Are all Lagrangians translationally invariant? I am rather stumped by David Tong's derivation of the energy-momentum tensor for a translationally invariant theory because it appears it doesn't assume any type of Lagrangian at all.
A Lagrangian $\mathcal{L}(\phi,\partial_\mu \phi)$ has a symmetry $\phi \rightarrow \phi + \delta \phi $ if the off-shell variation is given to first order by a total derivative
$$ \delta \mathcal{L} = \partial_\mu F^\mu(\phi). $$
Tong argues  that if we substitute in a particular field configuration into the Lagrangian then we can define a function $\mathcal{L}(x)\equiv \mathcal{L}(\phi(x),\partial_\mu \phi(x))$. Under a translation of the fields $\phi(x) \rightarrow \phi'(x) = \phi(x-\epsilon) $, we have 
$$ \mathcal{L}(x) \rightarrow \mathcal{L}(x-\epsilon) = \mathcal{L}(x) - \epsilon^\mu \partial_\mu\mathcal{L}(x)$$
which is a total derivative, as shown in (1.40). This did not assume any form of the Lagrangian so is this telling me that all Lagrangians are translationally invariant? This seems to apply to Lorentz transformations too, despite not assuming a Lorentz invariant Lagrangian, as seen in equation (1.53).
If I take this idea further and suppose I performed a conformal transformation described by a transformation $ x^\mu \rightarrow x^\mu + \epsilon^\mu(x)$, where 
$$ \epsilon^\mu(x) = a_\mu + b_{\mu \nu}x^\nu + c_{\mu \nu \rho} x^\nu x^\rho $$
as given by equation (2.7) of "Intro to CFT by Blumenhagen and Plauschinn", then I would say the off-shell variation is, from the arguments above, given by
$$  \delta \mathcal{L} = - \epsilon^\mu(x) \partial_\mu \mathcal{L}(x) $$
which can't be massaged into a total derivative. Is this telling me that no theories are conformally invariant? I know this is not true but I do not know how one could write this as a total derivative to fulfil the definition of a symmetry.
 A: *

*Yes they are all translationally invariant as long as they don't depend explicitly on $x^\mu$. As Tong assumes by writing $\mathcal{L}(\phi,\partial_\mu\phi)$.

*No, the argument wouldn't go through for Lorentz transformations as they also affect the indices $\mu$ in $\partial_\mu$ and in whatever other field that is not a scalar. That is, $\phi'_{\mu_1\cdots \mu_\ell}(x)$ is not $\phi_{\mu_1\cdots \mu_\ell}(x-\epsilon)$.

*The argument wouldn't go through for conformal transformations neither. Both for the reason above and for the fact that if $\epsilon$ depends on $x$ then it does not go through derivatives and so $\partial_\mu\phi$ transforms differently than $\phi$.

*That's not necessarily the form of the Lagrangian after a conformal transformation.
A: When you write $\mathcal{L}(x)=\mathcal{L}(\phi(x),\partial_\mu\phi(x))$, you're assuming translational invariance. A more general Lagrangian is written $\mathcal{L}(x)=\mathcal{L}(\phi(x),\partial_\mu\phi(x),x)$.
A: One approach to this is to invoke Nother's theorem.  For every symmetry of the system, there is a corresponding conserved quantity.  For spatial transnational symmetry, that conserved quantity is momentum (linear momentum).  So all systems which conserve momentum will have this symmetry.
One way you can avoid conservation of momentum is to design a system which couples to an external force, permitting momentum to be transferred out of the system.
