# A thought experiment paradox

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1. Oct 6, 2015

### TheScienceGuy

I am pretty familiar with the theories of special and general relativity. I know how to add velocities, I know the main postulates and the experimental confirmation. However, I thought of the following thought experiment:
There are 3 experimenters (at the same point in space), who adjust their clocks at time point 0.
Two of them get into their spaceships and travel in opposite directions at a constant speed close to the speed of light (for example 0.75c). Once they travel a certain distance measured by their clocks (both travel with a constant speed of 0.75c) they launch probes traveling back to the third experimenter - he is now exactly in the middle. Lets assume that each probe travels again with a speed of 0.75c towards the third experimenter left in the middle. Lets assume that at a given time he can observe only one of the probes traveling towards him (as if the other probe doesn't exist at all). Then he would measure that this one probe approaches him with 0.75c. Then, he does the same measurement with the other probe and finds out the same - that it approaches him with a speed of 0.75c. Why the hell then the two probes will approach each other with a speed of 288010km/s instead of 450000km/s? This system could be regarded either as two separate systems (experimenter 3 + probe 1 and experimenter 3 + probe 2) or as one system (experimenter + probes 1 and 2). Note that this experiment doesn't change the reference frame. It also takes care of the simultaneity and is completely symmetrical. Why the probes don't approach each other with 1.5c?
Further speculation: If the third experimenter stays where he is, he may get killed (and thus live a shorter life) by the two probes approaching him with a speed of 225000km/s (0.75c) when measured individually, compared to the case in which each of the probes approaches him with a half of their combined speed 288010/2=144005 (according to the theory of relativity).
Please help with a reasonable explanation. This for me is a paradox that I can't resolve.

Last edited: Oct 6, 2015
2. Oct 6, 2015

### DaveC426913

No observer (1, 2, or 3) observes any object moving at greater than c.

There's nothing wrong with observer 3 seeing two objects, each in motion, having a closing velocity greater than c. Nothing in SR forbids this.

3. Oct 6, 2015

### phinds

You are confusing reference frames and the "approach/separation" velocity of two bodies as seen by a third, as opposed to how they see each other. They each see the other approaching a more than .75c but less than c.

As I said, it's not a paradox at all, you simply have conflated things that cannot be conflated.

As I said, it's not a paradox at all, you simply have conflated things that cannot be conflated.

EDIT: just to be sure I'm clear here, what you are doing is taking the middle observers frame of reference and saying that it is valid for each of the approaching/receding objects. It is not.

4. Oct 6, 2015

### TheScienceGuy

Thanks for the answer, but I don't really see the rationale behind it. This is my basic assumption - experimenter 3 sees them each traveling with a speed of 0.75c. Their closing velocity should obviously be greater than c. When does the velocity addition rule kicks in? When would (according to which rule) the experimenter eventually die (assuming a collision with the two probes) ?

5. Oct 6, 2015

### phinds

Yes, their closing velocity IS greater than c, but so what. This is not proper motion. Each probe will reach him based on its velocity of .75c and the distance from him when it was launched. This has nothing to do with how fast the to objects see each other, which has nothing to do with what he sees.

EDIT: let's get more concrete: The three start out at the origin and one spaceship moves off in the -x direction and the other in the +x direction, both going at .75c. The guy left in the middle sees them each moving away from him at .75c and therefor moving away from each other at 1.5c. This has nothing to do with their proper motion relative to each other because it does not look at one of them from the other's frame of reference, which is how you get proper motion.

I say again, you are confusing frames of reference and conflating things that cannot be conflated.

6. Oct 6, 2015

### Staff: Mentor

In the reference frame of 3 they do approach each other at 1.5 c.

7. Oct 6, 2015

### Staff: Mentor

That's a speed of .75c relative to what? If you mean relative to Observer 3 in the middle (as the rest of your post suggests) then it would be a good exercise to calculate what the probe's speed is relative to the observer whose outgoing ship launches that probe.

You can start even further back in the problem, with the two experimenters heading in opposite directions at .75c. There's no need to further complicate things with the two probes until you have that first part down.

Observer three finds Observer one moving left at .75c and Observer two moving right at .75c. Neither one is moving faster than the speed of light relative to Observer three, so there's no relativity violation... But Observer two and Observer one are moving apart at 1.5c.

Now look it at from the point of view of Observer one (or Observer two - it's symmetrical). Observer one is at rest relative to himself, sees Observer two moving away from him at .75c, and sees Observer two ship somewhere beyond Observer three and moving away even faster, at speed $(.75+.75)/(1+.75^2)$ by the velocity addition rule.

Again, no one observes anyone moving faster than light relative to them.

8. Oct 6, 2015

### TheScienceGuy

I believe after all these answers I should make my post more clear: it is about the measurements and what happens in the reference frame of observer 3 + probe 1 + probe 2. I intentionally designed the experiment to be symmetrical. So, each probe can measure their distance and the speed towards the 3rd experimenter and use the symmetry to infer the speed of the other probe in their own reference frame (probe1 vs probe2 and vice-versa). If we isolate them (imagine universes with only experimenter 3 + probe 1 or experimenter 3 + probe 2) in each reference frame they will all have the same velocity - the experimenter 3 and the probes will each measure 0.75c. If we put them all together the story changes: experimenter 3 sees each probe approaching with a velocity of 0.75c. Simultaneously, the probes can measure their speed in relation to observer 3 (they can use reflected light, the parallax method or any other conceivable measurement), which will be again 0.75c and due to the symmetry of the experiment (knowledge about the initial experimental design) they can easily multiply their result by 2 (due to the symmetry of the system) to find the speed of the approaching probe in their own reference frame. However, if they measure that speed directly (as opposed to the above indirect method taking the symmetry into account) they will find out that in their own reference frame the other probe approaches with a speed of less than c. Why do these two methods of EXPERIMENTAL measurements give different results (they are conducted in the reference frames of the probes)? Would the experimenter live longer or shorter?

9. Oct 6, 2015

### phinds

Your flaw is in the \statement I bolded. Can you see why?

10. Oct 6, 2015

### TheScienceGuy

Can't really see that - in 2 separate "universes": i) universe 1 consists only of experimenter 3 + probe 1; and ii) universe 2 consists of experimenter 3 + probe 2. In universe 1 both the experimenter and the probe from the viewpoint of their own reference frames will measure 0.75c. The same holds true for universe 2 (both will measure 0.75c in their own reference frames).

11. Oct 6, 2015

### Orodruin

Staff Emeritus
But this is exactly where your problem lies. You cannot simply multiply the result by two if you want to know the relative velocity between the probes. In the central observer's frame, both probes are travelling at 0.75c. In order to find their velocities in a different frame you need to transform them according to the velocity addition formula.

12. Oct 6, 2015

### TheScienceGuy

Doesn't the symmetry of the system give me the right to multiply by 2?

13. Oct 7, 2015

### Orodruin

Staff Emeritus
No. You need to use relativistic velocity addition whenever you want to infer how fast something is moving in a reference frame different from that where you already know the velocity.

14. Oct 7, 2015

### Orodruin

Staff Emeritus
To be clear here, the middle observer will conclude that the closing speed of the probes is 1.5c, but the closing speed is not Lorentz invariant. It will be different in different inertial frames. In particular, it will be different in the probes' rest frames.

Nowhere is any observer finding an object travelling faster than c.

15. Oct 7, 2015

### TheScienceGuy

Both measurements are experimental - one of them just uses the symmetry of the system to infer the relative velocity of the other object. For example I can't apply the wave equation after the wave function collapses - the wave equation is deterministic, but the collapse is governed by a non-deterministic procedure. Here the situation seems identical - the relativistic velocities addition law can be applied to the "direct" measurement, but not to the indirect one (the one based on the symmetry of the system).

16. Oct 7, 2015

### Orodruin

Staff Emeritus
I am sorry, but this is just ramblings. There is no argument here for not using relativistic velocity addition. The probes must account for the fact that the observer is quoting velocities as measured in his frame and not the probe's frame.

17. Oct 7, 2015

### TheScienceGuy

So, what you imply is that the type of symmetry I used in my experiment (corresponding to simple reflection) doesn't allow me to infer the relative speed of one of the probes in the reference frame of the other one by multiplication (x2).

18. Oct 7, 2015

### Orodruin

Staff Emeritus
Your symmetry does not involve the necessary Lorentz transformation. Therefore it tells you absolutely nothing about how things appear in reference frames different from the middle observer's.

If you consider the symmetry to be the fact that the observer and probe agree on their relative velocity, the transformation associated with this symmetry is broken by the introduction of the second probe.

19. Oct 7, 2015

Staff Emeritus
Do you believe you can use the Lorentz transformations here?

If so, what do you get when you use them?

20. Oct 7, 2015

### Janus

Staff Emeritus
It kicks in when you work out how fast the probes move relative to each other when measured by either probe. This works out to being 0.96c Thus each probe sees the other probe moving at 0.96c and experimenter 3 moving at .75c
No, because for either probe, the symmetry that exists for the 3rd experimenter doesn't exist. For him, the third experimenter has a speed of 0.75c and the other probe has a velocity of 0.96c. Also, at any given moment, the distance between the 3rd experimenter and himself will be greater than the distance between the other probe and the 3rd experimenter. Remember, the two probes started simultaneously according to the 3rd experimenter, but don't do so in the frame of either probe due to the Relativity of Simultaneity Each probe will say that the other probe started before it did. And this makes up for the fact that the closing speed between the experimenter and the other probe is less than his relative speed with respect to the experimenter. As a result, both probes still agree that all three will meet up together.