I Galilean relativity for 2 frames

[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

 

[jbriggs444]

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

It was too verbose. I was left wondering what the point was. The part about something moving backward, but not really but yes, moving backward relative to the train frame was very off-putting. Here is that passage:

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.

There is no such thing as a "real sense". All motion is relative. There is no need to waffle. Pick a frame and describe the ball's motion in that frame. Don't handwave about other frames that you could have used but did not.

Do you mind explaining what you mean here ?

Your conclusion was that:

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.

But a conclusion about inertial frames has nothing to do with reasoning that involves accelerating frames. So your talk about accelerating frames was irrelevant.

Reasoning that only addresses the $x$ component of position is one dimensional. You only reasoned about one dimension. So your reasoning was restricted to one dimensional inertial frames of reference that may be in relative motion in that single dimension.

You've dealt with the case of a single object in force-free motion as described in two different one dimensional inertial frames and reasoned that the same force law (no force) applies in both frames.

That is a weak basis on which to conclude that the principle of [Galilean] relativity holds good for all physical force laws and does so in three dimensions.

 

[gionole]

There is no such thing as a "real sense". All motion is relative. There is no need to waffle. Pick a frame and describe the ball's motion in that frame

Ok then we say in train’s frame, ball moves backward and that’s it. Correct ? Hope everything else is correct too ?

That is a weak basis on which to conclude that the principle of [Galilean] relativity holds good for all physical force laws and does so in three

True, but I had to start somewhere and learn. But it holds true for even 3 dimensions and multiple particles right ? I mean Landaus argument

 

[A.T.]

Ok then we say in train’s frame, ball moves backward and that’s it. Correct

We describe the motion of the ball in a frame quantitatively.

In inertial frames that quantitative description matches all 3 of Newton's Laws of Motion.

In non-inertial frames it doesn't. But we can introduce inertial-forces, which violate Newton's 3rd Law, but allow us to match Newton's Laws 1 & 2.