What exactly is centrifugal force

[stevendaryl]

I very much doubt that they were as naive as to think they were actually near a conclusion. You cannot have any knowledge of their motives but you must know that anyone who breaks ground in any of this (since the concept of God given laws has ceased to be taken for granted, at least) can only hope to improve on existing scientific models. Models are not 'truth'; they are statements that can be shown to predict the outcomes of certain experiments.

I'm not sure what the relevance of this is to what I've said. Claiming that the goal of physics is understanding doesn't imply how close our current understanding is to the truth.

 

[A.T.]

I gave plenty of examples of what I meant. If you didn't understand what I meant after that, you could ask follow-up questions.

I understand that it is not relevant for the predictions.

Anyway, when Einstein, or Newton, or Schrödinger, or just about any other physicist was engaged in doing physics, it certainly wasn't coming up with formulas that make predictions.

Yes it was.

They were engaged in the struggle to understand the world.

Define "understand the world".

 

[Dale]

I don't think that's true. The fact that you and I might disagree about specific cases whether something is or is not "physics" or a "sport" or "music" does not get in the way of communication if there is substantial overlap.

The current discussion between you and AT seems to contradict this claim of yours. There is surely substantial overlap, and yet the remaining differences in the definitions are getting in the way of communication.

 

[Studiot]

How does any of this help or encourage the OP?

 

[Dale]

I think that there is no such thing as centrifugal force .

Am I right ? is this force fictitious ?

A centrifugal force is a fictitious force. That doesn't mean that there is no such thing as a centrifugal force.

 

[Andrew Mason]

But the other astronaut (sitting in the frame of the wheel) will see the departing astronaut accelerating, initially (during the first 90 degrees of motion, at least) and due to the geometry of the situation.

He would think the departing astronaut was accelerating only if he forgot that he was in a non-inertial (rotating) reference frame. To an inertial observer, the departing astronaut is simply continuing the motion he had when the bolts were cut.

Would he not conclude that there is a force still operating? This perceived force will also be making the departed astronaut perform a spiral outward path - so it would (might) not just be a centrifugal force that he would need in order to explain the guy's path.

I think the departing astronaut would prescribe a cycloidal outward spiral in the non-inertial reference frame of the astronaut on the rotating space station.

AM

 

[stevendaryl]

The current discussion between you and AT seems to contradict this claim of yours. There is surely substantial overlap, and yet the remaining differences in the definitions are getting in the way of communication.

I don't think that's a correct diagnosis.

 

[stevendaryl]

I understand that it is not relevant for the predictions.

Indirectly, it is. Getting straight the difference between connection coefficients and forces is an important step in understanding physics in curved spacetime. So if you want to go on to advanced topics, then understanding this is important.

Define "understand the world".

Come on. You know what the word "understand" means.

The idea that physics is exclusively about making quantitative predictions is just wrong.

We can go through many examples. Before Einstein developed Special Relativity, the equations of Special Relativity were already developed. That's why they're called the "Lorentz transformations" rather than the "Einstein transformations". Einstein's new theory didn't change the equations, it provided a new way of understanding those equations--a new way of deriving them. In the long run, this was a tremendous advance, making most of the physics since then possible. But the motivation wasn't new predictions, it was to understand things that people already knew about, but didn't understand.

Similarly for quantum mechanics. The Balmer series already gave a good quantitative prediction for the energy levels of hydrogen. It was just a guess. The steps that Bohr took, which opened up further developments by Schrödinger and Heisenberg, was an attempt to derive those energy levels from some kind of first principles. Of course, quantum mechanics ended up having enormous consequences and great predictive value. But the initial steps were an attempt to understand already existing information.

When Feynman developed his path integral formulation of quantum mechanics, at first it was simply a reworking of the Schrödinger--it was just a different way of understanding how to derive quantum amplitudes. It turns out that the ideas developed by Feynman in working on his path integral formulation had applicability beyond quantum mechanics, and could be applied in quantum electrodynamics and elsewhere.

The idea that there is nothing to physics other than quantitative predictions is a philosophical position, and in my opinion, it's very BAD philosophy. The attempt to understand data and formalisms has always been the most direct route to coming up with new theories that do make quantitative predictions. So even if, at the end, all you care about is quantitative predictions, trying to understand the mathematics, the data, and the models is a much more effective way to get to that point.

 

[Dale]

The idea that there is nothing to physics other than quantitative predictions is a philosophical position, and in my opinion, it's very BAD philosophy.

No, it is a definition of the term "physics", and in your opinion it is a very BAD definition.

 

[stevendaryl]

Some other examples: General Relativity. The motivation was to make a theory of gravity that was compatible with relativity. There was also a philosophical goal, which was to generalize the principle of relativity so that all coordinate systems were treated equivalently, not just inertial coordinate systems. The goal wasn't to make new predictions about bending of starlight, or whatever. The goal was reconciling two theories: Special Relativity and gravity. It certainly turned out that the project produced new predictions, and if hadn't produced new predictions, it would been considered a failure, or at best an interesting exercise. However, the goal of making predictions really didn't drive the development of the theory at all. Einstein was trying to understand the nature of gravity in a way that made sense in light of relativity.

I just think that the idea that there is nothing to physics other than quantitative predictions is just a severely claustrophic notion of what science is about. It is true that at the end of the day, your ideas have to have empirical consequences, but a lot of the development of a new theory is about clarifying concepts.

 

[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.