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Scientists Confirm Nuclear Decay Rate Constancy

timothy posted about 4 years ago | from the was-just-the-one-time-I-swear dept.

Science 95

As_I_Please writes "Scientists at the US National Institute of Standards and Technology and Purdue University have ruled out neutrino flux as a cause of previously observed fluctuations in nuclear decay rates. From the article: 'Researchers ... tested this by comparing radioactive gold-198 in two shapes, spheres and thin foils, with the same mass and activity. Gold-198 releases neutrinos as it decays. The team reasoned that if neutrinos are affecting the decay rate, the atoms in the spheres should decay more slowly than the atoms in the foil because the neutrinos emitted by the atoms in the spheres would have a greater chance of interacting with their neighboring atoms. The maximum neutrino flux in the sample in their experiments was several times greater than the flux of neutrinos from the sun. The researchers followed the gamma-ray emission rate of each source for several weeks and found no difference between the decay rate of the spheres and the corresponding foils.' The paper can be found here on arXiv. Slashdot has previously covered the original announcement and followed up with the skepticism of other scientists."

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Tremendous Fun (-1, Troll)

Anonymous Coward | about 4 years ago | (#33695652)

Im betting the author of the original piece is great fun at parties

Re:Tremendous Fun (1)

jack2000 (1178961) | about 4 years ago | (#33695822)

I love talking with such people. :3

Re:Tremendous Fun (2, Funny)

JustOK (667959) | about 4 years ago | (#33696008)

well, he's not the life of the party...he's the HALF-life!

Not a certain conclusion yet (4, Informative)

BSAtHome (455370) | about 4 years ago | (#33695658)

From the paper (emphasis mine):

In summary, the present experiment is the first direct precision test of whether the decay rate of a radioactive source depends on its shape. Our results in Table 1 indicate a 2.3 deviation of the foil/sphere ratio in experiment 1 from unity. From Table 2, based on the initial 30 spectra, the foil/sphere ratio for experiment 2 deviates from unity by 2.6. These results thus leave open the possibility that the half-life of a radioactive nuclide could in fact depend on its shape (due to the internal flux of neutrinos, photons, or electrons), and hence suggests that additional experiments are necessary.

So, there still is a chance that there is a deviation.

Re:Not a certain conclusion yet (2, Insightful)

Frequency Domain (601421) | about 4 years ago | (#33695720)

Yup, that's the nature of statistics. In order to "call" the result with a reasonable level of certainty the sample size requirements increase with smaller signal/noise ratios.

Re:Not a certain conclusion yet (2, Interesting)

Anonymous Coward | about 4 years ago | (#33695862)

"So, there still is a chance that there is a deviation."

Yes, of course. People have been saying that all along. And even with more experiments in the future it will always be the case that a deviation from constancy is possible. However, if there is a deviation, it's vanishingly small and what remains possible is getting smaller as the experiments are refined.

People have been testing the constancy of radiometric decay rates for many decades. Those experiments always have limits in terms of their resolution of decay rates, but there are so many tests by now that we're usually talking about deviations on the order of 1% or less in most cases. It says something that any deviation that might be there is so close to constancy that is difficult to measure experimentally. Variations in decay rate on that scale are certainly a scientifically interesting possibility worth pursuing, but a big "so what?" in terms of their implications for many practical applications such as radiometric dating, where measurement uncertainties are on that order anyway.

Re:Not a certain conclusion yet (3, Interesting)

ceoyoyo (59147) | about 4 years ago | (#33696786)

Those are some pretty big deviations to go with the headline "Scientists Confirm Nuclear Decay Rate Constancy." In any field except physics they would be considered significant evidence of a difference.

Untrue (2, Informative)

Kupfernigk (1190345) | about 4 years ago | (#33696910)

It seems you do not understand the nature of statistics. "significant evidence" is a statistical measure for which there are well-defined measurements. In any field of science, including social science like polling, "significant" has a precise meaning. In this case, the difference was not significant.

Journalism, by the way, is not science. In fact, it is usually the enemy of science.

Re:Untrue (2, Insightful)

ceoyoyo (59147) | about 4 years ago | (#33697000)

I think I have a pretty good grasp of statistics, thanks.

As I explicitly mentioned in my post, you're correct, there are different standards for "statistically significant" in different fields. Contrary to what you think, they're not particularly precise. They're basically rules of thumb and differ between fields, and even within fields, due to tradition, history, and sometimes experience. Note also that I was talking about the Slashdot headline and summary. In fact, I quoted the former. Sorry, I thought the quote made it obvious.

The headline claims "Scientists Confirm Nuclear Decay Rate Constancy." In fact, they did no such thing. They found (weak) evidence of change. In order to confirm constancy, they would have to find significant evidence of no change.

Why don't you go grab a cup of coffee, or maybe a nap? You're awfully cranky for a Saturday morning.

Re:Untrue (1)

jegerjensen (1273616) | about 4 years ago | (#33697376)

They found (weak) evidence of change. In order to confirm constancy, they would have to find significant evidence of no change.

This is not accurate. In the paper, the authors state that they are not able to conclude, so they suggest that more precise measurements are needed. Moreover, it is not possible to prove that there is no change by experiment. One can only demonstrate the effect, or determine an upper limit for it.

Re:Untrue (1)

ceoyoyo (59147) | about 4 years ago | (#33697600)

It is accurate.

You can't prove that their is a change either. You CAN show that there is certain, arbitrarily small chance that there is (or is not) a change. You are correct, the latter case has to assume a certain minimum magnitude of change but that can also be made arbitrarily small.

The evidence from their four experiments is somewhat contradictory, but two of them show a a p-value of 0.02 or less (that's in the paper, if you care to go look). That's (roughly) a 2% chance that there is no change, and a 98% chance that there is. In physics that's not sufficient to claim there is a change, but it's certainly not sufficient to conclude there isn't one.

Re:Untrue (2, Insightful)

jd (1658) | about 4 years ago | (#33698652)

I'm not going to bother RTFAing, but this would depend on whether the test is two-tailed or one-tailed, in terms of how to interpret the p value. Because this kind of statistical test ultimately boils down to a simple measure of the observed versus a given expected, it is also only really good at testing against H0, the Null Hypothesis. There are more complex tests which allow you to measure against multiple variables in a single test (a good approach as it, in theory, allows you to examine the interactions between variables).

However, these all make one core assumption: that variables you're not testing out are randomly distributed. In other words, as the sample size goes up, the errors will all balance out and so the noise approaches zero. Which is entirely fair if it is indeed noise, which is why we were taught at University that you should do an analysis of variance that allowed you to test to make sure that there weren't any untested factors.

Like I said, I've not RTFA'ed, so I can't say if they did this here, but the experiment seems to have been designed with some measure of care - better than anything I've seen from earlier stories on the subject.

Re:Untrue (0)

Anonymous Coward | about 4 years ago | (#33702254)

I will not bother creating an account for this. You can not prove that there is no change. You can only prove that change if happens is less than whatever (i did not read TFA).

About your claim about the different standards of statistical significance being "rules of thumb", well, you have no idea what you are talking about and more dangerously you are not aware of that.

Statistically significance is a well defined concept for any given test (t-test, chis square... ) You define the level at which you want to be confident to accept or reject the null hypothesis at 90, 95, 99%... to obtain an statistic then based on the sample size and the measured results you compute the new statistic and based on the measured vs the calculated you are ready to go.

You sound like my old firm (management consultancy) when doing some analytical work.

Short of telling other people to go and grab a coffee, you should go grab a can of cookies and a good book on statistics, probably you will need some basic calculus before that.

As a friend used to say, there are two kind of people, the ones who can understand that the estimate of the average of sample has a distribution and the ones who don't.

Re:Untrue (1)

Jane Q. Public (1010737) | about 4 years ago | (#33697086)

It doesn't matter anyway. This experiment does not show what the headlines of both OP and TFA claim. It does not reaffirm "constancy"; rather (if there is any significance to it at all) it only indicates that neutrinos are probably not the cause of observed fluctuations.

All in all, however, it seems to me that this experiment used far too little mass to involve many neutrino interactions at all, and thus is very questionable.

I agree... (3, Informative)

Grog6 (85859) | about 4 years ago | (#33697420)

This experiment covered only the decay of Gold-198; The ones that were found to be changing were exhibiting electron capture decays, a completely different mechanism.

For such a limited experiment, the claims are grandiose, IMHO.

Neutrinos also oscillate forms; perhaps the emitted form doesn't interact the same way.

.

Re:I agree... (1)

jd (1658) | about 4 years ago | (#33699038)

In principle, the emission of a neutrino with some energy E and momentum M should be identical to the absorption of an anti-neutrino of identical energy and momentum. BUT, and this is important, it would have to be absorbed in such a way as to alter the angular momentum of the nucleus by the correct amount. Because ALL of the equations have to match up exactly, it's not merely a matter of a neutrino being in the general vicinity. It has to impact in a way that makes the symmetry complete.

Because the system is quantized, if the system of equations would not result in a valid change of state, no interaction would take place. It is if and only if the end result is perfectly valid that the nucleus would absorb the neutrino, and if multiple such solutions exist, it is only those solutions for which a decay event exists that a decay event would occur, assuming neutrino absorption affected decay.

Although a lot of care was put into the experiment, there are some problems with it. First, there's no critical mass for a thin film of radioactive material, but there is one for a sphere. By implication, shape is going to change what you observe. Second, because ONLY valid quantized interactions occur, ALL invalid interactions will result in the neutrino passing by/through the nucleus. This is already known to be virtually all such interactions. Thus, the shape won't significantly alter the number of neutrinos any given gold nucleus would interact with. It might alter it a little, but if the variation in neutrino flux is smaller than the variation in decay events due to shape, your signal just got swamped.

Ok, mr smartypants, if you're so much better at this, you design an experiment!

I'll design one for you, sure, but I'd be an idiot to think that it would ever be utilized. Let us take some of this Gold-198*, but only thin films of it, all of equal size. Let's place them in three locations that we can expect to have different neutrino fluxes. Let's have one at high altitude, say a passenger jet that's going to make a fair number of transatlantic journeys. The second can be in a laboratory. The third, let's put that in a box and have an ROV place it in some deep sea trench. There are some highly accurate crystal clocks, or at least accurate to within the margin of error we're concerned with here. One should go along for the ride with each film. Gold-198 has a half-life of 2.69 days. So after 5.38 days, three-quarters should have decayed. That's sufficient for our purposes. The half-life only matters here in that we want enough decay events to make the experiment sufficiently sensitive.

Ok, so after however long the experiment runs for, you take an identical-sized chunk out of each film and run it through an AMS to count the number of daughter nucleotides there are. All radioactive nucleii decay into lighter nucleii, these are termed daughter nucleii. Since they can form no other way, they are the most sensitive measure of the number of decay events of the original gold atoms. Radioactive daughter nucleii will themselves produce further daughter nucleii when they decay. By counting the individual atoms of each kind, you can know precisely how many events have occurred. There is no question about it. There isn't a more sensitive test possible.

Because decay events are going to occur between retrieval and measurement for the two outside units, if you're using the same lab for all measurements, you can use the clocks to subtract out events after the official finish of the experiment, so that only events in the different environments are considered. If you like, note the timestamp on each clock at the start and at the point of retrieval and you can adjust for relativistic effects as well.

This should be about the most sensitive measure of external influence on decay that you can do on Earth.

Re:I agree... (1)

Jane Q. Public (1010737) | about 4 years ago | (#33699770)

"ALL invalid interactions will result in the neutrino passing by/through the nucleus. This is already known to be virtually all such interactions. Thus, the shape won't significantly alter the number of neutrinos any given gold nucleus would interact with. It might alter it a little, but if the variation in neutrino flux is smaller than the variation in decay events due to shape, your signal just got swamped."

Yes, exactly my point. It appears to me that it would be difficult to measure the differences of something that must be so near zero in the first place. They would have to use a vastly larger amount of mass (think: those massive neutrino detectors elsewhere) to get a sample of any significance.

Re:I agree... (1)

jd (1658) | about 4 years ago | (#33707412)

Yes, a larger mass would definitely help, as does the sensitivity of the measurements. You can also lengthen the time of the experiment. This is not dependent on the half-life as documented anywhere, except insofar as there has to be enough radioactive material left in all three samples that you can draw useful conclusions. Now, a larger mass only helps to a degree. Remember, after one half-life, half of that mass is gone as far as the experiment is concerned. You have to double the amount of mass to add a single half-life's duration to the maximum duration of the experiment, to stay above whatever threshold you've set as the minimum amount of radioactive material left to be useful.

At some point, it becomes impractical to make the mass larger. In order to get a mass onto an aircraft as a film with a well-defined decay rate, you can only cover the total floorspace. Nothing on the walls, no extra layers anywhere, the floor area is it. In order to get an identical mass to the bottom of an oceanic trench, it's got to be light enough for an ROV. That's a significant limit. Fortunately, these are both far in excess of the mass of radioactive material most labs are allowed, let alone how much space they have available.

This isn't too catastrophic - Avagadro's number is very very large, so any sample you have will have a LOT of atoms. And because AMS is able to count down to the level of atoms, your threshold point is impressively low. (Weighing the mass, or measuring decay events, at the end of the experiment as a measure is infinitely less accurate. That would not be an unusual way of doing such a test though.)

Re:I agree... (1)

Alsee (515537) | about 4 years ago | (#33708200)

three locations that we can expect to have different neutrino fluxes. Let's have one at high altitude, say a passenger jet that's going to make a fair number of transatlantic journeys. The second can be in a laboratory. The third, let's put that in a box and have an ROV place it in some deep sea trench

It's possible I'm misunderstanding what you're trying to test, but your experiment appears to be broken. All three samples will have effectively identical neutrino flux. Being the same shape they will have the same self-flux from decaying atoms, and they will all have the same solar neutrino flux because you can't shield neutrinos. Even at midnight they will all get equal and full blast neutrino flux from from solar neutrinos passing through the entire planet and hitting them from below.

-

Re:I agree... (1)

jd (1658) | about 4 years ago | (#33713698)

The reason neutrino detectors are underground is that you don't want them to detect any old neutrino. You can indeed shield from -some- neutrinos, and it is my argument that the very fact that you can shield from them makes them interesting. If they are being absorbed, they must presumably do something. The question then becomes one of what do they do. The sorts of neutrinos that affect one chlorine atom per many thousands of moles of the stuff are less interesting. Any effect they have would be too small for this experiment to measure.

But what makes this difference? Well, my experiment rests on the underlying assumption that the interaction between a neutrino and matter depends on a valid quantum state being achieved with absolutely nothing left over, that weakly-interacting neutrinos have a state such that almost no possible "collision" between a nucleus and the neutrino could produce such a state, and that more-strongly interacting neutrinos have a state such that there are a number of possible valid interactions.

If this is true, then you would expect different decay rates between the bottom of an oceanic trench and ground level.

However, it isn't just neutrinos that are absorbed. Cosmic rays of all kinds will be. How do you differentiate between a neutrino and a cosmic ray? Well, the atmosphere absorbs quite a few cosmic rays but practically no neutrinos. So by measuring the level of cosmic rays at altitude vs. on the ground, and the change in decay rate (if any), you can calibrate your experiment accordingly. Rather more worrying would be if the decay rate is the same in both cases, precisely because high-energy particles blasting into unstable nuclei generally don't just bounce off. Further, you can't just look up the average values because the whole point of the experiment is that it is intended to be extremely sensitive. Direct measurement at the time of the experiment is the only way you can know the values and the impact of those values.

Why is this important? Because the signal you're looking for is extremely weak. You must be able to eliminate or filter out any/all noise stronger than that. There's no guarantee that you can, but if you don't filter it out, you'll certainly never see what you're looking for.

Re:I agree... (1)

Alsee (515537) | about 4 years ago | (#33726080)

The reason neutrino detectors are underground is that you don't want them to detect any old neutrino.

Neutrino detectors are put underground to shield out noise from stuff like cosmic rays.

You can indeed shield from -some- neutrinos

A solid lead wall one trillion miles thick would provide less than 10% shielding against neutrinos.
An entire planet will shield 0.000000000% of neutrinos. An entire star will shield 0.00000000% of neutrinos. Nothing short of a black hole will noticeably shield against neutrinos, and even then the black hole would just "vacuum up" the neutrinos rather than shielding them in any conventional sense.

-

Re:Not a certain conclusion yet (1)

jegerjensen (1273616) | about 4 years ago | (#33696962)

I doubt that these number would be considered "evidence of a difference" in any field, since the uncertainty is too large to confirm any difference. The headline is misleading as you say, but that claim is not made in the paper.

Re:Not a certain conclusion yet (1)

ceoyoyo (59147) | about 4 years ago | (#33697568)

Assuming the error is normally distributed, 2.3 * sigma is p = 10. They get p=0.02.

In physics I believe the custom is to wait for 3*sigma before you get excited, and 6*sigma before you announce a discovery. In most sciences it's 1.96 (for a reasonable dof) though.

Re:Not a certain conclusion yet (0)

Anonymous Coward | about 4 years ago | (#33697732)

This has been said many times but its worth repeating:

Extraordinary claims require extraordinary evidence.

Re:Not a certain conclusion yet (1)

MrKaos (858439) | about 4 years ago | (#33697860)

These results thus leave open the possibility that the half-life of a radioactive nuclide could in fact depend on its shape

Just going with the possibility *if* it does maybe the radionuclides radiate at a certain frequency and the shape helps it achieve a resonance where the decay rate is altered.

I'm not qualified to say whether that question even makes any sense, my brother has some relevant qualifications (I'll ask him tomorrow). Surely there is some physicists here who can tell if that is feasible?

Semantism (4, Insightful)

Kilrah_il (1692978) | about 4 years ago | (#33695660)

I think the proper phrasing should be "No evidence for inconsistency of nuclear decay found". It seems pedantic, but proper scientific methodology works this way. There
can still be inconsistency in nuclear decay, just not in this test scenario. You cannot prove consistency, you con only be very, very sure this is how nuclear decay works because you performed many studies that have failed to show something else. (Not that I despute their findings).

Re:Semantism (0, Redundant)

TapeCutter (624760) | about 4 years ago | (#33695704)

"It seems pedantic, but proper scientific methodology works this way. There can still be inconsistency in nuclear decay, just not in this test scenario. You cannot prove consistency"

To be uber-pedantic they are not claiming proof of consistency. They are claiming the same thing you are, ie: their test rules out nutrino flux as a possible cause for the observations.

Re:Semantism (2, Interesting)

Kilrah_il (1692978) | about 4 years ago | (#33695754)

Well, but our nice editors have used the incorrect phrase "Scientists confirm nuclear decay rate consistancy". Just responding to that.

Re:Semantism (2, Insightful)

maxwell demon (590494) | about 4 years ago | (#33696042)

They did confirm nuclear decay rate constancy. A confirmation is not a proof. It's just what the word says: A strengthening of the claim. It makes you more confident that the claim is true.

Re:Semantism (1)

John Hasler (414242) | about 4 years ago | (#33696098)

> They did confirm nuclear decay rate constancy.

No. They confirmed that nuclear decay rate is independent of shape.

Re:Semantism (4, Informative)

Frequency Domain (601421) | about 4 years ago | (#33695784)

To be uber-pedantic they are not claiming proof of consistency. They are claiming the same thing you are, ie: their test rules out nutrino flux as a possible cause for the observations.

Not quite, it doesn't rule it out. The observed changes are not large enough to be considered inconsistent with the hypothesis that neutrino flux has no role. With a larger sample or better control of variability, it's still possible that future experiments could reject the hypothesis.

Re:Semantism (1)

Ceriel Nosforit (682174) | about 4 years ago | (#33696034)

Since we're already being pedantic... and all good science takes it to the max...

What if the decay-rate inconsistency observed in the previous results is the more sensitive measurement of neutrino flux? Then this would be like saying that a measurement of 1cm doesn't exist because you used a ruler with inches.

tl;dr - OR IS IT??

Re:Semantism (1)

Demena (966987) | about 4 years ago | (#33695788)

No. They have no conclusion on consistency. Only that (they say) it is not caused by 'neutron flux'. However their study does not seem to do even that as there has been no defined mechanic. WIthout a hypothesised mechanic then not test is of much value as it only eliminates one possible mechanism.

Why wouldn't the scientists in this study... (0, Troll)

Anonymous Coward | about 4 years ago | (#33695674)

Use the same isotopes (manganese-54, silicon-32, and radium-226) that were in the original studies?

Because by using gold-198 they aren't validating their claim that radioactive decay itself is uninfluenced by neutrino exposure, only that gold-198's radioactive decay is likely unaffected by neutrino exposure.

It would seem then that they're purposely avoiding replicating the original studies parameters, perhaps to avoid observing the same phenomenon.

Re:Why wouldn't the scientists in this study... (5, Insightful)

Kilrah_il (1692978) | about 4 years ago | (#33695826)

You are right, they are purposely avoiding using the same isotopes to avoid observing the phenomenon that caused them to perform this research. "Truly, you have a dizzying intellect."
Call me naive, but maybe they had better reasons not to use the same material. I am not a physicist, so I don't know if it's correct, but here are some reasons I thought of, of the top of my head:
1) Gold may have more neutrino activity, so there was a better chance to observe said phenomenon.
2) The scientists involved have more experience working with gold, so they preferred using a material they are experienced with.
3) Gold may be easier to work with and this it is easier to construct thin foils.
4) They had a pile of unused gold and didn't know what to do with it :)

Again, I don't know if these are valid/correct reasons, but I'm somehow convinced there is a better reason than the one you stated.

Re:Why wouldn't the scientists in this study... (0, Troll)

Anonymous Coward | about 4 years ago | (#33695896)

Call me naive, but maybe they had better reasons not to use the same material.

All right, you're naive. As far as I can tell, the OP comment is still valid. How can they invalidate the orginal experimenter's experiment unless they try to replicate the original experiment?

First experimenter: "We observed this phenomenon in manganese-54, silicon-32, and radium-226."

Second experimenter: "We did not observe that phenomenon in gold-198 therefore you are wrong."

Of course, once they (potentially) validate the original experiment and observe the same phenomenon is gold-198 (showing that gold-198 is sensitive to the same phenomenon), then using foil and spheres made out of gold-198 to test sensitivity to neutrino flux is a clever idea.

Re:Why wouldn't the scientists in this study... (1, Informative)

Anonymous Coward | about 4 years ago | (#33695998)

How can they invalidate the orginal experimenter's experiment unless they try to replicate the original experiment?

They weren't trying to invalidate the findings of the original experiment. That was just something Slashdot made up for the heading (I've no idea why). They were investigating whether a particular mechanism could account for the findings of the original experiment.

Re:Why wouldn't the scientists in this study... (1, Insightful)

Anonymous Coward | about 4 years ago | (#33696250)

Yes, they're not replicating the original experiment, but they are testing the proposed, more general, mechanism -- i.e. that seasonal variations in neutrino flux from the Sun cause variation in decay rates. That's what the original experimenters were wondering about -- does this affect everything? I suppose followup studies could do the same experiment for every element in the period table and every radioactive isotope, including the ones used in the prior experiment, but the fact remains that the new experiment with gold has narrowed the possibilities for what might be going on, and they used a rather elegant experiment to do it. Is there an effect on all isotopes? Well, apparently not within measurement uncertainties for gold-198, although a smaller variation remains possible.

You can also turn the argument around: if the first experimenters found a real effect for Mg-54, Si-32, and Rn-226, then so what? That doesn't mean there's any effect at all on the decay rates for any other isotope, including the ones used for radiometric dating and other techniques where the observed variation (if taken at face value) would have a small effect. Maybe the process doesn't work for potassium, uranium, or rubidium, and until people test the specific isotopes in those methods, who cares if they find something odd in Mg, Si, or Rn?

In other words, you can't fault the followup study for not testing the same isotopes if you are simultaneously claiming that the original experiment has grand implications for the decay rates of all isotopes. Well, you can, but it wouldn't be particularly consistent.

Re:Why wouldn't the scientists in this study... (0)

Anonymous Coward | about 4 years ago | (#33696826)

I suppose if the effect was observed in some series of elements/isotopes and not others, then perhaps another hypothesis about neutrinos affecting the decay rate for certain orbitals might need to be tested. In which case they should be able to predict the next elements and isotopes in a series which would also exhibit the phenomena... Then its on to performing experiments for validation. Something like that might give some intricate detail about atomic structure that was overlooked previously.

But then again what do I know, IANANP.

Re:Why wouldn't the scientists in this study... (1)

Kilrah_il (1692978) | about 4 years ago | (#33696902)

you may not be a physicist, but you nailed the scientific methodology exactly. Kudos.

Re:Why wouldn't the scientists in this study... (0)

Anonymous Coward | about 4 years ago | (#33697548)

Yes. For example, perhaps it affects, say, electron capture modes of decay, for which a tiny variation in decay rate was noticed many years ago for extremely high pressures. This effect is well accepted, but it doesn't affect other decay modes. Actually, you can practically stop electron capture modes if you ionize the atoms. Same story -- other decay modes are unaffected.

If that kind of isotope-specific and decay-mode-specific mechanism the case, it would mean that the possibility of significant influence on radiometric dating results is wildly overblown by some people, because some isotopes used for such methods involve electron decay modes, some don't. Same for other decay mechanisms. The methods are pretty diverse in terms of the decay processes involved. Essentially, if it isn't some kind of general effect that influences all decaying isotopes similarly, then it's hard to come up with a scenario where it would matter in any significant way to radiometric dating methods and not be noticed already -- again, unless it is such a tiny effect that it is hard to measure.

Re:Why wouldn't the scientists in this study... (3, Informative)

Anonymous Coward | about 4 years ago | (#33696346)

3) Gold may be easier to work with and this it is easier to construct thin foils.

This. Gold is a phenomenally ductile metal -- ideal for making the thin foils typically used in preparing radioactive sources. If you want a radioactive source, the easiest thing to try (broadly speaking) is electroplating your nuclide of interest on a gold substrate. Then all your measurements require you to take the shielding properties of gold into account, but that's not usually too big a deal.

I am a nuclear physicist (grad student), and one of the key issues we have to deal with is sample preparation. The bleeding-edge is thin carbon foils, but that's expensive and tricky and takes a long time. If you want a sample quick, you use a gold substrate. No conspiracy here, folks.

Re:Why wouldn't the scientists in this study... (1)

BraksDad (963908) | about 4 years ago | (#33701192)

...4) They had a pile of unused gold and didn't know what to do with it :)...

You may have been trying to be funny, but perhaps it is simply the cheapest material to work with.

Re:Why wouldn't the scientists in this study... (2, Interesting)

sFurbo (1361249) | about 4 years ago | (#33695914)

It would seem that they started with the simple experiment. If there had been an effect here, no need to go and do the complicated experiment. Now that there weren't an effect, make the same experiment, but with an alloy of gold-198 and manganese-54, and only measure the decay of manganese. Of course, just the proximity of gold-198 should be enough, so perhaps layers of foil of gold and manganese? Anyway, these experiment are more complicated to carry out, so it makes perfect sense to start with the simple experiment.

How big? (3, Insightful)

srussia (884021) | about 4 years ago | (#33695690)

From TFA:“There are always more unknowns in your measurements than you can think of,” Lindstrom says.

How big were the foil and spherical samples? Neutrinos interact very weakly, so much so that neutrino detectors need to be on the order of 1 km^3.

Heck, if I had that much gold (whatever isotope) I'd have better ways to spend my time.

Re:How big? (5, Interesting)

sFurbo (1361249) | about 4 years ago | (#33695736)

The change in neutrino flux due to shape was bigger than the neutrino flux from the sun is, so it must be much bigger than changes in the solar neutrino flux (if I read the summary correctly). If that change in neutrino flux does not induce a measurable change in the rate of decay, then neither will the solar neutrino flux. I think it is a very elegant experiment, testing just what the hypothesis said.

The effect might be different for different decays, so the hypothesis isn't completely dead. Now, if they made an alloy of gold-198 and the isotopes that is claimed to change decay rate...

Re:How big? (0)

Anonymous Coward | about 4 years ago | (#33696654)

if you had gold-198 you wouldn't have it for very long, because in two and 3/4 days you would have mercury instead.

Re:How big? (0)

Anonymous Coward | about 4 years ago | (#33696944)

Neutrinos interact very weakly

You mean "all the known neutrino interactions are very weak", but we're talking about a new interaction here so it could have any strength.

Well maybe its something else coming from the sun (0)

Anonymous Coward | about 4 years ago | (#33695790)

Dark matter particles?

Re:Well maybe its something else coming from the s (4, Funny)

CarpetShark (865376) | about 4 years ago | (#33695878)

That's crazy talk. Everyone knows that the answer to all astrophysics problems is "11-dimensional dark matter particles".

Re:Well maybe its something else coming from the s (1)

Seeteufel (1736784) | about 4 years ago | (#33697756)

I didn't know that.

Re:Well maybe its something else coming from the s (1)

MK_CSGuy (953563) | about 4 years ago | (#33697844)

I call bull'... It's turtles, turtles all the way down!!

Re:Well maybe its something else coming from the s (1)

arthurpaliden (939626) | about 4 years ago | (#33696182)

Dark matter particles?
Only at night.

Only if it's neutrinos. (5, Insightful)

Israfels (730298) | about 4 years ago | (#33695816)

This, of course, is only true under the assumption that it's the neutrinos that are really causing the increase in radioactive decay. The article does mention that there were many unknowns in the measurements. It may be something else that causes this increase, or even a combination of two. It may also be the case that more neutrinos, the rate at which they're emitted, or other interacting fields alter the effect.

Re:Only if it's neutrinos. (2, Informative)

John Hasler (414242) | about 4 years ago | (#33696646)

Right. One must remember that the original article did not assert that solar neutrinos were cause. They merely speculated that they might be.

Variations in time rather than decay? (2, Interesting)

timepilot (116247) | about 4 years ago | (#33695840)

Could it be that there are local variations in time during the original solar flare observations rather than fluctuations in the actual decay rate, and that it is not related to neutrinos from the flare but from some other gravitational changes coupled with flares?

I know, my ignorance is showing. Sorry. IANASH (I am not a stephen hawking)

Re:Variations in time rather than decay? (3, Interesting)

maxwell demon (590494) | about 4 years ago | (#33696078)

I'd guess any variation in time so large that you can see it in decay time measurements would have created so many other clearly visible effects that it would not have gone unnoticed.

Re:Variations in time rather than decay? (1)

timepilot (116247) | about 4 years ago | (#33699282)

I suppose that's true if they were observable or more importantly if they were being searched for. But if nobody was looking for them they could have been missed.

What else would one look for?

Re:Variations in time rather than decay? (1)

timepilot (116247) | about 4 years ago | (#33699308)

And furthermore, this would be an excellent conversation to have with beer.

Re:Variations in time rather than decay? (1)

maxwell demon (590494) | about 4 years ago | (#33701968)

The atomic clocks which form the base of our official time are constantly monitored, and they are far more precise than any decay time measurement can be. I doubt any major disturbance in time would have gone unnoticed.

Moreover effects in time should go with gravitational effects. Note that the earth's gravity only has an effect on time of about 1e-16 per meter, and that already gives a clearly noticeable gravitational force. I couldn't find anything about the size of the effect, but the accuracy of the experiment described here was at 2*10^-3, which is extremely much larger. From that I conclude that any effect on time that could be detected by radioactive decay in relatively short time (the solar flare was just 43 minutes!) would cause gravitational effects of order larger than earths gravity; I guess that would have been noticed even if not specifically looking for it.

Now I admit that I'm not a GR expert (although I do know quite a bit about it), therefore there may be a flaw in that reasoning. However, the atomic clock argument is independent of that.

Re:Variations in time rather than decay? (1)

jasticE (196565) | about 4 years ago | (#33728906)

I'd guess any variation in time so large that you can see it in decay time measurements would have created so many other clearly visible effects that it would not have gone unnoticed.

What would be the most striking?

Re:Variations in time rather than decay? (1)

maxwell demon (590494) | about 4 years ago | (#33728962)

Objects mysteriously moving around due to the gravitational fields connected with such time variations.

Re:Variations in time rather than decay? (0)

Anonymous Coward | about 4 years ago | (#33699204)

It's an interesting question, and it is possible to grind out a full solution using General Relativity.

Without doing that, there are a couple things to consider:

1. We can select a pseudo-preferred reference frame in a laboratory here on Earth; we assume that the space-time in the 4-volume of the laboratory over the duration of the experiment is flat, that is 3-dimensional geometry will be Euclidean (if we build regular polyhedra we get the expected internal angles and edge lengths) and that a set of very precise clocks placed around the laboratory will agree on the frequency of a second across the duration of the experiment.

(This is a little synthetic and so we approximate it with cutoffs on the order of nanohertz (frequency error), microradians (angular error) and micrometres (length error), which are practical to detect in a lab setting, and which is practical to realize in a designed-as-flat lab).

2. We construct a conformal map in which we can safely ignore the rotation of the Earth about its axis and the revolution of the Earth about the sun, the rotation of the sun about its own axis, and the contribution of other bodies in the inner solar system. We assume that for the duration of the observation the proper motion of the Earth-sun system is fully inertial, so we don't have to correct for anything outside the immediate vicinity.

(In particular, we are ignoring the gravity potential gradients around the Earth and sun, and their contribution to gravitational time dilation. That means we have a means of treating the lab clocks as accurate throughout the Earth and sun system and the empty space in between (through which the flare moves) and have a systematic set of corrections for the changes in ruler-lengths that we observe in the system (doppler shifting, object foreshortening, etc.)).

(This is the hard part, and you really want a computer to do it rigorously).

3. We treat the neutrinos and other particles from the flare as a separate entity with its own kinetic energy. In reality the structure (shape) of the ensemble is more complicated, and subtracts a small amount of mass-energy from the sun. The difference to the sun is negligible and can be dealt with as per (2). We treat its movement as inertial. In reality, all the components of the flare must move against the sun's gravitational potential gradient, and that of the Earth when close enough to "feel" non-negligible gravitation from that. Assuming the flare gets an initial "kick" and then travels inertially, it means that the flare components will be moving more slowly at greater distances from the sun, since they behave as if they are losing energy rolling uphill from the sun.

Treating the different particles (photons, neutrinos, atomic nuclei, etc) as if they move in the same way between the sun and Earth is fairly reasonable; however, treating all the particles as if they were emitted simultaneously in a blob is not very realistic. In particular it *magnifies* the expected distortions by piling a more diffuse mass-energy into a smaller 4-volume of space-time.

4. Our "blob" has a lot of mass-energy in it; mass-energy creates a gravitational potential gradient around it, and that in turn causes "clocks" in it (anything that oscillates or decays with a statistical frequency, etc) to run more slowly (from the perspective of our lab reference timescale in 1). CMEs are large, this is not a stretch.

5. Our "blob" moves very quickly between the sun and the lab; this movement also distorts space-time, and by the equivalence principle can be seen in the same fashion as increasing the gravitational potential gradient around the "blob" (so clocks within it tick slower still, from the perspective of the lab timescale).

6. Let's eliminate the charged particles before they hit the lab; we cut down their mass-energy by having them scatter away in contact with the Earth's magnetic field, electrons in the solar wind, and so forth; the photons can bounce off things too, like the atmosphere.

Their mass-energy still influences the amount of time it takes to transit from the sun to the top of the Earth's atmosphere, but only the neutrinos really pass through the lab itself.

7. We now treat the neutrino residual as if it were interacting only gravitationally; neutrinos interact only very rarely and weakly with matter anyway.

8. Let's have the residual "wave" pass through the lab from east to west. The easternmost clock will briefly disagree with the westernmost clock in our lab because the "wave" causes it to be at a lower gravitational potential while it's in the "wave"; the east clock will tick slower. Other "rulers" will also be distorted; if we measure stuff in c-seconds (where c is the causality speed limit in General Relativity; the speed of light in vacuum moves is c) it's pretty obvious that a relatively slow second will make things shorter than a relatively long second. This distortion is measurable; gravity wave searches look for this sort of thing.

However, the wave will also pass through the westernmost clock, and will affect it the same way within the limits of our measurement cutoff.

(In practice, even with a large flare, the effects of the "wave" would be so small that we won't really be able to measure these distortions within the cutoff limit in 1, and at the limits of current technology it is very very hard to notice them at all)

9. The disagreement between clocks in 8 is critical. There is a (very brief!) period in which a researcher standing right at the easternmost clock and watching the decay rate of radioactive isotopes at the westernmost clock will see an increase in the frequency of decays. (The window is the proper time (of the researcher) between him or her being immersed in the "wave" and the "wave" passing through the isotopes).

10. We can lengthen the time in (9) by making the "lab" big. Let's have the easternmost clock on one side of the planet, and the westernmost clock on the other. The time-delay at c between the two clocks on opposite sides of the planet is on the order of 43 milliseconds. (We could always make it even bigger if we wanted).

11. Answer:

Yes, there are local variations in time during the flare observations that can influence remote observations fluctuations in the decay rate.

However:

*Local* observations of the decay rate will show much shorter fluctuations.

The amplitude of the change of decay rate depends on the amplitude of the change of neutrino mass-energy.

In practice, the decay rate is observed so locally that the fluctuations from General Relativity are completely negligible. This is normal; particle physics is almost always set against Minkowski space in order to use the much simpler Special Relativity (by ignoring accelerations due to gravitation).

Also in practice, the amplitude of the frequency difference is so small that it takes a lot of work coordinating different observers comparing the differences between their local precise atomic clocks and different GPS satellites (which are in-orbit atomic clocks) just to see any effect at all. More precisely still, they track the signal phase differences at particular UTC time stamps (i.e., they do comparative interferometry on the signal from orbit).

Unlike gravity waves from distant inspiralling binary star systems and the like, we do occasionally detect "blobs" of low-interacting mass-energy passing through the earth, mainly from the direction of the sun. (We see high-interacting stuff even more frequently -- that's what the aurorae are for example). These are expected to be neutrinos (General Relativity doesn't really care, but the standard model of particle physics offers those up as the most likely things).

We *may* detect gravity waves in a similar way; the major difference is directionality and the absence of even low-interacting effects (i.e., neutrino flares light up neutrino detectors; actual gravity waves would not).

So while the answer to your question is "yes", local time variations are negligible and are not expected to be able to account for longer term (or sharp) variations in isotope decay rates, because "local" rapidly and reliably smooths out in practice (otherwise we would not be able to agree on what time of day it is at Greenwich Observatory, for example!).

Re:Variations in time rather than decay? (0)

Anonymous Coward | about 4 years ago | (#33699558)

I need to clarify one important thing.

In point 3 I said the equivalent of "let's treat the ejected mass as a single cloud of gas comprising neutrinos and other particles".

Apart from the various other objections to this that I listed above, coronal mass ejections (flares) carry away material from at or near the surface of the sun. Neutrinos are produced deep within the sun where solar fusion is occurring. It's *unlikely* that there are lots of extra neutrinos in a solar flare, or in other words that the mechanisms of solar flares cause a noteworthy change in the omnidirectionality of neutrinos radiated from deep within the sun.

I call bullshit (-1, Offtopic)

Anonymous Coward | about 4 years ago | (#33695910)

There is no such thing as radio active gold as it has intrinsic value.

I want one. (1)

Rashdot (845549) | about 4 years ago | (#33695930)

A Gold-198 foil hat, to keep the neutrinos out...

Re:I want one. (1)

darthdavid (835069) | about 4 years ago | (#33695950)

Sounds like an expensive way to give yourself cancer...

So really... this means? (1)

Rainwulf (865585) | about 4 years ago | (#33695970)

I hate to be THAT person, but what does this mean for us normal humans? Does it mean anything at all?

Re:So really... this means? (1)

John Hasler (414242) | about 4 years ago | (#33696126)

"Normal humans"? This is Slashdot. If you don't find science intrinsically interesting you don't belong here.

Re:So really... this means? (1)

Rainwulf (865585) | about 4 years ago | (#33696210)

Oh i find it intrinsically interesting.
I was just wondering if it had any real world implications, which as i have read other people's comments, was noted as to the accuracy of dating methods.
Thats all. Science IS cool, i just wanted to know if this actually had any significance, or just one of those cool but non significant things science brings around ya know.

Re:So really... this means? (1)

kmcarr (1185785) | about 4 years ago | (#33698310)

Knowing whether radioactive decay rates are constant goes to our fundamental understanding of matter. How does that not have significance??

Re:So really... this means? (1)

maxwell demon (590494) | about 4 years ago | (#33696130)

It means there's no new physics at this point. So also no hope to exploit that new physics in new technology (e.g. to deal with nuclear waste).
On the positive side, it means that we don't have to expect nasty surprises from this new physics for our existing technologies (e.g. we don't have to expect that an extraordinary large solar flare suddenly makes a nuclear reactor fail, or something like that).

Re:So really... this means? (1)

Megaport (42937) | about 4 years ago | (#33696466)

On the positive side, it means that we don't have to expect nasty surprises from this new physics for our existing technologies (e.g. we don't have to expect that an extraordinary large solar flare suddenly makes a nuclear reactor fail, or something like that).

Umm, we didn't know that already? Oh dear. "New physics" had better not turn out to be an excuse for why we all suddenly glow in the dark, while it still isn't the nuclear power industry's fault.

--M

(PS - I love nuclear power, but I'm always a sceptic about our confidence when predicting unforeseen consequences.)

Re:So really... this means? (1)

John Hasler (414242) | about 4 years ago | (#33696706)

...I'm always a sceptic about our confidence when predicting unforeseen consequences.

I'm not. I'm quite confident of our inability to predict unforeseen consequences, because if we did predict them they would not be unforeseen. On the other hand I am also quite confident about our ability to predict that every act (or inaction) will have unforeseen consequences. Fortunately, most are not consequential.

Re:So really... this means? (0)

Anonymous Coward | about 4 years ago | (#33702690)

Current phsyics will deal with nuclear waste well enough. If we use modern reactor designs most of what is currently called waste becomes fuel, that will deal with 99% of the waste and the remains can be buried somewhere out of the way. I worry far more about all the crap spewed into the atmosphere by all the coal plants and by all the cars we drive, since we can't just take it and bury it, and even it we could it would take orders of magnitude more space to store than the equivalent waste from the same amount of power generated.

Re:So really... this means? (2, Insightful)

axedog (991609) | about 4 years ago | (#33696174)

It is of huge significance in radiometric dating. If we can show that the half-life of a radioisotope is constant, then it increases confidence in these dating methods. Conversely, if it can be shown that decay rates vary significantly, then accurate dating becomes more difficult and merits further research.

Re:So really... this means? (1)

jegerjensen (1273616) | about 4 years ago | (#33698358)

I may be wrong, but I would expect other uncertainties to play a more dominant role in radiometric dating. The initial concentration of the measured isotope, for instance, or the possibility of contamination.

Re:So really... this means? (2, Interesting)

jegerjensen (1273616) | about 4 years ago | (#33697200)

The constancy of nuclear decay rates is a fundamental assumption for our understanding of stellar processes, big bang theory and the like. If someone can prove a deviation, it will break just about every cosmological model. If they had found a deviation, the implications could be comparable to what Copernicus did, so I believe this does mean something for the "normal human"...

Not enough time (1)

spaceman375 (780812) | about 4 years ago | (#33696224)

The variation in decay rates is said to have two cycles; a yearly fluctuation, and a 33-day cycle (proposed first because that's the rotation of the sun's core, THEN found in the data). These experiments should have been run for at least 66 days, preferably for more than two years, before making claims that this has anything at all to do with the effects that have been observed so far. They can't even say that gold-198 displays evidence of the phenomenon they are trying to measure. This experiment cannot provide any useful information for investigating the possible connection between nuclear decay rates and distance to the sun. Nothing to see here but some attention grabbing with no real substance, gold-198 or not. Yawn.

Re:Not enough time (3, Insightful)

ceoyoyo (59147) | about 4 years ago | (#33696812)

They're investigating the hypothesis that it's neutrinos that cause the variation. Did you even read the summary, or just the sensationalistic headline?

Re:Not enough time (0)

Anonymous Coward | about 4 years ago | (#33699328)

No self respecting /.er RTFA!

The original "experiment" was moronic (0)

Anonymous Coward | about 4 years ago | (#33696268)

Nuclear decay rates do not vary seasonally. The thing is, the effects of the solar neutrino flux on the decay of an unstable nucleus is already known. It would be a not-very-difficult homework problem for any physics graduate student. It is effectively zero. You might as we postulate that the change in decay rate was caused by fairies, because that would be equally consistent with the known laws of physics. You have some guys trolling through some vast data set for weird blips, so of course they find some. Anybody with an undergraduate understanding of statistics knows that for a large enough set of hypotheses, you can always find one that is consistent with the statistical fluctuations in your data. Shame on the media for playing stupid. Shame on slashdot for being gullible.

color it blue (1)

noshellswill (598066) | about 4 years ago | (#33696322)

Well yes everybody **does** know you can't change nuclear decay rates with nancy-boyz neutrinos. Can't happen. Doesn't work. Not in a Millennia of Centuries. Everybody knows that the only way to change nuclear decay rates in some substance is to color it blue. Yep that's right. Robins egg blue. Make it sad and lonely and feeling (blue). Then the lil' bastards quarks & gluonz & E-fields slow down and ... **shit doesn't happen**.

This uFP for GNAA. (-1, Offtopic)

Anonymous Coward | about 4 years ago | (#33696436)

contact to see if Lagged behiTnd,

Radioactive decay variations (2, Interesting)

russotto (537200) | about 4 years ago | (#33696634)

This study provides strong evidence against solar neutrino flux being the reason for observed variations in radioactive decay. However, it does not provide evidence against those variations -- nor was it designed to. The measurements still need to be explained; there have been reports of changes in radioactive decay during solar flares, and also seasonal variations; most likely IMO they're some sort of systemic measurement error, but maybe not.

Also note that the idea that decay rates might be affected by particle flux or shape isn't all that farfetched. Fission rates in certain isotopes are, for instance.

Re:Radioactive decay variations (1)

ceoyoyo (59147) | about 4 years ago | (#33696828)

The process that produces fission is well understood, and the reasons why shape should play a role are well understood. If source shape affects decay rates it would be something very unexpected.

Neutrino oscillations? (2, Informative)

calidoscope (312571) | about 4 years ago | (#33696860)

The study overlooks neutrino oscillations, the neutrinos from the gold have had little chance to oscillate. While it is probable that neutrinos don't affect decay rates, the study isn't as conclusive as the summary makes it out to be.

The decay rate for electron capture is mildly affected by pressure.

How can they do such an experiment? (1)

istartedi (132515) | about 4 years ago | (#33697630)

IANAphysicist, but everything I've heard about neutrinos is along the lines of "they pass through the entire Earth with a very small chance of hitting anything". This makes me wonder how you can measure any kind of effect involving neutrinos, in a sample that isn't the size of an underground cavern full of water. Certainly they don't have a chunk of gold that big, or does gold have unusually high neutrino-interacting properties? How long does the experiment have to run? How sensitive is the whole setup and how do they isolate it from other neutrino sources, etc.

Re:How can they do such an experiment? (2, Informative)

Richard Kirk (535523) | about 4 years ago | (#33697920)

You are right - neutrinos can pass through a lot of matter without the matter affecting the neutrino, or the neutrino affecting the matter. Or so we think. A couple of people noticed that the apparent decay rates were different during a solar flare, which could mean there may be strange circumstances where the neutrinos had more effect than we expect. Or it could have been something other than neutrinos, if our understanding is that far off. I didn't think that was likely, but it doesn't hurt to test your assumptions once in a while.

The trouble with original measurements were that they could not easily be repeated. You could wait for the sun to have another significant outburst, and see whether the same detectors measure the same pulse, but you could never be sure whether that burst happened to put out a 'clump' of neutrinos in a particular direction. However, they designed a much neater experiment. This used a gold isotope that emits neutrinos when it decays. Some of it was made into a foil and some was made into a ball. You could then measure the foil with one detector and the ball with another, and then you could swap the detectors around. If this experiment is done properly then this could cancel out any difference in the detectors or the measuring environment or the background neutrino flux (if that is important).

If you design experiments, you get to appreciate the forms of really good experiments: they have a canon-like symmetry to balance out all the known and unknown effects you can imagine except for the one you are trying to measure. This feels like one of them. So, in the end, science wins. Yay.

FAQ (1)

bcrowell (177657) | about 4 years ago | (#33697856)

FAQ: Do rates of nuclear decay depend on environmental factors?

There is one environmental effect that has been scientifically well established for a long time. In the process of electron capture, a proton in the nucleus combines with an inner-shell electron to produce a neutron and a neutrino. This effect does depend on the electronic environment, and in particular, the process cannot happen if the atom is completely ionized.

Other claims of environmental effects on decay rates are crank science, often quoted by creationists in their attempts to discredit evolutionary and geological time scales.

He et al. (He 2007) claim to have detected a change in rates of beta decay of as much as 11% when samples are rotated in a centrifuge, and say that the effect varies asymmetrically with clockwise and counterclockwise rotation. He believes that there is a mysterious energy field that has both biological and nuclear effects, and that it relates to circadian rhythms. The nuclear effects were not observed when the experimental conditions were reproduced by Ding et al.

Jenkins and Fischbach (2008) claim to have observed effects on alpha decay rates at the 10^-3 level, correlated with an influence from the sun. They proposed that their results could be tested more dramatically by looking for changes in the rate of alpha decay in radioisotope thermoelectric generators aboard space probes. Such an effect turned out not to exist (Cooper 2009). Undeterred by their theory's failure to pass their own proposed test, they have gone on to publish even kookier ideas, such as a neutrino-mediated effect from solar flares, even though solar flares are a surface phenomenon, whereas neutrinos come from the sun's core. An independent study found no such link between flares and decay rates (Parkhomov 2010). Jenkins and Fischbach's latest claims, in 2010, are based on experiments done decades ago by other people, so that Jenkins and Fischbach have no first-hand way of investigating possible sources of systematic error. Laboratory experiments[Lindstrom 2010] have also placed limits on the sensitivity of radioactive decay to neutron flux that rule out a neutrino-mediated effect at a level orders of magnitude less than what would be required in order to explain the variations claimed in [Jenkins 2008].

Cardone et al. claim to have observed variations in the rate of alpha decay of thorium induced by 20 kHz ultrasound, and claim that this alpha decay occurs without the emission of gamma rays. Ericsson et al. have pointed out multiple severe problems with Cardone's experiments.

He YuJian et al., Science China 50 (2007) 170.

YouQian Ding et al., Science China 52 (2009) 690.

Jenkins and Fischbach (2008), http://arxiv.org/abs/0808.3283v1 [arxiv.org]

Jenkins and Fischbach (2009), http://arxiv.org/abs/0808.3156 [arxiv.org]

Jenkins and Fischbach (2010), http://arxiv.org/abs/1007.3318 [arxiv.org]

Parkhomov, http://arxiv.org/abs/1006.2295 [arxiv.org]

Cooper (2009), http://arxiv.org/abs/0809.4248 [arxiv.org]

F. Cardone, R. Mignani, A. Petrucci, Phys. Lett. A 373 (2009) 1956

Ericsson et al., Comment on "Piezonuclear decay of thorium," Phys. Lett. A 373 (2009) 1956, http://arxiv4.library.cornell.edu/abs/0907.0623 [cornell.edu]

Ericsson et al., http://arxiv.org/abs/0909.2141 [arxiv.org]

Lindstrom et al. (2010), http://arxiv.org/abs/1006.5071 [arxiv.org]

Gold 198 emits antineutrinos, (1)

scruffie (174124) | about 4 years ago | (#33698146)

...while the Sun, through proton-proton fusion, emits neutrinos. If solar neutrinos do affect radioactive decay, maybe it's because of the difference between neutrinos and antineutrinos?

What it really confirms... (1)

AJWM (19027) | about 4 years ago | (#33698576)

Ignoring the noted discrepancies (which may mean the experiments don't confirm anything), the experiment as designed confirms only that neutrino flux -- of the type of neutrinos emitted by Au-198 decay -- does not affect the decay rate of Au-198.

One could generalize this further to say that (Au-198) neutrino flux doesn't affect beta decay, but that's only one type of decay ... and one flavor of neutrino. (Neutrinos come in three flavors, plus their antiparticles. Beta decay actually produces electron antineutrinos, and while some observations suggest than neutrinos can oscillate between different flavors, the time scale for that is too long for it to occur in these small gold samples.) In fact, one (controversial, of course) proposed explanation for the decay-rate oscillations (in Pr-140 and Pm-142) observed at Darmstadt relates to neutrino oscillation.

Even with this generalization, they've only covered 1/9th of the possible combinations of decay type and neutrino flavor (or 1/18th counting antineutrinos). So it's an interesting experiment, but doesn't confirm what the summary says it does -- except for Au-198 beta decay and electron antineutrinos.

I'm not saying that there is a real effect seen in the long-term experiments that seem to suggest it, but considering that measurements of decay rates of different isotopes seem to have shown a different phase vs the position of Earth's orbit, and that some isotopes (eg plutonium in deep space RTGs) may not be showing an affect at all (hard to tell because the decay rate is inferred from a measurement about three levels of indirection removed from the actual decay rate), it leaves open the possiblity that the mechanism (if there is one) varies with the type of nucleus involved.

If there is such a decay rate change, it may have nothing to do with Earth's distance from the Sun, but rather Earth's distance from something else (eg galactic core) in which case interplanetary RTG decay rates wouldn't vary nearly as much as if distance from the Sun were a factor.

(Also, I note that observed decay rate changes so far seem to occur in isotopes of lighter elements (Pr, Pm and Si) and not in isotopes of heavier elements (Au, Pu) so perhaps the more complex nuclear structure of heavy elements masks or cancels the effect.)

So, interesting experiment, and it narrows down the places to look for a possible cause, but it doesn't prove that there's no effect.

Headline is wrong (1)

DCFusor (1763438) | about 4 years ago | (#33698634)

All they proved is that neutrinos created by decaying gold 198 don't seem to affect the decay rate of gold 198. The variable decay rate issue is still alive and kicking.

And with the short half life of gold 198, it's hard to believe they even proved that. I work with it on a daily basis, as an integrating neutron detector for my fusor (normal gold 197 picks up a neutron in a moderated neutron oven and becomes radioactive). It's fairly numb compared to say, Silver or Indium, but a little longer lived so the error due to the time it takes to stop a run and start the activity count is less. To make enough gold hot enough to do a statistically valid test, they must have had one heck of an intense source of medium-low energy neutrons. Gold picks them up best at a resonance energy somewhat above "thermal" which is what is found in most reactors -- it's more complex than that, of course, as reactors have a spectrum of neutron energies available if designed for that.

As another poster (calidoscope) pointed out, neutrinos seem to oscillate, and another poster (scruffie) also pointed out that there are also antineutrinos.

So this test was pretty limited in terms of what was actually tested, and how well it could be tested over the pretty short half life of radioactive gold.

Better than not testing at all, but just barely, it doesn't cover many of the bases at all.

Frame of reference, distribution of mass (0)

Anonymous Coward | about 4 years ago | (#33698682)

Neutrinos? Could it be that the distribution of mass in the local frame of reference plays a role? No "particle" interaction required!

I'd like to see the experiment that tests this. :)

Headline is wrong (1)

sjames (1099) | about 4 years ago | (#33698790)

The experiment doesn't address the larger question of variable decay rate, nor was it designed to. Instead, it indicates that if there is a variability, it probably isn't caused by neutrino flux. That is, in itself, a useful (non)result.

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