I Special relativity timebomb on distant planet 'paradox'?

[PeterDonis]

Imagine if one of these muons was used as the clock to detonate the bomb?

Sure, go ahead and imagine it. What do you come up with? I see at least one key difference in this case: the muons are moving in the Earth's frame, so they are more like Alice than the clock on the planet in your original scenario. So you'll need to clarify what scenario you are envisioning here.

 

[phyzguy]

It's very clear that the OP was in fact talking about what Alice actually sees. In the OP, the statement was:

She is on her way to destroy the clock, regardless of what Bob thinks. She only measures the planet as being 3 light years away, but also sees the clock that will detonate the bomb tick slower and slower.

Note that "the clock that will detonate the bomb" is on the distant planet. It is the incorrect statement that she sees this clock tick slowly that I have been attempting to correct.

 

[Matternot]

Whether

Sure, go ahead and imagine it. What do you come up with? I see at least one key difference in this case: the muons are moving in the Earth's frame, so they are more like Alice than the clock on the planet in your original scenario. So you'll need to clarify what scenario you are envisioning here.

The muons are traveling towards the Earth at a speed very close to the speed of light, in the same way the planet is moving towards alice at very close to the speed of light. The lifetime of the muons are extended according to people on earth. But I am certain where the confusion is and my next post should explain.

 

[Matternot]

Vitro is correct

The problem of whether she measures the clock run faster or slower depends on how she is measuring. When writing the original problem statement, the subtleties between the words see/view and measure were not immediately apparent to me. Typically when considering the time observers record events in special relativity, clocks are synchronised according to the Einstein synchronisation technique in which all observers moving at a particular velocity will agree on when an event occurs. Upon configuring these clocks and using them to measure the time between ticks, the ticks of the clock moving towards her will appear slower (τγ). There is a crucial subtlety in her looking at the clock, in particular whether or not she takes Doppler shift and the distance the light has to travel into account. If she instead measures the time between events from when the photons from the event reach her, then she will view the clock as running at a speed τ/(1+β). In setting up the problem I had assumed the Einstein synchronisation technique, and yet used words like "view the clock" due to this lack of appreciation of the subtleties of the words. I should have instead used "measured the time the clock had proceeded". This is where the crux of the disagreements appeared.

If using the Einstein synchronisation technique, the paradox is resolved as suggested by Dale, Perok, geogir and I, with Relativity of simultaneity (that the clock is far further proceeded as measured by Alice than it is bob)

If using the the time of the event as the time a photon from the event would reach you, the paradox can be resolved as suggested by Phyzguy and Phinds by the clock ticking far faster than Alice due to Doppler shift effects

Both synchronisation techniques are valid for resolving this paradox, however the Einstein synchronisation technique reveals the nature of time itself, that even if Alice is savvy and accounted for the Doppler effect. If she couldn't see the clock ticking at all and no photons passed between them, she still would not be able to make it in time, despite her being able to reach the planet in 3 years from the time she passes Earth.

The idea that muon lifetime is extended as muons move towards the Earth at very close to the speed of light is particularly interesting given you might start thinking about the lifetime being extended, yet the frequency at which you record muon decay increases if Doppler shift is ignored. In this case, because we cannot see the clock, we decide to account for the Doppler shift and calculate a lower frequency after taking the Doppler shift. Here the convention used is that of Einstein synchronisation, as opposed to the convention suggested by "seeing the events," otherwise we might even record a shorter lifetime (which makes it now harder and more confusing to think about how they can reach the Earth before they decay).

 

[PeterDonis]

the frequency at which you record muon decay increases if Doppler shift is ignored

I'm not sure what you mean by this. If you consider a series of clocks along the path of the muons, all at rest relative to the Earth and Einstein synchronized with Earth clocks, then these clocks will measure the frequency of muon decay to decrease (compared to clocks at rest relative to the muons). This has nothing to do with Doppler shift; it's just the clocks recording their times and the fraction of muons decayed as the muons pass by.

Here the convention used is that of Einstein synchronisation, as opposed to the convention suggested by "seeing the events," otherwise we might even record a shorter lifetime (which makes it now harder and more confusing to think about how they can reach the Earth before they decay).

It seems like you are thinking here of a single observer on Earth, watching the muons come towards him. This observer will in fact see (actually observe through his telescope, or muon viewer, or whatever) the muons to be decaying at a greater rate than if they were at rest relative to him--because what he sees includes the effect of the Doppler shift. But he will also see the start of the muons' flight time delayed, compared to a clock located in the upper atmosphere where the muons actually start their flight (because the light emitted by the muons when they start their flight is only traveling a little faster than the muons themselves, so it only reaches the Earth observer a little before the muons do). And this time delay more than compensates for the faster rate at which he sees the muons decay, so that there are still muons left when they reach him.

(As an instructive exercise, you might want to figure out what an observer in the upper atmosphere, at rest relative to the Earth, will see if he watches the muons flying away from him through a telescope.)

 

[DarioC]

Reductionist time: First, what Alice perceives has no effect on what actually happens on the far planet. Next, get rid of the bomb. Have a light source that flashes every second on the far planet. It has been flashing forever (OK 50 years). In the time that it takes Alice to get to the planet it will have put out ten years worth of flashes. (in the far planets time). Alice will see these ten years of flashes spread over her time (3 years). She will see roughly 3 flashes per second.
Seems pretty simple to me.
DC

 

[PeterDonis]

In the time that it takes Alice to get to the planet

You have just confused the issue again, because this statement is frame-dependent. The ten years' worth of flashes are not emitted "in the time it takes Alice to get to the planet" in Alice's frame, only in the Earth-planet frame.

 

[DarioC]

Then you disagree with my concept that Alice will intercept each and everyone of these flashes while in transit to the far planet?
DC

 

[phinds]

Then you disagree with my concept that Alice will intercept each and everyone of these flashes while in transit to the far planet?
DC

You have misunderstood Peter's post. Read it again carefully.

 

[PeterDonis]

Then you disagree with my concept that Alice will intercept each and everyone of these flashes while in transit to the far planet?

I have said no such thing.

I strongly suggest that you work out the specific (t, x) coordinates (in the Earth/planet frame) at which each flash is emitted, and at which each flash intersects Alice's worldline. Then you can Lorentz transform those coordinates into Alice's frame. Hopefully that will help you to understand my previous post better.

 

[DarioC]

OK, I can see the phrase I used is not the way to express what I meant. Between the planets there is already a "string" of light pulses. When Alice passes through these pulses she will intercept them at much higher rate, according to her clock, than one pulse per her second--because her clock, along with everything else on her ship is slowed down. Those pulses can serve as a yardstick of distance that she will perceive as being much shorter than the speed of light for one second.
But none of her perceptions have any effect on what is actually happening on the far planet except for how she perceives it.

On consideration perhaps I should have said "during her trip to the far planet" the pulsing light on that planet will have emitted a number of pulses, according to the far planet clock, that is the number that you get when you calculate the number of seconds in 10 years. I see no reason why she would not intercept at least that number of pulses in route.

 

[PeterDonis]

Between the planets there is already a "string" of light pulses.

"Already" is still frame-dependent; "already" according to whose frame? "Already" has different meanings in different frames because of relativity of simultaneity.

On consideration perhaps I should have said "during her trip to the far planet" the pulsing light on that planet will have emitted a number of pulses, according to the far planet clock, that is the number that you get when you calculate the number of seconds in 10 years

"During" according to whose frame? Same comment as above.

Here is a frame-invariant statement that you might be groping towards: when Alice is just leaving Earth, some pulse from the far planet is just arriving. Call that pulse "pulse 0". When Alice is just arriving on the distant planet, some pulse is just being emitted; call that pulse "pulse N". Alice sees every pulse between pulse 0 and pulse N, in order, during her trip. (Note that "during" here has an invariant meaning because we are only talking about events on Alice's worldline--we don't care "when" they were emitted from the planet, we only care about their intersections with Alice's worldline.)

Now work out now much time passes, on the planet's clock, between the emission of pulse 0 and the emission of pulse N (which is also Alice's arrival). You will find it is not 10 years.

 

[PeterDonis]

When Alice passes through these pulses she will intercept them at much higher rate, according to her clock, than one pulse per her second--because her clock, along with everything else on her ship is slowed down.

This is also not quite right, because the rate at which Alice intercepts the pulses, according to her clock, is affected by the Doppler shift as well as time dilation. (Actually, if you look at the math, the relativistic Doppler shift formula by itself gives the correct answer.)

Those pulses can serve as a yardstick of distance

No, they can't, because Alice is moving relative to the source of the pulses.

 

[DarioC]

Well I guess that pretty much takes care of that. DarioC backs quietly out of the room, closing the door behind him.