# I What does it mean: "up to total derivatives"

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1. May 9, 2017

### Ken Gallock

Hi.
I don't understand the meaning of "up to total derivatives".

It was used during a lecture on superfluid. It says as follows:

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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,
$$\mathcal{L}_{NR}\simeq \dfrac{i}{2}(\varphi^*\dot{\varphi}-\dot{\varphi}^*\varphi)-\dfrac{1}{2m}\nabla\varphi^*\cdot \nabla \varphi -\dfrac{\lambda}{4m^2}|\varphi|^4\\ =\varphi^*\left( i\partial_t+\dfrac{\nabla^2}{2m} \right) \varphi -\dfrac{\lambda}{4m^2}|\varphi|^4.$$
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I don't understand the last part of this. Drop total derivatives?

2. May 9, 2017

### Orodruin

Staff Emeritus
Any difference between the first expression and the final expression is a total derivative.

3. May 12, 2017

### ZealScience

It means the difference between the terms is a derivative of some function. (e.g. In 3-D, the gradient of something.) In the language of differential forms, an exact form. The point is that the total derivative (or exact form) in the action could be converted into a boundary integral by Stokes' theorem (in 3-D the Gauss' theorem).

In classical fields, you do variational derivatives fixing the boundary conditions, so the boundary variation is zero and does not affect the equation of motion.

In quantum fields of many-body physics, you path integrate coherent states. In any case, it will only contribute as a constant factor and can be absorbed into path integral measures which is non-dynamical.

Last edited: May 13, 2017
4. May 19, 2017

### Ken Gallock

Thanks.
So, we don't have to think about fields at the far away like $x^i \rightarrow \infty$.
And from Gauss' theorem, an integral can be calculated like:
$$\int_M d^3x \partial_i X^i \sim \int_{\partial M}(d^2x)_i X^i = 0.$$

5. May 28, 2017

### haushofer

Well, we do have to think about it, because we need to impose boundary conditions. Mathematically, often it is said that fields have compact support on spacetime. This is not always the case though; think about (A)dS spaces and holography.

6. May 29, 2017

### Ken Gallock

I'm not sure about holography, but do you mean that $d$-dimension AdS spacetime's boundary is $(d-1)$-dimension CFT spacetime?

7. May 30, 2017

### haushofer

If you mean that the conformal boundary of AdS is Minkowski spacetime: yes.