In our last post we simplified how the gas particles inside the engine behave, but how do we know how they move, what their velocity is, and if they even escape through the hole at the bottom of the engine?
We may not know the exact position of every individual particle, but seeing as there are so many of them, we can assume that they are uniformly distributed across the engine.
Can we assume the same of the velocity? Let’s think about it for a second. Imagine you're watching a race - How are the people spread out across the track? Usually there are a few people in the lead and a few people in the back, while most people are somewhere in the middle.
Turns out we can describe how fast the runners are by a normal distribution, sometimes reffered to as a bell curve for its iconic shape, or a Gaussian distribution for its iconic man. <3 The probability (P) can be described by the normal distribution as:
\(P(x) = \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma} \right)^2}\)
The probability depends on a variable, x, on the average value, \(\mu\), and on the standard variation, \(\sigma\), which is our average difference from the average. Because the normal distribution depends on average values, it is more accurate for higher values of x, and can often be used to describe plenty of things in nature, including our particles! Using the average distribution we can find the probability of a certain amount of particles having a certain velocity in a certain direction, namely the \(x-\), \(y-\) and \(z\)-direction. This is a very important distinction from \(P(v)\) , where we look at speed rather than velocity.
If we rewrite this equation with velocity as the variable, we can set \(\mu = 0\), because a particle is just as likely to move to the left as to the right along an axis, and velocity is a vector - it includes direction. Most particles are in fact not at a complete standstill, even if the velocity distribution has reaches its maximum at \(v_x = 0\) (if that is the direction we are looking at, but the same goes for \(v_y\) and \(v_z\)). We define \(\sigma = \sqrt{\frac{kT}{m}}\), where T is the temperature in kelvin, m is the mass of the particles and k is the Boltzmann constant. We rewrite as:
\(P(\vec{v}) = \left( \frac{m}{2\pi kT} \right)^\frac{1}{2} e^{-\frac{1}{2}\frac{m\vec{v}^2}{kT}}\)
This is a version of the normal distribution known as the Maxwell-Boltzmann distribution, and it's often used when looking at the velocities of particles.
Lets have a look at a velocity distribution between the interval \(-2.5\cdot10^4 \: m/s\) and \(2.5\cdot10^4 \: m/s\):
The particles have a higher probability of having velocities under the higher part of the graph, and are less and less likely to have velocities of higher magnitude. Negative velocities indicate that the object is moving the opposite way from the objects with positive velocities, but they still have the same speed.
The velocity distribution lets us know the probability density, which makes for a pretty graph, but if we want to know the probability, we need to sum up the values within the interval of velocities we are curious about. However if this interval is very large we have to integrate over the chosen interval. For example, if we want to know the probability of a particle having a velocity between the interval \([5,30]\cdot10^3\)m/s, we integrate over the interval:
\(P(5\cdot10^3,30\cdot10^3) = \int^{5\cdot10^3}_{30\cdot10^3} P(v) \: dv\)
The probability for this interval will be approximately 0.158, or 15.8%. The answer will always be between 0 and 1, where 1 means there is a 100% chance the particle has this velocity. If we multiply this integral by the number of particles, \(n\), we get the number density, or the number of particles with the velocities in the interval. This is written as \(n(v)\), meaning we can define the number density as:
\(n(v)\:dv = nP(v)\:dv\)
This may look familiar if If you've had a look at the derivation for \(P = nkT\) in our first post!
An interesting quality of the normal distribution is that if we look for the probability between an interval of the standard variation (from negative to positive) we get the same values every time because the equation already depends on \(\sigma\).
\(P(-1\sigma\leq x-\mu \leq 1\sigma) = 0.68\\ P(-2\sigma\leq x-\mu \leq 2\sigma) = 0.95 \\ P(-3\sigma\leq x-\mu \leq 3\sigma) = 0.997\)
These values corresopond with the colored areas in the graph above! The darkest color covers 68% of the area under the graph, the lighter covers 95% and the lightest covers 99.7%.
Knowing the standard variation we can get a sense of the width of our distribution without having to use the more complicated Gaussian function. This relation is refered to as the full width at half maximum, and is defined as FWHM \(=2\sqrt{2\ln2}\sigma\). We won't be using this particular property, but those who want to further their understanding of statistics a little further can have a look at the derivation of FWHM here.
The information so far tells us a lot about how the particles move, but what if we would like to know the average speed of a particle?
We can start by finding the absolute velocity distribution. Think of it as the probability of the particles having a certain speed (note how \(v\) is not a vector value)!
\(P(v) = \left( \frac{m}{2\pi kT} \right)^\frac{3}{2} e^{-\frac{1}{2}\frac{m\vec{v}^2}{kT}}4\pi v^2\)
Which may produce a graph like this one:
This is the probability of a particle having a speed within the given interval. Speed is a value without any direction, and so we see how all values are positive. The curve is no longer the iconic bell shape of the gaussian distribution, because for speed we have a lower probability of finding very slow values than we do for higher ones.
If we wanted to find the average speed, \(\langle v\rangle\), we multiply with \(v\) and simply integrate:
\(\langle v\rangle = \int^\infty_0 v\:P(v)\:dv\)
This expression actually holds for any average of a function:
\(\langle f(x) \rangle = \int^\infty_0 f(x) \: P(x) dx\)
For example, if we insert the expression for kinetic energy, \(E = \frac{1}{2}mv^2\), with a variable \(v\) as our function, we get:
\(\langle f(v) \rangle = \int^\infty_0 \frac{1}{2}mv^2 \: P(v) dv\)
Inserting the above expression for \(P(v)\) and solving gives us the expression for the average energy of a molecule in an ideal gas:
\(\langle E \rangle = \frac{3}{2}kT\)
Which may come in handy, but even if it doesn't, it's still pretty cool! It does show us that the average energy of a molecule in an ideal gas is only dependent on the temperature.
We now know that the particles inside the engine will have velocities that are normally distributed, and we know what this means in more than just name. Such deeper understanding is important, so we're not simply throwing equations around, hoping one of them will launch our rocket ?\_(ツ)_/?