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Consider a scalar field with the following lagrangian density:

$$\mathscr{L}=-\frac{1}{2} \partial_{\mu} \phi \partial^{\mu} \phi-V(\phi).$$

I want to find the corresponding Euler-Lagrange equation, so:

$$\partial_{\mu} \frac{\delta L}{\delta\left(\partial_{\mu} \phi\right)}-\frac{\delta L}{\delta \phi}=0.$$

The term $$\frac{\delta L}{\delta \phi} = V'(\phi),$$ with the prime symbol indicating derivatives with respect to the field itself. Why is this the case? In other terms, why exactly do the other two terms of the varied lagrangian yield a null contribution? Do they cancel each other somehow or are they zero?

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    $\begingroup$ Hint 1 : \begin{align} \dfrac{\delta L(t)}{\delta\phi(\boldsymbol x,t)} & =\dfrac{\partial \mathscr L}{\partial\phi(\boldsymbol x,t)}-\boldsymbol\nabla\dfrac{\partial \mathscr L}{\partial\left[\boldsymbol\nabla\phi(\boldsymbol x,t)\right]} \tag{01a}\\ \dfrac{\delta L(t)}{\delta\dot{\phi}(\boldsymbol x,t)} & =\dfrac{\partial \mathscr L}{\partial\dot{\phi}(\boldsymbol x,t)} \tag{01b} \end{align} $\endgroup$
    – Frobenius
    Commented Jun 14, 2022 at 14:52
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    $\begingroup$ Hint 2 : \begin{align} &\dfrac{\partial}{\partial t}\left[\dfrac{\delta L(t)}{\delta\,\dot{\!\phi}(\boldsymbol x,t)}\right]-\dfrac{\delta L(t)}{\delta\phi(\boldsymbol x,t)} =0 \tag{02a}\\ &\dfrac{\partial}{\partial t}\left[\dfrac{\partial \mathscr L}{\partial\,\dot{\!\phi}(\boldsymbol x,t)}\right]+\boldsymbol\nabla\dfrac{\partial \mathscr L}{\partial\left[\boldsymbol\nabla\phi(\boldsymbol x,t)\right]}-\dfrac{\partial \mathscr L}{\partial\phi(\boldsymbol x,t)} =0 \tag{02b} \end{align} $\endgroup$
    – Frobenius
    Commented Jun 14, 2022 at 15:06

1 Answer 1

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The Lagrange density is ($c=1$) \begin{equation} \mathscr L\left(\phi\left(\boldsymbol x,t\right),\boldsymbol\nabla\phi(\boldsymbol x,t),\dot{\phi}(\boldsymbol x,t)\right)=-\frac12\left(\dot{\phi}^2-\Vert\boldsymbol\nabla\phi\Vert^2\right)-V\left(\phi\right) \tag{01}\label{01} \end{equation} while the Lagrangian is \begin{equation} L(t)=\int\mathrm d^3\boldsymbol x\,\mathscr L\left(\phi\left(\boldsymbol x,t\right),\boldsymbol\nabla\phi(\boldsymbol x,t),\dot{\phi}(\boldsymbol x,t)\right) \tag{02}\label{02} \end{equation} The Euler-Lagrange equation in terms of the Lagrangian and its functional derivatives is \begin{equation} \dfrac{\partial}{\partial t}\left[\dfrac{\delta L(t)}{\delta\,\dot{\!\phi}(\boldsymbol x,t)}\right]-\dfrac{\delta L(t)}{\delta\phi(\boldsymbol x,t)} =0 \tag{03}\label{03} \end{equation} Between the functional derivatives of the Lagrangian $L(t)$ and the partial derivatives of the Lagrange density $\mathscr L$ we have the following relations \begin{align} \dfrac{\delta L(t)}{\delta\,\dot{\!\phi}(\boldsymbol x,t)} & =\dfrac{\partial \mathscr L}{\partial\,\dot{\!\phi}(\boldsymbol x,t)} \tag{04a}\\ \dfrac{\delta L(t)}{\delta\phi(\boldsymbol x,t)} & =\dfrac{\partial \mathscr L}{\partial\phi(\boldsymbol x,t)}-\boldsymbol{\nabla\cdot}\left[\dfrac{\partial \mathscr L}{\partial\left(\boldsymbol\nabla\phi(\boldsymbol x,t)\right)}\right] \tag{04b} \end{align} so the Euler-Lagrange equation in terms of the Lagrange density and its partial derivatives is \begin{equation} \dfrac{\partial}{\partial t}\left[\dfrac{\partial \mathscr L}{\partial\,\dot{\!\phi}(\boldsymbol x,t)}\right]+\boldsymbol{\nabla\cdot}\left[\dfrac{\partial \mathscr L}{\partial\left(\boldsymbol\nabla\phi(\boldsymbol x,t)\right)}\right]-\dfrac{\partial \mathscr L}{\partial\phi(\boldsymbol x,t)} =0 \tag{05}\label{05} \end{equation}

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  • $\begingroup$ If the OP has in mind the FRW metric of this of his/her own questions (not written here explicitly) then my equation \eqref{01} is not valid. $\endgroup$
    – Frobenius
    Commented Jun 17, 2022 at 13:03

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