# The Collision Clock Experiment

1. Jul 27, 2010

### calebhoilday

This is a thought experiment that best depicts my issue in understanding special relativity, particularly the congruency of the velocity addition and velocity subtraction formula with other aspects of the theory.

This experiment consists of two clocks; clock A and clock B.

Clock A consists of a cannon that fires a projectile every second at a wall. The distance between the wall and the clock is (0.75)^0.5 light-seconds. The velocity of a projectile is (0.75)^0.5 C. These parameters result in a collision between a projectile and the wall every second in the observing frame of reference.

Clock B consists of two cannons that fire a projectile every second at each other. The distance between the cannons is 2x(0.75)^0.5 light-seconds. The velocity of each projectile is (0.75)^0.5 C. These parameters result in a collision between projectiles every second.

The purpose of this experiment is to convert from the frame of an observer to that of a projectile and determine the duration between event 1 firing and event 2 collision.

I am lead to believe that duration in a moving frame, divided by the duration in a stationary frame is equal to (1- (V^2/C^2))^0.5 . If this is the case, a projectile from either clock can be expected to consider that 0.5 seconds pass between event 1 and event 2 as time is dilated for the projectile so that for every one second on an observer’s clock, 0.5 seconds pass on a projectile’s clock based on the velocity the projectiles have.

The process of determining the duration between event 1 and event 2 in the projectiles frame of reference consists of determining:
(1) The distance between projectile and object one that the projectile will collide with in the frame of the projectile.
(2) The velocity of the object that the projectile will collide with in the frame of the projectile.
(3) The duration between event 1 and event 2 in the projectiles frame of reference, by dividing the distance as determined in (1) by the velocity as determined in (2)

Clock A
(1)
D =Length in observers frame/ (length of projectile with velocity/ length of projectile when stationary) = 2(0.75)^0.5 light-seconds

(2)
U = (S-V)/(1-(SV/C^2)
= (0.75)^0.5

(3)
T=D/U
= 2(0.75)^0.5/(0.75)^0.5
= 2 seconds

Clock B

(1)
D =Length in observers frame/ (length of projectile with velocity/ length of projectile when stationary) = 4(0.75)^0.5 light-seconds

(2)
U = (S-V)/(1-(SV/C^2)
= 2(0.75)^0.5/(1+0.75)
=(8/7)*(0.75)^0.5

(3)
T=D/U
= 4(0.75)^0.5/(8/7)*(0.75)^0.5
= 3.5 seconds

As can be seen Clock A and Clock B produce different results for the duration between event 1 and event 2 in the projectiles frame of reference, when in both the duration for the observer is 1 second. Neither of them produced a result that was consistent with the formula derived duration of 0.5 seconds.
????

2. Jul 27, 2010

### starthaus

This is wrong, the correct ratio is the inverse of what you "are lead to believe"

3. Jul 27, 2010

### calebhoilday

If what you are saying starthaus is correct, then everything is good for Clock A. My only problem is that i always considered that it was the moving observe that dilated or considered less time to pass, what your saying is the reverse.

4. Jul 28, 2010

### Austin0

Hi ca;ebholiday

I am somewhat confused regarding your parameters here. Comparing A to B

If I am understanding just B then with the two opposite cannons it seems that the

Relative velocity would be .989c But this provides no information regarding the distance between the two frames. The only available information is relative to the rest frame i.e. .866 ls between the two cannons as calculated in cannonball frame.
SO if you use the v relative to the rest frame you get the t'=.5 s
If you use the relative v between cannonballs you get t'=.4378 s
If you do use the .989c gamma as applied to the distance you get t' = .1295 s but I dont think this has a reasonable basis because self evidently the v wrt the rest frame is not .989c

SO I am unsure about your projectile length and dont get the 3.5 s you arrived at????

Or maybe I am just not understanding

5. Jul 28, 2010

### calebhoilday

I’ve seen it depicted before that if one was travelling with velocity that outside objects appear expanded. Iv gone about it this way, to be consistent with literature, even though it is probably wrong. If you inverse my length manipulations then you get a result that is consistent with the formula for clock A, but not for clock B. The Problem still remains.

6. Jul 28, 2010

### Staff: Mentor

You must take great care in analyzing these "clocks" since event 1 and event 2 take place at different locations. You can't simply apply the time dilation formula; these are not simple clocks. (A simple clock is one where in its own frame the 'ticks' take place at a single location.)

This is a misleading way of putting it. A clearer statement might be: An observer in a 'stationary' frame will observe a moving clock to run slow by a factor of gamma = 1/(1- (V^2/C^2))^0.5. This time dilation formula applies to measurements of moving clocks--normal 'clocks' that have a single location in the moving frame--not to generic 'durations'. To analyze how 'durations' between spatially separated events will transform between frames, you must also consider length contraction and the relativity of simultaneity. (All of which is contained in the Lorentz transformations.)

I suggest you start over.

7. Jul 28, 2010

### calebhoilday

Im guessing it is a relativity of simultaneity issue.

Event 1 and event 2 may not be happening in the same location, but event 1 for clock A and clock B happen in the same position and event 2 happens in the same position for both clocks also.

This idea of a simple clock doesn't make any sense to me. Time is defined by motion and requires changes in location that are consistant to be measured.

Should I change "D =Length in observers frame/ (length of projectile with velocity/ length of projectile when stationary)" to D =Length in observers frame * (length of projectile with velocity/ length of projectile when stationary) ???

8. Jul 28, 2010

### JesseM

True in practice, but in relativity an ideal clock is basically what you get if you take a light clock and consider the limit as the distance between mirrors approaches zero.

9. Jul 28, 2010

### Staff: Mentor

Yes it is.
Stop calling them 'clocks'; you'll just continue to confuse matters.

It might be helpful to imagine actual clocks co-located with the two events. In the rest frame of the cannons, those clocks are synchronized. In the frame of the moving cannonball, they are not.

You don't have a wristwatch or a wall clock? That's the sort of clock I'm talking about.

It takes one second for the cannonball to travel from cannon to target (in the frame of the cannons). Whether there's a collision every second depends on the rate at which you fire the cannon. Fire it every 10 seconds and the collisions will be 10 seconds apart. (Again, in the frame of the cannons.)

10. Jul 28, 2010

### calebhoilday

Sorry, I left out that the cannons fire every second, but I think it really doesn’t matter a whole lot.

A Collision clock is the same as any other clock that can exists. A wrist watch isn't a simple clock, it relies on something to move, even if it is just electrons over very small distances.

If considering relativity of simultaneity, the difference between events from one frame to another is dependent on the relative velocity of these frames and nothing else according to the formula. All the projectiles have the same velocity, compared to the cannons or the observer.

The velocity addition formula is meant to be the outcome of combining length contraction, time dilation and relativity of simultaneity. I just don't get how it is and this is the best way I can depict why.

11. Jul 28, 2010

### Staff: Mentor

If all you want to do is figure out the time it takes for the cannonball to go from cannon to target in the frame of the cannonball, you can just use the time dilation formula. From the view of the cannon frame, the time as measured on a clock moving with the cannonball is dilated by a factor of 2. So if the cannon frame measures the travel time as 1 second, the cannonball will measure the trip to take 1/2 second.

You can also get this by taking the distance the cannonball frame measures between cannon and target (use length contraction) and dividing it by the speed (which is given). You'll get the same answer, of course.

12. Jul 28, 2010

### JesseM

No, there is only contraction, in every frame. If my ship is moving at relativistic speed relative to your ship, then in my rest frame your ship is shrunk relative to mine, and in your rest frame my ship is shrunk relative to yours. All inertial frames are equally valid, and the laws of physics work the same way in all of them.

In your example of clock A, in the frame of the cannon the projectile moves at 0.866c and the distance from the cannon to the wall is 0.866 light-seconds, so the time is one second. In the frame of the projectile, the distance between the cannon and the wall is shrunk by a factor of $$\sqrt{1 - 0.866^2}$$ = 0.5, so the distance is only 0.5*0.866 = 0.433 light-seconds. The projectile is at rest while the wall is moving towards the projectile at 0.866c, so the time for the projectile to reach the wall is distance/speed = 0.433/0.866 = 0.5 seconds.

For clock B, we have to consider the relativity of simultaneity. If two events are simultaneous and a distance x apart in one frame, then in a frame moving at v relative to that first frame, the time between the events is gamma*(vx/c^2), with gamma=1/sqrt(1 - v^2/c^2). In your example the two cannons fire simultaneously and a distance of 2*0.866 = 1.732 light-seconds apart in their rest frame, and the left projectile's frame is moving at 0.866c relative to the cannon frame (giving us gamma=2), so the right cannon fired 2*(0.866c*1.732) = 3 seconds earlier than the left cannon. The distance between the cannons at any moment is 0.5*1.732 = 0.866 light-seconds in the frame of the left projectile--so if we define x=0 as the position of the left projectile, then at the moment the left cannon fires the right cannon is at x=0.866 light seconds, and since the right cannon is moving towards the left projectile at 0.866c in this frame, that must mean that 3 seconds earlier (when the right cannon fired) the right cannon was at position x=0.866 + 3*0.866 = 3.464 light-seconds.

According to the velocity addition formula, the right projectile was traveling at (0.866c + 0.866c)/(1 + 0.866*0.866) = 1.732c/1.75 = 0.9897c. So, if the right projectile was fired at a distance of 3.464 light-seconds from x=0, it reaches x=0 after an interval of 3.464/0.9897 = 3.5 seconds. Since the right cannon fired 3 seconds before the left cannon in this frame, it must hit the left projectile 0.5 seconds after the left cannon was fired in this frame.

So, when you correctly factor into account length contraction, the relativity of simultaneity, and relativistic velocity addition, you do find that the time to collision in the projectile frame is 0.5 seconds for both clocks A and B.

13. Jul 28, 2010

### calebhoilday

Here are the true results without relativity of simultaneity factored in.
Clock A
(1)
D =Length in observers frame*(length of projectile with velocity/ length of projectile when stationary) = 0.5(0.75)^0.5 light-seconds

(2)
U = (S-V)/(1-(SV/C^2)
= (0.75)^0.5

(3)
T=D/U
= 0.5(0.75)^0.5/(0.75)^0.5
= 0.5 seconds

Clock B

(1)
D =Length in observers frame* (length of projectile with velocity/ length of projectile when stationary) = (0.75)^0.5 light-seconds

(2)
U = (S-V)/(1-(SV/C^2)
= 2(0.75)^0.5/(1+0.75)
=(8/7)*(0.75)^0.5

(3)
T=D/U
= (0.75)^0.5/(8/7)*(0.75)^0.5
= 0.875 seconds

How do you get 3.5 seconds as clock B time before factoring in relativity of simultaneity?

14. Jul 29, 2010

### JesseM

I didn't, that figure took into account the relativity of simultaneity which says that in the frame of the left projectile, the right cannon fired 3 seconds earlier than the left one.

15. Jul 29, 2010

### calebhoilday

In in my calculations my length is 0.866 light seconds not 3.464 light seconds. Could you please explain jesseM. sorry im having difficulty following your formatting.

16. Jul 29, 2010

### Austin0

Hi JesseM Working off your figures above, leads to some interesting questions.

If we posit the projectiles being clocks that set to 0 on leaving the cannon:

The above simultaneity result, of the right cannon firing 3 s before the left presents some pregnant implications. The situation is symmetrical so the same applies to the right cannonclock i.e. the left cannon fired 3 s earlier, yes??

We can assume the cannon frame observers would see the clocks emerging with proper time =0 so this can be taken as hypothetical empirical reality ,that all frames would agree upon.
Likewise a cannon frame observer proximate to collision would see .5 s on both clocks.
So the idea that from the frame of the left cannonclock the right clock passed 3 s of proper time before impact would appear to not be consistent with observed reality.
It could only be that it passed 3 s of the left frames coordinate time, yes?

This would seem to infer that although both cannonclock frames would agree that proximate observers in their respective frames would be 3 s out of synch with the opposite cannonclock, that would only be possible if it was actually their clocks that were out of synch.
I.e. If it is established that both cannonclocks actually, by observation , read 0 at firing then it makes no sense to say an opposite observer see's it reading -3 at that point.
There is no point where it reads -3 .

This leads to some interesting speculation wrt the reality and meaning of simultaneity.

I.e. Is it something that occurs basically within a frame perhaps as a result of the synch convention or does it have actual temporal meaning as some seem to think.

There are other questions here but I d like your take on this. SOmething I am missing maybe?
Thanks
BTW did your "how to" book help with Heidigger , hermeneutically speaking?

17. Jul 29, 2010

### JesseM

Remember that in the frame where the left projectile is at rest, both the left and right cannons are moving at 0.866c. At the moment the left projectile is fired from the left cannon, the right cannon is indeed 0.866 light-seconds away from the left projectile. But 3 seconds earlier, the motion of the right cannon means that it was an additional distance of 3*0.866c from the position where the left projectile would later come to rest in this frame, which I defined as being position x=0. So, the total distance of the right cannon from position x=0 at the moment the right cannon fired must be 0.866 light seconds + 3*0.866c = 3.464 light-seconds.

18. Jul 29, 2010

### JesseM

In the frame where the right projectile is at rest after being fired, yes.
If you want to assume that each projectile has an onboard clock which is set to T=0 at the moment it is fired from the cannon, that's fine.
All frames would agree on the time the two onboard clocks show at the moment they collide, since this each clock's reading will be at the same local point in spacetime when they collide, and different frames never disagree about local events.
Proper time is frame-invariant, all frames agree that the two projectiles experienced 0.5 seconds of proper time between being fired and colliding. And if you're talking about the time between the right cannon firing and the impact, in the frame where the left projectile is at rest (after being fired, it's moving along with the cannon beforehand) the coordinate time is 3.5 seconds, not 3 seconds. 3 seconds was the coordinate time between the right cannon firing and the left cannon firing.
3 seconds is coordinate time, yes.
Each frame uses a system of clocks and rulers at rest in that frame (with the clocks synchronized in that frame according to the Einstein synchronization convention) to define the position and time coordinates of each event. So, imagine a ruler moving inertially in such a way that the left projectile is at rest relative to the ruler between being fired and colliding with the right projectile, and also imagine that we have a series of clocks at rest relative to the ruler, attached to each ruler-marking. Then when I say the left cannon is fired at x=0 light-seconds and t=0 seconds in this frame, I mean that the event of the left cannon firing occurs right next to the x=0 light-second mark on the ruler, and the clock at that mark reads t=0 seconds when it fires. And when I say that the right cannon fires at x=3.464 light-seconds, t=-3 seconds, I mean that the event of the right cannon firing occurs right next to the x=3.464 light-second mark on this ruler, and the clock sitting at that mark reads t=-3 seconds when this happens.

For an illustration of this sort of thing, you could check out the pictures and description I gave of two ruler/clock systems moving alongside each other in the OP of this thread...
I don't remember mentioning Heidegger on these forums, are you thinking of another poster? Or maybe you saw a Heidegger book on my librarything page linked on my website?

19. Jul 29, 2010

### JesseM

Incidentally, note that it actually is possible to get this result without referring to the relativity of simultaneity, as long as we know that the two projectiles hit each other at the exact midpoint of the two cannons. Instead of having x=0, t=0 be the coordinates of the left cannon firing as I assumed earlier, let's redefine x=0, t=0 to be the coordinates of the right projectile hitting the left projectile, in the frame where the left projectile was at rest. If the two cannons are 0.866 light-seconds apart in this frame, and the two projectiles crash at the midpoint, then in this frame the left cannon must be at x=-0.433 light-seconds at t=0, while the right cannon must be at x=0.433 light-seconds at t=0. And we know in this frame the right cannon is moving towards the left projectile at 0.866c, so the right cannon's position as a function of time must be:

x(t) = 0.433 - 0.866*t

Meanwhile, because of relativistic velocity addition, the right projectile must be moving towards the left projectile at 0.9897c. And if the right projectile reaches x=0 at t=0, its position as a function of time must be:

x(t) = -0.9897*t

So, to figure out when the right projectile emerged from the right cannon, just set those two equal:

0.433 - 0.866*t = -0.9897*t

0.433 = -0.1237*t

Dividing both sides by -0.1237:

t = -3.5

So, the right projectile must have been shot from the right cannon 3.5 seconds before the two projectiles collided in this frame.

Meanwhile, the left projectile is at rest at x=0 in this frame, so for the left projectile x(t)=0. And the left cannon was at x=-0.433 at t=0, and it's moving in the -x direction at 0.866c, so its position as a function of time is:

x(t) = -0.433 - 0.866*t

So setting x(t) for the left projectile equal to x(t) for the left cannon to find out when the left projectile was fired gives:

0 = -0.433 - 0.866*t

0.433 = -0.866*t

Dividing both sides by -0.866:

t = -0.5

So, the left projectile must have been shot from the left cannon 0.5 seconds before the two projectiles collided.

20. Jul 31, 2010

### calebhoilday

it kinda makes sense at first glance, ill have to run through the math. You have to remember that if the cannon on the right was fired before, that the velocity of the projectile from the left is 0 and will have to consider the firing of both cannons as simultaneous. As the cannon on the right couldn't be fired on the right until the left and once fired the right had already been fired, you deal with the projectile on the left never witnessing the right cannon firing. Not sure if this is an issue or not.