Relationship of the Demagnetizing Energy to the Demagnetizing Field

[PotatoMerrick]

Hello,

I'm currently reading material on micromagnetics. In these papers, authors define a quantity called the demagnetizing energy ($E_d$) as
$$E_d = -\frac{1}{2} \int_V \vec{m} \cdot \vec{H}_d\;dV$$
where $\vec{m}$ is the internal magnetization of a material sample of volume $V$ and $\vec{H}_d$ is the demagnetising field. The demagnetizing field itself is defined as the negative derivative of the demagnetizing energy with respect to the material magnetisation, i.e.
$$\vec{H}_d = -\frac{dE_d}{d\vec{m}}$$
My problem is that I would like to know how to derive $\vec{H}_d$ by taking the derivative of $E_d$ with respect to $\vec{m}$. This is as far as I have got (and I'm not too sure that this is correct)
$$\frac{dE_d}{d\vec{m}} = \frac{d}{d\vec{m}} \left( -\frac{1}{2}\int_V \vec{m}\cdot\vec{H}_d\;dV \right)$$
$$\frac{dE_d}{d\vec{m}} = -\frac{1}{2}\int_V \frac{\partial}{\partial\vec{m}}\left(\vec{m}\cdot\vec{H}_d\right)\;dV=-\frac{1}{2}\int_V\frac{\partial\vec{m}}{\partial \vec{m}} \cdot \vec{H}_d + \vec{m}\cdot\frac{\partial \vec{H}_d}{\partial \vec{m}}\;dV=-\frac{1}{2} \int_V \vec{H}_d\;dV - \frac{1}{2}\int_V \vec{m}\cdot\frac{\partial \vec{H}_d}{\partial \vec{m}}\;dV$$
Could some kind soul please give me some pointers as to how to proceed and/or explain to me where I'm going wrong?

 

[Vailanor]

Thank you in advance!The integral of the first term is just -1/2 times the total volume V, since \vec{H}_d is not a function of \vec{m}, so this term evaluates to -V\vec{H}_d. To evaluate the second integral, you need to use the chain rule and substitute in the definition of the demagnetizing field: \frac{dE_d}{d\vec{m}} = -\frac{1}{2}V\vec{H}_d - \frac{1}{2}\int_V \vec{m}\cdot\left(\frac{\partial \vec{H}_d}{\partial \vec{m}}\right)\;dV = -\frac{1}{2}V\vec{H}_d - \frac{1}{2}\int_V \vec{m}\cdot\left(-\frac{\partial E_d}{\partial \vec{m}}\right)\;dV = -\frac{1}{2}V\vec{H}_d + \frac{1}{2}\int_V \vec{m}\cdot\vec{H}_d\;dV = -V\vec{H}_d + E_d.Therefore, the derivative of E_d with respect to \vec{m} is\frac{dE_d}{d\vec{m}} = -V\vec{H}_d + E_d,which is equivalent to the definition of the demagnetizing field, \vec{H}_d = -\frac{dE_d}{d\vec{m}}.