Invariants from the covariant derivatives of a scalar field

Question

I am reading Theoretical minimum: Special Relativity and Classical Field Theory where you construct a Lagrangian for the field by the argument that it would be invariant under the Lorentz transformation, so basically any scalar function involving $\phi(x^\mu)$ since $\phi$ transforms as: $$\phi(x) = \phi'(x')$$ But then he argues that the derivatives of the field can also be included in the Lagrangian by making a scalar out of them in the following manner: $$\left(\frac{\partial \phi}{\partial t} \right)^2 - \left(\frac{\partial \phi}{\partial x} \right)^2 - \left(\frac{\partial \phi}{\partial y} \right)^2 - \left(\frac{\partial \phi}{\partial z} \right)^2$$ But I dont understand how covariant components can combine in the above way like the magnitude of a vector to create an invariant quantity since: $$\vec A = A^1 \vec e_1 + A^2 \vec e_2 = A_1 \vec e^1 + A_2 \vec e^2$$ $(A^1)^2 + (A^2)^2$ is an invariant but $(A_1)^2 + (A_2)^2$ is not.

Is there an intuitive way to see how the above squares of the covariant derivatives are invariant under the Lorentz transformation ?

For example if I actively transform $\vec e_x \rightarrow \vec e'_x = 2\vec e_x$ then $\partial_x\phi$ also doubles and the others ($\partial_t\phi$, $\partial_y\phi$, $\partial_z\phi$) remain the same, then how can $(\partial_t\phi)^2 + (\partial_x\phi)^2 + (\partial_y\phi)^2 + (\partial_z\phi)^2$ remain invariant ?

Answer 1

Oh, that is just stereotypical Susskind. He knows all the correct stuff, but when teaching, he is so sloppy.

What is really happening is that the contravariant metric is used to compute the invariant scalar product of two covariant fields. Namely, $$ \left ( \partial_t \phi \right )^2 - \left ( \partial_x \phi \right )^2 - \left ( \partial_y \phi \right )^2 - \left ( \partial_z \phi \right )^2 = \left ( \partial_\mu \phi \right ) g^{\mu\nu} \left ( \partial_\nu \phi \right ) = \left ( \partial \phi \right ) \cdot \left ( \partial \phi \right ) $$ There, you see upper and lower indices contracting properly, and it should no longer be a wonder why this thing is invariant.

You have a really nice animation of the contravariant components changing on the right. It is very well connected to the 2$^\text{nd}$ fundamental confusion of calculus.

However, it is unlikely that such a nice animation for the covariant components case can be correct. Because covariant components are not vectors, but rather one-forms, and their characterisation is by thick slabs, not by arrows.