I Constructing Vector in Another Basis

[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 $y$.

 

[Orodruin]

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 $y$, since in this case $\zeta$ need to be a vector.

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

 

[davidge]

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

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 $x$?

 

[PeterDonis]

since $x$ and $y$ are coordinate points, the vector $V$ should not depend on them.

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.

we have the familiar transformation of a vector from one point to another (a prime denotes quantities in yy)

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?

 

[davidge]

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?

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 $P$.

 

[stevendaryl]

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 $y$.

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?

 

[davidge]

@stevendaryl Oh, I see. Thank you