# I Coordinate independent version of "gradient"?

1. Jul 25, 2016

### Stephen Tashi

Is the "gradient" vector a concept that that is coordinate independent ?

For example, the concept of a vector representing a force is independent of what coordinate system is used to represent the vector. So is a "gradient vector" such a physical vector ?

The web page http://www.mathpages.com/home/kmath398/kmath398.htm says that the temperature at a point in space is invariant under coordinate transformations, but the gradient of the temperature at a point in space is not. I can understand that view if "gradient" always implies a choice of coordinate system. Does it ? - or should we protest to the authors of that page ?

2. Jul 25, 2016

### Staff: Mentor

The gradient at a point points to the steepest incline.

3. Jul 25, 2016

### Stephen Tashi

Are you taking the position that "gradient" is a coordinate independent concept ?

4. Jul 25, 2016

### DrGreg

No, what it actually says is that each component of the gradient is not invariant (just like individual components of a velocity vector are not invariant). The gradient has covariant components which transform in a particular way under a change of coordinates, similar to the way that a velocity vector's contravariant components transform (in a different way) under a change of coordinates.

It seems that this author refers to velocity as a "vector with contravariant components" and a gradient as a "vector with covariant components". Other authors would use the terms "vector" and "covector" instead.

Either way, the gradient of a scalar is independent of the choice of coordinates.

5. Jul 25, 2016

### Stephen Tashi

Is a "component" of a vector also a vector ?

6. Jul 25, 2016

### DrGreg

No. Components are coordinate-dependent.

7. Jul 25, 2016

### Staff: Mentor

There is a beautiful coordinate free definition on Wiki (and they say that e.g. engineers often use it)
$$grad \; f = \lim_{V \rightarrow 0} \frac{\oint_{\partial \mathcal{V}} f d \vec{A}}{V}$$
with a scalar vector field $f : \mathcal{V} \longrightarrow \mathbb{R}$ and a piecewise smooth boundary $\partial \mathcal{V}$ where $V$ denotes the volume and $d \vec{A}$ a surface element (normal to it) of $\partial \mathcal{V}$.

8. Jul 25, 2016

### Stephen Tashi

I'll agree that's what the page actually shows, but what the page actually says is

which asserts that if a gradient vector is "invariant" then its components should be invariant under a coordinate transformation.

I agree.

What prompted my interest in this was the statement in another thread https://www.physicsforums.com/threads/su-2-and-su-3.879988/#post-5528633 that

If we regard an element of $SU(2)$ as a transformation from vectors to different vectors then that definition makes sense. If we regard an element of $SU(2)$ as a transformation from components of a vector in one coordinate system to components of it in a different coordinate system, then no vector is mapped to a different vector.

9. Jul 25, 2016

### Staff: Mentor

One of the first things here on PF I have been remembered on was to carefully distinguish between vectors and their presentation through coordinates. Of course does a transformation of coordinates not change the vectors themselves, only how we calculate with them. (And I did not say representation here on purpose.)

Invariant vectors of an operation however, indicate states that doesn't change under this operation, e.g. the colour-mixture of quarks or electric charges.

10. Jul 25, 2016

### Stephen Tashi

We can regard a "component" of vector V as a scalar c_i that is coefficient of the i-th basis vector v_i in some representation of V with respect to a set of basis vectors.

- or we can ( as physics texts do) draw a picture of vector V "resolved into components" and portray and discuss these "components" as vectors.

Perhaps "component" is an example of ambiguous terminology.

11. Jul 25, 2016

### Staff: Mentor

Of course it is, since you can write $\vec{v} = \vec{x}+\vec{y}=\vec{a}+\vec{b}$ in unbounded many ways and all summands are components if you will.
As soon as you speak of coordinates $\vec{v} = (v_1 , \dots , v_n)$ according to a basis $\mathfrak{e}_i$, you mean $\vec{v} = \sum_{î=1}^{i=n} v_i \mathfrak{e}_i$ and with the coordinates you simultaneously fixed the components $v_i \mathfrak{e}_i$. At least this is usually meant by component as soon as a basis is involved.

12. Jul 25, 2016

### Stephen Tashi

That illustrates that if we accept that a "component" of a vector is a vector then we have the ambiguity that a given vector can be expressed in components in different ways. What I'm discussing is the ambiguous use of the word "component" to mean something that is vector and also something that is not a vector.

The question "Is a component of a vector also a vector" isn't the same as asking "Is a component of a vector a unique vector".

That suits me. I don't have any axe to grind about whether the components of $\vec{v}$ are defined to be the $v_i$ or the $v_i \mathfrak{e}_i$.

But it seems that in the typical scenario in physics where people are classifying physical quantities according to "how their components transform under a change of coordinates", interpreting a "component" to be a vector gets contradictory. If a component is a vector then it doesn't become a different vector under a change of coordinates. If a component is a scalar in the i_th place of a coordinate representation of a vector then there is no contradiction in saying it changes to a different scalar in the i_th place of a representation using a different basis.

13. Jul 25, 2016

### Staff: Mentor

I'm built simple. As soon as physics lurks around the corner, I tend to view components of $v$ as forces which altogether result in $v$, i.e. a bunch of summands.

To call $v_i \mathfrak{e}_i$ the $i$-th component of $v$ is a bit of sloppy for saying
"$v_i$ is the $i$-th coordinate with respect to the basis $\{ \mathfrak{e}_j \, | \, j \in I\}$ of $V$ where $I \subseteq \mathbb{N} \cup \{ \infty \}$ is the index set of the basis and $|I|$ the dimension of $V$".
You see: such a precision isn't fun to write. The more since there is no advantage to do so. Therefore I can well accept someone referring to $v_i \mathfrak{e}_i$ or for short $v_i$ as the $i$-th component.

Last edited: Jul 25, 2016