Code indexing in gitaly is broken and leads to code not being visible to the user. We work on the issue with highest priority.

Skip to content

Changes

Page history
manual latexmath authored by snuverink_j's avatar snuverink_j
Hide whitespace changes
Inline Side-by-side
opalt.asciidoc
View page @ 1b1c7032
......@@ -177,20 +177,20 @@ where
[latexmath]
++++
\mathcal{S}=\left(\begin{array}{ccc}
\cos\Theta & 0 & n\Theta \\
\cos\Theta & 0 & \sin\Theta \\
0 & 1 & 0 \\
-n\Theta & 0 & \cos\Theta
-\sin\Theta & 0 & \cos\Theta
\end{array}\right),
\quad
\mathcal{T}=\left(\begin{array}{ccc}
1 & 0 & 0 \\
0 & \cos\Phi & n\Phi \\
0 & -n\Phi & \cos\Phi
0 & \cos\Phi & \sin\Phi \\
0 & -\sin\Phi & \cos\Phi
\end{array}\right),
\quad
\mathcal{U}=\left(\begin{array}{ccc}
\cos\Psi & -n\Psi & 0 \\
n\Psi & \cos\Psi & 0 \\
\cos\Psi & -\sin\Psi & 0 \\
\sin\Psi & \cos\Psi & 0 \\
0 & 0 & 1
\end{array}\right).
++++
......@@ -704,7 +704,7 @@ potential is an analytic function and can be expanded as a power series
[latexmath]
++++
\tilde{A} (z) = \sum_{n=0}^{\infty} \kappa_n z^n, \quad \kappa_n = \lambda_n + i \mu_n\]]
\tilde{A} (z) = \sum_{n=0}^{\infty} \kappa_n z^n, \quad \kappa_n = \lambda_n + i \mu_n
++++
with latexmath:[\lambda_n, \mu_n] being real constants. It is
......@@ -713,8 +713,8 @@ latexmath:[(r, \varphi, s)]
[latexmath]
++++
\begin{aligned}
x & = r \cos \varphi \quad y = r n \varphi \\
z^n & = r^n ( \cos n \varphi + i n n \varphi )
x & = r \cos \varphi \quad y = r \sin \varphi \\
z^n & = r^n ( \cos n \varphi + i \sin n \varphi )
\end{aligned}
++++
......@@ -723,8 +723,8 @@ From the real and imaginary parts of equation () we obtain
[latexmath]
++++
\begin{aligned}
V(r, \varphi) & = \sum_{n=0}^{\infty} r^n ( \mu_n \cos n \varphi + \lambda_n n n \varphi ) \\
A_s (r, \varphi) & = \sum_{n=0}^{\infty} r^n ( \lambda_n \cos n \varphi - \mu_n n n \varphi )
V(r, \varphi) & = \sum_{n=0}^{\infty} r^n ( \mu_n \cos n \varphi + \lambda_n \sin n \varphi ) \\
A_s (r, \varphi) & = \sum_{n=0}^{\infty} r^n ( \lambda_n \cos n \varphi - \mu_n \sin n \varphi )
\end{aligned}
++++
......@@ -735,8 +735,8 @@ components, respectively
[latexmath]
++++
\begin{aligned}
B_{\varphi} & = - \frac{1}{r} \frac{\partial V}{\partial \varphi} = - \sum_{n=0}^{\infty} n r^{n-1} ( \lambda_n \cos n \varphi - \mu_n n n \varphi ) \\
B_r & = - \frac{\partial V}{\partial r} = - \sum_{n=0}^{\infty} n r^{n-1} ( \mu_n \cos n \varphi + \lambda_n n n \varphi )
B_{\varphi} & = - \frac{1}{r} \frac{\partial V}{\partial \varphi} = - \sum_{n=0}^{\infty} n r^{n-1} ( \lambda_n \cos n \varphi - \mu_n \sin n \varphi ) \\
B_r & = - \frac{\partial V}{\partial r} = - \sum_{n=0}^{\infty} n r^{n-1} ( \mu_n \cos n \varphi + \lambda_n \sin n \varphi )
\end{aligned}
++++
Furthermore, we introduce the normal multipole
......@@ -752,8 +752,8 @@ b_n = - \frac{n \lambda_n}{B_0} r_0^{n-1} \qquad a_n = \frac{n \mu_n}{B_0} r_0^{
[latexmath]
++++
\begin{aligned}
B_{\varphi}(r, \varphi) & = B_0 \sum_{n=1}^{\infty} ( b_n \cos n \varphi+ a_n n n \varphi ) \left( \frac{r}{r_0} \right) ^{n-1} \\
B_r (r, \varphi) & = B_0 \sum_{n=1}^{\infty} ( -a_n \cos n \varphi+ b_n n n \varphi ) \left( \frac{r}{r_0} \right) ^{n-1}
B_{\varphi}(r, \varphi) & = B_0 \sum_{n=1}^{\infty} ( b_n \cos n \varphi+ a_n \sin n \varphi ) \left( \frac{r}{r_0} \right) ^{n-1} \\
B_r (r, \varphi) & = B_0 \sum_{n=1}^{\infty} ( -a_n \cos n \varphi+ b_n \sin n \varphi ) \left( \frac{r}{r_0} \right) ^{n-1}
\end{aligned}
++++
To obtain a model for the fringe field of a
......@@ -763,7 +763,7 @@ field is
[latexmath]
++++
\begin{aligned}
V & = \sum_{n=1}^{\infty} V_n (r,z) n n \varphi \\
V & = \sum_{n=1}^{\infty} V_n (r,z) \sin n \varphi \\
V_n & = \sum_{k=0}^{\infty} C_{n,k}(z) r^{n+2k}
\end{aligned}
++++
......@@ -778,7 +778,7 @@ coefficients
[latexmath]
++++
\[V_n = \sum_{k=0}^{\infty} C_{n,k}(z) r^{n+2k}
V_n = \sum_{k=0}^{\infty} C_{n,k}(z) r^{n+2k}
++++
[latexmath]
......@@ -805,13 +805,14 @@ Therefore
[latexmath]
++++
V_n = - \left( \sum_{k=0}^{\infty} (-1)^{k+1} \frac{n!}{2^{2k} k! (n+k)!} C_{n, 0}^{(2k)}(z) r^{2k} \right) r^n
++++
The transverse components of the field are
[latexmath]
++++
\begin{aligned}
B_r & = \sum_{n=1}^{\infty} g_{rn} r^{n-1} n n \varphi \\
B_r & = \sum_{n=1}^{\infty} g_{rn} r^{n-1} \sin n \varphi \\
B_{\varphi} & = \sum_{n=1}^{\infty} g_{\varphi n} r^{n-1} \cos n \varphi
\end{aligned}
++++
......@@ -834,7 +835,7 @@ expressed in a similar form
[latexmath]
++++
\begin{aligned}
B_z & = - \frac{\partial V}{\partial z} = \sum_{n=1}^{\infty} g_{zn} r^n n n \varphi \\
B_z & = - \frac{\partial V}{\partial z} = \sum_{n=1}^{\infty} g_{zn} r^n \sin n \varphi \\
g_{zn} & = \sum_{k=0}^{\infty} (-1)^{k+1} \frac{n!}{2^{2k} k! (n+k)!} C_{n,0}^{2k+1} r^{2k}
\end{aligned}
++++
......@@ -845,9 +846,9 @@ latexmath:[g_{rn}, g_{\varphi n}, g_{zn}] are obtained from
[latexmath]
++++
\begin{aligned}
B_{r,n} & = - \frac{\partial V_n}{\partial r} n n \varphi = g_{rn} r^{n-1} n n \varphi \\
B_{r,n} & = - \frac{\partial V_n}{\partial r} \sin n \varphi = g_{rn} r^{n-1} \sin n \varphi \\
B_{\varphi,n} & = - \frac{n}{r} V_n \cos n \varphi = g_{\varphi n} r^{n-1} \cos n \varphi \\
B_{z,n} & = - \frac{\partial V_n}{\partial z} n n \varphi = g_{zn} r^{n} n n \varphi
B_{z,n} & = - \frac{\partial V_n}{\partial z} \sin n \varphi = g_{zn} r^{n} \sin n \varphi
\end{aligned}
++++
......@@ -860,6 +861,7 @@ profile on the central axis is the k-parameter Enge function
C_{n,0}(z) & = \frac{G_0}{1+exp[P(d(z))]}, \quad G_0 = \frac{B_0}{r_0^{n-1}} \\
P(d) & = C_0 + C_1 \left( \frac{d}{\lambda} \right) + C_2 \left( \frac{d}{\lambda} \right)^2 + \dots + C_{k-1} \left( \frac{d}{\lambda} \right)^{k-1}
\end{aligned}
++++
where latexmath:[d(z)] is the distance to the
field boundary and latexmath:[\lambda] characterizes the fringe field
......@@ -875,50 +877,122 @@ We consider the Frenet-Serret coordinate system
latexmath:[ ( \hat{\mathbf{x}}, \hat{\mathbf{s}}, \hat{\mathbf{z}} )]
with the radius of curvature latexmath:[ \rho ] constant and the scale
factor latexmath:[ h_s = 1 + x/ \rho]. A conversion to these
coordinates implies that latexmath:[\[\begin{aligned}
coordinates implies that
[latexmath]
++++
\begin{aligned}
\nabla \cdot \mathbf{B} & = \frac{1}{h_s} \left[ \frac{\partial (h_s B_x )}{\partial x} + \frac{\partial B_s}{\partial s} + \frac{\partial (h_s B_z )}{\partial z} \right] \\
\nabla \times \mathbf{B} & = \frac{1}{h_s} \left[ \frac{\partial B_z}{\partial s} - \frac{\partial (h_s B_s )}{\partial z} \right] \hat{\mathbf{x}} + \left[ \frac{\partial B_x}{\partial z} - \frac{\partial B_z}{\partial x} \right] \hat{\mathbf{s}} + \frac{1}{h_s} \left[ \frac{\partial (h_s B_s)}{\partial x} - \frac{\partial B_x}{\partial s} \right] \hat{\mathbf{z}} \nonumber
\end{aligned}\]] To simplify the problem, consider multipoles with
\end{aligned}
++++
To simplify the problem, consider multipoles with
mid-plane symmetry, i.e.
latexmath:[\[b_z (z) = B_z (-z) \qquad B_x (z) = - B_x (-z) \qquad B_s (z) = - B_s (-z)\]]
The most general form of the expansion is latexmath:[\[\begin{aligned}
[latexmath]
++++
b_z (z) = B_z (-z) \qquad B_x (z) = - B_x (-z) \qquad B_s (z) = - B_s (-z)
++++
The most general form of the expansion is
[latexmath]
++++
\begin{aligned}
B_z & = \sum_{i,k=0}^{\infty} b_{i,k} x^i z^{2k} \label{eq:01} \\
B_x & = z \sum_{i,k=0}^{\infty} a_{i,k} x^i z^{2k} \label{eq:02}\\
B_s & = z \sum_{i,k=0}^{\infty} c_{i,k} x^i z^{2k} \label{eq:03}
\end{aligned}\]] Maxwell’s equations
\end{aligned}
++++
Maxwell’s equations
latexmath:[ \nabla \cdot \mathbf{B} = 0 ] and
latexmath:[ \nabla \times \mathbf{B} = 0 ] in the above coordinates
yield
latexmath:[\[\frac{\partial}{\partial x} \left( (1+x/ \rho) B_x \right) + \frac{\partial B_s}{\partial s} + (1+x/ \rho) \frac{\partial B_z}{\partial z} = 0 \label{eq:21}\]]
latexmath:[\[\begin{aligned}
[latexmath]
++++
\frac{\partial}{\partial x} \left( (1+x/ \rho) B_x \right) + \frac{\partial B_s}{\partial s} + (1+x/ \rho) \frac{\partial B_z}{\partial z} = 0 \label{eq:21}
++++
[latexmath]
++++
\begin{aligned}
\frac{\partial B_z}{\partial s} & = (1+x/ \rho) \frac{\partial B_s}{\partial z} \label{eq:22} \\
\frac{\partial B_x}{\partial z} & = \frac{\partial B_z}{\partial s} \label{eq:23} \\
\frac{\partial B_x}{\partial s} & = \frac{\partial}{\partial x} \left( (1+x/ \rho) B_s \right) \label{eq:24}
\end{aligned}\]] The substitution of ([eq:01]), ([eq:02]) and
\end{aligned}
++++
The substitution of ([eq:01]), ([eq:02]) and
([eq:03]) into Maxwell’s equations allows for the derivation of
recursion relations. ([eq:23]) gives
latexmath:[\[\sum_{i,k=0}^{\infty} a_{i,k} (2k+1) x^i z^{2k} = \sum_{i,k=0}^{\infty} b_{i,k} i x^{i-1} z^{2k}\]]
[latexmath]
++++
\sum_{i,k=0}^{\infty} a_{i,k} (2k+1) x^i z^{2k} = \sum_{i,k=0}^{\infty} b_{i,k} i x^{i-1} z^{2k}
++++
Equating the powers in latexmath:[x^i z^{2k}]
latexmath:[\[a_{i,k} = \frac{i+1}{2k+1} b_{i+1,k} \label{eq:11}\]] A
similar result is obtained from ([eq:24]) latexmath:[\[\begin{aligned}
[latexmath]
++++
a_{i,k} = \frac{i+1}{2k+1} b_{i+1,k} \label{eq:11}
++++
A similar result is obtained from ([eq:24])
[latexmath]
++++
\begin{aligned}
\sum_{i,k=0}^{\infty} \partial_s b_{i,k} x^i z^{2k} & = \left( 1+ \frac{x}{\rho} \right) \sum_{i,k=0}^{\infty} c_{i,k} (2k+1) x^i z^{2k} \\
\Rightarrow c_{i,k} & + \frac{1}{\rho} c_{i-1,k} = \frac{1}{2k+1} \partial_s b_{i,k} \label{eq:12}
\end{aligned}\]] The last equation from
\end{aligned}
++++
The last equation from
latexmath:[\nabla \times \mathbf{B} = 0] should be consistent with the
two recursion relations obtained. ([eq:22]) implies
latexmath:[\[\sum_{i,k=0}^{\infty} \left[ \frac{i+1}{\rho} c_{i,k} x^i + c_{i,k} i x^{i-1} \right] z^{k+1} = \sum_{i,k=0}^{\infty} \partial_s a_{i,k} x^i z^{2k}\]]
latexmath:[\[\Rightarrow \frac{\partial_s a_{i,k}}{i+1} = \frac{1}{\rho} c_{i,k} + c_{i+1,k}\]]
[latexmath]
++++
\sum_{i,k=0}^{\infty} \left[ \frac{i+1}{\rho} c_{i,k} x^i + c_{i,k} i x^{i-1} \right] z^{k+1} = \sum_{i,k=0}^{\infty} \partial_s a_{i,k} x^i z^{2k}
++++
[latexmath]
++++
\Rightarrow \frac{\partial_s a_{i,k}}{i+1} = \frac{1}{\rho} c_{i,k} + c_{i+1,k}
++++
This results follows directly from ([eq:11]) and ([eq:12]); therefore
the relations are consistent. Furthermore, the last required relations
is obtained from the divergence of *B*
latexmath:[\[\sum_{i,k=0}^{\infty} \left[ \frac{a_{i,k} x^i z^{2k+1}}{\rho} + i a_{i,k} x^{i-1} z^{2k+1} + \frac{i a_{i,k} x^i z^{2k+1}}{\rho} + \partial_s c_{i,k} x^i z^{2k+1} + 2kb_{i,k}x^i z^{2k-1} \right] = 0 \nonumber\]]
latexmath:[\[\Rightarrow \partial_s c_{i,k} + \frac{2(k+1)}{\rho} b_{i-1,k+1} + 2(k+1) b_{i,k+1} + \frac{1}{\rho} a_{i,k} + (i+1) a_{i+1,k} + \frac{1}{\rho} a_{i,k} = 0 \nonumber\]]
[latexmath]
++++
\sum_{i,k=0}^{\infty} \left[ \frac{a_{i,k} x^i z^{2k+1}}{\rho} + i a_{i,k} x^{i-1} z^{2k+1} + \frac{i a_{i,k} x^i z^{2k+1}}{\rho} + \partial_s c_{i,k} x^i z^{2k+1} + 2kb_{i,k}x^i z^{2k-1} \right] = 0 \nonumber
++++
[latexmath]
++++
\Rightarrow \partial_s c_{i,k} + \frac{2(k+1)}{\rho} b_{i-1,k+1} + 2(k+1) b_{i,k+1} + \frac{1}{\rho} a_{i,k} + (i+1) a_{i+1,k} + \frac{1}{\rho} a_{i,k} = 0 \nonumber
++++
Using the relation ([eq:11]) to replace the latexmath:[a] coefficients
with latexmath:[b]’s we arrive at
latexmath:[\[\partial_s c_{i,k} + \frac{(i+1)^2}{\rho (2k+1)} b_{i+1,k} + \frac{(i+1)(i+2)}{2k+1} b_{i+2,k} + \frac{2(k+1)}{\rho} b_{i-1,k+1} + 2(k+1) b_{i,k+1} = 0\]]
[latexmath]
++++
\partial_s c_{i,k} + \frac{(i+1)^2}{\rho (2k+1)} b_{i+1,k} + \frac{(i+1)(i+2)}{2k+1} b_{i+2,k} + \frac{2(k+1)}{\rho} b_{i-1,k+1} + 2(k+1) b_{i,k+1} = 0
++++
All the coefficients above can be determined recursively provided the
field latexmath:[B_z] can be measured at the mid-plane in the form
latexmath:[\[B_z(z=0) = B_{0,0} + B_{1,0}x + B_{2,0} x^2 + B_{3,0} x^3 + \dots\]]
[latexmath]
++++
B_z(z=0) = B_{0,0} + B_{1,0}x + B_{2,0} x^2 + B_{3,0} x^3 + \dots
++++
where latexmath:[B_{i,0}] are functions of latexmath:[s] and they
can model the fringe field for each multipole term latexmath:[x^n]. As
an example, for a dipole magnet, the latexmath:[B_{1,0}] function can
......@@ -939,7 +1013,10 @@ derivative latexmath:[\frac{\partial}{\partial s}] is applied to the
scale factor latexmath:[h_s]. If the radius of curvature is set to be
proportional to the dipole filed observed by some reference particle
that stays in the centre of the dipole
latexmath:[\[\rho (s) \propto B(z=0, x=0, s) = B_x (z=0,x=0) = b_{0,0}(s)\]]
[latexmath]
++++
rho (s) \propto B(z=0, x=0, s) = B_x (z=0,x=0) = b_{0,0}(s)
++++
[[fringe-field-of-a-multipole]]
Fringe field of a multipole
......@@ -948,7 +1025,13 @@ Fringe field of a multipole
_This is a different, more compact treatment_ The derivation is more
clear if we gather the variables together in functions. We assume again
mid-plane symmetry and that the z-component of the field in the
mid-plane has the form latexmath:[\[B_z (x, z=0, s) = T(x) S(s)\]] where
mid-plane has the form
[latexmath]
++++
B_z (x, z=0, s) = T(x) S(s)
++++
where
latexmath:[T(s)] is the transverse field profile and
latexmath:[S(s)] is the fringe field. One of the requirements of the
symmetry is that latexmath:[B_z(z) = B_z(-z)] which using a scalar
......@@ -956,37 +1039,93 @@ potential latexmath:[\psi] requires
latexmath:[\frac{\partial \psi}{\partial z}] to be an even function in
z. Therefore, latexmath:[\psi] is an odd function in z and can be
written as
latexmath:[\[\psi = z f_0(x,s) + \frac{z^3}{3!} f_1(x,s) + \frac{z^5}{5!} f_3(x,s) + \dots\]]
[latexmath]
++++
\psi = z f_0(x,s) + \frac{z^3}{3!} f_1(x,s) + \frac{z^5}{5!} f_3(x,s) + \dots
++++
The given transverse profile requires that
latexmath:[f_0(x,s) = T(x)S(s)], while latexmath:[\nabla^2 \psi = 0]
follows from Maxwell’s equations as usual, more explicitly
latexmath:[\[\frac{\partial}{\partial x} \left( h_s \frac{\partial \psi}{\partial x} \right) + \frac{\partial}{\partial s} \left( \frac{1}{h_s} \frac{\partial \psi}{\partial s} \right) + \frac{\partial}{\partial z} \left( h_s \frac{\partial \psi}{\partial z} \right) = 0\]]
[latexmath]
++++
\frac{\partial}{\partial x} \left( h_s \frac{\partial \psi}{\partial x} \right) + \frac{\partial}{\partial s} \left( \frac{1}{h_s} \frac{\partial \psi}{\partial s} \right) + \frac{\partial}{\partial z} \left( h_s \frac{\partial \psi}{\partial z} \right) = 0
++++
For a straight multipole latexmath:[h_s = 1]. Laplace’s equation
becomes
latexmath:[\[\sum_{n=0} \frac{z^{2n+1}}{(2n+1)!} \left[ \partial_x^2 f_n(x,s) + \partial_s^2 f_n(x,s) \right] + \sum_{n=1} f_n(x,s) \frac{z^{n-1}}{(n-1)!} = 0\]]
[latexmath]
++++
\sum_{n=0} \frac{z^{2n+1}}{(2n+1)!} \left[ \partial_x^2 f_n(x,s) + \partial_s^2 f_n(x,s) \right] + \sum_{n=1} f_n(x,s) \frac{z^{n-1}}{(n-1)!} = 0
++++
By equating powers of latexmath:[z] we obtain the recursion relation
latexmath:[\[f_{n+1}(x,s) = - \left( \partial_x^2 + \partial_s^2 \right) f_n(x,s)\]]
[latexmath]
++++
f_{n+1}(x,s) = - \left( \partial_x^2 + \partial_s^2 \right) f_n(x,s)
++++
The general expression for any latexmath:[f_n(x,s)] is then obtained
from the mid-plane field by latexmath:[\[\begin{aligned}
from the mid-plane field by
[latexmath]
++++
\begin{aligned}
f_n(x,s) & = (-1)^n \left( \partial_x^2 + \partial_s^2 \right)^n f_0(x,s) \\
f_n(x,s) & = (-1)^n \sum_{i=0}^n \binom{n}{i}T^{(2i)}(x) S^{(2n-2i)}(s)
\end{aligned}\]] For a curved multipole of constant radius
\end{aligned}
++++
For a curved multipole of constant radius
latexmath:[h_s = 1 + \frac{x}{\rho} \quad \text{with} \quad \rho = const.]
The corresponding Laplace’s equation is
latexmath:[\[\left( \frac{1}{\rho h_s} \partial_x + \partial_x^2 + \partial_z^2 + \frac{\partial_s^2}{h_s^2} \right) \psi = 0\]]
[latexmath]
++++
\left( \frac{1}{\rho h_s} \partial_x + \partial_x^2 + \partial_z^2 + \frac{\partial_s^2}{h_s^2} \right) \psi = 0
++++
Again we substitute with the functional form of the potential to get the
recursion latexmath:[\[\begin{aligned}
recursion
[latexmath]
++++
\begin{aligned}
f_{n+1}(x,s) & = - \left[ \frac{1}{\rho + x} \partial_x + \partial_x^2 + \frac{\partial_s^2}{(1+x/ \rho)^2} \right] f_n(x,s) \\
f_{n+1}(x,s) & = (-1)^n \left[ \frac{1}{\rho + x} \partial_x + \partial_x^2 + \frac{\partial_s^2}{(1+x/ \rho)^2} \right]^n f_0(x,s)
\end{aligned}\]] Finally consider what changes for
\end{aligned}
++++
Finally consider what changes for
latexmath:[\rho = \rho (s)]. Laplace’s equation is
latexmath:[\[\left[ \frac{1}{\rho h_s} \partial_x + \partial_x^2 + \partial_z^2 + \frac{\partial_s^2}{h_s^2} + \frac{x}{\rho^2 h_s^3} (\partial_s \rho) \partial_s \right] \psi = 0\]]
[latexmath]
++++
\left[ \frac{1}{\rho h_s} \partial_x + \partial_x^2 + \partial_z^2 + \frac{\partial_s^2}{h_s^2} + \frac{x}{\rho^2 h_s^3} (\partial_s \rho) \partial_s \right] \psi = 0
++++
The last step is again the substitution to get
latexmath:[\[\begin{aligned}
[latexmath]
++++
\begin{aligned}
f_{n+1}(x,s) & = - \left[ \frac{\partial_x}{\rho h_s} + \partial_x^2 + \partial_z^2 + \frac{1}{h_s^2}\partial_s^2 + \frac{x}{\rho^2 h_s^3} (\partial_s \rho) \partial_s \right] f_n(x,s) \\
f_{n}(x,s) & = (-1)^n \left[ \frac{\partial_x}{\rho h_s} + \partial_x^2 + \partial_z^2 + \frac{\partial_s^2}{h_s^2} + \frac{x}{\rho^2 h_s^3} (\partial_s \rho) \partial_s \right]^n f_0(x,s) \label{eq:40}
\end{aligned}\]] If the radius of curvature is proportional to the
\end{aligned}
++++
If the radius of curvature is proportional to the
dipole field in the centre of the multipole (the dipole component of the
transverse field is a constant latexmath:[T_{dipole}(x) = B_0] and
latexmath:[\[\rho(s) = B_0 \times S(s)\]] This expression can be
[latexmath]
++++
\rho(s) = B_0 \times S(s)
++++
This expression can be
replaced in ([eq:40]) to obtain a more explicit equation.