# What is the energy density of the nabla-term of the Higgs hamiltonian in SI units?

According my source, the Higgs field hamiltonian density is

$$H(\phi )={\frac {1}{2}}\left|{\dot {\phi }}\right|^{2}+\left|\nabla \phi \right|^{2}+V(\left|\phi \right|)$$

The first term is the kinetic energy of the field. The second term is the extra potential energy when the field varies from point to point. The third term is the potential energy when the field has any given magnitude.

I am interested on the second term, more clearly in the energy density of a Higgs field, which

1. Static in time.
2. Has $$\rm{|\phi|=246GeV}$$ value.
3. But, it is not constant in space.

It means that moving in space, the Higgs components are rotated to preserve $$\rm{|\phi|=246GeV}$$.

I think, such a Higgs field would have energy only from the second term, which is the square of a space-derivative tag; meaning that if the change is enough long (in space), maybe a significant, macro-sized, permanent change of the Higgs field would be possible without astronomical energy needs.

My question is the energy density of this, in SI units.

I think, the answer would be in $$\rm{\frac{J}{rad \cdot m^3}}$$, i.e. if the energy in an $$\rm{1 m^3}$$ cube, if the angle between the Higgs-components of its two opposite sides is 1 rad.

It is not a problem if the answer is valid only for infinitesimal changes.

• I don't understand what this question is asking for - you've already written down the expression for the energy density (your $H(\phi)$). If you know $\phi$ as a function of $x$, you can compute $H(\phi)(x)$ from that, if you don't, you can't. What do you expect an answerer to do here? Oct 15 '20 at 17:29
• @ACuriousMind The energy of a small cube (let it the coordinates of its opposing corners $(0;0;0)$ and $(\Delta r;\Delta r;\Delta r)$), where the $\phi(\underline{x},t)=246GeV \cdot e^{i \frac{\alpha x_1}{\Delta r}}$. In SI units, where $\Delta r$ and $\alpha$ are input parameters. $\Delta r$ and $\alpha$ are small. (Probably I used badly the common QFT notation, I hope it is better comprehensible.) Oct 15 '20 at 17:54
• @ACuriousMind related question Oct 15 '20 at 18:03

• Why? The $|\nabla \phi|^2$ looks to me very clear. Did you read the penultimate sentence? Oct 13 '20 at 18:35