# Derivatives in a non-trivial metric

• cj7529
In summary, the conversation discusses the calculation of a derivative in curved space with a metric. The question is whether (∇f)^2, where f is a scalar function, is a tensor or not. The conclusion is that it is a tensor as long as it transforms appropriately under a change of coordinates.
cj7529
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

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:

## 1. What are derivatives in a non-trivial metric?

Derivatives in a non-trivial metric refer to the rate of change of a function or variable in a space with a non-trivial metric. This means that the distance between two points is not simply given by the Euclidean distance formula, but rather by a more complex metric that takes into account the curvature and structure of the space.

## 2. How are derivatives in a non-trivial metric different from derivatives in a Euclidean space?

In a non-trivial metric, the concept of a tangent vector and the idea of taking a derivative at a point become more complicated. This is because the metric itself affects the definition of a tangent vector and the calculation of a derivative. In a Euclidean space, derivatives are calculated using the standard methods of differential calculus, while in a non-trivial metric, more advanced techniques such as differential geometry are necessary.

## 3. What are some examples of non-trivial metrics?

One example of a non-trivial metric is the Riemannian metric, which is used to describe the curvature of a surface. Another example is the Minkowski metric, which is used in special relativity to describe the geometry of spacetime. Non-trivial metrics can also arise in more abstract mathematical spaces, such as in functional analysis or in algebraic geometry.

## 4. How are derivatives in a non-trivial metric used in scientific research?

Derivatives in a non-trivial metric are used in various scientific fields, including physics, engineering, and economics. In physics, they are used to model the behavior of particles and fields in curved spacetime. In engineering, they are used to optimize designs and understand the behavior of complex systems. In economics, they are used to study the dynamics of financial markets and to make predictions about future trends.

## 5. What are some challenges in working with derivatives in a non-trivial metric?

One of the main challenges in working with derivatives in a non-trivial metric is the complexity of the calculations involved. Unlike in a Euclidean space, where standard methods of differentiation can be applied, derivatives in a non-trivial metric require more advanced mathematical techniques and can be much more difficult to compute. Additionally, the interpretation of derivatives in a non-trivial metric can be more abstract and may not always have a straightforward physical meaning.

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