu/courses/8-022-physics-ii-electricity-and-magnetism-fall-2004/07002b8b4ec36c93c70bfc4697be3789_lecture6.pdf |
Be creative and think of distribution of point charges that will create the
same filed lines:
(cid:132) Example:
Method of images
+
+
- - - - - - - - -
- - - - - - - - -
-
G. Sciolla – MIT
8.022 – Lecture 6
24
12
Capacitors
(cid:132) Capacitance
have capacitance C
(cid:132) Two oppositely charged conductors kept... | https://ocw.mit.edu/courses/8-022-physics-ii-electricity-and-magnetism-fall-2004/07002b8b4ec36c93c70bfc4697be3789_lecture6.pdf |
2 General results of representation theory
2.1 Subrepresentations in semisimple representations
Let A be an algebra.
Definition 2.1. A semisimple (or completely reducible) representation of A is a direct sum of
irreducible representations.
Example. Let V be an irreducible representation of A of dimension n. Then Y ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0721b66f53c3c196ce86d6d867514442_MIT18_712F10_ch2.pdf |
is isomorphic to
niVi given
�
by multiplication of a row vector of elements of Vi (of length ri) by a certain ri-by-ni matrix Xi
with linearly independent rows: θ(v1, ..., vri ) = (v1, ..., vri )Xi.
V is a direct sum of inclusions θi : riVi
ni, and the inclusion θ : W
m
i=1riVi, ri
⊃
⊃
∗
∗
�
by induction in ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0721b66f53c3c196ce86d6d867514442_MIT18_712F10_ch2.pdf |
the matrix Xi goes to Xigi, while Xj , j = i don’t change. Take gi
Gi such that
(q1, ..., qni )gi = (1, 0, ..., 0). Then W gi contains the first summand Vi of niVi (namely, it is P gi),
hence W gi = Vi
nmVm is the kernel of the projection
−
of W gi to the first summand Vi along the other summands. Thus the require... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0721b66f53c3c196ce86d6d867514442_MIT18_712F10_ch2.pdf |
subrepresentation by the induction assumption.
2.2 The density theorem
Let A be an algebra over an algebraically closed field k.
Corollary 2.4. Let V be an irreducible finite dimensional representation of A, and v1, ..., vn
V there exists an element a
be any linearly independent vectors. Then for any w1, ..., wn... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0721b66f53c3c196ce86d6d867514442_MIT18_712F10_ch2.pdf |
=1 End(Vi) is surjective.
r
i=1δi : A
the map
�
r
Proof. (i) Let B be the image of A in End(V ). We want to show that B = End(V ). Let c
v1, ..., vn be a basis of V , and wi = cvi. By Corollary 2.4, there exists a
B, and we are done.
Then a maps to c, so c
End(V ),
A such that avi = wi.
�
�
�
i be the imag... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0721b66f53c3c196ce86d6d867514442_MIT18_712F10_ch2.pdf |
7. (Dual representation) Let V be a representation of any algebra A. Then the
dual representation V ⊕ is the representation of the opposite algebra Aop (or, equivalently, right
A-module) with the action
a)(v) := f (av).
(f
·
Proof of Theorem 2.6. First, the given representations are clearly irreducible, as for any ... | https://ocw.mit.edu/courses/18-712-introduction-to-representation-theory-fall-2010/0721b66f53c3c196ce86d6d867514442_MIT18_712F10_ch2.pdf |
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