- #1

- 8

- 0

## Main Question or Discussion Point

I have some issues understanding the following thought experiment:

Suppose you are standing still, and two balls are moving towards you from opposite direction. From your own reference frame, Ball A is ##10^5## m away from you, moving towards you from the left with speed ##0.8c##, and Ball B is ##10^5## m away from you, moving towards you from the right with speed ##0.8c##. And then (of course), from YOUR reference frame, both balls will hit you at the same time, specifically after

## time = distance/velocity = \frac{10^5 m}{0.8c} = 0.0004167 s = 416.7 μs ##

From the reference frame of Ball A, IT is standing still, and YOU are moving towards it with speed ##0.8c## from the right, and Ball B is moving towards it with speed:

##v_{B} = \frac{0.8c + 0.8c}{1 + (0.8c)^2/c^2} = \frac{1.6c}{1.64c} = 0.97c## .

This means that, from the reference frame of Ball A, YOU collide into Ball A after

##t = \frac{10^5 m}{0.8c} = 416.7 μs## (like we found above)

and Ball B collide into Ball A after

##t = \frac{2 \cdot 10^5 m}{0.97c} = 687.3 μs##,

which means that, from the reference frame of Ball A, the three objects (you, Ball A and Ball B) didn't collide at the same time. And both you AND Ball A would be correct. Is this the accurate way to measure it? Can Ball A (and Ball B for that matter) really claim that the collision didn't occur with all three objects colliding at the same time?

Thanks in advance!

Suppose you are standing still, and two balls are moving towards you from opposite direction. From your own reference frame, Ball A is ##10^5## m away from you, moving towards you from the left with speed ##0.8c##, and Ball B is ##10^5## m away from you, moving towards you from the right with speed ##0.8c##. And then (of course), from YOUR reference frame, both balls will hit you at the same time, specifically after

## time = distance/velocity = \frac{10^5 m}{0.8c} = 0.0004167 s = 416.7 μs ##

From the reference frame of Ball A, IT is standing still, and YOU are moving towards it with speed ##0.8c## from the right, and Ball B is moving towards it with speed:

##v_{B} = \frac{0.8c + 0.8c}{1 + (0.8c)^2/c^2} = \frac{1.6c}{1.64c} = 0.97c## .

This means that, from the reference frame of Ball A, YOU collide into Ball A after

##t = \frac{10^5 m}{0.8c} = 416.7 μs## (like we found above)

and Ball B collide into Ball A after

##t = \frac{2 \cdot 10^5 m}{0.97c} = 687.3 μs##,

which means that, from the reference frame of Ball A, the three objects (you, Ball A and Ball B) didn't collide at the same time. And both you AND Ball A would be correct. Is this the accurate way to measure it? Can Ball A (and Ball B for that matter) really claim that the collision didn't occur with all three objects colliding at the same time?

Thanks in advance!