Weirdness with this relative velocity formula when defining coordinate systems differently

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etotheipi
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Wikipedia gives, "The relative velocity ##{\displaystyle {\vec {v}}_{B\mid A}}## is the velocity of an object or observer B in the rest frame of another object or observer A."

Suppose the coordinate system being used in the rest frame of ##A## is has its origin slightly displaced from ##A## itself, such that the vector from the origin to A is ##\vec{R}##. Suppose also that this coordinate system is translating and rotating such that ##\vec{R}## is a function of time.

We can write the relative position of B wrt. A as ##\vec{r_{BA}} = \vec{r_{B}} - \vec{r_{A}}##, such that the position of ##B## as measured by the coordinate system in the rest frame of ##A## is ##\vec{r_{BA}} + \vec{R}##. Finally, since ##\vec{R}## is dependent on time, differentiating this position will give the velocity of ##B## measured by the coordinate system used in A's frame as ##\vec{v_{BA}} + \dot{\vec{R}}##.

This appears to contradict with the statement from wikipedia. It seems as though if we set everything up in a certain manner, the velocity of B as measured by a chosen coordinate system not centred on A in A's rest frame is not necessarily ##\vec{v_{BA}} = \vec{v_{B}} - \vec{v_{A}}##, since there is a possibility of an extra term.

Everything works out fine if, when dealing with relative positions/velocities/accelerations, we centre the coordinate systems on all of the bodies in question. However, if the coordinate systems are not centred as such (which is permissible, since we can set up our coordinate system as we please in any given rest frame), these formulae start to no longer work. A more easily visualisable example would be the relative position of B in A's rest frame, which evidently can take on infinitely many values depending on where we choose the origin of A's coordinate system!

I was wondering if anyone could tell me what the problem with all of the above junk is, as I feel I am misinterpreting some key ideas! Thank you.
 
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  • #2
PeroK
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You do have to be careful with rotating reference frames. If an object is rotating about the origin of an inertial reference frame, then its distance to the origin is constant and, in one sense, the object and the origin are at rest with respect to each other. But, they do not share a rest frame, because the point at the centre is not accelerating. And, in general, points further from the origin have a greater acceleration.

And, clearly, the velocity of a point at rest in the inertial frame relative to a rotating object depends on the distance of the object from the centre of the rotation.

A rotating reference frame, therefore, is not a rest frame for all objects that are rotating. There's more here:

https://en.wikipedia.org/wiki/Rotating_reference_frame
 
  • #3
etotheipi
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You do have to be careful with rotating reference frames. If an object is rotating about the origin of an inertial reference frame, then its distance to the origin is constant and, in one sense, the object and the origin are at rest with respect to each other. But, they do not share a rest frame, because the point at the centre is not accelerating. And, in general, points further from the origin have a greater acceleration.
Thanks for the clarification, I can sort of visualise now two points rotating about each other (like a binary star system) that might occupy the same rest frame (rotating at some velocity), however their relative velocities to a fixed point somewhere will evidently be different depending on the phase of the orbit!

When we say we make a relative measurement in the frame of 'X', it is not a completely unambiguous statement due to where we place the coordinate system within the frame! It's like how the relative position vector between particles A and B is always well defined, however this doesn't necessarily denote the relative position of B in A's frame, since we could set A's origin to be a billion kilometres from A itself!

I wonder whether it is always implied that the coordinate systems in the frame of each particle are centred on the particle?
 
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PeroK
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Thanks for the clarification, I can sort of visualise now two points rotating about each other (like a binary star system) that might occupy the same rest frame (rotating at some velocity), however their relative velocities to a fixed point somewhere will evidently be different depending on the phase of the orbit!
If you are going to analyse these things, you need to tighten up your terminology.

A binary star system has a centre of mass frame of reference. That will be an inertial frame in which the centre of mass is at rest. As there are no external forces, Newton's second Law guarantees that the centre of mass moves inertially.

Each star follows an ellipse with the centre of mass at one focus. The refernce frame of each star is, therefore, non-inertial and the two stars do not share a common rest frame.

There is no complexity here other than the stars are in relative (accelerated, non-inertial) motion with respect to each other.

When we say we make a relative measurement in the frame of 'X', it is not a completely unambiguous statement due to where we place the coordinate system within the frame! It's like how the relative position vector between particles A and B is always well defined, however this doesn't necessarily denote the relative position of B in A's frame, since we could set A's origin to be a billion kilometres from A itself!
Changing the origin of a coordinate system results simply in changing every measurement by a fixed amount. If it doesn't then you have done more than change the origin.

This is where you have to be careful with a rotating reference frame, which is not the rest frame for any single observer. You can see this by noting that points at different distances from the centre of rotation have different forces on them and different (real) acceleration. A rotating reference frame may have uniform angular velocity, but it doesn't have uniform real acceleration.

You can see this on the surface of the Earth where the Coriolis effects depends on latitude. So, although an observer on the surface of the Earth could add or subtract numbers from position measurements (change of origin) that is different from making measurements at a different location where the local acceleration is different.

I wonder whether it is always implied that the coordinate systems in the frame of each particle are centred on the particle?
This is definitely not the case. If you have several particles, all in uniform motion, then they share a rest frame and the origin has to be somewhere.

The difficulty comes when you consider non-unifrom motion. If this is linear, then it is obvious that the particles do not share a rest frame, because the distance between them is changing. But, in rotational motion, particles may not share a rest frame even though the distance between them remains constant.

Consider the following example. You are at rest in some inertial reference frame. An object is orbiting you. It has a velocity relative to you. Now, you begin to rotate to keep an eye on the object. You have no linear velocity in the inertial reference frame - or, at least, you can make any velocity arbitrarily small. But, now the object is stationary in your frame of reference.

This shows the complexity of using a rotating reference frame. You and the object do not share a rest frame, but it has zero velocity in your reference frame.

If you imagine you have a Cartesian grid, which you hold so that the object lies along the x-axis. Let's say the object is now at a coordinate of ##x = 100m##. You could, however, move the origin of your grid ##100m## outwards. The object is now at coordinate ##x = 0m## and you are at coordinate ##x = -100m##. THis changes nothing significant.

The reason for the complexity was not where you chose to put the origin. The complexity arose from using a rotating reference frame.
 
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  • #5
etotheipi
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This is definitely not the case. If you have several particles, all in uniform motion, then they share a rest frame and the origin has to be somewhere.
If in an inertial frame there exist a system of particles A through Z which are not performing uniform motion, whose positions are all being measured relative to a coordinate system in this frame. When we denote the velocity of one particle relative to another, e.g. ##v_{BA}##, then I thought we interpreted this as the velocity of B as measured in A's frame of reference. In that sense, doesn't A need a coordinate system in its own rest frame?

The difficulty comes when you consider non-unifrom motion. If this is linear, then it is obvious that the particles do not share a rest frame, because the distance between them is changing. But, in rotational motion, particles may not share a rest frame even though the distance between them remains constant.

Consider the following example. You are at rest in some inertial reference frame. An object is orbiting you. It has a velocity relative to you. Now, you begin to rotate to keep an eye on the object. You have no linear velocity in the inertial reference frame - or, at least, you can make any velocity arbitrarily small. But, now the object is stationary in your frame of reference.

This shows the complexity of using a rotating reference frame. You and the object do not share a rest frame, but it has zero velocity in your reference frame.

If you imagine you have a Cartesian grid, which you hold so that the object lies along the x-axis. Let's say the object is now at a coordinate of ##x = 100m##. You could, however, move the origin of your grid ##100m## outwards. The object is now at coordinate ##x = 0m## and you are at coordinate ##x = -100m##. THis changes nothing significant.

The reason for the complexity was not where you chose to put the origin. The complexity arose from using a rotating reference frame.
This is a nice explanation; in your final example, I assume that the coordinate system is still rotating around with the observer rotating in the inertial frame, though now the coordinate axes are pivoted around ##x=-100##.

Thanks for taking the time to write such a detailed answer!
 
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  • #6
PeroK
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If in an inertial frame there exist a system of particles A through Z, whose positions are all being measured relative to a coordinate system in this frame. When we denote the velocity of one particle relative to another, e.g. ##v_{BA}##, then I thought we interpreted this as the velocity of B as measured in A's frame of reference. In that sense, doesn't A need a coordinate system in its own rest frame?
The underlying assumption is that each coordinate system shares a common definition of the ##x, y## and ##z## directions. Or, at least, the relationship between coordinate axes do not change over time.

If, however, the coordinate axes change orientation over time, then you lose the simple relationship between relative velocities. This is what happens in a rotating reference frame. Compare your set of axes on the rotating surface of the Earth with, say, a hypothetical set of axes defining an inertial reference frame. Or, a fixed set of axes on the ground, with a set of axes carried on a circular fairground ride.

One idea is to use gyroscopes to define your coordinate axes, as these will continue to point in the same physical direction (relative to an inertial reference frame). If you took a gyroscope on the fairground ride, then it should continue to point in the same physical direction. Using these coordinates, defined by your gyroscope, will maintain the relationship between relative velocities.

The "distant stars" as seen from Earth provide a similar reference. They are so far away that their physical direction remains almost constant during the Earth's orbit and rotations.

There must be something online about this. The Wikipedia page you looked at avoids these complexities of how an accelerating observer defines their coordinate system. Or, even if the observer is not accelerating, what happens if they move their coordinate axes with time.
 
  • #7
etotheipi
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Is there any way of defining what a reference frame actually is? I think I now have a better understanding of the distinction between that and a coordinate system, but am struggling to find a way to think about the reference frame itself.

Wikipedia seems to imply that the frame itself has consists of an origin and a coordinate system, such that neither a frame nor coordinate system can exist without the other. However, I see no reason as to why a frame of reference can't have more than one associated coordinate system (e.g. solving a pulley question with two different coordinate systems on each side, with the y-axes pointing in opposite directions!). We have already established that there are infinitely many ways we could set up a system of coordinate axes once a frame is defined.

What I'm going on at the moment is that the frame defines a set of rest/reference points, and an observer whose rest frame is a particular frame is neither translating nor rotating relative to these reference points (i.e. the corners of a room).

I'm thinking of the example of a merry-go-round, positioned at the origin of a coordinate system in an inertial frame. When the merry-go-round starts rotating, an observer standing on the edge of a merry-go-round will define every point on the merry-go-round to be at rest relative to them, and hence this observer's rest frame constitutes a rotating frame relative to the inertial frame.

However it is at this point I notice my understanding must be incorrect. Supposing there is another object closer to the centre of the merry-go-round, as you stated,

You and the object do not share a rest frame,
This does not happen in my picture, since that object would also define the same set of rest points (i.e. the horses on the merry-go-round, etc.).
 
  • #8
PeroK
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Is there any way of defining what a reference frame actually is? I think I now have a better understanding of the distinction between that and a coordinate system, but am struggling to find a way to think about the reference frame itself.
A reference frame is something defined physically. A person at the centre of a merry-go-round has a choice. If they remain facing a fixed object. Perhaps they occupy a hole in the centre and are not spinning, then that defines an inertial reference frame. If they rotating at the centre, then that defines a non-inertial reference frame.

They decide which one based on physical considerations: do they experience a real force? Does Newton's first law apply? Note that to determine whether an object is at rest and whether it experiences a force does not require coordinates, as such. It may require a measurement and, for practical purposes, some sort of coordinate system of lengths and times.
 
  • #9
etotheipi
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A reference frame is something defined physically. A person at the centre of a merry-go-round has a choice. If they remain facing a fixed object. Perhaps they occupy a hole in the centre and are not spinning, then that defines an inertial reference frame. If they rotating at the centre, then that defines a non-inertial reference frame.
Right, so the notion of a reference frame is always tied to an observer. Is it a good idea to think of the observer as a pair of eyes? An observer in an inertial frame can still look around in any particular direction, though if the observer is "rotating" then their eyes are always fixed toward perhaps a rotating object. There is a definite sense of directionality, though it's not easily defined!

I'll try and see if I can find any good resources online to get the hang of all of this!
 
  • #10
PeroK
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Right, so the notion of a reference frame is always tied to an observer. Is it a good idea to think of the observer as a pair of eyes?
Not really. A reference frame is defined by something physical: the rest frame of a infinite charged wire; the rest frame of the (centre of mass of the) solar system. A frame of reference moving with respect to another.

Physics is not dependent on electromagnetic radiation for measurements. In fact, the best way to think of a reference frame is a line, plane or 3D grid of local observers. Each is at a predefined location. Each observer knows his/her Cartesian or polar coordinates and they all have synchronised watches.

This is where inertial reference frames are important. In Newtonian physics, if these local observers are all inertial, then you have an inertial frame of reference and Newton's laws apply.

And, again, a rotating reference frame involves the complication that local observers at different points have different real acceleration. Imagine a large merry-go-round with local observers at strategic points. When the merry-go-round is at rest we have an inertial frame of reference. But, once it is rotating, the local observers experience different forces to keep them in position. But, they can still record anything that happens locally and what time it happens. That information is then a measurement using that frame of reference. The coordinate system determines the position label that each observer has. If it's Cartesian, then someone might be at ##x = 4m, y = 3m##; and, if it's polar, then they will be at ##r = 5m,\theta = \arctan(\frac 3 4)##.

Maybe think of it that way: the reference frame is all the local observers you need - a physical array of them. The coordinate system is the label you give each observer to determine where they are.
 
  • #11
etotheipi
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Physics is not dependent on electromagnetic radiation for measurements. In fact, the best way to think of a reference frame is a line, plane or 3D grid of local observers. Each is at a predefined location. Each observer knows his/her Cartesian or polar coordinates and they all have synchronised watches.
I really like this idea, sort of a sea of point-like observers with watches. Going all the way back to yesterday, with the object performing circular motion, we could imagine a space station far out in deep space watching one planet orbit a star. The army of observers remain at rest relative to the space station, and perhaps also the star (if it is arbitrarily large!), and each one agrees on the coordinate system being used.

For a rotating frame, I imagine a team of observers who are rotating around the axis of rotation, not unlike the visual picture of a tornado... Each member is accelerating wrt. an arbitrary inertial frame in order to perform the rotational motion, and the resulting rotational frame of reference is non-inertial.

Most importantly, all members of the observer army have the constraint that they are at rest relative to the coordinate system (i.e. if we are using a polar coordinate system, we must fix the horizontal according to some physical feature, e.g. the line from the centre of a clock to the number 12.). If the frame is rotating, then the position of any observer relative to the ##x## axis isn't going to change, they will simply move around with the axes. I know it's probably not a mathematically rigorous definition, however this does make the idea of a reference frame much easier to understand!

Thank you @PeroK for your patience!
 
  • #12
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I think I now have a better understanding of the distinction between that and a coordinate system, but am struggling to find a way to think about the reference frame itself.
A reference frame is a collection of coordinate systems which are all related by constant transformations.
 
  • #13
etotheipi
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I left this alone for a day or so since I thought I was developing a mental block over insignificant details, and just to summarise this with a slightly fresher outlook today, here's what I settled on:

"A reference frame consists of an n-dimensional grid of observers who share the same rigid body motion as a set of physical reference points. Each location in the frame can be uniquely defined by one or more coordinate systems.”
 
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  • #14
etotheipi
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Actually, I’ve reconsidered; I promise this will be my last post!

I’ve edited the definition so that the two concepts are distinct. Although the Wikipedia definition states that a reference frame consists of a coordinate system, that’s like saying apples consist of oranges so it can’t be right.

A reference frame doesn’t have an origin, though a coordinate system defined in a reference frame does!
 
  • #15
PeroK
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Actually, I’ve reconsidered; I promise this will be my last post!

I’ve edited the definition so that the two concepts are distinct. Although the Wikipedia definition states that a reference frame consists of a coordinate system, that’s like saying apples consist of oranges so it can’t be right.

A reference frame doesn’t have an origin, though a coordinate system defined in a reference frame does!
Strictly speaking, yes, but normally you choose a reference frame and a coordinate system at the same time.
 
  • #16
vanhees71
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I left this alone for a day or so since I thought I was developing a mental block over insignificant details, and just to summarise this with a slightly fresher outlook today, here's what I settled on:

"A reference frame consists of an n-dimensional grid of observers who share the same rigid body motion as a set of physical reference points. Each location in the frame can be uniquely defined by one or more coordinate systems.”
This doesn't make sense within relativity, because there are no rigid bodies (at least not in the sense there are rigid bodies in Newtonian physics).

In SR you can define global inertial frames though, and they consist of a congruence of parallel straight timelike lines and the corresponding 3D hypersurfaces (this is a special type of "time slicing" possible only in SR).

In GR everything is defined for local observers, i.e., you have an arbitrity congruence of time-like worldlines (covering some part of spacetime) and the corresponding 3D normal-hypersurfaces. You can consider these time-like worldlines as a set of pointlike observers each carrying his clock with him, which measures his proper time and can be used as a measure of time at the point of the observer. The best approximation of "inertial frames" (local inertial frames defining Galilean coordinates) are given by freely falling observers, defining a congruence of timelike geodesics, but note that these are indeed only local inertial frames, and at "large enough" distances there are "tidal forces".
 
  • #17
etotheipi
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In SR you can define global inertial frames though, and they consist of a congruence of parallel straight timelike lines and the corresponding 3D hypersurfaces (this is a special type of "time slicing" possible only in SR).
I’m trying to digest this but am finding it slightly difficult to understand. I understand a timelike line as representing a body moving only in space (I.e. vertical on a spacetime diagram) whilst I am not sure what a hypersurface is. Perhaps a congruence of timelike lines implies a set of objects not moving in the frame?

Does this improved definition bear any conceptual similarities to the one I wrote for classical mechanics?
 
  • #18
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I’m trying to digest this but am finding it slightly difficult to understand. I understand a timelike line as representing a body moving only in space (I.e. vertical on a spacetime diagram) whilst I am not sure what a hypersurface is. Perhaps a congruence of timelike lines implies a set of objects not moving in the frame?

Does this improved definition bear any conceptual similarities to the one I wrote for classical mechanics?
A timelike path is one that always stays within the past and future light-cones of every event on the path. Or, perhaps more simply -- it is a path that a massive particle could take. [Paths with sharp corners may still be allowed, even though that would take infinite acceleration]

Space-like -- it is outside the light cone.
Time-like -- it is inside.
Light-like -- it is on the light cone.

Hypersurface: In three dimensions, a surface is a two dimensional shape. In four dimensions, a "hyper-surface" is a three dimensional shape. You can visualize it by erasing one of the dimensions from your mind.

Edit: Reaching outside my area of competence and possibly fumbling toward the same understanding you are...

The Wiki page on "congruence" is daunting. But the idea here seems to be that you have a family of freely falling objects. So not only are their world lines timelike, they are also geodesics. The idea of a congruence appears to be, roughly speaking, that these geodesics are all parallel.

So we can set up a "hypersurface" that is perpendicular to all of them. The fact that the congruence of worldlines fills a region means that the angle of the hypersurface is known throughout the region. Pick one event on a hypersurface and the remaining points are uniquely determined. We can then assign a coordinate system so that every event on the hypersurface gets the same time coordinate. This amounts to a "foliation" of the region.
 
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  • #19
etotheipi
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Thank you all for your insights; I won't pretend that that I have done any general relativity (I haven't...), and indeed this is the classical part of the forum (I don't know what the protocol is if the topic changes half-way through the thread!), however here's what I've grasped from what you've said:

Considering a 2+1 dimensional space, so that I can picture it, I've drawn the hyper-surface as horizontal and put the worldlines of the observers are perpendicular to this surface. I imagine we could then define a coordinate system on the hypersurface (I've drawn in some arbitrary ##x## and ##y## axes).

1581527425993.png


The following is taken from Astronomy and Astrophysics, Recent Developments, José Lemos:
In our discussion, we find necessary to make a distinction between "reference frames" and "coordinate systems" . By a reference frame we shall mean an observer set by which measurements are directly made. For example, a set of radially moving geodesic observers would comprise a frame of reference. On the other hand, a coordinate system refers to a set of numbers assigned to each point in the space-time manifold. In Newtonian physics a reference frame is an imagined extension of a rigid body and a clock. We can then choose different geometrical coordinate systems or charts (Cartesian, spherical, etc.) for the same frame.

But what is precisely a reference frame in general relativity? To build a physical reference frame in general relativity it is necessary to replace the rigid body by a fluid or a cloud of point particles that move without collisions but otherwise arbitrarily. One can define a reference frame as a future-pointing, timelike congruence, that is, a three-parameter family of curves ##x_{a} (A, y^{i})##, where A is a parameter along the curve and ##y^{i}## is a set of parameters that 'labels' the curves, such that one and only one curve of the family passes each point. If specific parameters A and ##y^{i}## are chosen on the congruence, we define a coordinate system. Of course, this choice is not unique. In general, a given reference frame can give rise to more than one
If I understand correctly, a "fluid" of observers need not maintain fixed relative positions. Consequently, what role does the stipulation that the worldlines of the observers are parallel play? Or is the parallel worldline case restricted to only SR? Indeed I've flicked through a few introductory SR texts which make reference to a rigid lattice of measuring rods and clocks, implying that perhaps the notion of the observers (in SR) retaining fixed relative positions isn't too bad an idea.

One positive is that the constant amongst all of these different definitions is the notion that the reference frame consists of a network of observers. The difference between Newtonian, SR and GR appears to be the relationship between these observers.
 
  • #20
PeroK
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Thank you all for your insights; I won't pretend that that I have done any general relativity (I haven't...), and indeed this is the classical part of the forum (I don't know what the protocol is if the topic changes half-way through the thread!), however here's what I've grasped from what you've said:
I would focus on classical concepts for the time being. SR and GR considerations can come later.
 
  • #21
etotheipi
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I would focus on classical concepts for the time being. SR and GR considerations can come later.
Yes, GR at the moment is rightfully much too advanced for me. However there is a SR module in my current scheme of work, so it's not totally out of the question!

I did read that there are no rigid bodies in SR. However, in the (granted, fairly basic) questions I've done so far in SR, I can't see a noticeable difference between the actual physical frames used there as opposed to in classical physics (transformations are a different story...), in that for instance the reference frame of a train in SR seems as if it could also be defined as an imagined extension of point-like observers glued to the train, which we might call ##S'##.

In SR you can define global inertial frames though, and they consist of a congruence of parallel straight timelike lines and the corresponding 3D hypersurfaces (this is a special type of "time slicing" possible only in SR).
The only difference appears to be that whilst in classical physics observers in all reference frames would agree on the distances between "adjacent observers" in another particular frame, in SR the grid of observers and then also the associated coordinate system would appear contracted from another frame. In this sense the SR frame is not rigid - it can be 'squished'! However the grid system remains intact, which seems fundamentally different to the fluid of observers in GR.

Apologies if I'm rambling, there's a lot of stuff to try and internalise!
 
  • #22
PeroK
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Yes, GR at the moment is rightfully much too advanced for me. However there is a SR module in my current scheme of work, so it's not totally out of the question!

I did read that there are no rigid bodies in SR. However, in the (granted, fairly basic) questions I've done so far in SR, I can't see a noticeable difference between the actual physical frames used there as opposed to in classical physics (transformations are a different story...), in that for instance the reference frame of a train in SR seems as if it could also be defined as an imagined extension of point-like observers glued to the train, which we might call ##S'##.



The only difference appears to be that whilst in classical physics observers in all reference frames would agree on the distances between "adjacent observers" in another particular frame, in SR the grid of observers and then also the associated coordinate system would appear contracted from another frame. In this sense the SR frame is not rigid - it can be 'squished'! However the grid system remains intact, which seems fundamentally different to the fluid of observers in GR.

Apologies if I'm rambling, there's a lot of stuff to try and internalise!
For SR you definitely want to think of an inertial reference frame as a grid of local observers, all at known coordinates and with synchronised clocks. That's my advice anyway. Several of the main text books emphasise this point. The alternative, which is widespread (unfortunately IMHO) is the "Bob and Alice" show, which emphasises individual observers.

And, yes, there are three differences between SR and Newtonian physics. Let's assume frame ##S'## is "moving" wrt frame ##S##. The grid of observers in frame ##S'## when measured in frame ##S## differs in:

1) The observers in the direction of relative motion are closer together (length contraction). The other dimensions, perpendicular to the relative motion are unafffected.

2) All clocks run slow, by the same factor as the length contraction.

3) The clocks are not synchronised (again in the direction of motion): "Leading clocks lag". If we imagine ##S'## is moving in the positive x-direction, then the clocks show progressively earlier times the further they are in the positive x-direction. Note that this has nothing to do with the transmission of light signals. These times are measured and recorded by the local observers in ##S##.

All of 1) - 3) are bundled into the Lorentz Transformation, which transforms the coordiantes in ##S## into the coordinates in ##S'##. Note that, by convention, the Lorentz transformation assumes a common origin for the two coordinates systems: i.e. the event ##t = 0, x = y = x = 0## cooresponds to the event ##t' =0, x' = y' = z' = 0##.
 
  • #23
etotheipi
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Thank you for clarifying, this is a really nice summary.

I also could not agree more about Alice and Bob, since the idea of human observers in a frame just introduces so many new potential misconceptions.

For instance, we don't account for light travel time in SR unless it's specifically stated. This is evident if we think of the reference frame as having a grid of observers which measure events, however the notion of having a Bob who measures events makes it seem as though we need to take into the account the time it takes for light to reach his eye!

And another example, even earlier on this thread I got confused as to whether we should think of observers as having eyes, which most likely stemmed from thinking of observers far too physically!
 
  • #24
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And another example, even earlier on this thread I got confused as to whether we should think of observers as having eyes, which most likely stemmed from thinking of observers far too physically!
I quite like thinking about experiments in SR where things are recorded locally and then, after the excitement is over, all the information is collected and what happened, where and when is analysed. This takes real-time light signals out of the question altogether. The recorded data might be sent in by email!

The other advantage of this is that observers from both frames can sit down together and look at each other's data. Too much is made in SR, IMHO, about observers "disagreeing" about things. If they want they can sit down afterwards and happily agree about everything that was recorded by observers in more than one frame of reference.
 
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vanhees71
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Well, the point is that observers can never disagree about anything, because the Lorentz transformation is invertible, i.e., a one-to-one-mapping between descriptions with Lorentz-covariant objects (tensor components with respect to Minkowski bases).

The Alice-Bob strategy in teaching RT is not that bad too, because that's how experiments are usually done, i.e., we have an observer (which is of course usually not a necessarily a human being; it can be an electronic device like a detector or a photo screen).
 

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