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I have a very simple question, but strangely I cannot find any answer on the internet; maybe the answer is too simple that I don't notice. I go straight to the point: if I define a Lagrangian from a Lagrangian density, and so from a definite integral in the coordinate space, how there can be an explicit coordinate dependence on the Lagrangian? In the picture I try to calculate the action along a fixed $s$-parametrized trajectory in the $n+1$-dimensional space in which a scalar field $\phi$ is present ($y_i$ is the generical $i$-component of the $n+1$-dimensional vector $\boldsymbol{y}=(\boldsymbol{x},t)$ ).

\begin{equation} S\boldsymbol{=}\int\limits_{t\left(s_1\right)}^{t\left(s_2\right)}\!\!\!\!\mathrm dt\,L\left(\boldsymbol{x}\left(s\right),t\left(s\right)\vphantom{\tfrac{a}{b}}\right)\boldsymbol{\doteq}\!\!\int\limits_{t\left(s_1\right)}^{t\left(s_2\right)}\!\!\!\!\mathrm dt\!\!\int\limits_{\mathcal M_n}\!\!\mathrm d^n x\,\mathscr L\left(\phi,\{\partial_{y_{i}}\phi\},\boldsymbol{x},t\vphantom{\tfrac{a}{b}}\right) \tag{01}\label{01} \end{equation}

I don't know if it is a stupid question, but this indetermination on what actually is the nature of the quantities I've got in front of me, completely mess my conception up, about what I can and what I can't do, such as when is the case to treat total and partial derivatives in the coordinate, of the lagrangian density; so thanks really in advance to anyone that will answer!

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  • $\begingroup$ What's your source? What is exactly the manifold (???} $\mathcal M_{n}$ ??? $\endgroup$
    – Frobenius
    Jan 30, 2021 at 1:37
  • $\begingroup$ I know to be very unclear when I ask questions, sorry. In this case what I'm actually asking is "if I have a lagrangian $L$ depending on spacetime index $\boldsymbol{x}$, can it arise from a lagrangian density $\mathscr{L}$, or not?". So to me the point is to get an analogy with the general relativity, if I would check my notes again in the future: I was trying to be the more general possible, and I considered an $n+1$-dimensional manifold, instead of a global flat spacetime, that's all, but if you want, that's just notation in the context of the question. $\endgroup$
    – Rob Tan
    Jan 30, 2021 at 13:04
  • $\begingroup$ I think the relativistic discussion is really complicate when I want an integral from a point $\boldsymbol{x}_0$ to another point $\boldsymbol{x}$, so at the moment I'm just interested in the flat spacetime case, to understand if it is possible to define $L(\boldsymbol{x},...)=\int_{\boldsymbol{x}_0}^{\boldsymbol{x}}\text{d}^n\tilde{x}\mathscr{L}(\tilde{\boldsymbol{x}},...)$ $\endgroup$
    – Rob Tan
    Jan 30, 2021 at 13:12

1 Answer 1

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OP's equation is a bit unclear. In this answer we will for simplicity only consider the case where OP's $s$-parametrization is spacetime-filling, i.e. $s=n+1$. Also to be concrete, let $n=3$. Then we are back at standard field theory in 3+1D.

In that case the Lagrangian $$\begin{align}L_V[q(\cdot,t);v(\cdot,t);t]~:=~&\int_V\! d^3x~ {\cal L}\left(q(\vec{x},t);v(\vec{x},t),\frac{\partial q(\vec{x},t)}{\partial \vec{x}};\vec{x},t\right),\cr v(\vec{x},t)~:=~&\frac{d q(\vec{x},t)}{d t},\end{align}\tag{A}$$ can still depend explicitly on time $t\in [t_i,t_f]$. It can obviously not depend on the integration variable $\vec{x}$. However it can in principle depend on the spatial positions of the spatial boundary $\partial V$ of the spatial volume $V$. Here $$\underbrace{V\times [t_i,t_f]}_{\text{spacetime}}\quad\stackrel{q}{\longrightarrow}\quad \underbrace{M}_{\text{target space}}.\tag{B}$$

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  • $\begingroup$ First, thank you for the answer. Ok, I get what you say, but what happens when a lagrangian actually depends on the "spatial" coordinates? Does it mean that it cannot be conceived as arising from an integration of the lagrangian density? Or this integration is not on the actual spatial boundaries of the system, but stops at $\boldsymbol{x}$, so it is like an indefinite integral? I hope it is clear what I said and it's not too unreasonable $\endgroup$
    – Rob Tan
    Jun 30, 2020 at 11:50
  • $\begingroup$ Perhaps your "spatial" coordinates are living in the target space? $\endgroup$
    – Qmechanic
    Jun 30, 2020 at 12:06
  • $\begingroup$ With "target space" do you mean the image of $V\times[t_i,t_f]$ trough the scalar field $q$? Because, sorry, I don't understand your question; if so this target space should be $\mathbb{R}$ I think. What I imagine, is to be in a space with $n$ dimensions, plus $1$ temporal dimension and I just want to understand how a lagrangian can have an explicit dependence on $\boldsymbol{x}$ if it emerges from a definite integral. $\endgroup$
    – Rob Tan
    Jun 30, 2020 at 12:43
  • $\begingroup$ $\uparrow$ Yes to the first question in above comment. $\endgroup$
    – Qmechanic
    Jun 30, 2020 at 19:26
  • $\begingroup$ Ok, but sorry I continue to don't understand how a lagrangian can have an explicit dependence on $\boldsymbol{x}$ $\endgroup$
    – Rob Tan
    Jul 1, 2020 at 6:45

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