Undergrad Do partial derivatives commute in general?

[jk22]

Suppose we have to deal with the question : $$\frac{\partial}{\partial x}\frac{\partial}{\partial y}=?\frac{\partial}{\partial y}\frac{\partial}{\partial x}$$

This seems true for independent variables. But if at the end x and y are linked in some way like $$x=f(t),y=g(t)$$ this is no more the case, since : $$\frac{\partial}{\partial x}\frac{\partial}{\partial y}=\frac{d}{\dot{f}dt}\frac{d}{\dot{g}dt}=\frac{d^2}{\dot{f}\dot{g}dt^2}-\frac{\ddot{g}d}{\dot{f}\dot{g}^2 dt}\neq\frac{\partial}{\partial y}\frac{\partial}{\partial x}=\frac{d}{\dot{g}dt}\frac{d}{\dot{f}dt}=\frac{d^2}{\dot{f}\dot{g}dt^2}-\frac{\ddot{f}d}{\dot{g}\dot{f}^2 dt}$$.

Is this equal to the covariant derivative ?

For example can we then say that if we consider a curve on a sphere that those partial derivatives do not commute in general ?

 

[fresh_42]

https://en.wikipedia.org/wiki/Symmetry_of_second_derivatives

 

[haushofer]

I'm confused about your notation. In your particular example, I'd say that

<br /> \frac{\partial}{\partial x} = \frac{\partial t}{\partial x}\frac{d}{dt}, \ \ \ \ \ \ \frac{\partial}{\partial y} = \frac{\partial t}{\partial y}\frac{d}{dt}<br />

With this you can apply the commutator

<br /> [\frac{\partial}{\partial x}, \frac{\partial}{\partial y}]<br />

on a function $f(t)$. If I do this, I find

<br /> [\frac{\partial}{\partial x}, \frac{\partial}{\partial y}]f \neq 0<br />

in general. So

I'm not sure what this has to do with covariant derivatives; the commutativity of partial derivatives involves independent variables and differentiable functions.

(edit: silly mistake concerning chain rule and variables; corrected it)

 

[jk22]

I'm confused about your notation.
<br /> \frac{\partial}{\partial x} = \frac{dx}{dt}\frac{d}{dt} = \dot{x} \frac{d}{dt} , \ \ \ \ \ \ \frac{\partial}{\partial y} = \frac{dy}{dt}\frac{d}{dt} = \dot{y} \frac{d}{dt}<br />

Is it not that $\frac{\partial}{\partial x}=\frac{\partial}{dt}\frac{dt}{\partial x}$ but this has indeed no sens formally.

Anyway, if it does not commute on a curve parametrized by $t$ like on a geodesic, does this mean that all the general relativistic framework should be written in terms of non commutative derivatives ? This would mean that this typical quantum aspect were involved in gr ?

 

[haushofer]

Yes, you're right, this doesn't make sense what I wrote in post #3, I was totally sloppy there; I applied the chain rule wrongly. But I'm not sure about your question. Non-commutativity is not automatically linked to quantum mechanics, so I don't see the connection there.

 

[suremarc]

In general, it is not true that
\frac{\partial}{\partial x}=\frac{d}{\dot{f}\, dt};
What is true is that
\frac{d}{dt}=\dot{f}\frac{\partial}{\partial x}+\dot{g}\frac{\partial}{\partial y}.