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Natural Science Forum / Physics / Relativity / October 2005



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Matrix Mass

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WaiteDavid137@yahoo.com - 27 Oct 2005 18:11 GMT
In the relativity faq it is discussed briefly how one can relate
ordinary force to coordinate acceleration by a matrix mass
f = Ma.
For those interested in what the full matrix looks like with f and a
extended to four-components(not four-vectors) see:
http://s2.phpbbforfree.com/forums/modernscience-about22.html
The fourth element I added to each is
a^0 = 0,
and
f^0 = (1/c)(dE_R/dt)
otherwise they are the ordinary force and coordinate acceleration, not
the four-vector force or four-vector acceleration.
Bilge - 28 Oct 2005 12:20 GMT
WaiteDavid137@yahoo.com:
>In the relativity faq it is discussed briefly how one can relate
>ordinary force to coordinate acceleration by a matrix mass
[quoted text clipped - 8 lines]
>otherwise they are the ordinary force and coordinate acceleration, not
>the four-vector force or four-vector acceleration.

 Isn't it rather specious to call that a mass matrix, since the same
factor of m appears in every element and the remainder is just a funny
way of writing the lorentz transform of an acceleration without the
acceleration? It's also not very clear what your equation, f = m_i a,
means, since you have f and a designated as vectors, and m_i is a
matrix. I literally read that in component form as f_k = m_ij a_k,
which doesn't make sense.
WaiteDavid137@yahoo.com - 28 Oct 2005 16:06 GMT
The I subscript is for inertia, not for an element. I didn't intend for
the equation to be in element notation. The equation
f = Ma
as I had inteded it means that the ordinary force as a vector is equal
to the Mass as a matrix times the acceleration as a vector. There is
nothing wrong with doing this and I think it was Steve Carlip that I
first heard suggest that it could be done. An addendum was added to the
relativity faq demonstrating that it could be done but he didn't work
out what the Matrix actually looked like in terms of its components.
Someone asked me what the matrix looked like so I wrote it out. The
only difference between what I wrote out and what he suggested could be
done was that he used it to say three component force is 3X3 matrix M
multiplying three component acceleration whereas to be more general I
extended the definitions of ordinary force and coordinate acceleration
to include a fourth element(not four-vectors). So my equation reads
four component ordinary force is 4X4 matrix M mutltiplying four
component acceleration. To visualise the product take the force and
acceleration to be columb vectors. It works out fine. In element
notation his would read
f^i = M^i_J*a^J
where the Matrix M elements are M^i_J
I didn't want to do that because someone will confuse f and a with
four-vector force F and four-vector acceleration A, but in element
notation mine would read
f^mu = M^mu_nu*a^nu
Tom Roberts - 29 Oct 2005 01:11 GMT
> f = Ma.
> For those interested in what the full matrix [M] looks like with f and a
> extended to four-components(not four-vectors) see:
> http://s2.phpbbforfree.com/forums/modernscience-about22.html
> [...]

Sure, one can do that. One could also take the cube of your equation, or
the 14th root, or.... Mathemtical masturbation is sterile.

    Actually, it is worse than sterile, as this sort of thing
    HIDES the underlying physical phenomena in a fog of symbols.

Most physicists prefer that the symbols in their equations correspond to
physical quantities in the world being modeled (rather than being
arbitrary mathematical combinations of such quantities). An incredibly
remarkable side effect of this is that the equations usually turn out to
be simpler and more elegant[#]. That is, at base, what dictates that the
mass m in our equations be an invariant, and that force, velocity, and
acceleration all be 4-vectors, etc.

    [#] Compare "F = DP/d\tau" to your complicated mess.

I think that calling your M a "matrix mass" is an unacceptable PUN on
the word "mass" -- it is virtually unrelated to the original meaning of
the word as "amount of stuff".

bilge wrote:
>   Isn't it rather specious to call that a mass matrix, since the same
> factor of m appears in every element and the remainder is just a funny
[quoted text clipped - 3 lines]
> matrix. I literally read that in component form as f_k = m_ij a_k,
> which doesn't make sense.

Yes, I think it is indeed quite specious.

But his equation f = M_i a, written in components is:
    f_k = (M_i)_k^j a_j
The "i" in his "M_i" is merely a label (meaning "inertial", I believe),
not an index; he wrote in vector notation, not component notation.

Tom Roberts    tjroberts@lucent.com
WaiteDavid137@yahoo.com - 29 Oct 2005 02:45 GMT
Well then perhaps you should explain to Steve Carlip why you think the
addendum should be removed. All I am saying is that someone wanted to
know what the matrix referred to in it looked like once written out and
so I am showing it to people at the link. I called it matrix mass
specifically to draw attention to the post. I see it did that, good.
Now if you'll visit the link you'll see that I don't hide the
underlying physics, nor do I think of this matrix as the mass. If
you'll read it you will see that I start out with a four-vector
equation where the mass m is an invariant and then from that work out
the relation between ordinary force and coordinate acceleration and
then mention that this result can be written as a matrix equation and
write out the matrix M. It isn't a fog of symbols, I work it out step
by step explicitly defining each term as I go. If you'll read it, you
will see that I think of the invariant m as the mass, not the matrix M.
I didn't think the i sucscript would be confused for an idex because
the rest of the equation was obviously not in element notation. I just
meant for it to mean inertia.
Ken S. Tucker - 29 Oct 2005 20:38 GMT
> In the relativity faq it is discussed briefly how one can relate
> ordinary force to coordinate acceleration by a matrix mass
[quoted text clipped - 4 lines]
> The fourth element I added to each is
> a^0 = 0,

I'm worried about relative clock rates.

You set a^0 == d^2 x^0 /dt^2 = 0,

(x^0 and t are acculumated on separate clocks)

thenthat means dx^0 = dt, how do the clocks
come to differ?
Ken
WaiteDavid137@yahoo.com - 29 Oct 2005 21:04 GMT
There aren't two different clocks for a. It isn't four-vector
acceleration.
a^0 = d^2x^2/dt^2
a^0 = (d/dt)(dx^0/dt)
a^0 = (d/dt)(dct/dt)
a^0 = (d/dt)(cdt/dt)
a^0 = (d/dt)(c)
a^0 = 0
Those wanting details:
http://s2.phpbbforfree.com/forums/modernscience-about22.html
Ken S. Tucker - 29 Oct 2005 21:54 GMT
> There aren't two different clocks for a. It isn't four-vector
> acceleration.
[quoted text clipped - 6 lines]
> Those wanting details:
> http://s2.phpbbforfree.com/forums/modernscience-about22.html

Well you specified a^1...a^3 =/=0 meaning two
relatively accelerating CS's is that correct?

Use the clock in CS K' to have time t' and
the clock in K to have time t. Then

t' =/= t, dt' =/= dt and

d^2t'/dt^2 =/=0 if the clocks are relatively
accelerating.

Anyway, you'll need to explain "time dilation",
in your own terms to make your argument sound.
Regards
Ken S. Tucker
WaiteDavid137@yahoo.com - 29 Oct 2005 22:02 GMT
No, not correct. Only one coortinate system is used to define a. It is
NOT four-vector velocity.
My arguement is sound, like it or not. There is not t' that enters into
it anywhere. Go look at how it is defined. I outright stated that it
was coordinate acceleration and that it was not four-vector
acceleration and gave the definition for it. Go read it.
http://s2.phpbbforfree.com/forums/modernscience-about22.html
Ken S. Tucker - 29 Oct 2005 23:19 GMT
> No, not correct. Only one coortinate system is used to define a. It is
> NOT four-vector velocity.
[quoted text clipped - 3 lines]
> acceleration and gave the definition for it. Go read it.
> http://s2.phpbbforfree.com/forums/modernscience-about22.html

Are we in GR here?
Are your accelerations absolute or relative?
Ken
Ken S. Tucker - 30 Oct 2005 00:01 GMT
> No, not correct. Only one coortinate system is used to define a. It is
> NOT four-vector velocity.
[quoted text clipped - 3 lines]
> acceleration and gave the definition for it. Go read it.
> http://s2.phpbbforfree.com/forums/modernscience-about22.html

Well how do you get the a^1= d^2x/dt^2,
without a reference.
Do you use an accelometer?
WaiteDavid137@yahoo.com - 30 Oct 2005 03:41 GMT
Use a radar gun for all I care.
Ken S. Tucker - 30 Oct 2005 08:42 GMT
> Use a radar gun for all I care.

Well a radar gun would need 2 CS's, but you
have specified one, that's why I presumed
you are using accelometer readings within
a single FoR. If so then a^0 is consistent
with a^i <>0, but then other parts of the
article become unclear, but fixable.
Have you considered the Christoffel's?
Ken
WaiteDavid137@yahoo.com - 30 Oct 2005 23:09 GMT
>Well a radar gun would need 2 CS's

B.S.

>Have you considered the Christoffel's?

How many times do I have to tell you, it is NOT four-vector
acceleration!
Ken S. Tucker - 30 Oct 2005 23:40 GMT
> >Well a radar gun would need 2 CS's
>
[quoted text clipped - 4 lines]
> How many times do I have to tell you, it is NOT four-vector
> acceleration!

Ok so GR does not apply, and SR doesn't, so
what EXACTLY did you try to do?
What courses have you taken in college?
Ken
Dirk Van de moortel - 30 Oct 2005 23:44 GMT
> > >Well a radar gun would need 2 CS's
> >
[quoted text clipped - 8 lines]
> what EXACTLY did you try to do?
>  What courses have you taken in college?

David should have taken a course on Applied Village
Idiot Psychology, where they teach you to kick an
idiot once - real hard - and then turn your back.

Dirk Vdm
WaiteDavid137@yahoo.com - 31 Oct 2005 00:16 GMT
Yeah apparently so, lol. At this point, I think I'll just let him stay
confused.
Ken S. Tucker - 31 Oct 2005 01:12 GMT
> Yeah apparently so, lol. At this point, I think I'll just let him stay
> confused.

Well fine...evidence indicates,
*David Waite SUCKS Dirk VD's cock*

Guess our group can figure out a pair a queenies
since (Waite & VD) self-outed!

I'll need to dismiss Waite's analysis since he
is using his homosexual involvement with VD
obviously intimately involved with the Dirk VD.

I find CASE DISMISSED..........

David Waite is clearly a homosexual and has
a greater interest in his sex life than any
thing to do with with real physics, therefore
his queer testimony is questionable.

Judged honestly by...
Ken S. Tucker
Peter M. Brown - 31 Oct 2005 01:58 GMT
> > Yeah apparently so, lol. At this point, I think I'll just let him stay
> > confused.
>
> Well fine...evidence indicates,
> *David Waite SUCKS Dirk VD's cock*

Wow! What brought all this anger on Ken? Fill me in please. What is it
you're seeking to find out? Is this an SR or GR question? I'm too far behind
to start reading from scratch and I don't read David's posts for obvious
reasons i.e. I don't want to argue with him. Nothing against David regarding
that decision. I'm just too sick to argue anymore. Its not within me
anymore.

> Guess our group can figure out a pair a queenies
> since (Waite & VD) self-outed!
[quoted text clipped - 12 lines]
> Judged honestly by...
> Ken S. Tucker

Now that's what I call and angry Ken. Splain please? :)

Pete
Ken S. Tucker - 31 Oct 2005 12:11 GMT
> > > Yeah apparently so, lol. At this point, I think I'll just let him stay
> > > confused.
[quoted text clipped - 8 lines]
> that decision. I'm just too sick to argue anymore. Its not within me
> anymore.

Hey Pete, can I send my email address to you?

Waite posted an essay 95% good, 5% ambiguous.
When I questioned Waite for the intend of
clarification, Waite couldn't do it.
Then Dirk VD and Roberts have decided to suck
Waites cock, in place of the real physics that's
lacking. It's a typical 3 stooge fest, it's
predictable in this group from the adulles.
The downside is Waites physics is now f.cked
and is likely lost, because no one will correct
his malformed essay, certainly his 2 stooges
can't, so it's down the tubes...lost.

> > Guess our group can figure out a pair a queenies
> > since (Waite & VD) self-outed!
[quoted text clipped - 14 lines]
>
> Now that's what I call and angry Ken. Splain please? :)

Set up an emotional field to confine 3 quarks
(David,Tom,VD) to create a baryon of emotional
dependancy...it worked like a charm predictably,
their fused like a unit.
Those SOB's hate me more than the enemy, now
that's discipline! The down side is they love
to love instead of fight.
Ken
Tom Roberts - 31 Oct 2005 06:10 GMT
> Yeah apparently so, lol. At this point, I think I'll just let him stay
> confused.

You don't have any say in the matter -- he _will_ stay confused
regardless of what you do or don't do. <shrug>

Unfortunately, that remark applies to all too many people around here.

Ken S. Tucker wrote:
> [... trash]

Not only is he confused, he is also outrageously offensive in his writing.

Tom Roberts    tjroberts@lucent.com
Bilge - 31 Oct 2005 07:33 GMT
WaiteDavid137@yahoo.com:
>Yeah apparently so, lol. At this point, I think I'll just let him stay
>confused.

 That's sort of like letting einstein stay dead. Even if you made
the attempt to ressurect him, you won't succeed and you won't get any
help from einstein no matter how much effort you spend trying to convince
him that it's to his advantage to not be dead.
Sue... - 30 Oct 2005 09:54 GMT
> In the relativity faq it is discussed briefly how one can relate
> ordinary force to coordinate acceleration by a matrix mass
[quoted text clipped - 8 lines]
> otherwise they are the ordinary force and coordinate acceleration, not
> the four-vector force or four-vector acceleration.

<< These notes fill in some of the steps needed to
calculate the fields and instantaneous power radiated
by an accelerated charge. All results are fully
relatvistic except when explicitly stated that the
limit is for particle velocity much less than the
speed of light. >> --K.E. Schmidt
http://fermi.la.asu.edu/PHY531/larmor/index.html

Sue...
 
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