Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a question that came up in a discussion with friends. If I throw a ball straight up in an enclosed train car moving with constant velocity, I believe the basic physics books say it will land in the same spot. But will it really? I think I can say that the answer is "not in the real world".

Trivially, a train car is never enclosed. Fresh air is being allowed into the carriage or the passengers would all die. Thus there's currents of air that would affect the ball, agreed? If we remove the passengers and have a trusty robot (who does not need oxygen) throw the ball up in a carriage that really is completely air-tight, I'm still not sure it will land in the same spot. I would imagine that there must still be air circulation. The train had to start from a stop. It's true the floor and the roof will drag the air right at the boundary along with it, but just as an open convertible car does not drag all the air in the world with it, I assume that the air in the middle of the car will not be dragged along at the same speed. The air in the middle will remain stationary with respect to earth and pile up at the back of the car. Then it will be forced along. I further imagine that this "pile of air" will try to redistribute itself uniformally. Won't all this set up currents? Will the air come to be completely still in the reference frame of the car? [I'm guessing the answer is yet] How long would this take?

Bonus question: I believe if I'm sitting in a convertible car and throw a ball straight up it will land back in my hand as long as I don't throw it too far up. At some point, I'll throw it to high and will lose the ball out the back of the car. What's the relevant equation covering this in a car travelling at X miles per hour in still air? Put another way, I'm trying to get a feel for how extensive the "boundary" layer of air around the car is and how it dissipates with distance.

share|cite|improve this question
You haven't defined "straight up" or "the same spot", and this matters a lot (in physics speak we say that you have not defined your coordinate system). The topic you're trying to get at here is called "Galilean relativity", and is a very important concept. The short version is that (once you have correctly defined your terms) both the passengers and pedestrians will make the same predictions. – dmckee Sep 23 '11 at 15:41
Straight up relative to the human (or robot) on train car*. Same spot is defined as paint drop (dried) on floor of car. Again, I'm asking what will happen in "real world" where we do have air. It's impossible that all the air is dragged along with the train right from the very start. Perhaps though, it settles after a very short time? Put another way, would you want to do this experiment as soon as the train had got to 50 mph and was travelling at constant velocity? Or would you want to wait some time? How much time? Thanks for the comments! – Dave Sep 23 '11 at 15:47
*I suppose defining staight up will have to be done with some sort of right angle set on ground. If I'm right, you can't just drop a ball to define straight up/down. At least not right away. – Dave Sep 23 '11 at 15:48
If the train is moving at constant velocity, the air in the air-tight car would not 'pile up at the back of the car.' If it was moving at constant acceleration, then yes, it would. But we are talking about constant velocity here. – Greg P Sep 23 '11 at 15:49
Greg, that's part of my point. In the "real world" the train had to get to a constant velocity. Thus it did accelerate and air (presumeably, correct me if I'm wrong) was not uniformly distributed at some point in time. I'm guessing this sets up currents that continue even after the train reaches it's final velocity. I'm asking how long it will take for things to settle down. Thanks for the comments. – Dave Sep 23 '11 at 16:03
up vote 7 down vote accepted

Yes, the ball would land in exactly the same spot, whether robot or person. The air does not remember the original speed, and new air coming in does not keep its velocity, but settles down with the co-moving air. The speed it has is determined by the fan blowing it in, not by the speed of the train.

The reason is that the train pushes the air just as it pushes everything else. The air transmits the push by a pressure force, and there is no significant airflow inside the car when you start and stop, even at huge acceleration. Nothing is different from a stationary train, except during acceleration. The effect of acceleration will create a small pressure gradient in the air, and a density gradient, but these are insignificant, because the acceleration is slow.

This is counterintuitive to many people, but it is absolutely 100% true in the real world. Aristotle also confused things with air, despite the fact that Aristothenes, Archimedes, and other ancient scientists believed in some sort of inertia principle.and this type of thing

share|cite|improve this answer
Thanks Ron. I believe the part about "everything settles down", but what's the basis for saying "except for a brief period at the beginning". First of all, what is brief? Milliseconds? Minutes, Hours? It seems reasonable that the air would continue to move until brought to a stop by drag from the surfaces (walls, floor) of the car and other air molecules. I guess I don't have a "gut" feel that this would happen super fast. Also, are you sure about "a pressure force to all the air". You wouldn't argue that the air in front of the train is affected by the train once you get away from train. – Dave Sep 23 '11 at 16:15
To expand on the last comment, are boundary conditions the way to think about this? Again, it's the air immediately around a train or car that is affected. Similarly, I'd suppose that air in middle of train is less affected by the moving train as it accelerates. – Dave Sep 23 '11 at 16:17
@Dave: the pressure in air transmits the force to all the air at the speed of sound, which is always much greater than any speed of the train. The acceleration at the beginning is smooth, so the air settles to its moving profile adiabatically, without producing sound waves (no flow in the middle). If you accelerated the train with an atomic explosion, then you would get pressure waves inside the train from the air in the middle bunching up in the back, but this is incredible acceleration. – Ron Maimon Sep 23 '11 at 16:48
@Dave: Your everyday experience can be a guide here. Wouldn't you notice if there we on-going significant winds blowing around a car or buss or airplane every time it started or stopped? Also, think about length and time scales here. If the air was moving fast enough to effect the ball (order of m/s) how long could it possible go in the confines of a railroad carriage? – dmckee Sep 23 '11 at 17:09
@dmckee: you are absolutely right, of course, but I looked up the viscosity, and the naive decay rate due to viscosity of a size 100 ft circulation in air is 10^8 seconds! This cannot be right. I suppose the reason winds in cars decay is due to boundary layer friction, but I haven't sorted it out. – Ron Maimon Sep 23 '11 at 18:46

No, the ball does not land in exactly the same spot, even in the absence of any air at all. However, in the real world, the effects in the train are probably too small to observe without a careful experimental setup.

dmckee's answer describes physics in an inertial reference frame. However, because the Earth is spinning, the ball experiences a velocity-dependent Coriolis force. This force comes from the ball's motion and the Earth's rotation, so its magnitude is on the order of $mv/T$, with $v$ the ball's velocity, $m$ its mass, and $T$ the time it takes Earth to rotate - one day. The exact magnitude depends on the angle between the train's velocity and Earth's axis. The force is perpendicular to the ball's velocity.

If this force pushes the ball to the side, off its trajectory, the distance it goes off should be on the order of $at^2$ with $a$ the acceleration and $t$ the time it's in the air. If it's in the air for about a second, the deflection is small because a second is small compared to a day. You'll get something around $10^{-5}v\textrm{s}$, so even if the train is going $100 m/s$ you still only get millimeters of deflection.

If the train is traveling at constant velocity, any oscillations induced a while ago by accelerating should equilibrate, and the air should be like normal still air (except again for Coriolis forces on the air). As Ron Maimon mentioned, this happens pretty fast. How fast? Try singing in the shower. When you hit a good resonance, suddenly stop singing and see how long the note keeps sounding. Maybe a tenth of a second. Physically, that's roughly the same mechanism, so the oscillations of air set up by the acceleration of the train will die out on a similar timescale, plus or minus an order of magnitude to adjust for the size of the car and the boundary conditions at the walls.

How about bulk motion of the air? Try turning on a fan, then suddenly stopping it with a stick? How long do you still feel wind? Maybe a little longer, on the order of seconds this time, but experience still informs us that this dies off quickly, too. I think Ron's point in comments that there essentially are no oscillations or currents set up by the train's acceleration is right. The time scale on which the air equilibrates is fast compared to the time over which the train accelerates, so the air is essentially always in equilibrium.

share|cite|improve this answer
Sound attenuation happens by different mechanism (heat flow between adjacent adiabatically compressed regions) than velocity attenuation for bulk air circulation, and there is no connection between the time scale. If you set up an air motion in the shower, it will circulate far longer than a sound wave will sound. Regarding Coriolis force, it will vanish when the train is moving east-west on the equator, and I believe the poster meant to ignore this effect. – Ron Maimon Sep 24 '11 at 5:42
@Ron Coriolis vanishes when you move north-south on the equator, not east-west. When you move east-west, it is parallel to gravity. I will think more about what you said about sound attenuation. – Mark Eichenlaub Sep 24 '11 at 5:55
Coriolis force won't be parallel to gravity--- it will push the ball in the direction of the Earth's rotation as it falls, that's a forward force. The balls motion is up-down, and the coriolis force is perpendicular to the motion and to the axis of rotation. I was totally wrong. You would have to go in any direction at the poles. But this effect is not the intent of the question--- the air circulation is. – Ron Maimon Sep 24 '11 at 6:17
@Ron I was calculating Coriolis due to the motion of the train, not due to the up/down motion. The question was "If I throw a ball straight up in an enclosed train car moving with constant velocity, I believe the basic physics books say it will land in the same spot. But will it really?". Coriolis is relevant to that. The OP didn't discuss it, most likely because he didn't know about it. I agree that there are different mechanisms for dissipating energy in the air, but since I was directly addressing vibrations, I'll leave the analogy to sound, while adding a note about bulk motion. – Mark Eichenlaub Sep 24 '11 at 15:07
Thanks Mark. I had neglected Coriolis for this problem. I was more concerned with how air transmits force as Ron pointed out. – Dave Sep 29 '11 at 17:54

I'll discuss two situation:

  • The experimenter is standing on a "X" in the enclosed railcar as it rolls along with velocity $v$, and throws just as she passes a "Y" painted on the ground outside the train. In this case she throws the ball "straight up" in her own frame.

    She predicts that the ball lands on the "X", and is not disappointed.

    A pedestrian watching from the verge also predicts that the ball lands on the "X", but in his frame of reference, this will be some distance $v\Delta t$ along the tracks (thus not on the "Y") when it does that.

  • The experimenter carefully (very carefully) throws the ball such that the pedestrian see it go "straight up". This will require that she throws it toward the back of the train.

    Unsurprisingly she predict that it will not land on the "X", but back along the train by a distance $v \Delta t$.

    The watcher predicts correctly that the ball will land beside the "Y", which it does.

In neither case do the very modest air currents in the car make an appreciable difference.

In the event that the railcar is open to the air, the we would need to make a fairly complicated calculation involving wind resistance, but it is still true that both observes will make the same predictions.

share|cite|improve this answer
I believe the question is not about Galilean relativity, but about how air transmits force. – Ron Maimon Sep 23 '11 at 16:04

Even without regarding air issues, I think in a 'real world' scenario the ball will never drop at exactly the same spot because of minor accelerations during the ride.

When a car / train moves, I don't think it ever moves at a completely steady velocity (in fact, although I can't explain it, I have a feeling it's physically impossible). There will always be minor accelerations of some sort.

share|cite|improve this answer
In the real world the uncertainty principle guarantees that the ball can't ever fall in the same position since the position can't even be measured exactly. – Brandon Enright May 30 '13 at 4:40

I think this thread got spammed and floated to the top, but it reminded me of a similar demonstrations about frames of velocity. Watch this video where a ball is fired from a moving vehicle yet if conditions are just right it falls straight down. Ball drops straight down when fired from cannon

share|cite|improve this answer

protected by Qmechanic Dec 3 '14 at 23:29

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.