I Galilean relativity in terms of homogeneity

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Landau's discussion on Galilean relativity emphasizes the importance of choosing an inertial frame of reference, where space is homogeneous and isotropic, allowing a free body at rest to remain at rest. In a non-inertial frame, such as inside an accelerating car, fictitious forces must be considered to explain the motion of objects like a hanging ball, which behaves differently compared to an inertial frame. The conversation highlights that while Newton's laws apply directly in inertial frames, they require adjustments in non-inertial frames due to the presence of these fictitious forces. The distinction between real and fictitious forces is crucial for understanding motion in different reference frames. Ultimately, the discussion illustrates the complexities of applying classical mechanics in varying frames of reference.
  • #31
gionole said:
From the car frame: I got a problem here. I sit in a car. I understand that to me, when car moves with constant speed at first, everything is stationary - i look at ball or anything for that matter, and nothing moves. Now acceleration happened, but to me, ceiling is still stationary - to me, car/ceiling still feel like stationary. Now, if to me, if i feel that everything is still stationary, why did the ball swing back ? I think this is the scenario we try to explain and can't explain with the same logic as we did in ground frame. Since ceiling to me didn't accelerate, it couldn't have caused force in the tension the same way as in ground frame,
The ceiling didn't accelerate in the car frame, but the ball did. I don't think it's a good idea to start talking about "cause and effect" in a non-inertial frame, because it creates confusion.
gionole said:
but ball still swang backwards, so if no force acted on it, why did it move backwards ?
It moved backwards because of the fictitious force on it. The same fictitious force that applied to everything else in the car and cancelled out the real "accelerating" force.
gionole said:
It seems like newton's law broke - if no force, it should have stayed at rest or moving with the same speed, but the ball accelerated. The way you explain the swing of ball backwards is(i.e it's acceleration - because to me, it's not stationary during its backwards movement), some fictitious force must have happened.
The fictitious force applies as soon as you use the accelerating reference frame.
gionole said:
Where would you say I'm wrong ? Would appreciate to point out the exact things that I wrongly said.

Thanks already so much for bearing with me. Physics in terms of intuitivity is lot harder than I imagined.
With non-inertial frames you need to rely more on solid technique and mathematics. Your inituition is letting you down and you are drifting from one misconception to another. I suggest you start using free-body diagrams. In an non-inertial frame you add the fictitious force to everything.

Note that, in fact, you can combine the fictitious force with the "real" gravity to get an overall virtual-gravity in the non-inertial frame. In the car frame this acts at an angle, down and backwards. The ball, therefore, exhibits the motion of a simple pendulum under this virtual gravitational force.
 
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  • #32
@PeroK

Q1. Was everything correct in ground frame analysis of mine ?

Q2:
The ceiling didn't accelerate in the car frame, but the ball did. I don't think it's a good idea to start talking about "cause and effect" in a non-inertial frame, because it creates confusion.
Isn't this what I said ? Could you re-read my "car frame reference" analysis ?

Q3:
It moved backwards because of the fictitious force on it. The same fictitious force that applied to everything else in the car and cancelled out the real "accelerating" force.
I asked the question: "why ball moved backwards" to convey the problem, but I answer it by myself in there. I think re-reading my analysis would be so appreciated.
 
  • #33
gionole said:
Car is moving with constant velocity 20m/s and then accelerates in uniform mode.
This is precisely the non-uniform acceleration that I wished to avoid. Changing acceleration from 0 to a constant is non-uniform acceleration. I would do this as two separate scenarios

1) Car is moving with constant velocity
2) Car is accelerating uniformly

I am not getting involved in this one as it is, but good luck maybe others like @PeroK can help. There are too many complications for having a non-uniform acceleration that it is not worth it for me.
 
  • #34
Dale said:
This is precisely the non-uniform acceleration that I wished to avoid. Changing acceleration from 0 to a constant is non-uniform acceleration. I would do this as two separate scenarios

1) Car is moving with constant velocity
2) Car is accelerating uniformly

I am not getting involved in this one as it is, but good luck maybe others like @PeroK can help. There are too many complications for having a non-uniform acceleration that it is not worth it for me.
hey Dale. You got it wrong. I didn't mean to bring non-uniform acceleration. The reason I brought 20m/s was I wanted to convey the idea that before accelerating, with constant speed of car, ball also was moving with 20m/s and when acceleration happened, my analysis of why ball didn't instantly receive the force due to force transfer delay caused the ball to continue moving with 20m/s and car with higher speed which caused the perception of the ball moving backwards. I don't want to complicate this even more, believe me. If 1) car is moving with constant velocity, I don't know what we need to discuss at all. I want to discuss uniform acceleration. I know you mean when car has a = 2 all the time which I agree, but as I. said I only brought 20m/s in the beginning to convey the speed difference in the ball/car due to delay. with uniform acceleration, my analysis should still be right as ball receives the updated speed at a delay due to force delay while car instantly increases speed by 2 hence swinging.

Having said that, would it be hard for you to let me know in your opinion what I've explained wrongly in my above analysis ? @Dale
 
  • #35
gionole said:
I didn't mean to bring non-uniform acceleration. The reason I brought 20m/s was I wanted to convey the idea that before accelerating, with constant speed of car, ball also was moving with 20m/s and when acceleration happened,
When there is a "before accelerating" and a "when acceleration happened" then the acceleration is non-uniform. I don't want to deal with the transient mess of the swinging ball. Good luck.

gionole said:
would it be hard for you to let me know in your opinion what I've explained wrongly in my above analysis ?
Yes. That is why I don't want to do non-uniform acceleration. It is MUCH harder and provides no instructional benefit.
 
  • #36
Thanks for the help. You don't have to answer, but I will still mention the below.

Dale said:
When there is a "before accelerating" and a "when acceleration happened" then the acceleration is non-uniform. I don't want to deal with the transient mess of the swinging ball. Good luck.
I didn't know that it was that complicated. That means my analysis of force delay thing is wrong. That's just great. Were you describing the case when the ball is already swung backwards ?

In the car frame (assuming uniformly accelerating car) the ball does not accelerate. It hangs at rest at an angle. The tension should make it accelerate horizontally, but it does not. So the fictitious force must be pulling it backwards to prevent it from accelerating.
I believe, you mean, looking at it from inside the car, the ball is stationary. It's tilted, but it doesn't move, just stays tilted.

In the inertial frame the ball is accelerating horizontally. The real force is the tension acting on the ball.
Wouldn't the ball still stay tilted at the angle all the time the same way here as well ? it's just from the ground frame, the tilted string ball for sure moves past the observer because the car is moving.

I think the distinction is, from looking outside(ground frame), we see the ball which moves, because the car moves and movement of the tilted string/ball is because there's acceleration on it by the tension. but from inside the car, we look at the ball, we see it's stationary and you say that if it's stationary, why is it tilted backwards ? in that case, ok, fictitous force causes this, but why do we even ask this question: "why the ball is tilted backwards" - it is because previous event caused it or the question we're asking is why does it stay tilted ? well if it's stationary, then it will remain at rest then until force acts on it so what's the correct question we should be asking that newtonian law fails to explain ?
 
  • #37
To be honest, this is not difficult. In an linearly accelerating reference frame you add a fictitious force to everything. The fictitious force represents a common acceleration equal and opposite to the "real" acceleration of the frame. That's it. You just pretend there is a second gravitational force acting on everything, and apply Newton's laws.

It's no more complicated than that.
 
  • #38
PeroK said:
To be honest, this is not difficult. In an linearly accelerating reference frame you add a fictitious force to everything. That's it. You just pretend there is a second gravitational force acting on everything, and apply Newton's laws.

It's no more complicated than that.
Yes, but I still have hard time intuitively. I'd love to ask you last favor.

In the above analysis(#30), I use the constant velocity and then acceleration to explain the backward motion of the ball in both of the frames. At least would be happy to know what exactly is wrong in that analysis. @PeroK you pointed out some things, but I believe I had already written them that way as you pointed out. check #32.

Then for uniform acceleration, from inertial frame, I get the idea. We see the ball which moves, because the car moves and movement of the tilted string/ball is because there's acceleration on it by the tension and tension is because the ceiling is accelerating, so ceiling -> tension -> ball. But ball stays tilted all the time in uniform acceleration. For non-inertial frame, I'm not sure what is the question I should be asking. Why do we say newton law doesn't hold true and we have to bring frictitious force ? what is it that doesn't hold ? is it that because from inside the car, tilted string/ball seems stationary, but we should be able to explain why it's tilted backward - is this the question ? or what else ?
 
  • #39
gionole said:
Were you describing the case when the ball is already swung backwards ?
Yes. All of my posts were assuming a uniformly accelerating frame. The acceleration is constant over time, and the ball is not swinging but just hanging at an angle.

You can easily draw a free body diagram for this scenario.
 
  • #40
Dale said:
Yes. All of my posts were assuming a uniformly accelerating frame. The acceleration is constant over time, and the ball is not swinging but just hanging at an angle.

You can easily draw a free body diagram for this scenario.
@Dale And what’s the question we ask that netwonian laws fail to answer in car frame ? Is it :’why the ball and string stay tilted if no force acts on it ?’

I understand that from car frame, ceiling does not accelerate which means it does not give force to the string which means there is now no tension force. Also, the ball itself is tilted and is stationary. Since there is no tension force, what causes it to stay tilted and stationary ? and thats where we bring fictious force to say that this fictitous force is what holds it stationary and tilted ?

And in ground frame, we didnt have this problem because we did not have to explain why the ball stays tilted/stationary. In there, we had to explain why the ball is accelerating and we explain it such as ‘tension causes it to accelerate’.

Am I right ? 🙏
 
  • #41
gionole said:
what’s the question we ask that netwonian laws fail to answer in car frame ? Is it :’why the ball and string stay tilted if no force acts on it ?’
No. The question is why does it not accelerate due to the real forces acting on it? Draw the free body diagram and you can see the real forces are unbalanced. And yet it remains stationary in the car’s frame.

gionole said:
it does not give force to the string which means there is now no tension force
The tension is a real force. It is directly measurable with a force gauge. It must exist in all frames, inertial or non-inertial.
 
  • #42
Here's a diagram of the forces on the ball while the car is under steady acceleration (and any oscillations have died out). There is tension ##T## at angle ##\theta## to the vertical and weight ##mg##
4c2d3749-68b6-4c1e-ac71-e1e20e2670e8.png


Obviously we can resolve components and there's an unbalanced force ##T\sin\theta## to the left, so the ball ought to accelerate to the left.

In an inertial frame that's fine. The ball does indeed accelerate to the left along with the car.

In the non-inertial frame where the car is at rest, we want to say that the ball is not accelerating. But there's still an unbalanced force! The solution is to introduce a so-called "inertial" or "fictitious" force pointing to the right. Its magnitude is ##m\alpha##, where ##\alpha## is the acceleration of the car, and this will turn out to be equal to ##T\sin\theta## if you work through the maths. Thus the pendulum hangs at an angle to the ceiling because the "inertial force" balances the horizontal component of the tension.

Mathematically, what happens is that in the inertial frame we have a net force causing an acceleration: ##F_\mathrm{net}=m\frac{d^2x}{dt^2}##. Then we switch to a coordinate frame accelerating to the left. If its coordinate is denoted ##x'## then we have ##x'=x+\frac 12\alpha t^2## where ##\alpha## is the car's acceleration. That means that ##\frac{d^2x'}{dt^2}=\frac{d^2x}{dt^2}+\alpha## and therefore ##F_\mathrm{net}=m\frac{d^2x'}{dt^2}-m\alpha##. But if we want to keep our intuitive idea that any unbalanced force causes an acceleration we can't have that second term on the right. So we just shift it over to the left and call it a "inertial force": ##F'_\mathrm{net}=m\frac{d^2x'}{dt^2}##, where ##F'_\mathrm{net}=F_\mathrm{net}+m\alpha## adds the inertial force to the real ones.
 
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  • #43
Thanks so much. Now it is finally clear. Though, alternatively, I could now ask that in an inertial frame, why the string/ball stays tilted if they are accelerating and if Tsin0 acts as horizontal force in the car’s movement direction? Why is it not getting back to complete equilibrium position ?

In a non inertial frame, I get it, you say that horizontal force is balanced out by fictious and it stays tilted. But how do you explain it in an inertial frame ? The only explanation I could come up with car accelerates faster than ball due to force delay transfer so car’s speed increases faster than ball’s and because of that ball stays tilted/behind. Is not this the right way to look at it ?
 
  • #44
gionole said:
Why is it not getting back to complete equilibrium position ?
Because both the car, to which the top of the string is attached, and the ball are always moving at the same (increasing) speed. That's the point of the equilibrium position - it's where the horizontal component of ##T## causes the ball to accelerate at the same rate as the car. So it never catches up nor falls behind, and the string stays at that angle.
 
  • #45
gionole said:
Thanks so much. Now it is finally clear. Though, alternatively, I could now ask that in an inertial frame, why the string/ball stays tilted if they are accelerating and if Tsin0 acts as horizontal force in the car’s movement direction? Why is it not getting back to complete equilibrium position ?

In a non inertial frame, I get it, you say that horizontal force is balanced out by fictious and it stays tilted. But how do you explain it in an inertial frame ? The only explanation I could come up with car accelerates faster than ball due to force delay transfer so car’s speed increases faster than ball’s and because of that ball stays tilted/behind. Is not this the right way to look at it ?
I would question to your whole approach to this subject. You seem to be treating physics as an inituitive, philosophical subject, where wordy arguments are used to justify things. Perhaps that's Landau's style. I don't know his books. Perhaps someone who does can comment.

We have 44 posts on this thread, where you have repeatedly posted the same questions over and over. You have never used any sort of formal analysis of these questions, by using a free-body diagram or by doing specific calculations. This is a bad sign.

If you learned mechanics from, say, Kleppner and Kolenkow, you would be doing physics by drawing free-body diagrams and using force-based calculations. If you attach a string to a ball and pull the string, then a tension in the string arises and the ball gets pulled. That would be an "obvious" starting point for most students, I would say. I wouldn't expect a page of supporting text to explain what's going on.

That said, if forces and motion are really something of a mystery to you, then classical mechanics is going to be tough to learn.

It's possible that you should put to one side the whole concept of non-inertial reference frames until you have a grasp of Newton's laws and have solved all different problems, static and dynamic, involving forces, acceleration and equilibrium - using inertial reference frames.
 
  • #46
In this case the description of motion in an accelerated frame is pretty simple, if we assume the car moves along a straight line, i.e., that its motion is described by an arbitrary function ##\vec{x}(t)=x_0(t) \vec{n}## with ##\vec{n}=\text{const}## wrt. an inertial reference frame. An observer at rest wrt. the car will then describe the motion of an arbitrary body by the position vector ##\vec{r}'(t)##. The position vector with respect to the inertial frame then obviously is
$$\vec{r}(t)=x_0(t) \vec{n} + \vec{r}'(t),$$
and in this frame of reference the usual Newtonian equations of motion hold true,
$$m \ddot{\vec{r}}=\vec{F}(\vec{r})$$
Now
$$m \ddot{\vec{r}}=\ddot{x}_0(t) \vec{n} + \ddot{\vec{r}}' \; \Rightarrow \; m \ddot{\vec{r}}'=\vec{F}(x_0 \vec{n}+\vec{r}')-m \ddot{x}_0(t) \vec{n},$$
i.e., in addition to the "true force" you have an inertial force (what some posters here call fictitious force, which can be kind of misleading, because these inertial forces are not fictitious, but the point is that you can transform them away by going the way backwards to the original inertial frame, and they are thus not "true forces" in the sense of some real interactions like the electromagnetic or gravitational interaction).
 
  • #47
gionole said:
I could now ask that in an inertial frame, why the string/ball stays tilted if they are accelerating and if Tsin0 acts as horizontal force in the car’s movement direction? Why is it not getting back to complete equilibrium position ?
It must accelerate with the car, so it needs the horizontal component of T, which is 0 if the string is vertical.
 
  • #48
Everything looks clear now. I want to thank everyone included in the discussion for your big help. It's not hard to be honest and drawing free-body diagram is easy. The whole idea of physics being hard is sometimes it's a game of playing words. Well, I understood Lagrange and euler quicker than this, so it's full of surprises. Some topics are easy, some are not. It's all relative to the person.
 
  • #49
gionole said:
I could now ask that in an inertial frame, why the string/ball stays tilted if they are accelerating and if Tsin0 acts as horizontal force in the car’s movement direction? Why is it not getting back to complete equilibrium position ?
Draw the free body diagram in the non-tilted position. What is the acceleration?

gionole said:
It's not hard to be honest and drawing free-body diagram is easy.
Then why didn’t you do that? FYI, it is kind of rude to ask people to answer 50 posts and then say “it’s easy”. Please start your next question with your best personal effort, particularly when the important part is easy for you.
 
  • #50
Dale said:
Draw the free body diagram in the non-tilted position. What is the acceleration?

Then why didn’t you do that? FYI, it is kind of rude to ask people to answer 50 posts and then say “it’s easy”. Please start your next question with your best personal effort, particularly when the important part is easy for you.
Well, Dale, you misunderstood my part. The easy part I meant was about free body diagram. I drew it before even I asked the question, but the hardship I had was somehow I couldn't make myself sure why acceleration in car frame of the ball was 0 and everything just added it up to this and I couldn't even distinguish what i understood and what not.

I don't think anything that I have done is rude. The physics sometimes is a play of words and once I understood all words in a good way, then I finally understood. Drawing free body diagram is again easy for this problem and I needed all of your help regarding the mixed up things in my head which you all guys solved it and I now get it.

Thanks again for you help
 
  • #51
You are right, I shouldn’t have said rude.

Your last question just struck me poorly since I don’t understand how it could be asked by someone who has done a free body diagram. But that shouldn’t have led me to assert rudeness in you. My apologies
 
  • #52
Dale said:
You are right, I shouldn’t have said rude.

Your last question just struck me poorly since I don’t understand how it could be asked by someone who has done a free body diagram. But that shouldn’t have led me to assert rudeness in you. My apologies
All good and thank you !
 

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