# Are vector expressions and vector operator expressions independent of coordinates

We encounter expressions for vectors and tensors in Euclidean space, such as

$$\vec{F}=\vec{A}+\nabla\phi,$$ or

$$\vec{H} = \nabla\vec{u}\cdot\vec{n}+\nabla\times(\nabla\times\vec{B}) + \frac{\partial \vec{B}}{\partial t},$$ or

$$\nabla\cdot\vec{E}=\vec{C}\cdot\hat\Phi\cdot[(A\cdot\nabla)\vec{B}]-\vec{A}\cdot\vec{B}.$$

and other similar expressions, which include in themselves vectors, tensors and differential operators. Then we are told that these expressions "are independent of coordinates". Does this mean these expressions are the same in ANY coordinate system in Euclidean space? What about non-orthogonal coordinates? Curvilinear coordinates?

Can someone explain why are they the same. If changing the basis, what does one need to know to arrive at the "same" expression in the new coordinates?

If one writes the expression in the new coordinates, and it is not the same, what went wrong?

• A vector, as element of a vector space, is by definition independent of its description. Coordinates are, more or less, maps from the vector to a set of numbers and as such they depend on the representation. – gented Jun 23 '17 at 15:43
• @GennaroTedesco, I hear what you are saying and read much about it on the net... I am concerned about the math , however. Don't want to miss anything. Sure, a vector must be independent of its mathematical description, but does this always work? What about non-orthogonal frames? Because we need to preserve $\nabla$, $\nabla\times$, dot product, $(\vec{A}\cdot\nabla)\vec{B}$ etc... When the math gets complicated, it doesn't seem obvious it will retain its form. Does it? – user142523 Jun 23 '17 at 15:48
• The fact that it must be the same is a definition (it needs not be proven). Then, the gradient changes its expression accordingly so that eventually the vector stays the same. Details can be found in any textbook on tensor calculus. – gented Jun 23 '17 at 16:07
• @GennaroTedesco, how do we maintain the form. $\nabla$ is covariant, $\vec{E}$ (or such) is contravariant, perhaps we should transform them differently to maintain the form of the equation, correct? – user142523 Jun 23 '17 at 16:12
• By definition the action of a co-form on a vector gives back a scalar, so what is it that is worrying you? – gented Jun 23 '17 at 16:16

Yes!

A vector is the same in every coordinate system, only the way the vector is written out varies.

For example, consider this expression: $$\nabla \cdot \vec E=\frac{\rho}{\epsilon_o}$$

This is true in any Euclidean space, and holds the form.

The only thing that changes is the form of the $\nabla$ operator. (and the resulting $\rho$, that undergoes a variable transformation)

For example, in rectangular coordinates, it is: $$\nabla=\hat{\pmb e}_{x}\frac{\partial}{\partial x}+\hat{\pmb e}_{y}\frac{\partial}{\partial y}+\hat{\pmb e}_{z}\frac{\partial}{\partial z}$$

Whereas in spherical polar coordinates, it is: $$\nabla =\hat{\pmb e}_{r}\frac{\partial }{\partial r}+\hat{\pmb e}_{\theta}\frac{1}{r}\frac{\partial }{\partial \theta}+\hat{\pmb e}_{\phi}\frac{1}{r\sin\theta}\frac{\partial }{\partial \phi}.$$

• thanks and +1 for your reply. You have $\nabla$, dot product, divergence operator and a scalar in your example. The $\nabla$ is covariant, but $\vec{E}$ must be contravariant, right? To keep the same expression under ANY change of basis, we must transform $\nabla$ and $\vec{E}$ differently then, correct? – user142523 Jun 23 '17 at 15:55
• I don't understand some of the terminology you're using there but as long as you change the basis properly and consistently there should be no issues. Why don't you try working it out on some simple examples? – Hritik Narayan Jun 23 '17 at 15:58
• Say, hypothetically, in another system I got $\alpha\nabla\cdot\vec{E}=\frac{\rho}{\epsilon}_0$. What went wrong? It must be the way I transformed $\nabla$ or $\vec{E}$, correct? – user142523 Jun 23 '17 at 16:03
• Yes, there might be something wrong with the transformation or maybe you gained the constant from how $\rho$ changed? – Hritik Narayan Jun 23 '17 at 16:07
• The so-called metric coefficients as written by Hritik Narayan for a spherical coordinate system will make sure that you can add those otherwise transforming covariant and contravariant vectors; this takes you to the subject of the Hodge dual and some such. When the vectors are given in a metric space you can associate with each covariant vector (1-form) such as the gradient a contravariant vector through these metric coefficients. – hyportnex Jun 23 '17 at 22:48