Velocity addition and relativity of simultaneity

  • #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!
 

Answers and Replies

  • #2
PeterDonis
Mentor
Insights Author
2019 Award
29,616
8,897
You are only transforming the time coordinates of events, not the space coordinates. To do this analysis properly, you need to define both ##x## and ##t## coordinates for all events of interest (there would be four: you at time ##t = 0## in your rest frame, Ball A at time ##t = 0## in your rest frame, Ball B at time ##t = 0## in your rest frame, and the event where all three of you intersect, at time ##t = 416.7 \mu s## in your rest frame).

Then, to find out how things look in the other frames, you need to Lorentz transform the coordinates for all four events. You should find that, in the other frames, you and the two Balls start out at different times (relativity of simultaneity), but still all meet at the same event (because Lorentz transforming one event in your rest frame gives one event in the other frames).
 
  • #3
robphy
Science Advisor
Homework Helper
Insights Author
Gold Member
5,627
885
Here's a diagram I made with my GeoGebra tool: https://www.geogebra.org/m/HYD7hB9v# .
It's an ordinary Minkowski diagram with light-clock diamonds supplementing worldlines (and other segments on a spacetime diagram)
(See my Insight for details: https://www.physicsforums.com/insights/relativity-rotated-graph-paper/ )

I then marked it up with Paint....
(I might have switched A and B... but you can resolve that.)

Let M be the collision event.
For simplicity, let the distance be PQ="8 light seconds" (instead of 10^5 meters) in your frame.
Then the time to collision is PM=##\frac{8\rm {l.s.}}{.8c}=10\rm\ s##. So, PQ/PM=0.8c.

In A's frame, by symmetry, we have P'Q'= -8 and P'M'=10... so P'Q'/P'M'=-0.8c.
(Note PQM and P'Q'M are similar triangles in Minkowski spacetime geometry.)

At that instant in A's frame where you are 8 light-seconds away, where is Ball B according to A?
Locate event Z as the intersection of Ball B's worldline and A's x-axis (which is parallel to the spacelike diagonal of A's diamond).
Count the light-clock diamonds along A's negative x-axis....
graphically, P'Z is about 9.75... light-seconds [not 2*8=16 light-seconds],
in accord with the relative speed of Ball B being about .975c.


upload_2017-3-9_17-55-20.png
 
Last edited:
  • #4
446
159
This means that, from the reference frame of Ball A ... Ball B collides into Ball A after

##t = \frac{2 \cdot 10^5 m}{0.97c} = 687.3 μs##
I believe your mistake is here.

From YOUR perspective in the middle, Ball A and Ball B are both ##10^5## meters away from you at the same time. Indeed, at any given time, the distance between YOU and Ball A is equal to the distance between YOU and Ball B.

But that's not true from Ball A's perspective. If Ball A says that YOU are ##10^5## meters away, then Ball A does not say that Ball B is ##2 * 10^5## meters away at that moment.

To figure out how far away Ball B was from Ball A (from Ball A's perspective) at precisely the moment when YOU were ##10^5## meters away, make use of the fact that Ball A, Ball B, and YOU must all meet at a single event in spacetime when you collide. Simply dial back Ball A's clock ##416.7## microseconds, to when YOU were ##10^5## meters away from Ball A, and use Ball B's speed (relative to Ball A) to find the distance ##\Delta x## that Ball B traveled during that time:

##\Delta x = v \Delta t = \left( \dfrac{1.6c}{1.64} \right) \left( \dfrac{10^5 m}{0.8c} \right) = \dfrac{50}{41} \cdot 10^5 m##

So from Ball A's perspective, when YOU are 100,000 meters away, Ball B is only about 122,000 meters away.
 

Related Threads on Velocity addition and relativity of simultaneity

Replies
19
Views
2K
Replies
13
Views
243
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
8
Views
2K
Replies
2
Views
1K
Replies
4
Views
2K
Replies
60
Views
4K
Replies
16
Views
2K
Replies
3
Views
6K
Replies
29
Views
2K
Top