# Given a translation of coordinates (and thus of fields) how does the lagrangian (density) transform?

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.

• 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. Feb 24 '20 at 16:03
• @LopeyTall apologies, I somehow never saw this comment. I've made an edit to my second point. Does this help? Apr 3 '20 at 19:10