Partial Derivative of x^2 on Manifold (M,g)

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Discussion Overview

The discussion revolves around the computation of the partial derivative of the expression ##x^2## on a manifold ##(M,g)##. Participants explore the implications of this operation in the context of differential geometry, particularly concerning the nature of coordinates and the distinction between partial and covariant derivatives.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant suggests that the partial derivative ##\partial_\mu x^2## should yield ##2x_\mu## but questions this, proposing instead that it might be ##2x_\mu + x^\nu x^\lambda \partial_\mu g_{\nu\lambda}##.
  • Another participant emphasizes the need to specify the coordinate system when discussing ##x^2##, arguing that it is a coordinate-dependent statement and that there is no position vector on a manifold.
  • A participant questions whether ##x^2## could be interpreted as the distance between two points, specifically between ##x^\mu## and ##x^\mu_0=(0,0,\dots)##.
  • There is a disagreement on the interpretation of ##x^2##, with one participant asserting that it cannot be considered as a distance, while another later clarifies that they meant the square of the distance, referencing the Euclidean distance as ##(x^\mu x_\mu)^{1/2}##.
  • Another participant asserts that the concept of distance in the manner described does not exist in a general manifold.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of ##x^2## and its implications on a manifold. There is no consensus on whether it can be considered as a distance or how to properly compute its derivative.

Contextual Notes

The discussion highlights the dependence on coordinate systems and the distinction between partial and covariant derivatives, which remains unresolved. The nature of distance in the context of a general manifold is also questioned without a definitive resolution.

shooride
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How can I figure out ##\partial_\mu x^2## on the manifold ##(M,g)##? I thought that it should be ##2x_\mu##, but I think I'm wrong and the answer is ##2x_\mu+x^\nu x^\lambda \partial_\mu g_{\nu\lambda}##, right?! In particular, it seems to me, we can't write ##\partial_\mu=g_{\mu\nu}\partial^{\nu}##. However, we can raise or lower the indices of the covariant derivative with metric, I mean ##\nabla^\mu=g^{\mu\nu}\nabla_\nu##. Is this true because partial derivative is not a tensor but covariant derivative is? Could you please explain it to me?
 
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What do you mean by ##x^2##? If ##x^\mu## are coordinates, then ##x^2## is a coordinate dependent statement and you must specify the coordinate system in which this is true in order to make any sense of it. Generally, there is no such thing as a position vector on a manifold.
 
Orodruin said:
What do you mean by ##x^2##?.
Can one consider ##x^2## as the distance between two points ##x^\mu## and ##x^\mu_0=(0,0,\dots)##?
 
shooride said:
Can one consider ##x^2## as the distance between two points ##x^\mu## and ##x^\mu_0=(0,0,\dots)##?
No.
 
Orodruin said:
No.
Sorry, I meant square of distance! ##(x^\mu x_\mu)^{1/2}## is just the Euclidean distance?!
 
There is no such concept in a general manifold.
 

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