Which inertial frame is used when resolving toward the centre

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  • #1
etotheipi
For simple circular motion it is common to resolve forces toward the centre and relate this to centripetal acceleration, provided that the current reference frame is inertial.

However where would such a coordinate system be positioned? Since it requires one axis passing through the line containing the rotating particle and the centre of rotation (in order to resolve the forces into this basis), I can only think of perhaps a series of instantaneous inertial reference frames all around the circle. And that it is implied that we'd apply NII in one of these frames at any given position during the motion.

For more complicated examples a rotating frame is perhaps more useful, however I'm asking in the context of inertial frames. Thank you.
 

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  • #2
PeroK
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For simple circular motion it is common to resolve forces toward the centre and relate this to centripetal acceleration, provided that the current reference frame is inertial.

However where would such a coordinate system be positioned? Since it requires one axis passing through the line containing the rotating particle and the centre of rotation (in order to resolve the forces into this basis), I can only think of perhaps a series of instantaneous inertial reference frames all around the circle. And that it is implied that we'd apply NII in one of these frames at any given position during the motion.

For more complicated examples a rotating frame is perhaps more useful, however I'm asking in the context of inertial frames. Thank you.
You have to distinguish between a reference frame and a coordinate system. An inertial reference frame is defined physically. One where Newton's first law applies.

You can use a polar coodinate system to describe motion in an inertial reference frame. In this case the straight lines followed by objects with no unbalanced force are described by the appropriate equations in polar coordinates.

If you have an object in uniform circular motion relative to an inertial reference frame, then you can describe the forces on it and its motion using the polar coordinate system. In which case the changing force remains in the changing ##-\hat r## direction.
 
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Since it requires one axis passing through the line containing the rotating particle and the centre of rotation

There is no need for the coordinate system to have an axis constantly aligned with the object in circular motion. You should describe the circular motion in a fixed coordinate system. All interesting quantities can be written as functions of time in the fixed coordinates.
 
  • #4
etotheipi
If you have an object in uniform circular motion relative to an inertial reference frame, then you can describe the forces on it and its motion using the polar coordinate system. In which case the changing force remains in the changing −^r−r^ direction.

Right, so if we setup a polar coordinate system centred on the centre of rotation, then the centripetal force is Fr^rFrr^ and the centripetal acceleration is still ar^r=−v2r^rarr^=−v2rr^?

You have to distinguish between a reference frame and a coordinate system. An inertial reference frame is defined physically. One where Newton's first law applies.

Yes, this might be the problem. Before closer inspection, a fixed polar coordinate system and a rotating Cartesian coordinate appear similar, though actually they are quite different; in this context, one applies to an inertial frame, the other to an accelerating frame. I was thinking of the latter, and hadn't considered that a polar coordinate system is still fine in an inertial frame and that there is no need for fictitious contributions and the like.

As far as I'm aware, an observer in any particular reference frame can come up with their own arbitrary coordinate system, with the only constraint that it remains at rest relative to them in that frame.
 
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PeroK
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Yes, this might be the problem. Before closer inspection, a fixed polar coordinate system and a rotating Cartesian coordinate appear similar, though actually they are quite different; one applies to an inertial frame, the other to an accelerating frame. I was thinking of the latter, and hadn't considered that a polar coordinate system is still fine in an inertial frame.

Here's the difference:

Imagine you build a set of physical Cartesian coordinate axes. Three meter sticks glued together, say. That is the basis of a Cartesian co-ordinate system. You can sit on the surface of the Earth and use that as the basis of your co-ordinate system. You can get into a car going round a circular race-track and use that system etc.

Whether you are using a Cartesian coordinate system is unrelated to you state of motion.

Similarly, whether you are using a polar, spherical or cylindrical (or general curvilinear) coordinate system is independent of your state of motion.
 
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PeroK
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Right, so if we setup a polar coordinate system centred on the centre of rotation, then the centripetal force is Fr^rFrr^ and the centripetal acceleration is still ar^r=−v2r^rarr^=−v2rr^?

Assuming the motion is being measured from an inertial reference frame, yes.
 
  • #7
etotheipi
Here's the difference:

Imagine you build a set of physical Cartesian coordinate axes. Three meter sticks glued together, say. That is the basis of a Cartesian co-ordinate system. You can sit on the surface of the Earth and use that as the basis of your co-ordinate system. You can get into a car going round a circular race-track and use that system etc.

Whether you are using a Cartesian coordinate system is unrelated to you state of motion.

Similarly, whether you are using a polar, spherical or cylindrical (or general curvilinear) coordinate system is independent of your state of motion.

That's interesting. Whenever I've encountered non-inertial frames, my thought process has been something like:

In an inertial frame (and some coordinate system in this frame), a particle's motion is described by ##\vec{F} = m\vec{a}##

In a non-inertial frame accelerating at ##\vec{a_{f}}## wrt. the inertial frame (the coordinate system is then also accelerating at ##\vec{a_{f}}## wrt the inertial coordinate system) the relative acceleration of the particle to the coordinate system being used in the non-inertial frame is ##\vec{a} - \vec{a_{f}}##, and since the frame is non-inertial we include a fictitious term, resulting in

##\vec{F} + \vec{F_{fict}} = m(\vec{a} - \vec{a_{f}}) \implies \vec{F_{fict}} = -m\vec{a_f}##

I suppose the key thing to remember is that any coordinate system exists within a certain frame, and if that frame happens to be non-inertial we obtain fictitious contributions. I hope I've not confused frame with coordinate system anywhere!!
 
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PeroK
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I suppose the key thing to remember is that any coordinate system exists within a certain frame ...

It's better to say that the choice of coordinate system is not dependent on your reference frame. How that coordinate system relates to someone else's is important.

Note that we are free to use any frame of reference and any coordinate system. For example, if you describe a scenario to me, I can choose any reference frame and any coordinate system to study it.
 
  • #9
etotheipi
Note that we are free to use any frame of reference and any coordinate system. For example, if you describe a scenario to me, I can choose any reference frame and any coordinate system to study it.

So if I understand correctly, any observer has their own reference frame associated with their motion, and in this frame the observer can setup a coordinate system of their choosing (the choices may be translated/rotated in space, though all remain at zero relative velocity to them).

When transforming between different coordinate systems, which may exist inside different frames of reference, we also need to consider whether the frame associated with a particular coordinate system is inertial or not to decide whether fictitious forces are required.
 
  • #10
etotheipi
You have to distinguish between a reference frame and a coordinate system. An inertial reference frame is defined physically. One where Newton's first law applies.

Another thought; if reference frames are associated with states of motion, do two observers at relative rest occupy the same reference frame? They will likely choose different coordinate systems within this frame, though we have since established that this is not an issue!
 
  • #11
PeroK
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So if I understand correctly, any observer has their own reference frame (the space they 'carry' with them), and in this frame the observer can setup a coordinate system of their choosing (which may be translated/rotated in space, though remain at zero relative velocity to them).

When transforming between different coordinate systems, which may exist inside different frames of reference, we also need to consider whether the frame associated with a particular coordinate system is inertial or not to decide whether fictitious forces are required.

In general, we make measurements using our specific reference frame. But, we can do physics and make predictions using any reference frame. For example, we are compelled to measure the position of the Sun, Moon and planets relative to Earth; but, we can switch to a heliocentric frame of reference to study the motion of the solar system.

In both cases, we can use Cartesian or polar coordinates as we wish.

I wouldn't say that coordinate systems "exist inside different frames of reference". It's perhaps better to say you can define a coordinate system using a frame of reference.

I wouldn't say that you "carry space around with you". This thread is under classical physics, which simplifies matters because you have global inertial reference frames, to which everything can be related.

You may have been reading about relativity, which does complicate the issue.

In general, you can transform between any two coordinate systems. A simple transformation could be simply a rotation of the coordinate axes; or a different time origin: there are (at least) 24 times zones in the world.

As fas as kinematics is concerned all this is purely mathematical. When you invoke forces and Newton's laws, these reflect physical considerations. An inertial reference frame is a physical, not a mathematical thing.
 
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PeroK
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Another thought; if reference frames are associated with states of motion, do two observers at relative rest occupy the same reference frame? They will likely choose different coordinate systems within this frame, though we have since established that this is not an issue!

There is probably a variety of conventions on this point. You could try:

https://en.wikipedia.org/wiki/Frame_of_reference

For what it's worth, I would tend to say that (in Classical Physics and SR) a reference frame is defined by a state of motion. In the sense that if two objects or observers are at rest with respect to each other, then their rest frames are the same frame of reference. And two reference frames differ when there is relative motion between them: i.e. an object at rest in one frame is not at rest in the other.

PS you can check that the wikipedia page is saying what I've been saying on this thread - hopefully!
 
  • #13
etotheipi
You may have been reading about relativity, which does complicate the issue.

It's a little confusing, since a text might define frames ##S## and ##S'## and talk about the ##(x,y,z,t)## coordinates as measured within that frame, such that the two concepts of frames and coordinate systems are mushed together.

I suppose what is actually meant is having coordinate systems defined with ##S## and ##S'##, and transforming between those. Though the distinction is generally not made clear at all.
 
  • #14
Mister T
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Another thought; if reference frames are associated with states of motion, do two observers at relative rest occupy the same reference frame?

You might be confusing things here. Every object, including every observer, can exist in every reference frame. I think the terminology you're looking for is rest frame. Two observers at rest relative to each other do indeed occupy the same rest frame. Rest frame being a frame of reference where the object or observer under consideration is at rest.

Also, they need not be located at the origin of the coordinate system to be at rest in a frame of reference.

When beginning a study of uniform circular motion we usually place the origin of the ##xy##-plane at the center of the circle of radius ##r## and measure ##\theta## counter-clockwise from the positive ##x##-axis. Then we can write ##x=r \cos \theta## and ##y=r \sin \theta##.
 
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vanhees71
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Right, so if we setup a polar coordinate system centred on the centre of rotation, then the centripetal force is Fr^rFrr^ and the centripetal acceleration is still ar^r=−v2r^rarr^=−v2rr^?
Vectors are invariant, only their components change with a change of the basis. To take up your example, take a circular motion with constant angular velocity in the ##12## plane around the origin of a inertial reference frame. Written with respect to a Cartesian basis ##(\vec{e}_1,\vec{e}_2,\vec{e}_3)## the position vector of the particle is described as
$$\vec{x}(t)=a \cos(\omega t) \vec{e}_1 + a \sin(\omega t) \vec{e}_2.$$
Since the Cartesian basis is time-independent you simply have
$$\vec{v}(t) =\dot{\vec{x}}(t) = -a \omega \sin(\omega t) \vec{e}_1 + a \omega \cos(\omega t) \vec{e}_2$$
and for the acceleration
$$\vec{a}(t)=\dot{\vec{v}}(t)=-a \omega^2 \cos(\omega t) \vec{e}_1 - a \omega^2 \sin(\omega t) \vec{e}_2.$$
In polar coordinates you have
$$\vec{x}=r \vec{e}_r=a \vec{e}_r = a \cos \varphi \vec{e}_1 + a \sin \varphi \vec{e}_2.$$
In polar coordinates the motion is described as ##r=a=\text{const}##, ##\varphi=\omega t##. The basis vectors are (given in terms of the Cartesian basis)
$$\vec{e}_r=\cos \varphi \vec{e}_1 + \sin \varphi \vec{e}_2, \quad \vec{e}_{\varphi} = -\sin \varphi \vec{e}_1 + \sin \varphi \vec{e}_2.$$
Since these basis vectors are orthonormal (though position dependent!) we simply find the components of the position vector wrt. this basis as
$$x_r=\vec{e}_r \cdot \vec{x}=a, \quad x_{\varphi}=\vec{e}_{\varphi} \cdot \vec{x}=0,$$
those of the velocity
$$v_r = \vec{e}_r \cdot \vec{v}=0, \quad v_{\varphi}=\vec{e}_{\varphi} \cdot \vec{v}=a \omega,$$
and of the acceleration
$$a_r = \vec{e}_r \cdot \vec{v}=-a \omega^2, \quad a_{\varphi}=\vec{e}_{\varphi} \cdot \vec{a}=0.$$
 
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  • #16
etotheipi
There are a few things I can think of which become a little more complicated with this new picture.

For instance, the relative position of ##B## wrt. ##A##, if ##A## and ##B## are at the origins of their own coordinate systems, is ##\vec{r_{BA}} = \vec{r_{B}} - \vec{r_{A}}##, as measured from a third frame of reference, that of ##C##.

If perhaps ##A## defines the origin of his coordinate system to be ##5## metres in front of him, the position of ##B## in his frame will not be ##\vec{r_{B}} - \vec{r_{A}}##, but we will instead need to add a correction somehow.

Since this shift in position is constant wrt. time, when we differentiate at least once this term disappears and we are left with relative velocity and acceleration as usual. However the point still stands for the relative position!

It seems convenient in a lot of cases to denote the origin of the coordinate systems at the positions of the bodies under consideration, despite the fact that it is not necessary!
 
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