Resolving the Three Colliding Observers Paradox in Special Relativity

[Meatlofe]

I have a random paradox I seem to have found with three colliding observers, though to explain it properly I'm going to use a bit of visual aid and a bit of abstracting it from the math (or setting the math so that it works according to my example).
For anyone who wants the TLDR, read this paragraph, for anyone who wants the precise details, read the next paragraph. We have 3 observers, A B and C that are some distance apart and moving towards each other. From B's point of view they all collide at the same time but from A's point of view they collide with B first, then C and then C hits B.
It starts with 3 observers, A, B and C, which are each in a line and 6 units away (86542628m according to B, if you care about the math). (A->B = B->C = 6 units) I will represent each unit as a "-"; So to start with our scene looks like this: (A------B------C). From B's frame of reference, A and B are traveling towards B at c/sqrt(3) or 173085256 m/s. This number is special because when you put it through relativistic velocity addition the final velocity is 1.5 the original. This example would work with any speed but is easiest to demonstrate like this. Any observer can provide the time steps as long as they are the same time steps used across each instance, so each time step is going to be how long it takes for the observed distance from A->B to change by 2 units (1 second according to either A or B). These time steps don't effect any math or events, they are just freeze-frames for us to observe the system at and compare them to each other.
Alright, so that's the groundwork. It's a little less complicated than the second paragraph makes it sound, a little more so than the first one does. So now I'm going to go through the example from the observer B's point of view. This is pretty straight forward.

0(A------B------C)

1(--A----B----C--)

2(----A--B--C----)

3(------ABC------)

It is perfectly symmetrical, A and C hone in on B and collide at the same time. That is what we should expect to happen from all frames of reference. So now let's look at the point of view from observer A. Observer A will see the space between them all as contracted and will think that time is going slower than observer B did, but Lorentz contraction is homogeneous and I have chosen my phrasing and units of measurement carefully so that different clocks won't effect the problem.
When adding velocities relativistically we use a different system of adding them. (Final V= (V1+V2)/(1+(V1*V2/c^2))) I'm sure everyone knows, but to be thorough; relativistic velocity addition will be less than V1+V2, especially for larger numbers. In my example relativistically adding the 2 velocities is 1.5 times the original, so C will travel 3 units of distance rather than 2 or the Galilean 4. Now let's see what happens.
0 (A------B------C)

1 (A----B-----C---)

2 (A--B----C------)

3 (AB---C---------)

4(B--AC------------)First A and B collide, then Observers A and C collide and then observers B and C eventually will collide on time step 6. Observer C will see the order in reverse. In special relativity it is not necessary for all observers to see the same order of events as other observers, but that only applies to things that are outside the cone if influence. We can find circumstances quite easily where a paradox occurs here, such as if they were 3 particles with known energy states, which one's scatter? Because of this it is clear that my proposed paradox does not have an effect in reality. I am looking for a potential resolution as to why.

I have had this problem for a while now and have realized some things and ruled some others out. On principle this problem feels similar to the twins paradox, but as no acceleration ever occurs so general relativity never comes into play. The most promising sounding solution I've been offered as of yet is that observer A on creation (whenever that was) should see them self as being in the past (or seeing the others as being in their future) so should actually have their position shifted backwards by 3 seconds/time-steps. Unfortunately this falls apart because if you double the distance you need 6 time steps, and double it again and you need 12...etc which would imply that space time knows prophetically when it's next interaction will be and holds the particle in place for that long (infinity for light), which we have not observed. This reveals that instead the speed or distance traveled by observer C in comparison to observer B (according to observer A) must be altered.

 

[Dale]

We have 3 observers, A B and C that are some distance apart and moving towards each other. From B's point of view they all collide at the same time but from A's point of view they collide with B first, then C and then C hits B.

This is not possible. Starting with this premise will undoubtedly lead to contradictions.

To do the math correctly just start with the scenario in B's frame, since that is easiest. Then apply the Lorentz transform to go to A's or C's frame. You will get a single collision event in both cases.

 

[Nugatory]

This problem will be be much easier to visualize if you start by drawing a spacetime diagram of the situation.

However, the key issue here may be that you've overlooked the relativity of simultaneity: if A and C are the same distance from B at the same time in the frame in which B is at rest, they are not in the frame in which A (and likewise for C) is at rest.

Try writing down the initial coordinates of A and C in a frame in which B is at rest at the origin, and then use the Lorentz transforms to find the coordinates of those events in the frame in which A is at rest. Look at the speeds of B and C relative to A, and it will be clear that all three objects come together at the same point in that frame as well as the frame in which B is at rest.

 

[Ibix]

You seem to have forgotten the relativity of simultaneity. As Dale says, start in B's frame and use the full Lorentz transforms instead of just the length contraction and time dilation formulae. They don't apply to this situation because there are no shared rest frames to work from.

 

[Meatlofe]

I do understand relativity of simultaneity. That is why this post was so long and carefully phrased. Relativity of simultaneity however only applies at distances greater than zero, because observer A interacts with object B from it's own perspective and thus can change object B's course of action so it never interacts with object C, even though the same goes for object C, we can rule out relativity of simultaneity as the cause.

As for the Lorentz transformations I am doing them now.

 

[Nugatory]

this problem feels similar to the twins paradox, but as no acceleration ever occurs so general relativity never comes into play.

It's a bit of a digression here, but general relativity is irrelevant to the twin paradox and in general to problems involving acceleration. GR is needed only in situations in which spacetime is not flat and the tidal effects caused by the curvature are significant. As long as the spacetime is flat, special relativity works just fine even when accelerations are involved; google for "Rindler coordinates" to see how.

(The misconception that SR only works for inertial motion and constant speeds comes about because that's all that most introductory treatments cover.)

 

[Jorrie]

I do understand relativity of simultaneity. That is why this post was so long and carefully phrased. Relativity of simultaneity however only applies at distances greater than zero, because observer A interacts with object B from it's own perspective and thus can change object B's course of action so it never interacts with object C, even though the same goes for object C, we can rule out relativity of simultaneity as the cause.

I think part of your problem is not taking into account that the collision is a single event in all three inertial frames. The event will have different spacetime coordinates in different frames, but that does not take away the fact that it is one event in all inertial frames.

 

[Nugatory]

I do understand relativity of simultaneity.

That may be, but you didn't consider it when you drew the two time-0 initial states in the two sequences of snapshots above. Because of the relativity of simultaneity, they can't both look the same... you'll see this when you do the Lorentz transformations.

 

[Janus]

When adding velocities relativistically we use a different system of adding them. (Final V= (V1+V2)/(1+(V1*V2/c^2))) I'm sure everyone knows, but to be thorough; relativistic velocity addition will be less than V1+V2, especially for larger numbers. In my example relativistically adding the 2 velocities is 1.5 times the original, so C will travel 3 units of distance rather than 2 or the Galilean 4. Now let's see what happens.
0 (A------B------C)

1 (A----B-----C---)

2 (A--B----C------)

3 (AB---C---------)

4(B--AC------------)

The problem here is that you are assuming that according to A, there is a moment( other than when they are together) when A, B and C are equally distant from each other. This never happens when the set up is such that B always measures and C to be equal distances from him. According to A, C will always be closer to B than he is to B.

A good way to examine this is to run the example backwards; A, B and C start at the same place and then separate.
As far as B is concerned, A and B travel at equal speeds away, and thus running backwards we can recreate the situation were A and C start at equal distances from B.

As far as A is concerned B heads of at a set speed, and C heads off in the same direction at at something less than twice that speed. Thus the distance between A and B grows at a faster rate than the distance between B and C. You can never recreate the situation shown in your line 0 above where A, B, and C are equally spaced from each other.
C sees the same thing, with A being closer to B than he is.

So your basic problem is assuming that the initial equal spacing between A, B and C as seen by B ever[i/] occurs according to A or C.

 

[Meatlofe]

Using Lorentz transformation I find Object A and B to be 9 units away by A's perspective and for A and B to be 48 units away. That means that now I have the problem extenuated. It takes A and B 4.5 seconds to collide (from A's perspective) and 16 seconds journey away from C (from A's perspective again) and it would take 48 seconds for B and C to collide.

 

[Nugatory]

Using Lorentz transformation I find Object A and B to be 9 units away by A's perspective and for A and B to be 48 units away. That means that now I have the problem extenuated. It takes A and B 4.5 seconds to collide (from A's perspective) and 16 seconds journey away from C (from A's perspective again) and it would take 48 seconds for B and C to collide.

Post your calculations so we can see where you went wrong.

 

[Meatlofe]

Turns out I forgot the square root, but the new answers still don't make sense; 7.35 and 24 where the farther one is more than double the closer. If I get the right answer A->C should be 1.5 * A->B

With A as location zero on a one dimensional line at time zero:
Location B = (6-173085256*0)/SQRT(1-(173085256*173085256)/(299792458*299792458))
Location C = (12-259627884.25*0)/SQRT(1-(259627884.25*259627884.25)/(299792458*299792458))

As copy pasted functionally to and from my spreadsheet.

 

[Ibix]

Work in more convenient units! E.g. decide that the distances are in light seconds and the times in seconds. Then c is one light second per second...

Am I right in thinking that you mean the things you've labelled "Location B" and "Location C" to be the starting positions of B and C as measured in the rest frame of A?

 

[Meatlofe]

Work in more convenient units! E.g. decide that the distances are in light seconds and the times in seconds. Then c is one light second per second...

Am I right in thinking that you mean the things you've labelled "Location B" and "Location C" to be the starting positions of B and C as measured in the rest frame of A?

Yes using the rest frame of A and the speeds calculated previously, and the reason I chose these numbers is because they worked well for the example I chose. When sculpting this question I did not anticipate them as being used for alternate equations. If you prefer they are equivalent to c/sqrt(3) or 1.5c/sqrt(3) and the last is of course c, but spreadsheet doesn't accept c as a valid input.

 

[Ibix]

OK. Then I have two questions:
1 - why are you using different $\gamma$ factors?
2 - have you done the time transformations too?

 

[Meatlofe]

γ

OK. Then I have two questions:
1 - why are you using different $\gamma$ factors?
2 - have you done the time transformations too?

So I am using different gamma functions because the different observers are traveling at different speeds/velocities (not that there's much difference in 1d) and I have done the time transformations but they come out to a negligable -1.22 and -6(yes that's using the distance in meters not units)

 

[Ibix]

But the velocity in the Lorentz transforms is the velocity of the frame you are transforming into. The velocity of an object doesn't enter into it - you are only transforming some coordinates, which happen to be where the object is at a given time.

You've got a time difference of around 1.2s (assuming your answers are correct - did you make the same mistake with the $\gamma$ factors?). How many meters does an object traveling at $c/\sqrt{3}$ travel in 1.2s? Do you think that's negligible compared to your 6m separation?

 

[Meatlofe]

But the velocity in the Lorentz transforms is the velocity of the frame you are transforming into. The velocity of an object doesn't enter into it - you are only transforming some coordinates, which happen to be where the object is at a given time.

You've got a time difference of around 1.2s (assuming your answers are correct - did you make the same mistake with the $\gamma$ factors?). How many meters does an object traveling at $c/\sqrt{3}$ travel in 1.2s? Do you think that's negligible compared to your 6m separation?

My separation is 6 *86542628m (and respectively 12 * 86542628m), which actually answers your question. That number is half the distance $c/\sqrt{3}$ in 1 s and one third the distance in 1.5 $c/\sqrt{3}$ which is why I used it. In the initial equation I edited out the larger number because it was was a linear scale with increase in distance, which caused the output answer to be in the previously used units (1 * 86542628m) rather than some horrible number of meters.

 

[Ibix]

This is why you should use more manageable units. And if you must work in SI units, at least use standard form and sensible precision (3 s.f.).

Let me make sure I've got this straight. In the rest frame of B, A and C are each a little over 10⁹m away from B, in opposite directions. They are both traveling towards B at $c/\sqrt{3}$, which is about 1.7x10⁸m/s. Is that right?

 

[Meatlofe]

This is why you should use more manageable units. And if you must work in SI units, at least use standard form and sensible precision (3 s.f.).

Let me make sure I've got this straight. In the rest frame of B, A and C are each a little over 10⁹m away from B, in opposite directions. They are both traveling towards B at $c/\sqrt{3}$, which is about 1.7x10⁸m/s. Is that right?

Yup. That seems about right.

 

[Ibix]

Then why are you using just 6 and 12 in your calculations in post #12? You're mixing units. You'll actually get away with it for these transformations, as a special case of an error not coming back to bite you, but the time transformations where you multiply x by $v/c^2$ will be wrong unless you've done something funky with the $c^2$ to fix it.

You need to use consistent units. Either SI all the way or your funny units.

Please post your calculations and values for the x' and t' coordinates of the start positions of A, B and C. Remember the $\gamma$ factor to use and consistent units.

 

[Nugatory]

OK, since I can't resist...

Let's measure all distances in light seconds and times in seconds, to get rid of those annoying factors of $c$ that clutter up the arithmetic.

Just for the sake of simplicity, let's take the the initial distance between B and the other two to be $1/\sqrt{3}$ light-seconds, exactly one second (in the frame in which B is at rest) before the three-way collision. Now we have the following events, with coordinates in the frame in which B is at rest at the origin and all labeled as $(x,t)$:
1: initial position of A: $(-1/\sqrt{3}, 0)$
2: three-way collision: $(0,1)$
3: initial position of C: $(1/\sqrt{3}, 0)$
4: initial position of B: $(0,0)$

Now transform the coordinates of these events into coordinates in a frame in which A is at rest and whose origin is the same as the frame above. In that frame the coordinates of these events are:
1: $(-1/\sqrt{2}, \gamma/3)$ where $\gamma=\sqrt{3/2}$
2: $(-1/\sqrt{2}, \gamma)$
3: $(1/\sqrt{2}, -\gamma/3)$
4: $(0,0)$
(Wise people will check these results, as I've made errors keying in Latex in the past)

Note the relativity of simultaneity at work: in this frame A and C are the same distance from B at different times, while in the first frame they were the same distance at the same time.

You can see that B is $1/\sqrt{2}$ light-seconds away from A at time $\gamma/3$ seconds before the collision; moving at speed $1/\sqrt{3}$ B gets there right on time. C is farther away but starts sooner and moves faster, so also arrives right on time.

 

[Meatlofe]

Then why are you using just 6 and 12 in your calculations in post #12? You're mixing units. You'll actually get away with it for these transformations, as a special case of an error not coming back to bite you, but the time transformations where you multiply x by $v/c^2$ will be wrong unless you've done something funky with the $c^2$ to fix it.

You need to use consistent units. Either SI all the way or your funny units.

Please post your calculations and values for the x' and t' coordinates of the start positions of A, B and C. Remember the $\gamma$ factor to use and consistent units.

I already explained why I used my funky units in that equation in post 18. It was for my output answer to work in my funky unit format as well, however for the t' equation the same method would not work because it was compared in a non-linear fashion to c (which I could have just converted into funky units but that wasn't the first solution I thought of). Additionally, I have posted my calculations for the locations of the starting x' coordinates for B and C according to A in post 12. The coordinates for positions that I was transforming them from were shown implicitly both by the same post and by the question its self, but for clarity I will display them here.
From Observer A's Perspective

From Observer B's Perspective; as told be original post
A Time:0s Location= -6 units or -5.2*10^8 meters away Speed = 2 units/s, c/sqrt(3) or 1.7 * 10^8m/s
B Time:0s Location = 0 Speed = 0
C Time:0s Location= 6 units or 5.2*10^8 meters away Speed = -2 units/s, -c/sqrt(3) or -1.7 * 10^8m/s

From observer A's perspective
A Time:0s Location = 0 Speed = 0
B Time:-1.22 Location = 7.35 units or 6.4 *10^8m, Speed = -2 units/s, -c/sqrt(3) or -1.7 * 10^8m/s
C Time:-6 Location = 24 units or 2*10^9m, Speed = -3 units per second, 1.5c/sqrt(3) or 2.6 * 10^8
This is the math for location in meters, similar to comment 12
Location B = (6*86542628-173085256*0)/SQRT(1-(173085256*173085256)/(299792458*299792458))
Location C = (12*86542628-259627884.25*0)/SQRT(1-(259627884.25*259627884.25)/(299792458*299792458))
This is the math for time in meters, newly posted here
Time B = (0-173085256*(6*86542628)/(299792458*299792458))/SQRT(1-(173085256*173085256)/(299792458*299792458))
Time C = (0-259627884*(12*86542628)/(299792458*299792458))/SQRT(1-(259627884*259627884)/(299792458*299792458))

C's perspective is A's with all instances of A and C swapped.
All of my work is out in the open and all units freely convertible.

 

[Ibix]

From Observer B's Perspective; as told be original post
A Time:0s Location= -6 units or -5.2*10^8 meters away Speed = 2 units/s, c/sqrt(3) or 1.7 * 10^8m/s
B Time:0s Location = 0 Speed = 0
C Time:0s Location= 6 units or 5.2*10^8 meters away Speed = -2 units/s, -c/sqrt(3) or -1.7 * 10^8m/s

I'm still confused about what you call "a unit". You agreed A was ~10⁹m away from B when I last asked; now you're saying A and C are that far apart.

Never mind - you gave the SI values, so I'll work with them.

A Time:0s Location = 0 Speed = 0

This is the location and time of A in A's frame at the start of the experiment. Agreed.

B Time:-1.22 Location = 7.35 units or 6.4 *10^8m, Speed = -2 units/s, -c/sqrt(3) or -1.7 * 10^8m/s

This is the location and time of B in A's frame at the start of the experiment. Agreed.

C Time:-6 Location = 24 units or 2*10^9m, Speed = -3 units per second, 1.5c/sqrt(3) or 2.6 * 10^8

This is presumably supposed to be the location of C in A's frame at the start of the experiment, but it is wrong. You're still using the wrong $\gamma$ factor - see the first paragraph of post #17. The same applies to your time calculation for C.

 

[Meatlofe]

OK, since I can't resist...

Let's measure all distances in light seconds and times in seconds, to get rid of those annoying factors of $c$ that clutter up the arithmetic.

Just for the sake of simplicity, let's take the the initial distance between B and the other two to be $1/\sqrt{3}$ light-seconds, exactly one second (in the frame in which B is at rest) before the three-way collision. Now we have the following events, with coordinates in the frame in which B is at rest at the origin and all labeled as $(x,t)$:
1: initial position of A: $(-1/\sqrt{3}, 0)$
2: three-way collision: $(0,1)$
3: initial position of C: $(1/\sqrt{3}, 0)$
4: initial position of B: $(0,0)$

Now transform the coordinates of these events into coordinates in a frame in which A is at rest and whose origin is the same as the frame above. In that frame the coordinates of these events are:
1: $(-1/\sqrt{2}, \gamma/3)$ where $\gamma=\sqrt{3/2}$
2: $(-1/\sqrt{2}, \gamma)$
3: $(1/\sqrt{2}, -\gamma/3)$
4: $(0,0)$
(Wise people will check these results, as I've made errors keying in Latex in the past)

Note the relativity of simultaneity at work: in this frame A and C are the same distance from B at different times, while in the first frame they were the same distance at the same time.

You can see that B is $1/\sqrt{2}$ light-seconds away from A at time $\gamma/3$ seconds before the collision; moving at speed $1/\sqrt{3}$ B gets there right on time. C is farther away but starts sooner and moves faster, so also arrives right on time.

To me this feels like you've solved the problem that I had been having by doing the math right. Correct me if I'm wrong, but I believe that you have correctly implemented Lorentz transformation, drawing a space-time graph that actually translates C forwards in time in order to cause it to arrive at the right time. However this translation of time effect is only caused by the objects speed and distance where if speed were lower then the time would be slower (I have the calculator for it hand built in front of me). Couldn't we just as easily see two slow moving particles collide to form this faster one which would break all the math and require the faster moving particle to appear before the event that caused it to be created? That's why I ruled this possibility out (perhaps prematurely) to begin with.

 

[Nugatory]

However this translation of time effect is only caused by the objects speed and distance where if speed were lower then the time would be slower (I have the calculator for it hand built in front of me).

Try it with different speeds... if the speeds are such that objects collide in one frame, then they will collide in all frames; if the speeds are such that they don't collide in one frame, then they don't collide in all frames. If you don't get that result, then one way or another you messed up the coordinate transformation.

Couldn't we just as easily see two slow moving particles collide to form this faster one which would break all the math and require the faster moving particle to appear before the event that caused it to be created?

Try drawing the spacetime diagram, and you'll quickly satisfy yourself that that can't happen.

Wherever you're learning your relativity from, you might want to give "Spacetime Physics" by Taylor and Wheeler a try.

 

[Janus]

To me this feels like you've solved the problem that I had been having by doing the math right. Correct me if I'm wrong, but I believe that you have correctly implemented Lorentz transformation, drawing a space-time graph that actually translates C forwards in time in order to cause it to arrive at the right time. However this translation of time effect is only caused by the objects speed and distance where if speed were lower then the time would be slower (I have the calculator for it hand built in front of me). Couldn't we just as easily see two slow moving particles collide to form this faster one which would break all the math and require the faster moving particle to appear before the event that caused it to be created? That's why I ruled this possibility out (perhaps prematurely) to begin with.

As per Nugatory's suggestion, here are space-time diagrams for the scenario. For better clarity I added two other references which represent points equal distances from B and at rest with respect to B. I also added clocks to B and these two reference points.

The first diagram is for B's frame

The dark blue line is B. The light blue lines are the reference points. The green line is A coming in from one direction and the red line is C coming in from the other direction. A and B pass the reference points when the clocks there and the clock at B read 0, and meet when all three read 5.

Same scenario from A's frame.

A still passes the reference point when the clock there reads 0, However, the clock at B and the other reference point do not read 0 at this moment. B's clock reads between 1 and 2 and the reference reads between 3 and 4. C passes its reference point when the clock thee reads 0, just like above, but according to A, this happened before A passes its own reference point. Again A B and C meet up when B's clock reads 5

In C's frame, things look like this

It is basically a mirror image of what happened according to A, as according to C, A passes its reference point before C passes its reference point.

 

[Dale]

This is one example where it is probably easier to do it analytically rather than numerically. If an object's position is $x=v t$ then it goes through the origin, regardless of $v$. If you use the Lorentz transform, then you will still get $x'=v' t'$ so it will still go through the origin. So if multiple objects collide at the origin in one frame then they collide in all frames. And with a simple translation this generalizes to a collision anywhere.

 

[sweet springs]

Hi. For B, A And C are equidistant anytime. But for A, B and C are equidistant in one moment. They are not equidistant in all the other time. Same for C. This difference assures that event A,B and C meet altogether is shared among the three.

(A------B------C) is, for B, anytime. Anytime is OK to start.
(A------B------C) is for A in very special moment to start in good arrangement of distance and speeds so that the three meet altogether.

Best.