Can you find the length of a pencil without a ruler or clock?

Question

Edit: Assume you do have access to all sorts of instruments, but they are all shrunk in proportion. My real question is:

If you are shrunken (or expanded) by a constant factor and put in a room shrunken by that constant factor, can you be fooled into thinking you are in a regular-size world, by altering your time perception by a constant factor?

Suppose a magic fairy comes up to you, and transports you into some opaque box. You are given a potion that alters your time perception, and changes the size of your body.

Assume that the effects of the potion are constant: each second feels like $x$ seconds, and your height has been multiplied by a factor of $y$. You do not know the values of $x$ and $y$.

Helpfully, the fairy has left you a suitably shrunken/expanded copy of a textbook, listing basically every single constant and formula in physics. Unfortunately for you, all the dimensionful constants are measured in SI units.

There is no clock or ruler in sight. You can measure the ratios of sizes or time intervals because the potion doesn't have variable effects.

The fairy gives you a pencil, and says that once you tell him how long the pencil is, in centimeters, you are released.

There are some ways to do this task of course. The ones I thought of involve building a microscope and attempting to eyeball the radius of, say, a carbon-12 atom, and then multiplying that by the ratio between the radius of a carbon-12 atom and a meter, found in the textbook. However, this is close to impossible to do. We might be able to try to diffract light to measure objects in multiples of, say, the wavelength of 500 nm green light.

However, neither of these methods are particularly accurate or useful. Besides, without any measurement equipment we can't be sure of the wavelength of the light. Would there be some obvious, classical way of deriving the length of a meter without access to any ruler?

Answer 1

OK, time to burn thru all my mod points:

for (j  in 0:1e6 cm,by=1Angstrom) {
    print("Hi, magic fairy, the pencil is" ,j, "cm long")
}

Sooner or later you'll get it right and off you go. (With apologies to the world-famous "If you tell me the height of the building I'll give you this barometer" story)

EDIT: for those who never ran across it, the "barometer" story is an example of lateral thinking in the classroom. I'm sure it's apocryphal. Here goes:

Teacher finishes explaining to a Physics class about variation of air pressure with altitude and then asks 'How can you use a barometer to find the height of a tall building?" Well, you could try to measure the variation in barometric pressure and dig up enough reference data to guesstimate the altitude, or you could 1) drop the barometer off the top and, knowing gravitational force, use the measurement of elapsed time to estimate the distance travelled, 2) knock on the door of the building supervisor and say "I'll give you this really nice barometer if you'll tell me from the construction plans how tall the building is." 3) [fill in your own]

Answer 2

transports you into some opaque box.

We are in a box, therefore we don't have additional measuring devices such as dynamometers or electrometers.

each second feels like x seconds, and your height has been multiplied by a factor of y. You do not know the values of x and y.

We can measure length and duration accurately albeit scaled. (For narrative purposes: use the pencil as a ruler, and count (1 Mississippi 2 Mississippi ) as a chronometer.)

Method

Drop the pencil from some height. At some point, Measure its velocity, the height it traveled, and duration since you dropped it.

You have

$v_0 (ym / xs) = g * t_0 (xs)$

and

$\frac{1}{2}v^2 (ym / xs)^2 = g h_0 (ym)$

solve for y and use it to give pencil length.

EDIT

After all your galilean experiments, you can only measure g in $pencil/Mississippi^2$, which gives you just $y/x^2$.

Cop out: Measure pencil's length while it moves at constant velocity. Solve for $c$ using length contraction formula $L=L_{0}\sqrt{1-v^{2}/c^{2}}$. Which gives you $y/x$. Now, solve for $y$, (for real this time).

Answer 3

Assuming all laws of physics are the same in this magical new world, and taking you at your word that everything is exactly proportional (so an atom would be larger/smaller depending on your new universe), a meter would still be defined as "the length of the path traveled by light in vacuum during a time interval of 1/299 792 458 of a second". So, a centimeter "there" is just like a centimeter "here". Figure out the dimensions of your book, or find some scale inside of the book, and measure the pencil as normal.

Answer 4

Strip out the graphite from the pencil and make an estimate of the diameter of the "lead" compared to the length of the pencil. e.g. If the pencil is height $n$ metres, and the pencil "lead" has a diameter of, say, $n/100$, then the volume of graphite you have is $V = \pi n^3/400$. Weigh the graphite and this tells you what mass of graphite you have and then use the known density of graphite to tell you what $V$ and hence $n$ is.

How accurate you would be depends on how well you can measure how many pencil "lead" diameters make up the height of the pencil.

In terms of whether you could be fooled into not realising your size had changed; there is a simple test.

You didn't say whether the transformation you proposed was mass conserving. If it were, then you should weigh the same, but be $y^{-3}$ times denser than you were (assuming it wasn't just your height that was altered!). If $y<1$, then you would not be able to float. If $y>1$ then you would float abnormally well.

If the process somehow modified your mass to keep your density the same, then the simple act of weighing yourself would reveal the value of $y$.

Answer 5

The condition to not have access to a ruler or a clock is actually not very restrictive.

Use 2 electrons and use a very precise forcemeter.

$$F = k\frac{q_aq_b}{r^2}$$

Measure F and use the formula to deduce r, the distance between the 2 electrons.

Put one electron at each end of the pencil to answer the riddle.

Now if your question is can we determine a one meter distance without making any kind of measurement, I don't have the answer.

EDIT: here's a force-meter

Answer 6

After reviewing the statements more carefully, I have deviced a method to determine the "real length" of a pencil (stick). If I can determine the values of the y (length) and x (time) factors, it will be simple to determine the pencil's length. In order to determine the y & x values, the following information is assumed/given: $g_1 = g_2 = g = 9.8 m/s^2$ ; $c_1 = c_2 = c$
I, and everything in the room shrunk by y = 1/2.
My thumb width tw, defined as 2cm long.

Solution: find out the shrink factor and thereby be able to find the length of the pencil.

Method: create a "pendulum clock" who's period is 1 second (local time). The length of a string for such pendulum is given by $L_2 = \frac{g{t_2}^2}{4{\pi}^2}$. For $T_2$ = 1 second, $L_2 = 24.85cm$. Using my thumb, I measure the string and create a "pendulum clock". Keep in mind that since I am shrunk by 1/2, $L_2$'s "real" length ($L_1$) is $\frac{L_2}{2}$, or 12.42cm. So,outside my shrunk room, this pendulum would have a period of $T_1 = 2 \pi$ $(\frac{L_1}{g})^{1/2} = \frac{1}{2^{1/2}}sec$. Using a fiber optic cable of length (d) of 3 x 10^10; an LED; and a light detector; the time delay should be $t_2$ = d/c = 1sec. However, the "real time" $t_1$ is = d/2c = 1/2 sec (since the real length is d/2), but since I am measuring time with my pendulum, then the pendulum period ($T_2$) required is $\frac{1/2}{\frac{1}{2^{1/2}}}$ = $\frac{1}{2^{1/2}}$. What must I do now to the string length so the pendulum's period is $\frac{1}{2^{1/2}}$ sec (instead of 1 sec)? I have to multiply its period by $\frac{1}{2^{1/2}}$, which means its length must be multiplied by 1/2, thus factor y = 1/2. So now, if you give me a stick that's 20cm long, I will determine it's 40cm long, but because I know the shrink factor is 1/2, I can say that the length of the stick is really 20cm.

Answer 7

A low-tech solution

Assuming that you're near or on the surface of the Earth, you construct a pendulum (of unknown length). Then determine the oscillation frequency, using your pulse as a time base. You can then determine the length of the pendulum, which you use as a ruler to calibrate all other lengths.

Your pulse isn't very accurate as a clock, but I would guesstimate that you can get the length better than 10%. Hopefully the fairy isn't too picky.

Answer 8

They do it with mirrors

Take an interferometer set up using a laser and where you can vary the distance between one mirror and the beam splitter.

By knowing how far you have gone between two points where the light was fully destructively interfering you can know the distance in the number of wavelengths.

Then you just put the pencil lined up with the mirror's starting point and move the mirror along a track until you get to the end of the pencil. You will know the length in terms of the wavelength of the light from the number of full destructive interferences you had.

Answer 9

The strict answer is: Count the atoms in the direction you want.

The answer by Dutch Brannigan is 100% correct. I will expand on it for those many that do not understand it.

All the measures we use are based in the atomic properties and counting.
For example the mass unit 1kg is the mass of a definite ensemble of atoms (in Paris prototype) and the mass of each atom is a little part of it.

Only pratical reasons prevent us to choose the usual mass unit ($m_u$) as being the mass of an electron.
The charge unit ($c_u$) is well known: the charge of an electron.
The length unit ($l_u$): we can agree to be the electron radius or the H Bohr radius.
We assume 'c' is constant and the time unit ($t_u$) is a derived one ($c=l_u/t_u$).

In fact when we measure distances we are counting time units.

It is now clear that we measure with atoms (like the ones we see around in the local universe - here and now - the 'regular-size world' in the question.

All the laws of physics are relations about the atomic properties and they are the same laws WHATEVER the actual 'size' of the atoms (mass, charge, Length, time). No law exists able to make a distinction between the atoms of now and the ones in the past.
No one can show with an experiment that the atoms are not changing as time goes by, because to make a measure one have to compare at the same instant the ones of now with their past version that are not available now.

Without a law and without an experiment it is improper to a physicist, and anyone, pretend that the atoms are not evolving. The consensus that the 'space expands', without any cause, assumes - it is an hidden axiom - that the atom is not a varying entity - without any kind of evidence.
To the extraordinary claim 'the space is expanding' an extraordinary demonstration is mandatory. It is the amount of space that is expanding or only 'the measure of it' with a shorter and shorter atom ?

I maintain with a mathematical proof (A self-similar model of the Universe unveils the nature of dark energy) that the atoms are evolving, contrary to the consensus, as seen in the redshifted radiation from the past emited by larger atoms.

An exercise on measure and units: suppose that inside the box we have a sample of radioactive Radon and we have a special eye to detect the radiation. If we use it as a clock unit - we can't know that the $t_u$ is decaying - and the laws of physics became awkward - think about the planets motion as they appear to be approaching us and speeding up - a longer trajectory in each $t_u$.

Using the scientific method one can never discover a new paradigm but one should used it to show that I'm claimning a falsehood. Give it a try.