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manual latexmath authored by snuverink_j's avatar snuverink_j
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......@@ -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.