How does the Gallilean Invariance Puzzle challenge our understanding of motion?

[Magatron]

TL;DR Summary

A thought experiment puzzle involving the Gallilean Invariance theory

Gallilean Invariance states that the laws of motion are the same in all inertial frames. One experiment involved being on a ship below deck with no frame of motion reference. Supposedly, there is no experiment which could show whether the ship is moving or in what direction or speed. I was thinking about this theory and I imagined a superball being bounced off a wall on a moving train. The results of this thought experiment, since I can't carry out the experiment in the real world currently, seemed contradictory to me. See if you can figure out this puzzle.

A person on a train moving at 100 mph throws a superball in the direction of the train's motion at a speed which is to him 100 mph. When it hits the wall and bounces, assuming 100% bounce efficiency and straight paths, how fast will it appear to return to the thrower? For it to appear to him to return at 100 mph it would have to stop dead in relation to the train tracks at the moment it hit the wall, because he's moving at 100 mph toward it so his motion alone would provide the full 100 mph.

However, from the viewpoint of the train tracks, or an observer on a bench by the tracks, it would appear that the ball when thrown was moving at 200 mph, the train speed plus the throw speed. Since the wall is only moving at 100 mph, the ball should bounce back at 100 mph relative to the observer and the thrower should see it returning to him at 200 mph, rather than 100. The stationary observer would not see the ball stop dead, they would see it bounce back at 100 mph, would they not? So how can this be? Does the ball stop dead and wait for the thrower to catch up to it at 100 mph or does it bounce back at 100 mph relative to the track and close in on the thrower at 200 mph? I don't know the answer myself, it just seems contradictory.

 

[Dale]

It only seems contradictory because you simply assumed contradictory results without actually working it out quantitatively. If you use the conservation of energy and momentum and actually work it out then there is no contradiction.

 

[PeroK]

However, from the viewpoint of the train tracks, or an observer on a bench by the tracks, it would appear that the ball when thrown was moving at 200 mph, the train speed plus the throw speed. The stationary observer would not see the ball stop dead, they would see it bounce back at 100 mph, would they not?

If you do the calculations for this collision using conservation of energy and momentum, then the trackside observer will see the ball stop dead.

Elastic collisions are characterised by the same separation speed before and after collision. In the frame of the tracks, that separation speed is $100 mph$ - before and after. The horizontal speed of the ball after the collison is, therefore, $0 mph$.

 

[A.T.]

I don't know the answer myself, it just seems contradictory.

Your main error: If the wall is moving, a 100% elastic collision doesn't preserve the speed of the ball.

A general problem of your post. You fail to stick to one reference frame in either paragraph, and keep mentioning velocities relative to the other frame. This approach is bound to confuse you.

 

[Magatron]

It only seems contradictory because you simply assumed contradictory results without actually working it out quantitatively. If you use the conservation of energy and momentum and actually work it out then there is no contradiction.

So which is it, stop dead or bounce back at 100 mph from the a stationary observer's viewpoint?

 

[PeroK]

So which is it, stop dead or bounce back at 100 mph from the a stationary observer's viewpoint?

Stop dead.

 

[Magatron]

Stop dead.

So it will go from 200 mph to zero just because it hit a 100 mph wall? Wouldn't a 200 mph ball hitting a 100 mph wall be the same as a 100 mph ball hitting a stationary wall? What would happen in that case, the ball would just stop and stay stuck to the wall?

 

[jbriggs444]

So it will go from 200 mph to zero just because it hit a 100 mph wall?

Yes. The ball rebounds so that the separation velocity is 100 mph, all of which is accounted for by the motion of the wall.

Wouldn't a 200 mph ball hitting a 100 mph wall be the same as a 100 mph ball hitting a stationary wall?

Yes. The ball rebounds so that the separation velocity is 100 mph, all of which is accounted for by the motion of the ball.

 

[PeroK]

So it will go from 200 mph to zero just because it hit a 100 mph wall?

If you have conservation of energy and momentum, then yes. Do the math, as they say.

But, of course, the easiest way to solve that problem is to transform to the frame where the train is at rest. Then it's what you calculated for the train frame. Then transform back. And, when you transform back then $- 100mph$ in the train frame becomes $0 mph$ in the ground frame.

Wouldn't a 200 mph ball hitting a 100 mph wall be the same as a 100 mph ball hitting a stationary wall?

It's the same in the sense that it's the same physical event viewed from two different frames. That's why you can do the calculations in the rest frame of the train/wall, then transform back.

But, if you do the calculations in all the ground frame, then it comes out the same. The ball stops in the ground frame. Which is equivalent to rebounding in the wall frame.

You raised this as a paradox. But, I get the feeling you are disappointed that there is no paradox. When you do the maths, the paradox disappears. I'm sorry!

 

[Ibix]

So it will go from 200 mph to zero just because it hit a 100 mph wall?

Yes

Wouldn't a 200 mph ball hitting a 100 mph wall be the same as a 100 mph ball hitting a stationary wall?

Yes

What would happen in that case, the ball would just stop and stay stuck to the wall?

In both cases the wall and ball move apart at 100mph. You've just got two frames, one where the wall moves away from a stationary ball at 100mph and one where the ball moves away from the stationary wall at 100mph.

 

[A.T.]

Wouldn't a 200 mph ball hitting a 100 mph wall be the same as a 100 mph ball hitting a stationary wall?

Same frame invariants. Different frame dependents, like the final velocity of the ball

What would happen in that case, the ball would just stop and stay stuck to the wall?

No, that is a different frame invariant outcome.

 

[Magatron]

Yes

Yes

In both cases the wall and ball move apart at 100mph. You've just got two frames, one where the wall moves away from a stationary ball at 100mph and one where the ball moves away from the stationary wall at 100mph.

It doesn't make sense that something moving at 200 mph could be stopped cold by hitting something moving half that speed. Common sense says that it will bounce back at 100 mph.

 

[PeroK]

It doesn't make sense that something moving at 200 mph could be stopped cold by hitting something moving half that speed. Common sense says that it will bounce back at 100 mph.

That just shows that if we were dependent on your common sense there wouldn't be much science to talk about!

Anyway, the proof is to do the experiment. Not that easy to set up, though. But, experiments don't care about your common sense either.

This is the nearest I could find:

 

[jbriggs444]

Anyway, the proof is to do the experiment. Not that easy to set up, though.

Last time I checked, the surface of the Earth where I live was moving eastward at about 600 or 700 miles per hour. Yet I've never had problems bouncing a super-ball off either east-facing or west-facing walls.

Nor have I ever had to account for time of day and the Earth's motion around the sun.

 

[A.T.]

Common sense says that it will bounce back at 100 mph.

Sounds more like nonsense. Why would tennis players swing the racket towards the incoming ball, if that wouldn't increase the final speed of the ball? Conversely if the racket is moving away, the ball's final speed will decrease.

 

[Dale]

So which is it, stop dead or bounce back at 100 mph from the a stationary observer's viewpoint?

Stop dead in the embankment frame.

 

[Ibix]

It doesn't make sense that something moving at 200 mph could be stopped cold by hitting something moving half that speed. Common sense says that it will bounce back at 100 mph.

It's what happens, at least assuming that the wall is hugely more massive than the ball. So I would suggest you need to adjust your notion of "common sense".

It's worth noting that it took us until the 1600s to figure this out, and we'd been walking the Earth for the better part of 10,000 years at that point. But it's trivial to work out. In the frame where the wall is stationary the ball approaches at +100mph and rebounds at -100mph. If you switch to a frame where the wall is doing +100mph, you need to add 100mph to all velocities. The zero of the wall becomes +100mph. The incoming ball velocity becomes +200mph and the rebound velocity becomes (-100+100)=0mph.

 

[Magatron]

It's what happens, at least assuming that the wall is hugely more massive than the ball. So I would suggest you need to adjust your notion of "common sense".

It's worth noting that it took us until the 1600s to figure this out, and we'd been walking the Earth for the better part of 10,000 years at that point. But it's trivial to work out. In the frame where the wall is stationary the ball approaches at +100mph and rebounds at -100mph. If you switch to a frame where the wall is doing +100mph, you need to add 100mph to all velocities. The zero of the wall becomes +100mph. The incoming ball velocity becomes +200mph and the rebound velocity becomes (-100+100)=0mph.

I guess you're right, it just seems odd. It really is something where you would have to see it to believe it. What about this though? If the wall was stationary and the ball hit it at 200 mph it would bounce from that point in space at 200 mph. So if the wall is moving at 100 mph then the ball should bounce back from that point in space at 100 mph. How could it not? The only thing that changed was that 100 mph was added in one direction, thus 100 mph should be subtracted from the other direction, leaving the ball moving at 100 mph from that point in space. Why would adding 100 mph in one direction result in subtracting 200 mph in the other?

 

[Ibix]

Why would adding 100 mph in one direction result in subtracting 200 mph in the other?

I'm having trouble parsing this. It doesn't seem to correspond to anything in the maths, which you look to have done correctly. Perhaps you could rephrawe your question?

 

[Mister T]

It really is something where you would have to see it to believe it.

The next time you're cruising in an airplane bounce a ball off the back wall. Also do it off the front wall.

 

[Jeff Root]

Common sense says that it will bounce back at 100 mph.

Your common sense is correct. The ball will bounce back at
100 mph relative to the wall, since that is the speed at which
it hit the wall. Its speed relative to you will be zero. That is a
change in the ball's speed relative to you of 200 mph. It was
going toward the wall at 100 mph, now it is going away from
the wall at 100 mph, and that is 0 mph relative to you.

That is just common sense, plus not conflating reference frames.

-- Jeff, in Minneapolis

 

[Vanadium 50]

The next time you're cruising in an airplane bounce a ball off the back wall.

The airlines do not care for this. Or <cough> so I've been told.

 

[jefferywinkler2]

We assume person A is on board the train throwing the ball towards the front of the train, or in the direction the train is traveling, if it is traveling. Person A is throwing the ball towards a wall which could be either inside the train, at the front of the car person A is on, or outside the train, next to track, perpendicular to the track, in which such case, person A throws it outside the window. We assume the angle between the path of the ball and the track is negligible. Tje ball impacts the wall, and rebounds in an elastic collision. Person B is an observer standing near the track. v is the velocity of the ball from the point of view of the person A, shortly after person A threw it. Velocity in the direction the train is moving is positive. Velocity opposite from that the train is moving is negative. When the train is moving, it goes 100 mph. We discuss three scenarios, the wall inside the train while the train is motionless, the wall is inside the train while the train is moving, and the wall is outside the train while the train is moving. For each, I give the velocities for the following reference frames, person A, person B, the ball while traveling from person A to the wall, and the ball while traveling from the wall to person A.

I. Wall inside train which is not moving

A. person A

person A = 0
wall = 0
ball traveling from person A to wall = v
ball traveling from wall to person A = -v

B. person B

person A = 0
wall = 0
ball traveling from person A to wall = v
ball traveling from wall to person A = -v

C. ball traveling from person A to wall

person A = -v
wall = -v
ball traveling from person A to wall = 0

D. ball traveling from wall to person A

person A = v
wall = v
ball traveling from person A to wall = 0

II. Wall inside train which is moving

A. person A

person A = 0
wall = 0
ball traveling from person A to wall = v
ball traveling from wall to person A = -v

B. person B, add 100 to above values for person A

person A = 0 + 100 = 100
wall = 0 + 100 = 100
ball traveling from person A to wall = v + 100
ball traveling from wall to person A = -v + 100, which may be either positive or negative, depending on the value of v

C. ball traveling from person A to wall

person A = -v
wall = -v
ball traveling from person A to wall = 0

D. ball traveling from wall to person A

person A = v
wall = v
ball traveling from person A to wall = 0

II. Wall outside train which is moving

A. person A

person A = 0
wall = -100
ball traveling from person A to wall = v
ball traveling from wall to person A = -v

B. person B, add 100 to above values for person A

person A = 0 + 100 = 100
wall = -100 + 100 = 0
ball traveling from person A to wall = v + 100
ball traveling from wall to person A = -v + 100, which may be either positive or negative, depending on the value of v

C. ball traveling from person A to wall

person A = -v
wall = -v + (-100) = -(v + 100)
ball traveling from person A to wall = 0

D. ball traveling from wall to person A

person A = v
wall = v + (-100) = v - 100, which may be either positive or negative, depending on the value of v
ball traveling from person A to wall = 0

Someone mentioned tennis rackets. If you swing a tennis racket, from the reference frame of the racket, the racket is motionless, so if the ball hits at v, it rebounds at -v.

The original poster wrote "So it will go from 200 mph to zero just because it hit a 100 mph wall? Wouldn't a 200 mph ball hitting a 100 mph wall be the same as a 100 mph ball hitting a stationary wall? What would happen in that case, the ball would just stop and stay stuck to the wall?" When they say "zero", they are not implying that the ball "would just stop and stay stuck to the wall". The velocity is only 0 mph for an instant. Before that, it is v, and after that it is -v.

Somebody said, "we'd been walking the Earth for the better part of 10,000 years at that point". The human species evolved 200,000 years old.

 

[Magatron]

I guess you really are right, even though it seems weird, because if the thrower moved out of the way so the ball bounced off the rear wall after bouncing off the front wall, the wall would hit the stationary ball at 100 mph and make it move at 200 mph toward the front wall in relation to the track. So an observer by the track would see the ball repeatedly go forward at 200 mph, stop dead, and go 200 mph again. One side of the bouncing being 200 mph and the other being zero mph, they average out to the same as 100 mph each side, so the math is consistent even though the results look weird when observed.

 

[A.T.]

... even though the results look weird when observed.

It doesn't look that weird, because even if you watch it from the ground, the brain often picks up the relative motions of the interacting objects (ball & train), rather than the motions relative to you. It also depends on what you are tracking with your eyes.

 

[Ibix]

It doesn't look that weird, because even if you watch it from the ground, the brain often picks up the relative motions of interacting object (ball & train), rather than the motions relative to you. It also depends on what you are tracking with your eyes.

Agreed. I think this is actually why the results of frame changes are so unintuitive. Your brain is so good at mixing frames that the results of formal analysis in a single frame seems weird. Just look at the Mythbusters video in #13. They don't just shoot a cannon out of a truck - they are extremely careful to do it in front of a strongly patterned background, right in front of a vertical line in the pattern. Physically, that yellow/black chessboard is irrelevant, but it seriously cuts down on the tricks your brain can play on you.

 

[A.T.]

So an observer by the track would see the ball repeatedly go forward at 200 mph, stop dead, and go 200 mph again.

Note that this is similar to what an outer point on the wheel of a driving car does, in the frame of the ground. Its horizontal velocity oscillates between zero and twice the car speed. But intuitively people just interpret the wheel as "spinning", for the reasons I gave in post #25.

 

[jbriggs444]

II. Wall outside train which is moving

A. person A

person A = 0
wall = -100
ball traveling from person A to wall = v
ball traveling from wall to person A = -v

This is not correct. A ball rebounding elastically from an approaching wall will increase in speed.

A natural way to work the correct results out is to shift to the ground frame. The ball is now traveling at v + 100 and rebounds elastically from the stationary wall at -v - 100.

Shifting into the train frame, the new velocity becomes -v - 100 - 100 = -v - 200.

One might feel uncomfortable with this. In the train frame we are told that this is an elastic collision but it is clear that energy has increased. The ball is going faster than it started. However, a careful accounting for energy reveals that all is well. The collision with a moving wall does work on the wall. The moving wall slows down. No matter how massive the wall is, the work done is roughly the same.

In the train frame, the ball gains energy and the wall loses energy.
In the track frame, no energy changes hands.

This is acceptable because energy is not an invariant quantity -- it changes depending on your frame of reference.

 

[A.T.]

A ball rebounding elastically from an approaching wall will increase in speed.

Right, otherwise tennis wouldn't work. The elastic collision in different frames is visualized at the beginning of this video:

 

[sophiecentaur]

It doesn't make sense

it just seems odd.

This applies to many situations and it's why confidence tricksters still regularly relieve people of their money. You just can't trust intuition; you have to do the (right) sums.

 

[Jeff Root]

It looks to me that what led the original poster astray was
the math. The situation is simple and intuitively obvious.
Applying the math to that situation without understanding
exactly how the math connects to it is the problem. If the
situation is described in one frame of reference only, the
problem does not arise. Only when one tries to work out
mathematical descriptions of both frames together do the
descriptions begin to seem contradictory.

The math is often essential to work out what happens,
especially when a quantitative result is needed, but it can
also be counterproductive to understanding what happens.

Ooo! You brought to mind something from a college class.
Public speaking or some such. It was a demonstration of
persuasion, and it happened to work out such that the impact
on me must have been greater than on anyone else in the
class of about 20 students. We were told this story, as best
I can recall now:

A farmer bought a calf for $30. A few weeks later he sold it
for $40. A few weeks after that he bought it back for $50.
Then a few weeks later he sold it again for $60. A few weeks
later yet he bought it back again for $70, and then after a
few weeks more he sold it again for $80.

How much money did the farmer make or lose?

We were not told not to write anything down, or do any
calculations. (This was just before the first pocket calculators
came out.) That felt like cheating, though, so I didn't do it.
I didn't notice anyone else writing or doing calculations.

Students volunteered answers. I believe that five different
answers were put forward. I raised my hand and offered
mine: There were three iterations, and the farmer made
$10 each time, so I figured he made $30. Others suggested
other amounts ranging from $10 on up to maybe $80.
I wasn't the first to give an opinion; I may have been the last.

I think we assumed zero costs for feed. I don't recall that
being specified. Maybe the calf just ate grass.

We were asked to group together with whoever expressed
a money amount we agreed with. I remember that there were
five different answers because I clearly remember one group
in each corner of the room and one in the center. I think three
others joined the group I started. Two gals and a guy.

The sizes of the groups were remarkably uniform: About four
people in each of the five groups. I still think that is interesting.

We discussed the problem for a few minutes. The other guy
who joined my group said he was an accounting major. He
started explaining the math. He went through the process
and came up with a different number than I had. He went
through it again, and by the time the instructor told us to
quit, all four of us appeared to be convinced that my guess
was mistaken. He went to the chalkboard and tried to show
the whole class his analysis. But something went wrong.
The numbers didn't work out. Of course it turned out in the
end that -- to my astonishment -- my original rather reckless
guess was right, and the accounting major's math was wrong.
I was persuaded by a good-sounding argument in which the
math seemed to work, but actually didn't. My analysis was
correct, that the farmer made $10 on each transaction, and
over three transactions he made $30. But I was persuaded
that I had made a mistake. I am not proficient at math, even
simple arithmetic, and I knew it. I was not surprised to be
shown I was wrong. I was very surprised to learn that I was
(apparently) the only one in the class who had been right.

As an example of one way persuasion can work, that exercise
has to have impressed me more than anyone else.

-- Jeff, in Minneapolis

 

[jefferywinkler2]

Let’s say you were standing on the ground, and a wall was moving towards you, and you threw a rubber ball at the wall which bounced off. The wall moving towards you would transfer some of its momentum to the ball so the ball would come back towards you with higher velocity than when you threw it.

delta p_wall = delta_ball
p = mv
m_wall delta v_wall = m_ball delta v_ball

Both m_wall and m_ball are fixed so an increase in velocity of the ball results in a decrease in velocity of the wall.

Now if imagine you are on a moving train which is passing a wall perpendicular to the train. The wall is stationary with respect to the ground. As the train approaches the wall. you throw the ball out the window to the wall, and the ball bounces off the wall. Assume that the angle between the path of the ball and the track is negligible. From the reference frame of the train, it is identical to the earlier example, where you are motionless, and the wall is moving towards you.

delta p_wall = delta_ball
p = mv
m_wall delta v_wall = m_ball delta v_ball

Both m_wall and m_ball are fixed so an increase in velocity of the ball results in a decrease in velocity of the wall.

However, the wall is only moving from the reference frame of the train. In the reference frame of the ground, the wall is stationary, and the train is moving. In order for the wall to slow down in the reference frame of the train, the train would have to slow down in the reference frame of the ground.

Does the ball bouncing off the wall cause the train to slow down?

 

[Ibix]

Does the ball bouncing off the wall cause the train to slow down?

I think you are describing a scenario where you have a wall stationary beside the track and your train is approaching it. You lean out of the window and throw a ball at the wall and catch it.

Let's call momentum positive when it is directed towards the front of the train. There are three interactions.

First you throw the ball. Its momentum increases and the train's decreases.

Second the ball bounces off the wall. The ball's momentum decreases and the wall's increases.

Third you catch the ball. Its momentum increases and the train's decreases.

So yes, the train is slowed by this process, both when you throw and when you catch the ball. Note that I haven't specified the frame I'm working in, and I can get away with this because the momentum changes are the same in any one.

 

[A.T.]

Does the ball bouncing off the wall cause the train to slow down?

If you consider wall+earth as movable and ignore the train's propulsion, then the relative velocity between ground and train will decrease, when the ball hits the wall. But due to the mass ratio the wall+earth are usually considered immovable in a collision with a ball.

 

[A.T.]

So yes, the train is slowed by this process, both when you throw and when you catch the ball.

And when the ball bounces from the wall, the train's groundspeed also decreases, because the wall accelerates forward.

Note that I haven't specified the frame I'm working in, and I can get away with this because the momentum changes are the same in any one.

In any inertial one.