5
$\begingroup$

Measurement units are defined in relation to some amount of physical quantity or physical phenomena. I have a doubt if certain amounts of a certain physical concept can never be measured by a given physical quantity.

For instance, suppose I have a rational amount of time written in the form:

$$ \frac{p}{q},$$

Then measuring that through periodic physical process is simple, I run the cycle $p$ times and I see how much is the $q$th part of it..., but how would I see the amount of seconds for an irrational amount of time? By the very definition it can't be written as "a part of some number of cycles" anymore.

So, is it impossible to ever be able to 'detect' or 'measure' an irrational amount of stuff?

Edit: my question is only loosely related to the dupe because mine is about dectability/measurability rather than existence.

$\endgroup$
11
  • 3
    $\begingroup$ I don't understand the question. This is a metrological definition of a second, not a prescription that you must only measure time by counting these cycles. The clock on your wall (if any) certainly doesn't count cesium cycles, yet it certainly measures time. $\endgroup$
    – ACuriousMind
    Jun 9 at 12:24
  • 4
    $\begingroup$ You can get arbitrarily precision since the rationals are dense. Does this work for you? $\endgroup$
    – DanDan0101
    Jun 9 at 12:29
  • 2
    $\begingroup$ I'm confused what this has to do with the specific definition of time - yes, if your measurement apparatus only measures the quantity in discrete steps, you'll have trouble measuring something that doesn't exactly hit one of these steps. How is this different from e.g. a ruler with certain markings where you can't really get a perfect reading of measurements that fall between the marks? Can you measure $\sqrt{2}$ centimeters with a ruler marked in centimeters? $\endgroup$
    – ACuriousMind
    Jun 9 at 12:33
  • 2
    $\begingroup$ Rational numbers are dense in the reals, so the question sounds a bit irrelevant or ill-posed. See also: physics.stackexchange.com/q/52273/226902 (possible to measure irrational length?). The opposite of your question would be: physics.stackexchange.com/q/2010/226902 $\endgroup$
    – Quillo
    Jun 9 at 12:40
  • 4
    $\begingroup$ And my point is that all physical measurements are like this (i.e. discrete/not an irrational number) even if only because all measurement apparati have finite precision. Both the SI definition of the unit of a second and the focus on time seem entirely irrelevant to me in this context. See physics.stackexchange.com/a/76823/50583 for a generic discussion of (ir)rational measurements $\endgroup$
    – ACuriousMind
    Jun 9 at 12:41

5 Answers 5

4
$\begingroup$

is it actually possible to measure irrational time through such a definition of a second?

Sure, simply use a different frequency standard. There is no reason that you must use caesium as your only possible frequency standard.

Any measurement will have some uncertainty. Within that region of uncertainty there will be an infinite number of rational numbers and an infinite number of irrational numbers. So choosing to use an irrational or a rational number is entirely a matter of choice. Either will fit the data just as well.

So the process would go like this: choose any different frequency standard whose period, in seconds, is $T\pm \delta T$. So then let $T-\delta T < T_{ rational} < T + \delta T$ be any rational number in that range and let $T-\delta T < T_{ irrational} < T + \delta T$ be any irrational number in that range. Then if you use the new standard to measure some time interval $nT$ you can equally claim that it is a rational number of seconds, $nT_{rational}$, or an irrational number of seconds, $nT_{irrational}$. Both numbers are equal to within the experimental uncertainty so both are equally valid and equally supported by the data.

$\endgroup$
2
  • $\begingroup$ I updated but to be honest I don't understand how adding diff frequency fixes it. Could you please explain in more details $\endgroup$ Jun 9 at 21:04
  • $\begingroup$ @Aplateofmomos hopefully that edit helps $\endgroup$
    – Dale
    Jun 9 at 21:25
2
$\begingroup$

Yes it’s impossible to get a direct measurement whose result is irrational. But it’s not so much that you’re comparing to a reference quantity but rather because every measurement has finite precision and is therefore rational.

On the contrary, if you do an indirect measurement and use a theory calculation based on your direct measurement then your indirect result can be irrational. For example if you measure the radius of a circle and infer it’s circumference using $\pi$ or if you use a clock which is know to tick at an irrational multiple of the cesium clock transition frequency (though I don’t know how you would come by such a clock in practice).

$\endgroup$
1
$\begingroup$

You can not measure anything irrational, if you for example measure the circumference of a circle, you will never have a multiple of pi. The concept of measuring is allways comparing with a unit and find the rational parts or multiples of it. you would not even know exactly what pi s or e kg are.Since e and pi are just names for something you can never know in our decimal system.

$\endgroup$
1
  • 1
    $\begingroup$ Actually you can, using appropriate units: take a circular ruler larger than your circle, with ticks marked from the top around and back to the top like e.g. 0, 0.1, ..., 0.9, 1. Use it to measure your circle, and you'll be able to tell its circumference in multiples of $\pi$ (within measurement uncertainty). Moreover, you can use it to also measure the diameter, it'll also be in multiples of $\pi.$ $\endgroup$
    – Ruslan
    Jun 9 at 21:45
0
$\begingroup$

You can certainly take a measurement and end up with an irrational number.

Here's an example of one way to do that. Suppose that you have a lump of a clay-like material and you want to determine its volume. One option is to roll it into a cylinder, and measure the radius and height of that cylinder. Suppose that you measure this cylinder and find that it has a height of $\def\cm{\ \mathrm{cm}} 2.66 \cm$ and a radius of $1.1 \cm$. Then you've measured its volume as $2.66 \cm \cdot \pi \cdot (1.1 \cm)^2 = 3.2186 \pi \cm^3$, which is an irrational number.

Now you can use this cylinder to measure irrational amounts of time. Suppose that you have a tiny faucet that's calibrated to put out $1 \cm^3$ of water per second. Then you take your cylinder and use it to displace some water into a test tube, and then you open your faucet into a second test tube and close the faucet again once the amount of water in the two test tubes is equal. You just measured out $3.2186 \pi$ seconds.

Of course, these methods of measuring aren't exact, and arguably, they're pretty silly. But the measured values that you end up with are definitely irrational numbers.

$\endgroup$
4
  • $\begingroup$ You say "the measured values that you end up with are definitely irrational numbers." This is definitely not true. The measured values have a finite number of valid digits, multiplying by pi does not change it, written pi as a multiplier give a inexperienced kid the impression of "irrational" which say infinitely many digits. True is 3.2186$\ pi $ is a irrational number but 3.2186$\ pi $s is not a measurable time. $\endgroup$
    – trula
    Jun 12 at 10:43
  • $\begingroup$ @trula The way I would describe it is that the numeric result of the measurement is an irrational number, but the measurement only has a finite amount of validity. So, maybe you take a measurement like I describe and you conclude that the faucet was turned on for $3.2186 \pi \pm 0.1 \pi$ seconds. The center point of that interval is still $3.2186 \pi$, an irrational number. The phrase "the measured value is an irrational number" does not imply that the measurement has an infinite amount of precision. $\endgroup$ Jun 12 at 13:19
  • $\begingroup$ Again no! your measurement gives you 3.2186±0.00005 and you can only multiply it by 3.1416 giving you 10.1115±0.00015 .Myschoo kid often give me answers coming from their calculators giving all the digits the calculator gives them if they divide some measurements by 3 or seven. I try to teach them that most of the digits are just " calculator garbage" $\endgroup$
    – trula
    Jun 12 at 14:05
  • $\begingroup$ @trula That's true. Any measurement that's recorded digitally involves a digitization step, and as far as I can think of, the result of digitization is always a rational number. $\endgroup$ Jun 12 at 15:52
0
$\begingroup$

Yes

Since both rational numbers $\mathbb{Q}$ and irrational numbers $\mathbb{I}$ are dense in the reals $\mathbb{R}$, you can measure in either rational or irrational quantities. That is, for any measurement, there will be both rational and irrational numbers arbitrarily close to the center of the measurement error region. Rational or irrational measurements only make sense if one has zero measurement error.

Example:

Define a piinch as $\pi$ inches. If we lathe a round dowel down to an inch thick, then we can ink the top and bottom of the dowel. Then we can make a measuring tape with demarcations of piinches by rolling the dowel over a blank tape measure. We can similarly mark fractional piinches on a tape measure.

Now you have the ability to measure length in piinches or fractions of a piinch. Any measurement with a piinch measuring tape (number of piinches and fractional piinches) will be an irrational number of inches.

For any length $l$ for an arbitrarily small error $e$, there will be some fractional values $x$ in inches and $y$ in piinches within $e$ of $l$.

$\endgroup$
2
  • $\begingroup$ NO!Multiplying a known measurement with $\pi$ is not a measurement of an irrational number. $\pi*r^2 $ gives you the area of a mathematical circle , not a physical object which is considered a good approximation of a mathematical circle. Multiplying $\pi$ with any measurements you made to a precision of n digits, gives you number which has n digits, all the rest is just garbage. $\endgroup$
    – trula
    Jun 10 at 15:49
  • $\begingroup$ Who says I have to use a rational number for unit conversion? $\endgroup$ Jun 10 at 17:01

Not the answer you're looking for? Browse other questions tagged or ask your own question.