2

My main source of confusion is the following.

Suppose I have a scalar potential $V(x,y,z)$. The electrostatic field for this potential is $ -\vec{E} =\vec{\nabla}V = \frac{\partial{V}}{\partial{x}}\hat{x} + \frac{\partial{V}}{\partial{y}}\hat{y} + \frac{\partial{V}}{\partial{z}}\hat{z}$. This is a covariant vector.

The electric field can also be expressed as $\frac{m}{q}\vec{a}$ where $\vec{a}$ is the acceleration of a charged particle of charge $q$ and mass $m$ placed in the electric field. Since acceleration is a contravariant vector in writing $-\vec{\nabla}V = \frac{m}{q}\vec{a}$, aren't we equating a covariant vector on the L.H.S to a contravariant vector on the R.H.S when expressed in the same basis. So if we apply a transformation of the basis on both sides of the equation the two sides should behave differently. What am I missing?

Qmechanic
  • 201,751
Abhishek Banerjee
  • 167
  • https://physics.stackexchange.com/questions/126740/gradient-is-covariant-or-contravariant#:~:text=A%20covariant%20vector%20is%20commonly,the%20contravariant%20vector%20x%CE%BC. – tryst with freedom May 26 '22 at 16:44

1 Answers1

5

That's because you're using the 'fake' version of gradient. The true version for a scalar field $F$ is given as:

$$ (\nabla F)= \frac{\partial F}{\partial Z^i } e^i$$

Here $e^i$ is the dual basis and not the basis. To get the contravariant form that you use, you contract both side with the upper index metric tensor:

$$ \text{grad} (F)^{\alpha}= (\nabla F)_i g^{i\alpha} = g^{i \alpha} \frac{\partial F}{\partial Z^i} $$

The idea is:

$$ \nabla F = \frac{\partial F}{\partial Z^i} e^i =( \frac{\partial F}{\partial Z^i} g^{i \alpha}) e_{\alpha}$$

The bracketed term you identify as the component of the gradient as a vector.

In Cartesian coordinate none of this matters as the $g^{i\alpha} = \delta_{i \alpha}$ so everything looks the same but the difference shows for real when you work in Curvilinear coordinates.

tryst with freedom
  • 7,940
  • So when we talk about covariant vectors, we are actually talking about vectors in the dual space? Is that correct? – Abhishek Banerjee May 26 '22 at 16:54
  • Could you suggest me any book or articles that illustrate these with a few concrete examples? – Abhishek Banerjee May 26 '22 at 16:55
  • 1
    I somewhat liked Pavel Grinfeld's tensor analysis book , it is simple and approchable but I found it too calculative. There is another book by Tristan Needham on Differential Geometry which pretty much goes on about this tensor stuff at the very final chap but with a focus on a certain class of such tensor entites, namely the anti symmetric differential forms. You can also check out Physicist books like Anthony Zee's fly by night physics which contains some discussion of this. I didn't find Zee's book to stick with me tho. – tryst with freedom May 26 '22 at 16:57
  • 1
    One pretty great book which goes on a whirlwind tour of all of this stuff in very high detail is Roger Penrose's Road to Reality. It it has discussion to every big idea imaginable on these things. @AbhishekBanerjee – tryst with freedom May 26 '22 at 16:59
  • Thanks a ton, is the statement in my first comment about covariant vectors being in the dual space correct? – Abhishek Banerjee May 26 '22 at 17:02
  • 1
    Sounds right to me @Abhishek Banerjee – tryst with freedom May 26 '22 at 17:08
  • Let us continue this discussion in chat. – Abhishek Banerjee May 26 '22 at 17:09