# Inertial reference frames vs. non-inertial reference frames

I have a problem concerning the motion in different frames of reference. This example confused me:

We've got a train which is moving and we declare the rail was our inertial frame of reference (with velocity $v=0$).

1. In the first case, the velocity of the train is constant. If we're on the train resting in one place and we throw a ball in the air straight up, it lands again in our hand (so we're in the inertial frame of reference of the with constant speed moving train). This is because as we throw the ball in the air we don't only give the ball a tangential speed (regarding the floor of the train), but also a parallel speed (which equals the speed of the train). That's why it lands in our hand again. Our friend how's outside sees as we throw the ball, but to them, the ball seems to move parabolically (they're in the inertial frame of reference of the rail). This is because the train moves with a velocity parallel to the rails and with the train the ball does. This velocity is going to be added to the velocity the ball has as it's thrown in the air.

This makes all sense to me, but the next example confused me:

1. Now our train has an acceleration. As we are in a state of rest on the train and throw the ball up, it accelerates (if a is the acceleration of the train and g is the gravitational acceleration, then the acceleration of the ball is the vectorial sum of the two accelerations). But to our friend, it moves linear (up and down). In the book that I read (Paul A.Tippler and Gene Mosca: Physik, 7.Auflage, Seite 154) it stated that this is because the motion of free fall is independent of the motion of the train.

But didn't we state in the first example the opposite? Didn't the ball land in our hand because the motion of the train gave the ball a velocity parallel to the ground of the train? Why is that not the case in the accelerated frame of reference?

Did you realize that in the case of the accelerating train you wouldn't catch the ball?

You would no longer be in an inertial frame (the train is accelerating) so your previous findings would no longer apply!

Didn't the ball land in our hand because the motion of the train gave the ball a velocity parallel to the ground of the train?

No, that is not exactly the reason. A frame of reference is a mathematical construct used to describe, it's a way to look at things. The physics are not concerned with frames of references, only laws of dynamics (Newton), and conservation of physical invariants (like mass, energy, momentum etc.). Note that the frame of reference is not mentioned in those laws.

Here is where your problem lies: The first example is correct, but the stated reason is wrong. You said (emphasis mine):

If we're on the train resting in one place and we throw a ball in the air straight up, it lands again in our hand (because we're in the inertial frame of reference of the train moving with constant speed)

THe frame of reference is for nothing in the analysis. In fact the actual reason the ball lands in our hand is that it has kept its momentum. The momentum is the velocity multiplied by mass (but we can forget about the mass, which doesn't change in this case). The horizontal component of the ball's momentum is the same as your hand's horizontal component - and that is the speed of the train $v_{train}$.

• In the case of the uniformly-moving train, the ball and your hands move at the same speed $v_{train}$ before you throw it, so after a time $\Delta t$ they are both $v_{train} \times \Delta t$ meters away and you can catch it again. Now you can describe this motion using the frames of reference you mentioned:
• In the frame attached to the train, that horizontal component is zero, so the parabola contracts to a line up-down.
• In the frame attached to your friend the horizontal component is non-zero (it's equal to $v_{train}$) so it's a true parabola.
• In the case of the accelerating train, the ball and our hands are both motionless before you throw it, so after a time $\Delta t$ the ball will not have moved horizontally ($0 \times \Delta t$), however you will have travelled with the train by $\frac{a_{train} \times {\Delta t}^2}{2}$ meters away and you won't be able to catch it. Now you can describe this motion using the frames of reference you mentioned:
• In the frame attached to the train which is no longer inertial, the horizontal component is zero initially. Then as the train accelerates, the ball picks up a negative horizontal speed due to the non-inertiality. When it falls back down of the parabola, it will have travelled $\frac{a_{train} \times {\Delta t}^2}{2}$ backwards from you, and you will not catch it.
• In the frame of reference of your friend which is inertial, things are much simpler: the ball starts with zero horizontal speed plus the vertical speed, and proceed with its free-fall. So your friend does see the ball move up-down vertically. It also sees you accelerating with the train and moving forward by $\frac{a_{train} \times {\Delta t}^2}{2}$ meters forward, so it also sees you missing the ball - by the same distance, so it all checks out.

Recap:

Don't base your computations of physical phenomenon on reference frames, but on elementary principles. Only use frames as a way to describe what happens under those principles.

Tip:

If changing of reference frame changes anything on what you are calculating - then somthing is wrong in your calculations!