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.

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

1 Answer 1


To be sure, the four-gradient of a scalar field yields a (covariant) four-vector field

$$\mathbf{\tilde d} \phi(x^\mu) = \partial_\mu\phi\, \tilde\omega^\mu = \frac{\partial \phi}{\partial x^\mu}\tilde\omega^\mu = \frac{\partial \phi}{\partial x^0}\tilde\omega^0 + \frac{\partial \phi}{\partial x^1}\tilde\omega^1+\frac{\partial \phi}{\partial x^2}\tilde\omega^2+\frac{\partial \phi}{\partial x^3}\tilde\omega^3$$

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.