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The probability for a gas molecule to have any velocity $v$ is $0$ since there are infinite possibilities for the velocity for the gas molecule to have. If that's so then how come the gas molecule has any velocity at all ,since according to probability there is $0$ probability for every velocity.

More confusion: How can we even say that the probability of any velocity is zero,since that implies $$1/\infty=0$$which isn't true.

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I think there is a more intuitive argument:

Say your average molecule speed in a gas cloud is $20\:\mathrm{\tfrac{m}s}$. Okay, so set up your apparatus and track one particle. And your apparatus shows indeed $20.000\:\mathrm{\tfrac{m}s}$. Now we have a small problem. Does it really have a speed of $20\:\mathrm{\tfrac{m}s}$? Afterall, we only measure to 5 decimal places. But no worries, just get a better apparatus, maybe the Magic-Molecule-Speed-Measurer-9000! Surely that is a great device. We have now determined the velocity to be $20.00000000000\:\mathrm{\tfrac{m}s}$. At the same time, we found lots of other particles with velocities of maybe $20.0000001\:\mathrm{\tfrac{m}s}$ or $19.99999817\:\mathrm{\tfrac{m}s}$. What? There's an even more precise apparatus? Take my money! ...

It may become clear, that there is no such thing as an "exact" speed (excluding $c$ of course). Disregarding quantum uncertainty arguments, you would need to measure to infinite decimal places, to exactly know the speed, because, in fact, $20\:\mathrm{\tfrac{m}s} = 20.00000...\:\mathrm{\tfrac{m}s}$. Thus, you can only say, a particle's velocity lies in an interval. There is a certain probability, that a particle is faster than $19.995\:\mathrm{\tfrac{m}s}$ and slower than $20.005\:\mathrm{\tfrac{m}s}$. And that can be calculated via the area under the curve of the probability density function (PDF) of the velocity distribution.

Addendum: For the following elaborations, let's assume a mean velocity of $\langle v \rangle = 20$.

We assume the following:

  • All velocities are between $19.5$ and $20.5\:\mathrm{\tfrac{m}s}$
  • All velocities are uniformly (evenly) distributed, although this doesn't matter.

If we were to graph this, it would simply look like this:

enter image description here

This is called the probability density. As I've said before, a probability is given by the area under a curve, which in this case, is just a straight line. Since this is just a rectangle, we can calculate the probability that a gas molecule has any velocity in the given interval, which is just width$\times$height so $1\times 1 = 1$, which checks out. Likewise, for a particle to be slower than $20\:\mathrm{\tfrac{m}s}$, we get 50%. The same is true for a particle to be faster than $20\:\mathrm{\tfrac{m}s}$. The probability is again 50%. Which is nice, because the probability to find a particle which is slower or faster than $20\:\mathrm{\tfrac{m}s}$ is then 50%+50%=100%. This would mean, that the probability to find a particle at exactly $20\:\mathrm{\tfrac{m}s}$ must be $0$. Otherwise, the total probability would be higher than 100%. But we can show this mathematically!

Let's now divide this rectangle into $n$ equal smaller rectangles. Each then has a width of $\frac{1}{n}$. E.g. if you split it into 10 rectangles, each has a width of $\frac{1}{10}$. If you split it into 1000 rectangles, each has a width of $\frac{1}{1000}$. But we don't want to know, if the speed is in a small rectangle. What we want is a line, of zero width. The exact speed. It turns out, this is not possible without calculus. By applying the limit, we find

$$\lim_{n\to\infty}\frac{1}{n}=0$$

That we need an infinite amount of smaller rectangles, to know the exact speed. But each rectangle then has a width of $0$. In the language of statistics, this means, that the probability for a particle to have any particular speed is $0$, while the probability for it to have a certain speed within an interval may be non-zero.

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    $\begingroup$ Your point is beautiful as to how it's impossible to 'know' the exact speed. But that doesn't mean it is impossible for a particle to have a particular speed. $\endgroup$ – Kashmiri Aug 11 '20 at 13:08
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    $\begingroup$ Don't particles have exact speeds,Irrespective of if we know or not? $\endgroup$ – Kashmiri Aug 11 '20 at 13:19
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    $\begingroup$ @MikeScott Can't the particle have a speed, even if we don't measure it? (If a tree falls...) $\endgroup$ – Dave Aug 11 '20 at 21:13
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    $\begingroup$ This question is answerable within the realms of classical mechanics. No need to complexify it with quantum mechanics. As far as I understand, this question is about limits. $\endgroup$ – infinitezero Aug 11 '20 at 21:18
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    $\begingroup$ @MikeScott I'm not sure how that clarifies anything. OP's objection is that just because we don't know an exact speed, doesn't mean there doesn't exist an exact speed. Your response is to talk about the inability to know the speed, which just brings us back to OP's original objection again. $\endgroup$ – JBentley Aug 12 '20 at 14:27
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This is actually two questions rolled into one. One is about the mathematics of probability, and how probability can be used with continuous spaces. The other is about physics and how we use mathematics to model it.

I'll focus on the former.

People tend to think of probability as something that is assigned to specific outcomes - and if you want the probability of a combination of outcomes, you just add up the individual probabilities. For example, if you roll a fair die, the probability to get 1 is $1/6$, the probability to get 2 is $1/6$, so the probability that the outcome will be in the set $\{1,2\}$ is $1/6+1/6=1/3$.

That last example happened to be a uniform distribution, with all elementary outcomes having the same probability - but you can just as well talk about probability spaces where outcomes have different probabilities. For example, you can have a loaded die, where the probability of 1 is $1/8$ and the probability of 2 is $1/4$ (and some probabilities for the other options). The the probability the result is in $\{1,2\}$ is $1/8+1/4=3/8$.

This all works when the space of possible outcomes is finite; and with a bit more effort, when it is countably infinite.

But if we want to use probability theory for continuous variables (in a space such as $\mathbb{R}$ or $[0,1]$) - and we do, since that is an extremely useful thing to do - we have to step back from such an approach. There is no way to assign a probability to each $X\in[0,1]$, such that uncountably many values have a positive probability, and the sum of probabilities is 1.

What we do is - instead of thinking of probabilities of specific outcomes, we think of probabilities of sets of outcomes. A set is no longer simply a collection of self-sufficient outcomes - a set is the fundamental object we use to define our probability space.

So when we have a variable in $[0,1]$, we can't talk about the probability that it will be $1/3$ or $\pi/4$ (well, we can, but the probability will be 0, which is not very interesting). What we can say is that its probability to be in $[1/2,2/3]$ is $7/36$ and the probability to be in $[1/10,1/7]$ is $51/4900$. If we specify the probability for every set we care about, we have defined our probability distribution.

The mathematical branch of assigning a size to every set, which satisfies a few intuititive properties, is called "measure theory". This is a generalization of the concepts of length, area, volume, integrals, and so on. Probability theory is basically measure theory when we require that the measure of the entire space is 1.

Note that it is actually impossible to assign a measure to every set. There are too many subsets of our space, and they are too weird.

But we don't have to. For the purposes of random variables on a subset of $\mathbb{R}$, it is enough to define a non-decreasing function $F(x)$, which specifies the probability that $X\le x$. From this we can calculate the probability of $X$ to be in any reasonable set we choose. This function is called "cumulative distribution function"

If $F(x)$ happens to be differentiable, we can talk, about its derivative $f(x)=F'(x)$, which we call "probability density function". We can also define a distribution by its PDF, but this is less general, since not all CDFs are differentiable.

By the way, the probabilities I gave above were for the distribution $f(x)=2x$ and $F(x)=x^2$, for $0\le x\le 1$.

So we can't meaningfully talk about the probabilities for specfic outcomes of a continuous random variables, we can talk about their probability densities, and this tells us which regions are more likely. While we will never find a molecule with a speed of exactly $1 m/s$ or $2 m/s$, we are more likely to encounter a speed of around $1m/s$ than around $2m/s$, if that's what the densities tell us.

I didn't go into the question of whether gas molecules exist, whether they are tiny billiard balls or quantum wave functions, whether they have velocities, whether the velocity is continuous or discrete, etc. I'm just assuming they have an unknown velocity which is modeled as a continuous random variable. That's a useful model for many applications.

Regarding the bit at the end:

First, having a probability of 0 does not really imply that $1/\infty=0$. Probabilities of things can be 0 without implying anything. It is true, though, that you can't have uncountably infinitely mutually exclusive events each with a positive probability.

Second, it is definitely not not true that $1/\infty=0$. $\infty$ is not the mysterious beast high-school teachers want you to believe it is. There are perfectly legitimate toplogic/algebraic structures, such as the Riemann Sphere, where $\infty$ is a full-fledged member, and $1/\infty=0$.

Finally: Note that the word "model" appeared several times in the answer. This is no accident. The truth is that we have no idea how physics actually works. What we do have is mathematical models of it. All models are wrong, but some models are useful. Gas molecules don't actually have probability densities to have certain velocities, we just model them as such.

This is something that should be kept in mind when asking questions like this. As I mentioned at the start - understanding the model, and understanding how we can use the model for a physical situation, are two different things.

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    $\begingroup$ People assume that probability is a strict progression from cause to effect, but actually from a non-linear, non-subjective viewpoint, it’s more like a big ball of wibbly-wobbly, sety-measury... stuff. $\endgroup$ – Meni Rosenfeld Aug 12 '20 at 10:15
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In this case I think the answer is given by the velocity density. In fact you cannot have a precise value of a velocity because $|\boldsymbol{v}|\doteq v\in\mathbb{R}$ and so is a continuous space; clearly you have to set probability $0$ for a point in that space. But the probability to have a velocity in an interval is non-zero once you consider $v\text{d}v$.

This concept may become useful in a statistical, pre-quantum mechanical description, because you cannot observe a velocity value of a particle with arbitrary precision; you just have a probability of catching it inside an interval.

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    $\begingroup$ What do you mean by "the value of a velocity" being a "dense space"? $\endgroup$ – Jannik Pitt Aug 11 '20 at 20:09
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    $\begingroup$ That's not what "dense" means in mathematics. Also there is no absolute requirement that a distribution of a real-valued quantity will be continuous. $\endgroup$ – Ian Aug 11 '20 at 20:10
  • $\begingroup$ According to you is 0 probability impossibility? $\endgroup$ – Kashmiri Aug 12 '20 at 6:03
  • $\begingroup$ The velocity modulus is a real number and so belongs to a dense set, because every point is a limit point. "Also there is no absolute requirement that a distribution of a real-valued quantity will be continuous" sure, but there's not absolute requirement for a distribution of a real-valued quantity not to be continuos. So just assume it's real and has no particular limitations, that's all; it's mathematics, the physics is in the concept of density. @YasirSadiq 0 probability it's not impossibility, but this is explained very well, more than I possibly can, in the links under your question $\endgroup$ – Rob Tan Aug 12 '20 at 7:40
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    $\begingroup$ @Rob Really, that's not what dense means. The word you want here is just "continuous". Density is always defined relative to an ambient topological space, and in particular a topological space is always dense in itself, so it makes no sense to just say a space is "dense" unless context specifies what it is dense in. $\endgroup$ – Ian Aug 13 '20 at 20:53
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There's a few ways to answer this question.

One way is to revert to quantum mechanics. The momentum states of particles in a box with finite size are quantized, so there are a finite and countable number of states below the energy of the system that the particles can take.

But this doesn't really get at the heart of the question, which is how can we observe events that have probability measure 0. I am not an expert, but I believe mathematicians answer this by saying essentially

you are misinterpreting what 0 means in probability theory.

e.g. probability 0 does not mean impossible, 0 is a sort of limit. But, personally, this doesn't sit very well with me.

Another more physical way to get around this is to remind yourself that "you never have infinite precision," (this answer was given elsewhere in this post) so you must consider intervals of possibilities. Mathematically, this is totally sufficient, however it does leave something to be desired philosophically.

My preferred way of resolving this is to remember that these models are just that, they are models. Just because a mathematical model yields numerical answers that are aligned with experiments, it does not mean that the quanitities you are describing mathematically are ontologically real. (e.g. see the debate in QM about the nature of the wavefunction)

To make this POV more concrete consider the example: In statistical thermodynamics, we often take the "thermodynamic limit" of a system by taking the number of particles to infinity. Obviously there are not infinite particles in any system, but there are so many that the limit still yields a good enough description of the situation for the bulk properties of the material.

So you can see that this issue basically arises from using the continuum to describe the physical world. What to make of this fact is still debated amongst physicists and mathematicians. However, this does not mean our current models do not constitute perfectly consistent mathematical systems that can be used to make predictions.

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Your question essentially applies to almost all physical equation or formulas. The problem is that physical equations and their solutions provide mathematical models for the real world and are always an approximation. As an example: a circle, meaning a pure mathematical and infinitely precise circle, does not exist in the real world. But for real world applications it provides a very handy model to perform calculations, e.g. to calculate the area if you are only given the radius. For that to work, however, you must assume that the mathematical model is a good model for your real world object. A wheel that doesn’t wobble for example.

In your question specifically, the trillions and trillions of molecules in the gas behave macroscopically as one continuous material, which justifies the use of mathematical smooth statistical model. Applying it to real world situations you can only measure how many molecules were found in a finite interval of a range of velocities, as explained in other answers.

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Others have given answers as to why the velocity of a single particle is unattainable, but I'll try to explain the errors in your question.

You state that there are an infinite number of speeds thus a 0 probability of any one speed, yet this assumes we don't live in a discreet universe where there is a minimum distance like the Plank length. I'm not sure if the universe is discreet or not, maybe this is another unknowable. But without first proving the universe is continuous, we do not have the foundation to answer your question.

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    $\begingroup$ A good point but wrt the 'question' at hand quite moot: Even with an infinite number of possible speeds, the speeds would not disappear. $\endgroup$ – TaW Aug 11 '20 at 21:03
  • $\begingroup$ This is true, we define speed of particles using distance traveled in a unit of time so "freezing" time and getting a speed is meaningless. $\endgroup$ – GnomeChompskee Aug 11 '20 at 21:09
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Think about it this way. Let the probability distribution of velocity be $P(v)$. Then the number of particles moving at exactly $100 m/s$ is $\int^{100}_{100} P(v) dv = 0$, since the integration limits are the same.

However if your measurement device is sensitive to the 3rd decimal place only - and all experimental devices have some measurement error - then you aren't measuring $100 m/s$, but rather $99.995 m/s$ to $100.005 m/s$ - and now the equation above becomes $\int^{100.005}_{99.995} P(v) dv \neq 0$!

Therefore you can find particles moving at some velocity even though the probability of moving at exactly that velocity is zero.

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The probability for a gas molecule to have any velocity v is 0

No, it's not. Probability distribution follows Maxwell–Boltzmann law :

enter image description here

This speed distribution probability law is of Gaussian type, so there is a central speed, where we most likely find random molecules with it. And checking other speeds farther from central - gives monotonically decreasing probability to find molecule with such speed.

there are infinite possibilities for the velocity for the gas molecule to have

Not true again. Molecule speed is bound to system temperature, so it can't have infinite possibilities for velocity. Lets say i cooled down gas to absolute zero, so in that case situation will be almost opposite to your claim - all molecules will stop moving in $0K$ temperature, so we can say that there's no any possibility of molecule having speed other than $0 ~\text{m/s}$.

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  • $\begingroup$ Imho, what you've drawn is the 'probability density function' .From this we get the probability of a 'range of velocities but not a particular velocity' . $\endgroup$ – Kashmiri Aug 13 '20 at 12:08

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