Rotational relativity problem

1. May 20, 2010

Jonnyb42

Ok I have had this problem for many years I have still not come to a satisfactory conclusion. I have asked my physics teacher and others I still am confused.

Imagine two particles revolving about one another. They attract one another via gravity (gravity ONLY for simplicity) and remain in equilibrium.

From a third distant observer that does not affect the system, the two particles continue to revolve about a common center and do not collide with each other.

This is where I get confused: Let's observe as from one of the particles, in other words, analyze the system relative to one of the particles. Now we can observe the gravitational force towards one another, but yet the distance between is constant as time goes on. What, in terms relative to one of the particles, is preventing the two from colliding?

How is the system revolving about the center, without an external observer? To the system, the system is not revolving about a center.

Another way to put it, what is the difference between a revolving system and a non-revolving system when viewed within the system.

I made the problem simple by only considering two particles, not a binary star system for example, which is where my confusion started. This problem assumes that these particles are spherically symmetrical.

2. May 20, 2010

Kynnath

Not sure if this is what you mean, and it's been a while since my last physics course so you'll have to excuse if I'm sloppy with the language, but I'll give it a shot.

So your reference frame is one of the particles, with the y axis looking 'up' at the other particle. According to this reference frame, the other particle is at a static location. It's coordinates in this reference frame would be (0,j). It's position is constant. Yet you know physics, and you know there should be a gravitational force acting on the system, and the other particle should be accelerating unless an equal and opposite force was keeping it (and you) in place.

That would be the case, at least, in an inertial reference frame. Because the particle where your reference frame is anchored is rotating around the center of mass, it is accelerated. If it is accelerated, its a non-inertial reference frame. Non-inertial reference frames are wonky.

This is like the example of a person standing on a bus when the bus stops. From within the bus, the passenger suddenly feels pulled forward, even though no identifiable force is acting on him, and he must brace himself to not get thrown around. To an outside observer in the bus station, however, the person is trying to slow down together with the bus, and the 'force' pulling him forwards is just the change in velocity, from speeding to stationary.

So, back to your example. There is a fictitious 'force' holding up the particle from the reference frame of the second particle. This force appears because you are in a non-inertial frame. In fact, the existence of these forces is what allows you to know that you are within a non-inertial reference frame.

The system knows it is revolving around a center because otherwise the other particle would be falling straight down. The difference are these 'fake' forces that appear from nowhere and keep things apart or throw them in other directions.

3. May 20, 2010

Jonnyb42

Thank you Kynnath, I read on wikipedia that fictitious forces must be considered in non-inertial reference frames, (such as the centrifugal force.) I am still confused though. Why must there be fictitious forces in non-inertial reference frames? Shouldn't things make sense in any reference frame?
Also, only a third observer would be able to tell that viewing from one of the particles is a non-inertial reference frame.

There still must be some explanation for why the particles do not collide as viewed from the non-inertial reference frame of one of the particles.

4. May 21, 2010

Staff: Mentor

Newton's laws are only valid for an inertial reference frame. To use them in a non-inertial frame, fictitious forces must be introduced.
They do!
No. An observer can certainly tell that he is in an non-inertial frame. Just check Newton's laws.

5. May 21, 2010

Kynnath

The first thing to understand is that your reference frame does not alter the system itself. It merely alters how you perceive it. If two particles do not collide in one reference frame, they will not collide in any reference frame. Think of it as watching a play (while you are not one of the actors, just a spectator) from different positions on stage, even while running around. It doesn't matter where you are or what you are doing, the play doesn't change. Romeo and Juliet don't get a happily ever after just because you choose to run circles around them.

So, once we are clear on that, why are there fictitious forces in non-inertial reference frames? There's probably a correct definition of this somewhere, but I don't know it, instead I'm going to try to explain it in a way that makes sense to me. Say you are in a drag race. The cars accelerate all the way from start to finish. You're watching from the sidelines. What do you see? You see a lot of spectators at rest, the world is at rest under your feet, and the cars are at rest at the start before suddenly accelerating and blasting away. They continue to accelerate until they hit the finish line, and then they decelerate (accelerate in the opposite direction) until they stop.

Since everything is at rest, the sum of the forces on each of these things must be 0.
F = ma, a = dv/dt, v = dp/dt, we know the positions don't change so velocity is 0 all the time, which means acceleration is 0, which means the sum of forces is 0. There may be forces like gravity and the floor holding them up which cancel each other, but we won't worry about these.

The velocity of the cars however changes. It starts at 0 at the start, then increases, peaks at the finish line, then decreases again to 0. Since dv/dt is not 0, a is not 0, and F = ma is not 0. So we can say that the sum of forces on the car is not 0.

Now, we move our reference frame to inside the car. For this reference frame, the car never moves. When you determine that the car's position is always (0,0), by definition you can't tell if it is moving or not, because it's position is always (0,0). As far as you can tell, it is always stationary. But what about everyone else? From your new vantage point, you see everyone else start accelerating away from you. Because F = ma, if they are accelerating away, the sum of forces on them must not be 0. These forces are not real forces, in the sense that they simply there to balance equations. In fact, you'll notice that the forces applied on them are exactly the opposite of the forces you are unable to feel due to being the reference frame.

That's really all there is to it. When you are in a non-inertial reference frame, the accelerated movement that makes you non-inertial has to be applied to everything else, and that's where the fictitious forces come from.

6. May 21, 2010

Jonnyb42

Linearly accelerated reference frames make sense, if you were in a rocket, you would easily feel a force corresponding to whenever the rocket supplied thrust, and everything is explained. But in rotational systems, it is continuous and in equilibrium, and does not change with time.

So when I think about a two particle system, one revolving about each other and another not revolving as observed from somewhere else, I do not see the difference between them, and do not understand what causes one to collide and not the other.

(When you observe from elsewhere however, it makes sense then, but I am talking about internal observation)

7. May 21, 2010

Kynnath

I think the problem you're having is wth forces that are perpendicular to your movement.

Acceleration is a vector. Vectors have a magnitude and a direction. A change in either magnitude or direction means a change in acceleration. When you are rotating, acceleration is perpendicular to your velocity, aimed towards the center. The direction of your movement is continuously changing though, and because the acceleration is always perpendicular, the direction of your acceleration changes as well. In fact, your acceleration at the start of a rotation is of equal magnitude but opposite direction to your acceleration after you've completed half a circle.

More importantly, velocity is a vector as well, and any change in the velocity of the reference frame means your reference frame is non-inertial.

Let's take your example again. Two spheres in space rotating around each other. The rule is the same as for linearly accelerated frames. You apply the opposite of the acceleration you're feeling to everything else to balance the equations. Your reference frame is constantly accelerating towards the other sphere, in consequence you need to compensate by adding the opposite acceleration to the other sphere, away from you. This acceleration, however, is of equal magnitude to the acceleration the other sphere was feeling towards you. When you add them together they cancel each other. Thus the sum of forces it is feeling is 0, and since it was at rest it will continue to be at rest.

8. May 21, 2010

Cleonis

Let me put your question in a wider context.

I think that underneath your question there is the following supposition: that relativistic physics is a relational theory.

With 'relational theory' I mean the following: a theory in which the interaction between a pair of objects depends only on internal properties of those two objects, and their relation to each other.

In the example that you give you are trying to find a way in which the fact that the two object are orbiting each other is not relevant in the frame of reference that is co-rotating with the two orbiting objects. But there is no such way.

Relativistic physics is not a relational theory.

(The name 'relativity' is a very wrong name. It was coined by Max Planck, 1906, or around that time. A couple of years later Einstein realized what the best name is: Theory of Invariance. But by that time the name 'relativity' had already taken root.)

It is not possible at all to formulate any relational theory of motion.
There is the phenomenon of inertia; objects are subject to inertia. The laws of motion describe the phenomenon of inertia:
- In the absence of a force an object will keep moving in the same direction, covering equal distances in equal intervals of time.
- To change the velocity of an object a force is required. If you double the amount of force you get double the acceleration.

Even if you do not concern yourself with the origin of inertia, you have to recognize that in the case of two objects orbiting each other the phenomenon of inertia is a third player in the field. Inertia is always there. Hence no possibility to formulate any relational theory of motion.

The name that does the best job in capturing the essence is 'Theory of Invariance'. That emphasizes that the substance of relativistic physics is properties that are invariant (under coordinate transformations).
So I recommend that everytime you see the name 'Theory of relativity' you replace that mentally with 'Theory of invariance'.

Last edited: May 21, 2010
9. May 22, 2010

Jonnyb42

Well that is very strange, I will have to think about that some more. Yeah I was thinking that the interaction within a system only depends on internal properties, and I don't understand how it wouldn't...

So it seems to me that it is the idea of inertia that is messing me up. I almost feel like when a mass accelerates it is smeared across spacetime, (like resistance to change velocity.)

Anyways, I have another question. The revolving two particle system and the non-revolving two particle system both have the same acceleration inward, so why does an observer within one view a fictitious force and not the other?

10. May 22, 2010

Kynnath

Actually, there are fictitious forces in both cases, but the effects are different.

In the first case, the fictitious force is what is keeping the other body aloft. Now, let's say the bodies are not rotating, but falling towards each other.

The real forces in play are one force from a to b on body a, and a force from b to a in body b. Both forces are of equal magnitude, since gravity has the same impact on the pair of particles, and the bodies are accelerating at the same rate since they are of equal mass.

The distance between the two bodies is going to decrease then at twice the rate of acceleration of the bodies, until they meet at the middle.

Now, place the reference frame at body a. What you see is body b falling with twice the acceleration it should have. So now the fictitious force is making the body fall faster than it shuld be falling.

11. May 22, 2010

Cleonis

In general I think you are wrongfooting yourself by trying to frame things in terms of 'fictitious force'.

The straightforward view is to think in terms of inertia.

We are always aware of the effects of inertia. When you're in a car you don't need to look outside to tell whether the car is pulling up or braking. If the car accelerates you feel the seat pressing against your back, if the car decelerates you feel the seat belt pressing against your chest. If the car is following a bend in the road you have to brace yourself to keep yourself from slipping sideways. In every direction the sensation is down to one thing: inertia.

Trying to wield an expression such as 'fictitious force' is awkward because inertia is as real as it gets: inertia is one of the fundamental properties of Nature. There is no point in contorting yourself to avoid using the word 'inertia'.

Last edited: May 22, 2010
12. May 22, 2010

Jonnyb42

Right, thanks. I had thought of that and forgot.

Well, I'm not really thinking about fictitious forces too much, I just don't understand the difference between the non-revolving system and the revolving system. You said there was no relational theory of motion, could you explain this to me please?

13. May 22, 2010

Tomsk

I think this question requires general relativity for a full answer, I'm no expert but I'll try and answer it.

Both the particles in question are following geodesics, and have no other forces acting on them so they don't feel any proper acceleration. So looking at it from a co-rotating frame, they should collapse. But in GR binary systems lose energy in the form of gravitational waves and do collapse, so I think that is what will happen. The same thing should happen regardless of coordinates, because of general covariance.

Another thing that's relevant from GR is frame dragging, which is where a rotating system drags nearby objects around with it. I think that stops a third observer from being able to determine whether the rotation is absolute in the way that they can in Newtonian gravity.

Also: I read that GR is relational. This is in Carlo Rovelli's Quantum Gravity book: http://www.cpt.univ-mrs.fr/~rovelli/book.pdf (page 54).

I think the two cases you mentioned are actually equivalent because of general covariance, but it requires GR to really explain it.

14. May 22, 2010

Cleonis

Well, I already explained as well as I could.

Perhaps the following is the stumbling block:
I asserted that the phenomenon of inertia is what shows/defines the difference between a revolving system and and a non-revolving system.

The thing about inertia is that while the effects of inertia are clear, we have no clear view on the source of inertia.
We assume the existence of inertia from observing the effects. What we have to go on is the effects.

Let's even take gravity out of the picture: imagine a region in space, so far away from any galaxy that gravitational effects are negligable.
Let two space capsules be connected by a tether. If the two capsules have no velocity relative to each other then there won't be any tension in the tether. If the two capsules are circumnavigating their common center of mass then the tension in the tether is providing the required centripetal force.

A passenger in one of the capsules will feel the same sensation as you do when you're a passenger in a car that's going round a bend; you're feeling the effect of inertia.

As far as we know, inertia is the same everywhere in the Universe. If it wouldn't be then astronomers would spot the difference.
Let me give an example: if an orbit is very eccentric (such as the orbit of Halley's comet) then the long axis of that ellipse-shaped orbit has a clear direction. That direction is unchanging.
Astronomers have observed distant double stars orbiting each other in highly excentric orbits. Like in the case of Halley's comet, the long axis of those orbits doesn't change direction.

15. May 22, 2010

Jonnyb42

Since inertia is the source of this problem, then is it true that there is no known solution?

When I think about it, it really is inertia that confuses me.

Also, this leads me to conclude that there IS a sort of "absolute" reference frame because we can refer to inertial reference frames while disregarding the reference frame WE are in when considering them. So it could be called the inertial reference frame.

I have not studied special nor general relativity enough (although you can bet I am going to!), so could you explain what is meant by general covariance, as it pertains to this case?

Last edited: May 22, 2010
16. May 22, 2010

Kynnath

There isn't an absolute reference frame, but the idea is close.

There are inertial reference frames, and non-inertial reference frame. Inertial reference frames are the ones we care about, because they have the handy property of all inertial reference systems being in a sense equivalent. In all inertial frames, you always see the same forces acting on the particles on the system.

17. May 22, 2010

Cleonis

Yeah, inertia is very peculiar. But since inertia is always and everywhere it tends to be overlooked.

We can think of the set of all inertial frames of reference as a class of frames. For describing motion all members of that set are equivalent. This is usually referred to as 'the equivalence class of inertial frames of reference'.

Now, we can choose to think of the equivalence class of inertial frames of reference as a single, unique reference for acceleration. While we do have relativity of inertial motion we don't have relativity of acceleration.

18. May 22, 2010

Jonnyb42

that's what I was thinking about.

Is it inertia that makes it impossible to formulate a relational theory of motion?

Does my problem simplify to "What is inertia?"

Last edited: May 22, 2010
19. May 23, 2010

Cleonis

In effect yes. As far as we know the phenomenon of inertia is uniform throughout the Universe. The effects of inertia act as a universal reference of acceleration. Obviously that excludes a relational theory of motion.

Currently we have no clear view on what the origin of inertia is, but that is not necessarily a problem. In order to proceed we assume the existence of inertia, with it's known properties.

In general that is how science makes progress. For example, Newton recognized that if he assumed the existence of gravity, with an inverse square law, then he could account for all gravitational effects. Some of his contemporaries criticized him for not explaining exactly how gravity acts over the vast distances of space. But if Newton would have struggled to explain gravity in an exhaustive way he would just have bogged himself down. Newton probed as deep as was possible at the time, making huge progress, by assuming the newtonian inverse square law of gravity.

The introduction of the General Theory of Relativity moved the theory building to a deeper level. In GTR describing the effects of inertia and describing the effects of gravitation are unified in a single theory.

20. May 23, 2010

Phrak

I'm not so sure you're question has been properly understood.

I'll try to use other words, and you can say if it's what you mean, or not:

Let's say for a moment that we place a coordinate system around your two spheres. That is, we attach to space, a coordinate system independent of the two spheres. Upon these coordinates we see that each sphere is rotating around the other sphere. We can even do the calculations and see that each sphere follows a circular path.

Now let's choose another coordinate system so that it rotates with respect to the first one. By our insightful choice, in this coordinate system neither sphere moves in a circle but sits at one point.

What in nature picks-out a particular coordinate system that is correct over any other?​

How's that?