How Does Speed Influence Time and Particle Decay in Einstein's Theory?

[SiennaTheGr8]

My interpretation is that all inertial observers will agree on the spacetime interval between two events.
So if something moves more in space from a particular observer's point of view, it must also move less in time for that observer.

That's not quite right. When it comes to the spatial distance and time elapsed between two events, it's the difference of their squares that everyone always agrees on.

Consider an equation like $a^2 - b^2 = c^2$ (for positive $a$ and $b$). What happens if $b$ increases while $c$ remains the same? Does $a$ decrease? Absolutely not! The opposite is true: $a$ increases also.

In special relativity, we have $(\Delta s)^2 = (\Delta t)^2 - (\Delta \mathbf{r})^2$. The invariance of the spacetime interval guarantees that if Observer A measures a greater spatial distance between two events than Observer B does, then Observer A likewise measures a greater elapsed time between them.

I think some confusion on this point arises from the fact that the formulas for time dilation and length contraction are "opposites": $dt = \gamma dt_0$ vs. $L = L_0/\gamma$.

 

[puzzled fish]

meni ohana, how do you explain the fact that a particle on the surface of the Earth decays slower than another put away 10 km above it say? They are not moving and are stationary relative to one another, right? But their time difference is a direct consequence of accelerating frames of reference, exactly the way SR predicts.

 

[PeterDonis]

their time difference is a direct consequence of accelerating frames of reference

No, it's a direct consequence of them being at different heights, i.e., different gravitational potential. That is independent of any choice of coordinates or reference frame.

Locally, i.e., over a small enough change of height (which 10 km is not given our current accuracy of measurement--see below), you can treat spacetime as flat and correctly predict the time dilation by treating the two particles as following worldlines of constant proper acceleration in flat spacetime, i.e., using SR. But if the height change is large enough that you can detect the effects of tidal gravity (which we can over a 10 km height change), you can't treat spacetime as flat, so you can't use SR.

 

[FactChecker]

No, it's a direct consequence of them being at different heights, i.e., different gravitational potential. That is independent of any choice of coordinates or reference frame.

Locally, i.e., over a small enough change of height (which 10 km is not given our current accuracy of measurement--see below), you can treat spacetime as flat and correctly predict the time dilation by treating the two particles as following worldlines of constant proper acceleration in flat spacetime, i.e., using SR. But if the height change is large enough that you can detect the effects of tidal gravity (which we can over a 10 km height change), you can't treat spacetime as flat, so you can't use SR.

Good point. In GPS satellites, both SR and GR must be taken into account to eliminate large GPS position errors. The relative velocity of the satellites make their time go slower (SR), but the orbital altitude makes their time go faster (GR). At that speed and altitude, the GR effect is over six times larger than the SR effect. (see https://en.wikipedia.org/wiki/Error...tioning_System#Special_and_General_Relativity )

 

[puzzled fish]

Locally, i.e., over a small enough change of height (which 10 km is not given our current accuracy of measurement--see below), you can treat spacetime as flat and correctly predict the time dilation by treating the two particles as following worldlines of constant proper acceleration in flat spacetime, i.e., using SR. But if the height change is large enough that you can detect the effects of tidal gravity (which we can over a 10 km height change), you can't treat spacetime as flat, so you can't use SR.

Agreed. I thought 10 km was enough for only SR to be taken into account. Please read 1 km distance.

 

[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.

 

[Herman Trivilino]

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.