What exactly is centrifugal force

[D H]

TIME OUT! Thread temporarily closed.

For the last few pages you all have been bickering over the "reactive centrifugal force", which is (1) quite distinct from the "(fictitious) centrifugal force", (2) a concept of limited applicability, (3) something physicists don't quite like.

I'm splitting this off-topic discussion of the reactive centrifugal force into a separate thread. This will take some time ...

 

[D H]

The off-topic discussion on the reactive centrifugal force has been moved to [thread=668756]this new thread[/thread]. Please limit the discussion in this thread to the fictitious centrifugal force.

Thread reopened.

 

[D H]

Surely, the last step isn't doing anything for you.

This is just an opinion, not a fact. Quite a few people find the concept of inertial forces (or fictitious forces) quite useful.

They are different from other forces you're likely to encounter, because they don't depend on the substance an object is made of, and they don't have an equal and opposite reactive force.

That's why they're called inertial forces. They don't have third law counterpart. So?

The real confusion that is at the heart of discussions of "inertial forces" is the assumption that, if \stackrel{\rightarrow}{U} is a vector (say, a velocity vector) with components U^i, then \frac{\stackrel{\rightarrow}{dU}}{dt} must be a vector with components \frac{dU^i}{dt}. That's just bad mathematics.

You have a non-standard concept of what constitutes a vector. There's not one thing in the mathematical definition of a vector that says how they transform. Inertial forces aren't covariant or contravariant tensors, but tensors are something different from vectors. Don't confuse the two concepts.

 

[stevendaryl]

This is just an opinion, not a fact. Quite a few people find the concept of inertial forces (or fictitious forces) quite useful.

Well, give an example of how it might be useful.

You have a non-standard concept of what constitutes a vector.

I don't think that's true. What do you think a vector is?

There's not one thing in the mathematical definition of a vector that says how they transform. Inertial forces aren't covariant or contravariant tensors, but tensors are something different from vectors. Don't confuse the two concepts.

I think you're confused about what vectors and tensors are, yourself. The usual notion of a tensor includes vectors as a special case.

The issue, as I said, is what does it mean to take the derivative of a vector quantity. If \vec{V(t)} is a vector quantity, such as velocity, then what is the meaning of \frac{\vec{dV}}{dt}?

However we define this derivative, we want it to obey the usual rules of calculus, such as the chain rule and the product rule. So if we decompose a vector \vec{V} as a linear combination of basis vectors \vec{e_\mu}, we have:

\vec{V} = \sum_\mu V^\mu \vec{e_\mu}
\dfrac{\vec{dV}}{dt} = \sum_\mu (\dfrac{dV^\mu}{dt} \vec{e_\mu} + V^\mu \dfrac{\vec{de_\mu}}{dt})

So if we expect derivatives to work in their normal way, we can't blithely assume that
(\dfrac{\vec{dV}}{dt})^\mu =\dfrac{dV^\mu}{dt} unless we assume that
\dfrac{\vec{de_\mu}}{dt} = 0

But if we assume that the basis vectors for Cartesian coordinates, \vec{e_x}, \vec{e_y} are all constant, obeying \dfrac{\vec{de_x}}{dt} = \dfrac{\vec{de_y}}{dt} =0, then when we switch to polar coordinates \vec{e_r}, \vec{e_\theta}, those basis vectors CANNOT be constant, because, for instance:

\vec{e_r} = cos(\theta) \vec{e_x} + sin(\theta) \vec{e_y}

\dfrac{\vec{de_r}}{dt} = -sin(\theta) \dfrac{d\theta}{dt} \vec{e_x} + cos(\theta) \dfrac{d \theta}{dt} \vec{e_y}

So when using polar coordinates, there are additional terms in computing the time derivative of a vector \vec{V} arising from \dfrac{\vec{de_r}}{dt} and \dfrac{\vec{de_\theta}}{dt}. These additional terms are not forces, they are just derivatives of basis vectors.

 

[Dale]

Well, give an example of how it might be useful.

E.g. to design a turbine blade that will not break during operation.

Non-inertial coordinate systems are also often useful in solving problems where the equations become numerically unstable in inertial coordinates. E.g. in calculating orbits in multi-body gravitational fields.

 

[stevendaryl]

E.g. to design a turbine blade that will not break during operation.

How is that helped by the concept of inertial forces? Inertial forces don't break things. You can see this by considering a ball at rest in an inertial Cartesian coordinate system. Now switch to a rotating coordinate system. In this coordinate system, there are huge "inertial forces" at work on the ball. Do they deform the ball, or stretch it, or anything else? No, they don't. Nobody was ever hurt by inertial forces. If you exert a centripetal force on the ball in an attempt to make the ball's coordinates constant in the rotating coordinate system, then you will definitely distort the ball. But it's the addition of the centripetal force that causes stresses on the ball, not the "inertial" force.

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.

 

[stevendaryl]

Non-inertial coordinate systems are also often useful in solving problems where the equations become numerically unstable in inertial coordinates. E.g. in calculating orbits in multi-body gravitational fields.

I think you added this paragraph after I responded to your post.

I perfectly well agree that it can be useful to use noninertial coordinates. In noninertial coordinates, the equations of motion for a point mass m are:

m (\dfrac{dU^i}{dt} + \Gamma^i_{jk} U^j U^k) = F^i_{int}

where \vec{F_{int}} is the "interaction" forces (noninertial), and \Gamma^i_{jk} are the connection coefficients for the coordinate system. If you're going to use noninertial coordinates (and certainly there are times when they are the most convenient way to go), you have to take the connection coefficients into account. However, the disagreement is over whether they are considered force terms (and belong on the right, with the forces) or are considered acceleration terms (and belong on the left, with the derivatives of the velocity). My point is that these extra terms arise whenever you want to take a derivative of a vector. It doesn't matter whether that vector is velocity, or what. So mathematically, they really should be considered part of the derivative, rather than forces. If you're computing acceleration, then it certainly doesn't matter what side of the equation you put the terms on, but in other cases, you are computing derivatives that may have nothing to do with forces.

 

[Dale]

How is that helped by the concept of inertial forces? Inertial forces don't break things.

If you are designing a turbine blade you will have some mathematical model of the shape and material. This is going to be a complicated shape so you will use some sort of numerical approximation, typically a finite element model. Then, to analyze the strain you simply go to the rotating frame and add a nice fictitious centrifugal force. That centrifugal force allows you to easily calculate the strain throughout the blade without any numerical problems such as you might encounter in an inertial frame.

As to whether or not inertial forces break things, I never claimed that they do. The important thing is that using them allows us to relatively easily determine if the blade breaks. That is one example how they are useful.

You can see this by considering a ball at rest in an inertial Cartesian coordinate system. ...

I never claimed they were useful for analyzing balls at rest in an inertial frame.

 

[Dale]

I perfectly well agree that it can be useful to use noninertial coordinates.

Then that makes your question above a little odd.

the equations of motion for a point mass m are:

m (\dfrac{dU^i}{dt} + \Gamma^i_{jk} U^j U^k) = F^i_{int}

where \vec{F_{int}} is the "interaction" forces (noninertial), and \Gamma^i_{jk} are the connection coefficients for the coordinate system. If you're going to use noninertial coordinates (and certainly there are times when they are the most convenient way to go), you have to take the connection coefficients into account. However, the disagreement is over whether they are considered force terms (and belong on the right, with the forces) or are considered acceleration terms (and belong on the left, with the derivatives of the velocity).

The terms are there, and they are called "fictitious forces". What is the problem? The fact is that even if you prefer to think of them as acceleration terms you can still use a lot of the standard machinery for forces to analyze them and their effects in the non-inertial frame.

 

[stevendaryl]

I think you added this paragraph after I responded to your post.

I perfectly well agree that it can be useful to use noninertial coordinates. In noninertial coordinates, the equations of motion for a point mass m are:

m (\dfrac{dU^i}{dt} + \Gamma^i_{jk} U^j U^k) = F^i_{int}

I've been working with the 4D notion of Galilean spacetime. If you prefer to keep your space and time separate, then this can be expressed as:

m (\dfrac{dU^i}{dt} + C^i_{jk} U^j U^k + D^i_j U^j + E^i) = F^i_{int}

where the coefficients C^i_{jk}, D^i_j and E^i are zero in an inertial Cartesian coordinate system, and can be computed in any other coordinate system via:
C^i_{jk} = \dfrac{\partial x'^i}{\partial x^u}\dfrac{\partial }{\partial x'^j}\dfrac{\partial }{\partial x'^k} x^u

D^i_j = \dfrac{\partial x'^i}{\partial x^u}\dfrac{\partial }{\partial x'^j}\dfrac{\partial }{\partial t} x^u

E^i = \dfrac{\partial x'^i}{\partial x^u}\dfrac{\partial^2}{\partial t^2} x^u

where x'^i are the noninertial, curvilinear coordinates, and x^u are and Cartesian, inertial coordinates.

 

[stevendaryl]

Then that makes your question above a little odd.

The terms are there, and they are called "fictitious forces". What is the problem?

Because they arise whenever one takes a derivative of a vector, whether or not that vector is a velocity. They don't have anything to do with "forces", they have to do with computing the changes in vector quantities.

 

[stevendaryl]

As to whether or not inertial forces break things, I never claimed that they do. The important thing is that using them allows us to relatively easily determine if the blade breaks. That is one example how they are useful.

I never claimed they were useful for analyzing balls at rest in an inertial frame.

I was using the ball as an example to show that inertial forces don't come into play in analyzing balls in rotating coordinate systems, either.

 

[Dale]

I've been working with the 4D notion of Galilean spacetime.

Btw, do you have any good internet references for this topic? I would like to learn a little more about it, but not enough to purchase anything. I still feel like the doubly-degenerate metric in 4D Galilean spacetime is an indication that it may be a bad fit between the math and the physics, but your recent comments have intrigued me enough to give it a second look.

 

[Dale]

Because they arise whenever one takes a derivative of a vector, whether or not that vector is a velocity. They don't have anything to do with "forces", they have to do with computing the changes in vector quantities.

For something that doesn't have anything to do with forces you sure can use a lot of the machinery of forces to work with them. I mean, you can treat them the same as real forces in Newton's 2nd law, Hooke's law, the work-energy theorem, etc.

 

[stevendaryl]

Btw, do you have any good internet references for this topic? I would like to learn a little more about it, but not enough to purchase anything. I still feel like the doubly-degenerate metric in 4D Galilean spacetime is an indication that it may be a bad fit between the math and the physics, but your recent comments have intrigued me enough to give it a second look.

Sorry, my understanding of this topic developed over many years through internet discussions, not from reading a textbook. I don't know of a good textbook that teaches it, although it is implicit in the Newton-Cartan theory of gravity, which is discussed in Misner, Thorne and Wheeler's Gravitation.

Whether the fact that there is no nondegenerate metric for 4D Galilean spacetime causes a problem or not depends on what you're doing. If you're trying to do equations of motion, then you don't need a metric, you just need a connection, which Galilean spacetime has.

The really nice thing about the 4D view is that changes to noninertial coordinates, or to curvilinear coordinates, or to another rest frame are all just coordinate transformations. Things like velocity transform like ordinary vectors: In going from coordinates x^\mu to x'^\u, you just use:

V^u = V^\mu \dfrac{\partial x^u}{\partial x^\mu}

Changing to boosted, accelerating, curvilinear, rotating, etc., coordinates are all instances of this general transformation scheme. But that's only true if you view vectors as having 4-components, not just three. Without using 4-vectors, transformations involving time aren't really coordinate transformations in the normal sense.

Also, with 4-vectors, "g" forces, Coriolis forces, Centrifugal forces all show up in exactly the same way, as connection coefficients associated with nonconstant basis vectors.

 

[Dale]

I don't know of a good textbook that teaches it, although it is implicit in the Newton-Cartan theory of gravity

Yes, that was the source of my exposure to it.

 

[stevendaryl]

The really nice thing about the 4D view is that changes to noninertial coordinates, or to curvilinear coordinates, or to another rest frame are all just coordinate transformations.

Obviously, it's equivalent to the usual way of doing things, but it has certain unifying advantages, in sort of the same way as viewing the electric and magnetic fields as components of a tensor can unify equations in electromagnetics.

 

[Dale]

Whether the fact that there is no nondegenerate metric for 4D Galilean spacetime causes a problem or not depends on what you're doing. If you're trying to do equations of motion, then you don't need a metric, you just need a connection, which Galilean spacetime has.

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?

 

[stevendaryl]

For something that doesn't have anything to do with forces you sure can use a lot of the machinery of forces to work with them. I mean, you can treat them the same as real forces in Newton's 2nd law, Hooke's law, the work-energy theorem, etc.

In what sense is any of that "the machinery of forces" as opposed to the machinery of accelerations? To me, Newton's 2nd law is a tautology if you don't also have his 3rd law. As for the work-energy theorem, how are you defining energy in a noninertial coordinate system?

 

[stevendaryl]

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.

 

[Dale]

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.

 

[AlephZero]

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).

 

[D H]

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.

 

[stevendaryl]

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.

 

[Dale]

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.

 

[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.

 

[stevendaryl]

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.

 

[Dale]

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.

 

[stevendaryl]

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"