Sunday, September 30, 2018

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?





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