Problem set 1¶
Problem 1.1¶
The probability density function is assumed to be constant, so \(P(\mathbf{x}) = 1/V\), where \(V\) is the volume of the classroom (\(P\) needs to be normalised). So, the probability of a single particle to be inside the box \(p\) is
\[ p = \frac{V_\mathrm{box}}{V} = 0.001 \]The probability that both are inside the box is therefore \(p^2 = 1\cdot10^{-6}\), i.e. one in a million.
That probability would be, under the same assumptions as before, \(P_\mathrm{in}\equiv p^N\) where \(N\) is the number of particles, so
\[ P_\mathrm{in} \equiv p^N = \frac{1}{10^{3\cdot10^{25}}}, \]that is, 1 divided by a 1 with \(3\cdot 10^{25}\) zeroes.
The total number of grains of sand on Earth and the total number of elementary particles in the Universe are completely negligible compared to this number.
Given that the distribution randomises every hour, and that one would sample \(M \equiv 10^{14}\) configurations, the probability that it would not occur is \((1-P_\mathrm{in})^M\). The probability to sample at least one situation where all the molecules are inside the box is
\[ P_\mathrm{occur} = 1 - (1-P_\mathrm{in})^M \approx M P_\mathrm{in}, \]which is still a fantastically small number, so there is no need to be afraid. (If you believe in the assumptions.)
Even if \(M\) were to be increased by a hundred orders of magnitude, it would still not be noticeable, except in the \((10^{25}-114)^\mathrm{th}\) significant digit.
According to the above assumptions that probability would be
\[ P_\mathrm{in} = p^N = \frac{1}{10^{3000}}, \]which is still unimaginably small. However, one needs to be careful with the assumption that the proteins be everywhere with equal probability. Proteins are constantly produced at certain locations in the cell, and they also interact with each other and the environment, changing the probability density distribution. To answer that, we need to know more about the Boltzmann distribution and other important concepts in Statistical Physics.
Problem 1.2¶
The expectation value \(\mu = 3.5\), and all outcomes are equally probable. The standard deviation is \(\sigma = \sqrt{\langle x^2\rangle - \langle x \rangle^2} \approx 1.708\), \(\sigma/\mu = 0.489\), and the distribution is constant, \(P_n = \frac{1}{6}\), where \(n\) is the outcome.
The expectation value is \(\mu_2 = 7\), the most probable outcome is \(7\), and the distribution is triangular, peaked at 7
\[\begin{split} P = \begin{cases} a(n-1) & n\leq 7 \\ a(n-13) & n \geq 7 \end{cases} \end{split}\]with \(a\) a normalisation constant. The standard deviation is close to \(2.415\), \(\sigma_2/\mu_2 = 0.345\).
The expectation value is \(\mu_3 = 10.5\), and the distribution consists of two parabolas, peaked at \(n=10\) and \(n=11\),
\[\begin{split} P = \begin{cases} a(n-2)^2 & n\leq 10 \\ a(n-19)^2 & n \geq 11 \end{cases} \end{split}\]with \(a\) a normalisation constant. The standard deviation should be close to \(\sqrt{3}\sigma \approx 2.96\), \(\sigma_3/\mu_3 = 0.282\).
One can use the central limit theorem here. The mean would be \(100\mu\), and the standard deviation \(10\sigma\), \(\sigma_{100}/\mu_{100} = 0.049\), which is visible as a very narrow peak at \(x = 350\).
Similar as d) except that the distribution is extremely peaked at the mean. The mean is \(\mu_N=N\mu_1 = 10^{12}\mu_1\) and the standard deviation is \(\sigma_N=\sqrt{N}\sigma = 10^6\sigma\), \(\sigma_N/\mu_N \approx 5\cdot 10^{-7}\), with \(N=10^{12}\). The figure of this distribution cannot be distinguished from a delta distribution (which is zero everywhere, except at the mean), except if you would use a microscope and zoom in at the position of the mean.
Problem 1.3¶
These are estimates of the order of magnitude.
Taking the water molecule to be roughly 0.01 nm\(^3\), and the cell to be \(10 \mu \mathrm{m}^3\), would add up to \(10^{12}\) molecules per cell, which is a number computer simulation algorithms cannot handle.
Taking the human body to be roughly 0.1 m\(^3\) and a cell to be 10 \(\mu\)m\(^3\), and assuming the body to consist completely of densely packed cells, this would add up to \(10^{16}\) cells. This is a slight overestimation, by about two orders of magnitude.
Taking a lipid membrane to be 10 \(\mu\)m\(^2\) and a single lipid molecule to occupy 0.1 nm\(^2\), this would add up to \(10^8\) lipid molecules.
Taking the skin to be 1 m\(^2\) and the cell to occupy 1 \(\mu\)m\(^2\) would add up to \(10^{12}\) microbes, if the skin were densely packed with a monolayer of cells.
Problem 1.4¶
All phenomena can be seen as emergent. The final points would deserve a much longer discussion, and have been discussed throughout the centuries, around the globe. These short sketches add a nanobiological flavor.
Temperature from the motion of molecules
Brownian motion from the collisions with other particles
The elasticity of a cell membrane from the interactions between the molecules in the membrane and the average kinetic energy of the molecules (their temperature)
Cell division from multiple processes including protein interaction networks that regulate the binding of filamentous proteins that shape and contract the cell at the right location. Underneath is still the mechanics of all the molecules, having very specific interactions. The expression of genes into proteins should also be distinguished.
Thoughts could be (speculatively) the consequence of a specific firing pattern of neurons, coupled to sensorial patterns and physical stimuli, which relate to the functioning of the individual cells involved, their components and their molecules. Many layers…
Music as emerging from individual notes, or a delicate superposition and succession of sounds, coupling to thoughts and internal processes via aural stimuli.
Awareness could be a collection of thoughts and other neuronal patterns, coupled to the physical infrastructure and environment, and seems difficult to define.