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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