COW

[ajb]

Thus KE is useless to determine direction (mathematically).

Which I think was swansont's point.

 

 

Can energy be vectorized? i.e. be made into a vector.

You can think of 4-mometum... maybe that is the closest notion.

 

 

And with the temperature of a single particle, you could just say that that the temperature is equal to the kinetic energy. But then you should realise that this is not a invariant notion, it depends on the relative velocity. Usually, we do not assign a notion of a temperature to a single particle.

Edited June 2, 2016 by ajb

[Capiert]

Which I think was swansont's point.

Oh (yes), I agree full heartedly there with Swansont. His point backward or foreward, was very enlightening. Very effective (=efficient).

I was so convinced, I advanced right into the next step, forgetting to mention it.

You can think of 4-momentum... maybe that is the closest notion.

What is 4-momentum?

And with the temperature of a single particle, you could just say that the temperature is equal to the kinetic energy. But then you should realise that this is not a invariant notion, it depends on the relative velocity.

Aren't all speeds relative? They are always a comparison, always a difference, from what we say is zero.

That would be a comparison to the container, e.g. a jar's surface, where it would strike, or a "stationary" thermometer's surface.

The molecule hits an area (surface). But that is too closely related to pressure. How do I distinguish?

I suspect part of the speed (per area, & numbers of collisions) represents temperature & the other part represents pressure. ?

 

Usually, we do not assign a notion of a temperature to a single particle.

There I would need to assign, temperature & pressure, & have to know the size (& shape) of the container.

Otherwise for me impossible to conceive of correctly. =Too little info.

An alternative, would be to determine the missing values, from a few assumptions.

 

(I can not even prepare to answer your previous posts, until I know enough about vectors.

E.g. can we vectorize energy? (=Can we turn KE into a vector with some sort of formula or operation, so we can deal with that result, as a vector?) Are vectors simply a math construct (tool) that we have invented?

Or are we bound to only what they (vectors) can deliver because they are a natural phenomena (that we cannot modify, as impossible to change, in any way, because it (the concept of vector) is not a tool)?

 

Y/N's + etcs.

 

Studiot may answer there (too) if he wants because he seems to put it in my words, better.

Edited June 2, 2016 by Capiert

[ajb]

What is 4-momentum?

Loosley, in special relativity you combine the momentum with energy to build a vector. This is the 4-momentum.

 

Aren't all speeds relative? They are always a comparison, always a difference, from what we say is zero.

In short, yes.

 

... represents temperature & the other part represents pressure. ?

The basic kinetic theory of gases tells us that the temperature is the average kinetic energy as measured in the frame for which the total momentum of the gas is zero. So if we have only one particle then this definition has little meaning.

 

 

 

E.g. can we vectorize energy? (=Can we turn KE into a vector with some sort of formula or operation, so we can deal with that result, as a vector?) Are vectors simply a math construct (tool) that we have invented?

I guess one can do lots of things to construct vectors, but that does not mean they all have any physical meaning.

[Capiert]

Loosley, in special relativity you combine the momentum with energy to build a vector. This is the 4-momentum.In short, yes.The basic kinetic theory of gases tells us that the temperature is the average kinetic energy as measured in the frame for which the total momentum of the gas is zero. So if we have only one particle then this definition has little meaning.

If I travel in my car at 100km/h

& I'm the only 1 on the highway, at all in (=on) the whole earth,

then my average speed is 100km/h

including for the whole earth.

The statistic(s) does not change when I discuss only 1.

The method including the math remains the same.

The meaning, thus, has not changed.

Often it's a lot easier to deal with only 1 thing

(vs a background, environment or container etc.).

 

I guess one can do lots of things to construct vectors, but that does not mean they all have any physical meaning.

Maybe not for you? It might have meaning for me? There are limits to how far I can go, though.

 

Is a vector multiplied by a vector, a(nother) vector?

I assume a vector multiplied by a scalar (e.g. just a number) is a(nother) vector. ?

Edited June 2, 2016 by Capiert

[imatfaal]

If I travel in my car at 100km/h

& I'm the only 1 on the highway, at all in (=on) the whole earth,

then my average speed is 100km/h

including for the whole earth.

The statistic(s) does not change when I discuss only 1.

The method including the math remains the same.

The meaning, thus, has not changed.

Often it's a lot easier to deal with only 1 thing

(vs a background, environment or container etc.).

 

 

 

How are you measuring your speed - you are measuring with respect to the road. There is no such thing as intrinsic velocity- ie velocity which can be measured without reference to something else.

[ajb]

Is a vector multiplied by a vector, a(nother) vector?

At the risk of seeming rude, you should really learn some mathematics.

 

If we just have a vector space, then there is no notion of multiplying vectors together. If we have further structure, say that of an algebra then we can multiply them. The context of the structures is important. If you just mean vectors on R^n then we do not have a canonical method of multiplying vectors.

 

 

I assume a vector multiplied by a scalar (e.g. just a number) is a(nother) vector. ?

This is one of the axioms of a vector space... you should know this.

[Capiert]

How are you measuring your speed - you are measuring with respect to the road. There is no such thing as intrinsic velocity- ie velocity which can be measured without reference to something else.

 

Yes, I think you are only repeating what was agreed upon in #52 & #53.

All speeds are relative, no exceptions.

I do not know why you are doing that repeat. Did I miss something you wanted to say? Please explain. Maybe I missed something? Do you know what it is?

At the risk of seeming rude, you should really learn some mathematics.

I will try to take it to heart. With all you say below you make me curious. My problem is finding a book .pdf that will maintain my interest. Perhaps I need a fly over & then details, to maintain perspective.

If we just have a vector space, then there is no notion of multiplying vectors together. If we have further structure, say that of an algebra then we can multiply them. The context of the structures is important. If you just mean vectors on R^n then we do not have a canonical (rule) method of multiplying vectors.

(Why not A^n factor as well? So you can combine A with R before exponentiating? Or am I too far off, naive?) Is there any reason why you use the church word canonical? Or am I astray again? Referring to a canon (e.g. ball) instead?.

This is one of the axioms of a vector space... you should know this.

If you say, I wish I did. I just haven't made the connection. I know too little.

 

Even your word vector space confounds me because I think of lines or arrows filling a volume.

I begin thinking about filling a volume with points instead.

I do not bring the math vocabulary in connection with reality.

Edited June 3, 2016 by Capiert

[ajb]

Is there any reason why you use the church word canonical? Or am I astray again?

Just because we do have finite dimensional real algebras... But if I give you just the space R^n then you do not have - by itself - multiplication of two vectors.

 

There are canonical algebras associated with R^n, but they do not correspond to naive multiplication of two elements of R^n to get another element of R^n.

 

If you say, I wish I did. I just haven't made the connection. I know too little.

 

Look up the definition of vector spaces and affine spaces. And from there metric spaces and also algebras.

 

I begin thinking about filling a volume with points instead.

That won't help. A smooth space contains an infinite number of points.

 

I do not bring the math vocabulary in connection with reality.

You seem not to have grasped some basic mathematical concepts. You will need to do this before caliming things are wrong and before trying to calculate things. You are making life hard on yourself.

Edited June 3, 2016 by ajb

[studiot]

 

Even your word vector space confounds me because I think of lines or arrows filling a volume.

 

 

Most mathematics starts with set theory.

 

A set is the posh mathematical name for a collection of (mathematical) objects, although there is some argument about this and the words class and type have also been used.

 

By itself a random collection of objects is pretty uninteresting and useless, mathematically.

 

So we place restrictions on the objects.

 

Usually we require that all the objects are of the same sort (eg all numbers, all vectors all angles )

We often require that there is only one appearance (instance) of any particular object.

 

But this is still pretty boring

 

So we set down a list certain properties that all our object must possess.

Obviously there are many possible lists so centuries of thought has gone into what should be on a particular list.

 

We call such a list 'the axioms of our set'

 

and a space is the set of all objects that obey they axioms.

 

So a vector space is the set of all objects that obey the (8 or 9 ) vector space axioms.

We call these objects vectors.

 

This gives the set some useful structure and justifies introducing a new word (space).

 

This structure allows us to perform mathematical operations on the vectors in the vector space, which is why we go to all that trouble to spout all that terminology.

 

There are many different vector spaces and that part of mathematics devoted to the study of these is called linear mathematics.

This is still probably the largest part of mathematics since most applications are linear.

 

 

You have a knack for digression by introducing new words that you do not fully understand.

So it was good that you acknowledged this.

This post was designed to help there.

 

You also made some very good comments on my post 43 (your post 46)

 

 

Studiot may answer there (too) if he wants because he seems to put it in my words, better.

 

 

If you like this post I will return to the subject of momentum, conservation and invariance and those replies.

Edited June 3, 2016 by studiot

[Capiert]

Most mathematics starts with set theory.

 

A set is the posh mathematical name for a collection of (mathematical) objects, although there is some argument about this and the words class and type have also been used.

 

By itself a random collection of objects is pretty uninteresting and useless, mathematically.

 

So we place restrictions on the objects.

 

Usually we require that all the objects are of the same sort (eg all numbers, all vectors all angles )

We often require that there is only one appearance (instance) of any particular object.

 

But this is still pretty boring

 

So we set down a list certain properties that all our object must possess.

Obviously there are many possible lists so centuries of thought has gone into what should be on a particular list.

 

We call such a list 'the axioms of our set'

 

and a space is the set of all objects that obey they axioms.

 

So a vector space is the set of all objects that obey the (8 or 9 ) vector space axioms.

What are those axioms? Are they written only in sentences like Euclid did?

Perhaps with helping formulas too?

We call these objects vectors.

Can those vectors be more than just arrows? length & direction.

Or is the distinguishing idea, the name of the (group) "space"? with only length & direction?

That brings up the question: Is my concept of vector outdated?

Has the definition evolved into something more that only direction, & magnitude?

(E.g. We can't rely on definitions, they change each year? or few (decades)?)

 

This gives the set some useful structure and justifies introducing a new word (space).

Yes, but isn't the word space old, & was already taken (in physics?) meaning something like volume?

E.g. Outer space.

Or is it more intended, like a (e.g. mailbox) slot, where we store info? E.g. Something reserved for info (list). ?

 

This structure allows us to perform mathematical operations on the vectors in the vector space, which is why we go to all that trouble to spout all that terminology.

 

There are many different vector spaces and that part of mathematics devoted to the study of these is called linear mathematics.

This is still probably the largest part of mathematics since most applications are linear.

 

 

You have a knack for digression by introducing new words that you do not fully understand.

Yes, too curious.

 

So it was good that you acknowledged this.

I try.

This post was designed to help there.

Thank you. The vocabulary (syntax) unsures me a bit, that mathematicians can call things the way they like, ignoring previous definitions, from other subjects. Its almost as bad as me? e.g. space, unless I've misunderstood. ?

 

You also made some very good comments on my post 43 (your post 46)

Thank you. It's nice to hear some encouragement, once & while, while the lions roar (=threat) & the sheep say bah! (=disagree, scoff).

 

If you like this post I will return to the subject of momentum, conservation and invariance and those replies.

Oh yes, please do!

 

At the risk of seeming rude, you should really learn some mathematics.

Theme: Just to confirm.

If we just have a vector space, then there is no notion of multiplying vectors together.

I'll assume you mean, empty as a vacuum, in (something like) x,y,z, coordinates. (I don't use polar coordinates much, because they are in angles. ? The idea there is I don't have a simple formula for angle, only a conversion exists &/or Taylor's series or something like it.) Since it's empty (=no structure, yet) no algebra applies. Then it gets interesting. We suddenly talk about something (e.g. algebra, no longer empty space).=

 

If we have further structure, say that of an algebra then we can multiply them.

The context of the structures is important.

If you just mean vectors on R^n

n is normally something like 3 for 3D. ?

My idea was to use a scalar A that also had an exponent ^n.

In other words take a simple scaler B & root it to the n, to get A.

In that way A could be merged with R, so both could be together with 1 common exponent.

An art or kind of scaling, together, if you know what I mean?

B^(1/n)=A, B*(R^n)=(A^n)*(R^n)=(A*R)^n.

It was just a guess (no rigor) if some of that could apply?

Naturally dot product must also recognize the angles between with cos (angle) factor(s).

E.g. That syntax probably does not apply?

Swansont probably wanted to see

the vector dot product as

U*V=u*v*cos (theta)=U1*V1+U2*V2+U3*V3

for the 2 vectors U & V, with lengths u & v

& the angle theta between them

& their components 1,2,3 (for ruffly x,y,z similar).

Vector cross product as

UxV

(U2*V3-U3*V2)*i +

(U3*V1-U1*V3)*j +

(U1*V3-U2*V1)*k

which is a vector

perpendicular to the vectors U & V,

with components postscripted 1,2,3,

& length u*v*sin (theta),

unit vectors i,j,k (again for ruffly x,y,z).

then we do not have a canonical

Please define canonical in a few words please.

My mind flips to different meanings, confusing me.

Please limit the definition to what you mean.

method of multiplying vectors.

This* is one of the axioms of a vector space... you should know this.

Are you saying yes, scalar multiplied by vector gives a vector?

I simply need a bit of confirmation, to the things I heard long ago, to get started.

That's why I say assume. I seemed to have lost track, what this* was, & got a glimse later what it could be.

Edited June 7, 2016 by Capiert

[ajb]

I'll assume you mean, empty as a vacuum, in (something like) x,y,z, coordinates.

I just mean a bog standard vector space, say over the reals. Nothing to do with a vacuum. We may choose a basis or a coordinate system, depending on how we want to proceeed.

 

(I don't use polar coordinates much, because they are in angles. ? The idea there is I don't have a simple formula for angle, only a conversion exists &/or Taylor's series or something like it.) Since it's empty (=no structure, yet) no algebra applies. Then it gets interesting. We suddenly talk about something (e.g. algebra, no longer empty space).=

 

All I mean is that we have no notion, by itself, of what vu means for vectors u and v.

 

n is normally something like 3 for 3D. ?

Of course.

 

My idea was to use a scalar A that also had an exponent ^n.

You mean AAA...A n-times or something else.

 

Vector cross product as

UxV

(U2*V3-U3*V2)*i +

(U3*V1-U1*V3)*j +

(U1*V3-U2*V1)*k

which is a vector

perpendicular to the vectors U & V,

with components postscripted 1,2,3,

& length u*v*sin (theta),

unit vectors i,j,k (again for ruffly x,y,z).

But be careful ... the dot product is not canonical. Moreover, the vector product, as you have given it only works in 3D.

 

Please define canonical in a few words please.

Made with the least number of choices and when choices are made we make the obvious ones.

 

Are you saying yes, scalar multiplied by vector gives a vector?

This is one of the axioms of a vector space. I am amazed that you are asking about this. If you do not know this, then you should.

[Capiert]

 

Please allow me to show my spontaineous "reaction" here, we can discuss reasonably later. What is 4-momentum?

Loosley, in special relativity you combine the momentum with energy to build a vector.

That is (for me) crazy, it looks so backwards if combine means something like multiply instead of divide. KE=mom*va.

(Momentum mom=m*v, m=mass, v=v1-v0, va=(v0+v1)/2, v0=initial_speed, v1=final_speed.)

The basic kinetic theory of gases tells us that the temperature is the average kinetic energy as measured in the frame for which the total momentum of the gas is zero.

"Moment!" "Who" came up with that 1 on momentum? It's got to be "somebody's" .. idea (assumption).

That is obviously an error for me to see. To err is human. It doesn't work & everybody is happy with it.

It's trying to say the average momentum (+ & -) adds to zero. But does not apply for only a single particle,

otherwise it would illegally rob that particle of its momentum. "Somebody" didn't think that statement out to the end. They goofed. Or do I see things wrong?

 

So if we have only one particle then this definition has little meaning.

Correct, so we can dismiss that definition (as Strange would say), & start again to reconstruct a more accurate description.

 

I guess one can do lots of things to construct vectors, but that does not mean they all have any physical meaning.

Pity. If we're loosing meaning, then we're off the right track (for physics). Edited June 8, 2016 by Capiert

[Strange]

That is (for me) crazy, it looks so backwards if combine means something like multiply instead of divide.

 

So because you don't understand what 4-momentum is, you assume it must be wrong. This seems typical of your approach.

 

 

"Moment!" "Who" came up with that 1 on momentum? It's got to be "somebody's" .. idea (assumption).

That is obviously an error for me to see.

 

This, again, seems to be an example of "don't understand therefore wrong".

 

There is a clear connection between thermodynamics and the kinetic energy of the particles in the material. This was worked out mainly by Maxwell and Boltzmann I think (this is not a part of history of science that I have studied) and is not an "assumption".

 

 

It doesn't work

 

Please show in appropriate mathematical detail that it does not work.

 

 

Or do I see things wrong?

 

Yes. I would suggest that you stop taking your lack of knowledge as the "litmus test" for whether science is correct or not. It doesn't really matter if it makes sense to you or not.

 

 

Correct, so we can dismiss that definition (as Strange would say), & start again to reconstruct a more accurate description.

 

I think everyone agrees that the definition does not apply in the case of a single particle. So I am not sure what your point is.

[swansont]

Or do I see things wrong?

You see things wrong.

[Capiert]

So because you don't understand what 4-momentum is, you assume it must be wrong. This seems typical of your approach.

 

This, again, seems to be an example of "don't understand therefore wrong".

If you think so. ? This time you are making the conclusion from your assumption, not me.

Btw. Did I say 4-momentum is wrong? Did I say I assume 4-momentum must be wrong?

I'll assume your 1st of 3 assumptions is correct. That's 2 against you.

 

There is a clear connection between thermodynamics and the kinetic energy of the particles in the material.

I don't disagree (=I agree).

This was worked out mainly by Maxwell and Boltzmann I think (this is not a part of history of science that I have studied) and is not an "assumption".

Bolltzmann hung himself, while his family went swimming. Probably too many enemies. They didn't like his stuff (probably) as bad as you don't like mine. Maybe a lot worse.

What did Maxwell do?

 

Please show in appropriate mathematical detail that it does not work.

mom=m*v

only 1 particle in the container has mass m=6.646*10^-27 kg

& speed v=100 m/s moving right, from middle in the stationary cubic container 1 m^3.

The particle's momentum is thus

mom=m*v=6.646*10^-27 kg*100 m/s=6.646*10^-25 kg*m/s.

That is not zero momentum.

Thus the total momentum of the particle cannot be zero.

Proven.

 

Normally we prove something works;

but here for scientists we have to prove they don't work

even when they admit they don't work. (See below.)

That sounds a little crazy to me.

It hints however, the task is possible (=not impossible).

 

Even if I had 2 particles, there could be times when 1 particle could be moving verical(ly up, as example),

while the other moving mostly sideways (horizontally).

Even there in detail we cannot always expect all momentum cancels to zero.

 

Yes. I would suggest that you stop taking your lack of knowledge as the "litmus test"

for whether science is correct or not.

What? Do you mean to turn the the scientists from red to blue? (pun intended).

It doesn't really matter if it makes sense to you or not.

It matters to me. (It's a lousey job, but somebody has got to do it.) So again you are not correct. Sorry.

 

I think everyone agrees that the definition does not apply in the case of a single particle. So I am not sure what your point is.

That was my point. It does not hold for that case, thus the definition is inferior. Throw it out. Don't try to hide its mistake as an exception just to maintain the nonsense. The definition is faulty. You can't win them all. The chain is as strong as the weakest link.

If it doesn't hold, the chain breaks. I don't know why you take pride in keeping an error? It indicates the premiss (assumption) is wrong.

How valueable is that if it leeds you the wrong way (without knowing)?

[Strange]

I don't disagree (=I agree).

 

Then why claim it is wrong?

 

Bolltzmann hung himself, while his family went swimming. Probably too many enemies. They didn't like his stuff (probably) as bad as you don't like mine. Maybe a lot worse.

What did Maxwell do?

Do try and stick to the subject.

 

That is not zero momentum.

Thus the total momentum of the particle cannot be zero.

Proven.

 

Let me repeat the sentence you claim is incorrect: "The basic kinetic theory of gases tells us that the temperature is the average kinetic energy as measured in the frame for which the total momentum of the gas is zero."

 

There is a frame where the momentum of that particle is zero. There is a frame where the total momentum of two particles is zero. There is a frame where the total momentum of all the particles is zero.

 

You need to read and understand what is said before leaping to claim it is wrong (because you haven't even bothered to read it).

 

It matters to me. (It's a lousey job, but somebody has got to do it.) So again you are not correct. Sorry.

 

If it matters to you, then you need to make an effort to understand.

 

Until you either show an error in the mathematics (which despite your wild claims, you have not done) or some evidence that contradicts current theory, it doesn't matter to science.

 

 

That was my point. It does not hold for that case, thus the definition is inferior. Throw it out.

 

Don't be ridiculous. Temperature is a property of bulk materials, collections of particles. That is the definition. It doesn't apply to single particles.

 

If you want to define some other property that does apply to single particles, then go ahead. But I assume that will just be kinetic energy (in some frame of reference).

[Capiert]

Then why claim it is wrong?

Because it's incomplete, its only an exceptional observation, not the full story for a basis of analysis.

My beef is stated further below with what bugs me there. It's incomplete.

 

You cut off the quotes too short to make any sense, sometimes.

So I repeat, I agree that there is a connection between thermodynamics & the kinetic energy of the particles .. of a material. There on that statement I agree with you.

For me that is obvious. Kinetic energy influences their temperature.

You said it, a "clear" connection. =Very obvious.

 

Do try and stick to the subject.

Yes, I think that was a fair question. What did Maxwell work on there concerning thermodynamics, so I can orient? I mentioned Boltzmann because I don't think too many people know how ruthless science can be.

We're dealing with people, representing it. Boltzmann is spelt with 1 l, & 2 n's. (Gosh! I almost spelt it wrong too.)

 

Let me repeat the sentence you claim is incorrect: "The basic kinetic theory of gases tells us that the temperature is the average kinetic energy as measured in the frame for which the total momentum of the gas is zero."

 

There is a frame where the momentum of that particle is zero. There is a frame where the total momentum of two particles is zero. There is a frame where the total momentum of all the particles is zero.

 

You need to read and understand what is said before leaping to claim it is wrong (because you haven't even bothered to read it).

If you want to know what I don't like there (=disagree with) is, first it's pure theory, somebody's idea, it didn't come from thin air, & I begin to start looking for the practical real side of things.

You've stated those frames in which the total momentum is zero, but that does not happen all the time.

You can describe those exceptional circumstances, but that's not enough for me. I can't close my eyes for the other times when the total momentum is not zero. There is more to the story than just the perfect ideal situation. The (almost rare) frames when total momentum is zero is not a basis for me, because it's not the whole picture. It doesn't happen all the time. (That definition was an attempt to eliminate all other motion except random motion for heat & maybe pressure, but it didn't work, it failed, falling on its face in the very first step.) You can do your thermodynamics the way you have been taught, but it's not acceptable for me so. Sorry. It can not be used all the time, as was shown. Zero total momentum, is not always true, so generally it is false. It only has a limited theoretical application, where it is true, & so there it must stay. If you want to do more than that, then you must use something else. Are we making any progress with (some sort of) agreement between you & me? I'm interested in the bigger picture? Are you?

 

 

If it matters to you, then you need to make an effort to understand.

 

Until you either show an error in the mathematics (which despite your wild claims, you have not done)

I demonstrated the math above,

but (ashamed?) you did not quote it.

 

or some evidence that contradicts current theory, it doesn't matter to science.

Your kinetic gas hypothesis has failed from the very beginning for me.

I start with nothing, & then begin with 1 particle,

& work my way up.

That it cannot support the very 1st instance is a total collapse for me.

There is no basis of support. It's a matter of good luck that it works so long.

Pisa's leaning tower was also built on weak foundation, but it still stands!

What a mesterpiece.

 

Water's boiling temperature has been fit to

Tbc=(Pb*Vb/(mol*Rb))^(1/4), [degrees Celsius)

Pb=1 atm pressure

Vb=0.02214 m^3 volume

mol=6.0221367*10^-23 Avogadro's number

Rb=2.2711*10^-5 [Pa*(m^3)/(mol*degrees_C^4)]

 

You might be more familiar with it so

P=mol*Rb*(T^4)/V.

 

The energy

P*V=mol*Rb*(T^4)

is related to the temperature

with 4th exponent;

unlike the ideal gas.

 

Don't be ridiculous.

I'm quite serious. Consevation of energy & momentum.

I don't throw them away before the last particle,

like you want to. I "try" to be consistent.

Temperature is a property of bulk materials, collections of particles. That is the definition. It doesn't apply to single particles.

I don't see why not. The total is added, from the sum.

0+1=1 sum.

The math should be exact.

Anything else is an approximation.

 

If you want to define some other property that does apply to single particles, then go ahead. But I assume that will just be kinetic energy (in some frame of reference).

Exactly. Why reinvent the wheel, when it already exists?

KE & momentum are sufficient, enough.

I need no other property or significance that I know of yet.

Thus it will do.

Edited June 8, 2016 by Capiert

[imatfaal]

!

Moderator Note

 

I will lock this thread unless the OP starts to listen to criticism, stops asserting made-up nonsense, and tries to work within current physics.

 

Do not respond to this moderation

 

[Strange]

If you want to know what I don't like there (=disagree with) is, first it's pure theory, somebody's idea, it didn't come from thin air, & I begin to start looking for the practical real side of things.

 

As you say: it is theory which, in science, means that it is based strongly on evidence. It didn't come from thin air. It came from measurements of energy and temperature and their relationship. It is already practical.

 

You've stated those frames in which the total momentum is zero, but that does not happen all the time.

You can describe those exceptional circumstances, but that's not enough for me.

 

Such a frame is guaranteed to exist, so it is not exceptional. The use of that frame is part of the definition.

 

You can do your thermodynamics the way you have been taught, but it's not acceptable for me so. Sorry. It can not be used all the time, as was shown.

 

You have not shown any such thing.

 

Zero total momentum, is not always true, so generally it is false.

 

It is always true in one frame of reference.

 

I demonstrated the math above,

but (ashamed?) you did not quote it.

 

Because, as explained, it is irrelevant.

[Klaynos]

Can you please provide an example when there is not a frame in which the total momentum of an ensemble of massive particles is zero?

 

Your example of two particles both with orthogonal momentum, the zero total frame is moving at 45 degrees to them both relative to the original rest frame.

 

Also, when giving speeds you need to state what they're relative to.

[swansont]

 

That was my point. It does not hold for that case, thus the definition is inferior. Throw it out. Don't try to hide its mistake as an exception just to maintain the nonsense. The definition is faulty. You can't win them all. The chain is as strong as the weakest link.

If it doesn't hold, the chain breaks. I don't know why you take pride in keeping an error? It indicates the premiss (assumption) is wrong.

How valueable is that if it leeds you the wrong way (without knowing)?

 

 

Most if us deal with this by learning when rules apply by understanding the assumptions that go into the rule. That way, it's not an error. The downside is you can't blithely apply a rule or definition all circumstances, but there are none which would work that way anyway, so nothing is really lost in this approach.

[Capiert]

As you say: it is theory which, in science, means that it is based strongly on evidence.

Yes, it's based on evidence, (that's nothing really remarkable, it's rather common, we observe (often=most of, all the time) how things go, thus also get an idea, of how they don't go too. Putting formulas to them helps reassure us even more,) it doesn't have to be perfect as can be seen but it is an attempt to fit things correctly into the picture of reality. How close we get to an accurate picture is another thing, qualitatively.

It didn't come from thin air. It came from measurements of energy and temperature and their relationship. It is already practical.

Yes, to a fair degree (=mostly). But sometimes you have to iron the bugs out. Or can you say it's 100% accurate, everywhere, all the time. The devil is in the details. If you're 95% accurate, most people are happy. 60% is a bit ugly. If 1 extreme of a curve looks good but the other end or the middle doesn't, then it's back to the old drawing board, trying to work the bugs out. It bulges & heaves if the assumptions are wrong, or off. It's a real world, little seems completely (=100%) perfect. It's always better to have some sort of theory than none. But some let sure question, considering the accuracy. I think that's obvious though. We're not perfect, how can our theories be expected to be? It's obvious, we can see (=notice) trends, & we try to describe them as accurately as possible. Theory is intended to help, most of the time (it does).

 

The formula I gave you, for water's boiling temperature also has problems (T^4 vs 3D volume?).

(E.g. Why 2 different exponents ^4 vs ^3?)

I can only guess, (=suspect, & this is a "speculation" website):

Maybe concerning 0 K, although it's (my formula is) in Celsius, having a limited temperature & pressure range (that were not mentioned, i.e. critical point 373.946 C, at 217.7 atm (22.06 MPa) ref Wiki). IAPWS-95, & http://www.iapws.org/faq1/boil.html

It's such a large website, but I find the frequently asked questions, most explainitory, & easiest to follow.

Further menus are available, for depth, if you want.

Thomson calculated the earth's age to be about 1 million years old based on the temperature difference,

wrt depth down into the crust (vs cooling effects), if I remember correctly.

Didn't that premiss affect the calculation for determining 0 K?

Anyway, we all know the earth's age is much older, because of dino fossils, sediments etc.

& Lord Kelvin was no dummy, but his calculation doesn't fit our facts.

(There are lots of paradoxes in science, you or I only have to stubble on them, to find them. Those are the bugs I mean.)

That formula does show my interest in thermodynamics, since I've achieved a simpler formula fit, that is easier than your scientists have. We can squabble about the accuracy, & typos if that is all you want to do.

I used their data to help derive it.

 

Such a frame is guaranteed to exist, so it is not exceptional.

It's not so much the frame's uniqueness, as it is to "how" you want to use it for all cases that is the error in your deduction. You've already excluded, certain things, by choosing only that (kind of) frame. You can not expect me to include those (exclusions) for further applications. That would be breaking the rules.

 

The use of that frame is part of the definition.

Please don't put the cart before the horse. Your priorities are reversed. You're taking a special example (=exception case), having success with it, & then falsely believing you can use the same method for all cases, because it worked well in the exceptional case.

It's like let's look at families with boy girl twin pairs no other extra children (balanced relations for momentum analogy), we conclude they run well (kinetic energy analogy). Conclusions are fine. Now we try to apply it to any type of family with unbalanced boy girl ratios. Doesn't work. The balanced pair was the exception, not the rule. Is that too abstract?

 

You have not shown any such thing (math example).

Yes, I have shown you a math example,

1 of the simplest

with only 1 particle,

but it is "so obvious"

that you ignore it as trivial

& thus wrongly argue that I have not shown it.

You do not acknowledge my calculation, at all,

although you do acknowledge (1 particle failure) independent of me.

That is "Strange" Strange.

That's not something you do not know,

you know it already,

& reject that I remind you.

That's a mind game, against me.

 

It is always true in one frame of reference.

If you mean the special cases that you mentioned,

where total momentum is balanced to zero

& I suspect you do,

I find it difficult to believe "always".

However, to give you the benefit of the doubt,

let's say they did (even if they didn't).

You are still only dealing with a special case.

What are you going to do about other frames that don't have that momentum balance?

Ignore them?

e.g. 1 particle, &/or 2 particle orthogonal.

 

 

Because, as explained, it is irrelevant.

I'll assume you mean by "as explained":

needed effort to understand,

math error momentum was shown to be not (balanced) to zero

for an alone single particle.

There is an effort to understand,

& maintain reasonability.

I'm not biased for no reason.

I've tried to show my reasons for my decisions.

(They are no secret.)

You can attempt to change them (my conclusions) from there.

Only the anti_thesis can change a person's decision,

stuck in their ways.

E.g. You tell them what it isn't.

(But why present, what you do not trust?

Why should I either?)

Can you please provide an example when there is not a frame in which the total momentum of an ensemble of massive particles is zero?

Good (=sly) question. I never assumed a frame had zero total momentum. It's (=zero total momentum is) a nice (neat) idea,

but it's an assumption. I can't say whether the total momentum is zero or is something else, at any time, because I can't measure either, yet. The concept of using large massive particles makes the idea (task) easier. I'll have to give it some thought. But at present, I have no other idea to become an extra example; other than a room of spinning tops, or the the solar systems' spinning planets, or a glass of water from a microwave oven, before the water molecules slow down spinning, from banging into each other, producing random linear motion heat temperature speed.

 

Your example of two particles both with orthogonal momentum, the zero total frame* is moving at 45 degrees to them both relative to the original rest frame.

I'm sorry, I didn't quite follow that*. Maybe I missed a word there?

Are you saying you derive a common frame which is zero speed relative to all particles, in the container?

In that (case), can then, that common frame be moving relative to the stationary container,

& as you said be moving at 45 degrees to both particles?

That's an interesting concept, but it tells me the net momentum in the container is not zero, it is the sum of those 2 particles,

in other words that common frame itself (is the total momentum) (relative to the stationary container).

1st, I consider the earth as the rest (reference) frame, unless otherwise stated.

That's obvious, & redundant (rebose) for me (to state).

Maybe you scientists have other problems I don't know about,

but I'm happy to have my feet on the earth (as ref)

& keep the language simple, if possible,

because it (often) gets too complicated & long.

E.g. If I'm moving in a car, we have a different story,

but if we have a beaker or container sitting in the lab,

what's the bother?

 

Also, when giving speeds you need to state what they're relative to.

Dito, above. Relative to earth('s surface as ref) unless otherwise stated.

I like to think simply, not complicated like you.

(Whether I succede, being so simple is another question. I'm informal.)

Is that ok?

Most of us deal with this by learning when rules apply by understanding the assumptions that go into the rule. That way, it's not an error. The downside is you can't blithely apply a rule or definition all circumstances, but there are none which would work that way anyway, so nothing is really lost in this approach.

 

That's not a bad arguement. I mean we try to do our best, & improve where we can, if possible (rare).

E.g. Zero (e.g learning) to hero (e.g. all), but maybe the gray (steps in) between them too.

Edited June 9, 2016 by Capiert

[Klaynos]

You seem to miss understand momentum. It is not a fundamental property, but a property of the system that depends on the frame in which it's measured.

 

In any situation you can sum the vectors of the linear momentum of the particles in an ensemble to give a total momentum. Your zero momentum frame would be the frame in which there is zero. This frame comoving with the ensemble in the lab frame.

 

I'm frankly agast you can state something is wrong without any basic understanding of what you're talking about. The lack of integrity is astounding, but not uncommon.

[Strange]

I think this is beyond even Dunning and Kruger's wildest imaginings.

[imatfaal]

!

Moderator Note

Thread locked - failed to comply with moderation. In essence - you need to get a grounding in science before you start asserting that bits of it are wrong.