## What exactly is centrifugal force

 Quote by DaleSpam Hmm, I guess this is due to my experience with GR, but I always think of the metric and the connection being tightly wound up together. In 4D Galilean spacetime is the connection not the unique metric compatible torsion free connection, like in GR?
It might be, but the easiest way for me to think of it (this actually applies to GR as well as Galilean spacetime) is that the connection coefficients $\Gamma^\mu_{\nu \lambda}$ for any coordinate system can be computed via

$\Gamma^\mu_{\nu \lambda} = \dfrac{\partial x^\mu}{\partial x^i}\dfrac{\partial^2 x^i}{\partial x^\nu \partial x^\lambda}$

where $x^i$ is any inertial, Cartesian coordinate system. I know that's cheating, because you'd like a characterization that doesn't involve special coordinate systems, but in practice, the noninertial coordinate systems people are interested in are related in a known way to some inertial Cartesian coordinate system.

Mentor
 Quote by stevendaryl In what sense is any of that "the machinery of forces" as opposed to the machinery of accelerations?
Forces and accelerations are closely related, so I wouldn't consider there to be much opposition between "the machinery of forces" and "the machinery of accelerations". In the end you have a quantity in units of force that goes into the equations in exactly the same place as the real forces go.

Here are the facts:
The equations of motion have terms containing the connection coefficients
In some valid coordinate systems those terms are non-zero
Those terms have units of force and are called "fictitious forces" or "inertial forces"
You cannot correctly represent the physics without those terms

Here is my opinion:
It is reasonable to call them "forces" and treat them as forces where appropriate

I think that we agree on the facts. Given the facts, I have a hard time seeing my opinion as being in need of reconsideration.

Recognitions:
 Quote by stevendaryl How is that helped by the concept of inertial forces? ... The condition that there are no stresses on the ball is a fact about the real (noninertial) forces on the various parts of the ball.
Maybe you forgot the fact that the ball on the table is at rest in the first system, but accelerating in second (rotating) system. In the rotating system, the inertia forces just happen to be the right value to cause the accelerations, so there is no stress. Now, isn't that an amazing coincidence? (/irony).

If you don't want to believe DaleSpam's example, try reading the users guides and theory manuals for a few standard commercial structural analysis programs like NASTRAN, ABAQUS, ANSYS, etc. Maybe they are all doing this the wrong way. Or maybe there is a difference between doing something practical, and writing some nice vector equations (whatever today's definition of "vector" happens to be).

Mentor
 Quote by stevendaryl Well, give an example of how it might be useful.
Docking of spacecraft. Modeling the weather. The restricted three body problem. Explaining the tides. Rotating machinery. The concept of "inertial force" is useful in a rather large number of very disparate applications.

 I don't think that's true. What do you think a vector is?
A vector is a member of a vector space. Vectors add per a commutative and associative operation. There's a zero vector, and every vector has a unique additive inverse. Vectors can be multiplied by a scalar, and this multiplication is associative. That they have a magnitude and a direction? Neither concept is part of what constitutes a vector. That they transform according to certain rules? That's even more foreign to the generic concept of a "vector" than are the concepts of magnitude and direction.

 I think you're confused about what vectors and tensors are, yourself.
Uh, no. You are. You have shown this confusion multiple times. Vectors and tensors are different things. Raising and lowering indices and operators are a tensor concept, not a vector concept.

You don't need to use tensors to understand inertial forces.

 Quote by D H Docking of spacecraft. Modeling the weather. The restricted three body problem. Explaining the tides. Rotating machinery. The concept of "inertial force" is useful in a rather large number of very disparate applications.
But all those are equally well described using connection coefficients. You need connection coefficients for whenever you are dealing with changes of vectors, not just when forces are involved.

 A vector is a member of a vector space. Vectors add per a commutative and associative operation. There's a zero vector, and every vector has a unique additive inverse. Vectors can be multiplied by a scalar, and this multiplication is associative. That they have a magnitude and a direction?
That's the definition of a "vector space". The vectors used by physics to represent velocities, accelerations, etc. are a very specific vector space, which is known as "tangent vectors", where "tangent" means a linear approximation to a parametrized path.

 Uh, no. You are. You have shown this confusion multiple times. Vectors and tensors are different things. Raising and lowering indices and operators are a tensor concept, not a vector concept.
I think you are confused about this point. The vectors that are consideration definitely are a special case of tensors.

 You don't need to use tensors to understand inertial forces.
I disagree. If you don't understand tensors, then you are really handicapped when it comes to understanding forces in noninertial coordinates.

Mentor
 Quote by stevendaryl But all those are equally well described using connection coefficients.
Of course they are equally well described using connection coefficients. "Fictitious forces" is just a name for the terms with connection coefficients.

That is like someone talking about the uses of water and someone else saying that all of those uses are equally well fulfilled by H2O.

 Quote by stevendaryl I think you are confused about this point. The vectors that are consideration definitely are a special case of tensors.
Take a look at http://en.wikipedia.org/wiki/Tensor#Examples.

A tensor is classified by a pair of numbers $(n,m)$ called its type, or rank.
• A (0,0) tensor is just a scalar.
• A (0,1) tensor is a vector.
• A (1,0) tensor is a covector, or 1-form.
• In general, a (n,m) tensor is a multlinear function that takes n vectors and m covectors and returns a real number.

 Quote by DaleSpam Of course they are equally well described using connection coefficients. "Fictitious forces" is just a name for the terms with connection coefficients.
But that's not true. "Fictitious forces" are described equally well using connection coefficients, but not the other way around, because connection coefficients are more general--they are used whenever one takes a derivative of a vector, regardless of whether it involves forces or not.

Mentor
 Quote by stevendaryl But that's not true. "Fictitious forces" are described equally well using connection coefficients, but not the other way around, because connection coefficients are more general--they are used whenever one takes a derivative of a vector, regardless of whether it involves forces or not.
Sorry, I should have been more clear. The terms in the equation of motion with the connection coefficients are fictitious forces. I didn't mean to imply that such terms in other equations were called fictitious forces.

In the equations of motion "fictitious forces" is just a name for the terms with connection coefficients.

 Quote by DaleSpam Sorry, I should have been more clear. The terms in the equation of motion with the connection coefficients are fictitious forces. I didn't mean to imply that such terms in other equations were called fictitious forces.
Okay. Yes, I certainly agree that the expression

$- m \Gamma^i_{jk} U^j U^k$

can unambiguously be called "the fictitious (or inertial) forces"