Tennis Ball Cannon

Made as part of a team-building contest at my company.

Some stills:

not bad for accuracy!

 The tennis ball is top center left, its shadow is just left of target! At 25 m distance, the precision was 3m in length and 1m in width. The pieces of bark and stones mark the "hits".

multi-launch

 Immediate post-launch: Elsa wanted to launch as many as possible simultaneously, we got up to 4!

Four in the air

Piston at full post-launch extension.

Looking down the mouth of a loaded cannon - see how much I trust Maya.

A 2 -minute video compilation

Why I am sad it is not a re-collapsing universe

Why I am soooooo sorry that we are apparently (meaning according to the latest astronomical data and relativistic cosmologies) no longer in a closed (re-collapsing) universe. According to the latest, the expansion of the universe is decelerating, but at an ever lower deceleration which will cumulatively not be enough to turn the expansion around. 

How terribly sad! Suppose that there were to have been enough mass density in the universe to cause the universe to stop expanding and then to contract. Then there are theorems in GR which state that within a finite time everything will recollapse to a final singularity. 

Now, listen closely: We would then already be INSIDE the event horizon of the final singularity. Yes, you, me, the earth the sun the solar system the galaxy the local cluster the supercluster etc. You don't notice anything different do you? No "the earth shook" nor nothing, nada? 

Being inside the event horizon of a sufficiently large, isolated Black Hole would feel exactly the same.

"But if I was inside the event horizon of a Black Hole wouldn't I be able to see the Black Hole I was falling towards!?"

Nope! A Black Hole, like any final singularity, is like next Monday. You can't see "next Monday" can you?

Physics Phriday - It is a Time Machine

From Jack van Ryswyck:
You know it's funny you should mention that. The other day I suddenly saw Mike vanish before my very eyes, desk chair and all. Left in his place was a box shaped area of seemingly totally empty space, filled with absolutely nothing but some strange looking radiation.

Two seconds later Mike suddenly reappeared, looking very normal. In fact, looking rather exactly like he had at the moment he vanished. He then looked at me as if I had done something totally strange.

He told me that he had not seen anything vanish at all, but to him it looked like I instantaneously had shifted position and facial expression, as if I were an old time movie that had skipped a few frames. We then noticed that the atomic clocks that we had coincidentally been carrying were inexplicably off by 2 seconds. We had just synchronized them earlier that morning, but now his was 2 seconds behind mine.

Some time later a very similar thing happened, but this time a part of his desk chair stayed behind when Mike and the rest of his chair vanished. The remaining part of his chair just got cut clean off and fell to the floor. Mike and the rest of the chair reappeared two seconds later, though the chair did not reassemble. He said that he saw part of his chair suddenly teleported to the floor instantly.

Then the other day Josh and I were passing juggling balls when the same thing happened again. Suddenly everything in the box shaped area vanished, and many copies of a juggle ball appeared, all moving in a parallel direction. I then found myself tossing a juggle ball into the box. Oddly enough it happened to have the same colour as all the copies, and by some mysterious coincidence it was moving in just the same direction. Two seconds later all but one of the balls disappeared, and everything else reappeared. At the far side of the box, one juggle ball flew off. The net effect was just as if I had thrown the ball across the box incredibly fast. Impossibly fast really.

The next time it happened I saw the empty box appear filled with many copies of about one-third of a juggle ball. All but one of them were moving in a parallel direction away from me, but the nearest copy seemed to be moving towards me. In self defense I grabbed the juggle ball nearest to me, which happened to be the same colour, and as a makeshift missile defense system I threw it against the partial juggle ball that was moving menacingly towards me. They collided. The partial ball got knocked into a trajectory parallel to all the others. My defense missile ball got nudged back towards me. Then all the partial balls disappeared, except one that flew off at the far side. My defense missile ball was suddenly missing about a third of it. The remainder fell to the floor. The part that fell to the floor on the other side of the box seemed to fit it exactly.

Weird.

Physics Phriday

Take your time on this one, as many times as you wish.

The first image is a model of a 1+1 dimensional spacetime, the time axis is vertical, the x axis is horizontal, the speed of light is 1, the orange cone is the light cone and the blue line is the world line of any physical object. At any point on it the world line is confined to the future (+) and past (-) of the light cone at that point, and the point always moves forward in time.
 

In the second image, I've done some "manifold surgery": I've made a cut each at t = -1 and t = +1 each of which extends from x=1 to x=3. Now I restitch them, but stitch the bottom edge of the lower cut to the top edge of the upper cut, and also the top edge of the lower cut is stitched to the bottom edge of the upper cut.

What have I built?

Answers for "Man from the South" probability questions

Links to Dahl's  "Man from the South"

The questions are repeated in this post.

Preliminary answers

Q1. What does American Boy think is the probability of his lighter lighting during any single attempt?

Most people accept a bet when they think the odds are at least 50-50. Since Man from the South and American Boy have had time to negotiate the terms and think about it, both think they can win it. So as neutral observers we can consider the probability that the American Boy will win the bet to be 0.5.

So what does this tell us about what he thinks is the probability of the lighter lighting on any one attempt? Let this probability be p. The probability that he will win the bet, i.e. that it will light N times out of N trials, is p^N. We just decided that this is ½, for N=10. So,

p¹⁰ = ½

I was driving and couldn't very well use my laptop (my phone is not a very smart phone), leave alone the log tables, slide rule or abacus in my back seat. And I can't do powers of 1/10 in my head, not in general. But I can calculate powers of numbers near 1! So let's calculate q=1-p instead, which is bound to be small since the boy is so sure of success.

½ = (1-q)^N ~ 1-qN, so for N= 10, q=0.5/10 = 0.05.

Which gives us our first answer: p = 95%, which we know is a bit of an overestimate. (The exact answer is 93.3%.)

But this is like focusing on conversion ratio and not on cost-per-action.

Q4. Were there to have been a 9^th attempt, and were American Boy to have failed in it, what would he have lost?

A: His left pinky, those were the terms of the bet. So American Boy is staking his pinky every time! All but the last time, if he wins, all he wins is the right to stay in the game, if he loses, chop-chop (and the right to play!)

So this is kinda sorta like conversion attribution: every bid-request we (RFI) win and then every impression we serve only gives us the right to stay in the game, in the sense that if we don't serve the impression we certainly won't get attribution credit.

Q2. How many fingers to a Cadillac?

On the face of it, it would seem to be one finger to a Cadillac, C = F, since if he loses, he loses a finger, and if he wins, he wins a Cadillac. But as we saw above, the American Boy stakes his one finger 10 times and thus faces 10 opportunities to lose his finger vs. one opportunity to win the Cadillac. So,

10F = C

After the eighth successful attempt, 

Man from the South's wife surprises them and ends the game. She sends American Boy away empty handed.

Q3. How should he have been compensated, if at all?

Aha! “How to distribute the spoils in an interrupted game?” The very question that my colleague Jack pointed out was the leading cause of the rise of probability theory!

The way I think of this is in terms of “vesting”. Each time they play and American Boy wins, he wins a 10^th of a Caddy, but, BUT, he only gets to keep his winnings (the entire Cadillac) if he keeps his head (figuratively) and finger (literally) for 10 trials. American Boy can't get cold feet and walk out after say 5 trials and demand half a Cadillac – so really he can get cold feet and walk out but he can't drive off. So there are two possibilities. Under one, the Man from the South gets cold feet and decides not to play anymore. In this case he forfeits his Cadillac, which was held in escrow by the narrator. Under the other possibility, the “authorities” intervene. Since neither party has broken their terms, in this case I think American Boy gets to keep his “unvested” winnings, which would be 8/10 of a Cadillac. Now since the Cadillac wasn't the possession of the Man to begin with, but belonged to the “authorities”, … this isn't a math problem anymore. (By the way, who intervened and broke up Cardano and Pascal's poker game?)

Legal opinions? Aji, Joanne?

Q1 (Re-evaluated). What does American Boy think is the probability of his lighter lighting during any single attempt?

Most people accept a bet when they think it is a game with at least a zero sum in their favor. Since Man from the South and American Boy have had time to negotiate the terms and think about it, both think they can win it. So as neutral observers we can consider the game to be a zero-sum game.

What does it mean for this to be a zero-sum game? Winnings * probability of winning – losses * probability of losing = 0! Which yields:

p = 1/(W/L + 1). (Check: if W are high, p is low; if W are nearly 0, p is nearly 1 and if W=L, p =1/2.)

In our case the winnings are the Cadillac C, the losses are the finger F and the probability P of winning the entire game is P = p^N, where p is the probability of the lighter lighting in a single trial. Combining things we have

C*P – F*(1-P) = 1, or P = 1/(C/F + 1)

Putting in C/F = 10,

p¹⁰ = 1/11

1/11 = (1-q)^N ~ 1-qN, so for N= 10, q=1/11 = 0.1.

Which gives us 78.7%.

Really? Would you play that game with a lighter which only lights less than 80% of the time? I think that the Man from the South has fuddled the American Boy into undervaluing his finger, by making him think he is wagering a finger vs. a Cadillac, whereas really he is wagering a finger against a 10^th of a Cadillac.

Some preliminaries

If you aren't interested in the nuts and bolts, skip them, but this is so anyone can check my work.

Notation: Sum[i, 0, Infinity] f(i) is to be interpreted as the sum of the function or series f(i) over the index i from i=0 to i= Infinity. Then,

Sum[n,0,Infinity] pⁿ = 1/(1-p)

Sum[n,0,N] pⁿ = (1-p^N+1)/(1-p)

and

Sum[n,1,N] pⁿ = (1-p^N)*p/(1-p)

Look at the following table of outcomes of consecutive tosses and the overall probabilities

W ← 1 → L

1: p (1-p) (End)

2: p² p(1-p) (End)

3: p³ p²(1-p) (End)

…

So after n trials, the probability of winning all is pⁿ and the probability of losing any is

(1-p)* Sum[i,0,n-1] pⁱ = (1-p) * (1-pⁿ)/(1-p) = 1-pⁿ = 1- prob(Winning). Which is good since it indicates I can still sum correctly.

Back to the problem, to get a handle on 

what if anything AB deserves when the game is stopped.

From making the last, Nth, trial a zero-sum game, we know that

p^N = 1/(C/F + 1). We've assumed that AB wins 1/N th of a Cadillac (virtually) when his lighter lights. So assuming the first trial is also a 0-sum game, we have:

p*C/N = (1-p) *F, which resolves as

p = (C/NF + 1) ^(-1).

Can these two equations be solved simultaneously for both C/F and p?

Yes, but the solution is i) independent of N and ii) meaningless:

We have (1+C/NF)^N = (1 + C/F), but the RHS is simply the first two terms in the binomial expansion for the LHS, so the equality holds only when C/F = 0 and p =1.

So one of our assumptions above is wrong. 

Let's try another tack. 

Assume that at the nth trial, AB wins some unknown portion of the Cadillac a(n)*C. (When Ari and I were talking about this last week, Ari guessed, “Wouldn't it be some quadratic or increasing portion that he wins?” Ari's motivation was to take into account the wearing out of the flint, the gas running out, the thumb getting tired, AB getting nervous etc. Dahl, spends an entire paragraph describing the care and attention to detail taken by AB, after each light, he blows on the lighter, closes the lighter, waits a few seconds perhaps for gas pressure to build up again, re-opens it and then flicks it once. ) What we know is that

Sum[n,1,N] a(n) = 1, over the course of the entire game, if he survives, he wins the entire Cadillac. So at every trial, AB stands to lose not just his finger and the right to play, but also the “won but not vested” portion of the Cadillac, and he stands to win some portion of the remaining.

At the last, Nth trial: W : L

p : (1-p)

a(N)C : F + (1-a(N))C

Using the 0-sum equation, we get:

a(N) = (1 +F/C)(1-p).

Great, so now we have ONE equation and 3 unknowns: a(N), F/C and p. But we also know that if the game is 0-Sum over all: p^N * C = F*(1-p^N),

which yields:

0 < p = (1+C/F)^(-1/N) < 1. So now at least we have two equations for three unknowns, and the solution for p is valid. This doesn't guarantee that 0

If we knew C/F, we could solve the problem. However, clearly, C=F is no longer valid. C = 10F could be used for AB's assumption. We also have another source of information: The Man from the South's wife explains that he has lost eleven cars and taken forty-seven fingers. Assuming that he considers these equivalent, we have 11C = 47F

so,

The fraction C/F is the ratio of the value of one Cadillac to the value of one Finger.

But all this hasn't answered the question of 

what AB deserves when the game is interrupted. 

Also, we've assumed that the entire game is 0-sum and that the last trial is 0-sum. Can't we make use of the assumption that 

every trial is also 0 sum 

to see if we can figure out the intermediate non-vested winnings?

Recall that at the nth trial, AB stakes his previous winnings and his finger for a chance to win a(n)*C portion of the Cadillac. So the 0-Sum equation for the nth trial is:

a(n)*C = (1/p -1) *(F + C* Sum[i, 0, n-1] a(i)). The resulting recursion relation is for a geometric series!

a(n+1) = (1/p)* a(n),

whose solution is

a(n) = a(0)/p^n.

Note immediately that a(0) != 0, so AB has to stake (even if only virtually) something more than just his finger. 

We find a(0) 

by using the fact that the total portion of the Cadillac gained over 10 trials is 1:

1 = Sum[i,1,N] a(i) = a(0) * Sum[i,1,N] (1/p)^i = a(0) * (1/p^N -1)/(1-p), or

a(0) = (1-p)/(1/p^N – 1)

For the first trial (note that this is independent information since so far we have used the recursion relation and established a(0) using the “normalization”, but we haven't yet used any 

“initial conditions”):

a(1)*C = (1/p – 1) * ( F + a(0)*C), which yields

C/F = ((1/p)^N -1)/p, which is a different relationship between C/F, N and p than we had previously. I am not sure I can invert this to yield p(C/F), but it can certainly be numerically solved.

Let us also calculate the virtual winnings after each trial:

W(n) = C*Sum[i,1,n]a(i) = C*a(0)*(1/p^n -1)/(1-p), which simplifies to

W(n) = ((1/p)ⁿ - 1)/((1/p)^N - 1)

So given p we could calculate C/F (or vice versa), a(0), a(n) and W(n)

Working on the “wife's numbers”, we see that the American Boy should be compensated with 66% of the Cadillac when the wife interrupts the game after the 8^th trial.

How do the winnings increase as the trials proceed:

Winnings in units of "Cadillacs"

 

which look like

Levenfeld curve

Conclusion and final answers:

Q1. What does American Boy think is the probability of his lighter lighting during any single attempt?

About 80%.

Q2. How many fingers to a Cadillac?

In American Boy's valuation based on his behaviour, 10 fingers to a Cadillac.

After the eighth successful attempt, Man from the South's wife surprises them and ends the game. She sends American Boy away empty handed.

Q3. How should he have been compensated, if at all?

With 60 or 66% of a Cadillac. I would go with 66%, which is based on the Man from the South's experienced equivalence between Cadillacs and fingers.

Q4. Were there to have been a 9^th attempt, and were American Boy to have failed in it, what would he have lost?

Ah, his finger, of course, and, his virtual stake, which is 3.9 % of a Cadillac. How could he have lost something he never had to begin with? Well, for the bet to proceed, AB would have had to ask the narrator to spot him 4% of a Cadillac, or its cash equivalent, or its (OUCH) finger equivalent, which is 17% (3.9% * 4.3 F/C).

If AB's lighter fails during the game, he loses 1.17 fingers since he would have to sell 0.17 fingers to the Man from the South to pay off the debt to the narrator. If AB's lighter doesn't fail during the game, he simply returns the cash or Cadillac equivalent from whoever he borrowed it, and is ahead one Cadillac. 

Why did I ever think of approaching the problem this way, with a “virtual stake”? In particle physics, one can borrow virtual particles from the vaccuum in order to simplify calculations. It is all halal as long as the virtual particles don't violate any conservation laws for quantum numbers and the mass-energy of the particles exists for a short enough duration of time that Heisenberg's Uncertainty principle is not violated. The really interesting thing is that these virtual particles have real effects: A pair of uncharged conducting plates will attract each other because a virtual charged particle – anti-particle pair will come into existence from the vacuum for a brief time, and the effective dipole and its images will cause the plates to experience an attractive force. Don't believe me, look up the Casimir Effect.

What happens if you grab those particles and forcibly separate them from each other and prevent them from annihilating each other as any decent particle-anti-particle pair should do? You end up creating a Black Hole-White Hole pair, which you can then use for superluminal transportation and as a time-machine! (Okay, I just made that up, but is it really any crappier than “The Secret” or Deepak Chopra?)

Back to reality: The Man from the South's wife explains that he has lost eleven cars and taken forty-seven fingers.

Q5. What does Man from the South think is the probability that American Boy's lighter will light during a single attempt?

86%.

Q6. How many fingers to a Cadillac does Man from the South figure?

Th ratio of the values is C/F = 47/11.

Q7. Do your answers to Q3 and Q4 change?

Yes.

Added on 16th Nov. 2012

What was Fermat and Pascal's approach? Instead of looking backwards, they looked forward and calculated the probability (on the condition of the current circumstances) of winning or losing the game and divided the spoils accordingly. So if AB has a probability of p of lighting the lighter and has already done so 8 times, the probability that he will then do so 10 times is simply p^2 and the probability that he will lose is (1-p^2). According to this approach, AB wins p^2 of the Cadillac and loses 1- p^2 of his pinky. I think this is close to Jon's suggestion, who strongly felt that since the game hadn't finished AB would have to lose part of his finger in exchange for part of the Cadillac.

Postscript to "Nature in Art"

In response to various comments from both close friends and from people who've known me for a couple of hours, which were along the lines of
A not-so-young physicist, un Indio
Posted on his blog about VIBGYOR.
"The passion is clear,
The physics, 'No fear!',
But the tone, ahh that is a failure!"


After what RS, ARM, DST, SP, AP, SK, PSK, JB etc have put me through over the last few weeks in the "Home Truths Dept.", I think a Maoist re-education camp will be a piece of cake!


...


One of the above friends: "Do you acknowledge that you have been personally responsible for suppressing millions of hard-working peasants?"


RST: "Yes."


Same friend: "Wait, you've already self-confessed? I don't get to torture you?"


RST: "The other friends in the list above have said pretty much the same thing about me, so it must be true."


Some friend!: "So what are you going to do about it?"


RST: " 'Do about it?' ? What do you mean? That would require me to be different from what I am! ... Oh, I see your point!"


...

So here goes: I want that post to be seen as a learning opportunity about observing nature, not art criticism nor artist criticism. To be very very explicit: The artistic representations aren't incorrect representations of nature - they don't have to be, reality is not the only domain of art. They're just not realistic/scientific representations of nature. True, which doesn't prevent them from sometimes being spot-on in capturing some essence of nature.

In exchange, I would like acknowledgment that
1)  scientists can appreciate beauty in nature, in spite of "having an equation for it", and that
2)  scientists can appreciate art, in spite of being very analytical (Read any art reviews lately?).

In that post "wrong" should be interpreted not as a commentary on the moral nature of the artist, but rather as "observationally incorrect representation of nature, but of course the artist has artistic license to do what she or he wants, just use this as an excuse to step outside and look, really look, at a rainbow!"


A painter friend of mine in Madrid (Javier F L)  put it best when I asked him why he painted, "I paint because it gives me an excuse to gaze for long periods of time." I couldn't disagree less about why I do science, or data analytics for that matter. (This isn't helping me get a job, is it?)

Observations of Nature in Art

May 23rd 2012, Please read this DISCLAIMER before proceeding. Today's edits are in red.

RAINBOWS:

A. What is wrong with this picture?

(Thanks to Shailesh S for motivating the following.) 

Have you ever seen an elliptical rainbow? No? Let's ask ourselves why not? Imagine looking at a circle, a child's hula, or a hoop-earring in a 3/4 profile photo of a beautiful woman looking out over a seascape (just saying!). The hula-hoop appears as a circle only from one particular angle, out of 2 Pi steradians. To anyone else looking at it, it appears as an ellipse. This becomes clear when they are as far to one side of you as you are from the hula-hoop. Now imagine that there is a big evening rainfront about 20 miles to your East, and you see a rainbow (in the East, why?), and your friend who lives 20 miles to your North texts you, "i c byu t ful rnbo, do u?". You respond, "Yes, i c it 2, as circle, so u must c it as ellipse." She txts bak, "No way, mine is circle 2! ;-?"

Now, the only way the same real object can appear as a circle to all viewers is if it is a ... sphere! Let's accept that a rainbow is not a sphere (in four dimensional space, it would be! Isn't just that a reason one would wish String Theory to be true?). Then we have to conclude that a rainbow is not real! What one means by "not real" is that you can't project it on a screen. Such images are called "virtual images", like the image you see through a magnifying glass. 

We see the rainbow, because the lens of our eye collects the light reflected and refracted from all the raindrops and projects a real image onto our retina. If we consider "seeing an object" to mean sensory perception of light emanating or reflected or otherwise having interacted with that object, then what we are seeing when we "see a rainbow" is all the raindrops!

So the only real image of the rainbow is the one that is projected onto your retina and which is then processed by your brain. Now since that is just as true for your friend as it is for you, what that means is that each and every person has their own personal rainbow, and she or he is at the center of it!   That is a pretty good analogy for consciousness of the universe.

You'll never see a rainbow as an ellipse, less so as co-axial elliptical bands. If we insist on imagining what it would look like to another observer far to the left of us, it would still not look like the above where the red arc remains on the left edge and the blue arc remains on the right edge. The colors form concentric annular discs, so the red arc should be imagined to stay on the outside edge of the elliptical form, and the blue arc on the inside edge.

B. What is wrong with this picture?

 The three rainbows have three different centers, implying three sources of light! On a planet with three or more spectrally identical suns, this landscape is possible. However, since rainbows are virtual images, they would not obscure one another.

Ignoring the rainbow on the right, non-concentric double rainbows are in principle possible. The lower one (whose center is the shadow of your head and is below the horizon) is the normal one formed by light from the sun refracting and reflecting off the raindrops. The upper rainbow, with its center above the horizon, is formed by sunlight which is first reflected by the body of water in the foreground, and then impinges at an upward angle on the raindrops! The center of the reflection rainbow is where the shadow of your head would appear to be after reflection from (a continuation of) the expanse of water. Because of the landforms, the upper rainbow should be incomplete except where there is water directly below the lower rainbow. What a thought-provoking painting!

Now, just as we can see multiple shadows due to multiple lights, for example at night in an illuminated parking lot, with some luck and lots of practice with a hose spraying water, one might be able to see three rainbows as well. 

C. What is wrong with this picture?

The sun is under the rainbow? Whatever the artist is smoking, I want some! This was seen by some of my readers as being derisive towards the artist. By no means! I really really do want some! A rather prosaic way of achieving the same effect would be to have a filtered mirror in front and just to the side of you while looking at the rainbow. By the way, even if the sun was above the rainbow it would still be wrong. Here is an excerpt from the page of an observant artist : "On a showery day, one may be blessed with the appearance of a rainbow. It is visible in an area of the sky opposite the light source."

D. What is wrong with this picture?

The order of the colors is inverted, as the artist could have ascertained by umm ...  looking. But hey! Why look at nature when you can do an art project? Or go online to check whether it is raining? What I think bothers me here is that young children, even when they see a rainbow, are perhaps missing the time with an adult who could help them look at the rainbow. By which I mean observe and mindfulness, which is a first step towards both science and art.

The angles for the first maximum, or the primary rainbow, are proportional to the wavelength. A rainbow, or anything else with a spectrum, requires dispersion, i.e. wavelength dependence of the relative refractive index of the material of the drops (water) in the atmosphere (air). In water, shorter wavelengths (blue) are dispersed more than longer wavelengths (red). Hence one might expect that red be on the inside and blue be on the outside, as in the above painting.  However, the light is also reflected from the back wall of the droplet and purely due to the geometry of this reflection, the colors cross each other and are inverted. See the wikipedia rainbow article.

E. What is wrong with this picture?

This is more subtle! Again, all the artist need have done is look! From inside to outside, the colors in the secondary rainbow go from red to blue! The secondary rainbow is formed when the light is reflected inside the rainbow twice! This causes the order of the colors to be inverted yet again and back to the original "blue is dispersed more". I'll confess that I haven't figured out the geometry of this, and neither has wikipedia.

F. What is i) right and ii) wrong with this picture?

i) The two rainbows are concentric, and the shadows cast by objects are consistent with the position of the sun. That is, the line joining the observer to the center of the rainbow is parallel to the lines joining objects to their shadows.
ii) However, I would have expected to see the shadow of the artist's head at the geometric center of the rainbow, near the bottom, but by my reckoning within the frame of the painting. This is a minor quibble.

G. In contrast, what is right with this picture?

The order of the colors of the secondary rainbow. An artist actually looked at nature! (Okay, the last sentence is unneccesary.)

G. What are two things wrong with this picture?

First, the only way the rainbow can be seen in front of the background trees is if there is rain or mist between the artist and the trees, which, from the clarity of appearance of the sheep and houses near the "pot-of-gold" point, does not appear to be the case. Second, from the artist's viewpoint one can see close to 180 degrees worth of the rainbow, implying that the sun is at the horizon. This is inconsistent with the rather short length of the children's shadows, which indicate the sun to be about halfway to the zenith, or about 45 degrees above the horizon, in which case the rainbow shouldn't be visible at all! (The sun has to be less than 42 degrees above the horizon, since that is the outer conical angle of the rainbow.)

 SNOWFLAKES: The next time I see non-C6 symmetry snowflakes I'll scream.

Scream 1

Scream 2

Scream 3

The USPS, which tends not to be particularly science friendly, did use actual photos of snowflakes for their stamps.

To which one can only say,
"Naught immortal hand nor eye
  could frame thy C6 symmetry"

Next up: descriptions of nature in Turgenev.

response to Pandurang nayak's blogpost on Greene's "elegant universe"

http://pandunayak.blogspot.com/2007/05/elegant-universe.html

Hmm ... "string theory", maybe it should be classified under meta-physics?

I haven't read Brain Greene's book. But I have been hearing about strings since about 1988, from people of the likes of Gary Horowitz (with whom I did a postdoc later), Ed Witten, David Gross, Strominger etc.

Early strings: In relativity, the action (which is later extremized to find the equations governing its motion) for a free particle is simply the path integral of the infinitesimal distance ds. So Nambu-Gottu proposed that the action governing the motion of a string is its surface area. Now if you just extremize this the solutin would lead to the collapse of the string to a point, so you add in a linear tension term. The resulting solutions to the Eqn of Motion have the usual oscillatory modes etc.

Now, fundamental theories are field theories and particles are thier quanta. So far so good. But when you put in interacting fields (EM field (photons) and Dirac field (charged particles)) and quantize them you can't do it exactly. So you do perturbations about the 0 interaction, or the vaccuum. It turns out that even QED is perturbatively "non-renormalizable", sort of like a divergent series. However, unlike a divergent series, it turns out that removing certain infinities and adding the perturbation term by term does lead to converging answers that agree with experiment to astounding degree of 10^-12 etc!

So why are interacting particle field theories singular? Hand waving, it goes like this, any paricle interaction involves a node in spacetime (the "collision" or disassociation) where 3 worldlines come together. Now this is no longer a "line" and leads to trouble. Also, there are (solved) causality problems and reference frame problems for multiple interactions. However, a string interaction is visualized as a smooth manifold, the trousers topology - imagine two circles coming closer and merging to form one: while the sequence of two dimensional slices of this event appear to have a singularity (non-hausdorff ness or a cusp formation) the spacetime manifold (surface) is smooth!

So lots of excitement, particularly for string field theory. Claims and expectations:

1. string interactions are just smooth 2 dimensional surfaces
2. lack of point interactions means should be renormalizable
3. critical dimensions for string FT to be consistent, either 26 or 10, lots of yummy math, extra dimensions are compact manifolds, could be beautiful Calabi-Yau manifolds, topological invariants etc.
4. Then things got even better, solutions to string field equations (which themselves arise from requiring SFT to be “perturbatively renormalizable”) include general relativity with matter, physical matter field equations etc. Hence the claim it a TOE – Theory of everything!
5. Lots of connections between different areas of math discovered, by Witten and others.
   However,

1. No 2D surface admits a non-singular, non-degenerate Lorentzian metric everywhere. Metric = infinitesimal distance tensor, which is the geometry of the manifold. Lorentzian means one spacelike and one timelike dimension, I.e, one positive and one negative eigenvalue. In other words, somewhere on that trouser there is a singularity. This is not very important.
2. “renormalizable” means “perturbatively renormalizable”. What this means is one that the background spacetime in which the strings live is flat (why should that be so?) and this flat metric is used to setup a measure of distance for the perturbative expansions. Now: no one has actually done any perturbative calculations in full string theory, unlike in QED. Two, assuming a flat background spacetime is senseless for a self-proclaimed TOE (either GR is assumed and hence the string tension will cause curvature, or the background metric is a parameter whose value should be determined by the string EOM – see later. Three, a TOE should be exact, not perturbative, about some again arbitrarily chosen homogeneous solution.
3. In the perturbative expansion, in order for the first order term to be finite, certain conditions have to be met. These are the string field equations. One solution to the string field eqn is that the background metric used satisfy Einstein's field eqn for the metric, G mu nu = 8 Pi Tmu nu. However, there are other solutions, how many (not “countably how many”, rather how many dimensions and are these countable)? One doesn't even know the structure of the solution space. Other than knowing that GR is right, there is no way to pick that particular solution and hence claim that GR arises from string theory. Also, all this can only take place in 26 (or 10 or 11) dimensions of the background spacetime. Then one is left with explaining the apparent 4 dimensionality of the physical world – how do the extra dimensions compactify, why are they small and not say large. Note that the extra dimensions are physical (like a very tightly rolled up sheet of paper appears to be 1D but is actually 2D on the small scale), not extra dimensions in the fibre-bundle sense (where the extra dimension is just a property holder, say e.g. a sphere to represent the spin of a point particle).
4. See above. So string theory is not just a TOE, it is a TOEPUCI+ – Theory Of Everything Possible U Can Imagine and then some.To counter this and patrt of the previous criticism, let me make an analogy: suppose that immediately after Maxwell's EM and the Lorentz transformations somebody (the analog of a string theorist) said I have this wonderful theory of a thing called a metric that explains everything, I have an action principle (very important in physics then and now), I extremize it, the eqn says the ricci curvature of this metric thing is 0. A solution is that the world is flat! See, I have proved the world is flat. The analog of Lee or myself (in a very small way) or Woit would say, wait a minute, but there are numerous other solutions to your “metric equation” which say the metric is curved (what we now call the weyl curvature can be non-zero), how do you pick your flat metric from all the other solutions without knowing in advance that the world is flat? The answer is that you can't, the world is not flat and the metric theory is correct! So might String theory!
5. There have been a series of String bandwagons, all but the most recent few started by Ed Witten: matrix theory, conformal field theory, String BHs (most notably the CGHS black hole - a 2 D mini-micro nanomodel and symmetry reduction of string theory then with the sign of the exponent in the potential term arbitrarily reversed because everything else is too hard to solve), M-theory, membrane theory, Maldacena etc. The lifetime of each of these (as measured by publications) is about 5-6 years. Every two years there is a big string conference, some major breakthrough is announced, stringers dance the macarena in the aisles, the fever subsides, it proves a physics dead end and one moves on.
6. Now, contrary to Voit's claim that ST is “not even wrong”, a Weyl-square theorist whose seminar I attended in 1996 pointed out that string theory does actually make one very concrete testable prediction, and that never in the history of science has a theory been so wrong! Roughly speaking, in a non-perturbatively quantized ST, the string tension (actually stress) would be quantized, its quanta would be the planck tension (appropriate dimensional combination of G, c and h), which will also be the leading contributor to the cosmological constant (which is an energy density and has the same dimensions as tension/ area). There are strong astrophysical observational upper limits on the value of the CC, the planck tension is greater by about 40 orders of magnitude!
   Now, I have written much less than I could write, but already much more than you are interested in reading, so I'll just sto     

In response to: Gamma-ray photon race ends in dead heat...

In response to:

ScientificBlogging Gamma-ray photon race ends in dead heat; Einstein wins this round. We'll get you yet, though, Al.

www.sciencecodex.com

Source: www.sciencecodex.com

Racing across the universe for the last 7.3 billion years, two gamma-ray photons arrived at NASA's orbiting Fermi Gamma-ray Space Telescope within nine-tenths of a second of one another. The dead-heat finish may stoke the fires of debate among physicists…

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Kevin D Wieland wrote:

How do you know they started at the exact same time and not 9/10th a second apart?

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Here are my collected comments:

The article itself says that the difference in arrival times is "… likely due to the detailed processes of the gamma-ray burst…", i.e. in the source of the γ rays.

Addressing Kevin’s question in more detail:

The telescope was looking at the same part of the sky, presumably small enough to contain only one source of γ s. A Gamma ray burst was observed (bunch of γ s with energies between the two extremes mentioned in the article.). Gamma ray bursts last on the order of a couple of seconds (See above article.). On a macroscopic scale, the photons traverse the same spatial trajectory and intervening objects, but one is a few seconds behind the other. Nothing changes cosmologically on a scale of two seconds - except of course for gamma bursts and supernovae, which weren't observed. Macroscopic dispersive effects (change in refractive index due to frequency) in large intervening nebulae etc. cannot in principle be ruled out, but what do I know about astrophysics? Presumably they carefully chose a very empty part of the sky.

So, in order to reject the null hypothesis (that there are no differences in the speeds of the photons of different energies) any observed difference has to be >~ 2 secs. If there are hypothesized physical processes that do predict larger differences in arrival times, the observed 0.9 sec difference will put very tight constraints on the viability of those theories.

Now, watch the string theorists squeeze themselves through those very tight constraints! They have had 30 + years of practise!

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Bo Asciu wrote:

Kevin good point, keen mind.

Space/Time can't be that "frothy" since the laws of the universe are so precise. Too much looseness and the whole thing falls apart.

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Briefly addressing Bo’s comments:

There is classical, deterministic chaos, there are classical uncertainties due to stochastic processes, and we haven't even got to talking about QM! -in terms of causing frothiness in spacetime.

The point is that there are viable and very precise theories - "laws of the universe" - that separately either hypothesize or predict a "frothy" universe on a scale somewhere between 10^-18m and the Planck scale (10^-35m). The observations discussed put presumably very tight constraints on the free parameters in the theories that distinguish them from Einsteinian General Relativity.

If say the two photons had arrived more than 8 secs apart, the observations would tell us precisely how "frothy" spacetime actually is – how big are the hyopothesized microstructures, on what time scales do they change, how do they interact with different photons.

Even Einstein's very beautiful, very precise etc etc General Theory of Relativity CANNOT rule out frothiness of the universe on some heretofore unobserved, untested scale, though very good arguments can be made that the GTR will break down at some scale > Planck. If frothiness is observed at some larger scale, it will put a lower bound on the domain of applicability of the GTR (Since it hypothesizes a smooth (or at least twice differentiable) geometry for spacetime, GTR is valid above but not below the scale at which those “violations” are observed.).