1
I.
3D PISTONS WITH MIXED BOUNDARY
CONDITIONS
August 21st By Zhonghai Liu
2
II.
3D RECTANGULAR PISTONS WITH MIXED BOUNDARY
CONDITIONS——SCALAR FIELD
Lemma 1: Define two operators N and D as below
N xa f (x, x0 ) = f (2a − x, x0 )
(1)
Dax f (x, x0 ) = − f (2a − x, x0 )
(2)
then based on method of image we have (by number of reflections):
f N (x, x0 ) = f +(N xa f +N xb f )+(N xa N xb f +N xb N xa f )+(N xa N xb N xa f +N xb N xa N xb f )+. . . (3)
satisfies Neumann boundary conditions at both x = a and x = b—–NN;
f D (x, x0 ) = f +(Dax f +Dbx f )+(Dax Dbx f +Dbx Dax f )+(Dax Dbx Dax f +Dbx Dax Dbx f )+. . . (4)
satisfies Dirichlet boundary conditions at both x = a and x = b—–DD;
f M (x, x0 ) = f +(Dax f +N xb f )+(Dax N xb f +N xb Dax f )+(Dax N xb Dax f +N xb Dax N xb f )+. . . (5)
satisfies mixed boundary conditions: Dirichlet at x = a and Neumann at x = b or
Neumann at x = a and Dirichlet at x = b—–DN or ND.
3
III.
CYLINDER KERNEL OF 3D RECTANGULAR CAVITY WITH
MIXED BOUNDARY CONDITIONS
The cylinder kernel in free space without any boundary is:
T (t, r, r0 ) = −
1
1
2π2 (t2 + |r − r0 |)2
(6)
For the cavity with boundary condition (NN)(NN)(DN), cylinder kernel can be
written as:
T
NN M
=T + (N x0 T + N xa T + Ny0 T + Nyb T + D0z T + Nzc T )
+ (N x0 N xa T + N xa N x0 T + Ny0 Nyb T + Nyb Ny0 T + D0z Nzc T + Nzc D0z T ) + . . .
From the point of view of classical paths, all paths above fall into 4 kinds:
Periodic paths—undergoing even number reflections at the walls;
Side paths—–undergoing odd number reflections at the walls;
Edge paths—–paths involving reflections off edges;
Corner paths—–paths involving reflections off corners;
(7)
4
IV.
ε1 ε2 ε3
Ur,r
0
4 KINDS OF CLASSICAL PATHS
∞
1 X
(−1)n
=− 2
2π l,m,n=−∞ t2 + (2la + x + ε1 x0 )2 + (2mb + y + ε2 y0 )2 + (2nc + z + ε3 z0 )2
(8)
then cylinder kernel for boundary conditions (NN)(NN)(DN):
T
NN M
−−−
+−−
−+−
−−+
++−
−++
+−+
+++
(t, r, r0 ) = Ur,r
0 + U r,r0 + U r,r0 + U r,r0 + U r,r0 + U r,r0 + U r,r0 + U r,r0
(9)
−−−
+++
Periodic paths—– (P ∼ Ur,r
0 );Corner paths—–(C ∼ U r,r0 )
−−+
−+−
+−−
Side paths—–(S x ∼ Ur,r
0 )( S y ∼ U r,r0 )(S z ∼ U r,r0 )
++−
−++
+−+
Edge paths—–(E xy ∼ Ur,r
0 )(E yz ∼ U r,r0 ) (E xz ∼ U r,r0 )
For (NN)(DD)(DN):
T
NDM
−−−
+−−
−+−
−−+
++−
−++
+−+
+++
(t, r, r0 ) = Ur,r
0 + U r,r0 − U r,r0 + U r,r0 − U r,r0 − U r,r0 + U r,r0 − U r,r0
(10)
For (DD)(NN)(DN):
T
DN M
−−−
+−−
−+−
−−+
++−
−++
+−+
+++
(t, r, r0 ) = Ur,r
0 − U r,r0 + U r,r0 + U r,r0 − U r,r0 + U r,r0 − U r,r0 − U r,r0
(11)
For (DD)(DD)(DN):
T
DDM
−−−
+−−
−+−
−−+
++−
−++
+−+
+++
(t, r, r0 ) = Ur,r
0 − U r,r0 − U r,r0 + U r,r0 + U r,r0 − U r,r0 − U r,r0 + U r,r0
(12)
5
V.
ENERGY DENSITY AND TOTAL ENERGY
1
∂2
ε = − lim 2 T (t, r, r)
2 t→0 ∂ t
(13)
Z
E=
εdV
(14)
Periodic paths
E −−−
M
0
∞
1 X
(−1)n abc
=− 2
2π l,m,n=−∞ [(2la)2 + (2mb)2 + (2nc)2 ]2
=−
(15)
abc
[2Z3 (a, b, 2c; 4) − Z3 (a, b, c; 4)]
32π2
Side paths
E
+−−
∞ 0
X
(−1)n
1
= − 2 bc
2π m,n=−∞ [(2mb)2 + (2nc)2 ] 23
=−
E −+−
(16)
bc
[2Z2 (b, 2c; 3) − Z2 (b, c; 3)]
64π
∞ 0
X
(−1)n
1
= − 2 ac
2π l,n=−∞ [(2la)2 + (2nc)2 ] 32
=−
(17)
ac
[2Z2 (a, 2c; 3) − Z2 (a, c; 3)]
64π
E −−+
=0
M
where Zd (a1 , .., ad ; s) is the Epstein zeta function.
(18)
6
VI.
ENERGY DENSITY AND TOTAL ENERGY
Edge paths
E
++−
∞ 0
c X (−1)n
π
=−
=
32π n=−∞ n2
192c
(19)
E +−+ = 0
(20)
E −++ = 0
(21)
E +++ = 0
(22)
Corner paths
E NN M = E −−− + E +−− + E −+− + E −−+ + E ++− + E −++ + E +−+ + E +++
(23)
E NDM = E −−− + E +−− − E −+− + E −−+ − E ++− − E −++ + E +−+ − E +++
(24)
E DN M = E −−− − E +−− + E −+− + E −−+ − E ++− + E −++ − E +−+ − E +++
(25)
E DDM = E −−− − E +−− − E −+− + E −−+ + E ++− − E −++ − E +−+ + E +++
(26)
Till now, we can deal with the rectangular cavity with any combination of D
and N. Easily to be extended for pistons.
7
VII.
T
AN ALTERNATIVE EXPRESSION FOR CYLINDER KERNEL
NN M
T
T
T
NDM
DN M
DDM
∞
X
|Dlmn |2
=−
cosk1 xcosk2 ysink3 zcosk1 x0 cosk2 y0 sink3 z0 e−ωlmn t (27)
ωlmn
l,m,n=−∞
∞
X
|Dlmn |2
cosk1 xsink2 ysink3 zcosk1 x0 sink2 y0 sink3 z0 e−ωlmn t
=−
ωlmn
l,m,n=−∞
(28)
∞
X
|Dlmn |2
=−
sink1 xcosk2 ysink3 zsink1 x0 cosk2 y0 sink3 z0 e−ωlmn t
ωlmn
l,m,n=−∞
(29)
∞
X
|Dlmn |2
sink1 xsink2 ycosk3 zsink1 x0 sink2 y0 cosk3 z0 e−ωlmn t
=−
ωlmn
l,m,n=−∞
where ω2lmn = k12 + k22 + k32 and k1 =
lπ
a ,k2
=
mπ
b ,k3
=
(n+ 12 )π
c .
(30)
It follows from the
normalization condition that:
|Dlmn |2 =
0l 0m 0n
abc
(31)
0i = 1 for i = 0 and 0i = 2 otherwise.
If we stick to this way with the help of Poisson Sum Formula, we can reproduce
the previous results (9 − 12).
8
VIII.
3D RECTANGULAR PISTONS WITH MIXED BOUNDARY
CONDITIONS——EM FIELD
Hertz Potential:
→
−
→
−
→
−
→
−
(Φ, A) = (−∇ · Π e , ∂t Π e + ∇ × Π m )
(32)
→
−
−
−e and →
−e ,then
If we choose the Hertz potentials as Π e = ϕ→
Π m = ψ→
3
3
→
−
(Φ, A) = (−∂3 ϕ, ∂2 ψ, −∂1 ψ, +∂0 ϕ)
(33)
By Maxwell Equation:
→
−
E = (∂1 ∂3 φ − ∂0 ∂2 ψ, ∂2 ∂3 φ + ∂1 ∂0 ψ, ∂23 φ − ∂20 φ)
→
−
B = (∂0 ∂2 φ + ∂1 ∂3 ψ, −∂0 ∂1 φ + ∂2 ∂3 ψ, ∂23 ψ − ∂20 ψ)
(34)
Conducting Boundary Condition—Et = 0, Bn = 0
Permeable Boundry Condition—En = 0, Bt = 0
We investigate the cavity with mixed boundary conditions: permeable boundary
at z = c and conducting boundary at other walls. The appropriate Hertz potentials
are:
φlmn (x, y, z) =
∞
X
l,m,n=−∞
∞
X
Dlmn
sink1 xsink2 ycosk3 ze−iωlmn t
√
⊥
|klmn
| 2ωlmn
cosk1 xcosk2 ysink3 ze−iωlmn t
ψlmn (x, y, z) =
Dlmn
√
⊥
|k
|
2ωlmn
l,m,n=−∞
lmn
(35)
9
IX.
ENERGY DENSITY AND TOTAL ENERGY
The energy density ∼ 12 (E 2 + B2 )
iωlmn (t − t0 ) ∼ −ωlmn τ
(k12 , k22 , k32 ) ∼ (−∂21 , −∂22 , −∂23 , )
E32 (t, x, x0 )
∞
X
|Dlmn |2 (k12 + k22 )
0
=
sink1 xsink2 ycosk3 zsink1 x0 sink2 y0 cosk3 z0 eiωlmn (t−t )
2ωlmn
l,m,n=−∞
(36)
E32 (τ, x, x0 )
∞
X
|Dlmn |2 (k12 + k22 )
=
sink1 xsink2 ycosk3 zsink1 x0 sink2 y0 cosk3 z0 e−ωlmn τ
2ωlmn
l,m,n=−∞
1
DDM
= (∂21 + ∂22 )T
2
(37)
1
ε5C1P (τ, r, r) = (E32 + B23 + E22 + B22 + E12 + B21 )
2
1
DDM
NDM
DN M
= [(∂21 + ∂22 )T
+ (∂22 + ∂23 )T
+ (∂21 + ∂23 )T
4
+ (∂21 + ∂22 )T
NN M
+ (∂22 + ∂23 )T
DN M
+ (∂21 + ∂23 )T
NDM
(38)
]
1
DDM
NDM
DN M
NN M
DN M
NDM
ε5C1P (τ, r, r) = − [∂2τ (T
+T
+T
+T
+T
+T
)
4
+ ∂21 (T
NDM
+T
DN M
) + ∂22 (T
DN M
+T
NDM
) + ∂23 (T
DDM
+T
NN M
)]
(39)
10
X.
ENERGY DENSITY AND TOTAL ENERGY
Substitute (9-12) into (39)
ε5C1P (τ, r, r) = −∂2τ U −−− − ∂2τ U −−+ − ∂23 U ++− − ∂23 U +++
(40)
Notice that:
∂23 U ++−
∂23 U +++
=
∂2τ U ++−
∞
2
2
4 X
n (2nc) − t
− 2
(−1)
E
π l,m,n=−∞
(d xy )6
lmn
2
∞
X
2
4
2 +++
n (2nc) − t
− 2
= ∂τ U
(−1)
C 6
π l,m,n=−∞
(dlmn
)
(41)
Then:
ε5C1P (τ, r, r) = − ∂2τ [U −−− + U −−+ + U ++− + U +++ ]
∞
∞
2
2
4 X
4 X
n (2nc)
n (2nc)
(−1) E xy + 2
(−1) C 6
+ 2
π l,m,n=−∞
(dlmn )
(d )6 π l,m,n=−∞
(42)
lmn
−−+
++−
+++
E 5C1P =2[E −−−
M + EM + EM + EM ] −
π
π
+
48c 72c
π
144c
0
∞
abc X
π
π
(−1)n
=−
+
−
16π2 l,m,n=−∞ [(la)2 + (mb)2 + (nc)2 ]2 96c 144c
++−
=2[E −−−
M + 0 + E M + 0] −
=−
POSITIVE!
abc
π
[2Z
(a,
b,
2c;
4)
−
Z
(a,
b,
c;
4)]
+
3
3
288c
16π2
(43)
11
XI.
ENERGY DENSITY AND TOTAL ENERGY
To compare with the rectangular cavity with conducting boundary at every wall,
which has energy
π
π
π
+
+
24a 24b 24c
π
π
π
+
+
= E DDD + E NNN +
24a 24b 24c
abc
π 1 1 1
=−
Z
(a,
b,
c;
4)
+
( + + )
3
48 a b c
16π2
E Con =2[E −−− + E −++ + E +−+ + E ++− ] +
(44)
Next to form a piston. E Con is x − y − z symmetrical, the force on the partition
could be either attractive or repulsive to the nearest wall, depending on the ration
a : b : c.
However, for E 5C1P , x − y − z symmetry is destroyed but x − y symmetry stays.
E 5C1P = −
π
abc
[2Z
(a,
b,
2c;
4)
−
Z
(a,
b,
c;
4)]
+
3
3
288c
16π2
(45)
We have two ways to form a piston:
1. with the permeable wall (z=c) movable;
Fc5C1P = −∂c [E 5C1P (a, b, c) + E 5C1P (a, b, L − c)]|L−→∞
(46)
2. with the conducting wall movable.
Fa5C1P = −∂a [E 5C1P (a, b, c) + E 5C1P (L − a, b, c)]|L−→∞
For both cases, the force on the partition is always repulsive.
(47)