riddle - Desert Wanderings bring…Ciphers? Part 1

^{Part 2 can be found here
Part 3 can be found here}

You should’ve know not to listen to that sketchy peddler. Now you’re lost in the middle of the desert with nothing to eat and hardly any water left. If you don’t somehow manage to find an oasis, you’re going to die.
As you continue walking you see something in the distance. Is that what you think it is? Apparently so, you’ve found a lush oasis! One slight downside however, between you and the oasis is a Sphinx and she’s noticed you, so sneaking around is no longer an option, nor is going the other way. So, resolutely you walk up to her and she greets you in the usual fashion of Sphinxes:

 Greetings potential lunch, my name is Azerty. Answer these 3 questions

 correctly or I eat you. If you get all 3 correct, I’ll let you go to
 the oasis behind me. I’ll even go so far as to tell you how to leave
 this desert. Although, that last part is simply because I don’t want
 you to wander over here again.

As you ready yourself for try and figure out some riddle this Sphinx starts to write instead of speaking her puzzle. As you look at her writings you realize that this isn’t a riddle but a cipher! Who ever heard of a Sphinx writing ciphers?

39 64 32 411 64 410 39 46 47 36 59 64 57 38 36 64 58 310 36 64 311 32 35 36 39 55 410 34 64 310 35 64 53 32 56 34 510
39 64 55 32 58 64 57 34 64 44 310 53 58 59 64 57 38 36 64 47 34 410 311 64 37 310 38 64 43 36 32 37 64 38 311 510

53 39 36 47 64 411 32 58 37 64 310 45 64 411 34 59 64 37 310 38 35 64 410 39 45 34 64 39 64 43 32 56 34 510

37 310 38 64 36 47 32 36 64 47 32 56 34 64 411 34 64 32 35 34 64 58 310 36 64 32 64 311 38 311 510

39 58 64 310 410 44 34 58 64 44 32 37 43 59 64 53 39 36 47 64 53 310 310 44 64 32 58 44 64 43 36 34 34 410 59 64 39 64 53 32 43 64 38 43 34 44 64 36 310 64 55 32 38 43 34 64 43 38 55 47 64 44 34 32 36 47 510

310 35 64 43 310 411 34 36 39 411 34 43 64 38 43 34 44 64 57 34 45 310 35 34 64 32 64 43 34 32 410 64 36 310 64 55 32 35 35 37 64 53 310 35 44 43 59 64 57 38 36 64 58 310 36 64 57 37 64 57 35 34 32 36 47 510

 This is your first question. All you should need to know is mines’
 location and location. Once you have that, give me my answer.

Author notes: The riddle portion of this puzzle is Rand Al'Thor's so, he'll be determining if you're correct once we get to that.

The solution to the cipher and the riddle must both be posted to get credit.

Answer

In the cipher it is clear that

64 is space and 510 is some kind of punctuation

which with a brief trip to quipqiup and some fiddling yields the plaintext

I AM LIGHT, BUT NOT PARTICLE OR WAVE.
I CAN BE DOWN, BUT HELP YOU STAY UP.
WITH MANY OF ME, YOUR LIFE I SAVE.
YOU THAT HAVE ME ARE NOT A PUP.
IN OLDEN DAYS, WITH WOOD AND STEEL,
I WAS USED TO CAUSE SUCH DEATH.
OR SOMETIMES USED BEFORE A SEAL

TO CARRY WORDS, BUT NOT BY BREATH.

whose answer seems clearly to be

a feather (light in weight; down feathers; not entirely sure about the life-saving -- artificial wings? filling for blankets etc. to keep you warm?; once you've grown feathers you are literally fledged, a term used metaphorically to mean not being a novice; feathers used to stabilize arrows; quill pens used for writing).

The cipher key appears to be

the letters of a French keyboard (as one might have guessed from the Sphinx's name): AZERTYUIOP etc. So e.g. 37 means row 3, key number 7, which is Y just as it is on a QWERTY keyboard. (The function keys are row 1, so this is a computer rather than typewriter keyboard. Key numbers bigger than 9 become 10 etc., which is why some numbers have three digits.)

which I guess (and the questioner kinda confirms in comments) should have been deduced somehow from "mines' location and location" but I'm not really sure what that means or how I was meant to deduce it...

Perhaps "location and location" means

row and column?

symmetry - How to apply Noether's theorem

Say I have a point transformation:

$$x' ~=~ (1 +\epsilon)x,$$

$$t' ~=~ (1 +\epsilon)^2t,$$

and Lagrangian

$$ L ~=~ \frac{1}{2}m\dot{x}^2 - \frac{\alpha}{x^2}.$$

How do I go out about showing that this transformation is a symmetry of the Lagrangian?

When I plug in $x'$ and $t'$ I do not get the same Lagrangian at all, it is off by factors of $(1+\epsilon)^2$ in both terms. I am not sure how I to calculate this. What sort of "symmetry" is this?

Answer

1) In this answer we provide more details for David Bar Moshe's correct answer. The action reads

$$\tag{1}S[q; t_i,t_f]~:=~\int_{t_i}^{t_f} \!dt~ L, \quad L~:=~T-V ,\quad T~:=~\frac{m}{2}\dot{q}^2,\quad V~:=~\frac{\alpha}{q^2}.$$

It is not hard to check that the action has an exact symmetry

$$\tag{2} S[q; t_i,t_f] \quad \longrightarrow \quad S[q^{\prime}; t^{\prime}_i,t^{\prime}_f] ~=~S[q; t_i,t_f]$$

under the following scaling

$$\tag{3}t \quad \longrightarrow \quad t^{\prime}~=~\lambda^2 t, \qquad q(t) \quad \longrightarrow \quad q^{\prime}(t^{\prime})~=~\lambda q(t),$$

with a non-negative parameter $\lambda\neq 0$, if we also scale the initial and final time integration limits in the same way as the time parameter $t$:

$$\tag{4}t_i\quad \longrightarrow \quad t^{\prime}_i~=~\lambda^2 t_i, \qquad t_f\quad \longrightarrow \quad t^{\prime}_f~=~\lambda^2 t_f.$$

Interestingly, the transformation (3) is not a symmetry of the Lagrangian

$$\tag{5}L(t) \quad \longrightarrow \quad L^{\prime}(t^{\prime})~=~\frac{L(t)}{\lambda^2}, $$ as OP already observes in the question(v1). This is a good opportunity to recall that Noether's theorem is about (quasi)-symmetries of the action rather than the Lagrangian.

2) Next let us consider the corresponding infinitesimal transformation. Assume that $\lambda=1+\epsilon$, where $\epsilon$ is infinitesimal, i.e. neglect higher-order terms in $\epsilon$. The so-called horizontal infinitesimal variation is

$$\tag{6}\delta t ~:=~t^{\prime}-t ~=~ \epsilon \cdot 2t. $$

The infinitesimal variation of the dynamical variable $q$ is

$$\tag{7} \delta q(t)~:= ~q^{\prime}(t^{\prime})-q(t)~=~\epsilon \cdot q(t), $$

so the vertical infinitesimal variation is

$$\tag{8} \delta_0 q(t)~:= ~q^{\prime}(t)-q(t)~=~\epsilon \cdot(q(t)-2 t\dot{q}(t)).$$ In other words, the transformation (3) has horizontal generator $2t$ and vertical generator $q-2 t\dot{q}$.

The bare Noether current (=charge) $Q$ is defined as the momentum times the vertical generator plus the Lagrangian times the horizontal generator:

$$\tag{9} Q~:=~ \frac{\partial L}{\partial \dot{q}}\cdot(q-2 t\dot{q})+ L\cdot 2t ~=~mq\dot{q} -2t(T+V).$$

[In general, if the infinitesimal transformation $\delta S$ of the action is only invariant up to boundary terms, it is called a quasi-symmetry, and the full Noether current would then receive boundary contributions. However in our case, the symmetry (3,4) is actually exact (2), i.e. without any boundary terms, so the full Noether current is just the bare Noether current (9).]

It is easy to check that the Noether charge (9) is conserved on-shell

$$\tag{10}\frac{dQ}{dt} ~=~-(q-2 t\dot{q})\frac{\delta S}{\delta q}~\approx~0, $$

where the $\approx$ sign means equality modulo equation of motion

$$\tag{11} 0~\approx~ \frac{\delta S}{\delta q} ~=~\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}} ~=~\frac{2\alpha}{q^3}-m\ddot{q},$$

i.e. on-shell.

optics - Amateur moon laser ranging

Questions first, then my rough estimations:

1) Is it possible to perform moon laser ranging with amateur motorized 114mm telescope? My calculations suggest that for 1mJ laser it should receive ~2 photons per source 1mJ laser pulse.

2) Given that we already talking about individual photons, how it was possible to perform moon laser ranging BEFORE retro-reflectors were deployed to the moon? Retroreflector sends back light in ~1arcsecond angle, while bare lunar sufrace - in ~6 archours, which means we supposed to receive signal ~(6*60*60)^2 = 4.5*10^8 weaker, i.e. even with 2.5 meter telescopes we are talking 1 photon per 250 1J pulses.

My rough estimations: Given that atmosphere turbulence limits telescope resolution to ~1 arcsecond (adaptive optics was not available when laser ranging experiments started, nor it is available now for amateurs), if we use telescope with diameter larger than ~150mm (so that we are limited by atmosphere, not diffraction) to expand the laser beam we will get ~1939x1939 meter illuminated area on the moon surface (tan(1arcsec)*400'000km). Which means only 1/(1939*1939) part of our energy will reach reflector.

Retroreflector is ~1x1 meter in size. It will reflect the light with same beam divergence - 1arcsecond. Too sad, as diffraction limit for retro-reflector of such size is ~0.2 arcsecond.

So, if our receiving telescope has area of ~1 meter^2, we will receive again 1/(1939*1939) part of what reached the moon, so total attenuation is ~ 1.4*10^13.

If we use 532nm pulse laser with 1mJ pulse energy, it will emit 2.67*10^15 photons, which means we are going to receive ~190 photons per pulse. Sounds realistic.

These calculations suggests that 114mm amateur telescope should be able to detect 2 photon per pulse - again should be detectable statistically.

Answer

Per BarsMonsters's request I'll expand my comment into an answer.

---

This same calculation has been done in a post on the blog Built on Facts. His conclusion is that amateur lunar ranging isn't feasible with a the lasers at his disposal. (Specs are given in the quote.)

“Amateur Lunar Ranging? Hmm.”

Typical values for the lasers we use might be somewhere in the 1 milliradian range. The distance to the moon is roughly 400,000 km, so the portion of the lunar surface illuminated by our laser will be about (0.001)*(400,000 km) = 400 kilometers in diameter. This is an area of about 125 billion square meters. If the retroreflector is one square meter, only about 1 part in 10^11 of our emitted light even makes it to the reflector. Now the reflected light has to make it back to the earth. If we’re extremely optimistic, we might say that the reflector introduces no extra angular spread, and its reflected light is spread over a 400 km diameter of the earth’s surface. To a first approximation this just means the total there-and-back efficiency is about 1 part in (10^11)^2, or one photon in 10^22.

...

The energy of one photon of wavelength λ is:

$$\frac{1}{A}\frac{dp}{dt}=\frac{S}{c}$$

The laser we’d likely use has a wavelength of 532 nanometers, and plugging into the equation we find that each photon has an energy of about 3.7 x 10^-19 joules. Therefore we’d need about 1000 joules per pulse to get up to the neighborhood of 10^22 photons per pulse. And we need 10^22 photons just to get one photon back per pulse on average.

We have a few compact 15 mJ/pulse q-switched Nd:YLF lasers with 1 KHz rep rates, and we even have a few not-so-compact 2 J/pulse with a 10 Hz rep rate. With a 2 second round trip time, the repetition rate isn’t so relevant since we can effectively only use one pulse per round-trip time. Even 2 joules won’t cut it unless we’re willing to do many thousands of shots worth of statistics. And that’s without even thinking about noise, which will not be inconsequential even with good filtering.

---

I can't answer your second question, but I can help with a few references. Linked from the above post, there is an article on The Lunar Ranging Experiment (PDF). The introduction gives some history of the retroreflector measurements, as well as the ranging experiments before the retroreflectors were placed.

In 1962, Smullin and Fiocco (3) at Massachusetts Institute of Technology succeeded in observing laser light pulses reflected from the lunar surface using a laser with millisecond pulse length. Additional measurements of this kind were reported by Grasyuk et al. (4) from the Crimean Astrophysical Observatory, and later Kokurin et al. reported successful results (5) using a Q-switched ruby laser.

...

(3): L. D. Smullin and G. Fiocco, Institute of Electrical and Electronics Engineers Proceedings 50, 1703 (1962).

(4): A. Z. Grasyuk, V. S. Zuev, Yu. L. Kokurin, P. G. Kryukov, V. V. Kurbasov, V. F. Lobanov, V. M Mozhzherin, A. N. Sukhanovskii, N. S. Chernykh, K. K. Chuvaev, Soviet Physics Doklady 9, 192 (1964).

(5): Yu. L. Kokurin, V. V. Kurbasov, V. F. Lobanov, V. M. Mozhzherin, A. N. Sukhanovskii, N. S. Chernykh, Journal of Experimental and Theoretical Physics Letters 3, 139 (1966). (link)

string theory - How much can AdS/CMT tell us?

I will begin my research on AdS/CMT, however I find AdS/CMT is only a phenomelogical method, so I want to know can AdS/CMT give some results the condensed matter physicists can not give, or even predict some unseen exotic matter states?

newtonian mechanics - Estimating atmospheric friction by measuring the change in velocity of a ball thrown straight upwards

Imagine I throw a ball straight upwards with some velocity $v_1$, and filming the ball with a camera, I can estimate a velocity $v_2$ (along the same vector) after the ball has moved a distance $D$. Using the difference between $v_1$ and $v_2$, and assuming constant friction due to air, how well can I estimate the initial velocity necessary to toss the ball some height $H$?

For fun - provided some $v_1$, $v_2$, and $D$, can we estimate an upper-bound for the Earth escape velocity with air friction/drag? Or is there unpredictable scaling of friction with velocity?

word problem - Guess the object - What Am I Riddle (Part 1)

In one form I can stretch to infinity, in another I have a clearly defined limit.
In one form I am very flexible, in another I am exceedingly rigid.
In one form intersections are catastrophic, in another intersections are common and can be helpful.
In one form I am a physical object, in another I am a conceptual object
In both forms I am used daily by millions of people all round the world.

What Am I?

Don't forget the spoiler tags.

If you get the correct answer, you must then think of a thing whose first letter begins with the end letter of the answer to this problem. So if the answer was apple, you need to think of an object that starts with 'e' and post a riddle about that. We can keep going until people get bored. However it does rely on getting the answer to this right first. If you really can't be bothered to think up a new riddle, say so in your answer. First person who comments something like 'I'll do it' can have the honor.

_{No hate, this is my first entirely made up word based riddle. Have Fun!}

Answer

Network. The virtual/social vs road networks.

In one form I can stretch to infinity, in another I have a clearly defined limit.

The virtual/social can cross boundaries reaching the endless depths of connections between people. Roads are physically bounded.

In one form I am very flexible, in another I am exceedingly rigid.

The digital can be instantly modified and twisted (eg. changing physical layout of a network but regaining functionality, or people changing locations not interfering with their friendship). Roads are very difficult to change.

In one form intersections are catastrophic, in another intersections are common and can be helpful.

Car accidents vs. meeting people / transferring data

In one form I am a physical object, in another I am a conceptual object In both forms I am used daily by millions of people all round the world.

self explanatory

general relativity - What is the relationship between $a$ and $m$

$a$ is defined in HERE

What is the relationship between the length-scale $a$ and the mass $m$?

Answer

According to Schutz (chapter 11) for the Schwarzschild metric $dr/d\phi$ is given by:

$$ \left(\frac{dr}{d\phi}\right)^2 = \frac{E^2}{L^2}r^4 - \left(1 - \frac{r_s}{r}\right)\left(\frac{r^4}{L^2} + r^2\right) $$

where:

$$\begin{align} E &= \frac{p_0}{m} \\ L &= \frac{p_\phi}{m} \end{align}$$

Comparing this with the equation in the Wikipedia article we find:

$$ a = L = \frac{p_\phi}{m} = r^2\frac{p^\phi}{m} $$

Though this makes $a$ an angular velocity rather than a length scale.