# I Constructing a vector in another basis

1. Mar 11, 2017

### davidge

Suppose we have defined a vector $V$ at a point $x$, so it has components $V^\mu(x)$ at $x$. Let $y$ be another point, such that $y^\mu = x^\mu + \epsilon \zeta^\mu(x)$, $\epsilon$ a scalar. Now, since $x$ and $y$ are coordinate points, the vector $V$ should not depend on them. So we have the familiar transformation of a vector from one point to another (a prime denotes quantities in $y$):
$$V'^\mu (y) = V^\nu (x) \frac{\partial y^\mu (x)}{\partial x^\nu}$$

We would end up with $V'^\mu (y) = V^\mu (x) + \epsilon V^\nu (x)\partial_{\nu}\zeta^{\mu}(x)$. My question is if it's correct to assume that relation between $x$ and $y$.

2. Mar 11, 2017

### Orodruin

Staff Emeritus
Your function $\zeta$ is not a vector. It is a collection of functions that describe the relation between two coordinate systems.

3. Mar 11, 2017

### davidge

Thanks. I have edited my post, because this has become not relevant to me
Is it ok to take $\zeta^\mu(x)$ such that $\zeta^\mu(x) = 0$ but its first derivatives non-zero at $x$?

4. Mar 11, 2017

### Staff: Mentor

Why not? You've only defined $V$ at the point $x$; you haven't defined it at the point $y$. Vectors are "attached" to particular points.

Now you're talking as though $y$ denotes the same point as $x$, just in a different coordinate chart. This is not the same as $y$ being a different point from $x$, expressed in the same chart. Which is it?

5. Mar 11, 2017

### davidge

Oh yes. I'm thinking of a point $P$ on a manifold, where the vector $V$ depends on $P$. That is why I assumed the vector to be the same. $x$ and $y$ are different maps to $P$.

6. Mar 11, 2017

### stevendaryl

Staff Emeritus
I think you're mixing up two different concepts:
1. The components of a vector in two different bases
2. The value of a vector field at two different points.
If you use a coordinate basis, and you change coordinate systems from $x^\mu$ to $y^\mu$, then you would use the transformation equation:

$V'^\mu = \frac{\partial y^\mu}{\partial x^\nu} V^\nu$

On the other hand, if you are talking about two different, nearby points, $\mathcal{P}$ with coordinates $x^\mu$ and $\mathcal{P'}$, with coordinates $x^\mu + \epsilon \zeta^\mu$ (you're keeping the same coordinate system), then you could use:

$V^\mu(\mathcal{P'}) \approx V^\mu(\mathcal{P}) + \epsilon \zeta^\nu \frac{\partial V^\mu}{\partial x^\nu}$

That only makes sense for a vector field (a function assigning a vector to each point in space), not a vector.

There is yet a third concept that you might be thinking of: parallel transport of a vector. If you have a vector $V^\mu$ at one point $\mathcal{P}$ with coordinates $x^\mu$), then what is the result of transporting that vector to a nearby point $\mathcal{P'}$ with coordinates $x^\mu + \epsilon \zeta^\mu$?

7. Mar 11, 2017