- #1
Ken Gallock
- 30
- 0
Hi.
I don't understand the meaning of "up to total derivatives".
It was used during a lecture on superfluid. It says as follows:
---------------------------------------------------------------------
Lagrangian for complex scalar field ##\phi## is
$$
\mathcal{L}=\frac12 (\partial_\mu \phi)^* \partial^\mu \phi - \frac12 m^2 |\phi|^2 -\lambda |\phi|^4.
$$
Take non-relativistic limit:
$$
\phi(x)=\dfrac{1}{\sqrt{2m}}e^{-imt}\varphi(t,x).
$$
Then, lagrangian for non-relativistic complex scalar field ##\mathcal{L}_{NR}## can be written as follows:
$$
\mathcal{L}_{NR}=\partial_t\phi^* \partial_t \phi - \nabla \phi^* \cdot \nabla \phi - m^2|\phi|^2 -\lambda|\phi|^4\\
=\dfrac{1}{2m}(im\varphi^*+\dot{\varphi}^*)(-im\varphi+\dot{\varphi})-\dfrac{1}{2m}\nabla\varphi^*\cdot \nabla \varphi -\dfrac{m}{2}|\varphi|^2-\dfrac{\lambda}{4m^2}|\varphi|^4.
$$
In non-relativistic limit,
$$
\partial_t\sim \nabla^2,
$$
therefore, we only consider first order of ##\partial_t##.
$$
\mathcal{L}_{NR}=\dfrac{1}{2m}[im(-im)\varphi^*\varphi+im\varphi^*\dot{\varphi}-im\dot{\varphi}^*\varphi]-\dfrac{1}{2m}\nabla\varphi^*\cdot \nabla \varphi -\dfrac{m}{2}|\varphi|^2-\dfrac{\lambda}{4m^2}|\varphi|^4\\
\simeq \dfrac{i}{2}(\varphi^*\dot{\varphi}-\dot{\varphi}^*\varphi)-\dfrac{1}{2m}\nabla\varphi^*\cdot \nabla \varphi -\dfrac{\lambda}{4m^2}|\varphi|^4\\
$$
Now, up to total derivatives,
$$
\mathcal{L}_{NR}\simeq \dfrac{i}{2}(\varphi^*\dot{\varphi}-\dot{\varphi}^*\varphi)-\dfrac{1}{2m}\nabla\varphi^*\cdot \nabla \varphi -\dfrac{\lambda}{4m^2}|\varphi|^4\\
=\varphi^*\left( i\partial_t+\dfrac{\nabla^2}{2m} \right) \varphi -\dfrac{\lambda}{4m^2}|\varphi|^4.
$$
-----------------------------------------
I don't understand the last part of this. Drop total derivatives?
I don't understand the meaning of "up to total derivatives".
It was used during a lecture on superfluid. It says as follows:
---------------------------------------------------------------------
Lagrangian for complex scalar field ##\phi## is
$$
\mathcal{L}=\frac12 (\partial_\mu \phi)^* \partial^\mu \phi - \frac12 m^2 |\phi|^2 -\lambda |\phi|^4.
$$
Take non-relativistic limit:
$$
\phi(x)=\dfrac{1}{\sqrt{2m}}e^{-imt}\varphi(t,x).
$$
Then, lagrangian for non-relativistic complex scalar field ##\mathcal{L}_{NR}## can be written as follows:
$$
\mathcal{L}_{NR}=\partial_t\phi^* \partial_t \phi - \nabla \phi^* \cdot \nabla \phi - m^2|\phi|^2 -\lambda|\phi|^4\\
=\dfrac{1}{2m}(im\varphi^*+\dot{\varphi}^*)(-im\varphi+\dot{\varphi})-\dfrac{1}{2m}\nabla\varphi^*\cdot \nabla \varphi -\dfrac{m}{2}|\varphi|^2-\dfrac{\lambda}{4m^2}|\varphi|^4.
$$
In non-relativistic limit,
$$
\partial_t\sim \nabla^2,
$$
therefore, we only consider first order of ##\partial_t##.
$$
\mathcal{L}_{NR}=\dfrac{1}{2m}[im(-im)\varphi^*\varphi+im\varphi^*\dot{\varphi}-im\dot{\varphi}^*\varphi]-\dfrac{1}{2m}\nabla\varphi^*\cdot \nabla \varphi -\dfrac{m}{2}|\varphi|^2-\dfrac{\lambda}{4m^2}|\varphi|^4\\
\simeq \dfrac{i}{2}(\varphi^*\dot{\varphi}-\dot{\varphi}^*\varphi)-\dfrac{1}{2m}\nabla\varphi^*\cdot \nabla \varphi -\dfrac{\lambda}{4m^2}|\varphi|^4\\
$$
Now, up to total derivatives,
$$
\mathcal{L}_{NR}\simeq \dfrac{i}{2}(\varphi^*\dot{\varphi}-\dot{\varphi}^*\varphi)-\dfrac{1}{2m}\nabla\varphi^*\cdot \nabla \varphi -\dfrac{\lambda}{4m^2}|\varphi|^4\\
=\varphi^*\left( i\partial_t+\dfrac{\nabla^2}{2m} \right) \varphi -\dfrac{\lambda}{4m^2}|\varphi|^4.
$$
-----------------------------------------
I don't understand the last part of this. Drop total derivatives?