Analytical calculation of the stray field

Since the phase shift of the cantilever attached to the sharp tip in a magnetic force microscope (Sáenz et al., 1987) when measured corresponds to the measured signal (McVitie et al., 2001), the second derivative of the stray field (Rugar et al., 1990, Hug et al., 1998) is proportional to the rate of change of the force on the tip (Barthelmeß et al., 2004).

If the second derivative is evaluated at a fixed height above a demagnetising energy dataset of micromagnetic simulation results then this is comparable to a magnetic force microscope. This derivative is presented in this appendix.

The dipolar energy between two points in a magnetic system, assuming each discrete cell is a dipole, can be considered to be

$$E_{\mathrm{pot}} = {\mu_0 } {\ensuremath{\mathbf{m}}\cdot \ensuremath{\mathbf{m'}} -... ...thbf{r'}}\vert^3} \forall \ensuremath{\mathbf{r}} \neq \ensuremath{\mathbf{r}}'$$ (A.1)

where $\ensuremath{\mathbf{e}}$ is the unit vector of $\ensuremath{\mathbf{r}}$.

For fixed $\ensuremath{\mathbf{m}}'$ (such as that in the magnetic tip in magnetic force microscopy), the effective field is

$$\ensuremath{\mathbf{H}}_d = {\ensuremath{\mathbf{m}} - 3(\ensuremath{\mathbf{m}}\cdot\ensurem... ...athbf{e}} \over \vert\ensuremath{\mathbf{r}} - \ensuremath{\mathbf{r'}}\vert^3}$$ (A.2)

because $E = -\mu_0\ensuremath{\mathbf{m}}\cdot\ensuremath{\mathbf{H}}_d$, noting that $\ensuremath{\mathbf{m}}$ is located at $\ensuremath{\mathbf{r}}$ and $\ensuremath{\mathbf{m}}'$ -- the tip -- is located at $\ensuremath{\mathbf{r}}'$. We define:

$$\ensuremath{\mathbf{e}} = {\ensuremath{\mathbf{r}} - \ensuremath{\mathbf{r}}' \over \vert\ensuremath{\mathbf{r}} - \ensuremath{\mathbf{r}}'\vert}$$ (A.3)

For the MFM data:

$${\partial^2E_{\mathrm{pot}} \over \partial z'^2} \propto {\partial^2\ensuremath{\mathbf{H}}_d^z \over \partial z'^2}$$ (A.4)

Expand A.1 by substituting A.3:

$$E_{\mathrm{pot}} = {\mu_0 } { {\ensuremath{\mathbf{m}}\cdot\ensuremath{\mathbf{m}}' ... ...f{r}}')) \over \vert\ensuremath{\mathbf{r}} - \ensuremath{\mathbf{r}}'\vert^5}}$$ (A.5)

We precompute some expressions:

$$g(\ensuremath{\mathbf{r}},\ensuremath{\mathbf{r}}') \equiv \vert\ensuremath{\mathbf{r}} - \ensuremath{\mathbf{r}}'\vert$$ (A.6)

$$= \sqrt{(x-x')^2 + (y-y')^2 + (z-z')^2}$$ (A.7)

$$\bigl(= \sqrt{(\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}')^2}\bigr)$$ (A.8)

$${\partial g(\ensuremath{\mathbf{r}}, \ensuremath{\mathbf{r}}') \over \partial z'} = {1 \over {2 \sqrt{(\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}')^2}}}\cdot{2(-1)(z-z')}$$ (A.9)

$$= {1 \over {\sqrt{(\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}')^2}}}\cdot{-1(z-z')}$$ (A.10)

$$= {-(z-z') \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert}$$ (A.11)

$$= {z'-z \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert}$$ (A.12)

$$f \equiv g^3$$ (A.13)

$$= \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^3$$ (A.14)

$${\partial f \over \partial z'} = {\partial f \over \partial g}\cdot {\partial g \over \partial z'}$$ (A.15)

$$= 3 \cdot g(\ensuremath{\mathbf{r}}, \ensuremath{\mathbf{r}}')^2\cd... ...artial g \over \partial z' }(\ensuremath{\mathbf{r}}, \ensuremath{\mathbf{r}}')$$ (A.16)

$$= 3 \vert\ensuremath{\mathbf{r}} - \ensuremath{\mathbf{r}}'\vert^2 \cdot {z' - z \over \vert\ensuremath{\mathbf{r}} - \ensuremath{\mathbf{r'}}\vert}$$ (A.17)

$$= 3 \vert\ensuremath{\mathbf{r}} - \ensuremath{\mathbf{r}}'\vert(z' - z)$$ (A.18)

$$\tilde{f} \equiv {1 \over f}$$ (A.19)

$$= {1 \over g^3}$$ (A.20)

$$= {1 \over \vert\ensuremath{\mathbf{r}} - \ensuremath{\mathbf{r}}'\vert^3}$$ (A.21)

$$= g^{-3}$$ (A.22)

$${\partial \tilde{f} \over \partial z'} = {\partial \tilde{f} \over \partial g} {\partial g \over \partial z'}$$ (A.23)

$$= -3g^{-4}\cdot {\partial g \over \partial z'}$$ (A.24)

$$= -3 {1 \over \vert\ensuremath{\mathbf{r}} - \ensuremath{\mathbf{r}... ...\cdot {z'-z \over \vert\ensuremath{\mathbf{r}} - \ensuremath{\mathbf{r}}'\vert}$$ (A.25)

$$= -3 {z'-z \over \vert\ensuremath{\mathbf{r}} - \ensuremath{\mathbf{r}}'\vert^5}$$ (A.26)

$$h \equiv g^5$$ (A.27)

so

$${\partial h \over \partial z'} = {\partial h \over \partial g} {\partial g \over \partial z'}$$ (A.28)

$$= 5g^4\cdot {z'-z \over \vert\ensuremath{\mathbf{r}} - \ensuremath{\mathbf{r}}'\vert}$$ (A.29)

$$= 5 \vert\ensuremath{\mathbf{r}} - \ensuremath{\mathbf{r}}'\vert^3\cdot (z'-z)$$ (A.30)

$$\tilde{h} \equiv g^{-5}$$ (A.31)

so

$${\partial \tilde{h} \over \partial z'} = {\partial \tilde{h} \over \partial g} {\partial g \over \partial z'}$$ (A.32)

$$= -5g^{-6} {z'-z \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert}$$ (A.33)

$$= -5 {z'-z \over \vert\ensuremath{\mathbf{r}} - \ensuremath{\mathbf{r}}'\vert^7}$$ (A.34)

Looking at the mixed terms in A.5:

$$\Psi(\ensuremath{\mathbf{r}}, \ensuremath{\mathbf{r}}') \equiv \bigl[\ensuremath{\mathbf{m}}\cdot (\ensuremath{\mathbf{r}} - \en... ...ath{\mathbf{m}}'\cdot(\ensuremath{\mathbf{r}} - \ensuremath{\mathbf{r}}')\bigr]$$ (A.35)

$$= \bigl[m_x(x-x') + m_y(y-y') + m_z(z-z')\bigr]$$

$$\cdot \bigl[m'_x(x-x') + m'_y(y-y') + m'_z(z-z')\bigr]$$ (A.36)

$${\partial \Psi \over \partial z'} = \bigl[\ensuremath{\mathbf{m}}(\ensuremath{\mathbf{r}}-\ensuremath... ...dot{\partial \over \partial z'}\bigl(m'_x(x-x') + m'_y(y-y') + m'_z(z-z')\bigr)$$

$$+ \bigl[\ensuremath{\mathbf{m}}'(\ensuremath{\mathbf{r}}-\ensurem... ...igr]\cdot {\partial \over \partial z'}\bigl(m_x(x-x')+m_y(y-y')+m_z(z-z')\bigr)$$

$$= \ensuremath{\mathbf{m}}(\ensuremath{\mathbf{r}} - \ensuremath{\ma... ...emath{\mathbf{m}}'(\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}')\cdot(-m_z)$$ (A.37)

$$\phi \equiv {\Psi \over \vert\ensuremath{\mathbf{r}} - \ensuremath{\mathbf{r}}'\vert^5}$$ (A.38)

$$= \Psi \over h$$ (A.39)

$${\partial \phi \over \partial z'} = { h \cdot {\partial \Psi \over \partial z'} - \Psi \cdot {\partial h \over \partial z'} \over h^2}$$ (A.40)

$$= {g^5 \cdot {\partial \Psi \over \partial z'} - \Psi \cdot 5g^3(z'-z) \over g^{10}}$$ (A.41)

$$= { {\partial \Psi \over \partial z'} \over g^5} - {5 \Psi (z'-z) \over g^7}$$ (A.42)

$$= {-\ensuremath{\mathbf{m}}\cdot({\ensuremath{\mathbf{r}} - \ensure... ...{r}}')m_z \over \vert\ensuremath{\mathbf{r}} - \ensuremath{\mathbf{r}}'\vert^5}$$

$$- { 5 (\ensuremath{\mathbf{m}} \cdot (\ensuremath{\mathbf{r}} - \... ...'))(z'-z) \over \vert\ensuremath{\mathbf{r}} - \ensuremath{\mathbf{r}}'\vert^7}$$ (A.43)

Combining the above to give the first derivative of the dipolar interaction energy as A.5:

$$E_{\mathrm{pot}} = {\mu_0} \Bigl( {\ensuremath{\mathbf{m}}\cdot\ensuremath{\mathbf{m... ...}}')) \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^5}\Bigr)$$ (A.44)

With respect to $z'$:

$${\partial E_{\mathrm{pot}} \over \partial z'} = {\mu_0} \left ( {\ensuremath{\mathbf{m}}\cdot\ensuremath{\mathbf{... ...) \over \vert\ensuremath{\mathbf{r}} - \ensuremath{\mathbf{r}}'\vert^5} \right.$$

$$\left. + {3 \ensuremath{\mathbf{m}} \cdot (\ensuremath{\mathbf{r}... ...z \over \vert\ensuremath{\mathbf{r}} - \ensuremath{\mathbf{r}}'\vert^5} \right.$$

$$\left. + 3 {5(\ensuremath{\mathbf{m}}\cdot(\ensuremath{\mathbf{r}... ... \over \vert\ensuremath{\mathbf{r}} - \ensuremath{\mathbf{r}}'\vert^7} \right )$$ (A.45)

$$= {\mu_0} (\mathcal{A} + \mathcal{D} + \mathcal{E} + \mathcal{F})$$ (A.46)

$$\mathcal{A} \equiv {3\ensuremath{\mathbf{m}}\cdot\ensuremath{\mathbf{m}}'(z-z') \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^5}$$ (A.47)

$$\mathcal{D} \equiv {3\ensuremath{\mathbf{m}}(\ensuremath{\mathbf{r}}-\ensuremath{\ma... ...)\cdot m'_z \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^5}$$ (A.48)

$$\mathcal{E} \equiv {3\ensuremath{\mathbf{m}}'(\ensuremath{\mathbf{r}}-\ensuremath{\m... ...')\cdot m_z \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^5}$$ (A.49)

$$\mathcal{F} \equiv {15(\ensuremath{\mathbf{m}}\cdot(\ensuremath{\mathbf{r}}-\ensurem... ...}}'))(z'-z) \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^7}$$ (A.50)

$$\mathcal{A} = {\mathcal{B} \over \mathcal{C}}$$ (A.51)

To find ${\partial^2E_{\mathrm{pot}} \over \partial z'^2}$, we need the derivatives of $\mathcal{A}$, $\mathcal{D}$, $\mathcal{E}$ and $\mathcal{F}$:

$${\partial \mathcal{A} \over \partial z'} = { \mathcal{C}\cdot {\partial \mathcal{B} \over \partial z'} - \mathcal{B}\cdot{\partial \mathcal{C} \over \partial z'} \over \mathcal{C}^2}$$ (A.52)

$$= { \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^5 \c... ...^3(z'-z) \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^{10}}$$ (A.53)

$$= {-3\ensuremath{\mathbf{m}}\cdot\ensuremath{\mathbf{m}}' \over \ve... ...}}'(z-z')^2 \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^7}$$ (A.54)

$${\partial \mathcal{D} \over \partial z'} = {\mathcal{C}\cdot{\partial \over \partial z'}(3\ensuremath{\mathb... ...^3(z'-z) \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^{10}}$$ (A.55)

$$= {3m'_z {\partial \over \partial z'}(m_x(x-x')+m_y(y-y')+m_z(z-z')) \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^5}$$

$$+ {15\ensuremath{\mathbf{m}}(\ensuremath{\mathbf{r}}-\ensuremath{... ...)m'_z(z-z') \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^7}$$ (A.56)

$$= {3m'_z (-m_z) \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mat... ...\mathbf{m}} \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^7}$$ (A.57)

$${\partial \mathcal{E} \over \partial z'} = { 3 \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^5 ... ...cdot m_z \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^{10}}$$

$$-{3\ensuremath{\mathbf{m}}'(\ensuremath{\mathbf{r}}-\ensuremath{\... ...^3(z'-z) \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^{10}}$$ (A.58)

$$= {3(-m'_zm_z) \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\math... ...')m_z(z-z') \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^7}$$ (A.59)

$$r_0 = 15(\ensuremath{\mathbf{m}}\cdot(\ensuremath{\mathbf{r}}-\ensurema... ...ensuremath{\mathbf{m}}'\cdot(\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'))$$ (A.60)

$$= 15(m_x(x-x')+m_y(y-y')+m_z(z-z'))$$

$$\cdot (m'_x(x-x')+m'_y(y-y')+m_z(z-z'))$$ (A.61)

$$\mathcal{F}_1 = r_0(z'-z)$$ (A.62)

In the derivative, only the terms with $z'$ matter:

$$\mathcal{F}_0 = {\mathcal{F}_1 \over z'-z}$$ (A.63)

$${\partial \mathcal{F}_0 \over \partial z'} = {\partial \over \partial z'} \left ( 15 \left ( m_x(x-x')m'_z(z-z') \right. \right.$$

$$+ m_y(y-y')m'_z(z-z')$$

$$\left. \left. + m_z(z-z')m'_z(z-z')\right ) \right )$$ (A.64)

$${\partial \over \partial z'} = m_zm'_z\cdot 2(z-z')(-1)$$ (A.65)

$${\partial \mathcal{F}_0 \over \partial z'} = 15 (m_x(x-x')m'_z(-1)$$

$$+ m_y(y-y')m'_z(-1)$$

$$+ m_zm'_z\cdot 2(z-z')(-1)$$

$$+ m_zm'_x(x-x')(-1)$$

$$+ m_zm'_y(y-y')(-1))$$

$$= -15 (\ensuremath{\mathbf{m}}\cdot (\ensuremath{\mathbf{r}}-\ensur... ...h{\mathbf{m}}'\cdot(\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}')\cdot m_z)$$ (A.66)

$${\partial \mathcal{F}_1 \over \partial z'} = \mathcal{F}_0 {\partial \over \partial z'}(z'-z) + (z'-z){\partial \mathcal{F}_0 \over \partial z'}$$ (A.67)

$$= 15(\ensuremath{\mathbf{m}}\cdot (\ensuremath{\mathbf{r}}-\ensurem... ...th{\mathbf{m}}'\cdot (\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'))\cdot 1$$

$$+ (z'-z)(-15)(\ensuremath{\mathbf{m}}\cdot(\ensuremath{\mathbf{r}... ... \ensuremath{\mathbf{m}}'(\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}')m_z)$$ (A.68)

$$= 15((\ensuremath{\mathbf{m}}\cdot (\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'))$$

$$+\ensuremath{\mathbf{m}}\cdot (\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}')m'_z(z-z')$$

$$+\ensuremath{\mathbf{m}}'\cdot(\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}')m_z(z-z'))$$ (A.69)

$$q \equiv g^7$$ (A.70)

$$= \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^7$$ (A.71)

$${\partial q \over \partial z'} = {\partial q \over \partial g} {\partial g \over \partial z'}$$ (A.72)

$$= 7g^6 {\partial g \over \partial z'}$$ (A.73)

$$= {7g^6 (z' - z) \over g}$$ (A.74)

$$= 7 \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^5(z'-z)$$ (A.75)

$${\partial \mathcal{F} \over \partial z'} = {g^7\cdot {\partial \mathcal{F}_1 \over \partial z'} - \mathcal{F}_1\cdot {\partial g^7 \over \partial z'} \over {(g^7)}^2}$$ (A.76)

$$= { {\partial \mathcal{F}_1 \over \partial z'} \over g^7} - {\mathcal{F}_1 \cdot {\partial g^7 \over \partial z'} \over g^{14}}$$ (A.77)

$$= {15 \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\v... ...'))(\ensuremath{\mathbf{m}}'(\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'))$$

$$+ \ensuremath{\mathbf{m}}(\ensuremath{\mathbf{r}}-\ensuremath{\ma... ...remath{\mathbf{m}}'(\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}')m_z(z-z'))$$

$$- {15 (\ensuremath{\mathbf{m}}\cdot(\ensuremath{\mathbf{r}}-\ensu... ...5(z'-z)) \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^{14}}$$ (A.78)

$$= {15 \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\v... ...'))(\ensuremath{\mathbf{m}}'(\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'))$$

$$+ \ensuremath{\mathbf{m}}(\ensuremath{\mathbf{r}}-\ensuremath{\ma... ...remath{\mathbf{m}}'(\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}')m_z(z-z'))$$

$$+ {105 (\ensuremath{\mathbf{m}}\cdot (\ensuremath{\mathbf{r}}-\en... ...-z')(z'-z)) \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^9}$$ (A.79)

The second derivative is:

$${\partial^2E_{\mathrm{pot}} \over \partial z'^2} = {\mu_0} (\mathcal{A}' + \mathcal{D}' + \mathcal{E}' + \mathcal{F}')$$ (A.80)

Where possible, collecting terms and expanding gives:

$${\partial^2E_{\mathrm{pot}} \over \partial z'^2} = {\mu_0} \left ( {-3\cdot \ensuremath{\mathbf{m}}\cdot \ensuremath... ...}}' \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^5} \right.$$

$$+ {15 \cdot \ensuremath{\mathbf{m}}\cdot \ensuremath{\mathbf{m}}' (z-z')^2 \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^7}$$

$$- {6 \cdot m'_zm_z \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^5}$$

$$+ {30 (\ensuremath{\mathbf{m}}\cdot(\ensuremath{\mathbf{r}}-\ensu... ...)m'_z(z-z') \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^7}$$

$$+ {30 (\ensuremath{\mathbf{m}}'\cdot(\ensuremath{\mathbf{r}}-\ens... ...))m_z(z-z') \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^7}$$

$$+ {15 (\ensuremath{\mathbf{m}}\cdot(\ensuremath{\mathbf{r}}-\ensu... ...thbf{r}}')) \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^7}$$

$$\left. - {105 (\ensuremath{\mathbf{m}}\cdot(\ensuremath{\mathbf{r... ...^2 \over \vert\ensuremath{\mathbf{r}}-\ensuremath{\mathbf{r}}'\vert^9} \right )$$ (A.81)

This is the final second derivative, which should be proportional to the signal at the tip of the MFM -- assuming the MFM tip is a dipole:

$$\ensuremath{\mathbf{m}}' = \left( \begin{array}{c} 0\\ 0\\ C \end{array} \right )$$ (A.82)