Download 3. Derivation of the two-dimensional ideal dipole model

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Supplementary Text S1
2
Validity of the 2D ideal dipole approximation in shallow water
3
In the main text, we used the two-dimensional ideal dipole formula to predict the voltage
4
difference between a pair of electrodes. The 2D ideal dipole formula was derived by
5
assuming an infinite 2D space without boundaries. In this text, we will derive a more
6
realistic ideal dipole formula by including the depth dimension and the boundary
7
conditions to show that the 3D ideal dipole formula reduces to the 2D ideal dipole
8
formula for a shallow, circular body of water. Our analysis will include the effects of the
9
boundaries and the finite electrode dimension, which influence the voltage difference
10
between a pair of vertically oriented electrodes in the presence of a dipole. The dipole is
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situated in the circular body of water of a radius R and depth d; and the water is surround
12
by the top, bottom, and side boundaries.
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1. Effects of the top and bottom interfaces
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First, let us consider the effects of the top and bottom interfaces on the electric fields, and
15
assume an infinitely wide body of water of a depth d. The current dipole induces surface
16
charges on the two dielectric interfaces, one on the top between the water and the air, and
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another at the bottom between the water and the glass. The induced surface charges
18
contribute to the electrode potential, but it is difficult to directly determine the surface
19
charge distributions. The method of image charges simplifies our analysis to determine
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the electrode potential, by replacing the surface charges with image current sources on
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the opposite side of each boundary [57,58]. Each dielectric boundary acts as a mirror
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surface to generate an image source I’ on the opposite side of the boundary at an equal
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distance to the surface, having a magnitude:
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   k
I' w
 w  k

 I  k I ,

(S1.1)
 k
25
where  w is the permittivity of water,  k is the permittivity of the neighboring media (air
26
or glass), and  is the attenuation factor. Note that the image source expression does not
1
27
depend on the conductivity of the media to calculate the electric field inside of the
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circular region [58]. In our problem, the two parallel surfaces at the top and at the bottom
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create an infinite number of image sources by creating an infinite number of reflections.
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Each reflection by a surface k creates a new image source with an attenuation factor  k .
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For instance, the first order reflection creates two image sources: I0 above the top
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interface at z0  2d  h , and I1 below the bottom interface at z1   h (see Fig. S1A).
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I0 
w  0
I  0 I ,
w  0
 0
I1 
 w  1
I  1 I .
 w  1
1
(S1.2)
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where  0 is the permittivity of air, and  1 is the permittivity of glass. Similarly, the
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second order reflections create image sources of I 0 and I1 . For example, the bottom
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surface creates a second order image source I 01 from the first order image source I1 at a
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height of z01 , where:
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
 I 01  1 I 0  1  0 I    01 I ,


 z01   z0  h  2d .
(S1.3)
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We can describe the reflection operations for each surface recursively. The reflection
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operations for the top interface (air-water interface, k=0) are:
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
 Ik0  f 0  I k    0 I k ,


 zk0  g 0  zk   2d  zk .
(S1.4)
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Note that the top surface will always reflect an image charge located below the bottom
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surface ( zk  0 ), such that the distance to the new image source ( zk0 ) from the original
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current source will always increase: zk0  2d  zk . The reflection operations for the
45
bottom interface (glass-water interface, k=1) are:
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
 Ik1  f1  I k   1 I k ,


 zk1  g1  zk    zk .
(S1.5)
2
47
48
Now, we can express the potential at a field location r due to an image source I k :
V  r | Ik   
cI k
cI k

,
2
r  rk
r 2   z  zk 
(S1.6)
49
where c   4 w 
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water. The net potential at r due to the current source I, and its first order reflections is:
1
is the constant of proportionality, and  w is the conductivity of
Vnet  r   V  r | I   V  r | I 0   V  r | I1  ,
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
cI 0
cI1
cI


,
r  rI
r  rI1
r  rI0
(S1.7)
2 1/ 2
2 1/ 2
2 1/ 2

 2 
 2 
d 
d 
d  

2
 cI  r   z      0  r   z0     1  r   z1     .
2 
2 
2  







52
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The net potential up to the second order reflections is:
Vnet  r   V  r | I   V  r | I 0   V  r | I1   V  r | I 01   V  r | I10  .
first order reflections
(S1.8)
second order reflections
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Figure S1A shows the net potential calculated up to the nth reflection as a function of n.
55
Our numerical calculation indicates that the net potential converges. The convergence of
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the potential is expected since each reflection produces a weaker image source at a
57
further location from the electrode.
58
3
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2. Finite electrode dimension
60
Now, let us consider the effect of a finite electrode dimension. The potential at the
61
surface of an electrode is equal throughout its surface since the electrode is a conductor,
62
and the electrode measures the average of potentials at different heights. This could be
63
shown by extending the proof based on connecting two conducting spheres. Thus, the
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electrode potential can be determined by averaging the potentials at different heights h,
65
where 0  h  d :
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Ve  r | I   
1 d
cI
cI 
d h
1  h  
dz '    sinh 1 
  sinh    .

z
'

0
2
2
d
d
 r 
 r 
r   z ' h 
(S2.1)
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If the distance from the current source to the electrode (=r) is much greater than the depth
68
of water (=d): d
69
70
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r , we can apply a Taylor approximation up to the first order:

h h
1  h 
1,
sinh  r   r , r

 

sinh 1  d  h   d  h , d  h



r
r
 r 
(S2.2)
1.
Substitute (S2.1) into (S2.2) to obtain the potential of a rod-shaped electrode:
Ve  r | I   
cI
.
r
(S2.3)
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Therefore, the potential of a vertically oriented extended electrode due to a single current
73
source is approximately equal to the potential of a point-like electrode if the electrode is
74
sufficiently far from the current source relative to the depth of water.
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4
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3. Derivation of the two-dimensional ideal dipole model
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Let us derive an expression for the potential due to an ideal dipole in a two-dimensional
78
space. According to Gauss’ law, the electric field strength due to a current source (I) in
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2D is:
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
r ' r
E  r ' dr ' 
I

E
I
cI
 ,
  2 r  r
(S3.1)
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where c is a positive constant,  is the conductivity of the medium and r is the distance
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from the current source to the field location. We can obtain the electric potential (V) by
83
integrating the electric field from Eq (S3.1):
84
V  r    E  a  da  cI ln r.
(S3.2)
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The potential due to a dipole is equal to the sum of contributions from the positive and
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negative current sources:
87
Vdip  V  V  cI  ln r  ln r  ,
(S3.3)
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where r is the distance between the positive source to the field location, and r is the
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distance between the negative source to the field location. According to the cosine law,
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the distances from the sources to the field location are:
2
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93
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r  r 1
d
d 
cos     ,
r
 2r 
(S3.4)
where  is the angle between the vectors r and d (Fig. 1C). Taking the ln of Eq (S3.4):
2
1  d
d  
ln r  ln r  ln 1
cos      ,
2 
r
 2r  
(S3.5)
and substituting Eq (S3.5) in Eq (S3.3):
5
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96
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Vdip
2
2
 d
cI   d
d  
 d  
 ln 1  cos       ln 1  cos       .
 r
2   r
 2r  
 2r   

In the limit of d
Vdip 
(S3.6)
r , Eq (S3.6) can be approximated as:
cI  1   d / r  cos 
ln 
2  1   d / r  cos 

 .

(S3.7)
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We can apply a Taylor expansion formula:
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1  1 x 
x3 x5
ln 

x

 

2  1 x 
3 5
, x  1,
(S3.8)
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to Eq (S3.7) by setting x   d / r  cos , since  d / r  cos   1 from d
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(S3.8) to Eq (S3.7) yields the ideal dipole potential in the two-dimensional case:
102
103
Vdip  r ,  
cp cos 
, p  Id ,
r
r . Applying Eq
(S3.15)
where p is the current dipole moment.
104
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Now, let us compare the potentials due to a current dipole placed in shallow water
106
determined from the two-dimensional ideal dipole model approximation (S3.15) with
107
numerical solutions. The dipole potential can be numerically calculated by estimating a
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dipole as a pair of closely spaced current source and sink, and summing all potential
109
contributions due to the current sources and their image currents up to the 1000th order.
110
Figure S2D shows the potential of a vertically oriented electrode due to a current dipole
111
as a function of the normalized inverse distance. The two current sources were separated
112
by 0.01d at a depth of z = d/2, and oriented toward the electrode. Our numerical result
113
(blue curve in Fig. S2D) confirms that the 2D ideal dipole voltage approximation (red
114
line in Fig. S2) is valid for the potential of a vertically oriented extended electrode in
6
115
shallow water. The 2D ideal dipole voltage approximation worked very well even when
116
the dipole was located very close to the electrode (r~0.1d).
117
4. Effect of the circular boundary
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A current dipole accumulates a surface charge on the interface between the circular
119
plastic wall and the water, thus we also need to consider the effect of the circular
120
boundary on our measurements. Let us treat our tank as a circle in a two-dimensional
121
space, and use the method of image charges since the 2D approximation is shown to work
122
well in the previous section. Let us consider a current source I oriented inside of the
123
circular region. The inside region of the circular boundary has a conductivity of  1 and a
124
permittivity of  1 ; and the outside region has a conductivity of  2 and a permittivity of
125
 2 . The aim is to find the magnitudes and the locations of the image sources
126
corresponding to I. Since image sources cannot be located where the field is evaluated,
127
we must separately find the image sources inside and outside of the circular region. To
128
simplify our derivation, the length unit was normalized to the radius of the aquarium,
129
such that the radius of the aquarium was set to one.
130
Case 1: Field location inside of the circular region (r<1):
131
We must place an image source I1 outside of the region to compute the potential inside of
132
the circular region (see Fig. S2A). According to the cosine law, the distance from the
133
current sources to the field point r is:
134
r '  r 2  b2  2rb cos ,
r1  r 2  h2  2rh cos .
(S4.1)
135
From Eq (6) of the main text, the net potential due to the current source I and its image
136
source I1 is:
137
Vin  r   
I
21
ln  r ' 
I1
21
ln  r1 .
(S4.2)
138
7
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Case 2. Field location outside of the circular region (r > 1):
140
We must place an image source I3 and replace the original current source I with I2 inside
141
of the circular region [56] to compute the potential outside of the region (see Fig. S2B).
Vout  r   
142
I2
2 2
ln  r ' 
I3
2 2
ln  r   .
(S4.3)
143
The constant  is required to make the voltage continuous at the circular boundary such
144
that: Vin (r  1)  Vout  r  1 . Now the electric field must satisfy the boundary conditions
145
below:

d
d

Vout
,
 Ein  Eout  d Vin
d

r a
r a

d
 E    E    d V
  2 Vout
1 in
2 out
1
in

dr r  a
dr
146
(S4.4)
.
r a
147
First, let us find the parallel components of the electric field ( E ). From (S4.2) and
148
(S4.3):
149
150
 dVin
I h sin 
1 
Ib sin 


 2 1
 d
2

,
2

b

2
b
cos


1
h

2
h
cos


1


r a
1 

I 2 b sin 
1 

 dVout

.
2

 d r  a
2 2  b  2b cos   1 
(S4.5)
Equating the above two equations using (S4.4):
I h sin 
I b sin 
Ib sin 




 2 1
 1  2 2
2

,
 b  2b cos  1 h  2h cos  1
 b  2b cos  1
 2 Ib
 2 I1h
 1 I 2b
 2
 2
 2
.
b  2b cos  1 h  2h cos  1 b  2b cos  1
2 
151
152
153
154
(S4.6)
Rationalizing (S4.6) yields:
 2 Ib  h2  2h cos  1   2 I1h  b2  2b cos  1  1I 2b  h2  2h cos  1.
(S4.7)
Separating cosine dependent and independent terms:
8
 2b  h2  1 I   2 h  b2  1 I1  1b  h2  1 I 2   2bh cos    2 I   2 I1  1 I 2   0.


155
156
The equation above must be satisfied for any choice of  , thus:
 2 I   2 I1  1I 2  0  1I 2   2  I  I1 .
157
158
(S4.8)
(S4.9)
Substituting (S4.9) into (S4.8):
 2b  h 2  1 I   2 h  b 2  1 I1   2b  h 2  1  I  I1   0,
159
  2 I1  hb 2  h  bh 2  b   0  b 
1
1
h .
b
h
(S4.10)
160
The solution to (S4.10) can be found using the quadratic formula, yielding: h  b,1/ b .
161
However, image sources cannot be located in the same region of the real current source,
162
thus:
1
h .
b
163
(S4.12)
164
Now, let us find the perpendicular component of the electric field ( E  ). From (S4.2) and
165
(S4.3):
166
167
168
169
 dVin
I  h cos   1 
  I  b cos   1
 1  2
 21
1
,
 dr r  a 2 1  b  2b cos   1 h  2h cos   1 


  I  b cos   1
 dVout
 2  22
 I3  .
 2 dr
2 2  b  2b cos   1
r a


(S4.13)
Equating the above equations according to (S4.4):
I  h cos  1   2  I 2  b cos   1

1  I  b cos  1
 21
 I3  .
 2
  2
 1  b  2b cos  1 h  2h cos  1   2  b  2b cos  1

(S4.14)
Substituting (S4.12) into (S4.14):
9
 I  b cos   1
I1  b cos   b 2  
 I  b cos   1


 I3  ,
   1 2  22
2
2
 b  2b cos   1

 b  2b cos   1 b  2b cos   1 
 2 1 
170




  2 1 I  b cos   1   2 1 I1 b cos   b 2   1 2 I 2  b cos   1   1 2 I 3 b 2  2b cos   1 ,

(S4.15)

   2 1 I   2 1b 2 I1   1 2 I 2   1 2 b 2  1 I 3   b cos   2 1 I   2 1 I1   1 2 I 2  2 1 2 I 3   0.
171
Similarly, the equation above must be satisfied for any choice of  , thus:
172
 1 2

 I1    I 3 ,

2 1

 I   2 1  I  I  .
2
1

 1 2

173

 21  I  I1    1 2  I 2  2 I 3  ,


2
2
 21  I  b I1    1 2 I 2   b  1 I 3  0,



(S4.16)
Substituting (S4.16) into (S4.9) yields all the image sources:
1   2
I,
1   2
175
Observe that if  1
176
circular region, as expected from a body of water surrounded by a circular plastic wall.
177
5. Differential voltage due to an image dipole
178
Previously, we have found the potential due to a current source in 2D with the circular
179
boundary by using the method of image charges. We can apply our previous results to
180
determine the potential due to an ideal current dipole in 2D with the circular boundary
181
condition. The circular dielectric boundary creates an image current dipole outside of the
182
region, and we need to determine the potential difference between two electrodes due to
183
the ideal current dipole (see Fig. S2C). The potential due to a current dipole in 2D is:
Vdip  c
pr
r
2
I3 
 21 1   2
I.
 1 2 1   2
I1 
184
I2 
 2 2 2
I,
 1 1   2
174
(S4.17)
 2 , I 2 and I 3 vanishes and we obtain zero potential outside of the
 cp
r  
r
2
,
(S5.1)
185
where    cos  ,sin   is the unit vector of the dipole p . From (S5.1), the voltage
186
difference between the two electrodes e1 and e2 (see Fig. S2C) is:
10
187
 r   r   
Vdip , p  cp  1 2  2 2  ,
r2 
 r1
(S5.2)
2
188
where r1,2  1  2r cos1,2  r 2 . Now, let us compute the potential difference due to the
189
image dipole. The image dipole is created outside of the circular boundary due to the
190
reflection from the dielectric circular wall, and it has a magnitude of [56]:
191
192
193
 w   p 1 
p'  
p,
    r 2 
w
p


(S5.3)
with the unit vector      . The voltage difference due to the image dipole is:
r '   

 r2'       
Vdip , p '  cp  1

,
' 2
' 2


r
r
1
2


2
(S5.4)
2
194
where r1,2'  1  2cos1,2 / r  1/ r 2  r1,2 / r 2 . It can be shown that Vdip , p ' from (S5.4) can
195
be algebraically simplified to:
196
Vdip , p ' 
w   p
V ,
 w   p dip , p
(S5.5)
197
for any choice of a dipole location ( r , ) and the electrodes locations 1,2 . From (S5.5),
198
the net voltage difference is thus:
199
Vdip  Vdip , p  Vdip , p ' 
2 w
Vdip , p .
w   p
(S5.6)
200
In summary, the circular boundary simply rescales the differential dipole potential by a
201
constant factor. The voltage rescaling does not influence our dipole localization
202
algorithm, since it uses the relative signal intensities between multiple channels.
203
204
-end11
205
206
Figure S1. The method of image charges applied to the shallow body of water. (A) The
207
image currents of the current source I created by the top and the bottom dielectric
208
interfaces are shown up to the two first order reflections I0 and I1. (B) The net potential
209
(Vnet) due to the current source and its image currents is plotted as a function of the
210
number of reflections (nr). Vnet was measured at the electrode at a distance r=d, and
211
normalized to I / 4 w d (d: depth of water). The current source was located at the height
212
d/2. (C) The potential measured at the vertically oriented extended electrode was
213
determined by averaging the potentials measured at different heights. (D) The
214
numerically calculated potential of the vertically oriented electrode (Vdip) is plotted in
215
blue as a function of the normalized inverse distance (R/d)-1. The 2D ideal dipole voltage
216
approximation is shown in red.
12
217
218
Figure S2. The method of image charges applied to the side circular boundary. (A) The
219
current source (I) and its image source (I1) are shown for the field location inside of the
220
circular region. (B) Two image sources (I2, I3) are shown for the field location outside of
221
the circular region. (C) The image current dipole ( p ' ) location is shown to calculate the
222
differential potential between the electrode pair (e1, e2).
13