Given a spacetime translation,

$$ x^\nu \rightarrow x^\nu {'}=x^\nu-\epsilon^\nu $$ and the corresponding field transformation $$ \phi(x) \rightarrow \phi(x)^{'}= \phi(x) + \epsilon^\nu \partial_\nu \phi(x) $$

it is stated in a variety of sources (Tong, Wikipedia, Srednicki, Peskin/Schroeder, etc.) that the corresponding transformation of the Lagrangian is,

$$ \mathcal{L}(x) \rightarrow \mathcal{L}{'}(x) = \mathcal{L}(x) + \epsilon^\nu \partial_\nu \mathcal{L}(x) $$

I have two problems with this.

Firstly I cannot derive it by brute force. For instance, if I plug $\phi^{'}$ into the following Lagrangian,

$$ \mathcal{L} = \frac{1}{2} (\eta^{\mu\nu}\partial_\mu \phi \partial_\nu \phi -m^2\phi^2) $$

I get a factor of 2 on the second term in the expression for $\mathcal{L}'$ above.

Secondly, the variable of the Lagrangian density. Does the Lagrangian density not depend on both $\phi$ and $\partial_\mu \phi$?! Why is the Lagrangian written as only depending on spacetime?

The closest thing I can get to solving this puzzle is from Peskin/Schroeder, where they argue that since the Lagrangian is a scalar (which I agree with) it must transform as the scalar fields do.

This doesn't address my first concern regarding the calculation, but I'll take what I can get.

For reference (if it helps those formulating an answer work from a starting point) here is the wording a variety of authors use before stating the Lagrangian transformation above:

Tong — "... once we substitute a specific field configuration $\phi(x)$ into the Lagrangian, the Lagrangian itself transforms as,..."

Timo Weigand — "Because $\mathcal{L}$ is a local function of x it transforms as..."


Try $$\partial_\nu \mathcal{L} = \partial_\nu \phi \partial_\phi \mathcal{L} + \partial_\nu \partial_\mu \phi \partial_{\partial_\mu {\phi}}\mathcal{L}$$ and there will be no factor of 2.

$$\mathcal{L(x)} \equiv \mathcal{L}(\phi(x),\partial_a \phi(x))$$


1) The lagrangian density very rarely depends explicitly on $x$-if the derivative was indeed a spacetime derivative $\partial_\nu\mathcal{L}$ would be zero. In fact, this is sloppy notation for $\mathcal{L}$ depending IMPLICITLY on $x$ via $\phi$ and $\partial\phi$. This is like $dL/dt$ vs $\partial L/\partial t$ in point mechanics-the latter is almost always zero, the former isn't.

2) To actually see what's going on, note the transformation for a generic scalar function-$$\phi(x)\to \phi'(\Lambda x)=\phi(x)\implies \phi'(x)=\phi(\Lambda^{-1}x),$$ where $\Lambda$ is a generic coordinate transformation. Here, we have $$\phi'(x')=\phi'(x+\epsilon)=\phi(x)\implies \phi'(x)=\phi(x-\epsilon)=\phi(x)-\epsilon^\mu\partial_\mu\phi+..$$ Now, note that $$\mathcal{L}=\mathcal{L}(\phi(x),\partial_\mu\phi(x)))$$ so $\mathcal{L}'(\phi'(x'),\partial'\phi')$ will transform as a scalar function does. You will need the chain rule, to see the implicit variation wrt $x$. Properly, one should write $d_\mu\mathcal{L}$ and not $\partial_\mu\mathcal{L}$, because we are only studying implicit variation, via chain rule. @Qmechanic has a detailed post on this that I will link if I can find it.

EDIT: The question of $L$ not transforming like a scalar because $\partial_\mu\phi$ doesn't transform like a scalar; is the same as the saying $\phi(x^\mu)$ doesn't transform like a scalar because $x^\mu$ doesn't transform like a scalar. It is meaningless-$\phi$, and $L$, are DEFINED to transform as scalars when their arguments are lorentz-transformed.

3) Lastly, most references, such as Tong, hide this implicit dependence, and write $\mathcal{L}(x)$ which is to be understood as $\mathcal{L}(\phi(x))$. $\mathcal{L}(x)$ transforms as $\phi(x)$, as they write. This is an abuse of notation, but common. To re-iterate, there is NO explicit coordinate dependence here-the variations are implicit via the chain rule.

  • $\begingroup$ Thank you for #1 But #2 is still entirely lost on me. Notably, your statement "Now, note that $\mathcal{L}=\mathcal{L}(\phi(x),\partial_\mu\phi(x))$ so $\mathcal{L}'(\phi'(x'),\partial'\phi')$ will transform as a scalar function does." is precisely what I have a question about. Specifically, $\partial_\mu(\phi{}'(x))$ does NOT transform as $\phi$ does. $\endgroup$
    – Lopey Tall
    Feb 24 '20 at 16:03
  • $\begingroup$ @LopeyTall apologies, I somehow never saw this comment. I've made an edit to my second point. Does this help? $\endgroup$
    – GRrocks
    Apr 3 '20 at 19:10

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