Why do dual vectors extract components the same way as dot products?

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LagrangeEuler
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I found notation in the form
##\vec{e}_{\alpha} \cdot \vec{x}=x^{\alpha}##
and also
## \langle \vec{e}^{* \alpha}, \vec{x} \rangle =x^{\alpha} ##.
If I understand this ##\vec{e}_{\alpha}## is unit vector in direct space, and ##\vec{e}^{* \alpha}## is unit vector in dual space, and the bracket ##\langle, \rangle## is like dual vector on vector on direct space. But why result is ##x^{\alpha}##?
 
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The ##x^\alpha## is the ##\alpha##-component of ##\vec{x}## in the basis ##\{\vec{e}_\alpha\}##. So ##\vec{x}=x^\alpha\vec{e}_\alpha##, the rest is the definition of the inner product and what a dual basis is.
 
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LagrangeEuler said:
I found notation in the form
##\vec{e}_{\alpha} \cdot \vec{x}=x^{\alpha}##
This is true only in an orthonormal basis. In general,
$$
\vec e_\alpha \cdot \vec x = \vec e_\alpha \cdot x^\beta \vec e_\beta = g_{\alpha\beta} x^\beta.
$$
 
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Ok. Tnx. I am liitle bit confused with notation.
[tex]\frac{\partial}{\partial x^{\mu}}=\partial x_{\mu}[/tex]?
So if I write
[tex]\frac{\partial}{\partial x^{\mu}} x^{\nu}=\partial x_{\mu}x^{\nu}=\delta_{\mu}^{\nu}[/tex].
So if I have for example
[tex]\frac{\partial}{\partial x_{\mu}}=\partial x^{\mu}?[/tex]
Is it correct?
 
LagrangeEuler said:
Ok. Tnx. I am liitle bit confused with notation.
[tex]\frac{\partial}{\partial x^{\mu}}=\partial x_{\mu}[/tex]?
This is not standard notation. The standard notation is
$$
\frac{\partial}{\partial x^\mu} = \partial_\mu.
$$
So if I write
[tex]\frac{\partial}{\partial x^{\mu}} x^{\nu}=\partial x_{\mu}x^{\nu}=\delta_{\mu}^{\nu}[/tex].
It is not so much a question about writing as it is about performing the derivative, but yes, ##\partial_\mu x^\nu = \delta^\nu_\mu##.
So if I have for example
[tex]\frac{\partial}{\partial x_{\mu}}=\partial x^{\mu}?[/tex]
Is it correct?
This would be ##\partial/\partial x_\mu = \partial^\mu##.
 
Side-note: shoudn't this be I level and not B?
 
Orodruin said:
This is not standard notation. The standard notation is
$$
\frac{\partial}{\partial x^\mu} = \partial_\mu.
$$
Question: is that because

$$
\frac{\partial}{\partial x^\mu}$$

is an operator, while

$$\partial x_\mu.$$

would be a specific operation on xμ?

Trying to learn these terms. Thanks.
 
Sorcerer said:
Side-note: shoudn't this be I level and not B?
Yes, good point. Fixed. (And you can report these instead of asking in thread - generally gets a faster response)
 
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Sorcerer said:
while

$$\partial x_\mu.$$

would be a specific operation on xμ?
This notation does not make sense to me at all. You would first have to identify what you mean by just ##\partial## without the index.
 
Orodruin said:
This notation does not make sense to me at all. You would first have to identify what you mean by just ##\partial## without the index.
Well, I was seeing it as just a differential of xμ. Is there such thing as just a partial differential, in the sense of say, dx, dy, dz?

But my other question, is that right? It's an operator?
 
That operator is completely different and typically denoted with d, not ##\partial##. The one-forms ##dx^\mu## form a basis of the dual space.
 
Sorcerer said:
Question: is that because

$$
\frac{\partial}{\partial x^\mu}$$

is an operator, while

$$\partial x_\mu.$$

would be a specific operation on xμ?

Trying to learn these terms. Thanks.

To some extent you have to learn the specific notation and accept it. You can't necessarily deduce what a shorthand means from some absolute first principles.

Orodruin said:
The standard notation is
$$
\frac{\partial}{\partial x^\mu} = \partial_\mu.
$$

You have to accept this. It's just convention to drop the ##x## on the RHS. You're not going to find a deep mathematical reason why there's an ##x## in one and not the other. It's just the way people choose to write things. There's a certain logic in it, of course, but it could have been different.
 
Orodruin said:
This would be ##\partial/\partial x_\mu = \partial^\mu##.

Does ##x_\mu## have a standard meaning? ##x^\mu## means a coordinate, usually, but what is ##x_\mu##?
 
stevendaryl said:
Does ##x_\mu## have a standard meaning? ##x^\mu## means a coordinate, usually, but what is ##x_\mu##?
Typically ##x_\mu = g_{\mu\nu}x^\nu##, even if ##x^\nu## is a coordinate and not a vector. This is particularly true in Minkowski space where the concept of a spacetime vector that does have ##x^\nu## as its components (in standard coordinates) does make sense, since it is an affine space (just like a position vector makes sense in Euclidean space). Come to think of it, I would strongly advise against using this notation in anything other than Minkowski space.