| Topic: |
Science > Physics |
| User: |
"Jeany" |
| Date: |
29 Jan 2005 07:51:40 AM |
| Object: |
Who will stun the world as next Einstein? |
In article <3a9ce32c.0106230251.743f15f3@posting.google.com>, cybermind
<cybermind@163.com> writes:
Many Chinese people believe Dr. Cui Silong, Leifei Institute of New
Energy, is stunning the world to with his new physics theory
Analytical Space-Time, which helps develop most powerful
electromagnetic technics. The new theory fundamentally unified general
relativity and quantum mechanics seamlessly with principle of a string
and deflection of space-time. Professor Chuanjian commented it in a
special physics review: Now that the new theory has allowed us to see
the aureole of God through the general effect of deflection or
rotation of space-time, would it lead us to see the true face of God
or comprehend the mind of God?
You can refer to the new theory at: http://www.tastphysics.com
or http://cuisilong.itgo.com
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| User: "Franz Heymann" |
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| Title: Re: Who will stun the world as next Einstein? |
29 Jan 2005 09:31:40 AM |
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"Jeany" <ctiei@hotmail.com> wrote in message
news:b6843a93.0501290551.5bacd0ab@posting.google.com...
In article <3a9ce32c.0106230251.743f15f3@posting.google.com>,
cybermind
<cybermind@163.com> writes:
Many Chinese people believe Dr. Cui Silong, Leifei Institute of New
Energy, is stunning the world to with his new physics theory
Analytical Space-Time, which helps develop most powerful
electromagnetic technics. The new theory fundamentally unified
general
relativity and quantum mechanics seamlessly with principle of a
string
and deflection of space-time. Professor Chuanjian commented it in a
special physics review: Now that the new theory has allowed us to
see
the aureole of God through the general effect of deflection or
rotation of space-time, would it lead us to see the true face of
God
or comprehend the mind of God?
Balls.
I'm afraid your hero seems to be talkoing crap.
I only read as far as the first diagram, where he shows a total lack
of undestanding of the Lorentz transform.
Provided that the event described in the LH diagram lies within the
light cone of the observer in the LH frame, there must exist a frame
in which the area he talks about is zero.
So much for the invariance of that area of which he speaks.
There is no need to pursue reading that paper any further.
Franz
You can refer to the new theory at: http://www.tastphysics.com
or http://cuisilong.itgo.com
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| User: "Bill Hobba" |
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| Title: Re: Who will stun the world as next Einstein? |
29 Jan 2005 05:26:24 PM |
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"Jeany" <ctiei@hotmail.com> wrote in message
news:b6843a93.0501290551.5bacd0ab@posting.google.com...
In article <3a9ce32c.0106230251.743f15f3@posting.google.com>, cybermind
<cybermind@163.com> writes:
Many Chinese people believe Dr. Cui Silong, Leifei Institute of New
Energy, is stunning the world to with his new physics theory
Analytical Space-Time, which helps develop most powerful
electromagnetic technics. The new theory fundamentally unified general
relativity and quantum mechanics seamlessly with principle of a string
and deflection of space-time. Professor Chuanjian commented it in a
special physics review: Now that the new theory has allowed us to see
the aureole of God through the general effect of deflection or
rotation of space-time, would it lead us to see the true face of God
or comprehend the mind of God?
You can refer to the new theory at: http://www.tastphysics.com
or http://cuisilong.itgo.com
It is junk of the first order. Consider the supposed important questions
whose answers lie beyond relativity -
1. What is the physical meaning of the contraction factor in STR?
It represents a hyperbolic rotation.
2. In Lorentz transformation, on what basis is the judgement of y = y' and z
= z' (y and z are orthogonal to relative velocity)?
See
http://arxiv.org/abs/physics/0110076,
3. Is there possibly another theoretical explanation for red shift?
That another theory may exist and explain a phenomena is true of any theory.
4. Whether do observers of two different coordinates get the same result in
describing relative velocity, or not?
Obviously they do not - eg consider a frame in standard configuration. In
one frame the velocity is in the x direction in the other in the -x
direction.
5. Mercury orbital perturbation is due to its movement along a shortest
route (geodesic) on bent space, then, how does gravitation make space bent?
What is the essential reason for the phenomenon?
The essential reason for space time curvature is that gravity is space-time
curvature.
It is such obvious tripe I could not bring myself to address the rest.
Bill
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| User: "Jesse Mazer" |
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| Title: Re: Who will stun the world as next Einstein? |
29 Jan 2005 08:01:14 PM |
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Bill Hobba wrote:
5. Mercury orbital perturbation is due to its movement along a shortest
route (geodesic) on bent space, then, how does gravitation make space bent?
What is the essential reason for the phenomenon?
The essential reason for space time curvature is that gravity is space-time
curvature.
Also, he gets the basic notion of a geodesic in general relativity
wrong--Mercury's orbit is not the path through curved space with the
shortest length, it is the path through curved spacetime with the
*greatest* proper time.
Jesse
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| User: "Bill Hobba" |
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| Title: Re: Who will stun the world as next Einstein? |
29 Jan 2005 10:57:01 PM |
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"Jesse Mazer" <vze2ztqw@mail.verizon.net> wrote in message
news:41FC69E2.3020705@mail.verizon.net...
Bill Hobba wrote:
5. Mercury orbital perturbation is due to its movement along a shortest
route (geodesic) on bent space, then, how does gravitation make space
bent?
What is the essential reason for the phenomenon?
The essential reason for space time curvature is that gravity is
space-time
curvature.
Also, he gets the basic notion of a geodesic in general relativity
wrong--Mercury's orbit is not the path through curved space with the
shortest length, it is the path through curved spacetime with the
*greatest* proper time.
I did not even read that far. But that is a very common crank
misconception. You will find some cranks even say you do not understand
Einstein when is pointed out to them. I even have given them the following
link where it is explained in detail -
http://www.eftaylor.com/pub/FmaAJPguest5.pdf. They even have the gall to
say Taylor (that is the Taylor of Taylor and Wheeler) does not understand
relativity. It never seems to occur to cranks what is so obvious to anyone
that has examined the issue - they basically do not understand the theory
they are criticizing.
Thanks
Bill
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| User: "Koobee Wublee" |
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| Title: Re: Who will stun the world as next Einstein? |
30 Jan 2005 12:40:10 AM |
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"Bill Hobba" <bhobba@rubbish.net.au> wrote in message
news:xOZKd.140842$K7.43402@news-server.bigpond.net.au...
I did not even read that far. But that is a very common crank
misconception. You will find some cranks even say you do not understand
Einstein when is pointed out to them. I even have given them the
following
link where it is explained in detail -
http://www.eftaylor.com/pub/FmaAJPguest5.pdf. They even have the gall to
say Taylor (that is the Taylor of Taylor and Wheeler) does not understand
relativity. It never seems to occur to cranks what is so obvious to
anyone
that has examined the issue - they basically do not understand the theory
they are criticizing.
You still have not explained why the following equation would indicate
gravity as a repulsive force.
D(dE/dv)/Dt - dE/dr = 0, where
* D = total derivative
* d = partial
* E = Lagrangian
* v = Dr/Dt
With
E = m v^2 / 2 - G M m / r
We have
* dE/dv = m v
* dE/dr = G M m / r^2
So,
D(m v)/Dt - G M m / r^2 = 0, or
m dv/dt = G M m / r^2
Thus, that equation should be
D(dE/dv)/Dt + dE/dr = 0
Only it should be in special conditions where the system only has
* Only one state variable, q(s), in the Lagrangian, L(Dq/Ds, q), or
* All state variables q(s)_i are independent of each other
[From now on, I am using 'd' both as total derivatives and partial
derivatives.]
That means the geodesic equation of motion has to be scrutinized further.
We have the following simplified spacetime equation with Swarzschild metric
ds^2 = c^2 (1 - 2 U) dt^2 - dr^2 / (1 - 2 U) - r^2 dH^2, where
* U = G M / c^2 / r
The derivation of choice is to locate the follwing state variables, q(s)_i
* dt/ds and t
* dr/ds and r
* dH/ds and H
And perform the Lagrangian operation mentioned above.
However, only dt/ds and t are independent from the others. Thus, only the
following applies.
d(dL/d(dt/ds))/ds + dL/dt = 0
Notice it does not matter if the sign on the 2nd is positive or negative in
this case because dL/dt = 0. The other equation fondly used to establish
the advance of Mercury's orbit is
d(dL/d(dH/ds))/ds + dL/dH = 0
Where the conservation of angular momentum is explained away through by this
equation. However, when one uses this equation above, one should not play
with dr/dH later on. Amazingly, the advance of Mercury's orbit is done with
dr/dH on the assumption that r and H are independent of each other.
Now, how does one calculate the advance of Mercury's orbit? To me, if we
can observe anything we should be able to write down the observed energy
even if the observed event takes place in curved spacetime. So, I am going
to divide the spacetime equation by dt instead of ds because dr/dt and r
dH/dt are the observed radial speed and the observed tangential speed. We
get
0 = c^2 - (dr/dt)^2 / (1 - 2 U)^2 - r^2 (dH/dt)^2 / (1 - 2 U) - (ds/dt)^2 /
(1 - 2 U) = Lagrangian, L
So,
d(dL/d(ds/dt))/dt + dL/ds = 0
Yields
ds/dt = (m^2 c^6 / E^2) (1 - 2 U), where
* m^2 c^6 / E^2 = integration constant
Plugging it back to the spacetime equation, we get
(m^2 c^6 / E^2) (1 - 2 U) = c^2 - (dr/dt)^2 / (1 - 2 U)^2 - r^2 (dH/dt)^2 /
(1 - 2 U)
The right hand side of the equation above is very simlar to the gamma term
in Lorentz Transformations except the radial speed is scaled with (1 - 2
U)^2 and the tangential speed is done with (1 - 2 U). Under strong
gravitational influence where 2 U is very close to 1, all the observed
speeds slow down as they should be under General Relativity.
After re-arranging the equation above, we get
E = m c^2 sqrt(1 - 2 U) / sqrt(1 - (dr/dt)^2 / (1 - 2 U)^2 / c^2 - r^2
(dH/dt)^2 / (1 - 2 U) / c^2), where
* E = observed energy
* m = rest mass in flat space
* E / c^2 = observed mass in spacetime
Applying the conservation of observed energy, we have
dE/dt = 0, total derivative
With the following constraint applied to the equation above,
* r^2 dH/dt d^2(H)/dr/dt = - 2 r (dH/dt)^2
* d^2(r)/dt^2 = 0, in near circular orbit
Where the 1st equation describes the conservation of angular momentum which
can be derived through simple vector derivatives in polar coordinate and
should be found readily available in the internet. If not, I can post the
derivation of it. The, we get
B^2 = U (1 + 1.5 U), where
* B = r dH/dt / c, with dr/dt = 0
* In every revolution, the orbit is advanced by 1.5 U not 3 U as we
observed.
I may have made a mistake in my calculation. If so, I will post a follow
up.
On a scale of 1 to 10, where do I rank in the crankiness?
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| User: "Bill Hobba" |
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| Title: Re: Who will stun the world as next Einstein? |
30 Jan 2005 06:31:49 AM |
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Koobee Wublee
You still have not explained why the following equation would indicate
gravity as a repulsive force.
m dv/dt = G M m / r^2
BTW still equating vectors to scalars I see. Learn some basic math then
repost.
Rest of junk mercifully snipped - well I actually did not read it to see if
it is junk but making basic errors such as equating vectors to scalars does
not inspire confidence.
Bill
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| User: "G=EMC^2 Glazier" |
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| Title: Re: Who will stun the world as next Einstein? |
30 Jan 2005 06:44:45 PM |
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ME
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| User: "hanson" |
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| Title: Re: Who will stun the world as next Einstein? |
30 Jan 2005 08:12:43 PM |
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"G=EMC^2 Glazier" <herbertglazier@webtv.net> wrote in message
news:3468-41FD7F7D-95@storefull-3174.bay.webtv.net...
ME
Yeap, that would *really stun* the world, Herbie... ahahaha
ahaha..... ahahanson
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| User: "John C. Polasek" |
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| Title: Re: Who will stun the world as next Einstein? |
30 Jan 2005 01:02:56 PM |
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On Sun, 30 Jan 2005 12:31:49 GMT, "Bill Hobba" <bhobba@rubbish.net.au>
wrote:
Koobee Wublee
You still have not explained why the following equation would indicate
gravity as a repulsive force.
m dv/dt = G M m / r^2
BTW still equating vectors to scalars I see. Learn some basic math then
repost.
Rest of junk mercifully snipped - well I actually did not read it to see if
it is junk but making basic errors such as equating vectors to scalars does
not inspire confidence.
Bill
Bill you are all wet. Both terms are vectors, the RHS sometimes
expressed as G M m r/r^3, quite obviously a radial vector. A cheap
shot where none is warranted.
John Polasek
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| User: "Bill Hobba" |
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| Title: Re: Who will stun the world as next Einstein? |
30 Jan 2005 04:29:08 PM |
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"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:dnbqv0drurl9hhcal20b375apulm5rijds@4ax.com...
On Sun, 30 Jan 2005 12:31:49 GMT, "Bill Hobba" <bhobba@rubbish.net.au>
wrote:
Koobee Wublee
You still have not explained why the following equation would indicate
gravity as a repulsive force.
m dv/dt = G M m / r^2
BTW still equating vectors to scalars I see. Learn some basic math then
repost.
Rest of junk mercifully snipped - well I actually did not read it to see
if
it is junk but making basic errors such as equating vectors to scalars
does
not inspire confidence.
Bill
Bill you are all wet. Both terms are vectors, the RHS sometimes
expressed as G M m r/r^3, quite obviously a radial vector. A cheap
shot where none is warranted.
What can I say - G M m r/r^3 (where r - usually written as rhat - rhat
being a vector of length r in the direction of the line joining the particle
m and M) is not he same as G M m / r^2. rhat/r^3 does not give 1/r^2. That
is assuming that is what you mean - not using standard terminology is a well
known crank standby ranking up there with Koobee Wublee's technique of
making wild silly claims then when such is pointed out simply accusing the
poster of not knowing what they are talking about. However in this case the
facts are far too wall known for it to work. You like Koobee Wublee should
learn some basic vector analysis. But knowledge of basic math or even basic
physics history is not something Mr Dual Space has ever demonstrated.
Bill
John Polasek
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| User: "John C. Polasek" |
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| Title: Re: Who will stun the world as next Einstein? |
30 Jan 2005 05:04:55 PM |
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On Sun, 30 Jan 2005 22:29:08 GMT, "Bill Hobba" <bhobba@rubbish.net.au>
wrote:
"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:dnbqv0drurl9hhcal20b375apulm5rijds@4ax.com...
On Sun, 30 Jan 2005 12:31:49 GMT, "Bill Hobba" <bhobba@rubbish.net.au>
wrote:
Koobee Wublee
You still have not explained why the following equation would indicate
gravity as a repulsive force.
m dv/dt = G M m / r^2
BTW still equating vectors to scalars I see. Learn some basic math then
repost.
Rest of junk mercifully snipped - well I actually did not read it to see
if
it is junk but making basic errors such as equating vectors to scalars
does
not inspire confidence.
Bill
Bill you are all wet. Both terms are vectors, the RHS sometimes
expressed as G M m r/r^3, quite obviously a radial vector. A cheap
shot where none is warranted. JP
What can I say - G M m r/r^3 (where r - usually written as rhat - rhat
being a vector of length r in the direction of the line joining the particle
m and M) is not he same as G M m / r^2. rhat/r^3 does not give 1/r^2.
SNIP
Bill
John Polasek
No, Bill, "rhat/r^3 does not give 1/r^2", it specifically makes the
RHS into a vector with the same magnitude as the scalar 1/r^2. The
radial vector nature of the RHS is well understood by the cognoscenti.
In gravity, there is no other vector.
Give it up. If you would read more of the literature you would see
this as very common practice with a bold r in the numerator, or fixing
it with u_r.
Your gripe is an obvious case of "piling on", where, in fact, you had
nothing to contribute, and really puts the tarnish on anything else
you post here.
And regarding your tweedle note, you shoud think twice about staying
on those funny cigarettes.
John Polasek
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| User: "Franz Heymann" |
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| Title: Re: Who will stun the world as next Einstein? |
31 Jan 2005 06:03:05 AM |
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"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:8bpqv096tu6ccmupbbvjm02qh813h40jmp@4ax.com...
On Sun, 30 Jan 2005 22:29:08 GMT, "Bill Hobba"
<bhobba@rubbish.net.au>
wrote:
[snip]
What can I say - G M m r/r^3 (where r - usually written as rhat -
rhat
being a vector of length r in the direction of the line joining the
particle
m and M) is not he same as G M m / r^2. rhat/r^3 does not give
1/r^2.
No, Bill, "rhat/r^3 does not give 1/r^2",
That's what Bill said. So what does the "No" do there?
[snip]
Franz
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| User: "John C. Polasek" |
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| Title: Re: Who will stun the world as next Einstein? |
31 Jan 2005 09:01:27 AM |
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On Mon, 31 Jan 2005 12:03:05 +0000 (UTC), "Franz Heymann"
<notfranz.heymann@btopenworld.com> wrote:
"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:8bpqv096tu6ccmupbbvjm02qh813h40jmp@4ax.com...
On Sun, 30 Jan 2005 22:29:08 GMT, "Bill Hobba"
<bhobba@rubbish.net.au>
wrote:
[snip]
What can I say - G M m r/r^3 (where r - usually written as rhat -
rhat
being a vector of length r in the direction of the line joining the
particle
m and M) is not he same as G M m / r^2. rhat/r^3 does not give
1/r^2.
No, Bill, "rhat/r^3 does not give 1/r^2",
That's what Bill said. So what does the "No" do there?
[snip]
Franz
I should have said:
No,(I AGREE) Bill, "rhat/r^3 does not give 1/r^2", BECAUSE it
specifically makes the
RHS into a vector (WHILST NOT DISTURBING) the same magnitude as the
scalar 1/r^2.
Gosh, physics is exciting!
John Polasek
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| User: "Tom Capizzi" |
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| Title: Re: Who will stun the world as next Einstein? |
31 Jan 2005 11:16:46 AM |
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"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:3shsv0peutkaif59larqomdfqvrc8adacu@4ax.com...
On Mon, 31 Jan 2005 12:03:05 +0000 (UTC), "Franz Heymann"
<notfranz.heymann@btopenworld.com> wrote:
"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:8bpqv096tu6ccmupbbvjm02qh813h40jmp@4ax.com...
On Sun, 30 Jan 2005 22:29:08 GMT, "Bill Hobba"
<bhobba@rubbish.net.au>
wrote:
[snip]
What can I say - G M m r/r^3 (where r - usually written as rhat -
rhat
being a vector of length r in the direction of the line joining the
particle
m and M) is not he same as G M m / r^2. rhat/r^3 does not give
1/r^2.
No, Bill, "rhat/r^3 does not give 1/r^2",
That's what Bill said. So what does the "No" do there?
[snip]
Franz
I should have said:
No,(I AGREE) Bill, "rhat/r^3 does not give 1/r^2", BECAUSE it
specifically makes the
RHS into a vector (WHILST NOT DISTURBING) the same magnitude as the
scalar 1/r^2.
It appears to me that your definitions of rhat are different, leading to
separate
interpretations of the formula. If "rhat/r^3" has the same magnitude as
1/r^2,
then this rhat must be the actual radius vector between two points. Bill has
clearly defined his version of rhat to be a unit vector in the direction of
the
line between two points. Why don't you just agree on one usage for the sake
of argument?
.
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| User: "John C. Polasek" |
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| Title: Re: Who will stun the world as next Einstein? |
31 Jan 2005 12:27:13 PM |
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On Mon, 31 Jan 2005 17:16:46 GMT, "Tom Capizzi"
<etianshrdlu@verizon.net> wrote:
"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:3shsv0peutkaif59larqomdfqvrc8adacu@4ax.com...
On Mon, 31 Jan 2005 12:03:05 +0000 (UTC), "Franz Heymann"
<notfranz.heymann@btopenworld.com> wrote:
"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:8bpqv096tu6ccmupbbvjm02qh813h40jmp@4ax.com...
On Sun, 30 Jan 2005 22:29:08 GMT, "Bill Hobba"
<bhobba@rubbish.net.au>
wrote:
[snip]
What can I say - G M m r/r^3 (where r - usually written as rhat -
rhat
being a vector of length r in the direction of the line joining the
particle
m and M) is not he same as G M m / r^2. rhat/r^3 does not give
1/r^2.
No, Bill, "rhat/r^3 does not give 1/r^2",
That's what Bill said. So what does the "No" do there?
[snip]
Franz
I should have said:
No,(I AGREE) Bill, "rhat/r^3 does not give 1/r^2", BECAUSE it
specifically makes the
RHS into a vector (WHILST NOT DISTURBING) the same magnitude as the
scalar 1/r^2.
It appears to me that your definitions of rhat are different, leading to
separate
interpretations of the formula. If "rhat/r^3" has the same magnitude as
1/r^2,
then this rhat must be the actual radius vector between two points. Bill has
clearly defined his version of rhat to be a unit vector in the direction of
the
line between two points. Why don't you just agree on one usage for the sake
of argument?
Your are muddying the waters. He has made no such clear definition.
Dimensionally, rhat must be a vector with the length r to go with
1/r^3. If rhat were a unit vector, then you go with rhat/r^2. I
mentioned that you could use a unit vector u_r/r^2.
mg/r^2 is commonly understood to be a vector just as the slovenly use
of z redshift = gh instead of gh/c^2 with c^2 being understood. This
is considered very "in" by techies.
I still say the original slur was just "piling on".
John Polasek
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| User: "Bill Hobba" |
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| Title: Re: Who will stun the world as next Einstein? |
31 Jan 2005 04:56:10 PM |
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"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:tbtsv0tcbte1n2n45gs6d8gto3nfvvrf3h@4ax.com...
On Mon, 31 Jan 2005 17:16:46 GMT, "Tom Capizzi"
<etianshrdlu@verizon.net> wrote:
"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:3shsv0peutkaif59larqomdfqvrc8adacu@4ax.com...
On Mon, 31 Jan 2005 12:03:05 +0000 (UTC), "Franz Heymann"
<notfranz.heymann@btopenworld.com> wrote:
"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:8bpqv096tu6ccmupbbvjm02qh813h40jmp@4ax.com...
On Sun, 30 Jan 2005 22:29:08 GMT, "Bill Hobba"
<bhobba@rubbish.net.au>
wrote:
[snip]
What can I say - G M m r/r^3 (where r - usually written as rhat -
rhat
being a vector of length r in the direction of the line joining the
particle
m and M) is not he same as G M m / r^2. rhat/r^3 does not give
1/r^2.
No, Bill, "rhat/r^3 does not give 1/r^2",
That's what Bill said. So what does the "No" do there?
[snip]
Franz
I should have said:
No,(I AGREE) Bill, "rhat/r^3 does not give 1/r^2", BECAUSE it
specifically makes the
RHS into a vector (WHILST NOT DISTURBING) the same magnitude as the
scalar 1/r^2.
It appears to me that your definitions of rhat are different, leading to
separate
interpretations of the formula. If "rhat/r^3" has the same magnitude as
1/r^2,
then this rhat must be the actual radius vector between two points. Bill
has
clearly defined his version of rhat to be a unit vector in the direction
of
the
line between two points. Why don't you just agree on one usage for the
sake
of argument?
Your are muddying the waters. He has made no such clear definition.
John you are tampering rather drastically with the truth - I wrote in a
previous response to you (and to Koobee Wublee as well) 'Then the equation
should be written in the correct form - which is - m dv/dt = G M m Rhat /
r^2 where Rhat is a unit vector pointing is the direction of the particle of
mass m to the particle of mass M.'. The sequence of events is plain for any
to see. Koobee Wublee wrote: 'You still have not explained why the
following equation would indicate gravity as a repulsive force. m dv/dt = G
M m / r^2' Because the equation is nonsense as it equates a vector with a
scalar I wrote (as I also replied in another thread where he tired the same
rubbish) 'BTW still equating vectors to scalars I see. Learn some basic
math then repost.' You responded 'Bill you are all wet. Both terms are
vectors, the RHS sometimes expressed as G M m r/r^3, quite obviously a
radial vector. A cheap shot where none is warranted.'. G M m / r^2 is not a
vector. G M m r/r^3 is a vector if the r in the numerator of that equation
is interpreted as a vector. But to avoid confusion with r^3 in the
denominator (where to make sense that r can not be a vector - it is the
length of the vector r - you are not allowed to divide a vector by a vector)
the r is usually written as Rhat - meaning a vector in the direction of the
line joining m to M and of length r. In my equation I defined Rhat a little
differently to have r^2 on the denominator because I believe it is clearer.
Now the whole claim of Koobee Wublee is that from the classical Hamiltonian
mv^2/2 - G M m/r you obtain gravity as repulsive by some obviously incorrect
manipulation. The reason it is not correct is that he ends up with a vector
equal to a scalar which nonsense. The correct derivation is as follows.
The lagrangain the previous Hamiltonian corresponds to is mv^2/2 + G M m/r.
Applying the PLA yields the well known equations of motion - m dvi/dt = part
deriv (G M m/r.)/part deriv xi where vi and xi are the components of
velocity and position respectfully. Now to simplify the problem we take the
origin at the particle M and the direction of the x axis along the line
joining M to m. Thus the equation becomes m dvx/dx = part deriv (G M m
/x)/part deriv. dvx is the component of the velocity in the x direction -
all other components are zero due to the coordinate system chosen. Hence we
end up with m dvx/dx = - G M m /x^2. Or going to vector notation m dv/dx =
G M m Rhat /r^2 where Rhat is a unit vector in the line joining m to M. The
reason the minus sign is removed is because it is in the opposite direction
of the line going from M to m. Thus we see - correctly - that gravity is
attractive. All I am asking of Koobee Wublee is to take similar care to
ensure vectors are equated to vectors and scalars to scalars with his
supposed derivation from the Hamiltonian. I am sure once he does that he
will also find that gravity is attractive as it should be.
You also stated 'The radial vector nature of the RHS is well understood by
the cognoscenti.' I am not a physicist but I am a mathematician and the
'cognoscenti' of that discipline are not so reckless as to equate a vector
to a scalar. Nor are physicists for that matter.
Bill
Dimensionally, rhat must be a vector with the length r to go with
1/r^3. If rhat were a unit vector, then you go with rhat/r^2. I
mentioned that you could use a unit vector u_r/r^2.
mg/r^2 is commonly understood to be a vector just as the slovenly use
of z redshift = gh instead of gh/c^2 with c^2 being understood. This
is considered very "in" by techies.
I still say the original slur was just "piling on".
John Polasek
.
|
|
|
| User: "Bill Hobba" |
|
| Title: Re: Who will stun the world as next Einstein? |
31 Jan 2005 11:25:08 PM |
|
|
Bill Hobba, being careless as usual, incorrectly wrote:
Thus the equation becomes m dvx/dx = part deriv (G M m
/x)/part deriv. dvx is the component of the velocity in the x direction -
all other components are zero due to the coordinate system chosen. Hence
we
end up with m dvx/dx = - G M m /x^2. Or going to vector notation m dv/dx
=
G M m Rhat /r^2 where Rhat is a unit vector in the line joining m to M.
Of course that should not be dvx/dx or dv/dx it should be dvx/dt and dv/dt.
Thanks
Bill
.
|
|
|
| User: "Bill Hobba" |
|
| Title: Re: Who will stun the world as next Einstein? |
31 Jan 2005 11:32:30 PM |
|
|
"Bill Hobba" <bhobba@rubbish.net.au> wrote in message
news:UoELd.143475$K7.17670@news-server.bigpond.net.au...
Bill Hobba, being careless as usual, incorrectly wrote:
Thus the equation becomes m dvx/dx = part deriv (G M m
/x)/part deriv. dvx is the component of the velocity in the x
direction -
all other components are zero due to the coordinate system chosen.
Hence
we
end up with m dvx/dx = - G M m /x^2. Or going to vector notation m
dv/dx
=
G M m Rhat /r^2 where Rhat is a unit vector in the line joining m to M.
Of course that should not be dvx/dx or dv/dx it should be dvx/dt and
dv/dt.
Thanks
Bill
It just occurred to me that I make this silly stupid obvious error and these
guys who recon I have no idea what I am talking about miss it and argue
about things like the derivative of a scalar function being a vector. It
really makes you wonder, it really does.
Bill
.
|
|
|
| User: "John C. Polasek" |
|
| Title: Re: Who will stun the world as next Einstein? |
01 Feb 2005 08:42:10 AM |
|
|
On Tue, 01 Feb 2005 05:32:30 GMT, "Bill Hobba" <bhobba@rubbish.net.au>
wrote:
"Bill Hobba" <bhobba@rubbish.net.au> wrote in message
news:UoELd.143475$K7.17670@news-server.bigpond.net.au...
Bill Hobba, being careless as usual, incorrectly wrote:
Thus the equation becomes m dvx/dx = part deriv (G M m
/x)/part deriv. dvx is the component of the velocity in the x
direction -
all other components are zero due to the coordinate system chosen.
Hence
we
end up with m dvx/dx = - G M m /x^2. Or going to vector notation m
dv/dx
=
G M m Rhat /r^2 where Rhat is a unit vector in the line joining m to M.
Of course that should not be dvx/dx or dv/dx it should be dvx/dt and
dv/dt.
Thanks
Bill
It just occurred to me that I make this silly stupid obvious error and these
guys who recon I have no idea what I am talking about miss it and argue
about things like the derivative of a scalar function being a vector. It
really makes you wonder, it really does.
Bill
Even 3 consecutive messages to yourself are not enough to make a wrong
into a right.
It is well understood in gravity and potential theory that mg/r
represents an equipotential surface quite like that at the metal
surface of a charged sphere.
Grad mg/r has 3 partial derivative components each accompanied by the
unit vectors i, j and k in spherical coordinates. In this case by
symmetry it is obvious that d/dr will do the whole job and put a u_r
on it if you want, it's a matter of notation.
Grad mg/r = -mg/r^2 is a vector field extending like a porcupine from
the equipotential surfaces.
Why struggle with this? As you say, " It really makes you wonder, it
really does" and that's a waste. It might even make a person stop and
think.
The OP has been lost in the dust, but out of respect for him, why not
a tiny apology?
John Polasek.
.
|
|
|
| User: "Bill Hobba" |
|
| Title: Re: Who will stun the world as next Einstein? |
01 Feb 2005 05:55:39 PM |
|
|
"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:bh4vv0heo1cc9r74mq621ccpmnnlo76s6o@4ax.com...
On Tue, 01 Feb 2005 05:32:30 GMT, "Bill Hobba" <bhobba@rubbish.net.au>
wrote:
"Bill Hobba" <bhobba@rubbish.net.au> wrote in message
news:UoELd.143475$K7.17670@news-server.bigpond.net.au...
Bill Hobba, being careless as usual, incorrectly wrote:
Thus the equation becomes m dvx/dx = part deriv (G M m
/x)/part deriv. dvx is the component of the velocity in the x
direction -
all other components are zero due to the coordinate system chosen.
Hence
we
end up with m dvx/dx = - G M m /x^2. Or going to vector notation m
dv/dx
=
G M m Rhat /r^2 where Rhat is a unit vector in the line joining m to
M.
Of course that should not be dvx/dx or dv/dx it should be dvx/dt and
dv/dt.
Thanks
Bill
It just occurred to me that I make this silly stupid obvious error and
these
guys who recon I have no idea what I am talking about miss it and argue
about things like the derivative of a scalar function being a vector. It
really makes you wonder, it really does.
Bill
Even 3 consecutive messages to yourself are not enough to make a wrong
into a right.
It is well understood in gravity and potential theory that mg/r
represents an equipotential surface quite like that at the metal
surface of a charged sphere.
That is one way of looking at the surface traced out as the vector that
function depends on and mg/r remains constant. But again that does not
change facts - a scalar is a scalar and a vector is a vector.
Grad mg/r has 3 partial derivative components each accompanied by the
unit vectors i, j and k in spherical coordinates.
Grad is not derivative.
In this case by
symmetry it is obvious that d/dr will do the whole job and put a u_r
on it if you want, it's a matter of notation.
Your semantic ramblings will not change the facts - the derivative of a
scalar is a scalar. The grad of a scalar is a vector. Vectors must be
equated to vectors, scalars to scalars.
Grad mg/r = -mg/r^2 is a vector field extending like a porcupine from
the equipotential surfaces.
Grad mg/r = -mg/r^2 is a nonsense statement because you are equating a
vector to a scalar.
Why struggle with this?
Because it does not matter how you cut and dice it a vector equals a vector
and a scalar equals a scalar.
As you say, " It really makes you wonder, it
really does" and that's a waste. It might even make a person stop and
think.
The OP has been lost in the dust, but out of respect for him, why not
a tiny apology?
Because a vector equals a vector and a scalar equals a scalar. But what can
I expect from Mr Dual Space - you can not even get basic facts about the
history of the positron and Dirac correct. I notice you never apologized to
me about that. Like most cranks you live in a dream world and refuse to
face facts.
Bill
John Polasek.
.
|
|
|
| User: "Ken S. Tucker" |
|
| Title: Re: Who will stun the world as next Einstein? |
01 Feb 2005 06:24:07 PM |
|
|
Bill Hobba wrote:
"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
Grad mg/r has 3 partial derivative components each accompanied by
the
unit vectors i, j and k in spherical coordinates.
Grad is not derivative.
Say what???, last I learned, the Grad is the
maximum slope (rate of change == derivative)
at the point it's being done at. I was quite
impressed by the Grad for that reason, but I
was a HS punk at the time, and maybe missed
something, please explain...
TIA
Ken
Bill
John Polasek.
.
|
|
|
| User: "Tom Capizzi" |
|
| Title: Re: Who will stun the world as next Einstein? |
01 Feb 2005 08:26:41 PM |
|
|
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1107303847.690945.56720@f14g2000cwb.googlegroups.com...
Bill Hobba wrote:
"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
Grad mg/r has 3 partial derivative components each accompanied by
the
unit vectors i, j and k in spherical coordinates.
Grad is not derivative.
Say what???, last I learned, the Grad is the
maximum slope (rate of change == derivative)
at the point it's being done at. I was quite
impressed by the Grad for that reason, but I
was a HS punk at the time, and maybe missed
something, please explain...
TIA
Ken
Bill
John Polasek.
Grad is a vector and derivative is a scalar.
.
|
|
|
| User: "Ken S. Tucker" |
|
| Title: Re: Who will stun the world as next Einstein? |
02 Feb 2005 01:02:42 PM |
|
|
Tom Capizzi wrote:
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1107303847.690945.56720@f14g2000cwb.googlegroups.com...
Bill Hobba wrote:
Grad is not derivative.
Say what???, last I learned, the Grad is the
maximum slope (rate of change == derivative)
at the point it's being done at. I was quite
impressed by the Grad for that reason, but I
was a HS punk at the time, and maybe missed
something, please explain...
TIA
Ken
Grad is a vector and derivative is a scalar.
That can't be right, derivatives don't
automatically generate invariants.
Tom, me thinks you've been Hobbatated.
Ken
.
|
|
|
| User: "Tom Capizzi" |
|
| Title: Re: Who will stun the world as next Einstein? |
02 Feb 2005 01:15:49 PM |
|
|
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1107370962.408865.4520@o13g2000cwo.googlegroups.com...
Tom Capizzi wrote:
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1107303847.690945.56720@f14g2000cwb.googlegroups.com...
Bill Hobba wrote:
Grad is not derivative.
Say what???, last I learned, the Grad is the
maximum slope (rate of change == derivative)
at the point it's being done at. I was quite
impressed by the Grad for that reason, but I
was a HS punk at the time, and maybe missed
something, please explain...
TIA
Ken
Grad is a vector and derivative is a scalar.
That can't be right, derivatives don't
automatically generate invariants.
Tom, me thinks you've been Hobbatated.
Ken
Maybe I should have been more explicit. The Grad of
a scalar is a vector, but the derivative of a scalar is a scalar.
.
|
|
|
| User: "Ken S. Tucker" |
|
| Title: Re: Who will stun the world as next Einstein? |
02 Feb 2005 01:52:45 PM |
|
|
Tom Capizzi wrote:
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1107370962.408865.4520@o13g2000cwo.googlegroups.com...
Tom Capizzi wrote:
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1107303847.690945.56720@f14g2000cwb.googlegroups.com...
Bill Hobba wrote:
Grad is not derivative.
Say what???, last I learned, the Grad is the
maximum slope (rate of change == derivative)
at the point it's being done at. I was quite
impressed by the Grad for that reason, but I
was a HS punk at the time, and maybe missed
something, please explain...
TIA
Ken
Grad is a vector and derivative is a scalar.
That can't be right, derivatives don't
automatically generate invariants.
Tom, me thinks you've been Hobbatated.
Ken
Maybe I should have been more explicit. The Grad of
a scalar is a vector, but the derivative of a scalar is a scalar.
Perhaps I was not clear.
A scalar is an invariant ok?
An invariant is not necessarily a constant ok?
Denote "A" to be such a variable invariant.
Then dA/dt =/= invariant because dt is not
invariant and hence dA/dt is not a scalar.
Something that's interesting though, is the
relativistic fact that dA/dt is not a scalar.
Maybe that's a good way to present dt as being
a vector component, specifically with ct=x^0,
then, dA/dt == dA/dx^0, and is also a vector
component, but not a scalar (invariant).
I just wanted to caution a derivative only
in special cases produces a scalar, and
certainly not generally in physics.
Regards
Ken S. Tucker
.
|
|
|
| User: "Tom Capizzi" |
|
| Title: Re: Who will stun the world as next Einstein? |
02 Feb 2005 04:12:04 PM |
|
|
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1107373965.215881.192060@o13g2000cwo.googlegroups.com...
Tom Capizzi wrote:
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1107370962.408865.4520@o13g2000cwo.googlegroups.com...
Tom Capizzi wrote:
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1107303847.690945.56720@f14g2000cwb.googlegroups.com...
Bill Hobba wrote:
Grad is not derivative.
Say what???, last I learned, the Grad is the
maximum slope (rate of change == derivative)
at the point it's being done at. I was quite
impressed by the Grad for that reason, but I
was a HS punk at the time, and maybe missed
something, please explain...
TIA
Ken
Grad is a vector and derivative is a scalar.
That can't be right, derivatives don't
automatically generate invariants.
Tom, me thinks you've been Hobbatated.
Ken
Maybe I should have been more explicit. The Grad of
a scalar is a vector, but the derivative of a scalar is a scalar.
Perhaps I was not clear.
A scalar is an invariant ok?
An invariant is not necessarily a constant ok?
Denote "A" to be such a variable invariant.
Then dA/dt =/= invariant because dt is not
invariant and hence dA/dt is not a scalar.
Something that's interesting though, is the
relativistic fact that dA/dt is not a scalar.
Maybe that's a good way to present dt as being
a vector component, specifically with ct=x^0,
then, dA/dt == dA/dx^0, and is also a vector
component, but not a scalar (invariant).
I just wanted to caution a derivative only
in special cases produces a scalar, and
certainly not generally in physics.
Regards
Ken S. Tucker
Interesting semantic quibble. I use the definition of scalar
as an entity with magnitude but no direction, and a vector
has both. To require a scalar to be an invariant is overly
restrictive, but I won't rule out the possibility that this is
customary usage in some circles. However, your argument
above is invalid because d/dt is not a component of the
Grad operator. Grad is a spatial vector derivative. And it
is a vector. The derivative of a scalar apparently may or
may not be a scalar (in the restricted sense of the word),
but it is never a vector. I am curious what you call an entity
with magnitude and no direction that is not an invariant.
.
|
|
|
| User: "Ken S. Tucker" |
|
| Title: Re: Who will stun the world as next Einstein? |
02 Feb 2005 04:44:16 PM |
|
|
Tom Capizzi wrote:
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1107373965.215881.192060@o13g2000cwo.googlegroups.com...
Tom Capizzi wrote:
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1107370962.408865.4520@o13g2000cwo.googlegroups.com...
Tom Capizzi wrote:
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1107303847.690945.56720@f14g2000cwb.googlegroups.com...
Bill Hobba wrote:
Grad is not derivative.
Say what???, last I learned, the Grad is the
maximum slope (rate of change == derivative)
at the point it's being done at. I was quite
impressed by the Grad for that reason, but I
was a HS punk at the time, and maybe missed
something, please explain...
TIA
Ken
Grad is a vector and derivative is a scalar.
That can't be right, derivatives don't
automatically generate invariants.
Tom, me thinks you've been Hobbatated.
Ken
Maybe I should have been more explicit. The Grad of
a scalar is a vector, but the derivative of a scalar is a scalar.
Perhaps I was not clear.
A scalar is an invariant ok?
An invariant is not necessarily a constant ok?
Denote "A" to be such a variable invariant.
Then dA/dt =/= invariant because dt is not
invariant and hence dA/dt is not a scalar.
Something that's interesting though, is the
relativistic fact that dA/dt is not a scalar.
Maybe that's a good way to present dt as being
a vector component, specifically with ct=x^0,
then, dA/dt == dA/dx^0, and is also a vector
component, but not a scalar (invariant).
I just wanted to caution a derivative only
in special cases produces a scalar, and
certainly not generally in physics.
Regards
Ken S. Tucker
Interesting semantic quibble. I use the definition of scalar
as an entity with magnitude but no direction, and a vector
has both. To require a scalar to be an invariant is overly
restrictive, but I won't rule out the possibility that this is
customary usage in some circles. However, your argument
above is invalid because d/dt is not a component of the
Grad operator. Grad is a spatial vector derivative. And it
is a vector. The derivative of a scalar apparently may or
may not be a scalar (in the restricted sense of the word),
but it is never a vector. I am curious what you call an entity
with magnitude and no direction that is not an invariant.
.
|
|
|
| User: "Tom Capizzi" |
|
| Title: Re: Who will stun the world as next Einstein? |
03 Feb 2005 01:13:48 PM |
|
|
This is the complete text of your last post. Is something missing?
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1107384256.335012.211010@z14g2000cwz.googlegroups.com...
Tom Capizzi wrote:
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1107373965.215881.192060@o13g2000cwo.googlegroups.com...
Tom Capizzi wrote:
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1107370962.408865.4520@o13g2000cwo.googlegroups.com...
Tom Capizzi wrote:
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1107303847.690945.56720@f14g2000cwb.googlegroups.com...
Bill Hobba wrote:
Grad is not derivative.
Say what???, last I learned, the Grad is the
maximum slope (rate of change == derivative)
at the point it's being done at. I was quite
impressed by the Grad for that reason, but I
was a HS punk at the time, and maybe missed
something, please explain...
TIA
Ken
Grad is a vector and derivative is a scalar.
That can't be right, derivatives don't
automatically generate invariants.
Tom, me thinks you've been Hobbatated.
Ken
Maybe I should have been more explicit. The Grad of
a scalar is a vector, but the derivative of a scalar is a scalar.
Perhaps I was not clear.
A scalar is an invariant ok?
An invariant is not necessarily a constant ok?
Denote "A" to be such a variable invariant.
Then dA/dt =/= invariant because dt is not
invariant and hence dA/dt is not a scalar.
Something that's interesting though, is the
relativistic fact that dA/dt is not a scalar.
Maybe that's a good way to present dt as being
a vector component, specifically with ct=x^0,
then, dA/dt == dA/dx^0, and is also a vector
component, but not a scalar (invariant).
I just wanted to caution a derivative only
in special cases produces a scalar, and
certainly not generally in physics.
Regards
Ken S. Tucker
Interesting semantic quibble. I use the definition of scalar
as an entity with magnitude but no direction, and a vector
has both. To require a scalar to be an invariant is overly
restrictive, but I won't rule out the possibility that this is
customary usage in some circles. However, your argument
above is invalid because d/dt is not a component of the
Grad operator. Grad is a spatial vector derivative. And it
is a vector. The derivative of a scalar apparently may or
may not be a scalar (in the restricted sense of the word),
but it is never a vector. I am curious what you call an entity
with magnitude and no direction that is not an invariant.
.
|
|
|
| User: "Ken S. Tucker" |
|
| Title: Re: Who will stun the world as next Einstein? |
03 Feb 2005 05:11:37 PM |
|
|
Tom Capizzi wrote:
This is the complete text of your last post. Is something missing?
I don't know, sometimes google seems to send an
echo or doesn't post at all. It's a new system
the're trying and figured I'd see how it evolves.
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1107384256.335012.211010@z14g2000cwz.googlegroups.com...
Tom Capizzi wrote:
Perhaps I was not clear.
A scalar is an invariant ok?
An invariant is not necessarily a constant ok?
Denote "A" to be such a variable invariant.
Then dA/dt =/= invariant because dt is not
invariant and hence dA/dt is not a scalar.
Something that's interesting though, is the
relativistic fact that dA/dt is not a scalar.
Maybe that's a good way to present dt as being
a vector component, specifically with ct=x^0,
then, dA/dt == dA/dx^0, and is also a vector
component, but not a scalar (invariant).
I just wanted to caution a derivative only
in special cases produces a scalar, and
certainly not generally in physics.
Regards
Ken S. Tucker
Interesting semantic quibble. I use the definition of scalar
as an entity with magnitude but no direction, and a vector
has both. To require a scalar to be an invariant is overly
restrictive, but I won't rule out the possibility that this is
customary usage in some circles. However, your argument
above is invalid because d/dt is not a component of the
Grad operator. Grad is a spatial vector derivative. And it
is a vector. The derivative of a scalar apparently may or
may not be a scalar (in the restricted sense of the word),
but it is never a vector.
It's a semantic quibble.
I am curious what you call an entity
with magnitude and no direction that is not an invariant.
It's a "relative tensor" such as
the determinant of g_uv, g=det g_uv.
Ken
.
|
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|
|
|
|
|
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|
|
|
|
| User: "John C. Polasek" |
|
| Title: Re: Who will stun the world as next Einstein? |
01 Feb 2005 09:07:11 PM |
|
|
On Tue, 01 Feb 2005 23:55:39 GMT, "Bill Hobba" <bhobba@rubbish.net.au>
wrote:
"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:bh4vv0heo1cc9r74mq621ccpmnnlo76s6o@4ax.com...
On Tue, 01 Feb 2005 05:32:30 GMT, "Bill Hobba" <bhobba@rubbish.net.au>
wrote:
"Bill Hobba" <bhobba@rubbish.net.au> wrote in message
news:UoELd.143475$K7.17670@news-server.bigpond.net.au...
Bill Hobba, being careless as usual, incorrectly wrote:
Thus the equation becomes m dvx/dx = part deriv (G M m
/x)/part deriv. dvx is the component of the velocity in the x
direction -
all other components are zero due to the coordinate system chosen.
Hence
we
end up with m dvx/dx = - G M m /x^2. Or going to vector notation m
dv/dx
=
G M m Rhat /r^2 where Rhat is a unit vector in the line joining m to
M.
Of course that should not be dvx/dx or dv/dx it should be dvx/dt and
dv/dt.
Thanks
Bill
It just occurred to me that I make this silly stupid obvious error and
these
guys who recon I have no idea what I am talking about miss it and argue
about things like the derivative of a scalar function being a vector. It
really makes you wonder, it really does.
Bill
Even 3 consecutive messages to yourself are not enough to make a wrong
into a right.
It is well understood in gravity and potential theory that mg/r
represents an equipotential surface quite like that at the metal
surface of a charged sphere.
That is one way of looking at the surface traced out as the vector that
function depends on and mg/r remains constant. But again that does not
change facts - a scalar is a scalar and a vector is a vector.
Grad mg/r has 3 partial derivative components each accompanied by the
unit vectors i, j and k in spherical coordinates.
Grad is not derivative.
In this case by
symmetry it is obvious that d/dr will do the whole job and put a u_r
on it if you want, it's a matter of notation.
Your semantic ramblings will not change the facts - the derivative of a
scalar is a scalar. The grad of a scalar is a vector. Vectors must be
equated to vectors, scalars to scalars.
Grad mg/r = -mg/r^2 is a vector field extending like a porcupine from
the equipotential surfaces.
Grad mg/r = -mg/r^2 is a nonsense statement because you are equating a
vector to a scalar.
Why struggle with this?
Because it does not matter how you cut and dice it a vector equals a vector
and a scalar equals a scalar.
As you say, " It really makes you wonder, it
really does" and that's a waste. It might even make a person stop and
think.
The OP has been lost in the dust, but out of respect for him, why not
a tiny apology?
Because a vector equals a vector and a scalar equals a scalar. But what can
I expect from Mr Dual Space - you can not even get basic facts about the
history of the positron and Dirac correct. I notice you never apologized to
me about that. Like most cranks you live in a dream world and refuse to
face facts.
Bill
John Polasek.
Hobba you are a consummate fool.
You say you are a mathematician; if you want to dip into physics study
some physics. There is just a whole bunch of field theory that hinges
on equipotentials and the gradient thereof and of which you are
totally unaware. You keep coming back with your insipid arguments.
I pointed out to you the 3 vector components of grad, and you should
by now also be congizant of the fact that with spherical symmetry, the
only derivative is with r, the other three are nil. So by not
mentioning them does not demote my gradient to your profane
deriviative.
Don't go on with your sophomoric rantings about derivatives not
making a vector. The derivative with r is what's left of the operator.
Remember what you are doing with your pustulations. You are defending
an insult you made to an inquirer 3 days ago. Nothing has been gained
by this exchange. Write this down; the gradient of a scalar potential
is a vector no matter how denominated.
John Polasek
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