Procrastinating, super-powers and caloric intake

Abstract

Things that keeps me up at night, how much calories in
a short plasma burst, and things I am
doing instead of writing articles.

If you are a member of the male caste of the human race, it is very probable that at some point or the other you have dreamt or even wished that you had super powers. Some of us (gee I wonder who...) still do. By the way in the spirit of equality, this is not to say that being a female precludes you from having the same super-aspirations, it's just that men are basically over-grown kids, thus we tend to sustain this dream longer into adulthood.

Who among us haven't mused about having superman's awesome super-power arsenal. But most of us would settle even for the modest powers of, eh, let's say, Cypher which is essentially that he is a polyglot. 

At any rate, this past week I was a bit under the weather, and one of my computers crashed, which gave me a good excuse to procrastinate. One of the things I did this week was binge on the first season of 'Titans' - the DC young superhero team comprised of Dick Grayson aka NightWing, Raven, StarFire, and Beast Boy. 

Burning daddy to a crisp at a gas station, with super-heated plasma bursts

The thing about being both a geek AND a physicist is that they sometime interfere. destructively.

This happened to me last night, as after watching two episodes of the series, I got to thinking, other than being a full on badass, does StarFire really work? I mean physics-wise? 

Her superpowers are (at this point in the plot) having the ability to discharge super heated plasma (I guess?) from her body, and a slight healing factor. It is also implied that her powers are dependent on some inner reservoir of energy. Which got me thinking. What might be the energy budget of such a thing, I mean at some point we see her burning a bad guy to a crisp, and at no point in the series do we actually see her eat anything. She drinks some hard liquor sometimes (tequila, whiskey shots etc.) but she doesn't seem to eat anything. This begs the question - since she exerts bursts of energy, how does she replenish? 

Well Done, extra crispy.

Supposedly she gets her powers from the sun, like good ol' Kal, and and some point she says she needs several hours in the sun to fully recharge, but that math doesn't really work at all.

<Warning: Physics ahead>

At one point we see her showcasing her powers by burning the body of a tractor. Usually cars and other heavy machinery's body is made out of steel. In fact the average automotive contains about 55% steel, which breaks down to chassis and body, the latter is the majority of the percentage. Some high end cars use aluminum for the body, but this is both rare and expensive. This is mostly done to decrease total weight in favor of energy conservation.

So she probably burnt through steel or iron, and since we don't see any spill-over it makes sense to assume she actually vaporized the iron. The latent heat for the vaporization of iron from room temperature is roughly given by 350 kJ/mol. From the picture we can assess the vaporized area as about \(\pi\;(m^2)\) and the average sheet metal used in cars bodies is about 1.6 mm and mass per square meter of 11.84 \(\frac{kg}{m^2} \) so the total mass of the evaporated material turns out to be about \(37.2\; kg\) which is about 666 moles of Iron. This means that this short burst of about ~3 seconds delivered at least 233 MJ, which is a whopping 55712 kcal. 

Let me illustrate this by using food.

This is the equivalent of 24 Large cheesy bite pies from Pizza Hut. Not slices. Whole pies.

This is the equivalent of 30 Big Mac meals, complete with a large order of fries and a big shake.

In 3 seconds.

And she never eats a darn bite.

So let's examine the story of solar charging.

The solar constant, which is the mean solar energy density flux on earth's surface is roughly \(1.362 \frac{kW}{m^2}\), let's call the human body surface at about \(2\; m^2\) thus it will take StarFire about \[  \frac{233 000}{2\cdot 1.362}\simeq 85,500\; sec \simeq 24\; hrs \]

</Warning: Physics ahead>

Which means that for that paltry 3 seconds of showing off she should have sun-bathed for that whole day. Assuming obviously she absorbs everything. With 100% efficiency. And a 100% efficiency discharge.

Lightly toasted tractor anyone?

Anyway, with a steady diet of maybe 15 large burger meals a day and a strict sunning regiment of say all the available sun per day, she might actually be dangerous for more than 3 seconds a day.

Now I get why she's such a devout martial arts student... 

She only has so much juice in her.

By the way, this makes me curious about scenes of female super-heroes eating. For some odd reason I can't seem to recall to many of those. Most female super-beings that were shot eating are either villains or straight up horribly mutated beings. hmmm....

Having said that, it was fun to watch the first season. And as an added bonus we get to see Deanna Troy smashed by a car.... which is positive in a way. I never cared for her, nor Riker for that matter. 

Be that as it may, the last Purim party my wife and I attended, we went as Riker and Troy, we had to be a couple. She was pregnant. No way was she pulling anything but the infamous Troy dress. 

Plus, I wasn't nearly as Picard-esque as I wanted to be, and I don't think she would have gone for the Crusher red hair look. Oh, right, and being that tool's mom. Awful.

Anyway, now you know what keeps me awake at night. Energy calculations and super heroes. I know. Lame.

 

I need to go back to work. Damn.

Oh right, another instance of superheroes charging up....

Physics of overeating and spaghetti (well, actually strings)

 
Jewish holidays are a horrible thing. Horrible. Especially so for the spatially challenged (fatso-es, like yours truly). Among those of Jewish origins, it is a known fact that every Jewish holiday could basically be described thus: "Hey look! they tried to kill us, we survived, let's eat!!!". Even in the Jewish day of atonement, where you are supposed to fast, if for any reason you are not allowed to - guess what: you need to EAT!

Before the holidays

After the holidays

Well, I promised at one point to elaborate on at least one way of how being fat can save your life. This might be actually a reference to string theory (!!). Let's look at the energy that is present in a standing wave. Oh shucks!, some of us might need an explanation of standing waves - so here goes: 

A game of DoubleDutch - standing waves in action!

Take a rope, tie one end to something very heavy (say a refrigerator handle), and wiggle the other end. You will notice that as you wiggle the rope's end a "hump" in the rope is created, it travels to the refrigerator door, and back again.

In purely ideal conditions, the hump would travel back and forth forever. However the world, much like my bank account situation, is NOT ideal. This means that dispersion and energy loss occurs, which is a main theme in our story. 

So, for the potentially stray physicist that may have come across my blog, here is the full derivation of string dynamics and (average) energy in standing waves.

< warning - physics ahead> 

We can do it in several ways, but we'll take a shortcut:

We will first assume we have a wave, derive the wave equation, and find the associated energy. Here goes:

The wave equation reads:

\[\frac{\partial^2 \psi}{\partial t^2}=v^2 \frac{\partial^2 \psi}{\partial x^2} \]

Which could be solved by method of variable separation or by ansatz. Let's go with variable separation:

\[\psi=X(x)\cdot T(t) \Rightarrow \frac{\ddot{T}}{T}=v^2\frac{X''}{X}\]

 Well, now we have two a-priori uncorrelated functions of different variables, which are equal regardless of time and space, thus they both equal a constant.

And so, we get a general solution: \[T=\tilde{A}e^{Ct}\;\; ; \;\; X=\tilde{B}e^{Dx},\] with the connection C^2=v^2 D^2. Now we \(C\) has dimensions of reciprocal time thus we may call it \(\omega\) up to a numeric constant. Also \(D\) has dimensions of reciprocal length, thus we can call it \(k\) such that \(\omega^2\propto v^2 k^2\).

Now by process of eliminating the constants being purely real, or zero, we are left with a solution of the form: \[\psi=Ae^{i(wt-kx)}+Be^{i(wt+kx)}\] which correspond to right propagating waves and left propagating waves. 

In order to completely solve for \(\psi\) we now need to apply boundary conditions. For people interested in string theory this is basically the whole story behind "Branes" (i.e. D-Branes etc...), meaning the applying of boundary conditions forces some interesting physical behavior of these objects.

BUT! in order to get what we want which is the average energy per wave solution. We can get that by realizing that in a wave dynamic energy is interchanged between kinetic and potential terms, so if we are only interested in a rough estimate we need only to consider the \("\frac{mv^2}{2}"\) term. Meaning, for a small section of the rope with \(\mu_0\) mass density per unit length we have:

\[E_k=\mu_0 \frac{v_{\perp}^2}{2}=\frac{\mu_0}{2}\left(\frac{\partial \psi}{\partial t}\right)^2\] 

This means that the total energy for a string vibrating with a frequency \{\omega\) is roughly:

\[E_k\propto  \mu_0 L A^2\omega^2\]

For an open or closed string there is a connection between the length of the string and the frequency: \(\omega_n \propto n\) in other words: 

\[E_n\propto \mu_0 L A^2 n^2\]

Now, there is a nifty principle in physics called equipartition. It's not always true, the theorem that actually holds is called the "virial theorem", but equipartition is a close and inaccurate relative.

Equipartition states that given a total quantity of energy, the partition of energy to the different constituents is basically equal. This means that plucking a guitar string will start all possible frequencies.

This means that if all active modes have the same amount of energy \(\epsilon\), the amplitude of each mode goes like the reciprocal of the wave number. This gives us a nice "tool" - If we know the amplitude of the "slowest" mode activated, we only need to count modes until the mode amplitude is comparable to "noise".  The mean free path of water is around 3 angstrom. So assuming we can treat human bodies like "ugly bags of mostly water" ,given the initial amplitude is \(A\) the number of activated modes are about \[N=\frac{A}{3\cdot 10^{-8} cm} \].

< /warning - physics ahead>

Suppose now, that you fall from a second floor window only to land squarely on your back. Let's assume the displacement of the compressed body is about a \(cm\). this means that by the previous "back of the envelope" assessment we have about \(10^7 \sim 10^8\) activated modes. the weight of a water molecule is about \(3\cdot 10^{-26}\; kg\), the speed of sound in water is about 1500 m/s, so the basic frequency is given by \(\omega=\frac{c_s}{2\pi \lambda}\simeq \frac{236}{\lambda}\) where we can substitute the basic wavelength \(\lambda\) with the length of the body \(L\). Thus the basic energy quote \(\epsilon\) is given by:

\[\epsilon = \frac{3\cdot 10^{-26}}{2}\cdot L \cdot 10^{-4}\left(\frac{236}{L}\right)^2 \simeq \frac{8.5}{L}\cdot 10^{-26} J\].

We will now suppose that \(L\simeq 2\), thus the characteristic energy quanta of energy is given by \[\epsilon \simeq 4 \cdot 10^{-26} J\]. So naively we get the energy in the form of wiggling to be comparable to \[E=4 \cdot 10^{-19} J\]. However! We are dealing with a 3 dimensional body, so mode degeneracy kicks in, as well as some other considerations.

I won't go into detail, rather I will state that the number of activated modes in a 3 dimensional body goes like \(N^3\)  (for high numbers). So let's get a minimal assessment for the total energy: \[E=4 \cdot 10^{-5} J\] This seems like a very small number.

A different assessment is given by understanding that in the above equation \(\mu_0 \cdot L\)

is the total string mass, we can replace it with the body mass ~ 100kg.

 Calculating this in this way we get: \[E \simeq 1100 J\] which is about a \(\frac{1}{6}\) of the energy deposited in that crushing interaction.

What I believe ACTUALLY happens is that the initial blow activates only the first mode, and then by a process of interference, the energy gets distributed to the other modes. This energy then is released into heat (and sound, and screams, and pain) which leaves the effective blow half as deadly.

How does being fat comes into all of this? Being fat means that the actual volume you have for containing wiggling movement is a lot larger, thus a lot more modes can reside in the same space. It also means that the initial displacement will be times 2 or 3 than if your a skinny dude. This means that the amount of energy that can go into that is bigger from the get go.

and that's my two cents about how being fat can actually protect you and delay your meeting with  John Cleese.

"The salmon mousse!!! Darling you didn't use canned salmon did you?

Prof. Avi Schiller's departure

A post will be dedicated to his memory,

this will take some time but I hope in a couple of days
something nice will be written.

This is by no means a eulogy, but I feel it is fitting to dedicate something to this great man.

It is customary to dub people "great" for their accomplishments and otherwise professional stature.

Avi Schiller was a great man, simply because he was a "mench". One of the few. 

He would also say: A boor cannot be sin-fearing, an ignoramus cannot be pious, a bashful one cannot learn, a short-tempered person cannot teach, nor does anyone who does much business grow wise. In a place where there are no men, strive to be a man.

                                                                       

                                                                                        -Old Hillel, Ethics of the Fathers, 2,5.

Vikings, Relativity and Couch potatoes

Abstract: It's not my fault I'm lazy,
Physics made me this way

dedicated to my sister-in-law and  her fresh hubby's mawwiege

Mazal Tov!!!

Recently I attended my sister-in-law's wedding...

After a full month of being forgotten, taken as granted and ultimately being treated as some ungainly babysitter to my own girl, it has arrived...

The event dreaded and anticipated, probably by the bride and groom, but also by yours truly -

THE WEDDING

Awesome and gruesome, with a multitude of family-related potential calamities, just waiting to happen, hanging on the off chance of someone sitting in the wrong seat or uttering the wrong sentence, incurring a death sentence, with the obvious exception of the crazy and otherwise fully-certifiable-bunkers pseudo-relative  (She can say whatever the hell she wants, nobody's home thus no-one really listens or cares what she says).

The food was good, the company better, and the music divine. well, not really... the first two statements are correct whereas the third - not so much....

The DJ was poor, probably not in the fiscal sense after the presumed hefty sum he was paid, but poor taste in music, poor know-how of his trade, poor execution, and ultimately, the only reason his music was anywhere near palatable, was this was their wedding and we were so damn happy for them, that everything else vanished, not  unlike the 10th order Taylor term for a quadratic potential function. 

At some point the music was pounding so hard, I literally saw hordes of vikings working at a huge smithy, pounding away with all their might to mold some unfathomably hard metal into some equally mysterious shape.

Now, if you follow my posts you know full well what happens to me when faced with potential demise - physics comes into play!

I immediately went on to contemplate vikings and physics, and stumbled upon Dichroism, which is basically when light polarized in different directions go through a certain special material, only one polarization survives (the otherwise polarized rays are reduced to a point of vanishing altogether).

Now since sunlight is non-polarized, this normally doesn't mean squat, but when there's heavy cloud cover, the light rays disperse over the clouds so you get a general bright haze, but no obvious direction for the sun.

Enter dichroism: by using dichroic glass the light that passes through the stone is brightest in the direction of the sun, because right in front of the sun (through the clouds) the light is differently polarized.

The trick is you have to find the right direction to hold the dichroic stone in, else you'll basically get nothing.

Anyway, this dichroism trick is how vikings were able to sail across wide stretches of open sea, since they didn't have land in sight for coastal navigation, and either they didn't know their celestial-nav basics, or the skies were constantly cloud-covered.

BOOM!!

One of the loud pounding beats threw me back into consciousnesses, in time to  realize how horribly out of shape I was...
You see, I was actually dancing (well, more like contorting) on the dance floor, sweating and panting, and  the all too familiar "why am I so lazy?" question kept burning in my mind...

Which again set a chain of events in my mind culminating with physics:

It's not my fault!!! Even physics is lazy!!!

And I will prove this immediately:

First off, remember old Newt's law: 

"Every body persists in its state of being at rest or of moving uniformly straightforward, except insofar as it is compelled to change its state by force impressed"

Translated: every body is basically a fat blob. you really have to exert force to change its current state.

I will now delve into a bit of Lagrangian physics to prove a point:

<Warning: Physics ahead>
This is a simplification of deriving Lagrangian Mechanics but here goes:
Consider Newton's 2nd Law:
\[F=ma\]
And if you're a physicist you know this is a simplification of:
\[\frac{dP}{dt}=F\]
So, now, applying D'Alambert pricnciple we can constrict ourselves to "external" forces, and taking F to always mean a conservative force (i.e. arising from some scalar potential function) we can write:
\[F=-\nabla \left(U\right)\] where U is just the potential function otherwise known by the name "Potential Energy". If we constrict ourselves to one dimension we can see:
\[F=-\frac{\partial U}{\partial x}\] 

Looking at the left hand side of the above equation where we treat \(P=mv\) let's explore the relations between the kinetic energy term - \(\frac{mv^2}{2}\) and the above term.
It's fairly easy to see that:
 \[ mv=\frac{\partial}{\partial v}\left(\frac{mv^2}{2}\right) \]

Let's call the kinetic energy term T and the potential energy term U, so we now get:
\[\frac{d}{dt}\left(\frac{\partial T}{\partial v}\right)=\frac{\partial (-U)}{\partial x}\]

Understanding kinetic energy to be non-dependant on location, and potential energy to be solely location dependant we can add expressions to both sides that vanish in derivation thus defining:
\[\mathcal{L}=T-U\]
to get the Euler-Lagrange equations
\[\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial \dot{q}}\right)=\frac{\partial\mathcal{L}}{\partial q}\] 
Where here I've also switched to some generalized coordinates \(q,\dot{q}\).

Now, defining the action \(\mathcal{A}\) as the time integral of the Lagrangian \(\mathcal{L}\), or in other words:
\[\mathcal{A}=\int\mathcal{L}dt\]

We have now successfully exchanged Newton's rules formulation by the "Principal of least action", since it follows that demanding minimal Action, actually yields Euler-Lagrange equations which are the equations of motion.

A word of caution though: 
First off this derivation is a sketch, the true process is a lot more rigorous, but I assure you, it works.

Secondly, Lagrangian Mech is a heavy canon, all you need to do is describe the problem in some sensible coordinates, find the kinetic energy and potential energy, perform some derivations...

and BAM!!! 

you get equations of motion, as well as conserved quantities and symmetries of the system like magic!!

But, since we use a couple of assumptions in deriving the Lagrangian, this actually covers LESS general physical situations. In reality there's ALWAYS friction, which is by no means a conserving force, thus adaptation to Lagrangian Mech are needed and they are not always as simple.

<Warning: Physics ahead/>

The obvious conclusion of all this mess, is that Newton laws of motion are equivalent (under some basic assumptions) to a first principle that states the following:

"Every physical system aspires to minimize the action taken"

for instance, water will flow in the path of least resistance, and light will travel the path of least (optical) distance, and I, ladies and gentlemen will aspire to do practically NOTHING if I can possibly get away with it.

This means I will absolutely abhor every single time I really have to get up, for instance to go to the fridge and get some grub, I'd much rather my lovely wife make lunch for me and serve it while I leisurely sit back and enjoy the latest mind-numbing episode ofsome stupid comedy show.

By the way, this is EXACTLY why light bends in the presence of gravitation. Not to watch stupid comedy shows, the other thing - being lazy.

See, gravitation can be thought of as a property of space(-time), rather than some force or field that occupies said space. by comrade Einstein's equivalency principle you can not tell whether you are being pulled by gravity, or being accelerated in the opposite direction.

And by that virtue, even light itself (having no mass is irrelevant here) bends to accommodate for this equivalency.

One might even say (and be wrong) that light is the submissive partner in the energy-gravitation relationship... But as I said, this is not correct, since light is a form of energy, and gravitation is a product of energy concentrates. So really, it's a marriage of energy and gravity, or in oriental philosophy terms, marriage of heaven and earth, or ying and yang, or horse and carriage (err... that's from a different philosophy methinks).

To conclude, marriage is such a blessing in one hell of a disguise, since left to our own devices we will do absolutely nothing until some external force propels us to action. When in a relatively successful marriage, the same is true, only the external force is called a wife, and hopefully, most of the times she kicks you in the gonads, it's for your own good.

This also enables two people to become a system of coupled equations, thus redefining the least action to actually produce some interesting dynamics, at least until a third little equation comes into play to dominate the hell out of their lives... if you're a parent, you know what I mean, right?

My own sweet third coupled equation 

Alternatively, if your marriage sucks, well, this explains why people stay in dysfunctional relationships for too long - it simply is to much ACTION to get out of one, better (by physical standards, not mine), to become mentally dislodged and otherwise apathetic to a fault.

But we know it's the former rather than latter. we really do.

Anyway we love you guys very much.

Mazal Tov!! 

Inflations, Oscillations, and Weight loss...

Both Cosmic and otherwise

To those who had the (mis)fortune of having me as a part of their lives,  weight gain, and weight loss are all too familiar. if not by first experience, they have  probably witnessed my own physique change, all across the range spanning from wirey-thin to brown dwarf ( a so called Failed-Star, or UltraCool Dwarf, which is a stellar body of roughly 13-85 Jupiter masses), and hopefully, one day, back again.
Since my travels through mass-space have been oscillatory in nature (for the most part), it is with pain-laced mirth I give you the following analysis and analogy of myself, and the Cosmos.

I know, I know, this seems a bit megalomanic on my part, but I assure you... I am. 

Well really, since my current field of study is Cosmology, and specifically inflation models, I found it might be funny to recount my thoughts and findings and relate it to the human condition, or rather, the fatso condition...

To any and all cosmologists out there, who might be reading this right now, be gentle, I am a mere neophyte to this field, and so I appeal to your sense of humor, and sympathy (enter violins please).

In order to first pique your interest, please listen to this number by Monty Python:

Which, by the way have some major faults in cosmological terms, for instance, the universe (acording to leading theories) actually does NOT expand at the speed of light. our EVENT-HORIZON (or simply horizon) does. It's meaningless to ascribe velocity to the expansion of the universe, since very close to us, the universe recedes from us in small velocities and far from us the universe seems to recede with greater velocity, as reflected by Hubble's law - velocity is proprtional to distance from us... \(\left(v=H_{0}D\right)\) .

By the way, this also implies that Galaxies which are far enough from us, recede at velocities greater than the speed of light (yes it IS possible), and thus they "drop off" our horizon, since they fade away faster than their light travels to us.

This also foretells a dark and lonely eventual demise, for our universe.

At any rate,  I would like at this juncture, to acquaint the reader with Friedmann's equation, which I will first write down and derive LATER (in two ways by the way...)
\[\left(\frac{\dot{R}}{R}\right)^{2}=\frac{8\pi G \rho}{3}-\frac{\kappa}{R^2}\]
Where \(R\) is the radius of an arbitrary sphere in space. this equation describes the evolution of \(R\) due to gravitational forces alone.

In layman terms:

The universe was born, got fat, tried dieting for a bit, got frustrated and ultimately gave up, becoming ever so fat.

For those who have some interest of keeping their sanity, you are invited to skip these derivations.
For those of brave soul and not so sound mind, please take special interest in the general relativity case...
(Important conclusions, in green)

<derivation - Newton style>
from Newt's 2nd law: \[F=ma\\
\Rightarrow F=m\ddot{R}\]
Gravitational force due to enclosed mass, on the perimeter of a sphere:
\[F=-\frac{GMm}{R^2}\]
Equating these yields:
\[\ddot{R}=-\frac{GM}{R^2} \Rightarrow \dot{R}\ddot{R}=-\frac{GM\dot{R}}{R^2}\]
Integrating, we get:
\[\dot{R}^2=\frac{2GM}{R}-\kappa\]
So, now, for matter in a sphere with a radius of  \(R\),   \(M=\frac{4\pi R^3 \rho}{3}\), thus:
\[\dot{R}^2=\frac{8\pi G \rho R^2}{3}-\kappa \Rightarrow \left(\frac{\dot{R}}{R}\right)^2=\frac{8\pi G \rho}{3}-\frac{\kappa}{R^2}\]
taking \(R=R_0\cdot a(t)\) we get:
\[\left(\frac{\dot{a}}{a}\right)^2=\frac{8\pi G \rho}{3}-\frac{\tilde{\kappa}}{a^2}\]
</derivation - Newton style>
It's fairly easy to see that, according to Newton, the term \(\frac{\kappa}{R^2}\), is simply a term that is a result of integration, i.e. an integration constant. And so according to this derivation \(\kappa \) is simply given by initial conditions, and given no other incentive, we can gauge it away.

Let's see what old Einei have to say about this...
<derivation - Einstein style>
I'll try to be brief yet informative:
The metric for a spherical symmetric curved space is given by:
\[ds^2=-dt^2+a(t)^2\left[\frac{dr^2}{1-\kappa r^2}+r^2d\Omega\right]\]
Or in matrix form:
\[g_{\mu\nu}=\left(\begin{array}{cccc}
-1&&&\\
&\frac{a^2}{1-\kappa r^2}&&\\
&&r^2&\\
&&&r^2\sin^2(\theta)
\end{array}\right)\]
From here it's relatively easy to find the Christoffel connections:
\[\Gamma^{0}_{11}=\frac{a\dot{a}}{1-\kappa r^2}\;;\;\Gamma^{0}_{22}=a\dot{a}r^2\\
\Gamma^{0}_{33}=a\dot{a}r^2\sin^2(\theta)\;;\;\Gamma^{i}_{0j}=\delta_{ij}\frac{\dot{a}}{a}\\
\Gamma^{1}_{11}=\frac{\kappa r}{1-\kappa r^2}\;;\;\Gamma^{1}_{22}=-r(1-\kappa r^2)\\
\Gamma^{1}_{33}-r(1-\kappa r^2)\sin^2(\theta)\;;\; \Gamma^{2}_{12}=\Gamma^{3}_{13}=\frac{1}{r}\\
\Gamma^{2}_{33}=-\sin(\theta)\cos(\theta)\;;\; \Gamma^{3}_{23}=\cot(\theta)\]
all other symbols vanish.

The Ricci scalar is then given by:
\[R=6\left[\frac{\ddot{a}}{a}+\left(\frac{\dot{a}}{a}\right)^2+\frac{\kappa}{a^2}\right]\]
The Einstein equation:
\[G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=8\pi G T_{\mu\nu}\]
or, equivalently :
\[R_{\mu\nu}=8\pi G \left(T_{\mu\nu}-\frac{1}{2}g_{\mu\nu}T\right)\]
Taking the 00 term we get:
\[3\left[\left(\frac{\dot{a}}{a}\right)^2+\frac{\kappa}{a^2}\right]=8\pi G\rho\Rightarrow \left(\frac{\dot{a}}{a}\right)^2=\frac{8\pi G\rho}{3}-\frac{\kappa}{a^2}\]
And, if we use the other notation also we get:
\[\frac{\ddot{a}}{a}=-\frac{4\pi G}{3}
(\rho+3p)\]


</derivation - Einstein style>
For now, we'll leave the second equation be, and look at the first.
We'll notice a couple of things first - Although it SEEMS as though the equations (Newt's, and Einei's) are the same, really the real Friedmann equation (derived from General Relativity) is the more general case. that is due to the former derivation relying on dynamics of MATTER only, and the latter does not, though, a careful massage of the former with reletavistic ideas, might give the correct equation for radiation, and other cosmic stuff...
The second thing I wish to emphasize is the existence of \(\kappa\) and it's meaning: In the Newtonian analysis, it was simply an integration constant, but in the Einsteinian analysis this is an intrinsic factor to the very fabric of the universe, this is the CURVATURE signature of the metric, that governs whether the cosmos is positively, negatively or null curved (i.e. flat) like in this picture:

Curvature types: positive, negative, and null.

<derivation - Gangnam style>
And this is just for fun:

</derivation - Gangnam style>

Admission of Guilt, errr... Ignorance

But wait!! What does all of this have to do with inflation?
Well, in fact, suppose space is sufficiently dillute, the matter density is sufficiently close to zero, and if it's sufficiently cold, radiation density drops to zero (almost), with an (almost) flat curvature, we are then left with some quantity we'll call \(\rho_0\), ("Rho naught"), and we get a dynamic equation, with an inflationary/deflationary solution:
\[\left(\frac{\dot{a}}{a}\right)=\pm\sqrt{\frac{8\pi G\rho_0}{3}} \Rightarrow a=A\exp\left(\sqrt{\frac{8\pi G\rho_0}{3}}t\right)+B\exp\left(-\sqrt{\frac{8\pi G\rho_0}{3}}t\right)\]
The second part drops off rapidely, and so we are left with an inflationarry solution.

"What is this witchcraft?!?!" you ask - where does this \(\rho_0\) come form?
well, suppose you're a fat guy, and you go into a fasting mode, note that you are actually GAINING weight (at least in the short run)... and by the time you've lost the battle against hunger, and went on a burger binge, guess what? you've now underwent inflation.

Well, actually this has nothing to do with \(\rho_0\), here's the REAL explanation for this:

WE DON'T KNOW!!!!

Oh my god, I can't believe I said that!!! this is the absolute NO NO for physicists!!!
To actually state that I don't know something? to recognize that everything we *THINK* we know is simply an approximate modeling of the awsome and complex reality we live in???!?

I should stop. NOW!!! I hear them knocking already... don't let them take me! NOOOOOOOOOOOOOOOOO!!!!

Ok, done with that gag, are we?

Moving on, there are several possible explanations for this, most are problematic to say the least, but hey, this is what it means to be in the front lines of science. you either get promoted or get diced....

At any rate, Inflation is a widely accepted theory of the adolescent universe, whatever mechanism manifests it.

What about oscillations?
In a nutshell - at the onset of the early universe, after inflation, some major ocillations occured in matter density, affected by ordinary (barionic) matter as well as dark matter, these oscillations are known as BARIONIC ACOUSTIC OSCILLATIONS (or modes) , and they could be seen quite nicely in analysis of the cosmic background radiation.

Moreover, when the universe was matter dominated, the same dynamics might have happened on a cosmic scale - Suppose matter is the dominant part of the universe, the dominant force then is gravitation, thus the universe itself, aspires to CONTRACT, offset only by radiation pressure, there MIGHT have been an epoch of slight contraction on the universe's part.

Sadly I didn't find a sufficiently fascinating animation to show, of the acoustic oscillations, but maybe some other time...

In conclusion, much like myself, the universe was "born", and began inflating.
Undergoing some oscillations, and at a certain point (just about.... now!) the universe moved from matter driven dynamics into  moderate inflationary epoch.

In layman terms:
The universe was born, got fat, tried dieting for a bit, got frustrated and ultimately gave up, becoming ever so fat.

In the words of our mutual friend:

Entropy and Einstein's turnover time

Abstract
A failed attempt at explaining Entropy,
and one zombie, coming right up...

And so it was, on a nice evening, much like this one, that we had all sat around the table, and a question popped up...

The question was along the lines of "what is the Entropic principle?", and it was asked by my brother, a brilliant man, and a science-fiction aficionado, who unfortunately for the physicist community, never had the chance to dabble with physics, and so they have to find a poor substitute in the image of your humble servant here...

At first I asked if he meant the Anthropic principle, but he just wanted to understand Entropy.

Thus I found myself trying to explain Entropy, and the 2nd law of thermodynamics to the uninitiated, in layman terms, and, after a fashion, follow the Einstein grandmother rule - "You do not really understand something unless you can explain it to your grandmother." (A.Einstein).

Incidentally , you could probably calculate the period time \(T\) of Einstein's turning over in his grave, by someone misquoting him, or otherwise deifying him, and justifying a falsehood or plain ol' stupidity by attributing something to old Einei that he never would have meant in a million light years.  

We'll start with some observational data - I have around 400 people in my human network, and on average I get an Einstein quote which falls under the aforementioned category, maybe once every two weeks.

Now, suppose only a third of the world's population leads a somewhat western lifestyle (either connected to facebook, google+, twitter etc. or alternatively reads the paper and or listens to the radio at least once a day), we have about 2.3 Billion people.

let's be harsh and assume each of the 400 people sub-networks are non-connected between them and so we neglect back-propagation we can put a lower limit of
\[\frac{2.3\cdot 10^{9}}{400}=5.75\cdot 10^{6}\,\text{instances in 2 weeks}\]
Divided by the number of seconds in a two-week period we get:
\[\frac{5.75\cdot 10^{6}}{14\cdot 24\cdot 60\cdot 60}\approx 4.75\,\text{times per second}\]

And so the period time of Einstein's turning in his grave would be \(T\approx 0.21 \,sec\).

So basically even the lower limit states that Einstein, by now, is a zombified Olympic athlete, even considering the initial rigor mortis....

By now, he would have a solid six-pack.

Anyway, I digress, I was going to explain Entropy and then a random rant stole my attention... sorry for that.

In a nutshell, Entropy is a measure of disorder, and I will explain.

<Failed attempt at an explanation : but is still worth a read>

Imagine a group of four coins, each with two sides - heads, and tails - right? (we'll have non of that Two-Face shenanigans here!)
Now suppose every coin is perfectly balanced so there's a fifty-fifty chance of getting heads or tails for each coin flip.

So, now, what are the chances of getting all 4 heads, when you flip 4 coins?
if you do the experiment enough times, you get an average of 1/16 chance.
the state of all heads, or equivocally all tails is the most "ordered" result, why?
because it is the most homogenous result (and we humans like homogeneity, symmetry and by the same token order).
Now, what's the most plausible result?
that's easy - the result where two coins are tails up, and two coins are heads up (regardless of their locations), happens ideally \(\frac{3}{8}\) of the times you flip (almost half of the times you flip the coins, you'll get this result).

That is the least "ordered" result, since we don't care about locations, and the coins show the most diversity in results.


Now, I won't go through the whole derivation, but if you'll try the same logic with 6 coins and then 8 coins you'll get a breakdown of \[\frac{1}{64},\frac{6}{64},\frac{15}{64},\frac{20}{64},\frac{15}{64},\frac{6}{64},\frac{1}{64}\,\text{for six coins}\]
and a breakdown of \[\frac{1}{256},\frac{8}{256},\frac{28}{256},\frac{56}{256},\frac{70}{256},\frac{56}{256},\frac{28}{256},\frac{8}{256},\frac{1}{256}\,\text{for eight coins}\]
And so on and so forth, the reason I'm sticking to even numbers is because it LOOKS more clear that way, but really, it makes no difference. You could go on and on until kingdom come, and you'll find the middle, most unorganized result will be the most common.

It turns out that this kind of dynamic is best approximated by a gaussian function called the \(g\) function (or the multiplicity function) and I'll spare you the details in favor of a graph:

Probable results graph

So, what you see here, is basically an overlay of 4 graphs that show the relative probability of results as they stray from the middle "disorganized" and probable result.
What is interesting, is the bigger the experiment is (i.e. instead of 8 coins, let's say a 100 or 1000 coins) the sharper the peak is, meaning it's narrower, and higher in respect to other possible results. that means by the way that the most probable result is highly probable, and the others highly improbable.  Now imagine an experiment of \(10^{23}\) coins, every result other then the most probable and it's immediate neighbors is SO improbable, it virtually is IMPOSSIBLE (in the sense that it would take a ludicrously impossible amount of experiments to perform to actually get a significant chance to get such a result).

A word of caution though - this is probability we're talking about, so in theory a highly organized result MIGHT happen, in actuality - yeah, not so much...

By the way, there are roughly no more than \(6\cdot 10^{14}\) coins in circulation today in THE WORLD, meaning even if you took all the coins in the world today you couldn't perform such an experiment, even once!

Incidentally the ridiculously high number of participants in a single experiment, makes all the difference between "hard sciences" even if they are statistically oriented, and "soft sciences".

Even if we take all the people in the world, and get them to participate in one of our experiments, the result will produce some correlation that may, or may not apply to a single participant.

In physics, while the same is true, you could say a statistic result applies and be absolutely correct on a macro level (with deviations so small as to be insignificant for most purposes), and be correct almost every time on a micro level as well!.

So anyway, Entropy is defined as the logarithm of the multiplicity function.

The reason for taking the logarithm is for the sake of defining a cumulative quantity, as opposed to multiplicative.

< /Failed attempt at an explanation : but is still worth a read> 

So anyway, obviously I failed at this attempt but let's try it in a simpler manner:

Entropy is a quantity that signifies how probable a result is.
by a fluke of chance, which isn't a fluke at all, more of a deep connection really, the most probable result is also the most diverse one, or differently put, the most disorganized.

Thus, Entropy becomes a measure of disorder of a system.

Entropy is a cumulative property in the sense, that when two non-interacting experiments are done the combined entropy is the sum.
However, when systems are allowed to interact, the combined entropy is typically larger than the sum of individual entropy.

it is by that sense, that entropy tends to increase over time (and interactions).

<example of entropy increase>

Suppose, we have two systems, each of 4 coins.

The most probable state is given by 2 heads, and 2 tails for a single experiment right?
as was explained in the above failed attempt, the chance for that happening is \(\frac{3}{8}\).

Now, what is the chance of each of the experiment to get the most probable state independently? you guessed it - the product of the two independent probabilities i.e. \(\frac{9}{64}\) right?

OK, but now, let's put all the coins in a single experiment, an flip all of them, the chance of hitting 4/4 division of heads/tails is given by \(\frac{70}{256}=\frac{35}{128}\) which is almost double the size of the product of individual probabilities.

So what happened here, really?
in essence, the combined system has more places to choose from, meaning more diversity of scenarios that lead to the same end result, thus the combined system is more "disordered" thus bigger probability that leads to bigger Entropy.

</example of entropy increase>

OK, that wraps it up for this time.
Obviously I don't understand Entropy enough, since I feel I have failed at explaining it,
but I will try again ("if at first you don't succeed" etc...).

Oh by the way, for all us \(\LaTeX\) geeks out there, isn't it cool that every time we use a fractional, we immediately get a reference to Battlestar Galactica?

Anyway, till next time...

P.S. some quick ideas to utilize Einei's incredible turnover time of 0.21 seconds :
1. Attach him to a turbine and generate electricity.
2. Display him as the 8th world wonder - the quickest man on the planet (faster than Usain Bolt).
3. Use him as the engine for a horse-ride carousel for my kid.
4. Use him to refute the 2nd law of Thermodynamics as he is both dead (Entropy supposed to get bigger), but in the greatest shape of his life (which requires work and decrease in local Entropy)...

Just sayin'...




 

Short Derivation of Archie's Law

 "captain's log: supplemental"

It's been quite a while since I wrote the previous post about Archie, his wife, a crown etc.
While I was writing that, I did not want to derive Archimedes' law on my own, so I looked around on the net, to maybe find someone who already posted that, sadly, I couldn't.

Also, this morning my wife and I had an argument about this law, and guess what, she was right, I was wrong... Isn't it weird these are the arguments my wife and I have?!? I mean, seriously, how nerdy can you get? 

So for the interest of completeness (which is actually a mathematical axiom, but let's leave that be for now), I give you a short derivation of Archie's law:

Deriving the law (I AM THE LAW)

Let us consider an infinitesimal volume element of a material with density \(\rho\) , and incident area and height \(A,h\) respectively.
So now let's suppose the material is lighter than the medium surrounding it we get a force equation that looks like this:
\[\Sigma F=A\cdot P_{up} -A\cdot P_{down}-Mg\]
Where the force equivalent (sum of all forces) is positive.

Now, let's wrap the forces that are the result of pressure and name it the Buoyancy force thus:
\[F_{Buoyancy}=A\cdot\left(P_{up}-P_{down}\right)\]
and with a little massage we get:
\[A\cdot\left(P_{up}-P_{down}\right)=A\cdot h \left(\frac{P_{up}-P_{down}}{h}\right)=-V\nabla P\]

So, now, let's take a small detour to understand who \(\nabla P\) is shall we?

suppose we are dealing with an element that is filled with the same material as the medium it's in, in that case we get a mechanical equilibrium and \(\Sigma F\) is simply zero. and so we get:
\[Mg=F_{Buoyancy}=-V\nabla P\]
and breaking up the \(Mg\) element we get:
\[V\rho_{medium}\cdot g=-V\nabla P\Rightarrow \nabla P = -\rho_{medium}\cdot g\]
and so we get quite simply:
\[F_{Buoyancy}=V\cdot \rho_{medium}\cdot g\]

Proper usage

What's useful with this representation, is that we get the force outright, thus we can make force calculations directly.

An important thing to understand is that the buoyancy force is directed not UPWARDS (which is a common misconception) but against the direction of the pressure gradient . which is quite different. that means for example that in a spinning tube of air, the buoyancy force will be mostly inwards as the pressure gradient is directed outwards (in cylindrical coordinates).

That's it.

this time it was really short.