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In David Tong Lecture Notes (page 14), it is written that

Proof of Noether's Theorem: We'll prove the theorem by working infinitesimally. We may always do this if we have a continuous symmetry. We say that the transformation $$ \delta \phi_a(x)=X_a(\phi)\tag{1.34} $$ is a symmetry if the Lagrangian changes by a total derivative, $$ \delta \mathcal{L}=\partial_\mu F^\mu\tag{1.35} $$ for some set of functions $F^\mu(\phi)$. To derive Noether's theorem, we first consider making an arbitrary transformation of the fields $\delta \phi_a$. Then $$ \begin{aligned} \delta \mathcal{L} & =\frac{\partial \mathcal{L}}{\partial \phi_a} \delta \phi_a+\frac{\partial \mathcal{L}}{\partial\left(\partial_\mu \phi_a\right)} \partial_\mu\left(\delta \phi_a\right) \\ & =\left[\frac{\partial \mathcal{L}}{\partial \phi_a}-\partial_\mu \frac{\partial \mathcal{L}}{\partial\left(\partial_\mu \phi_a\right)}\right] \delta \phi_a+\partial_\mu\left(\frac{\partial \mathcal{L}}{\partial\left(\partial_\mu \phi_a\right)} \delta \phi_a\right) \end{aligned}\tag{1.36} $$ When the equations of motion are satisfied, the term in square brackets vanishes. So we're left with $$ \delta \mathcal{L}=\partial_\mu\left(\frac{\partial \mathcal{L}}{\partial\left(\partial_\mu \phi_a\right)} \delta \phi_a\right)\tag{1.37} $$ But for the symmetry transformation $\delta \phi_a=X_a(\phi)$, we have by definition $\delta \mathcal{L}=\partial_\mu F^\mu$. Equating this expression with (1.37) gives us the result $$ \partial_\mu j^\mu=0 \quad \text { with } \quad j^\mu=\frac{\partial \mathcal{L}}{\partial\left(\partial_\mu \phi_a\right)} X_a(\phi)-F^\mu(\phi).\tag{1.38} $$

whereas in Schwartz (page 33), it is written that

"When there is such a symmetry that depends on some parameter $\alpha$ that can be taken small (that is, the symmetry is continuous), we find, similar to Eq. (3.13), that $$ 0=\frac{\delta \mathcal{L}}{\delta \alpha}=\sum_n\left\{\left[\frac{\partial \mathcal{L}}{\partial \phi_n}-\partial_\mu \frac{\partial \mathcal{L}}{\partial\left(\partial_\mu \phi_n\right)}\right] \frac{\delta \phi_n}{\delta \alpha}+\partial_\mu\left[\frac{\partial \mathcal{L}}{\partial\left(\partial_\mu \phi_n\right)} \frac{\delta \phi_n}{\delta \alpha}\right]\right\}\tag{3.22} $$ where $\phi_n$ may be $\phi$ and $\phi^{\star}$ or whatever set of fields the Lagrangian depends on. In contrast to Eq. (3.13), this equation holds even for field configurations $\phi_n$ for which the action is not extremal (i.e. for $\phi_n$ that do not satisfy the equations of motion), since the variation corresponds to a symmetry. When the equations of motion are satisfied, then Eq. (3.22) reduces to $\partial_\mu J_\mu=0$, where $$ J_\mu=\sum_n \frac{\partial \mathcal{L}}{\partial\left(\partial_\mu \phi_n\right)} \frac{\delta \phi_n}{\delta \alpha}\tag{3.23} $$ This is known as a Noether current."

I am not able to understand whether the two definitions are similar or not, if they are not similar how they are not similar. I will be glad if I can help regrading this.

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Comparing the two resulting currents you see that they only differ by the addition of the $-F^\mu$ term.

(The replacement of $X_a $ by $\delta \phi / \delta \alpha$ is only due to a different definition of $\delta \phi$ where the first formulation assumes this to only include a variation of $\phi$ induced by the symmetry, so these are actually the same.)

Now where does the term $F^\mu$ come from? This stems from the fact that actually, the lagrangian density does not have to be invariant but rather the action has to be and so it accounts for changes in the lagrangian that vanish under integration.

However, the second derivation does not allow for such terms by the assumption of your first equation i.e. that $\delta \mathcal{L}/ \delta \alpha =0$. Thus these formulas agree only if this is given. Other wise you can play the same game but varying the action instead of the lagrangian. Then the equation you get will include a term $\frac{\delta \mathcal{L}}{ \delta \alpha} + \partial \frac{\partial \mathcal{L} }{\partial \phi} \frac{\delta \phi }{\delta \alpha} + ...$

This extra term,$ \frac{\delta \mathcal{L}}{ \delta \alpha}$ accounts for this difference and will yield the same formula. (In other words the assumption for the second formulation you have is that the lagrangian does not directly depend on the variation of alpha but only does so through changes in the fields it contains.

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