Maxwellian velocity distribution and atoms

[grav]

The Maxwell-Boltzmann formula for the distribution of velocities is f(v)=4pi(m/2pikT)3/2v2/(emv2/2kT). This means (as far as I can tell) that if we have a total of N molecules per volume, that n molecules per volume will have velocities within a range of v to v' for f(v). In other words, f(v)*(v'-v)=n/N. If we square this, we get (n/N)2=(4pi)2(m/2pikT)3v4(v'-v)2/(emv2/kT). Now let's divide and multiply this by (v'+v)2. We now have (4pi)2(m/2pikT)3v4(v'2-v2)2/(v'+v)2(emv2/kt). For a very small velocity range for v to v', they are approximately equal, so (v'+v)=2v. Also, let's define mv2 as E and mv'2 as E'. We now have (n/N)2=E(E'-E)2/2pi(kT)3(eE/kt). If we then further define E/kT as y, then the entire formula reduces to (n/N)2=y(y'-y)2/(2pi)(ey). Now let's say that y'/y=(1+z). The formula is now (n/N)2=y3z2/2pi(ey). So for n/N, the formula is n/N=[y3z2/2pi(ey)]1/2. Now, if this is correct, then the its summation will equal one for infinitesimal values of z, where y is zero to infinity in multiples of (1+z), since y'=(1+z)y. It checks out. The summation is indeed equal to one. Just to make sure this wouldn't occur anyway, I tried other slight variations, such as for E2(E'-E) instead of E(E'-E)2, for instance. No others worked. However, one can change the definition of E, to E=mv2/2, for example, but the formula must change accordingly. In this case, it would become (n/N)2=(2y)[(2y')-(2y)]2/2pi(ey)=8y(y'-y)2/2pi(ey)2.

Okay. Now let's say we want to find the average energy over the entire distribution. That would be (E1n1+E2n2+E3n3+...)/N. This is simply equal to the energy times the distribution of energies for any particular range of energies, or Eaverage=[summation of] E*(n/N)=[summation of] (kT)*y*(n/N)=(kT)*[summation of] [y5z2/2pi(ey)]1/2=3kT. The average velocity of the molecules can also be found this way with vaverage=[summation of] v*(n/N)=[summation of] (E/m)1/2*(n/N)=[summation of] (kT)1/2(y/m)1/2*(n/N)=[(8/pi)kT/m)1/2, so mvaverage2=(8/pi)kT.

All of this is well and good, except for one thing. If the average energy is 3kT, then we can find the pressure with P=Eaverage/V=3kT*(N/V). But the gas law formula is the same thing except three times smaller. If both of these formulas are correct, then how can one be three times smaller than the other? Well, I believe the answer lies in the degrees of freedom of the molecules. When a material heats up, it expands. Linearly, it expands with (1+LT), where L is the expansion coefficient of the material. The expansion is therefore (1+LT)-1=LT linearly. If L is small, then the expansion of an area is (1+LT)2-1=2LT. For volume, it is (1+LT)3-1=3LT, which is where the difference in the factors come from. The energy formula (distribution of velocities) is linear and the gas law is for volume. However, there is another problem. The factor of three should be in the denominator, not the numerator. So the actual linear energy is expressed as 3kT, for area it becomes (3/2)kT, and for volume it is simply kT, as the formulas state. This can also be seen by multiplying the specific heat capacity of materials by the mass of the individual molecules. When they are in thermal equilibrium with the surrounding medium, they will have the same temperature, and this will be their heat capacity for that temperature. So csmT=energy. I have done this for eight elements- aluminum, copper, gold, iron, lead, mercury, silver, and zinc, and have found them to be almost exactly 3kT every time (csm=3k).

But this means that the temperature is really the additional energy applied. That is, the molecules already contain energy of their own, and the temperature merely adds a little more. This also means that molecules themselves would not freeze at absolute zero. They will still continue as normal with the same internal energies that support them as molecules in the first place, but they will just have no additional external energy, or kinetic energy. So where does this additional energy come from? The nuclei of the molecules themselves cannot collide, and so do not transfer energy directly, so it must lie with the outer electrons. This may also mean that the atoms themselves expand with heat added, not just materials. This certainly would explain how the materials themselves expand, if it is really all of the atoms, but I believe the present concept is that it is because the atoms collide more with each other due to the extra thermal energy and cause the material to expand in this way. It may be both. I am not sure. The extra energy given to the electrons could cause their shell to widen and the energy could also be transferred to the nucleus as it "chases" the electrons around. The nucleus is really free to move about and it and the electrons are only kept attached through their mutual attraction.

To see that the coefficient of linear expansion of materials may be related to the electrons, we can find an energy for this with simply E=3k/L. Dividing this by the mass of an electron and square rooting, we get a velocity which is comparable to that of the electron in a hydrogen atom, vH=ca (a=1/137.036). For the various materials I mentioned above, the velocity of the electrons comes out to about .5 to .9 of vH.

In addition to this, during electrolysis, ions are produced and mass is deposited at the plates. The amount of mass that is deposited is in accordance to m=QM/Fv, where Q is the total number of coulombs transferred, M is the atomic mass of the plating material, and v is the valence of the material. F is Faraday's constant, and according to my source, has been determined through careful experimentation to be F=9.65*107 C/kmol. Well, I have found that, much aside from careful experimentation, it can also be found with q*kA, where A is Avogadro's number. This means that for each ion which is deposited, one electron has been transferred. For a valence of two, two electrons will be tranferred per ion, and the mass is half as great. Seems pretty simple in retrospect, I know. But I just wanted to throw this in in case it wasn't already known, since my source didn't state it, but only said it was found through experimentation. Of course, after such a relationship is found, many such things probably become self-evident.

Anyway, these are just some things I've been working on. I wanted to see if I could get some feedback from anyone that is interested.

[tusenfem]

I think you should read up a little on the variation principle and work with v and dv and then expand the equations. I have no time at the moment to do the caculation, but it looks kind of shaky what you do up there in the small dv approximation.

[grav]

I think you should read up a little on the variation principle and work with v and dv and then expand the equations. I have no time at the moment to do the caculation, but it looks kind of shaky what you do up there in the small dv approximation.
Well, you're right of course. I sort of have my own way of performing calculus that I have developed through the years. It is simply the way I have learned to deal with calculations as seemed necessary to me to obtain results for whatever I was working on at any particular time. Although I have never taken the time to fully study calculus, I have nevertheless found it necessary and so have slowly developed my own way of doing it and now seem to be set in my ways. It's just the easiest way for me to think about it. It is also the easiest way to readily plug the numbers into a computer program to run similations or to find the limits of summations. The formulas are "computer-friendly" when expressed in this way. Plus I like to see it spelled out so I can quickly look back and see what I am dealing with. I assure you, however, that the math works out. I guess wherever I wrote something like (v'-v) would probably be dv. I'm not sure. If this is the case, then (v'-v)=v*z=dv, so where I wrote something like y5z2 would really be y3dy2 or y3d2y. Probably the latter, since it appears to show the square of the infinitesimal, or (y'-y)2, while dy2 would probably be something like y'2-y2. And since y=E/kT, it is really E3d2E/(kT)5. Please let me know if this is correct. (v'-v) or (E'-E) would in each case be a very small (infinitesimal) range of velocities or energies, as the case may be. The summation would, of course, be the sum of each using these infinitesimal ranges. The smaller I make the value of z, the smaller the range, and the more precise the result becomes for the limit. Also, the smaller the value of z, however, the longer it takes the computer to return a result, unfortunately, because it has to run through more calculations. For most computer language programs, this would also increase the margin of error, where the last digit might be off so the entire result runs further and further off with each calculation, but I have found the miracle of U-Basic, and so no longer have that dilemna. U-Basic is capable of storing numbers up to 2600 digits long, or a number long enough to fill about four computer screens. Hopefully, that is more than I should ever need. If you or anyone else out there knows how I can express these equations so that they can be better understood by the general public, please show me using one of my formulas above as an example if you would, please, and if it is not too complicated, I will attempt to use that formulation from now on to express them.

[grav]

See, now, this is why I like my method better. I am assuming first, however, that (v'-v)=dv, (v'2-v2)=dv2, and (v'-v)2=d2v. With these as my definitions, if one were to begin with (v'-v)=dv, and then multiply and divide by (v'+v), one obtains (v'2-v2)/(v'+v). Since the range between v' and v is infinitesimal, then v' is approximately equal to v, so we get dv=(v'2-v2)/2v=dv2/2v=mdv2/2mv=dE/2p (unless we define E as E=mv2/2, in which case we get simply dv=dE/p), where E is energy and p is momentum. So the relative velocities is equal to the difference in energy divided by twice the momentum, where the momentum is about the same between the two. Actually, considering I figured this out as I jotted it down, I guess I just learned something, so maybe this way of expression is indeed beneficial to some degree, but can still be found with my original method. Anyway, the point is, I still had to convert back to normal algebra a couple of times in order to obtain the formula dv=dE/2p. It may seem to relate simply now but I had to go through the step of dv=d2v/2v to get there, and it is not so easy to see how this relates, except through normal algebra. It's almost as if it would require a whole different form of mathematics to achieve it directly. I suppose mathematicians would have worked this out by now, so it is really that one must learn another entire form of mathematics. By using z, however, its value never changes. One can throw together as many variables as they wish, and z stays the same. It is always zero. Of course, one cannot perform a summation using z=0 precisely, since zero added to itself over and over still yields zero, or so it would seem. The actual infinite summation yields (1/0)*0=0/0=any real number, but that isn't readily seen using our present form of mathematics. We can, however, use some number that is very close to zero (infinitesimal) and we will achieve a result. The smaller the value of z, the more precise the result becomes.

[Nereid]

I think you should read up a little on the variation principle and work with v and dv and then expand the equations. I have no time at the moment to do the caculation, but it looks kind of shaky what you do up there in the small dv approximation.Well, you're right of course. I sort of have my own way of performing calculus that I have developed through the years. It is simply the way I have learned to deal with calculations as seemed necessary to me to obtain results for whatever I was working on at any particular time. Although I have never taken the time to fully study calculus, I have nevertheless found it necessary and so have slowly developed my own way of doing it and now seem to be set in my ways.
[snip]If this is the case, then you may have rather more work to do, to get your ideas established, than you might think.

For example, 'grav mathematics' may contain deep inconsistencies, or kinds that several centuries of mathematicians have uncovered, and resolved, in normal math (and it goes without saying that any of such inconsistencies could render useless any result derived using them). Or you might be lucky - your results may escape the inconsistencies because, somehow, they are rendered toothless by some curious quirk.

The point is, without a detailed investigation, you have no way of knowing.

Of course, it also means that the only* person who can check the results you post here in the ATM section of BAUT is yourself ... which kinda makes me wonder what the point of such posting is.

BTW, will you be re-working material you posted in other ATM threads? It seems, to me, somewhat odd for you to continue to build on at least some of that other work, knowing there are open questions and identified flaws in it.

*Unless someone else takes the trouble to get sufficiently familiar with 'grav mathematics' to perform an independent check.

[tusenfem]

I thought there was something fishy. I wanted to see what I could do with your idea, but by writing down the equation to start with there is a problem. The Maxwell(-Boltzmann) equation is:

f(v) = N (m/2 pi k T)3/2 exp-m v2/2 k T

(although the MB equation takes into account the total energy of the system, but that does not matter here I think). What you wrote down looks more like the second moment of the distribution function with a v2 extra.

[grav]

If this is the case, then you may have rather more work to do, to get your ideas established, than you might think.

For example, 'grav mathematics' may contain deep inconsistencies, or kinds that several centuries of mathematicians have uncovered, and resolved, in normal math (and it goes without saying that any of such inconsistencies could render useless any result derived using them). Or you might be lucky - your results may escape the inconsistencies because, somehow, they are rendered toothless by some curious quirk.

The point is, without a detailed investigation, you have no way of knowing.

Of course, it also means that the only* person who can check the results you post here in the ATM section of BAUT is yourself ... which kinda makes me wonder what the point of such posting is.

BTW, will you be re-working material you posted in other ATM threads? It seems, to me, somewhat odd for you to continue to build on at least some of that other work, knowing there are open questions and identified flaws in it.

*Unless someone else takes the trouble to get sufficiently familiar with 'grav mathematics' to perform an independent check.
Well, the mathematics I use is not really my own. It is just a simpler way of performing calculus (for me). That is to say, I did not invent a new mathematical method so much as I simply use the math that calculus is based on in its purest form, which is really just high school algebra. I would think that this would be easier to understand than calculus, since everything is basically "spelled out". I suppose that someone that is used to dealing with calculus with its present terms and form that it is presented in might have a tough time thinking about it in this way, however. I do not know. But the mathematics involved in my formulas is still just everyday mathematics. I even try to present it in a way that is comprehensible to the layman and some people appreciate that. However, for those that have moved on to higher levels of mathematics, I suppose I can try to translate it into those forms. I noticed you had a problem with the formulas for gravitational lensing, for instance. I could probably go back through those one more time if necessary and translate them into some form of calculus instead of just ordinary algebra if you like. I just think algebra would be easier to follow. As far as the infinite summations go, however, I have no idea how to translate these results except to just present the value of the summation being used and letting someone else that is interested perform it on their own computer to check the results. There are no other options open to me that I know of, except maybe some extensive proof that I have no idea how to perform. You are right, however. When it comes to physics, mathematics is the key. And when it comes to mathematics, communication is the key, after all.

[Nereid]

Well, the mathematics I use is not really my own. It is just a simpler way of performing calculus (for me). That is to say, I did not invent a new mathematical method so much as I simply use the math that calculus is based on in its purest form, which is really just high school algebra. I would think that this would be easier to understand than calculus, since everything is basically "spelled out". I suppose that someone that is used to dealing with calculus with its present terms and form that it is presented in might have a tough time thinking about it in this way, however. I do not know. But the mathematics involved in my formulas is still just everyday mathematics. I even try to present it in a way that is comprehensible to the layman and some people appreciate that. However, for those that have moved on to higher levels of mathematics, I suppose I can try to translate it into those forms. I noticed you had a problem with the formulas for gravitational lensing, for instance. I could probably go back through those one more time if necessary and translate them into some form of calculus instead of just ordinary algebra if you like. I just think algebra would be easier to follow. As far as the infinite summations go, however, I have no idea how to translate these results except to just present the value of the summation being used and letting someone else that is interested perform it on their own computer to check the results. There are no other options open to me that I know of, except maybe some extensive proof that I have no idea how to perform. You are right, however. When it comes to physics, mathematics is the key. And when it comes to mathematics, communication is the key, after all.So, from your own perspective, which is it?

"just high school algebra"?

or "I sort of have my own way of performing calculus that I have developed through the years"?

If it's the former, then anyone could approach what you've sketched in your posts from the POV of "high school algebra"; if it's the latter, then any reader of your posts must keep in mind that, whether stated explicitly or not, your derivations, conclusions, etc could well contain some aspect of 'grav mathematics' that differs from what all (other) mathematicians and physicists in the world use*, in some undefined, unspecified, non-standard, idiosyncratic way.

And if you, grav, can't take the time to be clear about this (and not flip-flop between posts that are only a few days apart), haven't you conveyed, in no uncertain terms, that nothing which you post should be taken at face value?

*or not; unless it's spelled out, clearly and explicitly, there's no way to tell!

[grav]

I thought there was something fishy. I wanted to see what I could do with your idea, but by writing down the equation to start with there is a problem. The Maxwell(-Boltzmann) equation is:

f(v) = N (m/2 pi k T)3/2 exp-m v2/2 k T

(although the MB equation takes into account the total energy of the system, but that does not matter here I think). What you wrote down looks more like the second moment of the distribution function with a v2 extra.
Well, I'm glad to see you edit that before I had a chance to get my hands on it. It looked like I really had my work cut out for me for a minute, there. My poor equations were suffering. I think they still feel the pain :sad: . But the formula you still present here doesn't look the same as the one I have found through many sources. For one thing, it should be multiplied by 4piv2. You mention that the v2 is extra, but I don't see how. You must be looking at the (m/2pikT)3/2 part of it as a pure number, but if you look again, you'll see it has units of 1/v3. After multiplying by v2, the entire formula has units of 1/v. And when multiplied by dv=(v'-v), we get a pure number value of f(v)dv=n/N, which is that fraction of the total number of molecules at any given time that have a velocity within a very small range of v to v'. If there is anything else you have problems with, please let me know. I am more than willing to help out with it and maybe I can learn some things from you as well, especially since I am just starting to get into all of this.

[grav]

So, from your own perspective, which is it?

"just high school algebra"?

or "I sort of have my own way of performing calculus that I have developed through the years"?

If it's the former, then anyone could approach what you've sketched in your posts from the POV of "high school algebra"; if it's the latter, then any reader of your posts must keep in mind that, whether stated explicitly or not, your derivations, conclusions, etc could well contain some aspect of 'grav mathematics' that differs from what all (other) mathematicians and physicists in the world use*, in some undefined, unspecified, non-standard, idiosyncratic way.

And if you, grav, can't take the time to be clear about this (and not flip-flop between posts that are only a few days apart), haven't you conveyed, in no uncertain terms, that nothing which you post should be taken at face value?

*or not; unless it's spelled out, clearly and explicitly, there's no way to tell!
Sorry about that. It looks like I still have some work to do communicating in English as well :silenced: . What I really mean to say is that I have never really gotten into the intricities of calculus all that much, so I have had to learn how to obtain the same results using the mathematics of just ordinary algebra. After having done so for so long, it is difficult for me to think about it in any other way, so this is the style I tend to use in all of my work.

[Bignose]

Well, I'm glad to see you edit that before I had a chance to get my hands on it. It looked like I really had my work cut out for me for a minute, there. My poor equations were suffering. I think they still feel the pain :sad: . But the formula you still present here doesn't look the same as the one I have found through many sources. For one thing, it should be multiplied by 4piv2. You mention that the v2 is extra, but I don't see how. You must be looking at the (m/2pikT)3/2 part of it as a pure number, but if you look again, you'll see it has units of 1/v3. After multiplying by v2, the entire formula has units of 1/v. And when multiplied by dv=(v'-v), we get a pure number value of f(v)dv=n/N, which is that fraction of the total number of molecules at any given time that have a velocity within a very small range of v to v'. If there is anything else you have problems with, please let me know. I am more than willing to help out with it and maybe I can learn some things from you as well, especially since I am just starting to get into all of this.

grav, before you get too far, you have to get yourself a copy of Chapman and Cowling's The Mathematical Theory of Non-Uniform Gases. This is a classic introduction to all things kinetic theory of gases.

In it you will find the answer to why you cannot just multiply by v2, as it really does take care of itself. The dv you keep writing is a really vector quantity, and it more properly written as dv which can be expanded as dv=dvxdvydvz, each dvi has units of velocity, so there are the 3 units of velocity you are looking for.

In the one-D Maxwell-Boltzman distribution, the exponent is 1/2, not 3/2, and consequently, you only integrate over one dimension, just dvx for example.

This is especially easily seen if you do the integration in spherical coordinates where instead of dv=dvxdvydvz, you write in therms of r,theta, and phi (I'm going to abbrevite this by r,t,p coordinates) In this case dv=r2 sin(p) dvpdvtdvr. And in this case, you see the squared term again coming out natually.

You have to be careful when dealing with these vector quantities, and really, you have to have more justification for putting an extra multiplier on the distribution that have been studied for over 100 years than 'It looks right to me.'

[grav]

Okay. I think I see the problem here. You are both considering the one dimensional formula of f(vx)=(m/2pikT)1/2/(emvx2/2kT). The formula I was working with was the three dimensional formula after all and is f(v)=4piv2(m/2pikT)3/2/(emv2/2kT). So I am incorrect with my thinking that it is greater by a factor of three because it is one dimensional. It is not. And so I still am unsure where that factor comes in. Maybe the pressure that pertains to the ideal gas law is one-dimensional, as measured in only one direction? The summation for n/N, however, works out beautifully. There is no other way to perform it. After going through the calculations, I come up with n/N=[yd2y/2pi(ey)]1/2. The summation for n for all possible velocities should, of course, add up to N. So the summation of n/N is 1. This means that the summation of [yd2y/2pi(ey)]1/2=1 for y equals zero to infinity (y is a pure number), and it does. The odds of something as chaotic looking as this equation coming out to be exactly one through pure chance must be tremendous. That is how I know it is correct.

[tusenfem]

Where do you get that equation with the v2?
You can take the Maxwell equation also for 3D if you want, but there will not be a squared velocity in the equation, because that one does not integrate to 1, if you take 3D you get the sum of the squares of the velocity components in the exponential.

Then about your interpretation of my "dv" which is the infinitesimal variation of v, and thus your (v'-v) would be dv when v' is only a very small amount bigger than v. This means that (v'-v)2 = (dv)2 (note that dv is one thing) and your idea that (v'2 -v2) = d2v is incorrect.

Now there may be some salvation here, as I have the idea that you want to do something with the (kinetic) energy of the particles. And there you might want to look at "second moment of the particle distribution" which is the integration of f(v) multiplied with the kinetic energy 0.5mv2. I would have to read more closely what you would like to do, before I start typing (wrong) equations again.

Another note on math, you write:

The actual infinite summation yields (1/0)*0=0/0=any real number, but that isn't readily seen using our present form of mathematics.

you want to sum something that is a function of z which is actually a function of y. Now, you cannot easily say what you did, the summation will not lead to any number, but has to be closely looked at and will give a real answer, but then first we would have to get the equations right. But taking infinite sums or infinite limits, although on the first eye they will give 0/0 do tend to go to a real answer. As an example:

take limit x goes to zero of (sin x) / x
this looks to go to 0/0, however for small x we find sin x = x + O(x3 and thus you see if you let x go to zero you find that (sin x) / x ~ (x + O(x3)/x = 1.

[grav]

Where do you get that equation with the v2?
My sources are:
Development of the Boltmann Distribution (http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html)
Kinetic Theory of Gases: A Brief Review (http://galileo.phys.virginia.edu/classes/252/kinetic_theory.html)
The Nature of Gases (http://www.ualberta.ca/-jplambec//che/p101/p01063.htm)
Maxwell-Boltzmann distribution (Wikipedia) (http://en.wikipedia.org/wiki/Maxwell-Boltzmann_distribution)

You can take the Maxwell equation also for 3D if you want, but there will not be a squared velocity in the equation, because that one does not integrate to 1, if you take 3D you get the sum of the squares of the velocity components in the exponential.
There is when we consider (m/2pikT)1/2 to the third power for 3D instead of 1D. In this case, (m/2pikT)1/2=1/vx becomes (m/2pikT)3/2=1/vx3, so it must be multiplied by 4piv2 to regain its 1/vx status in order to be multiplied by dv for a pure number ratio of n/N. It is interesting, however, that it is muliplied by 4pi for the surface area of a sphere instead of 4pi/3 for its volume. I wonder if that has anything to do with the factor of three. Also, it appears that the status of 3D as compared to 1D makes this formula a speed distribution instead of velocity distribution.

Then about your interpretation of my "dv" which is the infinitesimal variation of v, and thus your (v'-v) would be dv when v' is only a very small amount bigger than v. This means that (v'-v)2 = (dv)2 (note that dv is one thing) and your idea that (v'2 -v2) = d2v is incorrect.
Actually, I interpret d2v to be the square of the integral of v, or d2v=(dv)2=(v'-v)2. I figure dv2 to be the integral of v2, or (v'2-v2).

Now there may be some salvation here, as I have the idea that you want to do something with the (kinetic) energy of the particles. And there you might want to look at "second moment of the particle distribution" which is the integration of f(v) multiplied with the kinetic energy 0.5mv2. I would have to read more closely what you would like to do, before I start typing (wrong) equations again.
Well, I was actually trying to match energies with the speeds of photons from another recent thread I started, but I got sort of sidetracked, as I sometimes do. But I figure I should learn as much about how these formulas relate to molecules as possible, since this is what it was originally meant for, before I try to start applying them to other things.

Another note on math, you write:

The actual infinite summation yields (1/0)*0=0/0=any real number, but that isn't readily seen using our present form of mathematics.

you want to sum something that is a function of z which is actually a function of y. Now, you cannot easily say what you did, the summation will not lead to any number, but has to be closely looked at and will give a real answer, but then first we would have to get the equations right.
In the summation of [y3z2/2pi(ey)]1/2, I am taking z to be an infinitesimal number, and y is all numbers from zero to infinity in multiples of (1+z). In other words, if we start with some very small value for y and z, we plug them into the equation and find our result. We then multiply y by (1+z) and now we plug y(1+z) and z into the equation and add this to the former. Then we plug y(1+z)2 and z in and add it, then y(1+z)3 and z, and so on. It comes out to exactly one in this case. But I suppose the best way to show this formula is simply as the summation of [y(y'-y)2/2pi(ey)]1/2=(y'-y)*[y/2pi(ey)]1/2 for y=zero to infinity, and then anyone can perform the operation whichever way they want to.

[grav]

In my last post, I gave the summation as (y'-y)*[y/2pi(ey)]1/2=1. Well, I just realized that since y=E/kt, then this would be the formula for the distribution of energies, where f(E)dE=(y'-y)*[y/2pi(ey)]1/2=[(E'-E)/kt]*[E/2pikT(eE/kT)]1/2, so f(E)=[E/2pi(kT)3(eE/kT)]1/2. Here, the energy is defined as E=mv2.

[grav]

Okay. I have just looked up the distribution of energy, found at Maxwell-Boltzmann distribution, Distribution of the energy (Wikipedia) (http://en.wikipedia.org/wiki/Maxwell-Boltzmann_distribution) and found the formula, f(E)=2[E/pi(kT)3]1/2/(eE/kT). Apparently, then, they would be defining the energy as E=mv2/2 instead of E=mv2. In this case, since we are using the ratio, E/kT, and E is half as great, then kT must be half as great as well. So instead of f(E)=[E/2pi(kT)3(eE/kT)]1/2, my formula becomes f(E)=[E/2pi(kT/2)3(eE/(kT/2))]1/2=[4E/pi(kT)3(eE/kT)2]1/2=2[E/pi(kT)3]1/2/(eE/kT), which is exactly the same as what they show. So now I know without a doubt that it is correct, but obviously it is already known, which is usually the case when I come up with stuff like this :( . So I basically just performed a proof of the energy distribution based on the speed distribution formula. In any case, I would still like to know how the factor of three fits in, which I guess would actually be 3/2 if the energy is defined as E=mv2/2. So the average energy over all the molecules is now Eaverage=(3/2)kT.

[Bignose]

The distribution in the very first line of the first post is a distribution of speeds not velocity. The distinction is of course that speed is the magnitude of velocity, a scalar quantity, where velocity is a vector quantity. Working with speeds also effectively 'blur out' the 3-D nature of the problem since the speed could be in any of the three directions. This, in effect, turns the 3-D problem (velocity dist.) into a 1-D problem (speed dist.).

You have to be very careful in your use of the words, since precise definitions are implied.

Now, the reason you get 4*pi*v2 is because you are effectively looking at a spherical shell around each particle. The shell is 4*pi*v2*dv, which when integrated over v (from 0 to R) gives you (4/3)*pi*R3, like the volume of a sphere should. That gives you the volume, and if you want to know the expected number of particles in that shell, you multiply by the probability density and integrate over that as well.

The origin of all that is not magical, it comes from the change of variables from Cartesion to spherical just like I said. dv=r2 sin(p) dpdtdr. The limits change on the integral now include the integral of phi from o to pi, and the integrals on theta go from -pi to pi. Since the probability density is only a function of speed, and not angle, the integrals over theta and phi can be performed treating the f(v) as a constant.

Integral of phi (from 0 to pi) * Integral of theta (from -pi to pi) of [sin(phi) d(phi) d(theta)] = 4*pi ... exactly the constant that comes out.

These terms come from the math when done and understood properly.

I don't have my Chapman and Cowling with me, and I don't have the time to rederive all of it on my own, but later (few days) I can post how all these equations have been classically obtained. grav, it would be nice if you could get a copy and read along with it, especially since all the equations there are derived from first principles instead of trying to cobble together info from disparate websites and briefs like wikipedia. C&C's derivations start all the way from considering how two particles interact when they collide and go from there. All of these issues you are bringing up have been answered for over 50 years now.

[grav]

Thank you, Bignose. I think I understood most of what you said there. It should be helpful. I will continue researching as well, and see if I can eventually find a good book on the matter like the one you referred to.

[grav]

I have found what appears to be a wealth of information at Equilibrium Thermodynamics and Statistical Physics (http://www.phys.uri.edu/~gerhard/PHY525/topics525.html). Do you think this will do?

[tusenfem]

Okay. I have just looked up the distribution of energy, found at Maxwell-Boltzmann distribution, Distribution of the energy (Wikipedia) (http://en.wikipedia.org/wiki/Maxwell-Boltzmann_distribution) and found the formula, f(E)=2[E/pi(kT)3]1/2/(eE/kT). Apparently, then, they would be defining the energy as E=mv2/2 instead of E=mv2.

Any reason why kinetic energy would not be E=mv2/2?
Guess you are confusing here kinetic with rest mass energy which would be E=mc2.

Okay, now I understand your v2, my bad, I never work with speeds, I stand corrected.

[tusenfem]

In the summation of [y3z2/2pi(ey)]1/2, I am taking z to be an infinitesimal number, and y is all numbers from zero to infinity in multiples of (1+z). In other words, if we start with some very small value for y and z, we plug them into the equation and find our result. We then multiply y by (1+z) and now we plug y(1+z) and z into the equation and add this to the former. Then we plug y(1+z)2 and z in and add it, then y(1+z)3 and z, and so on. It comes out to exactly one in this case. But I suppose the best way to show this formula is simply as the summation of [y(y'-y)2/2pi(ey)]1/2=(y'-y)*[y/2pi(ey)]1/2 for y=zero to infinity, and then anyone can perform the operation whichever way they want to.

Now, here you want to do a sum for y = 0 to infty, and z is a very small number. And you also have defined y'/y=(1+z), so we know where the z comes from. But your idea of doing this summing is quite strange. First you will have to tell whether your y is all natural numbers or real numbers, I would assum you want real numbers as 1+z will be 1.something Unfortunately the summing as you like to do it does not work as y and z are not independent, so why would you multiply y with powers of 1+z and then plug it into the equation again? I will guess I will have to take your word for it that it comes out to be 1, but is that the correct answer? Naturally, as you already know the answer that integrated over all velocities the total number of particles N must come out it is easy to say this, but untill you show you have done the calculation I am not sure I am convinced that your method works.

[grav]

Any reason why kinetic energy would not be E=mv2/2?
Guess you are confusing here kinetic with rest mass energy which would be E=mc2.
Yes. That is it. I usually think of the energy as E=mv2, since E=mv2/2 is really, as far as I know, an average energy to or from zero, so that Eaverage=(Eoriginal+Efinal)/2, where each is defined as E=mv2 and one or the other is zero makes Eaverage=mv2/2. I have seen it used either way, however, so it must be defined. Otherwise the average energy to or from zero becomes E=mv2/4.

[grav]

Now, here you want to do a sum for y = 0 to infty, and z is a very small number. And you also have defined y'/y=(1+z), so we know where the z comes from. But your idea of doing this summing is quite strange. First you will have to tell whether your y is all natural numbers or real numbers, I would assum you want real numbers as 1+z will be 1.something Unfortunately the summing as you like to do it does not work as y and z are not independent, so why would you multiply y with powers of 1+z and then plug it into the equation again? I will guess I will have to take your word for it that it comes out to be 1, but is that the correct answer? Naturally, as you already know the answer that integrated over all velocities the total number of particles N must come out it is easy to say this, but untill you show you have done the calculation I am not sure I am convinced that your method works.
I do it this way because it is easy for me and that is how I plug the numbers into the computer. But I will do it here so you can see. I will use a rather large value for z so it will add up quickly for this demonstration. But a precise value only works for very small z, so the result will be slightly off. The summation is [y3z2/2pi(ey)]1/2=z*[y3/2pi(ey)]1/2. Let's use a very large value of z of .1 and start y originally at .0923. The summation is z*([y3/2pi(ey)]+[(y*(1+z))3/2pi(ey*(1+z))]1/2+[(y*(1+z)2)3/2pi(ey*(1+z)2)]1/2+...). So we get .1*(.01068+.01227+.01408+.01615+.018522+.02123+.02 431+.02781+.03180+.03633+.04146+.04726+.05381+.061 19+.06947+.07876+.08913+.10067+.11346+.12758+.1431 0+.16004+.17844+.19828+.21949+.24197+.26554+.28996 +.31489+.33989+.36445+.38793+.40961+.42869+.44431+ .45559+.46168+.46182+.45542+.44212+.42186+.39496+. 36213+.32448+.28350+.24090+.19853+.15821+.12149+.0 8958+.06316+.04239+.02695+.01614+.00905+.00473+.00 228+.00101+.00040+.00015+.00005+.00001+...)=.1*10. 42116=1.042116. With a very small starting value for y, it comes to .1*10.49205=1.049205. And with a very small x, one gets closer and closer to the value of one. Well, you get the idea.

[tusenfem]

Yes. That is it. I usually think of the energy as E=mv2, since E=mv2/2 is really, as far as I know, an average energy to or from zero, so that Eaverage=(Eoriginal+Efinal)/2, where each is defined as E=mv2 and one or the other is zero makes Eaverage=mv2/2. I have seen it used either way, however, so it must be defined. Otherwise the average energy to or from zero becomes E=mv2/4.

No! this just comes from taking moments, nothing averaged here.
First you have the mass of the particle: m
Then you get the impulse of the particle: m v
Then you get the kinetic energy of the particle: 0.5 m v2

You cannot just re-invent physics to fit your own ideas.

[grav]

No! this just comes from taking moments, nothing averaged here.
First you have the mass of the particle: m
Then you get the impulse of the particle: m v
Then you get the kinetic energy of the particle: 0.5 m v2

You cannot just re-invent physics to fit your own ideas.
Then how do you determine the average energy of a particle beginning at rest?

[tusenfem]

Oh for goodness sake Grav, read an introductory physics book, e.g. Allonso & Finn "Physics". I am not going to teach you basic physics.

And I never wrote that I calculate "the average energy" of a particle at rest. I just assumed a particle with mass m and a velocity v. There is no averaging, because take the derivative of the kinetic energy with respect to the velocity and then once more and you will find that it holds up as it should.

d Ek / dv = m v = P
d P / dv = d2 Ek / dv2 = m.

The only way you may want to calculate an "average energy" of "a particle" is when you have a distribution function of particle (Take your MB distribution) and then you will find that the average energy is 0.5 or 1.5 kT (depending if you calculate it in 1D or 3D), but then still if you want to know the velocity of this "average" particle you take 0.5 m v2 = 1.5 kT.

[grav]

d Ek / dv = m v = P

The formula you show here demonstrates my point (I think). You have dEk/dv=mv, so [Efinal-Eoriginal]/[vfinal-voriginal]=mv. Then [Ef-Eo]/[vf-vo]=[mvf2/2-mvo2/2]/[vf-vo]=[m(vf-vo)(vf+vo)/2]/[vf-vo]=m(vf+vo)/2. So m(vf+vo)/2=mv. Now, if we have an original velocity that is close to that of the final velocity, so that the change in velocity is very small, one gets m(vf+vo)/2=m(2vf)/2=mv, exactly as you say. But if the change in velocity is large, say from zero, then the change in energy is large as well, also from zero in this case, and we get m(vf+vo)/2=m(vf+0)/2=mvf/2=mv/2, which is only half as great as what you show. So dE/dv=mv or dE/dv=mv/2 depends not just on the final velocity, but on the starting velocity as well. What you are demonstrating is for a constant velocity, and so dE/dv doesn't apply. It is simply E/v=mv. But even then there is a problem, since if one defines E as mv2/2, then E/v=(mv2/2)/v=mv/2, so we actually still have the same dilemna.

I am not just making all of this up. I read it in a very old but vividly detailed applied physics book as a teenager. I wish I had it. I have not seen another with so much detail since. Now, that was indeed a long time ago, and it's possible I may have contorted some of the meaning since then, but it has stuck with me ever since. It explained it differently, however. I think it was something about the difference in mometum times the average velocity, which comes to [m(vf-vo)]*[(vf+vo)/2). For vf is very close to vo, we get (vf+vo)/2=(2vf)/2=vf. But is we break the equation up, it looks like [m(vf-vo)]*[(vf+vo)/2]=m(vf2-vo2)/2=mvf2/2-mvo2/2, so that the individual original and final energies look like mvo2/2 and mvf2/2. And since this is the formula for the difference in energies, it is convenient to look at it that way and define the energy as such, whereas Eo=mvo2/2 and Ef=mvf2/2, but it is really the difference in momentum times the average velocity.

But then again, if the original velocity is about the same as the final velocity, then the difference in momentum will be small, so it difficult to tell how that figures in anyway unless we use the previous formula for the energy. So I may have actually proved your point after all. I'm not sure. If so, there are still two things I wonder about. Would the energy as defined as E=mv2/2, then, now average out to or from zero (for a single particle with constant change in energy) to be Eaverage=(mvf2/2+mvo2/2)/2=(mvf2/2+0)/2=mvf2/4? And what about E=mc2? Why not E=mc2/2? In reality, though, the actual definition with or without a divisor of two doesn't matter, just as long as it is understood, and the rest of the equation must be made to accomodate it by supplying a multiple of two where applicable. All in all, then, none of this matters. What matters is the accepted definition of the energy, and according to what you are saying, that would be E=mv2/2, and so I suppose that is what I should use. Whether it may be based on a misconception that has caught on or not (or on my part), it does not matter when it comes to the communication of these ideas.

[tusenfem]

This may be so in your private universe, but I do hope you know the differnce between taking differences and taking a differential (indeed with the "difference in velocity very very small"). I would advise you not to not to rely on your teenage memory but to just take a book and read up on it, e.g. Allonson & Finn, Physics, where all these kind of things are discussed.

And the equation E = m c2 comes from a completely different perspective, and has NOTHING to do with kinetic energy.

Until you start reading a book, I guess I will stop this discussion. It is not my task to edumacate you in physics and math. And you cannot reinvent physics just to please your own worldview.

[grav]

Until you start reading a book, I guess I will stop this discussion. It is not my task to edumacate you in physics and math. And you cannot reinvent physics just to please your own worldview.
That's fine. You're right, it's not you're task, and I never expected you to. As a matter of fact, I read about every book I can get my hands on, at least when it is related to something I am working on at the time. But I don't learn these things overnight, and no book contains every detail, and even if it did, I can only comprehend so much, and in so much time, so I appreciate it when such things are pointed out, such as my confusion about why the energy is defined with a denominator of two when it could just as easily not be. I don't look at that as reinventing physics, since it does not effect the outcome of a formula as long as the terms are defined, but just a problem in semantics. I thought it demonstrated an average, but it doesn't matter, since I must use that definition that is in common use. In any case, thanks for setting me straight on that. I will use E=mv2/2 from now on. :)

[Bignose]

grav, even beyond the differential reasons given above, which are very compelling, there is another huge reason why the 1/2 is in there. So that all the units of energy work out to be the same. Specifically, so that when x Joules of work has been done on an object in free space, it should also have x J of energy. Among the main ideas, of course, is that if there was no dissipation, you could possible get that x J of work back out of the object. If the 1/2 wasn't in the definition of kinetic energy, it would have to be in a "work to kinetic energy conversion factor", which, and maybe this is because all the years I've known it the way it is currently defined, seems much more awkward.

I hope this does not sound condescending, but tus is right, this is basic first semester physics stuff. Before tackling the much more complicated issues, do yourself a favor and learn/review from the beginning. If you have questions please don't hesitate to ask, but these basic errors cannot show up when doing more difficult calculations.

[grav]

grav, even beyond the differential reasons given above, which are very compelling, there is another huge reason why the 1/2 is in there. So that all the units of energy work out to be the same. Specifically, so that when x Joules of work has been done on an object in free space, it should also have x J of energy. Among the main ideas, of course, is that if there was no dissipation, you could possible get that x J of work back out of the object. If the 1/2 wasn't in the definition of kinetic energy, it would have to be in a "work to kinetic energy conversion factor", which, and maybe this is because all the years I've known it the way it is currently defined, seems much more awkward.

I hope this does not sound condescending, but tus is right, this is basic first semester physics stuff. Before tackling the much more complicated issues, do yourself a favor and learn/review from the beginning. If you have questions please don't hesitate to ask, but these basic errors cannot show up when doing more difficult calculations.
Thanks, Bignose. In your example, the displacement would be d=at2/2+vt. So this is apparently where the value of 1/2 comes from, and why the value of 2 keeps popping in and out of equations. Since the work is defined simply as W=force * distance and the force is F=ma=m(v'-v)/t, then the work would be W=(ma)(at2/2+vt)=m(at)2/2+mv(at)=m(v'-v)2/2+mv(v'-v)=mv'2/2-mv'v+mv2/2+mv'v-mv2=mv'2/2-mv2/2. Now, if it were an average energy with E=mv2, it would be (mv'2+mv2)/2=mv'2/2+mv2/2, which it obviously is not. So if we define the work also as a difference in energy, I can see how E=mv2/2, then, would definitely be more convenient. It is not convenient to me, however, since I must now multiply all of my values for energy by two (mv2=2E), but then, only in my presentation. :neutral: I guess it's not that big of a deal, really. I've been doing things my own way for quite a while, and now it's time to conform. But I'm still new to a lot of this, and this forum is the first time that I've managed to communicate about any of my ideas. I hope that you, tusenfem, Nereid, and the others will continue to be patient with me and help me out. :) In the meantime, I will continue to read up on these things, as I am still quite eager to tackle the much more complicated issues, and I'm sure I will learn a lot by doing so and from all of you as well.

[grav]

And interestingly Mr. Crumb and Grav seem to think that E = m c2 seems to be a "speed of light" limit of the kinetic energy of a particle E = 0.5 m v2.

Oh, please keep me out of that mess (slop). I thought I'd post over here where I feel much more at ease. :)

As far as energy goes, it appears that E=mv2/2 is indeed the way it is defined, otherwise the work done would would be defined as W=E/2, if E=mv2. It could easily be defined either way, but it seems to be a matter of convenience to define it as such so that the amount of work performed is mathematically equal to the loss (or gain) of energy, as Bignose mentioned. In this case, the average energy to or from zero would be Eaverage=mv2/4, but that's just the breaks.

Now, it seems to me that a body can have any speed whatsoever, but it can never be greater than c relative to another. The greatest kinetic energy a body could have, then, relative to another at "rest", would be E=mc2/2. So now I must determine why it is E=mc2 instead. Well, my best guess is that the reason c is the maximum relative speed in the first place is because of the distortion of time in GR. Of course, I have different ideas about this, but it amounts to the same thing mathematically and I want to express this in a way everybody understands. So since the distortion of time is also a factor, then we must also account for this in the energy involved, which apparently brings it to twice the Newtonian value. I have seen this time and time again, so I am now sure this must be the reason. E=mc2 works out rather nicely in all of my formulas, so I have no reason to doubt it, and so it must necessarily be that all high energy particles (with low mass) experience time dilation. Thank you for bringing it to my attention, as I probably would not have thought about it to this extent otherwise.

I now wonder how GR would figure into the distribution of the speeds for very high energy molecules. :think: ;)

[tusenfem]

Grav

I did not want you to drag into the slop of Stevencrum, I just happened to see a similarity. Sorry bout that.

The maximum kinetic energy of a (relativistic) particle is not E = m c2/2, in relativistic physics the kinetic energy of a particle is given by:

Ekin = (gamma - 1) E0

where E0 is the rest mass energy of the particle m0c2 and gamma is the well know function:

gamma = 1 / sqrt(1 - v2/c2)

This means that the kinetic energy of a particle is the total energy of a particle with the rest mass energy subtracted (that is where the -1 comes from).

You really would make your scientific endeavors so much easier if you would just reas a basic book on physics.

[grav]

Grav

I did not want you to drag into the slop of Stevencrum, I just happened to see a similarity. Sorry bout that.

The maximum kinetic energy of a (relativistic) particle is not E = m c2/2, in relativistic physics the kinetic energy of a particle is given by:

Ekin = (gamma - 1) E0

where E0 is the rest mass energy of the particle m0c2 and gamma is the well know function:

gamma = 1 / sqrt(1 - v2/c2)

This means that the kinetic energy of a particle is the total energy of a particle with the rest mass energy subtracted (that is where the -1 comes from).

You really would make your scientific endeavors so much easier if you would just reas a basic book on physics.
Well, I assure you I do read physics books, and plenty of them, on all sorts of related topics, and I even collected some for a while until they all started to sound the same, but some of them are old, and other are very general, and I don't always know what I am looking for. I just read through them and hope something sticks. I did not know I should be looking for something like this until you pointed it out, but I do recall reading a couple of times about a similar equation, but it showed something like how
the kinetic energy and rest mass was related to the square of the momentum and c to the fourth power or something along those lines. I'll have to look it up.

I assumed that E=mc2 instead of E=mc2/2 came about when the time distortion is accounted for, which usually doubles the result. For the equation you show here, however, for a small v/c, we get Ekin=(1/[1-(v/c)2]-1) E0=([1+(v/c)2]-1)(m0c2)=(v/c)2(m0c2)=m0v2. So for v is small, Ekin=m0v2. Not that I doubt this or anything, but I'm curious, where is the value of 1/2 here? It seems like gamma should probably be 1/(1-v2/2c2) in this case.

Now, here the kinetic energy starts at E=m0v2 and climbs to infinity as one approaches the speed of light. Obviously one cannot really travel faster than light relative to anything else and gain an infinite energy, according to the theories of relativity, so it seems to me that the kinetic energy should be that of the kinetic energy as one would measure for flat space-time. In other words, that is the real energy, so since that cannot be translated into a speed greater than c, it becomes v, which is always less than c, and the rest would be converted into apparent mass.

[tusenfem]

you forgot the square root

[hhEb09'1]

you forgot the square rootYepgamma = 1 / sqrt(1 - v2/c2)Not that I doubt this or anything, but I'm curious, where is the value of 1/2 here? It seems like gamma should probably be 1/(1-v2/2c2) in this case.For small v/c, (1-v2/2c2) is approximately equal to sqrt(1 - v2/c2)

[grav]

That's true. :doh: That would do it. Thanks. :)

[Nereid]

That's true. :doh: That would do it. Thanks. :)So, grav, if your track record (so far) in the ATM section is "grav 0, mainstream {insert your estimate here}", do you have anything you could offer, in the way of re-assurance that your (quixotic?) approach has legs?

For example, I note that you seem to have not corrected any of the previous ATM ideas you floated here, addressing the inconsistencies, errors, etc that other BAUT members were kind enough to point out.

[grav]

So, grav, if your track record (so far) in the ATM section is "grav 0, mainstream {insert your estimate here}", do you have anything you could offer, in the way of re-assurance that your (quixotic?) approach has legs?

For example, I note that you seem to have not corrected any of the previous ATM ideas you floated here, addressing the inconsistencies, errors, etc that other BAUT members were kind enough to point out.
I don't see how my track record is zero. I have figured out many things as far as I can tell. Unfortunately, most all of those things turn out to be already known, except by me, of course, until after a lot of pain-staking calculations. Then again, I suppose that if it turns out to be already known, it becomes mainstream, so in a way, you're right. But what better way to learn these things than through first-hand experience? I still count each and every one as a personal victory. It shows me I know what I'm doing, at least, for the most part. Even if I am not as knowledgable as most of the other forum members, and my math level is rather low, I am still capable of achieving reasonable results if I stick with it. Take this thread, for example. I achieved an energy distribution for molecules from the speed distribution formula. Granted, I originally used the wrong formula for energy itself and was corrected, and that someone else that is experienced in this sort of thing may have done it easily, but I did in fact come up with the correct formula after all, depending on the definition for the energy that was used. It was the first time I had ever done this, as is the case for almost all of my posts, so I am still rather proud of them, even if they require some tweaking, which is often the case. The rest of the thread became a metaphysical discussion on why the definition of energy is what it is, from which I have learned very much. I also made some observations that have not been commented on yet. One of which I am confident is 3kB=mcs. I have tried this for 8 elements and it works out. Whether this will be the same for molecules, I am not sure. But it will probably turn out that this is already known as well, as I am only one man, and most all of these things have probably been thought about before, just not by me.

As for any inconstistencies left uncorrected in other threads, I am still working on them. I don't believe I have learned enough to satisfactorily answer them yet. Even if I did, it would just pose more questions. Did you have any particular ones in mind? I can only think of a couple of major ones off hand, though, which are the decay of binary star systems and the interactions of neutrinos, of which I would have to say I just don't know yet, but I am presently working on them. That is why I am so persevering in this forum. I am determined to learn as much as possible and then answer each and every one of them as soon as things turn full circle. I believe GR may be the key in doing this, perhaps not the way most are used to thinking about it, but only because I believe it requires a real, measurable mechanism, and I am trying to find it. For right now, it seems every answered question begs more, and not only on my part, but in the reality of physics itself. I am steadily learning more with the help of the forum members and yourself, and I am attempting to apply what I have learned to anything I can, to gain a better perspective of reality. Some of my methodology might seem quite different than what most are used to, I'm sure, but I must do these things in the best way I know how, and by whatever means I can conceive of them. Part of my problem so far, however, is that my mathematics language is not up to speed, and I am also having trouble grasping some of the concepts of GR, but I keep trying. I am stubborn that way.

[Bignose]

One of which I am confident is 3kB=mcs. I have tried this for 8 elements and it works out. Whether this will be the same for molecules, I am not sure. But it will probably turn out that this is already known as well, as I am only one man, and most all of these things have probably been thought about before, just not by me.



You are right in that this kind of relationship has been known for sometime, though only for ideal gas molecules. Your assumptions about Aluminum, for example, was treating it exactly like Neon gas, which just is not so.

Have you looked through, http://en.wikipedia.org/wiki/Heat_capacity, as just one example of the relationship between ideal gases and heat capacities? (I think that the 3 should be 3/2 because of the energy misdefinition you were using.)

[grav]

You are right in that this kind of relationship has been known for sometime, though only for ideal gas molecules. Your assumptions about Aluminum, for example, was treating it exactly like Neon gas, which just is not so.

Have you looked through, http://en.wikipedia.org/wiki/Heat_capacity, as just one example of the relationship between ideal gases and heat capacities? (I think that the 3 should be 3/2 because of the energy misdefinition you were using.)
So I'm off by a factor of 2 again, huh? :) After reading your post, I thought maybe the energy thing was the reason I came up with that, too. But then I realized that although kB is an energy constant, I did not actually use any definition for energy in finding this particular relationship, so that's not it. I then thought maybe I was considering it for diatomic molecules instead of monatomic, but I only used a single atom in each case, so that's not it either. I noticed that the link you referred to uses cV. So maybe it is the specific heat capacity for volume while mine is for surface area, or something along that nature. The relationship I discovered is a direct one, the specific heat capacity for a particular element times the mass of a single atom. I simply realized that the units were the same as for KB, except for the denominator of kg, so I tried it for atomic masses. I thought if this relationship was already known, it would have been stated in my source, but it's not. Also, it could easily just be replaced by 3kB/m, unless it is different for molecules for some reason. I did notice that the specific heat capacity for ice is just slightly greater than that for steam, but it seems that would have something to do with the initial temperature as well. I thought that the energy of kBT was directly proportional to the temperature in the first place, so this is a mystery to me. In any case, here are the values I used.

Element.....c (kcal/kg*K) *4186=c (J/kg*K) *m(atomic)*1.6606*10-27 kg= 3kB(4.14197436*10-23 J/K)
Aluminum.........(.22).................(920.920)*( 26.9815).................................(4.12623* 10-23)
Copper............(.093)...............(389.298)*( 63.546)...................................(4.10805 *10-23)
Gold................(.03).................(125.580 )*(196.967).................................(4.107 51*10-23)
Iron................(.113)................(473.018 )*(55.847)...................................(4.38 675*10-23)
Lead...............(.031)................(129.767) *(207.19)...................................(4.464 73*10-23)
Mercury...........(.033)................(138.138)* (200.59)...................................(4.6013 7*10-23)
Silver...............(.056)...............(234.416 )*(107.868)..................................(4.19 899*10-23)
Zinc................(.092)................(385.112 )*(65.37).....................................(4.1 8052*10-23)

I figured the small differences might be due to a margin of error for the values used (or isotopes).

[Bignose]

This is taken from Section 2.43 of Chapman and Cowling (which I referenced above):

Let a unit mass of a gas in equilibrium be enclosed in a constant volume. A certain (small) amount of heat is added to increase the temperature by deltaT; energy required to add will be written as cv*deltaT. The coefficient cv is the specific heat at constant volume (this is different from cp which is heat capacity as constant pressure. To do thermodynamics calculations correctly, you have to use the correct one!) B/c it is constant volume, the gas does no work against external pressure, and hence all the energy goes entirely into heat energy. The number of molecules in a unit mass is (1/m). Let the initial heat energy in the gas be denoted E/m; and this is increased by cv*deltaT when T is increased by deltaT. Therefore

deltaE/m = cv*deltaT

Or taking the limit as delta-->0

cv = (1/m)*(dE/dT)v.

where (dE/dT)v denotes the rate of increase in energy E with respect to T, when V (the volume) is kept constant.

E=(3/2)kbT from the Maxwellian distribution therefore these can be combined (do the derivative)

cv = 3kb/2m

Compare this with your relationship... its been known for quite some time. Pretty much only need the definition of constant volume heat capacity and temperature from kinetic theory.

Any molecule that you treat as an ideal gas will have this property. But, those metals are not ideal gases, you are only treating them as such. This formula only works for the noble gases, Helium, Neon, Argon, etc. Specifically, this comes from the statement the thermal energy consists only of energy of translation or in other words the energy of a given molecule is e=(1/2)m*c2 (c is the velocity), whereas real gases like hydrogen or CO2 or methane have energies of rotation and possibly energy of contraction or expansion of the bonds, e=(1/2)m*c2 + (1/2)m*w2 + h where w=angular momentum and h would be average elasticity energy of a molecule. Some of the simpler gases, like Hydrogen gas, cv = 5kb/2m because rotation energy is pretty much the only extra energy (the bonds on H-H are pretty much always the same length, so stretching energy can be assumed 0, and the spin temperature for hydrogen gas at moderate temperatures is pretty close the translational temperature. But, spin temperature is only equal to the translational temperature in exceptional cases, in most cases they are unequal), but the more complicated the gas, the farther away from ideal it will be.

[grav]

Thanks again, Bignose. I see what you mean on this (http://en.wikipedia.org/wiki/Specific_heat_capacity) Wikipedia link. cv is for constant volume and cp is constant pressure. I used cs (the {s} stands for specific heat) to express this in the beginning of the thread, by the way, because my source only used c and I didn't want it to get confused with the speed of light. Although the values were slightly different, the table it gives for cp is the same as my source. So apparently, cv gives half the value as cp when it comes to kB because P=E/V, where one is for pressure and one is for volume and the energy is 3kT/2 when considering the average energy of the molecules (or something like that, plus other factors, like you said). Very interesting.

[Bignose]

So apparently, cv gives half the value as cp

Actually, the relationship between cv & cp for gases is

cv + R = cp

where R is the gas constant.

E.g. (all units in calories per degree C)

Air: cp = 6.960 cv=4.965 cp-cv=1.995

compare that with R=1.9865 cal/degree C

This relationship between gas heat capacities holds really well, especially considering that air is a mixture of several gases.

[grav]

Actually, the relationship between cv & cp for gases is

cv + R = cp

where R is the gas constant.

E.g. (all units in calories per degree C)

Air: cp = 6.960 cv=4.965 cp-cv=1.995

compare that with R=1.9865 cal/degree C

This relationship between gas heat capacities holds really well, especially considering that air is a mixture of several gases.
Huh. That's also interesting. From you're earlier post and from the link you referred to, I guess it would depend on the number of degrees of freedom. 3R/2 plus 1R due to the rotation of the molecules. Is that correct?

[Bignose]

No, it's not rotation, it is the different definitions. cv has all the energy going into heat (which is then translation, rotation, etc.) because the volume of the gas is held constant. Whereas cp the pressure is kept constant. That means when energy is put in, the volume has to expand to keep the same pressure, and some of the energy being put in goes to heat, but some to work also. This is why it is so important to define and explicitly state what is happening in the system, because the definitions of these terms are exact, and using the wrong term results in big errors in thermodynic calculations.

[korjik]

Hey grav, how much calculus do you know? I seems like alot of the mistakes I see here are from trying to do physics with algebra only, when you should be doing calculus. The factor of 1/2 in 1/2mv^2 fairly leaps out at you when you derive KE with calculus.

[hhEb09'1]

I seems like alot of the mistakes I see here are from trying to do physics with algebra only, when you should be doing calculus. The factor of 1/2 in 1/2mv^2 fairly leaps out at you when you derive KE with calculus.I think the missing 1/2 was a result of an error in algebra (http://www.bautforum.com/showpost.php?p=831845&postcount=36) though

Even Newton himself tried to derive some of his famous results in other fashions, after finding them using his calculus

[grav]

I think the missing 1/2 was a result of an error in algebra (http://www.bautforum.com/showpost.php?p=831845&postcount=36) though

Even Newton himself tried to derive some of his famous results in other fashions, after finding them using his calculus
Thanks, hhEb09'1. But really it wasn't so much an error as an oversight. I forgot the square root, as tusenfem stated, which led me to twice the result.

Hey grav, how much calculus do you know? I seems like alot of the mistakes I see here are from trying to do physics with algebra only, when you should be doing calculus. The factor of 1/2 in 1/2mv^2 fairly leaps out at you when you derive KE with calculus.

Twice the value for the definition of energy was due to a misconception on my part, which tusenfem also pointed out. I seem to have a subconscious dislike for the number 2 or something (or maybe the other way around). Anyway, Bignose showed me that it was convenient to use the 1/2 in the value for the energy of work performed in order for both to come out the same, so instead of W=E/2, we use W=E (which I guess should really be W=dE). I did this with just algebra, but the factor of two still leaped out at me when all was said and done. I probably do need to learn calculus, though, as I am most likely missing out on some very important and useful formulas that I could utilize in my work if I could only understand them better. Although my way of doing it is much longer than it probably would otherwise need to be, I never thought anyone would have a hard time reading it. But if calculus is what they are used to, then it probably looks to them like something would look to me when written as (1+1+1)*(1+1)=(1+1+1+1+1+1) instead of the condensed version. It's true, and on the most basic level, but it would still take some getting used to, I'm sure. I found a calculus tutorial online some time ago, and printed off the first 80 pages of it, but the first half just seemed like common sense, and the second half ran off on a tangent and lost me. About the only thing I retained (barely) was right about in the middle when it expressed a derivitive as something like [f(a)-f(b)]/(a-b). That's it, and I'm not even sure how it should be applied, even though I came up with something similar to this formula on my own a while back when working with summations, and is probably the only reason I remember this part at all.

As for m*cp=3kB, it seems this formula is also trying its hardest to be off by a factor of 2 as well, since Wikipedia defines it as half that value for cv, at least. But the formula I came up with is not slop. It is factual (dangit, tusenfum, now you even have me talking like that other thread :) ). Just take the cp value for any element and multiply this by its atomic mass. It will come out to about 3kB. If it is then divided by 2 to obtain the energy or something, I'm not sure. I still don't quite get that part of it. If this is the case, I don't know why they wouldn't just use half the values for cp to begin with.

[grav]

Would the speed and energy distribution formulas work for neutral subatomic particles as well?

[Bignose]

grav, one more time, your 3*kb predictions you showed were for all solid phases, or a really dense liquid (mercury). But, if you go back to http://en.wikipedia.org/wiki/Heat_capacity again, and look under the section called solid phase, you'll see why it is working. The Dulong-Petit law (propsed in 1819) for heat capacities of solids says that the heat capacity is about equal to 3R (R the gas law constant again). 3R is on a energy/mole basis, when dividing back by the molecular mass, you get 3kb/m. Boltzmann's constant and the gas law constant are very related. The Dulong-Petit law was derived based on the molecular vibrational energy available to molecules in a solid lattice structure. It certainly isn't perfect, it fails miserably at low energies, and not every solid is the same, but it holds pretty well for some number of solids, especially the elemental ones like you did.

Your formula will not work for gas phase predictions. Do the rest of that table for Helium, Argon, Hydrogen, Oxygen, Nitrogen etc. Solid state molecules do not behave according to kinetic theory -- kinetic theory describes gases not molecules in a lattice! It is a completely different beast, and completely totally incomparable. Its not even apples to oranges, it is apples to orangutans.

[Bignose]

Would the speed and energy distribution formulas work for neutral subatomic particles as well?

There is a kinetic theory for electrons and the like, but there are additional complications what with the electromagnetic fields that can be generated. Also, there can be relativistic effects that have to be accounted for. That, and if the particles don't collide like billiard ball molecules do (depends on how the subatomic partciles interact with one another), there will be a need for additional modification.

[grav]

grav, one more time, your 3*kb predictions you showed were for all solid phases, or a really dense liquid (mercury). But, if you go back to http://en.wikipedia.org/wiki/Heat_capacity again, and look under the section called solid phase, you'll see why it is working. The Dulong-Petit law (propsed in 1819) for heat capacities of solids says that the heat capacity is about equal to 3R (R the gas law constant again). 3R is on a energy/mole basis, when dividing back by the molecular mass, you get 3kb/m. Boltzmann's constant and the gas law constant are very related. The Dulong-Petit law was derived based on the molecular vibrational energy available to molecules in a solid lattice structure. It certainly isn't perfect, it fails miserably at low energies, and not every solid is the same, but it holds pretty well for some number of solids, especially the elemental ones like you did.

Your formula will not work for gas phase predictions. Do the rest of that table for Helium, Argon, Hydrogen, Oxygen, Nitrogen etc. Solid state molecules do not behave according to kinetic theory -- kinetic theory describes gases not molecules in a lattice! It is a completely different beast, and completely totally incomparable. Its not even apples to oranges, it is apples to orangutans.
Thanks again, Bignose. That makes sense. I was wondering why the specific heat capacity of ice was slightly different than that for steam. It wouldn't make much sense, I don't think, although still possible, that it depended on the initial temperature, ice being much colder than steam, obviously, especially since it seems the specific heat capacity is meant to be temperature independent in itself. So I guess it is a phase dependency (plus other factors).

[grav]

There is a kinetic theory for electrons and the like, but there are additional complications what with the electromagnetic fields that can be generated. Also, there can be relativistic effects that have to be accounted for. That, and if the particles don't collide like billiard ball molecules do (depends on how the subatomic partciles interact with one another), there will be a need for additional modification.
I am attempting to define the interactions of neutrinos with each other. I am extremely lucky that they are electrically neutral, since it would indeed make for some terrible complications if they were not. But the relativistic effects concern me. Neutrinos should travel close to the speed of light. But I want to start with the basics to get a feel of it. That would be billiard ball like collisions of neutral particles without the relativistic effects for now.