Reference Frames in Simultaneity

[am2010]

Lets say that a person in a moving train throws a ball in the direction of motion from one end and hits the other end of the train at 10 mph (relative to the person on the train). According to special relativity, from the perspective of a platform observer though they would see the ball having to take marginally longer to reach the other end of the train thus according to the platform's frame, time is different.

However for objects that are traveling at 10mph wouldn't it be the case that (from the platform observer's frame) since the train is moving as well the velocities of the ball is NOT 10mph but instead 10 mph + the train's velocity (Law of Relative Motion). Wouldn't this addition in speed then compensate for the slowing down of time so that the platform observer would see the ball reaching the other end of the train at the same time as the person in the train.

 

[Doc Al]

However for objects that are traveling at 10mph wouldn't it be the case that (from the platform observer's frame) since the train is moving as well the velocities of the ball is NOT 10mph but instead 10 mph + the train's velocity (Law of Relative Motion). Wouldn't this addition in speed then compensate for the slowing down of time so that the platform observer would see the ball reaching the other end of the train at the same time as the person in the train.

You are correct that with respect to the platform, the speed of the ball is not 10 mph but faster due to the speed of the train itself. But it's not simply 10 mph + the train's velocity. One must add velocities using the relativistic law of velocity addition, which looks like this:

$$V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2}$$

Instead of this (which is only good for low speeds):

$$V_{a/c} = V_{a/b} + V_{b/c}$$

There are other factors to consider, such as length contraction: With respect to the platform, the train is shorter.

 

[am2010]

You are correct that with respect to the platform, the speed of the ball is not 10 mph but faster due to the speed of the train itself. But it's not simply 10 mph + the train's velocity. One must add velocities using the relativistic law of velocity addition, which looks like this:

$$V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2}$$

Instead of this (which is only good for low speeds):

$$V_{a/c} = V_{a/b} + V_{b/c}$$

There are other factors to consider, such as length contraction: With respect to the platform, the train is shorter.

Since we're dealing with low speeds (ball at 10 mph and the train at an even lower speed) then I would use the bottom equation. Since we add the velocities using this relativistic law, isn't it correct that the platform observer sees the ball hitting the other end of the train first BEFORE the person in the train. Thus time is speeding up for the platform observer?

 

[Doc Al]

Since we're dealing with low speeds (ball at 10 mph and the train at an even lower speed) then I would use the bottom equation. Since we add the velocities using this relativistic law, isn't it correct that the platform observer sees the ball hitting the other end of the train first BEFORE the person in the train. Thus time is speeding up for the platform observer?

I don't understand your point. If you are only talking about low, non-relativistic speeds, then there's no time dilation or length contraction to worry about. Time is not slowing down (or speeding up) for anyone. All observers will say that the ball takes the same amount of time to travel from one end of the train to the other.

What do you mean when you say that the "platform observer sees the ball hitting the other end of the train first BEFORE the person in the train"?

There's no problem in having everyone's clock synchronized. If the person on the train says the ball hit the end of the train at 1:00 pm, the platform observer will agree.