How Does Rotational Relativity Affect Particle Dynamics in a Two-Body System?

In summary: In short, fictitious forces are needed to make Newton's laws work in a non-inertial reference frame. Without them, the laws would be invalid.
  • #1
Jonnyb42
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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.
 
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  • #2
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.

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.
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
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
Jonnyb42 said:
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?
Newton's laws are only valid for an inertial reference frame. To use them in a non-inertial frame, fictitious forces must be introduced.
Shouldn't things make sense in any reference frame?
They do!
Also, only a third observer would be able to tell that viewing from one of the particles is a non-inertial reference frame.
No. An observer can certainly tell that he is in an non-inertial frame. Just check Newton's laws.
 
  • #5
Jonnyb42 said:
I am still confused though. Why must there be fictitious forces in non-inertial reference frames?

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
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
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
Jonnyb42 said:
How is the system revolving about the center, without an external observer? To the system, the system is not revolving about a center.

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'.
 
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  • #9
It is not possible at all to formulate any relational theory of motion.
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
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
Jonnyb42 said:
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?

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'.
 
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  • #12
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.
Right, thanks. I had thought of that and forgot.

In general I think you are wrongfooting yourself by trying to frame things in terms of 'fictitious force'.
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
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
Jonnyb42 said:
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?

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
...we have no clear view on the source of inertia.

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 think the two cases you mentioned are actually equivalent because of general covariance,
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?
 
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  • #16
Jonnyb42 said:
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.
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.

Non-inertial frames don't share this property.
 
  • #17
Jonnyb42 said:
When I think about it, it really is inertia that confuses me.

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

Jonnyb42 said:
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.

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
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'.
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?"
 
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  • #19
Jonnyb42 said:
Is it inertia that makes it impossible to formulate a relational theory of motion?

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
Jonnyb42 said:
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.

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?
 
  • #21
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?
Yeah, that is what I mean, thanks.

To me it would make sense that the laws of physics stay the same in any coordinate system, without the need for fictitious forces.
 
  • #22
Jonnyb42 said:
Yeah, that is what I mean, thanks.

To me it would make sense that the laws of physics stay the same in any coordinate system, without the need for fictitious forces.

Imagine that. This is what drove Einstein to hypothesize his theory of general relativity. You have the right question, which is a good part of finding answers. Keep it up.

I don't have the answer, but your question is far better than I've mangaged on this very topic. Maybe someone else has some background that can clarify things...
 
  • #23
Jonnyb42 said:
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?

Like you said a couple of posts ago, general covariance says that physics should be the same in any coordinate system. So if two bodies collide due to gravity in one coordinate system, they better had in all others! Because of the principle of equivalence, the transition between seeing orbiting and seeing no orbiting is just a coordinate transformation, so I think they should collide in both cases - which is why GR is necessary, since they need to lose energy as gravitational waves.

It doesn't work in Newtonian gravity because acceleration is absolute, and there's no frame dragging so an external observer can measure the angular momentum, which is what keeps them apart. In relativity angular momentum is tied into inertia, so the angular momentum isn't absolute, only the combined a.m+inertia. Actually it's even more complicated in GR, I don't really know how angular momentum works there, sorry!
 
  • #24
Imagine that. This is what drove Einstein to hypothesize his theory of general relativity.
awesome!
Does this mean that there is an answer in General Relativity?

, since they need to lose energy as gravitational waves.
Isn't that a problem though, because that would mean they lose enough energy to collide at the same rate as if they were not revolving, but I read the effects are more subtle than that.

Thank you everyone,
so far it looks like I need to study GR. (along with all the other physical theories!)
 
  • #25
Jonnyb42 said:
Does this mean that there is an answer in General Relativity?

No.

I mean, GR didn't and doesn't move physics in the direction that intrigues you: a relational theory.General Relativity is a deeper theory than Newtonian dynamics, it forces a truly profound rethinking of familiar concepts, but being firmly rooted in reality GR is not a relational theory.

Of course I do encourage you to learn as much about GR as you can. Whether or not it will be what you expect, it'll be a worthwile experience.Addendum:
Here's a resource that I would like to recommend, an inaugural lecture by Einstein from 1920.
Around that time Einstein accepted an honorary professorship appointment at the University of Leiden, Netherlands. It did not involve more than visiting Leiden more than a couple of days per year, but Einstein took the inaugural speech very serious indeed.
The lecture was titled http://www.tu-harburg.de/rzt/rzt/it/Ether.html"

In one respect the content of that lecture is outdated. Within years Einstein abandoned Mach's Principle, regarding it as distracting and unhelpful, but other than that the insights presented in that lecture are as valid today as they were then.
 
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  • #26
Addendum:
Here's a resource that I would like to recommend, an inaugural lecture by Einstein from 1920.
Around that time Einstein accepted an honorary professorship appointment at the University of Leiden, Netherlands. It did not involve more than visiting Leiden more than a couple of days per year, but Einstein took the inaugural speech very serious indeed.
The lecture was titled Ether and the Theory of Relativity
Thanks, I will look at that!

This is what drove Einstein to hypothesize his theory of general relativity.

Does this mean that there is an answer in General Relativity?
No.

Based on the above two quotes, does this mean that this problem drove him to hypothesize the theory of general relativity, but in fact never came to a solution for the problem, just came up with a theory for gravity?
 
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  • #27
This is arelatively simple problem to solve. Take A and B to be the two particles in orbit around each other and C to be the external reference frame. First take A to be the vantage point or rigid frame of reference. From here, C is irrelevant, but as far as A goes it is stationary while B revolves around it. since the true point of reference for A is B and for B, A, it is possible that A's refernce frame leaves it in motion around B. This depends on any unicluded factors that may assist A and or B in positioning themselves. This would also be the case for B. THe two are in motion relative to each other. IF you were on the Moon, both the Earth and Sun would appear to rotate around you.
 
  • #28
Yeah but then you have changed the coordinate system to explain the problem, where the problem is how to explain the particles not colliding in a coordinate system rotating about the center along with the particles.

Phrak re-worded the problem better than how I originally worded it:

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?
 
  • #29
The question essentially follows what I stated. The ultimate question involved here is the last statement asking what grants rights to which is correct. This is it, they are all correct. Also in teh case that both particles are of equal size and mass, you could place the origin in the center of the two such that each is the 180 degree opposite of the other upon a rotaing disk, or sphere if you wish. In my opinion, the ultimate solution to the entirety of your question is that no solution exists for correct solutions may be as false as the false solutioins may be correct. The only answers we can be sure of are those that are not correct in any regards to the matter...I would take my words lightly I am very "self-opinionated" and do not fully buy into our understanding as two senturies from now we wil be the fools who thought the world was flat.
 
  • #30
Jonnyb42 said:
[...] could you explain what is meant by general covariance, as it pertains to this case?

The name 'General covariance' refers to a systematic approach to dealing with the mathematics of coordinate transformation.

You can think of it as as superclass of coordinate transformations that includes all classes of coordinate transformations you might conceivably use.

To give an example: in the case of rotation around a fixed axis there is (for the rotating coordinate system) a term called the centrifugal term, that is proportional to the square of the angular velocity and the radial distance to the central axis of rotation.
If the physics is mapped in an inertial coordinate system then the centrifugal term evaluates to zero (since the angular velocity of the inertial coordinate system is zero). For consistency you can choose to write down that centrifugal term anyway. That way the notation will be always be the same.

Or maybe you want to postpone the decision as to whether the coordinate system is rotating, but you do want to explore some properties of the system.

General covariance acts as an abstraction layer. With the entire apparatus n place you can do math on the motion physics while remaining uncommitted as to which coordinate system you are going to use to map the motion.

In itself the mathematical apparatus of general covariance has no physical meaning. Any theory can be formulated in terms of the general covariance apparatus. As I understand it this has been actually done in the case of Newtonian theory of gravity: it's formulation in general covariant form has been worked out.

It is sometimes suggested that General Relativity is a deeper theory 'because it uses general covariance'. That's not why it's a deeper theory; any theory can be worked out in a general covariant formalism.

Abstraction layer
Let me return to the idea of general covariance formalism acting as an abstraction layer.
As long as you work on the level of the abstraction layer you are working with a single formalism. That is why it's sometimes suggested that when working with a general covariant formalism 'the laws are the same in all coordinate systems'.

Note that this "the same" is just an artifact of the mathematical formalism, it has no bearing on the physics taking place.

By contrast, the Lorentz invariance that is asserted by special relativity is a physical statement. That is, the assertion is that the members of the equivalence class of inertial coordinate systems are physically equivalent.
 
  • #31
As I see it, it is not a matter of whether a coordinate system is right or wrong. They are all right, in that they all describe what they each see.

A way to see it is, imagine that the universe is a play. And each coordinate system is a seat. It doesn't matter what seat you are in, you are always watching a single play. You might see it from the right, from dead center, or the left, and that may change your perspective, but it is still a single play. The events happening on stage are the same for all spectators.

Can you imagine a universe where, depending on your seat, a different play occurs? There would need to be infinite ramifications from each single act, each visible only to a specific reference frame. You could imagine a universe such as that, I suppose. But it wouldn't be the universe we are in. You could never transform from one coordinate system to another, either. You'd be stuck with infinite universes, each running it's own version of events.

In our universe, reference frames are arbitrary. They don't affect the events in any way whatsoever. A coordinate system is merely a set of numbers we choose to give to each position in space. They carry no meaning, and they don't affect what you are watching from that coordinate system.

Imagine you grab a wire frame, and put it in front of your screen. So you can plot each pixel in the screen to an (x,y) coordinate on the wire frame. Now, if you decide to turn the wire frame 45° clockwise, is your screen affected in any way? Does your switching the wire frame turn your screen as well? Would you expect it to? And yet, you now have a new coordinate system, and each pixel is now plotted to a different (x,y) coordinate.

That's what coordinate systems are. Wire frames we slap on top of physcal systems. We are free to move them around any way we want to, but they don't impact the actual event taking place. If two bodies are orbiting each other, then it doesn't matter if your reference frames are inertial, non-inertial, shrinking, expanding, moving randomly about or what have you. They can each show entirely different things.

The one thing they can never do, however, is show a collision. Because if it didn't happen in one reference frame, it can't ever happen in any other. Things either happen or they don't, and a reference frame can merely show us when and where. That's what they do. Slap timestamps and locations on events.

We like inertial frames, because even if they show differing speeds and locations, they agree on accelerations. That's all they have in special: the entire set of inertial systems agree with each other with regards to acceleration. Non-inertial systems do not agree with each other with regards to acceleration, because they themselves are accelerated, and they pass on this acceleration to everything plotted on them.

Because inertial systems agree with each other, we can convert from one to another trivially, which helps when we want to study some things. But that doesn't make them more right than non-inertial systems. Non inertial systems are more of a pain to work with, that is all.

So, back to the original example. You select a reference system that is centered on the center of mass. You see two particles rotating. Why should moving the grid from the center to one and keeping the y-axis level with the second body show anything other than the body hanging in midair?

Let's ignore stuff about reference systems. If you see something hanging in the air, you know the acceleration is 0. That is a definition. Acceleration is defined as dv/dt, and v is defined as dp/dt. If p doesn't change, v is 0 and a is 0.

From the law of gravitation, you know the two bodies must be attracting each other. There is a pair of forces pulling on each other, as Newton discovered. But a force adds an acceleration to a body, and neither body seems to be accelerating. Something must be keeping them apart. The one thing you absolutely cannot do is decide that because you don't know what is causing the bodies to stay apart, they must be falling together. Instead, you do the next best thing. You add a force to balance things out. You don't know where the force comes from, true, but you know it must be there.

When you then move to an inertial frame, things make a little more sense. You can compare what you were seeing in the rotating frame, and understand why the two bodies didn't fall together. Being in that particular reference frame had blinded you, having a second reference frame allows for a new perspective.

We slap the 'fictitious' label on the forces we didn't know the source of, but that doesn't mean the force, inside the rotating coordinate system, didn't exist. What is a force, after all? It is not a physical thing, it is more of an idea, a model we use to describe the universe. And if within the rotating frame you need an extra force to make sense of the universe, well, that is what forces are.

Neither frame is wrong. There is no requirement that all frames share the same forces. It is a requirement that all inertial frames share the same forces, but that is not the case here.
 
  • #32
Kynnath said:
Because inertial systems agree with each other, we can convert from one to another trivially, which helps when we want to study some things. But that doesn't make them more right than non-inertial systems. Non inertial systems are more of a pain to work with, that is all.

Let me discuss the example of a rotating coordinate system.

The centrifugal term in the equation of motion for the rotating coordinate system is proportional to the square of the angular velocity of the rotating coordinate system relative to the inertial coordinate system.

In other words, the rotating coordinate system can be put to use if and only if you use the inertial coordinate system as underlying reference.

For theory of motion it's not a matter of the inertial coordinate system being a more convenient referential system, it's the only referential system. Whenever a rotating coordinate system is used the equation of motion always contains the angular velocity of the rotating coordinate system with respect to the inertial coordinate system.
 
  • #33
Kynnath said:
A way to see it is, imagine that the universe is a play. And each coordinate system is a seat. It doesn't matter what seat you are in, you are always watching a single play. You might see it from the right, from dead center, or the left, and that may change your perspective, but it is still a single play. The events happening on stage are the same for all spectators.

I like this way of explaining things. No matter where you are sitting in the theater, you are watching the same play with the same characters interacting. In special relativity, for instance, the order of events may differ, but it’s difficult to find an example within special relativity where it does not apply. Boundary states are often a good place to look. Acceleration can place some interacting particles over the Rindler horizon but eventually, under any permissible acceleration leaving v<c, will they return, where all interactions are accounted for?
 
  • #34
Cleonis said:
In other words, the rotating coordinate system can be put to use if and only if you use the inertial coordinate system as underlying reference.

For theory of motion it's not a matter of the inertial coordinate system being a more convenient referential system, it's the only referential system. Whenever a rotating coordinate system is used the equation of motion always contains the angular velocity of the rotating coordinate system with respect to the inertial coordinate system.

I'll have to disagree. Imagine you want to plot the trajectory of a ball you launch inside a space shuttle. If you used any inertial reference system, you'd have to account for the force of gravity, and the ball you launched would follow a curved trajectory. It'd be a pain to do it that way, when if you used a reference frame that tracked the shuttle, and thus already took into account gravity keeping you in orbit, you can plot the straight trajectory the ball follow with respect to anything else inside the shuttle. If you wanted to aim the ball at a hoop on the opposite wall, would you rather plot the intercept of the two curves in an inertial reference frame, or just the striaght shot from the non inertial frame tracking the room already?
 
  • #35
Kynnath said:
Imagine you want to plot the trajectory of a ball you launch inside a space shuttle.

For simplicity, let me address this in terms of Newtonian physics.

We know from experience that gravitational mass and inertial mass are equivalent (There are no known exceptions. If there would be exceptions we would certainly have noticed them by now.)

The orbiting shuttle is in free fall, and in free fall a local measurement of the effects of gravity is not possible. A standard accelerometer onboard the orbiting space shuttle will register zero acceleration.

In effect we can treat the coordinate system that is co-moving with the orbiting shuttle as an inertial coordinate system. If the space shuttle has some spin around its own center of mass then the trajectory of a ball inside the space shuttle will be curvilinear with respect to the shuttle. Then to account for the motion of the ball the rotation of the space shuttle with respect to the local inertial coordinate system is used.

Of course, with sufficiently accurate measuring equipment you will also be able to discern effects arising from the fact that the space shuttle is orbiting the Earth.
Let's say you give the ball a minute velocity with respect to the shuttle, so that it takes the ball many minutes to make it to the other end. Then during that free motion the ball is in its own orbit around the Earth, with a corresponding curvilinear trajectory. With sufficiently accurate tracking you will be able to discern that.

Depending on how accurate your measurement devices are you need to set your scope differently. At first approximation just a local evaluation suffices. With more and more accurate tracking of individual objects you need to shift to a wider and wider perspective.

For each scale of perspective a different common center of mass will serve as the zero point of your coordinate system. If you need to take into account that the space shuttle is in orbit then the inertial coordinate system that is co-moving with the Earth's center of mass is the appropriate reference. If you need to take into account that the Earth is orbiting the Sun then the inertial coordinate system that is co-moving with the Solar System's center of mass is the appropriate reference. This goes on at ever larger scales.
 
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