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Did Homer Simpson calculate the correct Higgs boson mass?

No. But we may reveal the deepest secrets about Homer's mathematical brain, too.

For a while, I didn't understand where the story came from. But the number of articles pointing out that Homer Simpson discovered the Higgs boson (example) has exceeded 50 which is a lame reason why a TRF blog post may be appropriate. ;-)

All the excitement seems to be about this screenshot from the 1998 “The Wizard of Evergreen Terrace” episode. I can't realize that I have ever seen Homer Simpson who was this technically powerful but he must have been.

Now, all the media say that Homer Simpson either discovered the Higgs boson before CERN, or that he calculated the correct mass 14 years before it was measured at CERN (on the 2012 Independence Day).

The first interpretation is obviously completely wrong. One may only discover the Higgs boson experimentally and the collider was needed for that. (Even the recent claims that "the" Higgs boson was seen by condensed matter physicists are hype, an attempt of the squalid state physicists to suck some fame from their more catchy high-energy colleagues. It's not "the" Higgs boson they are seeing even though what they are seeing may be described by similar mathematics. The ideas flow between particle physics and condensed matter physics in both directions and it's very healthy and mutually beneficial but that doesn't mean that condensed matter physicists may really answer some of the particle physicists' most urgent questions – or vice versa, for that matter.)

If someone says that Homer Simpson discovered the Higgs boson theoretically, it is also crazy because the theory of the Higgs boson was developed by Peter Higgs (with the independent contributions of the other usual suspects) in 1964, thirty-four years before Homer Simpson allegedly did the same thing (and he didn't do it even in 1998).

OK, so those claims didn't survive. What about the most modest claim, namely that Homer Simpson knew about the correct Higgs boson mass in 1998?

Apparently, none of the journalists has worried whether the mass written by Homer is actually correct. They don't seem to care. Others say that Homer Simpson did it correctly so it must be right, mustn't it? Well, as you may expect, it is not right.

So let's start to analyze Homer's blackboard somewhat critically. First, it is not a blackboard, it is a blueboard. This correction was easy.

Now, another correction. The second equation says\[

3987^{12} + 4365^{12} = 4472^{12}

\] The sum of the twelfth powers of two integers equals the twelfth power of a third integer. Is that true? If you know what Fermat's Last Theorem is, you should be able to answer this question. The answer is No, it cannot be true. Fermat's Last Theorem proved rather recently says that\[

x^n + y^n = z^n, \quad \{x,y,z,n-2\}\in\{1,2,3,\dots\}

\] doesn't have any solutions. The exponents must either be "one" (simple addition of integers which works) or "two" (the Pythagorean theorem that has lots of integer-valued examples) but it doesn't work for any higher exponents, whether they're primes or not (it's enough to prove the theorem for exponents that are primes or 4). So the equation with the twelfth power just cannot be right!

If you remember the multiplication table and calculate the expressions numerically on the top of your head, you will see that\[

3987^{12} + 4365^{12} = \\
={\small 63976656349698612616236230953154487896987106}

\] while \[

={\small 63976656348486725806862358322168575784124416}

\] These are two 44-digit integers and they're very similar. But fortunately and unsurprisingly, they are not equal. You may see that only the first 10 digits agree – a large enough number for the generic calculator to fool you into thinking that it may work exactly.

I have to mention that it is not surprising that you may find triplets of integers for which the power-twelve Fermat's identity "almost works" at this accuracy. Note that we needed to specify about 12 digits for the numbers \(3987,4365,4472\), so there were about \(10^{12}\) ways to do it. (It was actually closer to "effectively 10 digits or so" only because some of the first digits are almost required to agree.) We only got an agreement in ten digits, so the probability of that agreement was approximately \(10^{-10}\). If you try to hit a target \(10^{12}\) times (or \(10^{10}\) times, after the correction explained in the previous parentheses) and the probability of a success is \(10^{-10}\), you will probably succeed. It is not a big deal that such approximate solutions exist.

OK, now the Higgs mass formula

Homer's formula for the mass reads\[

M_{\rm Homer}(H^0) = \pi \zav{ \frac{1}{137} }^8 \sqrt{ \frac{hc}{G} }

\] Does it agree with the observed mass – well, let us use the latent energy at rest \[

M_{\rm obs}(H^0) c^2 \approx 125\GeV

\] or not? First, let us notice that Homer's formula is dimensionally correct because \(m_P=\sqrt{\hbar c / G}\) is the Planck mass, and Homer multiplies it by a tiny dimensionless (pure) number \(\pi(1/137)^8\). OK, is the value right?

You may see that Homer's formula is equivalent to\[

M_{\rm Homer}(H^0)c^2 = \pi \zav{ \frac{1}{137} }^8 \sqrt{2\pi} E_{\rm Pl}

\] where \(E_{\rm Pl}\) is the Planck energy \(E_{\rm Pl} = \sqrt{\hbar c^5/G}\approx 1.22\times 10^{19}\GeV\). I had to add the factor of \(\sqrt{2\pi}\) because Homer used \(h\) rather than \(\hbar\) under the square root.

OK, so Homer had\[

M_{\rm Homer}(H^0)c^2 &= \pi \zav{ \frac{1}{137} }^8 \sqrt{2\pi} E_{\rm Pl}\approx\\
& \approx 3.1416 \times 8.06\times 10^{-18} \times \\
&\times 2.507 \times 1.22\times 10^{19}\GeV\approx\\
&\approx 774\GeV

\] He got over \(700\GeV\), close to the "a priori" upper bound for the Standard Model Higgs boson mass that people were aware of before the correct mass was measured.

So the formula as Homer wrote it down isn't producing the right result. The simplest way I see to produce a more accurate result is to replace Homer's initial prefactor \(\pi\) by \(1/2\). But keep \(h\) instead of the Planck-energy \(\hbar\) under the square root. With those numbers, one gets \(2\pi\) times less than Homer did, about \(123\GeV\) which is good enough but too little and "marginally" excluded by the most precise measurements that ATLAS and CMS have offered. (Whether you use \(1/137\) or \(1.137.036\) for the fine-structure constant only affects the result by 0.1% or so.)

Homer's numerical coefficients can't be quite right. Let me now discuss whether a formula like that might be correct.

Simpson sensibly chose the mass to be a numerical multiple of the Planck mass. That's a professional attitude because the Planck mass is the only natural and truly universal constant with the units of mass that appears in quantum gravity (described by effective Einstein's equations in 4 dimensions). But because the Planck mass – equal to 22 micrograms or so, some tiny insect – is so much heavier than the Higgs boson (and other known elementary particles), we have to multiply the Planck mass by a tiny numerical prefactor.

Homer chose a power of the fine structure constant \(1/137\), something that encodes the moderate weakness of the electromagnetic force, and he needed the eighth power to climb down from the Planck mass close to the electroweak scale. One could imagine that this contribution to the Higgs boson mass arises from a Feynman diagram with 16 electromagnetic vertices, something like an 8-loop diagram in QED with two external Higgs legs (yes, readers who independently think have noticed that it's the same number of legs that Peter Higgs has, too LOL).

It's rather implausible that there is a scheme in which the Higgs boson wouldn't get any non-cancelling corrections to its mass from the diagrams with 0-7 loops and only got some 8-loop diagrams to start with. Moreover, it seems unnatural to use the electromagnetic fine-structure in the first place. The Higgs boson is a part of the electroweak theory which replaces the electromagnetic \(\alpha\) by its \(SU(2)_W\) and \(U(1)_Y\) counterparts: electromagnetism and its \(U(1)_{\rm em}\) is just a "fundamentally awkward" mixture of the hypercharge \(U(1)_Y\) generator and the third component \(T_3\) of the \(SU(2)_W\) weak isospin.

Moreover, \(1/137\) is the value of the fine-structure constant at low energies which is obtained indirectly by some complicated "renormalization group flows" from the high-energy values, and it's the high-energy values of \(\alpha\) – and as we said, we really want the high-energy value of \(\alpha_{U(1),Y}\) and \(\alpha_{SU(2),W}\) to be used – that may play a fundamental role in the equations of physics.

So (with some extra discussion about other possible loopholes which would end negatively) I find it implausible that any simple formula resembling Homer's formula may be fundamentally right.

The last two things on the blueboard

Homer wrote that \(\Omega(t_0)\gt 1\) which probably says that the initial energy density of the Universe exceeded the critical density. Well, that's an interesting opinion but as far as the "solid enough" – experimental and inflation-based – arguments go, we have \(\Omega\approx 1\) today. This constant itself evolved in time – see this article about Alan Guth's greatest findings. During inflation, \(\Omega\) was exponentially converging ever closer to one, so that it had enough proximity to stay close to one when the post-inflation, ordinary big bang expansion drove it away from one.

Some refined physicists have interesting arguments on whether \(\Omega\) was greater than one or smaller than one but I don't understand these arguments perfectly and I would argue that this binary question – whether the Universe is open or closed, and so on – remains open. (Note that the state of the Universe and the state of the question differ: the question is open but the Universe is either open or closed LOL.)

The doughnut diagrams

At the bottom of the blueboard, you see some doughnut and sphere diagrams with the right arrow, \(\to\), in between them. You could think that they're perturbative diagrams in string theory – tree-level and one-loop world sheets. However, in that case, the more natural sign in between them would be plus, \({+}\).

I find it more likely that Homer Simpson was describing a topology change of the spacetime (because of the arrow). And the transition actually almost exactly agrees with Brian Greene's toy model for the topology change from Chapter 13 of The Elegant Universe (which I remember vividly, having been the translator of the book to Czech).

In fact, Homer's picture at the bottom of the blueboard is almost exactly identical to Figure 13.3 in The Elegant Universe (see Page 148/189 here and then buy a physical copy not to feel as a thief). A torus degenerates so that a nontrivial circle \(S^1\) on the torus shrinks to zero size. It's replaced by another sphere, \(S^0\), which happens to be a pair of points. Those blow up the holes on the two sides around the shrunk \(S^1\) and the genus-one torus changes into the genus-zero sphere.

This is a great toy model description of a more realistic conifold transition where one may shrink an \(S^3\) in a six-real-dimensional manifold and replace it by an \(S^2\) or vice versa.

Clarifying the origin of Homer's ideas as well as the recent hype

This Simpson blueboard is said to be aired in 1998 and The Elegant Universe was released in 1999 so that Homer apparently knew this cute explanation before the readers of Brian Greene's popular book. But the uniqueness of this idea is so strong that I actually believe that Homer Simpson – or whoever actually wrote this blueboard for him – got an advance copy of Brian Greene's book. That's my conjecture. Or maybe Homer Simpson attended a popular talk by Brian Greene that exposed a similar idea later included in the book.

Now, the question is: If we agree that it wasn't Homer Simpson himself (because of his virtual status), who was the person who got the advance copy of Brian Greene's The Elegant Universe in 1998 and who convinced Homer to have drawn what he drew?

I actually think that the person who convinced Homer to be smart in this particular way was the same person who is behind all the hype about this 1998 episode in recent days. And I believe that both of these men are no one else than Simon Singh who is being cited in these media reports about Homer Simpson's discovery.

What is my evidence?

Well, it boils down to two books that Simon Singh wrote. In 2013, Singh published The Simpsons and Their Mathematical Secrets arguing that the Simpson family are secretly a bunch of mathematical geniuses. I haven't read the book but I think that Singh is actually the person who is responsible for the hidden references to mathematics in The Simpsons.

This Singh's book gives its author a motive to promote secret discoveries made by Simpsons – because it will convince some extra readers to buy the book and bring some royalties to Singh.

Singh wrote several great books – The Code Book, or another book about secrecy from ancient Egypt to quantum cryptography. But Singh's brainchildren also include the excellent book Fermat's Enigma about Fermat's Last Theorem which would explain why Homer Simpson was obsessed with the approximate solution to this famous theorem, too.

So Mr Singh, are you behind all these clever ideas of Mr Simpson and all the hype that surrounds them? ;-)

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reader MikeN said...

The Simpsons probably had sources other than Mr Singh available to them. They were able to get Leonard Nimoy in an earlier episode(after Sulu said no because they were mocking public transportation). Perhaps some of them are just knowledgeable about math.
Having trouble remembering the episode. There was a 3D episode of the Simpsons, possibly a Halloween one. Homer is stuck in a 3D universe a la Flatland. Later on we see p=np floating around.

Since Homer is writing this, then maybe this is an episode where he listens to some vocabulary tapes at night and becomes brilliant, similar to the short story Flowers for Algernon. Later he is seen writing that he expects the effect to wear off.

reader Luboš Motl said...

LOL, Mike, your argument involving Leonard Nimoy is cute.

You realize that Nimoy wasn't a real science officer aboard the Starship Enterprise but just an actor (and poet and photographer etc.), don't you? ;-)

How does the proximity of the writers of Simpsons to Leonard Nimoy increases the odds that they had several informers about mathematics?

reader paul said...

"I haven't read the book but I think that Singh is actually the person
who is responsible for the hidden references to mathematics in The

I recommend reading the book, it is fun and rather light. If you do, you will find out that most of the authors of the Simpsons and Futurama are in fact trained mathematicians (masters and doctorats). that is why they try to crame the most mathematics inside their episodes.

reader Luboš Motl said...

Wow. How does it happen that "many trained mathematicians" end up producing the Simpsons?

reader papertiger0 said...

in season 6 episode 19 of TBBT
There's a particular scene this reminds me of. Link.

I would have never expected Lubos to unpack the donut, and that it actually meant something besides Homer's thoughts drifting off to donuts.

reader Troy said...

more fun than trying to get tenure

reader Liam said...

Hey it's another one of Uncle Al's (not really) "counterexamples" to Fermat's Last Theorem...

These things seem to be all the rage nowadays!

reader Uncle Al said...

3987^12 + 4365^12 = 4472^12 is good to its 10 most significant figures (no rounding). 10/44 = 22.7%

(3,472,073)^7 + (4,627,011)^7 = (4,710,868)^7 is good to its 21 most significant figures. 21/47 = 44.7% We call that "quality." "8^>)

reader Michael said...

Okay, I am starting to believe that the weather simulations supporting the butterfly effect (I mean in the atmosphere, an actual butterfly) in a classical weather simulation have not incorporated damping effects properly. So either it can make no substantial difference in the atmosphere because of damping effects, or its tiny perturbation of the initial state does matter, but then arbitrarily small perturbations matter and quantum mechanics then renders the butterfly unimportant anyway, as two initial states where only the butterfly is the difference will both give probabilities for hurricanes in the quantum case. I therefore think the flap of a butterfly's wings can't matter in the real weather system.

reader papertiger0 said...

Thanks for this buddy. You always go above and beyond what I imagine or expect. That's an exceedingly rare thing.

reader Luboš Motl said...

Thanks, Leandro, you clarified what your position is, and the answer is that you are just another one among the fundamentally wrong people.

1. Learning occurs by learning new data that are demonstrably i.e. experimentally right; and by generalizing them. That's why starting with the most general statement is usually a bad idea - it is pedagogically impossible.

2. It's a complete fallacy, if I avoid the word bullšit, that the assumption about the independence of the particles in the initial state strengthens the validity of Loschmidt's deluded objectsion.

The independence of the particles in the initial state is very well justified. When I (quantum mechanically, to be specific, but it is also true classically) measure all the particles' states, for example, I bring the system to a non-entangled state, a tensor product of states for each particle, and the particles *are* independent in this state.

This "collapse into a product state" occurs whenever one makes a measurement. But even if it is done at the "final moment" of the evolution, it doesn't change anything about the law because the measurement *means* that one knows some information about the physical system *after* the measurement, but not before the measurement (this asymmetry is the "psychological arrow of time", and it is completely unremovable, it's not up to a discussion - no laws of physics that operate within a time can violate it). By measuring, one can't make "oneself yesterday more informed". If this were possible, trading would be easier LOL - it would violate causality.

Loschmidt was full of šit and so you are because you are saying the damn same stupid thing.

reader Leandro said...

Luboš, I thought I said I agreed that the independence was a correct thing to assume?

reader Luboš Motl said...

But you just wrote the opposite, that the assumption "begs the question", and all this nonsense, too.

The correlations are not just "experimentally inaccessible". When one measures the state of atom 1 and then the atom 2, then he is sure that the correlations are *exactly* zero and the physical system is described by a product state. In the same sense classically, when one measures the positions and momenta of two particles, their states are uncorrelated after the measurement! Nothing "begs any question" here.

You are also wrong that the second law of thermodynamics isn't about dynamics. It is completely about dynamics. It's a fundamental law about the dynamics - evolution according to the laws of physical - that involves heat or temperature, and this dynamics is called thermodynamics. That's why the 2nd law is a key principle of thermodynamics and not thermostatics, for example. ;-)

The entropy only increases if the physical system changes over the time and one studies the changes. For the increase of the entropy, the evolution over a finite period of time is a necessary condition, and mathematically, the increase also boils down to something that happens in between the initial and final state - it is not a feature of the initial state separately or the final state separately. You are wrong about all these things.

reader Luboš Motl said...

LOL, I surely upvoted this Gentleman.

reader Anonymous said...

The planck mass appears to have no physical significance, certainly doesn't appear to be involved in setting the mass of any particles. It just seems like desperation, hoping to include it in a Higgs mass calculation.

reader John Archer said...

His name is Pat Condell. Short of killing them, he's very good at upsetting the type I despise.

reader phonon said...

reader JollyJoker said...

Doesn't the h to hbar change give a factor of sqrt(2 pi) instead of just 2 pi? Pointless comment, I just checked that possibility just before reading further and seeing you did too.

reader Dilaton said...

Thanks Lumo for this cool nice detailled discussion of the bluebord :-)!

If the universe is closed, I immediately vote to reopen ... ;-P !

reader Luboš Motl said...

Yes, it does! And I have sqrt(2*pi), don't I?

reader Ralph Hartley said...

The fact that the first formula was dimensionally correct, and consistent with what was known at the time, indicates that the person who wrote it probably knew exactly what he was doing. Making an inside joke.

Homer must have known perfectly well that Fermat's last theorem was proven a few years before. He wrote an equation that he knew was wrong, but that wouldn't be detected by a quick check with a calculator. It isn't an inside joke if just anyone can get it.

I don't quite get the joke about Omega. Maybe I'm not enough of an insider for that one, or maybe I just don't remember 1998 well enough. Reading all the cosmology papers published in the previous few years might help.

He may have known about the figure you refer to (or an earlier version of it), but what the formula actually *says* is that a doughnut changes topology as you eat it. he may be a "wizard" (at least in that episode, which I have not seen), but he is still Homer Simpson.

reader papertiger0 said...

Here's the value of the post. Go back up to the top where is says "But the number of articles pointing out that Homer Simpson discovered the Higgs boson (example) has exceeded 50 which is a lame reason why a TRF blog post may be appropriate. ;-)" . Click that link today, and you'll find all the usual hacks that beleive whole heartedly in global warming religion typing up storys such as 'How Homer Simpson discovered Higg's Boson' by the Nature World News;
'The Simpsons' made a scientific breakthrough years ago..." by the Business Insider; and

'D'oh! Homer Simpson beat scientists in race to find Higgs '
by the Los Angeles Times.

It says something about those outlets' credibility.
If even one mind is saved from those fraudsters with bylines, it's all worth it...

reader Uncle Al said...

Christopher Hitchens, YouTube. Hitch-slapping is reborn! Universities sell fantasies of being clever and able as "Inspected by" stamps.

reader MikeNov said...

You are correct, except I saw this episode on DVD, and from the audio commentary it is clear they are at least crazed Star Trek fans, with the exception of Conan O'Brien, who is the source for the Takei quote. It is probably his work in that environment that makes Conan mock Star Trek so much.

Another episode reveals one of the writers to be a hater of Jimmy Carter, to the other liberal writers there. He had to save his career by criticizing George Bush.

reader W.A. Zajc said...

Lubos, I commented on this a day or so ago (in response to papertiger0) on your post about Brian Greene's new audiobook

Unfortunately, my low-resolution eyes didn't notice on my high resolution monitor that Homer was using h rather than hbar. So Homer's formula nicely predicts the Higgs mass after he corrects some simple algebra mistakes: h -> hbar and pi -> 4/pi ;-)

reader lukelea said...

Well, here he is apologizing:

reader papertiger0 said...

"...just an actor (and poet and photographer etc.)"

I'm using the Vulcan ritual of Kolinahrto suppress my anger at that comment.

reader LM said...

The butterfly's wings cannot influence beyond a couple of metres on a very still day. So there is always a major separation between the butterfly and the confirmed hurricane. For the 'wings' influence to extend further, it has to coincide with a chain of intermediate also chaotic events. The eventual hurricane needs a heat difference, which the butterfly has no influence over. So that further coincidence is necessary. Due to the energetic nature of the hurricane. If there's a heat difference in play, of the parameters hurricaines typically resolve, then this is going to happen whether or not butterfly beats its wings. I always thought the butterfly, while possible in theory, has always been much more an illustrative metaphor, for the sensitivity of initial conditions

reader LM said...

If you look at the structure of what he says, he wasn't arguing for the literal butterfly. The reason it's easy to interpret that he was, is because the goal of his essay was to caste light on underpinnings of the sensitivity to initial conditions of chaotic systems. The name of that is The Butterfly Effect - if you follow the link to the Wikipedia. It's one of those things the butterfly often becomes an in-line metaphor, which to anyone really familiar with the theory is not confusing, but to people who are not familiar but intelligent to reason things through on the fly - such as yourself - it can obviously be confusing. I'm the same...not hugely familiar. NOTE: it's very possible you know all this and simply use the working metaphor already in play. I don't know but we need to buzz off before Motl gets back if we know what's good for us. Parting joke: What goes "Zzub ZZub" --> a bee flying backwards.

reader JollyJoker said...

Sorry, haven't checked the blog in a while.

"With those numbers, one gets 2π times less than Homer did, about 123GeV which is good enough but too little "

So you use the wrong factor in one place there, if I understand correctly.

reader Luboš Motl said...

Sorry, I don't have any errors in my calculation, Homer does. Many people have confirmed that Homer's figure is about 775 GeV.

reader TomVonk said...

If you pick sufficiently general final operators to measure - those that are sensitive to the relative motion of all the atoms etc., and not just some "overall average operators" (like the number of atoms in a cubic millimeter) that we often talk about, then these final operators will depend on the initial microstate.

This includes the dependence on the exact microstate of the butterfly wings. This exact pure state of the wings does affect the generic final-state operators that "feel" the microscopic correlations. And if there's a way for these operators to get translated to "big event" operators such as the existence of a hurricane, and there is such a translation in the real world, then this hurricane depends on the pure initial state of the wings, too.

This is perfectly right.
A practical demonstration has been given by the storm of the century in december 1999 and I recommend to everybody wondering about the butterfly effect to look at that.
The weather institutions computed then about a 100 "macrostates" by varying infinitesimally the values of a few "microstates".
The macrostates didn't gather in a sharp gaussian blob around some average macrostate.
They were all over the phase space and qualitatively extremely different.
In some cases a typical usual winter storm followed and in some cases a real monster appeared.
Of course taking only 100 values of V(0) among the uncountable infinity, doesn't allow to make any statistical prediction so the weathermen having observed that the "monsters" were in a relative minority, didn't worry much.
But the monster came.
The difference between a razed countryside and a regular winterstorm was literally a flap of a butterfly wing.
One thing is not clear to me Lubos.
I didn't read the paper (I know this may be bad) but it is already known for a long time that the exponential divergence of orbits in the Hilbert space is bounded in a chaotic system.
This is due to the fact that these orbits always belong to an attractor and the attractor is finite.
So as the orbits start to exponentially diverge, we get fast to a point where F(t,X0+dx) - F(t,X0) is the size of the attractor (usual L2 norm).
When this happens, the orbits "fold" and stop diverging.
The behaviour of a chaotic system is a succession of stretching and folding of orbits so that the idea of a bound for the dynamical orbits seems to be quite natural.
But L can't really be bounded for a general case especially because there is a whole Lyapounov Spectrum (depending on the detailed dynamics of the system) and there is no reason why the largest positive L (which governs the speed of divergence) should be less than some constant.
Now I may have misunderstood what Maldacena & al want to say and it would be due to my laziness and lack of time to read the paper.

reader onefrog said...

Looks like Le Monde newspaper has picked up this post. Funnily they call you Cech mathematician with family name Motl Pilsen.

reader Luboš Motl said...

LOL, yes, I noticed that, thanks to SiteMeter data. Not bad that Le Monde may guess pretty well even though they have apparently never heard my name - and they haven't read L'Equation Bogdanov, either. ;-)

French guests, you may buy

which is not only about the topic in the title of the book. I hope that the book is faithfully translated from the Czenglish original.

Sadly, I also wanted to add a special automatic translation button to French for the French readers who don't have it in Chrome etc. - but the URL-based Google Translate seems to return errors, even when one clicks at the flags near the title of the blog posts. That's bad!

reader Jitter said...

Mork did a better prediction. In one of the shows he said he had visited all the planets in the Solar system. His friend then asked what about pluto? Mork said no, it's a Micky Mouse planet anyway.