Maxwell distributions and average, RMS, and most probable speeds

[rtareen]

What I know about a Maxwell distribution is that an area under the distribution on an interval gives the fraction of the molecules with speeds on that interval. My first question is what does Maxwell's distribution represent? It is given by the formula
##P(v) = 4\pi \left(\frac{M}{2\pi RT}\right)^{3/2}v^2e^{-Mv^2/2RT}##

If we want to find the fraction of molecules that range from speeds ##v_1## to ##v_2## we can find that by using:

##\int_{v_1}^{v_2}P(v)dv##

So what does P(v) itself represent? It has units of s/m.

Secondly, the book says that to find the average speed of the molecules, we weight each value of v in the distribution. This means we multiply the integral by v and take the limits to go from zero to infinity. In other words:

##v_{av} = \int_0^{\infty}vP(v)dv##

Can somebody explain what is going on here because I am completely lost. How does this formula give the average speed? What is weighting?

Similarly, we can find the rms speed with the following formula

##v_{rms} = \sqrt{\int_0^{\infty} v^2P(v)dv}##

Again, what's with the ##v^2## and how does it help finding the mean squared speed? When we take the square root of this it becomes the root mean square speed.

Finally, we have the most probable speed, in which to find we set dP/dv = 0. I don't know how to take that derivative, and from what I've seen from my calculator its a really messy derivative. But anyways we then solve the equation for v and we get
##v = \sqrt{\frac{2RT}{M}}##

and this is the most probable speed. Why do we set the derivative to zero to find the most probable speed. I have a feeling all of this has to do with statistics and probability, which I have no experience with. Do statistics and probability come up a a lot in physics? If so would it be a good idea to take an upper level probability course? I want to major in physics.

 

[etotheipi]

Edit: This ended up being a lot longer than I "expected" (pun intended)

The function is showing the probability distribution of molecular speeds, ##v##, in an ideal gas. All of this is based on probability density functions, and you would be served well by doing some reading about them (Wikipedia is not a bad start). The general idea is this; the probability of a randomly selected particle having a speed between ##v## and ##v+dv## is ##P(v)dv##, i.e. ##P(v)## is the probability per unit increment of ##v##. If you integrate ##P(v)## between two chosen speeds, that's the probability that a randomly chosen particle will have a speed in that range.

You mentioned that ##P(v)## has units of ##\text{sm}^{-1}##, so it follows that ##P(v) dv## has units of ##\text{sm}^{-1}\text{ms}^{-1} = 1## i.e. it is dimensionless, which is what we expect of a probability!

For such a distribution, the mean of ##v## is ##\langle v \rangle = \int v P(v) dv##, where ##vP(v) dv## is integrated over the entire domain - in this case, zero to infinity, since speeds are strictly positive. If you were to measure the speeds of many many randomly chosen particles, and average them, you would approach ##\langle v \rangle##. To give some context, for a discrete distribution (with which you might be more familiar) the expectation of a random variable ##X## is defined as the sum of all the possible values of ##X## each multiplied (weighted) by their respective probabilities,$$\mathbb{E}(X) = \sum_i x_i p_i$$If you think of an integral as the limit of a sum, that formula for ##\langle v \rangle## above should seem a little more intuitive in the continuous case: very loosely, ##x_i \rightarrow v## and ##p_i \rightarrow P(v)dv##.

Similarly the mean of ##v^2## is ##\langle v^2 \rangle = \int v^2 P(v) dv## (do you notice a pattern?). The RMS - literally root mean square - is the square root of the mean of ##v^2##, denoted ##\sqrt{\langle v^2 \rangle}##. Another useful measure is the variance, which is the square of the standard deviation, and is calculated by ##\langle v^2 \rangle - \langle v \rangle^2##.

If you sketch the probability density function, the maximum of the function will correspond to the speed with the highest probability. You will be aware that a maximum of a function can be found via differentiation (here we don't need to worry about checking the 'endpoints' of the domain, since the function goes to zero at both ends).

Do statistics and probability come up a a lot in physics? If so would it be a good idea to take an upper level probability course? I want to major in physics.

You should probably wait for some advice from someone more experienced than I am, but I might add that there is a whole branch of Physics called Statistical Physics. If you are interested, maybe you will like some of David Tong's stuff

 

[rtareen]

Edit: This ended up being a lot longer than I "expected" (pun intended)

Thanks for your response. I appreciate your effort!

You say that P(v)dv is the probability of a random molecule having a speed between v and dv. As you mentioned before this probability is unitless. But what about the book's interpretation? It says that P(v)dv gives the fraction of molecules that have speeds in that range. I guess these could be the same thing. Actually now that I think about it, it probably is.

You say P(v) is called a probability density function. I did a little reading and learned that the dependent variable P always have reciprocal units of whatever it is a function of. This makes sense whenever you think of P as a fraction or a probability. But i still don't really know what P is. Let's say you plug in v = 200 m\s and you get P(v) = ##10^{-3}##s/m. What does P by itself actually represent? Does it have any meaning?

I don't know anything about the expectation function like you suggested I might have, but assuming it is true then the formulas for average and rms speed actually make sense! Thank you!

And as for the most probable speed, If we interpret P(v) as probability, whch i don't know if it correct, then I can see how the most probable speed occurs at the critical point. Maybe you or someone else can clear this up by explaining what P is.

 

[etotheipi]

You say that P(v)dv is the probability of a random molecule having a speed between v and dv. As you mentioned before this probability is unitless. But what about the book's interpretation? It says that P(v)dv gives the fraction of molecules that have speeds in that range. I guess these could be the same thing. Actually now that I think about it, it probably is.

Most probably . You can treat 'fraction' and 'probability' synonymously here.

You say P(v) is called a probability density function. I did a little reading and learned that the dependent variable P always have reciprocal units of whatever it is a function of. This makes sense whenever you think of P as a fraction or a probability. But i still don't really know what P is. Let's say you plug in v = 200 m\s and you get P(v) = ##10^{-3}##s/m. What does P by itself actually represent? Does it have any meaning?

You need to be a little careful with continuous distributions. The probability of getting an exact value, e.g. ##v=200 \text{ms}^{-1}##, is exactly zero (crazy, huh?!), since the limits of the integral are equal! As such, it only makes sense to ask what is the probability of the random variable being in a certain range, which is the area under the curve between those points, i.e. the integral of the probability density function.

What you can say is that the probability that the speed is within a small interval ##[200 \text{ms}^{-1}, 200 \text{ms}^{-1} + \delta v]## will be approximately ##P(200\text{ms}^{-1})\delta v = (10^{-3} \text{sm}^{-1})\delta v##, which becomes exact in the limit ##\delta v \rightarrow dv## (I haven't checked the numbers, so I am trusting you here!).

Loosely, you can often think of the value of the probability density function in a certain neighbourhood as a vague measure of the likelihood of the result being in that neighbourhood. But the value of the probability density function at any given point is not a probability. It is quite valid for the PDF to be greater than 1 in certain places. However, we must have that the integral of the PDF across the domain is exactly 1 (we say the PDF is normalised), which just means that the total probability is unity.