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I was surprised last week to find this value:
for a formula I had designed for a completely unrelated purpose.

$$\sum _{m=1}^{\infty } \frac{1}{\left(m^2+1\right){}_{2 m}}$$ where $\left(m^2+1\right){}_{2 m}$ is the Pochhammer symbol.

I posted over onPhysics.SE where they beat me up over the definition of mass. I'm not a physicist so it was a bit confusing. But, I don't want to discuss proton mass in this post. I want to discuss the odds of this happening at all!

We have $0.167$ as $3$ digits of accuracy for $1959$ and $0.1672621$ as $7$ digits of accuracy for $2012$. At $10$ numbers per each of the $4$ digits difference, we beat $9999$ to $1$ odds against that happening.

Is there a point after a number of successful steps that we would have some kind of confidence factor that would indicate that the number corresponds to the proton's mass or not?

1: Mathematica code:
NSum[10/Pochhammer[m^2 + 1, 2 m], {m, 1, [Infinity]}, WorkingPrecision -> 50]
2: CRC Handbook for Chemistry and Physics, 12th edition, (1959), Page 16.
3: http://pdg.lbl.gov/2012/reviews/rpp2012-rev-phys-constants.pdf


The number $m_{p_{0000}}$ is a percentage. When we multiply the length of the sides of the gram by 10 and then cube it to get kg, we don't cube the number. We change its exponent by 3. Therefore, the number is dimensionless.

I've added a bounty for the best answer that is back on topic.

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If there are planets with intelligent life, it is unlikely their unit of mass is the kilogram. So what you have observed would not be observed on other planets, and therefore cannot have physical significance. –  André Nicolas Nov 20 '12 at 2:10
This also uses the mass in kg. @Fred, please explain what you mean by the alien cube remark. André is right; his remark is not on a side track but goes to the core of your question. –  joriki Nov 20 '12 at 6:59
@Fred: I'm afraid I have no idea what you're talking about. What does "throw out the gram" mean? What are the "steps we choose"? You're asking about a quantity specified in units of kilograms, and I don't see any reason, including in any of what you've written, to assume that anyone other than Earthlings who happen to use kilograms would ever encounter this number. –  joriki Nov 21 '12 at 17:50
The wording of the comment was deliberate. Whether there is intelligent life on Earth is open, though there is some positive evidence in favour of dolphins. –  André Nicolas Nov 22 '12 at 0:06
@Fred: You must be very old! Since 1889, the kilogram has been defined as the mass of the International Prototype Kilogram, which is a lump of metal somewhere near Paris. –  TonyK Nov 23 '12 at 7:58

1 Answer 1

up vote 2 down vote accepted

The threshold in Physics is colloquially known as "five sigma." See, for example, this article, which might resolve some of your underlying confusion. Note that it is important to go into an experiment with a well-supported hypothesis; you should not (in general) first get the result and then go back and see if it corresponds to something known.

For example, your first number begins: $0.1672621822\ldots$

Next, let us consider $15357 \pi/288442 = 0.1672621822\ldots$

Neither mathematical nor physical reasoning allows us to conclude whether or not they are equal.

In any event, to answer your question directly: there is no chance the constant you have found corresponds to the physics term of interest, and with quantities that can be measured very well, you would need both (1) digits matching as far as they can be computed, and (2) a convincing reason [where the meaning of convincing depends on your discipline] for why they should be equal.

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The 'five sigma" article was interesting. So what would be the five sigma for the fact that testing (over the last 50 years) has added four significant digits that match to my digits? Or is this not enough to be significant? –  Fred Kline Nov 22 '12 at 3:40

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