# Special Relativity - Transformation from scalar to 4-vector

I'm studying Leonard Susskind's lectures on special relativity. In lecture 4, when he talks about transformations between scalars and vectors(which is around 1:10:00), he mentions that, to turn a scalar into a vector, you must differentiate the scalar's function with respect to the 4 spacetime dimensions...

My question is: When the referred differentiation is mentioned, does he mean to take the gradient of the scalar's function? Because that's what comes to my mind.

Thanks to anyone who answers.

• Yes, if $\phi$ is a scalar, the quantity $\partial_\mu \phi$ is a 4-vector that is the four-dimensional version of the gradient. Commented Sep 24, 2016 at 11:12

$$\mathbf{\tilde d} \phi(x^\mu) = \partial_\mu\phi\, \tilde\omega^\mu = \frac{\partial \phi}{\partial x^\mu}\tilde\omega^\mu = \frac{\partial \phi}{\partial x^0}\tilde\omega^0 + \frac{\partial \phi}{\partial x^1}\tilde\omega^1+\frac{\partial \phi}{\partial x^2}\tilde\omega^2+\frac{\partial \phi}{\partial x^3}\tilde\omega^3$$
where the $\tilde\omega^\mu$ are the one-form basis for the coordinates $x^\mu$.
That is to say, $\partial_\mu\phi$ are the components of a four-vector.