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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
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junfan02
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1 Answers1

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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.

Clemens Bartholdy
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