[Mr. Scientist]

Oh yoron, would you like my explanation, which is no more correct than his, but might help yo?

[yor_on]

"Here you are mixing at least three different concepts: invariant mass in special relativity, gravitational mass and general relativity."

Well thank you Mr Lightarrow.
Is it ok if I lay down to die now?

Awhh, just joking:)
I had hoped that the definition of 'invariant' would be more intuitive to me.
But it's not.

Yet it is thought as giving a better definition of 'mass' if I understood the intention rightly.
But don't give up on me yet Lightarrow, I will try to understand.

(Reducible, as I was thinking before, was to me relating to some smallest constituent of, and defining primary, 'matter' as invariant in all frames.)
Embarrassingly naive, as I now find out.

This coin, if rotating.
Won't its highest 'gravity'(angular momentum?) be at its rim?
I know, mixing again:)

It's just that I would have wanted gravity and mass to be a simpler relation.
Highly subjective of me I admit.



And...

Yes Mr S, if you please:)
Go for it, if you can pound some more sense in my head?
Lightarrow have tried:)

And partly succeded.
I know as I now have a headache..
But that can only become better

( bigger?

[lightarrow]

This coin, if rotating.
Won't its highest 'gravity'(angular momentum?) be at its rim?

If you treat it as a rigid body, then you can talk of the entire system's angular momentum, so you just have a unique one. If the system is not rigid, or you want to consider just a small portion of it (e.g. a single particle) then, yes, the angular momentum is higher for those particles at greater distance from the centre. However, it's not angular momentum which makes the mass in this case, but rotational kinetic energy and elastic potential energy of the stretched bonds between the particles (because of centripetal force). Usually the second one is neglected in ordinary (non-relativistic and however for not very high angular velocities ω) problems, so you are left with rotational kinetic energy. For a rigid body this energy is:
Er = (1/2)Iω²
where I = moment of inertia (for an homogeneus disk I = MR²/2, M = disk's mass, R = disk's radius).

[yor_on]

Awh.
Lightarrow you instill ever new definitions for me.
'Rigid body' you say?

(Would that have anything to do with dreaming of Christina Aguilera:)

But I like it very much.
Exact definitions help me see how things are defined.
So now I will have to read up on that.

----------

"but rotational kinetic energy and elastic potential energy of the stretched bonds between the particles (because of centripetal force)."

Nice.

'Rotational kinetic energy'.
How does that differ from 'angular momentum'?
Is it in the geometric direction of 'force'.

[lightarrow]

Awh.
Lightarrow you instill ever new definitions for me.
'Rigid body' you say?

(Would that have anything to do with dreaming of Christina Aguilera:)

ROTFL  []

But I like it very much.
Exact definitions help me see how things are defined.
So now I will have to read up on that.

----------

"but rotational kinetic energy and elastic potential energy of the stretched bonds between the particles (because of centripetal force)."

Nice.

'Rotational kinetic energy'.
How does that differ from 'angular momentum'?
Is it in the geometric direction of 'force'.

They are different concepts, even because they have different units and one is a scalar while the other is a (pseudo) vector.
Given a system of N particles, you have:

Angular momentum (pseudo vector) = K = ∑_i r_i Λ m_iv_i          units: Joule*second
r_i = position vector of the i-esim particle
m_i = mass of the i-esim particle
v_i  = velocity of the i-esim particle
Λ = vectorial product

Rotational kinetic energy (scalar) = Er = ∑_i (1/2)m_i(v_iG)²         units: Joule
where v_iG is the velocity of the i-esim particle with respect to the centre of mass G. For a rigid body such velocities v_iG are all perpendicular to the vector radius from the axis of rotation; for this reason it could be demonstrated that
Er = (1/2)Iω²

Edit: Indicating with K_u the component of angular momentum along the axis of rotation, the following equation is valid:

K_u = Iω

so you also have:

Er = (1/2)K_uω

[yor_on]

Thanks Lightarrow.
I will ask you more things, I'm sure.
But first I will need to assimilate what you already shown me.

Considering the average density of my head.
Quite thick I'm afraid.
It will take me some time:)

But, as they say in sunny California...
"I'll be back"

--------Quote-------

They are different concepts, even because they have different units and one is a scalar while the other is a (pseudo) vector.
Given a system of N particles, you have:

Angular momentum (pseudo vector) = K = ∑i ri Λ mivi          units: Joule*second
ri = position vector of the i-esim particle
mi = mass of the i-esim particle
vi  = velocity of the i-esim particle
Λ = vectorial product

Rotational kinetic energy (scalar) = Er = ∑i (1/2)mi(viG)2         units: Joule
where viG is the velocity of the i-esim particle with respect to the centre of mass G. For a rigid body such velocities viG are all perpendicular to the vector radius from the axis of rotation; for this reason it could be demonstrated that
Er = (1/2)Iω2

--------------------End - Quote --


Yes, I can see that one is scalar and that the other describing the whole 'system / coin' of scalar 'bonds' will be seen as a vector (magnitude and direction)

Is there anywhere one can download the most common expressions of 'symbolic' expressions and their definitions?
Or do they change depending on what you use them for?

Often when I see those mathematical symbolic formulas they don't get explained.
And sometimes they don't seem to be bound to any 'definite description'

So doing like you do makes me want to put numbers into your formulas to see if I really understands them.
And that's why I will need to go slow.
This is a new language for me.

I've done some boolean logic, Computers and Philosophy, but that is the nearest I've ever cometh
So I do enjoy the care you take in explaining your concepts and math here.

Cheers.
yoron.

[lightarrow]

If you google rotational kinetic energy you find many links, e.g.:
http://theory.uwinnipeg.ca/physics/rot/node6.html

About vectorial product, the rule is the following. Given the two vectors:

A = (Ax,Ay,Az)

B = (Bx,By,Bz)

then:

AΛB = (AyBz-AzBy,AzBx-AxBz,AxBy-AyBx)

[yor_on]

What?

Only joking:)

Thanks..
And I'm goggling.
Or should it be ogling?

:)

[lightarrow]

What?

Only joking:)

Thanks..
And I'm goggling.
Or should it be ogling?

Actually, there is a little picture which helps remembering that rule: write the symbols x, y, z in this order, clockwise in a circle. So you understand that the x component of the vectorial product is given by the product of the y component of the first vector by the z one of the second (from y to z you go clockwise) minus the z component of the first by the y one of the second (because now you go anti-clockwise): AyBz-AzBy.
The same rule for the other components.
 [ Invalid Attachment ]

[yor_on]

Ahem, perfectly understandable.
With some thought?
perhaps?

And yes, we are talking density here.
(between my ears:)

-----

The first one was easier on my mind than this second )
But that may have to do with my English?

There is only one conclusion can I draw after reading your explanation over, some times.
You are right no matter what sense I will make.

I will never forget that little picture:)