Hessian matrix of the Newtonian potential is zero?

AI Thread Summary
The discussion centers on the Hessian of the Newtonian potential, with the initial claim that it equals zero due to certain mathematical manipulations. The calculations involve the relationship between force, mass, and the second derivatives of position with respect to time. However, a critical error is identified in the assumption that mixed partial derivatives can be freely swapped, as acceleration is time-dependent and cannot be treated as a simple function of spatial coordinates. The correct approach emphasizes the need to account for the dependency of acceleration on time, leading to a non-zero result for the Hessian. The conclusion highlights the importance of careful differentiation when dealing with time-dependent variables.
Chain
Messages
35
Reaction score
3
So I'm looking at the hessian of the Newtonian potential:

\partial^2\phi / \partial x_i \partial x_j

Using the fact that (assuming the mass is constant):

F = m \cdot d^2 x / d t^2 = - \nabla \phi

This implies:

\partial^2\phi / \partial x_i \partial x_j = -m \cdot \frac{\partial}{\partial x_j} (d^2 x_i / d t^2) = -m \cdot \frac{\partial}{\partial x_j} (\partial^2 x_i / \partial t^2)

As we can swap the total derivatives for partial derivatives since for Cartesian coordinates:

\partial x_i / \partial x_j = \delta_{ij}

Using the fact that we can swap the order of differentiation for mixed partials (assuming continuity of the partial derivatives) we obtain:

\partial^2\phi / \partial x_i \partial x_j = -m \cdot \partial^3 x_i / \partial x_j \partial t^2 = -m \cdot \frac{\partial}{\partial t^2} \partial x_i / \partial x_j = -m \cdot 0 = 0

Hence I obtain the result that the hessian of the Newtonian potential is zero which can't possibly be correct but I can't find the error in my calculation.

Any help would be much appreciated :)
 
Physics news on Phys.org
What you wrote doesn't make too much sense and the mathematical manipulations are illegal. Acceleration depends on time, coordinate depends on time: a(x) = a(t(x)). Good luck reverting x(t) into t(x).
 
So the problem is in the last step where I swap the order of differentiation because it is not possible to find time as a function of position?

I guess the proper expression for the differential of acceleration with respect to a spatial coordinate is:

\partial a(t(x)) / \partial x = \frac{\partial a(t)}{\partial t} \cdot \frac{\partial t}{\partial x} = \frac{\partial a(t)}{\partial t} \cdot (\frac{\partial x}{\partial t})^{-1}

Which is clearly non-zero.
 
Thread 'Is 'Velocity of Transport' a Recognized Term in English Mechanics Literature?'
Here are two fragments from Banach's monograph in Mechanics I have never seen the term <<velocity of transport>> in English texts. Actually I have never seen this term being named somehow in English. This term has a name in Russian books. I looked through the original Banach's text in Polish and there is a Polish name for this term. It is a little bit surprising that the Polish name differs from the Russian one and also differs from this English translation. My question is: Is there...
I know that mass does not affect the acceleration in a simple pendulum undergoing SHM, but how does the mass on the spring that makes up the elastic pendulum affect its acceleration? Certainly, there must be a change due to the displacement from equilibrium caused by each differing mass? I am talking about finding the acceleration at a specific time on each trial with different masses and comparing them. How would they compare and why?
Back
Top