I Galilean relativity for 2 frames

[gionole]

TL;DR Summary

Just want to see if I'm correct

Learning Galilean transformation and just want to see if I understand the concept well.

both frames are moving relative to some other frame(me standing all the time, not moving). frame A moving 5m/s, frame B moving 7m/s, which in turn means frame B moving 2m/s relative to frame A. Galilleo says: x=x′+vt. As far as I understand, x′ coordinates starts from frame B and x starts from frame A(and not from my origin - I'm standing). This means that frame B's object is always at x′=0 in its own frame, whereas frame A's object is at x=0 in its own respective frame. The distance between the objects is definitely vt. Also x coordinate system goes and even covers x′ coordinate system. We can also choose some point in frame B which's coordinat will be x′ in its own frame and x from frame A. Then we can write x=x′+vt.

Including the image ! I just want to see whether my 2 statements are correct.

* that if 2 frames are moving, in the real world case, this means 2 objects are moving relative to each other and their coordinates in their own frames always will be 0. It's just gallilean's formula is much more general and choosing some other points in the frame B for example.
* If x coordinate starts from frame A as shown on the image.

 

[PeroK]

I think you have the right idea. A reference frame is essentially a system of coordinates. The principle of relativity says that the laws of physics are valid in all inertial reference frames. And, inertial reference frames differ from each other in terms of a uniform velocity between them. Also, generally we take the origin of the reference frames to coincide at time $t =0$.

That means that if $A$ and $B$ are inertial reference frames, then (for some velocity $v$):
$$x_A = x_B + vt$$where $t= t_A = t_B$ is the common time coordinate.

Now, for an object moving inertially, we can find define an inertial reference frame where the object remains at the origin. This is often described as the "rest frame" of the object. That said, once you have a specific object at the origin of a reference frame, you need to potentially allow an offset at time $t = 0$:
$$x_A = x_0 + x_B + vt$$Because $A$ and $B$ may not be at the same place at $t = 0$.

 

[gionole]

I think you have the right idea. A reference frame is essentially a system of coordinates. The principle of relativity says that the laws of physics are valid in all inertial reference frames. And, inertial reference frames differ from each other in terms of a uniform velocity between them. Also, generally we take the origin of the reference frames to coincide at time $t =0$.

That means that if $A$ and $B$ are inertial reference frames, then (for some velocity $v$):
$$x_A = x_B + vt$$where $t= t_A = t_B$ is the common time coordinate.

Now, for an object moving inertially, we can find define an inertial reference frame where the object remains at the origin. This is often described as the "rest frame" of the object. That said, once you have a specific object at the origin of a reference frame, you need to potentially allow an offset at time $t = 0$:
$$x_A = x_0 + x_B + vt$$Because $A$ and $B$ may not be at the same place at $t = 0$.

Thanks for the great analysis.

Whats the thing that seems wrong in my statements or in my image ? Just making sure I dont miss something

 

[PeroK]

Thanks for the great analysis.

Whats the thing that seems wrong in my statements or in my image ? Just making sure I dont miss something

There was nothing wrong. You don't need a third, baseline frame (your "me" frame). And, just to clarify the relationship between reference frames and objects. Note that several objects all at rest relative to each other share a rest frame (in one sense), but they can't all be at the origin. You just need to note that.

 

[gionole]

Note that several objects all at rest relative to each other share a rest frame (in one sense), but they can't all be at the origin. You just need to note that.

I think this is an example of the 2 frames moving at the same speed where frames’ origins are not the same. In my image, if frame A and B moving at the same speed then we can say they are at rest. But In that case, v would be 0. And x would be equal to x’ which would not be the case as they are separated by some distance from the beginning. I think you meant something else which I dont think I understood but is it that important ? I mean in my examples, if v=0 then the only time galilean works is if from the beginning, frames started at the same origin.

 

[PeroK]

I think this is an example of the 2 frames moving at the same speed where frames’ origins are not the same. In my image, if frame A and B moving at the same speed then we can say they are at rest. But In that case, v would be 0. And x would be equal to x’ which would not be the case as they are separated by some distance from the beginning. I think you meant something else which I dont think I understood but is it that important ? I mean in my examples, if v=0 then the only time galilean works is if from the beginning, frames started at the same origin.

More generally $x' = x_0 + x + vt$. You can't in general ignore the $x_0$. Only in cases where the origins coincide at $t = 0$.

 

[gionole]

Amazing. Thanks a lot

 

[gionole]

@PeroK After thinking some more, I realized 2 questions related to frames. If you don't mind, I'd appreciate it and thats pretty much it on this subject.

We know that inertial frame in a nutshell means:

1. the velocity is constant(no acceleration) or
2. there's an acceleration, but the acceleration can be explained by real force(no fictitious force needed). Newton's law will hold true in this case

It's said that galillean transform makes sense for only inertial frames.

Question 1: In my image, how do we define inertial frames ? should the inertial frame be in respect to their own ? for example: frame A can be inertial from its own point of view, but could be non-inertial from frame B's point of view. Which one should hold true ?

Question 2: Does galillean transform work for my (2) case above ? I wonder how, if frame B's acceleration is present, $vt$ won't be the distance anymore. Wouldn't this break ? can we then assume it only works for when both frame's velocities don't change ?

 

[PeroK]

@PeroK After thinking some more, I realized 2 questions related to frames. If you don't mind, I'd appreciate it and thats pretty much it on this subject.

We know that inertial frame in a nutshell means:

1. the velocity is constant(no acceleration) or
2. there's an acceleration, but the acceleration can be explained by real force(no fictitious force needed). Newton's law will hold true in this case

It's said that galillean transform makes sense for only inertial frames.

Question 1: In my image, how do we define inertial frames ? should the inertial frame be in respect to their own ? for example: frame A can be inertial from its own point of view, but could be non-inertial from frame B's point of view. Which one should hold true ?

A reference frame is either inertial or non-inertial. The test is whether Newton's first law holds. This is an intrinsic property of the reference frame. It's not relative to anything else.

Question 2: Does galillean transform work for my (2) case above ? I wonder how, if frame B's acceleration is present, $vt$ won't be the distance anymore. Wouldn't this break ? can we then assume it only works for when both frame's velocities don't change ?

The surface of the Earth is approximately an inertial reference frame. We have to ignore the rotation of the Earth and technically all objects are subject to the force of gravity. The rest frame of an elevator moving up at constant speed is also approximately an inertial reference frame. These two frames are related by a single relative velocity; and their coordinates are related by a single parameter representing this velocity.

The rest frame of an accelerating vehicle is not an inertial reference frame. Newton's first law does not hold (even approximately) in this frame. The rest frame of a merry-go-round is not an inertial reference frame.

Note the following subtle point.

An inertial reference frame is one where Newton's first law applies. But, there is no concept of a velocity associated with an inertial frame. Two inertial reference frames have a relative velocity, as above, that appears in the Galilean transformation of their coordinates.

Likewise, there is no concept of a velocity associated with a non-inertial frame. Instead, there is a real (measurable) acceleration associated with such a frame.

Note also that all velocities are relative. There are no absolute velocities in Newtonian physics. But, in general, acceleration is absolute - in the sense that acceleration is absolutely measurable and is the same in all inertial reference frames. This is encapsulated in Newton's second law:$$F = ma$$Where all three quantities are invariant, which means they are the same in all inertial reference frames. Mathematically, if we have two inertial reference frames whose coordinates are related by:$$x' = x_0 + x + vt$$Then$$\dot x' = \dot x + v$$This is the relation between velocities (of a particle, say) as measured in the two frames. And$$\ddot x' = \ddot x$$And we see that the same acceleration (of a particle) is measured in the two frames.

The invariance of Newton's second law is critical. It's fundamentally important that acceleration does not depend on the choice of inertial reference frame.

 

[gionole]

By first question, I think you didn't fully get my point. What I asked was car moving with acceleration is a non-inertial frame, but seeing it from road, the road becomes inertial frame. I guess, the frame itself should be inertial from its point of view and not from point of others(I also understand that frame seen from another frame doesn't say anything in terms of what the original frame is). accelerated car is thought to be the non-inertial frame for Galilleo.

So if an object accelerates, from its own frame, it's never going to be inertial frame. That's what I arrived to because in the object, we can always imagine something else for which newton's law won't hold true. (example - car=our object and pendulum=our imagined something in the object and for pendulum, newton's law fails from car's frame)

Let's say frame A and B are both accelerating by the same uniform acceleration. In their own frames, they're non-inertial. one accelerates by 2, second accelerates by 4. Now, we got relative acceleration = 2. So each second, relative velocity between these changes by 2. Writing $x' = x - vt$ seems not to work, because we don't have the relative velocity as it changes. I mean, if you come up with a function for relative velocity, $v = 2t$, then it works, but in terms of galillean, do we say this is enough or if not, why not ?

 

[PeroK]

By first question, I think you didn't fully get my point. What I asked was car moving with acceleration is a non-inertial frame, but seeing it from road, the road becomes inertial frame.

First, a car is an object, not a reference frame. Second, whether a reference frame is inertial or non-inertial does not depend on "who's looking at it". You can't look at a frame. A frame is a system of coordinates.

It's like asking if an apple is an apple from an orange's point of view. Or, from its own point of view. An apple is an apple, regardless of how you look at it.

I guess, the frame itself should be inertial from its point of view and not from point of others

No. You are misunderstanding the whole concept of a reference frame. It is NOT an object.

 

[gionole]

To me, reference frame is the point of view. What I mean by when I say accelerating car is an non inertial frame is that from car’s point of view, car is non inertial. If as you say car is not a reference frame, what do you call non inertial frame when car is in accelerating mode ? Lots of examples mention that when car accelerates, road is an inertial frame and car as non inertial. You can call an non inertial frame to be system of coordinates, but in nutshel, when we say car is non inertial, i mean that there exists a car frame(point of view seen from inside the car).

 

[vanhees71]

There was nothing wrong. You don't need a third, baseline frame (your "me" frame). And, just to clarify the relationship between reference frames and objects. Note that several objects all at rest relative to each other share a rest frame (in one sense), but they can't all be at the origin. You just need to note that.

One should add that there can nothing go wrong with using a third auxilliary frame, because the Galilei transformations form a group. That's why symmetries are (usually) described by transformations build a group.

Further Galilei symmetry follows from the spacetime model of Newtonian mechanics. In modern terms you can state the Lex I as that there exists an inertial frame. Together with the usually not explicitly mentioned demands that (a) time is absolute (i.e., up to arbitrary choices of an "initial time" (time-translation invariance, as one subgroup of the Galilei group)) independent of the choice of the inertial reference frame and (b) that space for an inertial observer obeys the laws of Euclidean geometry.

Choosing the spacetime model implies restrictions to the physical laws, i.e., that they must be consistent with the symmetries of this spacetime model, and this explains why Newtonian mechanics looks the specific way it looks.

Finally you can use any non-inertial frame of reference too. The distinction from inertial frames is that then the physical laws do not take the same form as in inertial frames (i.e., there occur additional terms when expressing the acceleration of the particles in the coordinates of the non-inertial frame, which usually a reinterpreted as additional "inertial forces"), but they are still the same physical laws, and according to Lex I you can always introduce an inertial frame of reference to describe the physics in this inertial frame of reference, and there the laws take the specific form as they must take in inertial frames.

 

[vanhees71]

First, a car is an object, not a reference frame. Second, whether a reference frame is inertial or non-inertial does not depend on "who's looking at it". You can't look at a frame. A frame is a system of coordinates.

It's like asking if an apple is an apple from an orange's point of view. Or, from its own point of view. An apple is an apple, regardless of how you look at it.No. You are misunderstanding the whole concept of a reference frame. It is NOT an object.

This is also a bit misleading. All the mathematics of Newtonian mechanics must be realized for their experimental observation in the real world, and thus you need to operationally construct a reference frame with real-world objects, e.g., three (rigid) rods fastened at a point pointing in three different directions, defining 3 coordinates for each point in space, and an arbitrary clock to define time (in Newtonian physics due to the absoluteness of time, a single clock is sufficient to define a synchronized time at all points in space). Only with such a operational realization of a reference frame (consisting of a clock and three non-coplanar rigid rods) you have a clear meaning of the four spacetime coordinates occuring in the math of Newtonian mechanics.

Given the fact that there is indeed no absolute space and time, it's not so trivial to a priori define an inertial reference frame. Usually in everyday work we just use a frame fastened at a point on Earth and work with the so defined coordinates as if this were an inertial frame of reference, which of course is not true, and we can nicely demonstrate this with the Foucault pendulum experiment, demonstrating that the frame is rotating against the class of inertial reference frames (due to the rotation of the Earth around its axis).

According to our best contemporary knowledge the best realization of a (local) inertial reference frame is to use a frame which is moving with constant velocity against the "thermal bath" of the cosmic microwave background radiation. With this, however, we enter the much more delicate issue of (inertial) reference frames according to General Relativity.

 

[PeroK]

To me, reference frame is the point of view. What I mean by when I say accelerating car is an non inertial frame is that from car’s point of view, car is non inertial. If as you say car is not a reference frame, what do you call non inertial frame when car is in accelerating mode ? Lots of examples mention that when car accelerates, road is an inertial frame and car as non inertial. You can call an non inertial frame to be system of coordinates, but in nutshel, when we say car is non inertial, i mean that there exists a car frame(point of view seen from inside the car).

An object moves either inertially (no external force) or non-inertially (real external force). If an object is moving inertially then you can define an inertial frame with that object permanently at the origin. This is called the object's rest frame.

We can start by assuming that the road is moving inertially (this is not quite true, but it's a good approximation for most purposes). More precisely, we take the surface of the Earth locally to be an inertial reference frame.

If a car is moving with constant velocity relative to the surface of the Earth, then we know that the car is moving inertially. And, the car's rest frame is an inertial reference frame - in which Newton's first law applies. [Note that you haven't mentioned Newton's first law regarding inertial reference frames, which is a problem.]

If a car accelerates relative to the surface of the Earth, then it must be moving non-inertially. And we know that there must be a measurable, real external force on the car. And, the car's rest frame is a non-inertial reference frame, where Newton's first law does not hold.

Taking about a car's "point of view" is a little sloppy, IMO. What is a "point of view"? If you are in an accelerating car, then you know you are accelerating: you can feel the external force. Are you accelerating or not from your poin of view? Instead, it seems clearer to me to say that your rest frame is non-inertial. You can define your rest frame relative to the surface of the Earth. For uniform acceleration you have:
$$x_{car} = x_0 + v_0t + \frac 1 2 at^2$$Where $a$ is the (real) acceleration of the car. And $x_0$ and $v_0$ are its initial position and velocity relative to the surface of the Earth.

You need to start thinking in terms of Newton's laws and basic, 1D coordinate systems. Introducing fuzzy terms like "point of view" may be leading you astray.

 

[gionole]

Let me give you the idea of what I mean by point of view.
as we know, there can be multiple frame of references that we can construct for the accelerating car. Frame of reference is the point of view so there is a ground frame that exists for a car which means things/point of view seen from the ground. Alternatively, there is a car frame which is the point of view seen from inside the car. I know car itself is not a reference frame, but we kind of need to imagine expressing it like that because we are curious whether laws in the car seen from the car’s point of view obey newton. That is why I said car is an non inertial frame. Ofc, I get that frames cant see each other in a way to say that if I am a ground frame, I cant just look at car and say it is a non inertial frame.

So when we say frames moving and having velocities and so on, we mean that each frame is such a frame seen from its own. You can say it is a system of coordinates, but to me, it seems easier that in galillean, we focus on the object’s frames itself and dont look for other frames such as ground frame. So in a nutshell, when 2 objects move, their frames that we talk about in galillean is the frames seen from their point of view respectively. This kind of makes things easier for me. I think what you call rest frame, I call own frame and we must be saying the same thing in the end.

Now what I arrived to now is for the galillean transform to work is that frames should be inertial. If car moves, it should be inertial frame and the only time this can be true is if its not accelerating because if it does, it becomes non inertial and I brought example as pendulum inside an accelerating car.

Let's say frame A and B are both accelerating by the same uniform acceleration. In their own frames, they're non-inertial. one accelerates by 2, second accelerates by 4. Now, we got relative acceleration = 2. So each second, relative velocity between these changes by 2. Writing $x' = x - vt$ seems not to work, because we don't have the relative velocity as it changes. I mean, if you come up with a function for relative velocity, $v = 2t$, then it works, but in terms of galillean, do we say this is enough or if not, why not ?

 

[PeroK]

Let me give you the idea of what I mean by point of view.

I give up. Perhaps someone else can help you.

 

[vanhees71]

Sure, if you have a car that is accelerating against the inertial reference frames and use its rest frame, that's not an inertial frame of reference. You can do that, but then to transform to the coordinates defined by this non-inertial frame you cannot use a Galilei transformation, but it's a different transformation, and then the natural laws take a different form than when using coordinates defined by an inertial reference frame. Nevertheless they are still the same physical laws.

 

[A.T.]

... you need to operationally construct a reference frame with real-world objects, e.g., three (rigid) rods fastened at a point pointing in three different directions, defining 3 coordinates for each point in space,...

One can define an infinite number of different reference frames for analysis based on a single physical object, therefore using the term "reference frame" for both is just confusing. We had this discussion before.

 

[vanhees71]

Why is it confusing? Any such construction with real-world "clocks and rods" defines a frame of reference. If you arrange them in different ways you just use different frames of reference. If you find this confusing, you have to think a bit longer about it. It's at the very beginning of any learning of physics. Without a clear understanding, how reference frames are operationally defined and relate to the coordinates used in theory, there's no clear understanding of physics. Physics is more than just an axiomatic scheme defining the math of some physical theory!

 

[gionole]

Why is it confusing? Any such construction with real-world "clocks and rods" defines a frame of reference. If you arrange them in different ways you just use different frames of reference. If you find this confusing, you have to think a bit longer about it. It's at the very beginning of any learning of physics. Without a clear understanding, how reference frames are operationally defined and relate to the coordinates used in theory, there's no clear understanding of physics. Physics is more than just an axiomatic scheme defining the math of some physical theory!

Perok had difficulties to what I was saying by "point of view" = "reference frame". Do you find this confusing as well ? I don't know. Physics use lots of complicated words sometimes and I'm just a beginner trying to understand it intuitively. I'm not sure why I shouldn't be trying to imagine things in a clearer way.

Funny thing is when I said: by point of view seen from the car, it appears to be the same as "rest frame" mentioned by you guys.

 

[gionole]

Taking about a car's "point of view" is a little sloppy, IMO. What is a "point of view"? If you are in an accelerating car, then you know you are accelerating: you can feel the external force. Are you accelerating or not from your poin of view? Instead, it seems clearer to me to say that your rest frame is non-inertial. You can define your rest frame relative to the surface of the Earth. For uniform acceleration you have:

If I'm in a car and the car accelerates, from my point of view, car didn't accelerate, just look at the ceiling, did it accelerate ? no. but i still felt the force even without acceleration and newton fails which is why from my point of view or car's point of view to itself, it's a non-inertial frame. I repeat that what you call "rest frame" seems to be the same what I call: "point of view of the object to itself".

 

[A.T.]

Why is it confusing?

Because it's not clear if by "reference frame" one means an abstract set of coordinates or an actual physical object.

Any such construction with real-world "clocks and rods" defines a frame of reference.

Any such real-world construction can be used to define infinitely many abstract frames of reference. There is no 1:1 relationship.

If you arrange them in different ways you just use different frames of reference.

I don't need to arrange the physical object differently. I can just specify how my abstract reference frame moves relative to the physical object.

 

[vanhees71]

Because it's not clear if by "reference frame" one means an abstract set of coordinates or an actual physical object.

A reference frame is an actual set of physical objects (something that defines the set of spatial coordinates and a (set of synchronized) clocks). Coordinates are coordinates, reference frames are reference frames. Nothing is confusing here.

Any such real-world construction can be used to define infinitely many abstract frames of reference. There is no 1:1 relationship.

If this were so, we had not well-defined physical theories, which obviously is not true. Of course, any properly constructed phsical reference frame leads to a 1:1 relationship (at least locally) between space-time points and space-time coordinates. That's implicit in the definition of what a reference frame is.

I don't need to arrange the physical object differently. I can just specify how my abstract reference frame moves relative to the physical object.

It's the other way round: To describe, how a physical object moves in terms of space-time coordinates, you need a reference frame. All motion is always relative to other objects, and to have a 1:1 correspondence (at least locally) between space-time coordinates of the object(s) under investigation you need a reference frame these space-time coordinates refer to.

I don't say, it's a simple concept, but it's also not confusing. You just need to think about it carefully!

 

[A.T.]

I repeat that what you call "rest frame" seems to be the same what I call: "point of view of the object to itself".

The "... to itself" part is not included in "rest frame". The rest frame of the car can be used to analyze objects on the ground too. Just say "rest frame". It's shorter and avoids confusion.

 

[A.T.]

A reference frame is an actual set of physical objects (something that defines the set of spatial coordinates and a (set of synchronized) clocks). Coordinates are coordinates, reference frames are reference frames. Nothing is confusing here.

We had this discussion before:
https://www.physicsforums.com/threa...nflict-of-two-definitions.990480/post-6359096

I think a key issue needs to be clarified for this discussion: the term "reference frame" can be used to mean multiple things in physics. In particular, it can be used to mean any of the following three things:

(1) A coordinate chart;

(2) A frame field (i.e., a mapping of points in spacetime to 4-tuples of orthonormal vectors, one timelike and three spacelike);

(3) A concrete measuring apparatus that physically realizes #1 or #2.

These are three distinct things that should not be conflated.

If this were so, we had not well-defined physical theories, which obviously is not true.

You seem to misunderstand what I mean:

Using a single physical rod-set one can define multiple abstract frames of reference:

Frame A, where the rod-set is at rest.
Frame B, where the rod-set moves at 1 m/s along the X-axis
Frame C, where the rod-set moves at 2 m/s along the X-axis
...

And for every of those frames one can define multiple coordinate-charts, for example by rotating the axes by 90°.

So there is no 1:1 relationship between "physical object" and "abstract reference frame defined based on that physical object".

 

[PeroK]

I'm with @AT on this one. One reason I was trying to get the OP to think in terms of specific coordinate systems is that "point of view" is vague and ambiguous, can't easily be generalised and isn't useful for specific calculations.

 

[PeroK]

I repeat that what you call "rest frame" seems to be the same what I call: "point of view of the object to itself".

And this is still wrong. A frame is represented by some system of coordinates that extends over space and time. It is not the point of view of a single object. No matter how many times you repeat this misconception, it's still wrong.

 

[Nugatory]

I repeat that what you call "rest frame" seems to be the same what I call: "point of view of the object to itself".

You can think of the rest frame of an object (better terminology would be "that frame in which the object is at rest") that way if you want, but not all frames are the rest frame of anything. The Galilean transformation relates any two frames, not just ones that happen to have some object at rest in them.

 

[PeroK]

You can think of the rest frame of an object (better terminology would be "that frame in which the object is at rest") that way if you want, but not all frames are the rest frame of anything. The Galilean transformation relates any two frames, not just ones that happen to have some object at rest in them.

I was starting to wonder whether "Niel" and "gionole" were one and the same.

 

[gionole]

I was starting to wonder whether "niels" and "gionole" were one and the same

It is. I had some problem with old account and finally team of this site helped me merge the accounts.

You can think of the rest frame of an object (better terminology would be "that frame in which the object is at rest") that way if you want, but not all frames are the rest frame of anything

Point 1. Does not this mean that every object(whatever it does, accelerates or moves with constant speed or is at rest), has a rest frame ? Car moves but we can take a reference frame in which car is at rest - such frame would be “inside being a car”. This logic will apply to any object.

Point 2. On the other day, I was told that inertial frame is not the frame where acceleration is 0, but it is a frame where acceleration can be explained by real force. With this logic still, if object accelerates, that object’s rest frame will always be non inertial - i.e - if car accelerates, in the car’s rest frame, there is no acceleration, but there are forces and F =ma breaks as this says F=m*0 but we actually got a non zero force.

Point 3. Does not saying: object’s rest frame seem pointless ? Why cant we just say object’s reference frame directly ? I think this question is valid depending on my point 1.

Point 4. If you dont like the reference frame as saying point or view , then hopefully I understand what you mean by system of coordinates. So If I got an object, and I am curious about that object’s rest frame, in that frame? Coordinate system starts at the origin where the object is and if object moves, the whole coordinate system moves with it(object always placed at the origin). But when I say: ground frame, then coordinate system starts as the origin to be earth. And even for car rest frame, coordinate system starts at the origin where the car is and extends to all the way to the whole space(whole universe) and time.

Point 5. check here from 0:12 to 0:20 and listen. Just a honest question: are you suggesting to stop listening to Khan academy ? The other day I had tremendous discussion with Dale where I was saying to him to imagine as if the electric field lines from a constant speed charge moves inertialy and then everyone told me that field lines dont move even though I was imagining it to be clearer. This I listened to as well on khan academy.

 

[Nugatory]

Point 1. Does not this mean that every object(whatever it does, accelerates or moves with constant speed or is at rest), has a rest frame ? Car moves but we can take a reference frame in which car is at rest -

Yes, but....

such frame would be “inside being a car”

Is not right. A frame is rule for assigning coordinates to the position of an object (so it can't "be" anywhere, saying it's "inside the car" is like saying that the Pythagorean theorem is in my cupboard). Let's say I'm using x, y, z Cartesian coordinates to locate the position of the car at any given time: $x(t)$, $y(t)$, and $z(t)$ are the coordinates of the car's location at time $t$. If all three are constants I am using the rest frame of the car.

Point 3. Does not saying: object’s rest frame seem pointless ? Why cant we just say object’s reference frame directly ?

We can call it anything we want, as long as we are clear on our mind that we are using a convenient linguistic shortcut because "that frame in which the object at rest" is a bit clumsy. But when we say "object's rest frame" we have to be careful not to let that wording trick us into thinking that the frame is somehow attached to the object, or that it is any way special, or that we need to use that frame to calculations involving the object, or that everything is always "in" all frames, whether they are at rest using that frame or not.

Coordinate system starts at the origin where the object is and if object moves, the whole coordinate system moves with it(object always placed at the origin).

Not necessarily. We can put the origin anywhere we want. It is often convenient to put it where there is no object at all, especially in problems involving multiple at objects at rest relative to one another. The requirement is that $x(t)$, $y(t)$, $z(t)$ are constant, not necessarily zero.

 

[gionole]

@Nugatory

such frame would be “inside being a car”

Is not right. A frame is rule for assigning coordinates to the position of an object (so it can't "be" anywhere, saying it's "inside the car" is like saying that the Pythagorean theorem is in my cupboard).

Well, if rest frame is a frame in which object is at rest and "inside being a car" is not necessarily a rest frame of the car, then what would be the car's rest frame as an example ? you might say I will take a ground and in it, when car doesn't move(is at some coordinate x=2, y = 3, z=4), that I can call rest frame, but when car starts moving, it seems like ground is not rest frame of the car anymore which kind of means: frames change depending on time which seems wrong to me as well.

 

[Nugatory]

which kind of means: frames change depending on time which seems wrong to me as well.

The frame doesn't change but the velocity, as described by the frame, of objects can change. This is one of the more important reasons that you should stop thinking in terms of "the frame of the car" and rest frames in general.

I can construct a frame in which the car's coordinates remain x=2,y=3,z=4 even as the car speeds up, slows down, and even stops relative to the ground and this will indeed be the rest frame of the car, but it will not be inertial and it will be very inconvenient to use in calculations so we usually don't.

 

[vanhees71]

We had this discussion before:
https://www.physicsforums.com/threa...nflict-of-two-definitions.990480/post-6359096You seem to misunderstand what I mean:

Using a single physical rod-set one can define multiple abstract frames of reference:

Frame A, where the rod-set is at rest.
Frame B, where the rod-set moves at 1 m/s along the X-axis
Frame C, where the rod-set moves at 2 m/s along the X-axis

So what, where's the problem? That simply defines different frames of reference with the same physical equipment.

...

And for every of those frames one can define multiple coordinate-charts, for example by rotating the axes by 90°.

So there is no 1:1 relationship between "physical object" and "abstract reference frame defined based on that physical object".

Yes, you have to choose a reference point and a basis of your spacetime vectors (e.g., a tetrad). This defines the local reference frame. Which coordinates you use doesn't matter at all. That's dealt with by general covariance of GR, which is however not a symmetry but a gauge principle!

 

[A.T.]

That simply defines different frames of reference with the same physical equipment.

Yes, and thus calling the physical equipment itself a "frame of reference" can be ambiguous, because one can define many different frames of reference based on it.

 

[vanhees71]

For me the physical equipment is the "frame of reference". Which coodinates you use if completely arbitrary. That's why we can use Cartesian or spherical coordinates or whichever coordinates are convenient.

 

[A.T.]

That simply defines different frames of reference with the same physical equipment.

For me the physical equipment is the "frame of reference".

You just said above that the same physical equipment can be used to define multiple different frames of reference, and now you say the physical equipment is the frame of reference.

 

[vanhees71]

There is not only one frame of reference but arbitrary many. I can use the same equipment like "three rods and clocks" to define different reference frames. Why should this be a problem?

To be concrete, let's start with Newtonian mechanics and thus Galilei spacetime, because this is the most simple case one can think of. Galilei space time by assumption defines that there exists the class of inertial reference frames, and it's impossible to distinguish any inertial reference frame from any other. So you have to pick a specific one, and this can be done by taking three rigid rods joined in a point (called the origin of the reference frame), defining a basis in 3D Euclidean vector space (for simplicity we take the rods forming a right-handed Cartesian basis, $\vec{e}_j$, $j \in \{1,2,3 \}$) and a clock, showing absolute time.

Then you can define the location of a particle as function of time in a one-to-one relation with the postition vector $\vec{r}(t)=\overrightarrow{OP(t)}$, where $O$ is the origin of the reference frame and $P(t)$ the location of the particle under observation. By definition $O$ is time-independent, i.e., the observer is at rest relative to the now defined inertial reference frame.

Of course, whether a such given reference frame is really (with sufficient accuracy) an inertial reference frame can be investigated by just checking, whether Newton's 1st Law holds good.

Now you can introduce coordinates. The most simple ones are just the three Cartesian components wrt. the chosen Cartesian basis given by the above introduced rigid rods. Then you have a one-to-one correspondence between these coordinates and the position vector (using the usual Cartesian Einstein summation convention)
$$(x_1,x_2,x_3)^{\text{T}} \leftrightarrow \vec{r}=x_j \vec{e}_j.$$
Of course, you can now introduce any other set of coordinates, e.g., spherical coordinates by expressing the $x_j$ in terms of these coordinates.

All such Cartesian or arbitrary "generalized coordinates" $q^k$ refer to the same frame of reference. A mere introduction of different coordinates does not introduce a new frame of reference.

So one must clearly distinguish arbitrary coordinates from a physically defined frame of reference. Then all this quibbles you seem to have go away.

I know that we had this discussion already at length, and I hope it won't again start a battle about semantics.

 

[A.T.]

I can use the same equipment like "three rods and clocks" to define different reference frames. Why should this be a problem?

This isn't a problem.

But using the same term: "reference frame", for two different things:
- a physical object
- an abstract reference frame
is ambiguous.

And it gives people the wrong idea, that defining a reference frame requires a physical object that is at rest in that frame

I hope it won't again start a battle about semantics.

It is purely about confusing semantics.

 

[vanhees71]

This isn't a problem.

But using the same term: "reference frame", for two different things:
- a physical object
- an abstract reference frame
is ambiguous.

The abstract reference frame must be realized in real-world experiments to test the theory. That's all. I still don't see, where your quibbles are.

And it gives people the wrong idea, that defining a reference frame requires a physical object that is at rest in that frame

That's not a wrong idea but just physics vs. pure math. To apply a physical theory to reality you have to operationally realize the notions within the theory with real-world equipment.

A reference frame in the lab is not an abstract set of coordinates but consists of real-world equipment.

This is related to the old discussion between Newton and Leibniz about Newton's "absolute inertial reference frame", which however is a mathematical fiction. All there is in reality is the possibility that you can build a reference body and a clock to describe the motion of any other body relative (!) to this reference body with the time given by the clock. Newton's Lex 1 in modern form just says that there exists a (global) inertial reference frame (and thus of course arbitrary many, i.e., there's an entire class with the reference bodies defining different inertial frames moving with constant velocity relative to each other). In this case Leibniz was right: There's no absolute space in reality, and motion must be described relative to other bodies.

It is purely about confusing semantics.

 

[A.T.]

I still don't see, where your quibbles are.

Is it so hard to understand, that using the same term for two different things can be confusing, especially for newbies?

That's not a wrong idea but just physics vs. pure math.

Nice distinction, which gets completely lost if you use the same term for both.

 

[vanhees71]

So how do you call the realization of a reference frame in the lab? A reference frame is a reference frame, isn't it?

 

[A.T.]

So how do you call the realization of a reference frame in the lab?

This has been discussed already:
https://www.physicsforums.com/threa...nflict-of-two-definitions.990480/post-6358619

 

[vanhees71]

I think we can leave it at that.

 

[A.T.]

The abstract reference frame must be realized in real-world experiments to test the theory.

We might be talking different purposes here:

- To directly test the frame transformation equations of your theory only, you have to physically realize at least two frames, and check if the physical measurements in them are related via the proposed transformation.

- To apply the whole theory, you don't have to physically realize every frame that you use in an analysis. For example, when analyzing a collision, you can transform the problem to the center-of-mass frame, solve there using conservation laws, and transform back to the frame where the problem is stated.

This is why I think it's important to distinguish physical objects from abstract frames which are not necessarily the rest frame of any physical object.

 

[vanhees71]

We might be talking different purposes here:

- To directly test the frame transformation equations of your theory only, you have to physically realize at least two frames, and check if the physical measurements in them are related via the proposed transformation.

- To apply the whole theory, you don't have to physically realize every frame that you use in an analysis. For example, when analyzing a collision, you can transform the problem to the center-of-mass frame, solve there using conservation laws, and transform back to the frame where the problem is stated.

This is why I think it's important to distinguish physical objects from abstract frames which are not necessarily the rest frame of any physical object.

Indeed, I have to do two experiments. For each of them I have to realize physically an inertial reference frame, and one of them should move with constant velocity against the other.

Just doing the experiment once and then do the Galilei transformation to a fictitious other one doesn't test the Galilei invariance at all. I just assume Galilei invariance. The question is whether this transformation to another inertial frame really describes correctly what is observed in the other inertial frame, for which I have to construct it. E.g., you could do an experiment measuring the velocity of light and then do a Galilei transformation and claim that this describes correctly the observations in the other inertial frame. Of course you have to check this with a real experiment, and then you'll see that the Galilei transformations are not correct for the dynamics of electromagnetic fields and that electromagnetism in fact does not define some preferred inertial frame of reference as thought before the development of special relativity.

This is why I thin it's important to stress that inertial reference frames are something to be operationally defined in the real world to give the mathematical formalism (e.g., "Newtonian mechanics" as a set of axioms) a meaning as a physical theory describing Nature.

 

[gionole]

Hi All. Lately, I needed to get back to this and realized I might have doubts whether my understanding is correct.

The text seems huge, but reads easily I think. I'd appreciate if you could exactly mention my wrong logic and not derive/explain the topic from scratch as this is more productive to me. Maybe "quoting" my text which you see wrong and then mentioning why.

If we have a train car that accelerates in which I'm standing and drop the ball at $t=0$, Would it move backwards ? (I'm the observer). It seems like it should because by some transformations, I arrive at:

$m\ddot{\vec{r}} = m\ddot{\vec{r'}} - m\vec a_{train}$ ($\vec{r'}$ is seen from ground frame(inertial), while $\vec{r}$ from train frame - non-inertial, hence $m\vec{r'}$ is the newton's law for the ball seen from ground frame).

While the ball is dropping, ground frame observer only can note $m\vec g$ and no other forces, so:

$\ddot{\vec{r}} = \vec{g} - \vec a_{train}$

In the $x$ direction, $g$ is 0, so: $\ddot{r_x} = 0 - a_{x-train}$

Train Frame: to me(observer in the train), while ball is dropping, it moves backward, but not in real sense, because while ball is in the air(dropping), train still accelerates, so it moves with increased speed, but ball in the air doesn't feel this increased speed. Though, I (observer in the train) move with the same acceleration as train, because I'm standing on the surface of the train.. Note that, in this frame, I get that, me and train are stationary, but I explained and mentioned all this to better make you understand why ball in the nutshell moves "backwards". Since before dropping the ball, I had it in my hand(hence, we were at the same place), after I let go of it, it will hit on the ground behind me and not at the same place that I'm standing. Whatever we wanna call this phenomena(whether ball moved backward or train and I accelerated even more than the ball), we still need to explain it in terms of forces. On the ball, no force acted on it in $-x$ direction, so question arises: if me and train are stationary, and I let go of the ball from my hand, why did it drop behind me ? surely, force must have acted on it in $-x$ direction, but there's no force we can think of, so fictitious force is needed. General formula is $x(t) = x_0 + vt + \frac{at^2}{2}$ and From the derivation(shown in this reply), We see that, in this non-inertial(train-frame), equation of motion of ball is: $x(t) = 0 + 0 - \frac{a_{x-train}t^2}{2}$ (assuming we dropped it from the origin of train frame).

For the ground frame:, we get: $x(t) = 0$. Wherever I let go of the ball, it will directly drop vertically to the observer standing on the ground.

Question 1: Is everything correct ?

Question 2: Does the equation of motion always have the same form in inertial frames for the same system ? For example, for the ground frame, above we got $x(t) = 0$. If we had the constant speed moving train, I also would get $x(t)=0$ from within that train's frame. or better asked: Correct ?

Question 3: If I got an accelerating car, let's say equation of ball in it has non-linear dependence/form on $t$ in that accelerating car's frame, but I can choose inertial frame and from that, ball would get different form of motion(linear dependence on $t$ or no dependence as I showed above $x(t)=0$) which though will be the same as if car were moving with constant speed and I had calculated the motion from within that constant moving car's frame. Let me give example.

• train1 - accelerating
• train2 - constant speed
• ground

I drop ball in both trains.

1. Calculating motion of ball in train1 from within train1's frame gives non-linear dependence on $t$. In short form of the equation includes $\frac{at^2}{2}$.
2. Calculating motion of ball in train1 from within ground's frame gives linear dependence on $t$ or no dependence at all as $x(t)=0$. In short form would be: $c_1t+c_2$
3. Calculating motion of ball in train2 from within train2's frame gives linear dependence on $t$ or no dependence at all. In this exact case we got $x(t)=0$. In short form of the equation is $c_1t+c_2$

We can see how in (2) and (3), equations of motion both yield the same equation form. Would this be correct ?

Question 4: Interestingly enough, how about the equation of motion of ball in train 1 from train 2’s frame ? We seem to be in inertial frame, so the answer must be of form $c_1t+c_2$

So Landau’s argument that in all inertial frames, equation of motion of the same system takes the same form seems to be correct if my analysis is correct.

Thank you.

 

[jbriggs444]

So Landau’s argument that in all inertial frames, equation of motion of the same system takes the same form seems to be correct if my analysis is correct.

Your analysis for objects subject to zero external forces in one dimensional inertial frames is consistent with Landau's claim, yes.

None of your verbiage about accelerating trains has anything to do with it.
The vertical dimension of falling balls also has nothing to do with it.

 

[gionole]

Your analysis for objects subject to zero external forces in one dimensional inertial frames is consistent with Landau's claim, yes.

@jbriggs444 So everything I said is correct ? Right ? Just to be sure.

None of your verbiage about accelerating trains has anything to do with it.
The vertical dimension of falling balls also has nothing to do with it.

Do you mind explaining what you mean here ?

Thank you