# Energy of the excitation in the superfluid

According to the chapter 1.3: Superfluidity, from the book "Methods of Quantum Field Theory In Statistical Physics by A.A Abrikosov, it is written that

Consider a Bose liquid at absolute zero, flowing with velocity $$\mathbf{v}$$ in a capillary. In the coordinate system fixed with respect to the liquid, the liquid is at rest and the capillary moves with velocity $$-\mathbf{v}$$. As a result of friction between the liquid and the wall of the capillary, the liquid begins to be "carried along" by the wall. This means that the liquid begins to have nonzero energy and momentum, which is possible only if elementary excitations appear in the liquid. As soon as a single such excitation appears, the liquid acquires momentum $$\mathbf{p}$$ and energy $$\epsilon(\mathbf{p})$$.

Now, suppose we go back to the coordinate system fixed with respect to the capillary. In this system, the energy of the liquid equals $$\epsilon + \mathbf{p} \cdot \mathbf{v}+\frac{1}{2}Mv^2$$

Thus the appearance of an excitation changes the energy by an amount $$\epsilon + \mathbf{p}\cdot\mathbf{v}$$.

However, I do not really see how the term $$\mathbf{p}\cdot\mathbf{v}$$ comes. Could somebody explain this to me, please?

• That term comes from approximating a Lorentz transformation. I'll post an answer later Nov 28 '17 at 16:42

There are two frames of reference. One in which the Bose fluid is flowing at speed $\textbf{v}$, the other one which is at rest; in this last frame, the fluid acquires momentum $\textbf{p}$ and energy $\epsilon(\textbf{p})$.
What is the energy $\epsilon'$ of the fluid in the original frame? The answer can be found by applying a Lorentz transformation with velocity $-\textbf{v}$. The speeds involved are much smaller than the speed of light, so it will be enough to approximate to first order. For ease of calculation, I'll only deal with the one dimensional case. The Lorentz transformation is given by:
$$\Lambda=\begin{pmatrix} \gamma & \gamma v/c & 0 & 0 \\ \gamma v/c & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \approx 1 + \begin{pmatrix} 0 & v/c & 0 & 0 \\ v/c & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} + \mathcal{O}(v^2 /c^2)$$
The four-momentum transforms like: $$\begin{pmatrix} \epsilon'/c \\ p' \\ 0 \\ 0 \end{pmatrix} = \Lambda \begin{pmatrix} \epsilon/c \\ p \\ 0 \\ 0 \end{pmatrix}$$
Therefore the energy of the excitations in the new frame is: $$\epsilon' = \epsilon + pv$$