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学习自旋电子学的笔记07：根据微磁学基本能量密度公式推导有效场
阅读说明：由于CSDN自带的Markdown编辑器目前对大篇幅KATEX公式的支持性不太好，导致文章内容有了字数限制，一旦超过字数限制，就不能正常保存和发布了。所以，我将笔记07的内容全部转换为了图片的形式，方便大家阅读！文末也有原始的.md文档，里面的公式使用的是KATEX格式，此外，若某些人没有合适的.md文档的阅读器，我也把本文打印了PDF版本，有需要的可以点击下载！
同理，可以根据结果的规律写出式子（2）的第二项：
∂ ∂ m x [ [ ( m x u 2 x + m y u 2 y + m z u 2 z ) ( m x u 3 x + m y u 3 y + m z u 3 z ) ] 2 ] \frac{\partial{}}{\partial{m_x}}[[(m_x u_{2x}+m_y u_{2y}+m_z u_{2z}) (m_x u_{3x}+m_y u_{3y}+m_z u_{3z})]^2] ∂mx∂[[(mxu2x+myu2y+mzu2z)(mxu3x+myu3y+mzu3z)]2]
= 2 [ ( m ⃗ ⋅ u 2 ⃗ ) ( m ⃗ ⋅ u 3 ⃗ ) 2 u 2 x + ( m ⃗ ⋅ u 2 ⃗ ) 2 ( m ⃗ ⋅ u 3 ⃗ ) u 3 x ] =2[(\vec{m} \cdot \vec{u_2}) (\vec{m} \cdot \vec{u_3})^2 u_{2x} +(\vec{m} \cdot \vec{u_2})^2 (\vec{m} \cdot \vec{u_3}) u_{3x}] =2[(m⋅u2)(m⋅u3)2u2x+(m⋅u2)2(m⋅u3)u3x]
同理，可以根据结果的规律写出式子（2）的第三项：
∂ ∂ m x [ [ ( m x u 3 x + m y u 3 y + m z u 3 z ) ( m x u 1 x + m y u 1 y + m z u 1 z ) ] 2 ] \frac{\partial{}}{\partial{m_x}}[[(m_x u_{3x}+m_y u_{3y}+m_z u_{3z}) (m_x u_{1x}+m_y u_{1y}+m_z u_{1z})]^2] ∂mx∂[[(mxu3x+myu3y+mzu3z)(mxu1x+myu1y+mzu1z)]2]
= 2 [ ( m ⃗ ⋅ u 3 ⃗ ) ( m ⃗ ⋅ u 1 ⃗ ) 2 u 3 x + ( m ⃗ ⋅ u 3 ⃗ ) 2 ( m ⃗ ⋅ u 1 ⃗ ) u 1 x ] =2[(\vec{m} \cdot \vec{u_3}) (\vec{m} \cdot \vec{u_1})^2 u_{3x} +(\vec{m} \cdot \vec{u_3})^2 (\vec{m} \cdot \vec{u_1}) u_{1x}] =2[(m⋅u3)(m⋅u1)2u3x+(m⋅u3)2(m⋅u1)u1x]
将以上三项带回式子（2）合并得到式子（2）的结果，即式子（1）的第一项：
∂ ∂ m x [ − K c [ [ ( m ⃗ ⋅ u 1 ⃗ ) ( m ⃗ ⋅ u 2 ⃗ ) ] 2 + [ ( m ⃗ ⋅ u 2 ⃗ ) ( m ⃗ ⋅ u 3 ⃗ ) ] 2 + [ ( m ⃗ ⋅ u 3 ⃗ ) ( m ⃗ ⋅ u 1 ⃗ ) ] 2 ] ] x ^ \frac{\partial{}}{\partial{m_x}}[-K_c [[(\vec{m} \cdot \vec{u_1}) (\vec{m} \cdot \vec{u_2})]^2 +[(\vec{m} \cdot \vec{u_2}) (\vec{m} \cdot \vec{u_3})]^2 +[(\vec{m} \cdot \vec{u_3}) (\vec{m} \cdot \vec{u_1})]^2]]\hat{x} ∂mx∂[−Kc[[(m⋅u1)(m⋅u2)]2+[(m⋅u2)(m⋅u3)]2+[(m⋅u3)(m⋅u1)]2]]x^
= − K c ( =-K_c \Bigg( =−Kc(
2 [ ( m ⃗ ⋅ u 1 ⃗ ) ( m ⃗ ⋅ u 2 ⃗ ) 2 u 1 x + ( m ⃗ ⋅ u 1 ⃗ ) 2 ( m ⃗ ⋅ u 2 ⃗ ) u 2 x ] + 2[(\vec{m} \cdot \vec{u_1}) (\vec{m} \cdot \vec{u_2})^2 u_{1x} +(\vec{m} \cdot \vec{u_1})^2 (\vec{m} \cdot \vec{u_2}) u_{2x}]+ 2[(m⋅u1)(m⋅u2)2u1x+(m⋅u1)2(m⋅u2)u2x]+
2 [ ( m ⃗ ⋅ u 2 ⃗ ) ( m ⃗ ⋅ u 3 ⃗ ) 2 u 2 x + ( m ⃗ ⋅ u 2 ⃗ ) 2 ( m ⃗ ⋅ u 3 ⃗ ) u 3 x ] + 2[(\vec{m} \cdot \vec{u_2}) (\vec{m} \cdot \vec{u_3})^2 u_{2x} +(\vec{m} \cdot \vec{u_2})^2 (\vec{m} \cdot \vec{u_3}) u_{3x}]+ 2[(m⋅u2)(m⋅u3)2u2x+(m⋅u2)2(m⋅u3)u3x]+
2 [ ( m ⃗ ⋅ u 3 ⃗ ) ( m ⃗ ⋅ u 1 ⃗ ) 2 u 3 x + ( m ⃗ ⋅ u 3 ⃗ ) 2 ( m ⃗ ⋅ u 1 ⃗ ) u 1 x ] 2[(\vec{m} \cdot \vec{u_3}) (\vec{m} \cdot \vec{u_1})^2 u_{3x} +(\vec{m} \cdot \vec{u_3})^2 (\vec{m} \cdot \vec{u_1}) u_{1x}] 2[(m⋅u3)(m⋅u1)2u3x+(m⋅u3)2(m⋅u1)u1x]
) x ^ \Bigg)\hat{x} )x^同理，可以根据第一项结果的规律写出式子（1）的第二项：
∂ ∂ m y [ − K c [ [ ( m ⃗ ⋅ u 1 ⃗ ) ( m ⃗ ⋅ u 2 ⃗ ) ] 2 + [ ( m ⃗ ⋅ u 2 ⃗ ) ( m ⃗ ⋅ u 3 ⃗ ) ] 2 + [ ( m ⃗ ⋅ u 3 ⃗ ) ( m ⃗ ⋅ u 1 ⃗ ) ] 2 ] ] y ^ \frac{\partial{}}{\partial{m_y}}[-K_c [[(\vec{m} \cdot \vec{u_1}) (\vec{m} \cdot \vec{u_2})]^2 +[(\vec{m} \cdot \vec{u_2}) (\vec{m} \cdot \vec{u_3})]^2 +[(\vec{m} \cdot \vec{u_3}) (\vec{m} \cdot \vec{u_1})]^2]]\hat{y} ∂my∂[−Kc[[(m⋅u1)(m⋅u2)]2+[(m⋅u2)(m⋅u3)]2+[(m⋅u3)(m⋅u1)]2]]y^
= − K c ( =-K_c \Bigg( =−Kc(
2 [ ( m ⃗ ⋅ u 1 ⃗ ) ( m ⃗ ⋅ u 2 ⃗ ) 2 u 1 y + ( m ⃗ ⋅ u 1 ⃗ ) 2 ( m ⃗ ⋅ u 2 ⃗ ) u 2 y ] + 2[(\vec{m} \cdot \vec{u_1}) (\vec{m} \cdot \vec{u_2})^2 u_{1y} +(\vec{m} \cdot \vec{u_1})^2 (\vec{m} \cdot \vec{u_2}) u_{2y}]+ 2[(m⋅u1)(m⋅u2)2u1y+(m⋅u1)2(m⋅u2)u2y]+
2 [ ( m ⃗ ⋅ u 2 ⃗ ) ( m ⃗ ⋅ u 3 ⃗ ) 2 u 2 y + ( m ⃗ ⋅ u 2 ⃗ ) 2 ( m ⃗ ⋅ u 3 ⃗ ) u 3 y ] + 2[(\vec{m} \cdot \vec{u_2}) (\vec{m} \cdot \vec{u_3})^2 u_{2y} +(\vec{m} \cdot \vec{u_2})^2 (\vec{m} \cdot \vec{u_3}) u_{3y}]+ 2[(m⋅u2)(m⋅u3)2u2y+(m⋅u2)2(m⋅u3)u3y]+
2 [ ( m ⃗ ⋅ u 3 ⃗ ) ( m ⃗ ⋅ u 1 ⃗ ) 2 u 3 y + ( m ⃗ ⋅ u 3 ⃗ ) 2 ( m ⃗ ⋅ u 1 ⃗ ) u 1 y ] 2[(\vec{m} \cdot \vec{u_3}) (\vec{m} \cdot \vec{u_1})^2 u_{3y} +(\vec{m} \cdot \vec{u_3})^2 (\vec{m} \cdot \vec{u_1}) u_{1y}] 2[(m⋅u3)(m⋅u1)2u3y+(m⋅u3)2(m⋅u1)u1y]
) y ^ \Bigg)\hat{y} )y^同理，可以根据第一项结果的规律写出式子（1）的第三项：
∂ ∂ m z [ − K c [ [ ( m ⃗ ⋅ u 1 ⃗ ) ( m ⃗ ⋅ u 2 ⃗ ) ] 2 + [ ( m ⃗ ⋅ u 2 ⃗ ) ( m ⃗ ⋅ u 3 ⃗ ) ] 2 + [ ( m ⃗ ⋅ u 3 ⃗ ) ( m ⃗ ⋅ u 1 ⃗ ) ] 2 ] ] z ^ \frac{\partial{}}{\partial{m_z}}[-K_c [[(\vec{m} \cdot \vec{u_1}) (\vec{m} \cdot \vec{u_2})]^2 +[(\vec{m} \cdot \vec{u_2}) (\vec{m} \cdot \vec{u_3})]^2 +[(\vec{m} \cdot \vec{u_3}) (\vec{m} \cdot \vec{u_1})]^2]]\hat{z} ∂mz∂[−Kc[[(m⋅u1)(m⋅u2)]2+[(m⋅u2)(m⋅u3)]2+[(m⋅u3)(m⋅u1)]2]]z^
= − K c ( =-K_c \Bigg( =−Kc(
2 [ ( m ⃗ ⋅ u 1 ⃗ ) ( m ⃗ ⋅ u 2 ⃗ ) 2 u 1 z + ( m ⃗ ⋅ u 1 ⃗ ) 2 ( m ⃗ ⋅ u 2 ⃗ ) u 2 z ] + 2[(\vec{m} \cdot \vec{u_1}) (\vec{m} \cdot \vec{u_2})^2 u_{1z} +(\vec{m} \cdot \vec{u_1})^2 (\vec{m} \cdot \vec{u_2}) u_{2z}]+ 2[(m⋅u1)(m⋅u2)2u1z+(m⋅u1)2(m⋅u2)u2z]+
2 [ ( m ⃗ ⋅ u 2 ⃗ ) ( m ⃗ ⋅ u 3 ⃗ ) 2 u 2 z + ( m ⃗ ⋅ u 2 ⃗ ) 2 ( m ⃗ ⋅ u 3 ⃗ ) u 3 z ] + 2[(\vec{m} \cdot \vec{u_2}) (\vec{m} \cdot \vec{u_3})^2 u_{2z} +(\vec{m} \cdot \vec{u_2})^2 (\vec{m} \cdot \vec{u_3}) u_{3z}]+ 2[(m⋅u2)(m⋅u3)2u2z+(m⋅u2)2(m⋅u3)u3z]+
2 [ ( m ⃗ ⋅ u 3 ⃗ ) ( m ⃗ ⋅ u 1 ⃗ ) 2 u 3 z + ( m ⃗ ⋅ u 3 ⃗ ) 2 ( m ⃗ ⋅ u 1 ⃗ ) u 1 z ] 2[(\vec{m} \cdot \vec{u_3}) (\vec{m} \cdot \vec{u_1})^2 u_{3z} +(\vec{m} \cdot \vec{u_3})^2 (\vec{m} \cdot \vec{u_1}) u_{1z}] 2[(m⋅u3)(m⋅u1)2u3z+(m⋅u3)2(m⋅u1)u1z]
) z ^ \Bigg)\hat{z} )z^将以上三项带回式子（1）合并得到磁晶立方各向异性能的有效场： H c u b i c → \overrightarrow{H_{cubic}} Hcubic：
= − 2 K c − u 0 M s ( ( m ⃗ ⋅ u 1 ⃗ ) ( m ⃗ ⋅ u 2 ⃗ ) 2 ( u 1 x x ^ + u 1 y y ^ + u 1 z z ^ ) + ( m ⃗ ⋅ u 1 ⃗ ) 2 ( m ⃗ ⋅ u 2 ⃗ ) ( u 2 x x ^ + u 2 y y ^ + u 2 z z ^ ) =\frac{-2K_c}{{-u_{0}}{M_s}}\Bigg( (\vec{m} \cdot \vec{u_1}) (\vec{m} \cdot \vec{u_2})^2(u_{1x}\hat{x} + u_{1y}\hat{y} + u_{1z}\hat{z}) +(\vec{m} \cdot \vec{u_1})^2 (\vec{m} \cdot \vec{u_2})(u_{2x}\hat{x} + u_{2y}\hat{y} + u_{2z}\hat{z}) =−u0Ms−2Kc((m⋅u1)(m⋅u2)2(u1xx^+u1yy^+u1zz^)+(m⋅u1)2(m⋅u2)(u2xx^+u2yy^+u2zz^)
+ ( m ⃗ ⋅ u 2 ⃗ ) ( m ⃗ ⋅ u 3 ⃗ ) 2 ( u 2 x x ^ + u 2 y y ^ + u 2 z z ^ ) + ( m ⃗ ⋅ u 2 ⃗ ) 2 ( m ⃗ ⋅ u 3 ⃗ ) ( u 3 x x ^ + u 3 y y ^ + u 3 z z ^ ) +(\vec{m} \cdot \vec{u_2}) (\vec{m} \cdot \vec{u_3})^2(u_{2x}\hat{x} + u_{2y}\hat{y} + u_{2z}\hat{z}) +(\vec{m} \cdot \vec{u_2})^2 (\vec{m} \cdot \vec{u_3})(u_{3x}\hat{x} + u_{3y}\hat{y} + u_{3z}\hat{z}) +(m⋅u2)(m⋅u3)2(u2xx^+u2yy^+u2zz^)+(m⋅u2)2(m⋅u3)(u3xx^+u3yy^+u3zz^)
+ ( m ⃗ ⋅ u 3 ⃗ ) ( m ⃗ ⋅ u 1 ⃗ ) 2 ( u 3 x x ^ + u 3 y y ^ + u 3 z z ^ ) + ( m ⃗ ⋅ u 3 ⃗ ) 2 ( m ⃗ ⋅ u 1 ⃗ ) ( u 1 x x ^ + u 1 y y ^ + u 1 z z ^ ) ) +(\vec{m} \cdot \vec{u_3}) (\vec{m} \cdot \vec{u_1})^2(u_{3x}\hat{x} + u_{3y}\hat{y} + u_{3z}\hat{z}) +(\vec{m} \cdot \vec{u_3})^2 (\vec{m} \cdot \vec{u_1})(u_{1x}\hat{x} + u_{1y}\hat{y} + u_{1z}\hat{z})\Bigg) +(m⋅u3)(m⋅u1)2(u3xx^+u3yy^+u3zz^)+(m⋅u3)2(m⋅u1)(u1xx^+u1yy^+u1zz^))
= 2 K c u 0 M s ( ( m ⃗ ⋅ u 1 ⃗ ) ( m ⃗ ⋅ u 2 ⃗ ) 2 u 1 ⃗ + ( m ⃗ ⋅ u 1 ⃗ ) 2 ( m ⃗ ⋅ u 2 ⃗ ) u 2 ⃗ =\frac{2K_c}{{u_{0}}{M_s}}\Bigg( (\vec{m} \cdot \vec{u_1}) (\vec{m} \cdot \vec{u_2})^2 \vec{u_1} +(\vec{m} \cdot \vec{u_1})^2 (\vec{m} \cdot \vec{u_2}) \vec{u_2} =u0Ms2Kc((m⋅u1)(m⋅u2)2u1+(m⋅u1)2(m⋅u2)u2
+ ( m ⃗ ⋅ u 2 ⃗ ) ( m ⃗ ⋅ u 3 ⃗ ) 2 u 2 ⃗ + ( m ⃗ ⋅ u 2 ⃗ ) 2 ( m ⃗ ⋅ u 3 ⃗ ) u 3 ⃗ +(\vec{m} \cdot \vec{u_2}) (\vec{m} \cdot \vec{u_3})^2 \vec{u_2} +(\vec{m} \cdot \vec{u_2})^2 (\vec{m} \cdot \vec{u_3}) \vec{u_3} +(m⋅u2)(m⋅u3)2u2+(m⋅u2)2(m⋅u3)u3
+ ( m ⃗ ⋅ u 3 ⃗ ) ( m ⃗ ⋅ u 1 ⃗ ) 2 u 2 ⃗ + ( m ⃗ ⋅ u 3 ⃗ ) 2 ( m ⃗ ⋅ u 1 ⃗ ) u 1 ⃗ ) +(\vec{m} \cdot \vec{u_3}) (\vec{m} \cdot \vec{u_1})^2 \vec{u_2} +(\vec{m} \cdot \vec{u_3})^2 (\vec{m} \cdot \vec{u_1}) \vec{u_1}\Bigg) +(m⋅u3)(m⋅u1)2u2+(m⋅u3)2(m⋅u1)u1)
= 2 K c u 0 M s ( ( m ⃗ ⋅ u 1 ⃗ ) [ ( m ⃗ ⋅ u 2 ⃗ ) 2 + ( m ⃗ ⋅ u 3 ⃗ ) 2 ] u 1 ⃗ =\frac{2K_c}{{u_{0}}{M_s}}\Bigg( (\vec{m} \cdot \vec{u_1})[(\vec{m} \cdot \vec{u_2})^2 + (\vec{m} \cdot \vec{u_3})^2] \vec{u_1} =u0Ms2Kc((m⋅u1)[(m⋅u2)2+(m⋅u3)2]u1
+ ( m ⃗ ⋅ u 2 ⃗ ) [ ( m ⃗ ⋅ u 3 ⃗ ) 2 + ( m ⃗ ⋅ u 1 ⃗ ) 2 ] u 2 ⃗ +(\vec{m} \cdot \vec{u_2})[(\vec{m} \cdot \vec{u_3})^2 + (\vec{m} \cdot \vec{u_1})^2] \vec{u_2} +(m⋅u2)[(m⋅u3)2+(m⋅u1)2]u2
+ ( m ⃗ ⋅ u 3 ⃗ ) [ ( m ⃗ ⋅ u 1 ⃗ ) 2 + ( m ⃗ ⋅ u 2 ⃗ ) 2 ] u 3 ⃗ ) +(\vec{m} \cdot \vec{u_3})[(\vec{m} \cdot \vec{u_1})^2 + (\vec{m} \cdot \vec{u_2})^2] \vec{u_3} \Bigg) +(m⋅u3)[(m⋅u1)2+(m⋅u2)2]u3)
总结
本文总结了微磁学中通过几个基本的能量密度公式推导出其有效场，内容核心就是求偏导，各种各样的偏导。至于微磁学中的其它公式推导，暂时还没学到皮毛，只有等以后再说了。以下内容是原始的.md文档
@[TOC](文章目录) <font color=#ff0000 > 雁来音信无凭，路遥归梦难成。李煜《清平乐别来春半》 font>
# 前言 在笔记06中为大家推荐了“ubermag”官网上的有关[微磁学公式的推导笔记](https://download.csdn.net/download/qq_43572058/86272578)，按照老辈人的说法，微磁学相关公式的推导应该在我们入门后两个月内学会，但并不是每一个人都有着基本的数学素养，有的人面对这些数学式子就是头晕，比如我，，，好在近期闲来无事想着动纸笔比划一下，遇到的一些陌生的名词，陌生的导数，卡住的推导步骤等都会去网上翻一翻，于是总归学会了一点三脚猫的功夫了。 本文的内容就是根据几个微磁学基本能量密度公式来推导其有效场，这几个微磁学基本能量包括：交换能，塞曼能，界面DMI能，体DMI能，单轴各向异性能，立方各向异性能。这几个能量的能量密度公式是已知的，利用有效场的公式： $$\overrightarrow{H_{eff}} =\frac{1}{{-u_{0}}{M_s}}\frac{\delta{E_{any}}}{\delta{\overrightarrow{m}}} $$ 即可推导出每一项能量对应的有效场。式中：$u_0，M_s$ 分别表示真空磁导率和材料的饱和磁化强度；$\overrightarrow{m}=\overrightarrow{m(\vec{r})}=\frac{\overrightarrow{M(\vec{r})}}{M_s}$ 表示单位磁化强度（矢量场）；$E_{any}$ 表示以上能量的能量密度（标量场）；符号 $\delta$ 表示变分算子。 这里涉及到“变分”这个概念，我自己目前也不是很明白，但从网上查了一些言论描述如下： 设 $F=F(y)，y=y(x)$，有所谓的 **函数的函数的导数** 表示为： $$F^{\prime}_y=\frac{\delta{F}}{\delta{y}}$$ 即 $$\frac{\delta{F}}{\delta{y}}=\frac{dF}{dy}或\frac{\partial{F}}{\partial{y}}$$ 于是，有效场的式子变为了能量密度（标量场）对单位磁化强度（矢量）求偏导： $$\overrightarrow{H_{eff}} =\frac{1}{{-u_{0}}{M_s}}\frac{\partial{E_{any}}}{\partial{\overrightarrow{m}}} $$ 大家对这个式子都非常熟悉了，在大部分微磁学文章中，一般这个式子会出现在LLG方程的后面。不过，此处又出现了有难度的步骤，即标量场如何对矢量场求偏导呢？其实结合“梯度”的定义来看： 设有一个标量场 $\phi=\phi(x,y,z)$，一个矢量线元 $d\vec{l}$ ，则定义 $\phi$ 的梯度为： $$grad\phi=\nabla{\phi}=\frac{d\phi}{d\vec{l}}=\frac{\partial{\phi}}{\partial{x}}\hat{x} + \frac{\partial{\phi}}{\partial{y}}\hat{y} + \frac{\partial{\phi}}{\partial{z}}\hat{z}$$ 梯度是一个标量场对矢量求导，从而不难得到“有效场等于能量密度对单位磁化强度的梯度”这个结论，这句话也经常出现在许多毕业论文第二章的开头，此时我才明白其中的含义。 于是借助梯度的概念将有效场的式子展开： $$\overrightarrow{H_{eff}}=\frac{1}{{-u_{0}}{M_s}}\nabla_{\vec{m}}{E_{any}}=\frac{1}{{-u_{0}}{M_s}}(\frac{\partial{E_{any}}}{\partial{m_x}}\hat{x} + \frac{\partial{E_{any}}}{\partial{m_y}}\hat{y} + \frac{\partial{E_{any}}}{\partial{m_z}}\hat{z})$$
# 一、塞曼能的有效场 已知塞曼能的能量密度公式： $$E_{zeeman}={-u_{0}}{M_s}(\vec{m} \cdot \vec{H})$$ 其中：$\vec{H}=\overrightarrow{H(\vec{r})}$ 为外加磁场（矢量场），单位 A/m。那么可以得到塞曼能的有效场： $\overrightarrow{H_{zeeman}}$ $=\frac{1}{{-u_{0}}{M_s}}\frac{\delta{E_{zeeman}}}{\delta{\overrightarrow{m}}}$ $=\frac{1}{{-u_{0}}{M_s}}\frac{\partial{E_{zeeman}}}{\partial{\overrightarrow{m}}}$ $=\frac{1}{{-u_{0}}{M_s}}(\frac{\partial{E_{zeeman}}}{\partial{m_x}}\hat{x} + \frac{\partial{E_{zeeman}}}{\partial{m_y}}\hat{y} + \frac{\partial{E_{zeeman}}}{\partial{m_z}}\hat{z})$ $=\frac{1}{{-u_{0}}{M_s}}[\frac{\partial{}}{\partial{m_x}}({-u_{0}}{M_s}(\vec{m} \cdot \vec{H})) \hat{x} + \frac{\partial{}}{\partial{m_y}}({-u_{0}}{M_s}(\vec{m} \cdot \vec{H})) \hat{y} + \frac{\partial{}}{\partial{m_z}}({-u_{0}}{M_s}(\vec{m} \cdot \vec{H})) \hat{z})]$ $=\frac{\partial{}}{\partial{m_x}}(\vec{m} \cdot \vec{H}) \hat{x} + \frac{\partial{}}{\partial{m_y}}(\vec{m} \cdot \vec{H}) \hat{y} + \frac{\partial{}}{\partial{m_z}}(\vec{m} \cdot \vec{H}) \hat{z}$ ***********************************************(式子 1） 其中 $\vec{m}$ 和 $\vec{H}$ 的点乘： $\vec{m} \cdot \vec{H} = m_xH_x + m_yH_y + m_zH_z$ 带回式子（1）的第一项： $\frac{\partial{}}{\partial{m_x}}(\vec{m} \cdot \vec{H}) \hat{x}$ $=\frac{\partial{}}{\partial{m_x}}(m_xH_x + m_yH_y + m_zH_z) \hat{x}$ $=H_x \hat{x}$ 同理，式子（1）的第二项： $\frac{\partial{}}{\partial{m_y}}(\vec{m} \cdot \vec{H}) \hat{y}$ $=\frac{\partial{}}{\partial{m_y}}(m_xH_x + m_yH_y + m_zH_z) \hat{y}$ $=H_y \hat{y}$ 同理，式子（1）的第三项： $\frac{\partial{}}{\partial{m_z}}(\vec{m} \cdot \vec{H}) \hat{z}$ $=\frac{\partial{}}{\partial{m_z}}(m_xH_x + m_yH_y + m_zH_z) \hat{z}$ $=H_z \hat{z}$ 最后合并这三项，式子（1）的结果为： $\frac{\partial{}}{\partial{m_x}}(\vec{m} \cdot \vec{H}) \hat{x} + \frac{\partial{}}{\partial{m_y}}(\vec{m} \cdot \vec{H}) \hat{y} + \frac{\partial{}}{\partial{m_z}}(\vec{m} \cdot \vec{H}) \hat{z}$ $=H_x \hat{x} + H_y \hat{y} + H_z \hat{z}$ $=\vec{H}$ 这个推导结果表示塞曼能的有效场： $\overrightarrow{H_{zeeman}}$ 就等于外加磁场 $\vec{H}$。 而且由塞曼能的能量密度表达式可知，若已知任意微磁能量的有效场 $\overrightarrow{H^{any}_{eff}}$ 的情况下，那么它对应的能量密度表达式即为： $$E_{any}={-u_{0}}{M_s}(\vec{m} \cdot \overrightarrow{H^{any}_{eff}})$$ # 二、交换能的有效场 已知交换能的能量密度公式： $$E_{ex}=A[\vec{m} \cdot (\nabla^2{\vec{m}})] =A[(\nabla{m_x})^2 + (\nabla{m_y})^2 + (\nabla{m_z})^2]$$ 其中 $A$ 表示交换作用强度，单位 J/m，式中 $\nabla^2$ 是拉普拉斯算子，可以直接作用在标量场或者矢量场，此处作用在矢量场 $\vec{m}$ ，表示： $\nabla^2{\vec{m}}=\frac{\partial^2{m_{x}}}{\partial{x^2}}\hat{x} + \frac{\partial^2{m_{y}}}{\partial{y^2}}\hat{y} + \frac{\partial^2{m_{z}}}{\partial{z^2}}\hat{z}$ 。 交换能的能量密度公式还有一个简单的，特殊的“写法： $E_{ex}=A(\nabla{\vec{m}})^2$ ，之所以特殊，是因为我们都知道矢量微分算子 $\nabla$ 只能直接作用在标量场，而不能直接与矢量场结合（即只能通过点乘或叉乘的方式来与矢量场结合），同时这种简单的写法在许多文章中也是最流行的。 那么可以得到交换能的有效场：$\overrightarrow{H_{ex}}$ $=\frac{1}{{-u_{0}}{M_s}}\frac{\delta{E_{ex}}}{\delta{\overrightarrow{m}}}$ $=\frac{1}{{-u_{0}}{M_s}}\frac{\partial{E_{ex}}}{\partial{\overrightarrow{m}}}$ $=\frac{1}{{-u_{0}}{M_s}}(\frac{\partial{E_{ex}}}{\partial{m_x}}\hat{x} + \frac{\partial{E_{ex}}}{\partial{m_y}}\hat{y} + \frac{\partial{E_{ex}}}{\partial{m_z}}\hat{z})$ $=\frac{1}{{-u_{0}}{M_s}}[\frac{\partial{}}{\partial{m_x}}(A[(\nabla{m_x})^2 + (\nabla{m_y})^2 + (\nabla{m_z})^2])\hat{x} + \frac{\partial{}}{\partial{m_y}}(A[(\nabla{m_x})^2 + (\nabla{m_y})^2 + (\nabla{m_z})^2])\hat{y} + \frac{\partial{}}{\partial{m_z}}(A[(\nabla{m_z})^2 + (\nabla{m_y})^2 + (\nabla{m_z})^2])\hat{z}]$ ***********************************************(式子 1) 其中式子（1）的第一项： $\frac{\partial{}}{\partial{m_x}}(A[(\nabla{m_x})^2 + (\nabla{m_y})^2 + (\nabla{m_z})^2])$ ***********************************************(式子 2） 式子（2）里面的 $(\nabla{m_x})^2$ $=(\frac{\partial{m_{x}}}{\partial{x}}\hat{x} + \frac{\partial{m_{x}}}{\partial{y}}\hat{y} + \frac{\partial{m_{x}}}{\partial{z}}\hat{z})^2$ $=(\frac{\partial{m_{x}}}{\partial{x}})^2 + (\frac{\partial{m_{x}}}{\partial{y}})^2+ (\frac{\partial{m_{x}}}{\partial{z}})^2$ 式子（2）里面的 $(\nabla{m_y})^2$ $=(\frac{\partial{m_{y}}}{\partial{x}}\hat{x} + \frac{\partial{m_{y}}}{\partial{y}}\hat{y} + \frac{\partial{m_{y}}}{\partial{z}}\hat{z})^2$ $=(\frac{\partial{m_{y}}}{\partial{x}})^2 + (\frac{\partial{m_{y}}}{\partial{y}})^2+ (\frac{\partial{m_{z}}}{\partial{z}})^2$ 式子（2）里面的 $(\nabla{m_z})^2$ $=(\frac{\partial{m_{z}}}{\partial{x}}\hat{x} + \frac{\partial{m_{z}}}{\partial{y}}\hat{y} + \frac{\partial{m_{z}}}{\partial{z}}\hat{z})^2$ $=(\frac{\partial{m_{z}}}{\partial{x}})^2 + (\frac{\partial{m_{z}}}{\partial{y}})^2+ (\frac{\partial{m_{z}}}{\partial{z}})^2$ 从三个式子的展开结果来看，显然， $(\nabla{m_x})^2$ 与变量 $m_x$ 有关，与 $m_y$ 和 $m_z$ 无关； $(\nabla{m_y})^2$ 与变量 $m_y$ 有关，与 $m_x$ 和 $m_z$ 无关； $(\nabla{m_z})^2$ 与变量 $m_z$ 有关，与 $m_x$ 和 $m_y$ 无关。 于是从式子（2）可以得到：$\frac{\partial{}}{\partial{m_x}}(A[(\nabla{m_x})^2 + (\nabla{m_y})^2 + (\nabla{m_z})^2])$ $=\frac{\partial{}}{\partial{m_x}}(A(\nabla{m_x})^2) + 0 +0$ $=A \frac{\partial{}}{\partial{m_x}}(\nabla{m_x})^2$ $=A \frac{\partial{}}{\partial{m_x}} [(\frac{\partial{m_{x}}}{\partial{x}})^2 + (\frac{\partial{m_{x}}}{\partial{y}})^2+ (\frac{\partial{m_{x}}}{\partial{z}})^2]$ ***********************************************(式子 3） 式子（3）中 $\frac{\partial{}}{\partial{m_x}} [(\frac{\partial{m_{x}}}{\partial{x}})^2]$ $=2\frac{\partial{m_{x}}}{\partial{x}} \frac{\partial{}}{\partial{m_x}} (\frac{\partial{m_{x}}}{\partial{x}})$ $=2\frac{\partial{m_{x}}}{\partial{x}} \frac{\partial{}}{\cancel{\partial{m_x}}} (\frac{\cancel{\partial{m_x}}}{\partial{x}})$ $=2\frac{\partial{m_{x}}}{\partial{x}} \frac{\partial{}}{\partial{x}}$ 式子（3）中 $\frac{\partial{}}{\partial{m_x}} [(\frac{\partial{m_{x}}}{\partial{y}})^2]$ $=2\frac{\partial{m_{x}}}{\partial{y}} \frac{\partial{}}{\partial{m_x}} (\frac{\partial{m_{x}}}{\partial{y}})$ $=2\frac{\partial{m_{x}}}{\partial{y}} \frac{\partial{}}{\cancel{\partial{m_x}}} (\frac{\cancel{\partial{m_x}}}{\partial{y}})$ $=2\frac{\partial{m_{x}}}{\partial{y}} \frac{\partial{}}{\partial{y}}$ 式子（3）中 $\frac{\partial{}}{\partial{m_x}} [(\frac{\partial{m_{x}}}{\partial{z}})^2]$ $=2\frac{\partial{m_{x}}}{\partial{z}} \frac{\partial{}}{\partial{m_x}} (\frac{\partial{m_{x}}}{\partial{z}})$ $=2\frac{\partial{m_{x}}}{\partial{z}} \frac{\partial{}}{\cancel{\partial{m_x}}} (\frac{\cancel{\partial{m_x}}}{\partial{z}})$ $=2\frac{\partial{m_{x}}}{\partial{z}} \frac{\partial{}}{\partial{z}}$ 将式子（3）的结果带回式子（2）得到：$\frac{\partial{}}{\partial{m_x}}(A[(\nabla{m_x})^2 + (\nabla{m_y})^2 + (\nabla{m_z})^2])$ $=A \frac{\partial{}}{\partial{m_x}} [(\frac{\partial{m_{x}}}{\partial{x}})^2 + (\frac{\partial{m_{x}}}{\partial{y}})^2+ (\frac{\partial{m_{x}}}{\partial{z}})^2]$ $=A[2\frac{\partial{m_{x}}}{\partial{x}} \frac{\partial{}}{\partial{x}} + 2\frac{\partial{m_{x}}}{\partial{y}} \frac{\partial{}}{\partial{y}} + 2\frac{\partial{m_{x}}}{\partial{z}} \frac{\partial{}}{\partial{z}}]$ $=2A \nabla{} \cdot \nabla{m_x}$ $=2A \nabla^2{m_x}$ 同理，式子（1）的第二项：$\frac{\partial{}}{\partial{m_y}}(A[(\nabla{m_x})^2 + (\nabla{m_y})^2 + (\nabla{m_z})^2])$ $=A \frac{\partial{}}{\partial{m_y}} [(\frac{\partial{m_{y}}}{\partial{x}})^2 + (\frac{\partial{m_{y}}}{\partial{y}})^2+ (\frac{\partial{m_{y}}}{\partial{z}})^2]$ $=2A \nabla^2{m_y}$ 同理，式子（1）的第三项： $\frac{\partial{}}{\partial{m_z}}(A[(\nabla{m_x})^2 + (\nabla{m_y})^2 + (\nabla{m_z})^2])$ $=A \frac{\partial{}}{\partial{m_z}} [(\frac{\partial{m_{z}}}{\partial{x}})^2 + (\frac{\partial{m_{z}}}{\partial{y}})^2+ (\frac{\partial{m_{z}}}{\partial{z}})^2]$ $=2A \nabla^2{m_z}$ 将以上三项带回式子（1）合并得到交换能的有效场：$\overrightarrow{H_{ex}}$ $=\frac{1}{{-u_{0}}{M_s}}[\frac{\partial{}}{\partial{m_x}}(A[(\nabla{m_x})^2 + (\nabla{m_y})^2 + (\nabla{m_z})^2])\hat{x} + \frac{\partial{}}{\partial{m_y}}(A[(\nabla{m_x})^2 + (\nabla{m_y})^2 + (\nabla{m_z})^2])\hat{y} + \frac{\partial{}}{\partial{m_x}}(A[(\nabla{m_z})^2 + (\nabla{m_y})^2 + (\nabla{m_z})^2])\hat{z}]$ $=\frac{2A}{{-u_{0}}{M_s}} ( \nabla^2{m_x}\hat{x} + \nabla^2{m_y}\hat{y} + \nabla^2{m_z}\hat{z})$ $=\frac{2A}{{-u_{0}}{M_s}} \nabla^2{\vec{m}}$ # 三、界面DMI能的有效场 已知界面DMI能的能量密度公式： $$E^{inter}_{DMI}=D(\vec{m} \cdot \nabla{m_z} - m_z \nabla \cdot \vec{m}) =D(m_x \frac{\partial{m_z}}{\partial{x}} - m_z \frac{\partial{m_x}}{\partial{x}} + m_y \frac{\partial{m_z}}{\partial{y}} - m_z \frac{\partial{m_y}}{\partial{y}})$$ 其中 $D$ 是界面DMI强度，单位 $J/m^2$。那么可以得到界面DMI能的有效场： $\overrightarrow{H^{inter}_{DMI}}$ $=\frac{1}{{-u_{0}}{M_s}}\frac{\delta{E^{inter}_{DMI}}}{\delta{\overrightarrow{m}}}$ $=\frac{1}{{-u_{0}}{M_s}}\frac{\partial{E^{inter}_{DMI}}}{\partial{\overrightarrow{m}}}$ $=\frac{1}{{-u_{0}}{M_s}}(\frac{\partial{E^{inter}_{DMI}}}{\partial{m_x}}\hat{x} + \frac{\partial{E^{inter}_{DMI}}}{\partial{m_y}}\hat{y} + \frac{\partial{E^{inter}_{DMI}}}{\partial{m_z}}\hat{z})$ $=\frac{1}{{-u_{0}}{M_s}}(\frac{\partial{}}{\partial{m_x}}(D(m_x \frac{\partial{m_z}}{\partial{x}} - m_z \frac{\partial{m_x}}{\partial{x}} + m_y \frac{\partial{m_z}}{\partial{y}} - m_z \frac{\partial{m_y}}{\partial{y}}))\hat{x} + \frac{\partial{}}{\partial{m_y}}(D(m_x \frac{\partial{m_z}}{\partial{x}} - m_z \frac{\partial{m_x}}{\partial{x}} + m_y \frac{\partial{m_z}}{\partial{y}} - m_z \frac{\partial{m_y}}{\partial{y}}))\hat{y} + \frac{\partial{}}{\partial{m_z}}(D(m_x \frac{\partial{m_z}}{\partial{x}} - m_z \frac{\partial{m_x}}{\partial{x}} + m_y \frac{\partial{m_z}}{\partial{y}} - m_z \frac{\partial{m_y}}{\partial{y}}))\hat{z})$ ***********************************************(式子 1) 其中式子（1）的第一项： $\frac{\partial{}}{\partial{m_x}}(D(m_x \frac{\partial{m_z}}{\partial{x}} - m_z \frac{\partial{m_x}}{\partial{x}} + m_y \frac{\partial{m_z}}{\partial{y}} - m_z \frac{\partial{m_y}}{\partial{y}}))$ $=D[\frac{\partial{}}{\partial{m_x}}(m_x \frac{\partial{m_z}}{\partial{x}} ) -\frac{\partial{}}{\partial{m_x}}(m_z \frac{\partial{m_x}}{\partial{x}}) +\frac{\partial{}}{\partial{m_x}}(m_y \frac{\partial{m_z}}{\partial{y}}) -\frac{\partial{}}{\partial{m_x}}(m_z \frac{\partial{m_y}}{\partial{y}})$ $=D[\frac{\partial{m_z}}{\partial{x}} -m_z \frac{\partial{}}{\partial{x}} +0 -0]$ $=D[\frac{\partial{m_z}}{\partial{x}} -m_z \frac{\partial{}}{\partial{x}}]$ 其中式子（1）的第二项： $\frac{\partial{}}{\partial{m_y}}(D(m_x \frac{\partial{m_z}}{\partial{x}} - m_z \frac{\partial{m_x}}{\partial{x}} + m_y \frac{\partial{m_z}}{\partial{y}} - m_z \frac{\partial{m_y}}{\partial{y}}))$ $=D[\frac{\partial{}}{\partial{m_y}}(m_x \frac{\partial{m_z}}{\partial{x}} ) -\frac{\partial{}}{\partial{m_y}}(m_z \frac{\partial{m_x}}{\partial{x}}) +\frac{\partial{}}{\partial{m_y}}(m_y \frac{\partial{m_z}}{\partial{y}}) -\frac{\partial{}}{\partial{m_y}}(m_z \frac{\partial{m_y}}{\partial{y}})$ $=D[0 - 0 + \frac{\partial{m_z}}{\partial{y}} -m_z \frac{\partial{}}{\partial{y}}]$ $=D[\frac{\partial{m_z}}{\partial{y}} -m_z \frac{\partial{}}{\partial{y}}]$ 其中式子（1）的第三项： $\frac{\partial{}}{\partial{m_z}}(D(m_x \frac{\partial{m_z}}{\partial{x}} - m_z \frac{\partial{m_x}}{\partial{x}} + m_y \frac{\partial{m_z}}{\partial{y}} - m_z \frac{\partial{m_y}}{\partial{y}}))$ $=D[\frac{\partial{}}{\partial{m_z}}(m_x \frac{\partial{m_z}}{\partial{x}} ) -\frac{\partial{}}{\partial{m_z}}(m_z \frac{\partial{m_x}}{\partial{x}}) +\frac{\partial{}}{\partial{m_z}}(m_y \frac{\partial{m_z}}{\partial{y}}) -\frac{\partial{}}{\partial{m_z}}(m_z \frac{\partial{m_y}}{\partial{y}})$ $=D[m_x\frac{\partial{}}{\partial{x}} -\frac{\partial{m_x}}{\partial{x}} +m_y\frac{\partial{}}{\partial{y}} -\frac{\partial{m_y}}{\partial{y}}]$ 将以上三项带回式子（1）合并得到界面DMI能的有效场： $\overrightarrow{H^{inter}_{DMI}}$ $=\frac{1}{{-u_{0}}{M_s}} [D[\frac{\partial{m_z}}{\partial{x}} -m_z \frac{\partial{}}{\partial{x}}]\hat{x} +D[\frac{\partial{m_z}}{\partial{y}} -m_z \frac{\partial{}}{\partial{y}}]\hat{y} +D[m_x\frac{\partial{}}{\partial{x}} -\frac{\partial{m_x}}{\partial{x}} +m_y\frac{\partial{}}{\partial{y}} -\frac{\partial{m_y}}{\partial{y}}]\hat{z}]$ $=\frac{D}{{-u_{0}}{M_s}} [(\frac{\partial{m_z}}{\partial{x}}\hat{x} -m_z \frac{\partial{\hat{x}}}{\partial{x}}) +(\frac{\partial{m_z}}{\partial{y}}\hat{y} -m_z \frac{\partial{\hat{y}}}{\partial{y}}) +(m_x\frac{\partial{\hat{z}}}{\partial{x}} -\frac{\partial{m_x}}{\partial{x}}\hat{z} +m_y\frac{\partial{\hat{z}}}{\partial{y}} -\frac{\partial{m_y}}{\partial{y}}\hat{z})]$ $=\frac{D}{{-u_{0}}{M_s}} [(\frac{\partial{m_z}}{\partial{x}}\hat{x} -m_z \cdot 0) +(\frac{\partial{m_z}}{\partial{y}}\hat{y} -m_z \cdot 0) +(m_x \cdot 0 -\frac{\partial{m_x}}{\partial{x}}\hat{z} +m_y \cdot 0 -\frac{\partial{m_y}}{\partial{y}}\hat{z})]$ ********************************（$\hat{x}，\hat{y}，\hat{z}$ 都是（方向）常矢量，它们的偏导数为0） $=\frac{D}{{-u_{0}}{M_s}} (\frac{\partial{m_z}}{\partial{x}}\hat{x} +\frac{\partial{m_z}}{\partial{y}}\hat{y} -\frac{\partial{m_x}}{\partial{x}}\hat{z} -\frac{\partial{m_y}}{\partial{y}}\hat{z})$ $=\frac{D}{{u_{0}}{M_s}}[(\nabla \cdot \vec{m})\hat{z} - \nabla{m_z}]$ # 四、体DMI能的有效场 已知体DMI能的能量密度公式： $$ E^{bulk}_{DMI}=D[\vec{m} \cdot (\nabla \times \vec{m})] $$ 首先展开 $\vec{m}$ 的旋度 $\nabla \times \vec{m}$ $=\begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ \frac{\partial{}}{\partial{x}} & \frac{\partial{}}{\partial{y}} & \frac{\partial{}}{\partial{z}} \\ m_x & m_y & m_z \end{vmatrix}$ $=(\frac{\partial{m_z}}{\partial{y}} - \frac{\partial{m_y}}{\partial{z}}, \frac{\partial{m_x}}{\partial{z}} - \frac{\partial{m_z}}{\partial{x}}, \frac{\partial{m_y}}{\partial{x}} - \frac{\partial{m_x}}{\partial{y}})$ 带回原式得到： $$ E^{bulk}_{DMI}=D[\vec{m} \cdot (\nabla \times \vec{m})] =D[(m_x,m_y,m_z) \cdot (\frac{\partial{m_z}}{\partial{y}} - \frac{\partial{m_y}}{\partial{z}}, \frac{\partial{m_x}}{\partial{z}} - \frac{\partial{m_z}}{\partial{x}}, \frac{\partial{m_y}}{\partial{x}} - \frac{\partial{m_x}}{\partial{y}})] $$ $$ =D[m_x(\frac{\partial{m_z}}{\partial{y}} - \frac{\partial{m_y}}{\partial{z}}) +m_y(\frac{\partial{m_x}}{\partial{z}} - \frac{\partial{m_z}}{\partial{x}}) +m_z(\frac{\partial{m_y}}{\partial{x}} - \frac{\partial{m_x}}{\partial{y}})] $$ 其中 $D$ 是体DMI强度，单位 $J/m^2$（注意：==**体DMI和界面DMI单位相同**==）。那么可以得到体DMI能的有效场： $\overrightarrow{H^{bulk}_{DMI}}$ $=\frac{1}{{-u_{0}}{M_s}}\frac{\delta{E^{bulk}_{DMI}}}{\delta{\overrightarrow{m}}}$ $=\frac{1}{{-u_{0}}{M_s}}\frac{\partial{E^{bulk}_{DMI}}}{\partial{\overrightarrow{m}}}$ $=\frac{1}{{-u_{0}}{M_s}}(\frac{\partial{E^{bulk}_{DMI}}}{\partial{m_x}}\hat{x} + \frac{\partial{E^{bulk}_{DMI}}}{\partial{m_y}}\hat{y} + \frac{\partial{E^{bulk}_{DMI}}}{\partial{m_z}}\hat{z})$ $=\frac{1}{{-u_{0}}{M_s}}\Bigg( \frac{\partial{}}{\partial{m_x}}[D(m_x(\frac{\partial{m_z}}{\partial{y}} - \frac{\partial{m_y}}{\partial{z}})+m_y(\frac{\partial{m_x}}{\partial{z}} - \frac{\partial{m_z}}{\partial{x}})+m_z(\frac{\partial{m_y}}{\partial{x}} - \frac{\partial{m_x}}{\partial{y}})]\hat{x} +\frac{\partial{}}{\partial{m_y}}[D(m_x(\frac{\partial{m_z}}{\partial{y}} - \frac{\partial{m_y}}{\partial{z}})+m_y(\frac{\partial{m_x}}{\partial{z}} - \frac{\partial{m_z}}{\partial{x}})+m_z(\frac{\partial{m_y}}{\partial{x}} - \frac{\partial{m_x}}{\partial{y}})]\hat{y} +\frac{\partial{}}{\partial{m_z}}[D(m_x(\frac{\partial{m_z}}{\partial{y}} - \frac{\partial{m_y}}{\partial{z}})+m_y(\frac{\partial{m_x}}{\partial{z}} - \frac{\partial{m_z}}{\partial{x}})+m_z(\frac{\partial{m_y}}{\partial{x}} - \frac{\partial{m_x}}{\partial{y}})]\hat{z}\Bigg)$ ***********************************************(式子 1) 其中式子（1）的第一项： $\frac{\partial{}}{\partial{m_x}}[D(m_x(\frac{\partial{m_z}}{\partial{y}} - \frac{\partial{m_y}}{\partial{z}})+m_y(\frac{\partial{m_x}}{\partial{z}} - \frac{\partial{m_z}}{\partial{x}})+m_z(\frac{\partial{m_y}}{\partial{x}} - \frac{\partial{m_x}}{\partial{y}})]$ $=D[(\frac{\partial{m_z}}{\partial{y}} - \frac{\partial{m_y}}{\partial{z}})+m_y(\frac{\partial{}}{\partial{z}} - 0) +m_z(0- \frac{\partial{}}{\partial{y}})]$ $=D(\frac{\partial{m_z}}{\partial{y}} - \frac{\partial{m_y}}{\partial{z}}+m_y\frac{\partial{}}{\partial{z}} - m_z \frac{\partial{}}{\partial{y}})$ 其中式子（1）的第二项： $\frac{\partial{}}{\partial{m_y}}[D(m_x(\frac{\partial{m_z}}{\partial{y}} - \frac{\partial{m_y}}{\partial{z}})+m_y(\frac{\partial{m_x}}{\partial{z}} - \frac{\partial{m_z}}{\partial{x}})+m_z(\frac{\partial{m_y}}{\partial{x}} - \frac{\partial{m_x}}{\partial{y}})]$ $=D[m_x (0 - \frac{\partial{}}{\partial{z}})+(\frac{\partial{m_x}}{\partial{z}} - \frac{\partial{m_z}}{\partial{x}}) +m_z(\frac{\partial{}}{\partial{x}}-0)]$ $=D(-m_x \frac{\partial{}}{\partial{z}} + \frac{\partial{m_x}}{\partial{z}} - \frac{\partial{m_z}}{\partial{x}} + m_z \frac{\partial{}}{\partial{x}})$ 其中式子（1）的第三项： $\frac{\partial{}}{\partial{m_z}}[D(m_x(\frac{\partial{m_z}}{\partial{y}} - \frac{\partial{m_y}}{\partial{z}})+m_y(\frac{\partial{m_x}}{\partial{z}} - \frac{\partial{m_z}}{\partial{x}})+m_z(\frac{\partial{m_y}}{\partial{x}} - \frac{\partial{m_x}}{\partial{y}})]$ $=D[m_x(\frac{\partial{}}{\partial{y}} - 0)+m_y(0 - \frac{\partial{}}{\partial{x}}) +(\frac{\partial{m_y}}{\partial{x}}- \frac{\partial{m_x}}{\partial{y}})]$ $=D(m_x\frac{\partial{}}{\partial{y}} - m_y\frac{\partial{}}{\partial{x}}+\frac{\partial{m_y}}{\partial{x}} - \frac{\partial{m_x}}{\partial{y}})$ 将以上三项带回式子（1）合并得到体DMI能的有效场： $\overrightarrow{H^{bulk}_{DMI}}$ $=\frac{1}{{-u_{0}}{M_s}}\Bigg( D(\frac{\partial{m_z}}{\partial{y}} - \frac{\partial{m_y}}{\partial{z}}+m_y\frac{\partial{}}{\partial{z}} - m_z \frac{\partial{}}{\partial{y}})\hat{x} +D(-m_x \frac{\partial{}}{\partial{z}} + \frac{\partial{m_x}}{\partial{z}} - \frac{\partial{m_z}}{\partial{x}} + m_z \frac{\partial{}}{\partial{x}})\hat{y} +D(m_x\frac{\partial{}}{\partial{y}} - m_y\frac{\partial{}}{\partial{x}}+\frac{\partial{m_y}}{\partial{x}} - \frac{\partial{m_x}}{\partial{y}})\hat{z}\Bigg)$ $=\frac{D}{{-u_{0}}{M_s}}\Bigg( (\frac{\partial{m_z}}{\partial{y}}\hat{x} - \frac{\partial{m_y}}{\partial{z}}\hat{x}+m_y\frac{\partial{\hat{x}}}{\partial{z}} - m_z \frac{\partial{\hat{x}}}{\partial{y}}) +(-m_x \frac{\partial{\hat{y}}}{\partial{z}} + \frac{\partial{m_x}}{\partial{z}}\hat{y} - \frac{\partial{m_z}}{\partial{x}}\hat{y} + m_z \frac{\partial{\hat{y}}}{\partial{x}}) +(m_x\frac{\partial{\hat{z}}}{\partial{y}} - m_y\frac{\partial{\hat{z}}}{\partial{x}}+\frac{\partial{m_y}}{\partial{x}}\hat{z} - \frac{\partial{m_x}}{\partial{y}}\hat{z})\Bigg)$ ********************************（$\hat{x}，\hat{y}，\hat{z}$ 都是（方向）常矢量，它们的偏导数为0） $=\frac{D}{{-u_{0}}{M_s}}\Bigg( (\frac{\partial{m_z}}{\partial{y}}\hat{x} - \frac{\partial{m_y}}{\partial{z}}\hat{x}+ 0 - 0) +(- 0+ \frac{\partial{m_x}}{\partial{z}}\hat{y} - \frac{\partial{m_z}}{\partial{x}}\hat{y} + 0) +(0 - 0+\frac{\partial{m_y}}{\partial{x}}\hat{z} - \frac{\partial{m_x}}{\partial{y}}\hat{z})\Bigg)$ $=\frac{D}{{-u_{0}}{M_s}}[ (\frac{\partial{m_z}}{\partial{y}} - \frac{\partial{m_y}}{\partial{z}})\hat{x} +(\frac{\partial{m_x}}{\partial{z}} - \frac{\partial{m_z}}{\partial{x}})\hat{y} +(\frac{\partial{m_y}}{\partial{x}}\hat{z} - \frac{\partial{m_x}}{\partial{y}})\hat{z}]$ $=\frac{D}{{-u_{0}}{M_s}}(\nabla \times \vec{m})$ # 五、磁晶单轴各向异性能的有效场 已知磁晶单轴各向异性能的能量密度公式： $$ E_{uniaxial}=-K_u (\vec{m} \cdot \vec{u})^2 或 E_{uniaxial}=-K_1 (\vec{m} \cdot \vec{u})^2 -K_2 (\vec{m} \cdot \vec{u})^2 （高阶形式） $$ 其中 $\vec{u}$ 是磁晶易轴方向（单位矢量）， $K_u$ 是磁晶单轴各向异性常数，单位 $J/m^3$。那么可以得到磁晶单轴各向异性能的有效场： $\overrightarrow{H_{uniaxial}}$ $=\frac{1}{{-u_{0}}{M_s}}\frac{\delta{E_{uniaxial}}}{\delta{\overrightarrow{m}}}$ $=\frac{1}{{-u_{0}}{M_s}}\frac{\partial{E_{uniaxial}}}{\partial{\overrightarrow{m}}}$ $=\frac{1}{{-u_{0}}{M_s}}(\frac{\partial{E_{uniaxial}}}{\partial{m_x}}\hat{x} + \frac{\partial{E_{uniaxial}}}{\partial{m_y}}\hat{y} + \frac{\partial{E_{uniaxial}}}{\partial{m_z}}\hat{z})$ 此处和推导塞曼能的有效场的步骤相同，将 $(\vec{m} \cdot \vec{u})$ 展开为 $m_x u_x+m_y u_y+m_z u_z$ 方便我们求导，带回上式： $=\frac{1}{{-u_{0}}{M_s}}\Bigg(\frac{\partial{}}{\partial{m_x}}[-K_u(m_x u_x+m_y u_y+m_z u_z)^2]\hat{x} +\frac{\partial{}}{\partial{m_y}}[-K_u(m_x u_x+m_y u_y+m_z u_z)^2]\hat{y} +\frac{\partial{}}{\partial{m_z}}[-K_u(m_x u_x+m_y u_y+m_z u_z)^2]\hat{z}\Bigg)$ ***********************************************(式子 1) 其中式子（1）的第一项： $\frac{\partial{}}{\partial{m_x}}[-K_u(m_x u_x+m_y u_y+m_z u_z)^2]$ $=-2K_u[(m_x u_x+m_y u_y+m_z u_z)][u_x + 0 + 0]$ $=-2K_u(\vec{m} \cdot \vec{u}) u_x$ 其中式子（1）的第二项： $\frac{\partial{}}{\partial{m_y}}[-K_u(m_x u_x+m_y u_y+m_z u_z)^2]$ $=-2K_u[(m_x u_x+m_y u_y+m_z u_z)][0 + u_y + 0]$ $=-2K_u(\vec{m} \cdot \vec{u}) u_y$ 其中式子（1）的第三项： $\frac{\partial{}}{\partial{m_z}}[-K_u(m_x u_x+m_y u_y+m_z u_z)^2]$ $=-2K_u[(m_x u_x+m_y u_y+m_z u_z)][0 + 0 + u_z]$ $=-2K_u(\vec{m} \cdot \vec{u}) u_z$ 将以上三项带回式子（1）合并得到磁晶单轴各向异性能的有效场： $\overrightarrow{H_{uniaxial}}$ $=\frac{1}{{-u_{0}}{M_s}}\Bigg(-2K_u(\vec{m} \cdot \vec{u}) u_x \hat{x} -2K_u(\vec{m} \cdot \vec{u}) u_y \hat{y} -2K_u(\vec{m} \cdot \vec{u}) u_y \hat{z}\Bigg)$ $=\frac{-2K_u}{{-u_{0}}{M_s}}(\vec{m} \cdot \vec{u}) (u_x \hat{x} +u_y \hat{y} -u_y \hat{z})$ $=\frac{2K_u}{{u_{0}}{M_s}}(\vec{m} \cdot \vec{u}) \vec{u}$ # 六、磁晶立方各向异性能的有效场 已知磁晶立方各向异性能的能量密度公式： $$ E_{cubic}=-K_c [(\vec{m} \cdot \vec{u_1})^2 (\vec{m} \cdot \vec{u_2})^2 +(\vec{m} \cdot \vec{u_2})^2 (\vec{m} \cdot \vec{u_3})^2 +(\vec{m} \cdot \vec{u_3})^2 (\vec{m} \cdot \vec{u_1})^2] $$ 其中 $\vec{u_1}，\vec{u_2}，\vec{u_3}$ 是磁晶易轴方向（单位矢量），而且 $\vec{u_3}$ 是通过 $\vec{u_1}，\vec{u_2}$ 的叉乘确定的：$\vec{u_3} = \vec{u_1} \times \vec{u_2}$ 。$K_c$ 是磁晶立方各向异性常数，单位 $J/m^3$。那么可以得到磁晶立方各向异性能的有效场： $\overrightarrow{H_{cubic}}$ $=\frac{1}{{-u_{0}}{M_s}}\frac{\delta{E_{cubic}}}{\delta{\overrightarrow{m}}}$ $=\frac{1}{{-u_{0}}{M_s}}\frac{\partial{E_{cubic}}}{\partial{\overrightarrow{m}}}$ $=\frac{1}{{-u_{0}}{M_s}}(\frac{\partial{E_{cubic}}}{\partial{m_x}}\hat{x} + \frac{\partial{E_{cubic}}}{\partial{m_y}}\hat{y} + \frac{\partial{E_{cubic}}}{\partial{m_z}}\hat{z})$ $=\frac{1}{{-u_{0}}{M_s}}\Bigg( \frac{\partial{}}{\partial{m_x}}[-K_c [(\vec{m} \cdot \vec{u_1})^2 (\vec{m} \cdot \vec{u_2})^2 +(\vec{m} \cdot \vec{u_2})^2 (\vec{m} \cdot \vec{u_3})^2 +(\vec{m} \cdot \vec{u_3})^2 (\vec{m} \cdot \vec{u_1})^2]]\hat{x}$ $+\frac{\partial{}}{\partial{m_y}}[-K_c [(\vec{m} \cdot \vec{u_1})^2 (\vec{m} \cdot \vec{u_2})^2 +(\vec{m} \cdot \vec{u_2})^2 (\vec{m} \cdot \vec{u_3})^2 +(\vec{m} \cdot \vec{u_3})^2 (\vec{m} \cdot \vec{u_1})^2]]\hat{y}$ $+\frac{\partial{}}{\partial{m_z}}[-K_c [(\vec{m} \cdot \vec{u_1})^2 (\vec{m} \cdot \vec{u_2})^2 +(\vec{m} \cdot \vec{u_2})^2 (\vec{m} \cdot \vec{u_3})^2 +(\vec{m} \cdot \vec{u_3})^2 (\vec{m} \cdot \vec{u_1})^2]]\hat{z}\Bigg)$ $=\frac{1}{{-u_{0}}{M_s}}\Bigg( \frac{\partial{}}{\partial{m_x}}[-K_c [[(\vec{m} \cdot \vec{u_1}) (\vec{m} \cdot \vec{u_2})]^2 +[(\vec{m} \cdot \vec{u_2}) (\vec{m} \cdot \vec{u_3})]^2 +[(\vec{m} \cdot \vec{u_3}) (\vec{m} \cdot \vec{u_1})]^2]]\hat{x}$ $+\frac{\partial{}}{\partial{m_y}}[-K_c [[(\vec{m} \cdot \vec{u_1}) (\vec{m} \cdot \vec{u_2})]^2 +[(\vec{m} \cdot \vec{u_2}) (\vec{m} \cdot \vec{u_3})]^2 +[(\vec{m} \cdot \vec{u_3}) (\vec{m} \cdot \vec{u_1})]^2]]\hat{y}$ $+\frac{\partial{}}{\partial{m_z}}[-K_c [[(\vec{m} \cdot \vec{u_1}) (\vec{m} \cdot \vec{u_2})]^2 +[(\vec{m} \cdot \vec{u_2}) (\vec{m} \cdot \vec{u_3})]^2 +[(\vec{m} \cdot \vec{u_3}) (\vec{m} \cdot \vec{u_1})]^2]]\hat{z}\Bigg)$ ***********************************************(式子 1) 此处和推导磁晶单轴各向异性能的有效场的步骤相同，将 $(\vec{m} \cdot \vec{u_{n}})$ 展开为 $m_x u_{nx}+m_y u_{ny}+m_z u_{nz}$ 方便我们求导，带回上式，计算式子（1）的第一项： $\frac{\partial{}}{\partial{m_x}}[-K_c [[(\vec{m} \cdot \vec{u_1}) (\vec{m} \cdot \vec{u_2})]^2 +[(\vec{m} \cdot \vec{u_2}) (\vec{m} \cdot \vec{u_3})]^2 +[(\vec{m} \cdot \vec{u_3}) (\vec{m} \cdot \vec{u_1})]^2]]$ $=-K_c\frac{\partial{}}{\partial{m_x}}\Bigg([(m_x u_{1x}+m_y u_{1y}+m_z u_{1z}) (m_x u_{2x}+m_y u_{2y}+m_z u_{2z})]^2$ $+[(m_x u_{2x}+m_y u_{2y}+m_z u_{2z}) (m_x u_{3x}+m_y u_{3y}+m_z u_{3z})]^2$ $+[(m_x u_{3x}+m_y u_{3y}+m_z u_{3z}) (m_x u_{1x}+m_y u_{1y}+m_z u_{1z})]^2\Bigg)$ ***********************************************(式子 2) 对于式子（2）的第一项，可以利用复合函数的求导法则（$(uv)^{\prime}=u^{\prime}v + uv^{\prime}$）来简化计算，于是： $\frac{\partial{}}{\partial{m_x}}[(m_x u_{1x}+m_y u_{1y}+m_z u_{1z}) (m_x u_{2x}+m_y u_{2y}+m_z u_{2z})]^2]$ $=2[(m_x u_{1x}+m_y u_{1y}+m_z u_{1z}) (m_x u_{2x}+m_y u_{2y}+m_z u_{2z})] \Bigg( (u_{1x}+0+0) (m_x u_{2x}+m_y u_{2y}+m_z u_{2z}) +(m_x u_{1x}+m_y u_{1y}+m_z u_{1z}) (u_{2x}+0 +0)$ $=2[(m_x u_{1x}+m_y u_{1y}+m_z u_{1z}) (m_x u_{2x}+m_y u_{2y}+m_z u_{2z})] \Bigg( u_{1x} (m_x u_{2x}+m_y u_{2y}+m_z u_{2z}) +(m_x u_{1x}+m_y u_{1y}+m_z u_{1z}) u_{2x}\Bigg)$ $=2[(\vec{m} \cdot \vec{u_1}) (\vec{m} \cdot \vec{u_2})^2 u_{1x} +(\vec{m} \cdot \vec{u_1})^2 (\vec{m} \cdot \vec{u_2}) u_{2x}]$ 同理，可以根据结果的规律写出式子（2）的第二项： $\frac{\partial{}}{\partial{m_x}}[[(m_x u_{2x}+m_y u_{2y}+m_z u_{2z}) (m_x u_{3x}+m_y u_{3y}+m_z u_{3z})]^2]$ $=2[(\vec{m} \cdot \vec{u_2}) (\vec{m} \cdot \vec{u_3})^2 u_{2x} +(\vec{m} \cdot \vec{u_2})^2 (\vec{m} \cdot \vec{u_3}) u_{3x}]$ 同理，可以根据结果的规律写出式子（2）的第三项： $\frac{\partial{}}{\partial{m_x}}[[(m_x u_{3x}+m_y u_{3y}+m_z u_{3z}) (m_x u_{1x}+m_y u_{1y}+m_z u_{1z})]^2]$ $=2[(\vec{m} \cdot \vec{u_3}) (\vec{m} \cdot \vec{u_1})^2 u_{3x} +(\vec{m} \cdot \vec{u_3})^2 (\vec{m} \cdot \vec{u_1}) u_{1x}]$ 将以上三项带回式子（2）合并得到式子（2）的结果，即式子（1）的第一项： $\frac{\partial{}}{\partial{m_x}}[-K_c [[(\vec{m} \cdot \vec{u_1}) (\vec{m} \cdot \vec{u_2})]^2 +[(\vec{m} \cdot \vec{u_2}) (\vec{m} \cdot \vec{u_3})]^2 +[(\vec{m} \cdot \vec{u_3}) (\vec{m} \cdot \vec{u_1})]^2]]\hat{x}$ $=-K_c \Bigg($ $2[(\vec{m} \cdot \vec{u_1}) (\vec{m} \cdot \vec{u_2})^2 u_{1x} +(\vec{m} \cdot \vec{u_1})^2 (\vec{m} \cdot \vec{u_2}) u_{2x}]+$ $2[(\vec{m} \cdot \vec{u_2}) (\vec{m} \cdot \vec{u_3})^2 u_{2x} +(\vec{m} \cdot \vec{u_2})^2 (\vec{m} \cdot \vec{u_3}) u_{3x}]+$ $2[(\vec{m} \cdot \vec{u_3}) (\vec{m} \cdot \vec{u_1})^2 u_{3x} +(\vec{m} \cdot \vec{u_3})^2 (\vec{m} \cdot \vec{u_1}) u_{1x}]$ $\Bigg)\hat{x}$ 同理，可以根据第一项结果的规律写出式子（1）的第二项： $\frac{\partial{}}{\partial{m_y}}[-K_c [[(\vec{m} \cdot \vec{u_1}) (\vec{m} \cdot \vec{u_2})]^2 +[(\vec{m} \cdot \vec{u_2}) (\vec{m} \cdot \vec{u_3})]^2 +[(\vec{m} \cdot \vec{u_3}) (\vec{m} \cdot \vec{u_1})]^2]]\hat{y}$ $=-K_c \Bigg($ $2[(\vec{m} \cdot \vec{u_1}) (\vec{m} \cdot \vec{u_2})^2 u_{1y} +(\vec{m} \cdot \vec{u_1})^2 (\vec{m} \cdot \vec{u_2}) u_{2y}]+$ $2[(\vec{m} \cdot \vec{u_2}) (\vec{m} \cdot \vec{u_3})^2 u_{2y} +(\vec{m} \cdot \vec{u_2})^2 (\vec{m} \cdot \vec{u_3}) u_{3y}]+$ $2[(\vec{m} \cdot \vec{u_3}) (\vec{m} \cdot \vec{u_1})^2 u_{3y} +(\vec{m} \cdot \vec{u_3})^2 (\vec{m} \cdot \vec{u_1}) u_{1y}]$ $\Bigg)\hat{y}$ 同理，可以根据第一项结果的规律写出式子（1）的第三项： $\frac{\partial{}}{\partial{m_z}}[-K_c [[(\vec{m} \cdot \vec{u_1}) (\vec{m} \cdot \vec{u_2})]^2 +[(\vec{m} \cdot \vec{u_2}) (\vec{m} \cdot \vec{u_3})]^2 +[(\vec{m} \cdot \vec{u_3}) (\vec{m} \cdot \vec{u_1})]^2]]\hat{z}$ $=-K_c \Bigg($ $2[(\vec{m} \cdot \vec{u_1}) (\vec{m} \cdot \vec{u_2})^2 u_{1z} +(\vec{m} \cdot \vec{u_1})^2 (\vec{m} \cdot \vec{u_2}) u_{2z}]+$ $2[(\vec{m} \cdot \vec{u_2}) (\vec{m} \cdot \vec{u_3})^2 u_{2z} +(\vec{m} \cdot \vec{u_2})^2 (\vec{m} \cdot \vec{u_3}) u_{3z}]+$ $2[(\vec{m} \cdot \vec{u_3}) (\vec{m} \cdot \vec{u_1})^2 u_{3z} +(\vec{m} \cdot \vec{u_3})^2 (\vec{m} \cdot \vec{u_1}) u_{1z}]$ $\Bigg)\hat{z}$ 将以上三项带回式子（1）合并得到磁晶立方各向异性能的有效场： $\overrightarrow{H_{cubic}}$： $=\frac{-2K_c}{{-u_{0}}{M_s}}\Bigg( (\vec{m} \cdot \vec{u_1}) (\vec{m} \cdot \vec{u_2})^2(u_{1x}\hat{x} + u_{1y}\hat{y} + u_{1z}\hat{z}) +(\vec{m} \cdot \vec{u_1})^2 (\vec{m} \cdot \vec{u_2})(u_{2x}\hat{x} + u_{2y}\hat{y} + u_{2z}\hat{z})$ $+(\vec{m} \cdot \vec{u_2}) (\vec{m} \cdot \vec{u_3})^2(u_{2x}\hat{x} + u_{2y}\hat{y} + u_{2z}\hat{z}) +(\vec{m} \cdot \vec{u_2})^2 (\vec{m} \cdot \vec{u_3})(u_{3x}\hat{x} + u_{3y}\hat{y} + u_{3z}\hat{z})$ $+(\vec{m} \cdot \vec{u_3}) (\vec{m} \cdot \vec{u_1})^2(u_{3x}\hat{x} + u_{3y}\hat{y} + u_{3z}\hat{z}) +(\vec{m} \cdot \vec{u_3})^2 (\vec{m} \cdot \vec{u_1})(u_{1x}\hat{x} + u_{1y}\hat{y} + u_{1z}\hat{z})\Bigg)$ $=\frac{2K_c}{{u_{0}}{M_s}}\Bigg( (\vec{m} \cdot \vec{u_1}) (\vec{m} \cdot \vec{u_2})^2 \vec{u_1} +(\vec{m} \cdot \vec{u_1})^2 (\vec{m} \cdot \vec{u_2}) \vec{u_2}$ $+(\vec{m} \cdot \vec{u_2}) (\vec{m} \cdot \vec{u_3})^2 \vec{u_2} +(\vec{m} \cdot \vec{u_2})^2 (\vec{m} \cdot \vec{u_3}) \vec{u_3}$ $+(\vec{m} \cdot \vec{u_3}) (\vec{m} \cdot \vec{u_1})^2 \vec{u_2} +(\vec{m} \cdot \vec{u_3})^2 (\vec{m} \cdot \vec{u_1}) \vec{u_1}\Bigg)$ $=\frac{2K_c}{{u_{0}}{M_s}}\Bigg( (\vec{m} \cdot \vec{u_1})[(\vec{m} \cdot \vec{u_2})^2 + (\vec{m} \cdot \vec{u_3})^2] \vec{u_1}$ $+(\vec{m} \cdot \vec{u_2})[(\vec{m} \cdot \vec{u_3})^2 + (\vec{m} \cdot \vec{u_1})^2] \vec{u_2}$ $+(\vec{m} \cdot \vec{u_3})[(\vec{m} \cdot \vec{u_1})^2 + (\vec{m} \cdot \vec{u_2})^2] \vec{u_3} \Bigg)$
# 总结 <font color=#999AAA > 本文总结了微磁学中通过几个基本的能量密度公式推导出其有效场，内容核心就是求偏导，各种各样的偏导。至于微磁学中的其它公式推导，暂时还没学到皮毛，只有等以后再说了。- 1
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- 原文地址：https://blog.csdn.net/qq_43572058/article/details/126234976