Derivatives in a non-trivial metric

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Discussion Overview

The discussion revolves around the calculation of derivatives in a non-trivial metric space, specifically focusing on the expression (∇f)² where f is a function. Participants explore the implications of working in curved space and the appropriate methods for differentiation, including covariant derivatives.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant seeks clarification on how to calculate (∇f)² in a curved space defined by a specific metric.
  • Another participant suggests that the question may involve a two-dimensional space with coordinates t and r, and emphasizes the need for covariant differentiation to obtain an invariant quantity.
  • There is a discussion about the distinction between the gradient (∇f) and the covariant derivative, with some participants expressing uncertainty about the notation used.
  • Some participants clarify that while partial derivatives of a scalar function are valid, the distinction between partial and covariant derivatives becomes significant when dealing with vectors or tensors.
  • A question is raised about whether the square of the partial derivatives can be considered a tensor in general, with some uncertainty expressed regarding the conditions under which this holds true.
  • One participant indicates that they believe they have resolved their confusion after reviewing the responses.

Areas of Agreement / Disagreement

Participants express differing views on the nature of derivatives in curved space, particularly regarding the distinction between partial and covariant derivatives. There is no consensus on whether the square of the partial derivatives constitutes a tensor in general, indicating ongoing debate.

Contextual Notes

Limitations include potential misunderstandings of notation and the implications of using different types of derivatives in curved space. The discussion does not resolve the conditions under which the square of partial derivatives may or may not be a tensor.

cj7529
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I'm trying to work out:
(∇f)^2 (f is just some function, its not really important)

While working in curved space with a metric:
ds^2 = α dt^2 + dr^2 + 2c√(α+1) dtdr

I'm not really sure how to calculate a derivative in curved space, any help would be appreciated

thanks
 
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Well, hmm...Your question is above my pay-grade. But it seems to me that you have a 2 dimensional space here.
you have t and r. ∇f is DEFINED as ∑ ∂f/∂qi over all i (where qi are the coordinates).
I am assuming you do really mean (∇f)² and not ∇²f which is completely different.
In general, you are not going to get an invariant (coordinate independent) quantity by simply differentiating a variable. You need to use covariant differentiation (which includes the Christofel symbol) and results in a tensor (invariant). This may or may not be important in your problem, IDK.
Oh my! In looking at the wikipedia article on covariant derivatives, I see that ∇ is also used to symbolize them! (I was familiar with using Df to symbolize them). So, now I'm not sure what your question is? Did you mean the square of the gradient or did you mean the square of the covariant derivative? Sorry, I almost never revisit a thread I've answered, but hopefully someone else will. Again, your question is above my competence level. Take it for what its worth.
 
cj7529 said:
I'm trying to work out:
(∇f)^2 (f is just some function, its not really important)
I'm not really sure how to calculate a derivative in curved space, any help would be appreciated
The gradient ∇f is ∂f/∂xi, just the partial derivatives wrt the coordinates.
∇f·∇f is where the metric comes in, ∇f·∇f = gij ∂f/∂xi ∂f/∂xj
 
Are you trying to express that as a tensor? partial derivatives are not tensors in curved space time.
 
HomogenousCow said:
Are you trying to express that as a tensor? partial derivatives are not tensors in curved space time.
Partial derivatives of a scalar are. The distinction between partial derivative and covariant derivative arises only when you're differentiating a vector or tensor quantity. I assume by "some function" the OP means a scalar.
 
Bill_K said:
Partial derivatives of a scalar are.
But is the square of the partials a tensor in general? To simplify, in Rn is Tμ=(∂f)^2 a true vector in general? I'd say it isn't in general, but I'm not sure exactly in which cases it is.
 
ah yeah thanks for the replies, think I've got it sorted now!
 
I guess as long as Ti=δij ∂f/Xj
 
Last edited:

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