Exploring the Effects of Speed on Time and Decay: Unraveling Einstein's Theory

[PeterDonis]

Please read 1 km distance

Even that might be too large. Here are some numbers for the variation in $g$ with height (all units are SI):

Newton's constant: $G = 6.67 \times 10^{-11}$

Mass of Earth: $M = 5.94 \times 10^{24}$

Radius of Earth: $R = 6.378 \times 10^6$

So $g = GM / R^2$ at Earth's surface is $9.73964772...$. And 1 km above the surface, i.e., at $R = 6.379 \times 10^6$, we have $g = 9.73659430...$

The difference is $0.00305...$, which seems like it should be well within our current technology's ability to measure.

Even if 1 km is too large, there will be some height change which is small enough that no effects of spacetime curvature are measurable, though.

 

[nitsuj]

Even that might be too large. Here are some numbers for the variation in $g$ with height (all units are SI):Even if 1 km is too large, there will be some height change which is small enough that no effects of spacetime curvature are measurable, though.

here is the referenced link from a wiki article that said laboratory tests have it to less than a meter.

 

[calinvass]

Time dilation has been observed with particles like muons that don't have any inner parts.

I don't understand how muons can decay if they have no inner structure.
I would expect something to change to them over in time. If nothing changes with time, how does time affect them?
Perhaps, like in QM, this doesn't need to be intuitive, it just happens.

 

[Dale]

I don't understand how muons can decay if they have no inner structure.
I would expect something to change to them over in time. If nothing changes with time, how does time affect them?
Perhaps, like in QM, this doesn't need to be intuitive, it just happens.

If something in them changed over time then the probability of decay could change over time. For instance, a human's inner structure changes over time such that their probability of decay is much greater from 78-79 years than from 28-29 years. For a muon the chance of decay from 28-29 us is exactly the same as the chance of decay from 78-79 us. This indicates that nothing about the muon changed.

 

[calinvass]

If something in them changed over time then the probability of decay could change over time. For instance, a human's inner structure changes over time such that their probability of decay is much greater from 78-79 years than from 28-29 years. For a muon the chance of decay from 28-29 us is exactly the same as the chance of decay from 78-79 us. This indicates that nothing about the muon changed.

Oh, now its even more confusing. If the mean lifetime of a n number of muons is 2.2 ns, then at time t=0 probability of one decay is very small. At t=1.5 s, the probability of one decay is the same, but approximately 50% of them have already decayed. This is like throwing a die every second and when it falls on 1 it means we have a decay. Although there is no internal clock for the die to trigger a decay, there is still a structure that changes randomly with time, until it gets to the final state where it decays.

 

[Mister T]

Oh, now its even more confusing. If the mean lifetime of a n number of muons is 2.2 ns, then at time t=0 probability of one decay is very small. At t=1.5 s, the probability of one decay is the same, but approximately 50% of them have already decayed.

The mean lifetime is 2.2 μs and the half-life is 1.5 μs. These are constants that don't change with time, thus they have the same value when t = 0 as they do at any other time t.

This is like throwing a die every second and when it falls on 1 it means we have a decay. Although there is no internal clock for the die to trigger a decay, there is still a structure that changes randomly with time, until it gets to the final state where it decays.

The chances of rolling a "1" on your first roll is the same as rolling a "1" on your tenth roll. There is no change in the internal structure with time.

But none of this directly addresses your initial query.

I don't understand how muons can decay if they have no inner structure.
I would expect something to change to them over in time. If nothing changes with time, how does time affect them?
Perhaps, like in QM, this doesn't need to be intuitive, it just happens.

Intuition is sometimes right and sometimes wrong. And what's intuitively right for one person may be intuitively wrong for someone else. There's no universal standard.

 

[Dale]

If the mean lifetime of a n number of muons is 2.2 ns,

No, I mean a single muon's probability of decay, although I see that I was not clear at all. My apologies for the confusion.

What I meant to focus on is the "memoryless" property. I.e. For a single muon the probability of decay in a given time period, t, follows: $P(t>t_0+\Delta t|t>t_0)=P(t>\Delta t)$. This means that the system has no internal memory about its age. This is what you would expect for something with no internal structure.

 

[jbriggs444]

If something in them changed over time then the probability of decay could change over time. For instance, a human's inner structure changes over time such that their probability of decay is much greater from 78-79 years than from 28-29 years. For a muon the chance of decay from 28-29 us is exactly the same as the chance of decay from 78-79 us. This indicates that nothing about the muon changed.

That does not read as correctly as one might have hoped.

The chance of a decay between 28 and 29 us [given that the muon had survived the first 28 us] is the same as the chance of a decay between 78 and 79 us [given that the muon had survived the first 78 us].

This sort of behavior is "memoryless". No matter how long you wait, the chance of decay in the next 1 us is always the same [assuming it hadn't already decayed before lasting that long]. The distribution for the decay time for a single particle under such a decay process is exponential.

 

[Nugatory]

Oh, now its even more confusing. If the mean lifetime of a n number of muons is 2.2 ns, then at time t=0 probability of one decay is very small. At t=1.5 s, the probability of one decay is the same, but approximately 50% of them have already decayed. This is like throwing a die every second and when it falls on 1 it means we have a decay. Although there is no internal clock for the die to trigger a decay, there is still a structure that changes randomly with time, until it gets to the final state where it decays.

You are misunderstanding the relationship between mean lifetime and decay probability.
If I throw a die once a second and interpret a one result as "the die is decayed" (maybe I'll smash it with a hammer when it first comes up one, so a decay event really does end its life), the probability that an undecayed die will decay on the next throw is always one in six. It doesn't matter if the die has already lived for one throw or one thousand before I start counting; it has no memory and no internal state that "remembers" its previous survival. I don't even need to think in terms of lifetime at all if I don't want to: The question "Will this die survive the next N throws" is mathematically equivalent to "If I throw N independent dice all at once, what is the probability that none of them will come up 1?"

Likewise, given an undecayed muon, the probability that it will decay in the next usec is the same whether the muon is one usec old or one year old - although the probability of me finding a one-year-old muon is very low, if I do find one it will be indistinguishable in every respect including its prospects for future survival, from one that is one usec old.

 

[FactChecker]

I don't think that the fact that it is a "memoryless" Poisson process can be used alone to imply that there is no internal structure and process. There are many Poisson processes where an internal process occurs over time but it was started in a Poisson way. To reach the conclusion that there is no internal structure, we would need some other logical or theoretical evidence.

 

[calinvass]

Actually it was confusing at first look, but then I gave the example with the die, because the probability for getting 2 is always 1/6. no matter the number of iterations. However, unlike muons, the state of the die changes every second, because its structure holds (memorizes) the last state. Something changes physically. For a muon that would be equivalent to continuously interrogating its state, with a probability of getting the final state being infinitesimal. But you are saying that such a memory doesn't exist and this I find difficult to understand.

 

[jbriggs444]

Actually it was confusing at first look, but then I gave the example with the die, because the probability for getting 2 is always 1/6. no matter the number of iterations. However, unlike muons, the state of the die changes every second, because its structure holds (memorizes) the last state. Something changes physically. For a muon that would be equivalent to continuously interrogating its state, with a probability of getting the final state being infinitesimal. But you are saying that such a memory doesn't exist and this I find difficult to understand.

You seem to want a mechanical explanation. So that the behavior of a muon (for example) is understandable if and only if there is a mechanical model that yields the observed behavior.

You posit a model in which the muon has some complex and unknown internal structure which evolves over time with some infinitesimal probability per unit time of of entering a final decay state. For instance a "salt shaker" model in which there is a salt shaker with a grain of salt and a hole that is exactly the right size and shape to permit the grain of salt to exit if it happens to hit exactly right].

As long as we have no visibility into the internal structure of the muon, it is easier to simply accept the model of an exponential decay distribution and not bother trying to postulate an explanation for it.

There is no need for a physical model as long as we have a mathematical model that matches the observed behavior.

 

[PeterDonis]

the state of the die changes every second, because its structure holds (memorizes) the last state

Does it? If it does, why is it that the probability of a given face coming up doesn't change from one throw to the next?

 

[Grinkle]

Does it? If it does, why is it that the probability of a given face coming up doesn't change from one throw to the next?

Possibly because the human hand cannot repeat the throw well enough to achieve any correlation.

There are potentially mechanical systems that can be controlled well enough that the initial condition of the die pre-toss will show up statistically in the result.

This is perhaps your point, that just looking at the die in isolation instead of the entire process involved in a toss can be misleading.

 

[FactChecker]

Does it? If it does, why is it that the probability of a given face coming up doesn't change from one throw to the next?

Certainly as it rolls, there is a state and a process. Between rolls, we like to assume that two rolls are independent, but if a machine picked it up after one roll and made the next roll exactly the same on a flat surface, there would probably be a pre-determined sequence of results.
I don't know about muons and wonder if there is a theoretical way of knowing that there is no internal process or structure of a muon.

 

[PeterDonis]

There are potentially mechanical systems that can be controlled well enough that the initial condition of the die pre-toss will show up statistically in the result.

Yes, but that's not the same as the die itself storing information about the results of previous tosses. The correlation between tosses is due to the throwing system, not the die itself.

 

[calinvass]

Does it? If it does, why is it that the probability of a given face coming up doesn't change from one throw to the next?

I'm not sure if it is this what you ask. The example was with a throw every second, and between seconds no throw, but this is only a mechanism that shows the probability for getting one state doesn't change over time, but the mean life of the die can be calculated as 6 seconds I guess.

 

[calinvass]

Possibly because the human hand cannot repeat the throw well enough to achieve any correlation.

There are potentially mechanical systems that can be controlled well enough that the initial condition of the die pre-toss will show up statistically in the result.

This is perhaps your point, that just looking at the die in isolation instead of the entire process involved in a toss can be misleading.

I'm not thinking of a real world object. In this example the die is a perfect system that gives completely random results although a real world one is also almost perfect I think.

 

[PeterDonis]

the probability for getting one state doesn't change over time

Yes, and my question was how that can be true if the structure of the die "memorizes" the last state, as you claimed it did in post #46. I don't think that claim is correct; a die, assuming it's fair, will not "memorize" any information from previous throws.

the mean life of the die can be calculated as 6 seconds I guess

Not quite. The half life (I use that because it's easier to conceptualize) will be the number of throws after which there is a 50% probability of throwing a 1 (or whichever face we treat as "destroying" the die when it comes up). The easiest way to calculate that is to find the smallest exponent $N$ such that

$$
\left( \frac{5}{6} \right)^N < \frac{1}{2}
$$

In other words, how many throws does it take before there is a less than 50% chance of not throwing a 1? (This method is easier because there is only one term on the LHS of the equation.) I get $N = 4$ as the smallest integer exponent (the corresponding probability, of not throwing a 1 in 4 throws, is about 48 percent). So the "half life" of a die is about 4 throws (a little less if we want an exact floating point answer).

 

[Dale]

I don't think that the fact that it is a "memoryless" Poisson process can be used alone to imply that there is no internal structure

I agree. The implication is the other way around. If it has no internal structure then its chance of spontaneously decaying must be memoryless. We postulate that muons have no internal structure, which implies a memoryless decay probability, we observe such a decay probability, which we take to be evidence supporting the postulate.

 

[calinvass]

Yes, and my question was how that can be true if the structure of the die "memorizes" the last state, as you claimed it did in post #46. I don't think that claim is correct; a die, assuming it's fair, will not "memorize" any information from previous throws.

Yes, you are right, but still its state changes every time. It doesn't make any progress over time but its state changes and that can be thought as a change in its structure I suppose.

 

[Dale]

its state changes and that can be thought as a change in its structure I suppose

Not its internal structure.

There are two memoryless distributions, one for continuous and one for discrete. Muons decay according to the continuous one, dice roll according to the discrete one.

 

[Grinkle]

I am finding the die analogy confusing. Each toss is self-contained and there is no toss-to-toss memory. Within a single toss there is obvious structure. The die tumbles, different faces are 'up' during the toss process, and finally the die comes to rest. Only some state transitions are possible, the die cannot flip from 6-dots to 1-dot without having moved through some other intermediate state, for instance.

What is the comparison to muon decay? Every half life is a toss, and we don't know what happens inside the half life interval, but we do know that each half life behaves as though it is the only half life to have ever expired? If that is the comparison, it does not lead me to any inclination that muon's have no internal structure, since a single die toss clearly does.

 

[FactChecker]

Whether something has an internal structure is not the same question as whether it has "memory" of its state. A die is in a state which includes its orientation and velocities. It retains that state till it is changed by some external force. That is "memory".
Suppose a die is sitting with the 5 side up and the next roll ended with the 5 side up. Now suppose the die had started with the 3 side up. With the exact same external influences, the roll would end with the 3 side up. That means that the die itself has a state and "memory" of it that influences the next result. The die having no internal structure does not prevent that. The randomness of rolling a die is due to external forces that are treated as random changes of its state . We just don't (or can't) keep track of the "random" external influences.

 

[PeterDonis]

but still its state changes every time

In the sense that which face is up changes, yes. (Or, if you want to keep track in more detail, the die's orientation and velocity changes with time.) But, as Dale and FactChecker have pointed out, this is not the same as the die's internal structure changing or the die having "memory" of previous throws stored in its internal structure.

 

[PAllen]

here is the referenced link from a wiki article that said laboratory tests have it to less than a meter.

Actually, the one meter tests are NOT detecting tidal gravity. They are detecting potential difference over one meter, and cannot distinguish the Earth surface observation from a uniformly accelerating rocket in empty space - the results of the experiment would be identical. Note, that Peter's computation shows you would need several significant digits in precision of time dilation measurement over a kilometer to distinguish tidal gravity. These experiments are at detection limit at one meter, e.g one significant digit. Instead they would need e.g 10 to detect tidal gravity.

[edit: well, a series of measurements from such clocks can verify tidal gravity. Separate such clocks radially versus tangentially, and do this on opposite sides of the earth, to take the extreme. Then only tidal gravity can explain the collection of observations. But the, again, we are expanding the scope of observations greatly. The key point being that if observations are sufficiently local, you cannot distinguish.]

 

[nitsuj]

Actually, the one meter tests are NOT detecting tidal gravity. They are detecting potential difference over one meter, and cannot distinguish the Earth surface observation from a uniformly accelerating rocket in empty space - the results of the experiment would be identical. Note, that Peter's computation shows you would need several significant digits in precision of time dilation measurement over a kilometer to distinguish tidal gravity. These experiments are at detection limit at one meter, e.g one significant digit. Instead they would need e.g 10 to detect tidal gravity.

[edit: well, a series of measurements from such clocks can verify tidal gravity. Separate such clocks radially versus tangentially, and do this on opposite sides of the earth, to take the extreme. Then only tidal gravity can explain the collection of observations. But the, again, we are expanding the scope of observations greatly. The key point being that if observations are sufficiently local, you cannot distinguish.]

I don't know enough about that distinction, but was replying to Peter's comment "Even if 1 km is too large, there will be some height change which is small enough that no effects of spacetime curvature are measurable, though."

I do know the distinction between flat spacetime and curved spacetime, and this experiment shows that we have this measurable to less than a meter. What of tidal gravity or difference in gravitational potential...when neither are flat.

 

[PAllen]

I don't know enough about that distinction, but was replying to Peter's comment "Even if 1 km is too large, there will be some height change which is small enough that no effects of spacetime curvature are measurable, though."

I do know the distinction between flat spacetime and curved spacetime, and this experiment shows that we have this measurable to less than a meter. What of tidal gravity or difference in gravitational potential...when neither are flat.

No, this measurement does not distinguish curvature at all over one meter. It would give the exact same result in flat spacetime in an accelerating rocket.

 

[nitsuj]

I am finding the die analogy confusing. Each toss is self-contained and there is no toss-to-toss memory. Within a single toss there is obvious structure. The die tumbles, different faces are 'up' during the toss process, and finally the die comes to rest. Only some state transitions are possible, the die cannot flip from 6-dots to 1-dot without having moved through some other intermediate state, for instance.

I'm finding the distinction between "memory" and causal connection confusing.

 

[nitsuj]

No, this measurement does not distinguish curvature at all over one meter. It would give the exact same result in flat spacetime in an accelerating rocket.

Sorry PAllen, I still fail to see the issue. You're mentioning the equivalence principal. So in a sealed free fall laboratory, this experiment doesn't work...neat physics.

 

[PAllen]

Sorry PAllen, I still fail to see the issue. You're mentioning the equivalence principal. So in a sealed free fall laboratory, this experiment doesn't work...neat physics.

I will try again to clarify. Uniform acceleration in flat spacetime is distinguishable from behavior near Earth over sufficient distances or time. Peter noted that for one kilometer, it should be possible to make the distinction. You claimed this experiment shows a distinction over one meter. This is simply false. This experiment could not distinguish the time dilation over one meter that would occur in an accelerating rocket from that on the Earth's surface. To make that distinction would require many order of magnitude greater precision. In, particular, a device that could measure presence of time dilation over e.g. one millimeter might stand a chance of detecting curvature over one meter.

 

[PeterDonis]

You're mentioning the equivalence principal. So in a sealed free fall laboratory, this experiment doesn't work

The correct comparison is not with a sealed free fall laboratory, but with a sealed rocket accelerating at 1 g in flat spacetime. And PAllen is saying, correctly, that the results of such an experiment would be indistinguishable from those of the experiment as it was performed on the Earth's surface. In other words, the experiment shows time dilation but does not show tidal gravity--it isn't precise enough for the latter.

 

[Mister T]

Yes, you are right, but still its state changes every time. It doesn't make any progress over time but its state changes and that can be thought as a change in its structure I suppose.

In the same way, the decay of a muon is a change in its state.

I am finding the die analogy confusing.

Each roll of the die is analogous to an observation of the muon. When you roll the die it either will or will not show "1". When you observe a muon it either will or will not decay.

You can find break-downs in the analogy. That's easy. Because it's an analogy.

Let's remember that the reason for the analogy was to try to understand why a particle such as a muon can decay despite the fact that it has no internal structure. The point was made that unlike humans, the decay rate stays the same over time, which is something we notice only after looking at large numbers of both humans and muons decaying. But that still doesn't explain how something with no internal structure can decay. It may be that there is no explanation, in which case we'd have to ask ourselves why we'd expect one to exist!

 

[nitsuj]

The correct comparison is not with a sealed free fall laboratory, but with a sealed rocket accelerating at 1 g in flat spacetime. And PAllen is saying, correctly, that the results of such an experiment would be indistinguishable from those of the experiment as it was performed on the Earth's surface. In other words, the experiment shows time dilation but does not show tidal gravity--it isn't precise enough for the latter.

Thanks PAllen and Peter, I see the accelerating rocket compared to Earth equivalence. And if I see that I got to say it's not measuring tidal gravity. The experiment simply determines the differential in gravitational potential, which is (this is where all my fog is) distinctly different from tidal gravity.I not good at math, or even keeping track of significant digits. but as far as the measurement accuracy in this experiment even if there were four measures taken, say one meter square area, The horizontal (to earth) comparatives wouldn't deviate, but the vertical ones do? Compared to "flat space", i intuit a trapezoid shape.

 

[PeterDonis]

The experiment simply determines the differential in gravitational potential, which is (this is where all my fog is) distinctly different from tidal gravity.

Yes, this is correct.

The horizontal (to earth) comparatives wouldn't deviate, but the vertical ones do?

Tidal gravity on Earth shows up in both the horizontal and vertical directions. But the horizontal change is in the direction of gravity, whereas the vertical change is in its magnitude. So you might use different methods, with different accuracies, to measure the two.