Integral problems with polar coordinates and variable substitution

[Edison Bias]

Not that informative, plaese explain the difference.

 

[haruspex]

Not that informative, plaese explain the difference.

If I walk 3km in one hour my average speed is 3km/h, but my average velocity depends on how far my finish point is from my start point (displacement). If the 3km was around a circle my average velocity is zero.

 

[Edison Bias]

Interesting! But I don't know if I grasp this one. Speed is then always measured "ahead" but velocity depends on how fast I have been moving from one point to another and if that point happens to coincide with my start point, velocity is zero, right? Velocity is then a measure of how fast you move from A to B with the destiction that you really need to have A separated from B otherwise velocity is zero, right? Strange how there can be a distiction of this in the english language, here in sweden we just call it "hastighet" (speed).

 

[haruspex]

Interesting! But I don't know if I grasp this one. Speed is then always measured "ahead" but velocity depends on how fast I have been moving from one point to another and if that point happens to coincide with my start point, velocity is zero, right? Velocity is then a measure of how fast you move from A to B with the destiction that you really need to have A separated from B otherwise velocity is zero, right? Strange how there can be a distiction of this in the english language, here in sweden we just call it "hastighet" (speed).

In physics and maths, as expressed in English, position, velocity and displacement are vectors. Displacement is the change in position, and velocity is the rate of change of position. Average velocity is the displacement divided by the elapsed time.
Displacement is $\Delta \vec r=\int \vec{dr}$, instantaneous velocity is $\vec v=\frac{d\vec r}{dt}$, average velocity is $\vec v_{avg}=\frac{\Delta \vec r}{\Delta t}$, where $\vec r$ is the position vector.
Speed is the magnitude of the velocity. Distance traveled is along the path taken, $\int |\vec {dr}|$, instantaneous speed is $\frac{|\vec {dr}|}{dt}$, and average speed is $\frac{\int |\vec {dr}|}{\Delta t}$

 

[Edison Bias]

Interesting, thanks for trying to teach me!

Integrals aren't that easy to understand for me but as we just have solved a difficult integral numerically I now understand better. Your integrals contain only the differential part, thus the "function" equal 1. This is equivalent to viewing your integral expressions as pure summations (with pretty small steps, dr). And by viewing the integrals as pure summations, they make sense. So to try to understand what you have written I will comment expressions from left to right (don't know how to write vectors i Tex).

First we have that the rough step movement (displacement) equals the summation of all tiny movements added with direction in mind (vectorial addition), instantaneous velocity seems to be how fast a single tiny movement is being done (and velocity has the movement direction), average velocity is then the whole step movement divided by the total time. Speed is not a vector, only a magnitude (scalar), distance traveled is a summation of tiny steps where direction is of no importance, instantaneous speed once again is a scalar that only describes the rate of tiny movements, average speed is a summation of tiny scalar movements over time.

Now, I have tried to interpret what you have said, am I even close to the truth?

Edison.

 

[haruspex]

Interesting, thanks for trying to teach me!

Integrals aren't that easy to understand for me but as we just have solved a difficult integral numerically I now understand better. Your integrals contain only the differential part, thus the "function" equal 1. This is equivalent to viewing your integral expressions as pure summations (with pretty small steps, dr). And by viewing the integrals as pure summations, they make sense. So to try to understand what you have written I will comment expressions from left to right (don't know how to write vectors i Tex).

First we have that the rough step movement (displacement) equals the summation of all tiny movements added with direction in mind (vectorial addition), instantaneous velocity seems to be how fast a single tiny movement is being done (and velocity has the movement direction), average velocity is then the whole step movement divided by the total time. Speed is not a vector, only a magnitude (scalar), distance traveled is a summation of tiny steps where direction is of no importance, instantaneous speed once again is a scalar that only describes the rate of tiny movements, average speed is a summation of tiny scalar movements over time.

Now, I have tried to interpret what you have said, am I even close to the truth?

Edison.

You have got it. Well put in plain language.

 

[Edison Bias]

Just to finish this, the density (n) may equal

n=2A\sqrt{\frac{2kT}{m}}\int_{0}^{\infty}e^{-u^2}du-2A\sqrt{\frac{2kT}{m}}\int_{0}^{\infty}e^{-u^2}\frac{v_0}{v_0+u\sqrt{\frac{kT}{m}}}du

which with

mv_0^2=kT

gives

n=2A\sqrt{\frac{2kT}{m}}\int_{0}^{\infty}e^{-u^2}du-2A\sqrt{\frac{2kT}{m}}\int_{0}^{\infty}e^{-u^2}\frac{1}{1+u}du

where the first integral equals

\frac{\sqrt{\pi}}{2}

according to our previous reasoning.

And the last integral equals 0,61 according to Charles Link's tedious numerical evaluation.

So we have that

n=A\sqrt{\frac{2kT}{m}}(\sqrt{\pi}-2*0,61)=A\sqrt{\frac{2kT}{m}}*0,55

which is almost half of the normally calculated particle density.

Edison

 

[haruspex]

Ok, I've just read your post #27, which I had not done before because although I understand your integral problem as a purely algebraic question I am not a thermodynamicist.
But I do understand why the average velocity should be zero. The mean velocity, as a vector, represents the bulk motion of the gas or body. This is mechanical energy. Thermal energy, by definition, comes from only velocities relative to this.
Of course, a body traveling through a gas experiences higher speed impacts, on average, and thus a higher effective temperature.

 

[Edison Bias]

Ok, I've just read your post #27, which I had not done before because although I understand your integral problem as a purely algebraic question I am not a thermodynamicist.
But I do understand why the average velocity should be zero. The mean velocity, as a vector, represents the bulk motion of the gas or body. This is mechanical energy. Thermal energy, by definition, comes from only velocities relative to this.
Of course, a body traveling through a gas experiences higher speed impacts, on average, and thus a higher effective temperature.

This is interesting, "the mean velocity, as a vector, represents the bulk motion of the gas or body". Well yes of course, the body stands still. Even a balloon stands still (when it is filled with air). But the air inside the balloon has a temperature and I have learned that that means that the body's molecules move. They will not move in such a way as to push the balloon in one direction, but they move. Now, there has to be a kind of huge difference between the bulk stand-still of the balloon and the thermal speed of the air molecules inside. I mean that the mean velocity is not zero but centers around a velocity that is equally probabable for +v_0 as it is for -v_0 because there has to be a thermal velocity and the sign doesn't matter because it is simply a speed (scalar) that gives the temperature.

Edison

 

[Charles Link]

This is interesting, "the mean velocity, as a vector, represents the bulk motion of the gas or body". Well yes of course, the body stands still. Even a balloon stands still (when it is filled with air). But the air inside the balloon has a temperature and I have learned that that means that the body's molecules move. They will not move in such a way as to "automatically" push the balloon in one direction, but they move. Now, there has to be a kind of huge difference between the bulk stand-still of the balloon and the thermal speed of the air molecules inside. I mean that the mean velocity is not zero but centers around a velocity that is equally probabable for +v_0 as it is for -v_0 because there has to be a thermal velocity and the sign doesn't matter because it is simply a speed (scalar) that gives the temperature.

Edison

If you look carefully through my post #28, it shows the results that come from using the speed "v" scalar parameter. Some of the mathematics getting there is semi-complex-it uses a coordinate system in velocity space($v_x, v_y, v_z$) and superimposes a spherical coordinate system representing speed "v".

 

[Edison Bias]

I actually kind of understand your mathematics, but I do not get your results. Would be nice if you could derive them for me. Because I'm kind of getting comfortable with advanced math, it is fun!

Edison

 

[Charles Link]

I actually kind of understand your mathematics, but I do not get your results. Would be nice if you could derive them for me. Because I'm kind of getting comfortable with advanced math, it is fun!

Edison

Simplest way might be to google HyperPhysics-Development of Maxwell Distribution. That article contains the results I mentioned, but there are most likely numerous other papers with the same result=google "Maxwellian Distribution for gas particles"", etc.