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Misner, Thorne and Wheeler, Exercise 8.5 (c)
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Hetware
Apr 7, 2013, 8:14:35 PM4/7/13
to
This is the geodesic equation under discussion:
d^2(r)/dt^2 = r(dp/dt)^2
d^2(p)/dt^2 = -(2/r)(dp/dt)(dr/dt).
r is radius in polar coordinates, p is the angle, and t is a path parameter.
The authors ask me to "[S]olve the geodesic equation for r(t) and p(t),
and show that the solution is a uniformly parametrized straight
line(x===r cos(p) = at+p for some a and b; y===r sin(p) = jt+k for some
j and k).
I tried the following:
(d^2(p)/dt^2)/(dp/dt) = -(2/r)(dr/dt)
f=dp/dt
(df/dt)/f = -(2/r)(dr/dt)
-1/2 ln(f) + k = ln(r)
a(f^(1/2)) = r
a(dp/dt)^(1/2) = r
And substitute for r in:
d^2(r)/dt^2 = r(dp/dt)^2
to get
d^2(r)/dt^2 = a(dp/dt)^(3/2)
But there I'm stuck.
How should the problem be handled?
d^2(r)/dt^2 = r(dp/dt)^2
d^2(p)/dt^2 = -(2/r)(dp/dt)(dr/dt).
r is radius in polar coordinates, p is the angle, and t is a path parameter.
The authors ask me to "[S]olve the geodesic equation for r(t) and p(t),
and show that the solution is a uniformly parametrized straight
line(x===r cos(p) = at+p for some a and b; y===r sin(p) = jt+k for some
j and k).
I tried the following:
(d^2(p)/dt^2)/(dp/dt) = -(2/r)(dr/dt)
f=dp/dt
(df/dt)/f = -(2/r)(dr/dt)
-1/2 ln(f) + k = ln(r)
a(f^(1/2)) = r
a(dp/dt)^(1/2) = r
And substitute for r in:
d^2(r)/dt^2 = r(dp/dt)^2
to get
d^2(r)/dt^2 = a(dp/dt)^(3/2)
But there I'm stuck.
How should the problem be handled?
Hetware
Apr 7, 2013, 11:13:47 PM4/7/13
to
T.M. Sommers
Apr 7, 2013, 11:55:56 PM4/7/13
to
On 4/7/2013 11:13 PM, Hetware wrote:
> On 4/7/2013 8:14 PM, Hetware wrote:
>> This is the geodesic equation under discussion:
>>
>> d^2(r)/dt^2 = r(dp/dt)^2
>>
>> [...]
> On 4/7/2013 8:14 PM, Hetware wrote:
>> This is the geodesic equation under discussion:
>>
>> d^2(r)/dt^2 = r(dp/dt)^2
>>
>>
>> How should the problem be handled?
>
> The silence is deadpanning!
While you're waiting for "Wilson" and Seto and the other crackpots to
>> How should the problem be handled?
>
> The silence is deadpanning!
answer, you could go to http://www.physicsforums.com/, where there are a
lot more people who could really help.
--
T.M. Sommers -- ab2sb
Lord Androcles, Zeroth Earl of Medway
Apr 7, 2013, 11:34:00 PM4/7/13
to
"Hetware" wrote in message
news:AuGdnVu7TbVslv_M...@megapath.net...
=============================================
What you have is a second order differential equation.
Unlike the solution to the general polynomial equation,
ax +bx^2 + cx^3 + ... + kx^n = 0, where you seek a value
for x given values for a,b,c etc., the solution for a
differential equation is a FUNCTION.
In other words you cannot obtain a numerical or algebraic
value (you don't have enough information and that is not
the idea anyway) but you can find functions r(t) and p(t) .
The authors have already told you the solution is a straight
line, which is of course a function.
http://search.snap.do/?q=solving+differential+equations&category=Web
HTH, because we don't do homework for you.
-- This message is brought to you from the keyboard of
Lord Androcles, Zeroth Earl of Medway.
When the fools chicken farmer Wilson and Van de faggot present an argument I
cannot laugh at I'll retire from usenet.
news:AuGdnVu7TbVslv_M...@megapath.net...
What you have is a second order differential equation.
Unlike the solution to the general polynomial equation,
ax +bx^2 + cx^3 + ... + kx^n = 0, where you seek a value
for x given values for a,b,c etc., the solution for a
differential equation is a FUNCTION.
In other words you cannot obtain a numerical or algebraic
value (you don't have enough information and that is not
the idea anyway) but you can find functions r(t) and p(t) .
The authors have already told you the solution is a straight
line, which is of course a function.
http://search.snap.do/?q=solving+differential+equations&category=Web
HTH, because we don't do homework for you.
-- This message is brought to you from the keyboard of
Lord Androcles, Zeroth Earl of Medway.
When the fools chicken farmer Wilson and Van de faggot present an argument I
cannot laugh at I'll retire from usenet.
Hetware
Apr 8, 2013, 12:05:19 AM4/8/13
to
zionazis. Never have, and never will.
Lord Androcles, Zeroth Earl of Medway
Apr 8, 2013, 12:07:05 AM4/8/13
to
"Lord Androcles, Zeroth Earl of Medway" wrote in message
news:Njr8t.334633$Ic.4...@fx07.fr7...
================================
Hetware's silence is deafening.
We should all be poised at the ready to answer his questions
immediately instead of sleeping in bed when he writes them.
news:Njr8t.334633$Ic.4...@fx07.fr7...
Hetware's silence is deafening.
We should all be poised at the ready to answer his questions
immediately instead of sleeping in bed when he writes them.
William Elliot
Apr 8, 2013, 12:20:01 AM4/8/13
to
On Sun, 7 Apr 2013, T.M. Sommers wrote:
> While you're waiting for "Wilson" and Seto and the other crackpots to
> answer, you could go to http://www.physicsforums.com/, where there are a
> lot more people who could really help.
That link gives an error message.
> While you're waiting for "Wilson" and Seto and the other crackpots to
> answer, you could go to http://www.physicsforums.com/, where there are a
> lot more people who could really help.
Hetware
Apr 8, 2013, 12:28:46 AM4/8/13
to
comparison to actual relevant content? Oh well, whatever,
nevermind....http://www.youtube.com/watch?v=pkcJEvMcnEg
Hetware
Apr 8, 2013, 12:50:22 AM4/8/13
to
intolerance explicitly excludes any thought congruent with that of
Albert Einstein and Erwin Schroedinger regarding quantum mechanics vs.
general relativity. It implicitly excludes the more subtle cognition of
Bohr and Heisenberg, as well. In short, it is a den of thieves and vipers.
Lord Androcles, Zeroth Earl of Medway
Apr 8, 2013, 12:58:09 AM4/8/13
to
"Hetware" wrote in message
news:EIednTlOU8ji2v_M...@megapath.net...
=====================================================
SR and GR are codswallop, so it is scarcely surprising if 99%
of posts to sci.physics.relativity are codswallop.
The video you cited is also codswallop. This is much better:
http://www.youtube.com/watch?v=t3217H8JppI
news:EIednTlOU8ji2v_M...@megapath.net...
SR and GR are codswallop, so it is scarcely surprising if 99%
of posts to sci.physics.relativity are codswallop.
The video you cited is also codswallop. This is much better:
http://www.youtube.com/watch?v=t3217H8JppI
T.M. Sommers
Apr 8, 2013, 1:09:23 AM4/8/13
to
On 4/8/2013 12:20 AM, William Elliot wrote:
Dirk Van de moortel
Apr 8, 2013, 9:27:57 AM4/8/13
to
Hetware <hat...@speakyeasy.net> wrote:
> This is the geodesic equation under discussion:
>
> d^2(r)/dt^2 = r(dp/dt)^2
>
> d^2(p)/dt^2 = -(2/r)(dp/dt)(dr/dt).
>
> r is radius in polar coordinates, p is the angle, and t is a path
> parameter.
>
> The authors ask me to "[S]olve the geodesic equation for r(t) and
> p(t),
> and show that the solution is a uniformly parametrized straight
> line(x===r cos(p) = at+p for some a and b; y===r sin(p) = jt+k for
> some
> j and k).
Normally we'd write dotted variables, but with quotes it's easier.
> This is the geodesic equation under discussion:
>
> d^2(r)/dt^2 = r(dp/dt)^2
>
> d^2(p)/dt^2 = -(2/r)(dp/dt)(dr/dt).
>
> r is radius in polar coordinates, p is the angle, and t is a path
> parameter.
>
> The authors ask me to "[S]olve the geodesic equation for r(t) and
> p(t),
> and show that the solution is a uniformly parametrized straight
> line(x===r cos(p) = at+p for some a and b; y===r sin(p) = jt+k for
> some
> j and k).
So write
r' = dr/dt
r'' = d^2(r)/dt^2
p = dp/dt
p'' = d^2(p)/dt^2
then you have
r'' = r (p')^2 [1]
p'' = -2/r p' r' [2]
Deriving [1] gives
r''' = r' (p')^2 + 2 r p' p''
which with [2] gives
r''' = -3 r' (p')^2
which again with [1] gives
r r''' + 3 r'' r' = 0
So if you find some r(t) that has r''(t) = r'''(t) = 0 for all t,
you're done. Easy. Take
r(t) = A t + B [3]
then
r'(t) = A
r''(t) = 0
r''"(t) = 0
so indeed
r r''' + 3 r'' r' = 0
Now, from [1] and [3] you get
0 = ( A t + B ) (p')^2
so
p' = 0
so
p(t) = C [4]
So you get
r(t) = A t + B
p(t) = C
Check it out with [1] and [2]. Trivial.
So
x(t) = (A t + B ) cos(C)
y(t) = (A t + B ) sin(C)
There's probably another solution, but seems to be the one they're
after.
Dirk Vdm
Lord Androcles, Zeroth Earl of Medway
Apr 8, 2013, 12:17:14 PM4/8/13
to
"Dirk Van de moortel" wrote in message
news:kjugkr$ns1$1...@speranza.aioe.org...
======================================
Pulled sin and cos out of your arse, Dork?
For A = 0, B = 1, C =(0 to 2pi) you've drawn a circle.
Is that the "probably another solution" you were
looking for?
exp(i.t) = cos(t) + i.sin(t) -- Euler.
> Indeed, writing LT and inverse:
> x' = g ( x - v t ) [1]
> t' = g ( t - v x ) [2]
> x = g ( x' + v t' ) [3]
> t = g ( t' + v x' ) [4] -- A fucking idiot called Dork.
news:kjugkr$ns1$1...@speranza.aioe.org...
Pulled sin and cos out of your arse, Dork?
For A = 0, B = 1, C =(0 to 2pi) you've drawn a circle.
Is that the "probably another solution" you were
looking for?
exp(i.t) = cos(t) + i.sin(t) -- Euler.
> Indeed, writing LT and inverse:
> x' = g ( x - v t ) [1]
> t' = g ( t - v x ) [2]
> x = g ( x' + v t' ) [3]
> t = g ( t' + v x' ) [4] -- A fucking idiot called Dork.
Wesley Burns
Apr 8, 2013, 8:00:43 PM4/8/13
to
Dirk Van de moortel wrote:
> Normally we'd write dotted variables, but with quotes it's easier.
> So write
> r' = dr/dt
> r'' = d^2(r)/dt^2
dr = (r1-r0) ?
> Normally we'd write dotted variables, but with quotes it's easier.
> So write
> r' = dr/dt
> r'' = d^2(r)/dt^2
dt = (t1-t0) ?
d^2(r) = ?
dt^2 = ?
d^2(r)/dt^2 = (r2-r1)(r1-r0)/(t1-t0)(t1-t0) ?
> p = dp/dt
p' ?
Alfred Einstead
Apr 8, 2013, 8:18:24 PM4/8/13
to
On Apr 7, 7:14 pm, Hetware <hatt...@speakyeasy.net> wrote:
> This is the geodesic equation under discussion:
> d^2(r)/dt^2 = r(dp/dt)^2
> d^2(p)/dt^2 = -(2/r)(dp/dt)(dr/dt).
Where the problem comes from is not important, since all you're asking
> This is the geodesic equation under discussion:
> d^2(r)/dt^2 = r(dp/dt)^2
> d^2(p)/dt^2 = -(2/r)(dp/dt)(dr/dt).
about is how this is solved.
Notice that it's independent of p and depends only on do/dt. So,
define v = dr/dt, f = dp/dt and write the system as
dv/dt = rf^2, df/dt = -2fv/r (along with dr/dt = v, dp/dt = f).
The second equation can be rewritten as
0 = 1/f df/dt + 2/r dr/dt = 1/(f r^2) d(f r^2)/dt
Therefore, f = K/r^2, for some constant K.
The first equation then becomes, dv/dt = K^2/r^3. The standard trick
is to turn this into a conservation of energy integral:
v dv/dt = K^2/r^3 v = K^2/r^3 dr/dt,
from which it follows that
d/dt (v^2/2) = d/dt (-K^2/2 1/r^2).
The solution is
v^2 = v_0^2 - K^2/r^2
for some constant v_0.
You can take it on from here; solving for r as a function of t, and
then putting this into the equation for f (i.e. dp/dt) to get p as a
function of t.
If you go back to the original problem, the following properties hold
true. The gravitational source (which I assume is the Schwarzschild
solution) has rotational symmetry and time translation symmetry. This
corresponds (by way of the Noether Theorem) to conserved quantities:
Angular Momentum for rotational symmetry and Energy for time-
translation symmetry. Therefore, the first things to look for in the
geodesic law are to extract out the integrals for angular momentum and
energy. The angular momentum part was already removed from the problem
by the time you brought the matter here in sci.math, so that left the
energy integral to take care of. That was the missing step.
Hetware
Apr 8, 2013, 8:53:47 PM4/8/13
to
I was getting close. I remembered what Wheeler told me: "Anytime you
are struggling to understand or explain something, draw a picture."
http://www.speakeasy.org/~hattons/chapter08.nb
I worked the problem in pencil today while away from my computer. Just
use p=arctan(t/r_0) where r_0 is a radial vector perpendicular to the
"curve". r=(r_0^2+t^2)=r_0/cos(p). The rest is just a matter of
plugging these into the geodesic equation.
rotchm
Apr 8, 2013, 9:08:01 PM4/8/13
to
> which again with [1] gives
> r r''' + 3 r'' r' = 0
>
> So if you find some r(t) that has r''(t) = r'''(t) = 0 for all t,
> you're done. Easy. Take
> r(t) = A t + B [3]
The general solution to this is
r = s(At2 + Bt + C), which can be ~ at + b if its a perfect square.
I do like Alfred's elegant approach!
Lord Androcles, Zeroth Earl of Medway
Apr 8, 2013, 10:03:15 PM4/8/13
to
"Hetware" wrote in message
news:CeSdneS1Wf0B-_7M...@megapath.net...
==========================================
Good. Note that what you are doing is strictly NOT anything
to do with relativity or gravitation or even physics, it is just
math.
Pity Einstein didn't work the examples or he'd have made fewer
blunders like this one:
http://www.androcles01.pwp.blueyonder.co.uk/DominoEffect.GIF
news:CeSdneS1Wf0B-_7M...@megapath.net...
Good. Note that what you are doing is strictly NOT anything
to do with relativity or gravitation or even physics, it is just
math.
Pity Einstein didn't work the examples or he'd have made fewer
blunders like this one:
http://www.androcles01.pwp.blueyonder.co.uk/DominoEffect.GIF
Hetware
Apr 8, 2013, 11:21:06 PM4/8/13
to
On 4/8/2013 8:18 PM, Alfred Einstead wrote:
> On Apr 7, 7:14 pm, Hetware <hatt...@speakyeasy.net> wrote:
>> This is the geodesic equation under discussion:
>> d^2(r)/dt^2 = r(dp/dt)^2
>> d^2(p)/dt^2 = -(2/r)(dp/dt)(dr/dt).
>
> Where the problem comes from is not important, since all you're asking
> about is how this is solved.
I don't fully agree with that view. The context of a problem often
> On Apr 7, 7:14 pm, Hetware <hatt...@speakyeasy.net> wrote:
>> This is the geodesic equation under discussion:
>> d^2(r)/dt^2 = r(dp/dt)^2
>> d^2(p)/dt^2 = -(2/r)(dp/dt)(dr/dt).
>
> Where the problem comes from is not important, since all you're asking
> about is how this is solved.
suggests a particular approach.
Clearly, what I posted as my effort to solve the problem suggested /an/
approach.
> Notice that it's independent of p and depends only on do/dt. So,
> define v = dr/dt, f = dp/dt and write the system as
> dv/dt = rf^2, df/dt = -2fv/r (along with dr/dt = v, dp/dt = f).
>
> The second equation can be rewritten as
> 0 = 1/f df/dt + 2/r dr/dt = 1/(f r^2) d(f r^2)/dt
> Therefore, f = K/r^2, for some constant K.
>
> The first equation then becomes, dv/dt = K^2/r^3. The standard trick
> is to turn this into a conservation of energy integral:
> v dv/dt = K^2/r^3 v = K^2/r^3 dr/dt,
> from which it follows that
> d/dt (v^2/2) = d/dt (-K^2/2 1/r^2).
> The solution is
> v^2 = v_0^2 - K^2/r^2
> for some constant v_0.
>
> You can take it on from here; solving for r as a function of t, and
> then putting this into the equation for f (i.e. dp/dt) to get p as a
> function of t.
hard time following math in ASCII. I appreciate your help. It did get
me further along than I was. Differential equations is not a strong
suit of mine. Unfortunately, I can't spend any more time on it tonight.
> If you go back to the original problem, the following properties hold
> true.
Lord Androcles, Zeroth Earl of Medway
Apr 8, 2013, 11:39:35 PM4/8/13
to
"Hetware" wrote in message
news:AMydnYGNKLq-FP7M...@megapath.net...
The exercise is titled "A sheet of paper in polar coordinates."
==============================================
That's a Mercator projection (cylinder around a sphere).
r_0 = Earth radius, r extends to touch the cylinder, p = latitude.
https://en.wikipedia.org/wiki/Mercator_projection
news:AMydnYGNKLq-FP7M...@megapath.net...
The exercise is titled "A sheet of paper in polar coordinates."
That's a Mercator projection (cylinder around a sphere).
r_0 = Earth radius, r extends to touch the cylinder, p = latitude.
https://en.wikipedia.org/wiki/Mercator_projection
Hetware
Apr 8, 2013, 11:50:53 PM4/8/13
to
http://www.youtube.com/watch?v=7XvlsM39C-U
Hey; don't get me wrong, I contribute $100s USD to WETA per year.
http://www.weta.org/fm
Hetware
Apr 9, 2013, 12:23:22 AM4/9/13
to
doesn't spend his time trying to prove how smart he is. He just shows
you what he things in the clearest manner available to him. The mark of
a true genius.
http://people.oregonstate.edu/~drayt/Courses/MTH437/2007/hw/geodesic.pdf
Lord Androcles, Zeroth Earl of Medway
Apr 9, 2013, 12:34:43 AM4/9/13
to
"Hetware" wrote in message
news:LKOdnaa-ddKCDf7M...@megapath.net...
===========================================
Ugh...
No more please, I want the barrel organ's designer, not
the monkey.
=========================================
I understand the obligation, mine went to WQED when I lived there.
In Britain the BBC is funded by a communist license fee which I do
NOT pay on principle. One can ask and I'll give, but a demand backed
by a huge database of every home in the country and threats of a
£1000 fine for watching broadcast TV without said license must be
resisted when the newsreaders alone are getting $150,000 salaries.
news:LKOdnaa-ddKCDf7M...@megapath.net...
Ugh...
No more please, I want the barrel organ's designer, not
the monkey.
=========================================
I understand the obligation, mine went to WQED when I lived there.
In Britain the BBC is funded by a communist license fee which I do
NOT pay on principle. One can ask and I'll give, but a demand backed
by a huge database of every home in the country and threats of a
£1000 fine for watching broadcast TV without said license must be
resisted when the newsreaders alone are getting $150,000 salaries.
Hetware
Apr 9, 2013, 12:40:52 AM4/9/13
to
On 4/8/2013 9:27 AM, Dirk Van de moortel wrote:
> Hetware <hat...@speakyeasy.net> wrote:
>> This is the geodesic equation under discussion:
>>
>> d^2(r)/dt^2 = r(dp/dt)^2
>>
>> d^2(p)/dt^2 = -(2/r)(dp/dt)(dr/dt).
>>
>> r is radius in polar coordinates, p is the angle, and t is a path
>> parameter.
>> The authors ask me to "[S]olve the geodesic equation for r(t) and
>> p(t), and show that the solution is a uniformly parametrized straight
>> line(x===r cos(p) = at+p for some a and b; y===r sin(p) = jt+k for
>> some j and k).
>
> Normally we'd write dotted variables, but with quotes it's easier.
> So write
Like this?
> Hetware <hat...@speakyeasy.net> wrote:
>> This is the geodesic equation under discussion:
>>
>> d^2(r)/dt^2 = r(dp/dt)^2
>>
>> d^2(p)/dt^2 = -(2/r)(dp/dt)(dr/dt).
>>
>> r is radius in polar coordinates, p is the angle, and t is a path
>> parameter.
>> The authors ask me to "[S]olve the geodesic equation for r(t) and
>> p(t), and show that the solution is a uniformly parametrized straight
>> line(x===r cos(p) = at+p for some a and b; y===r sin(p) = jt+k for
>> some j and k).
>
> Normally we'd write dotted variables, but with quotes it's easier.
> So write
<math xmlns='http://www.w3.org/1998/Math/MathML'
mathematica:form='TraditionalForm'
xmlns:mathematica='http://www.wolfram.com/XML/'>
<mrow>
<mrow>
<mover>
<mi>r</mi>
<mo>..</mo>
</mover>
<mo>=</mo>
<mrow>
<msup>
<mrow>
<mi>r</mi>
<mo>⁡</mo>
<mo>(</mo>
<mover>
<mi>ϕ</mi>
<mo>·</mo>
</mover>
<mo>)</mo>
</mrow>
<mn>2</mn>
</msup>
<mo>=</mo>
<mi>r</mi>
</mrow>
</mrow>
<mo>⁣</mo>
<msup>
<mi>ω</mi>
<mn>2</mn>
</msup>
</mrow>
</math>
<math xmlns='http://www.w3.org/1998/Math/MathML'>
<mtext>
</mtext>
</math>
<math xmlns='http://www.w3.org/1998/Math/MathML'
mathematica:form='TraditionalForm'
xmlns:mathematica='http://www.wolfram.com/XML/'>
<mrow>
<mrow>
<mi>r</mi>
<mo>⁢</mo>
<mover>
<mi>ω</mi>
<mo fontweight='bold'>·</mo>
</mover>
</mrow>
<mo>=</mo>
<mrow>
<mrow>
<mo>-</mo>
<mn>2</mn>
</mrow>
<mo>⁢</mo>
<mi>ω</mi>
<mo>⁢</mo>
<mover>
<mi>r</mi>
<mo fontweight='bold'>·</mo>
</mover>
</mrow>
</mrow>
</math>
<math xmlns='http://www.w3.org/1998/Math/MathML'>
<mtext>
</mtext>
</math>
<math xmlns='http://www.w3.org/1998/Math/MathML'
mathematica:form='TraditionalForm'
xmlns:mathematica='http://www.wolfram.com/XML/'>
<mrow>
<mrow>
<mfrac>
<mover>
<mi>r</mi>
<mo>..</mo>
</mover>
<mi>r</mi>
</mfrac>
<mo>=</mo>
</mrow>
<mo>⁣</mo>
<msup>
<mi>ω</mi>
<mn>2</mn>
</msup>
</mrow>
</math>
<math xmlns='http://www.w3.org/1998/Math/MathML'>
<mtext>
</mtext>
</math>
<math xmlns='http://www.w3.org/1998/Math/MathML'
mathematica:form='TraditionalForm'
xmlns:mathematica='http://www.wolfram.com/XML/'>
<mrow>
<mfrac>
<mover>
<mi>ω</mi>
<mo fontweight='bold'>·</mo>
</mover>
<mi>ω</mi>
</mfrac>
<mo>=</mo>
<mrow>
<mrow>
<mo>-</mo>
<mn>2</mn>
</mrow>
<mo>⁢</mo>
<mfrac>
<mover>
<mi>r</mi>
<mo fontweight='bold'>·</mo>
</mover>
<mi>r</mi>
</mfrac>
</mrow>
</mrow>
</math>
<math xmlns='http://www.w3.org/1998/Math/MathML'>
<mtext>
</mtext>
</math>
<math xmlns='http://www.w3.org/1998/Math/MathML'
mathematica:form='TraditionalForm'
xmlns:mathematica='http://www.wolfram.com/XML/'>
<mrow>
<mrow>
<mi>ln</mi>
<mo>⁡</mo>
<mo>(</mo>
<mi>ω</mi>
<mo>)</mo>
</mrow>
<mo>=</mo>
<mrow>
<mrow>
<mrow>
<mo>-</mo>
<mn>2</mn>
</mrow>
<mo>⁢</mo>
<mrow>
<mi>ln</mi>
<mo>⁡</mo>
<mo>(</mo>
<mi>r</mi>
<mo>)</mo>
</mrow>
</mrow>
<mo>+</mo>
<mi>k</mi>
</mrow>
</mrow>
</math>
<math xmlns='http://www.w3.org/1998/Math/MathML'>
<mtext>
</mtext>
</math>
<math xmlns='http://www.w3.org/1998/Math/MathML'
mathematica:form='TraditionalForm'
xmlns:mathematica='http://www.wolfram.com/XML/'>
<mrow>
<mrow>
<mi>ω</mi>
<mo>=</mo>
<mi>a</mi>
</mrow>
<mo>⁣</mo>
<msup>
<mi>r</mi>
<mrow>
<mo>-</mo>
<mn>2</mn>
</mrow>
</msup>
</mrow>
</math>
<math xmlns='http://www.w3.org/1998/Math/MathML'>
<mtext>
</mtext>
</math>
<math xmlns='http://www.w3.org/1998/Math/MathML'
mathematica:form='TraditionalForm'
xmlns:mathematica='http://www.wolfram.com/XML/'>
<mrow>
<mfrac>
<mi>ω</mi>
<mi>a</mi>
</mfrac>
<mo>=</mo>
<msup>
<mi>r</mi>
<mrow>
<mo>-</mo>
<mn>2</mn>
</mrow>
</msup>
</mrow>
</math>
<math xmlns='http://www.w3.org/1998/Math/MathML'>
<mtext>
</mtext>
</math>
<math xmlns='http://www.w3.org/1998/Math/MathML'
mathematica:form='TraditionalForm'
xmlns:mathematica='http://www.wolfram.com/XML/'>
<mrow>
<msqrt>
<mfrac>
<mi>ω</mi>
<mi>a</mi>
</mfrac>
</msqrt>
<mo>=</mo>
<msup>
<mi>r</mi>
<mrow>
<mo>-</mo>
<mn>1</mn>
</mrow>
</msup>
</mrow>
</math>
<math xmlns='http://www.w3.org/1998/Math/MathML'>
<mtext>
</mtext>
</math>
<math xmlns='http://www.w3.org/1998/Math/MathML'
mathematica:form='TraditionalForm'
xmlns:mathematica='http://www.wolfram.com/XML/'>
<mrow>
<mrow>
<mover>
<mi>r</mi>
<mo>..</mo>
</mover>
<mo>=</mo>
<msqrt>
<mfrac>
<mi>a</mi>
<mi>ω</mi>
</mfrac>
</msqrt>
</mrow>
<mo>⁣</mo>
<msup>
<mi>ω</mi>
<mn>2</mn>
</msup>
</mrow>
</math>
Lord Androcles, Zeroth Earl of Medway
Apr 9, 2013, 12:44:30 AM4/9/13
to
"Hetware" wrote in message
news:YsudnS3BrvtXCv7M...@megapath.net...
=======================================
Fine, but it has no more connection with physics than politics or music.
news:YsudnS3BrvtXCv7M...@megapath.net...
Fine, but it has no more connection with physics than politics or music.
Hetware
Apr 9, 2013, 12:55:08 AM4/9/13
to
On 4/9/2013 12:34 AM, Lord Androcles, Zeroth Earl of Medway wrote:
> "Hetware" wrote in message
> news:LKOdnaa-ddKCDf7M...@megapath.net...
>
> On 4/8/2013 12:58 AM, Lord Androcles, Zeroth Earl of Medway wrote:
>> The video you cited is also codswallop. This is much better:
>> http://www.youtube.com/watch?v=t3217H8JppI
>>
> The English translation:
>
> http://www.youtube.com/watch?v=7XvlsM39C-U
> ===========================================
> Ugh...
> No more please, I want the barrel organ's designer, not
> the monkey.
>
>
For you and the yack you rode in on.
> "Hetware" wrote in message
> news:LKOdnaa-ddKCDf7M...@megapath.net...
>
> On 4/8/2013 12:58 AM, Lord Androcles, Zeroth Earl of Medway wrote:
>> The video you cited is also codswallop. This is much better:
>> http://www.youtube.com/watch?v=t3217H8JppI
>>
> The English translation:
>
> http://www.youtube.com/watch?v=7XvlsM39C-U
> ===========================================
> Ugh...
> No more please, I want the barrel organ's designer, not
> the monkey.
>
>
http://www.youtube.com/watch?v=gPmSdTVXsdU
Lord Androcles, Zeroth Earl of Medway
Apr 9, 2013, 1:07:54 AM4/9/13
to
"Hetware" wrote in message
news:osWdndc0hb-wAv7M...@megapath.net...
===========================================
Doesn't he know the words?
http://www.youtube.com/watch?v=eh31j6L95Ok
news:osWdndc0hb-wAv7M...@megapath.net...
Doesn't he know the words?
http://www.youtube.com/watch?v=eh31j6L95Ok
Hetware
Apr 9, 2013, 1:15:45 AM4/9/13
to
On 4/9/2013 12:44 AM, Lord Androcles, Zeroth Earl of Medway wrote:
> "Hetware" wrote in message
> "Hetware" wrote in message
> I don't have time to follow up on this, but this is from a guy who
> doesn't spend his time trying to prove how smart he is. He just shows
> you what he thinks in the clearest manner available to him. The mark of
> doesn't spend his time trying to prove how smart he is. He just shows
> a true genius.
>
> http://people.oregonstate.edu/~drayt/Courses/MTH437/2007/hw/geodesic.pdf
> =======================================
> Fine, but it has no more connection with physics than politics or music.
>
This is most apropos and irrelevant:
>
> http://people.oregonstate.edu/~drayt/Courses/MTH437/2007/hw/geodesic.pdf
> =======================================
> Fine, but it has no more connection with physics than politics or music.
>
http://www.youtube.com/watch?v=dh0woT7NkKI
>
> -- This message is brought to you from the keyboard of
Lord Androcles, Zeroth Earl of Medway
Apr 9, 2013, 1:37:21 AM4/9/13
to
"Hetware" wrote in message
news:spGdnRP1stmXOf7M...@megapath.net...
On 4/9/2013 12:44 AM, Lord Androcles, Zeroth Earl of Medway wrote:
> "Hetware" wrote in message
> I don't have time to follow up on this, but this is from a guy who
> doesn't spend his time trying to prove how smart he is. He just shows
> you what he thinks in the clearest manner available to him. The mark of
> a true genius.
>
> http://people.oregonstate.edu/~drayt/Courses/MTH437/2007/hw/geodesic.pdf
> =======================================
> Fine, but it has no more connection with physics than politics or music.
>
This is most apropos and irrelevant:
http://www.youtube.com/watch?v=dh0woT7NkKI
===========================================
I don't have time for horse drawn seed drill inventors, it was only a
variation on the odometer anyway.
http://en.wikipedia.org/wiki/Jethro_Tull_(agriculturist)
http://www.mlahanas.de/Greeks/HeronAlexandria-Dateien/image002.jpg
Is the guitar the only instrument you know, Mickey?
-- This message is brought to you from the keyboard of
news:spGdnRP1stmXOf7M...@megapath.net...
On 4/9/2013 12:44 AM, Lord Androcles, Zeroth Earl of Medway wrote:
> "Hetware" wrote in message
> I don't have time to follow up on this, but this is from a guy who
> doesn't spend his time trying to prove how smart he is. He just shows
> you what he thinks in the clearest manner available to him. The mark of
> a true genius.
>
> http://people.oregonstate.edu/~drayt/Courses/MTH437/2007/hw/geodesic.pdf
> =======================================
> Fine, but it has no more connection with physics than politics or music.
>
This is most apropos and irrelevant:
http://www.youtube.com/watch?v=dh0woT7NkKI
I don't have time for horse drawn seed drill inventors, it was only a
variation on the odometer anyway.
http://en.wikipedia.org/wiki/Jethro_Tull_(agriculturist)
http://www.mlahanas.de/Greeks/HeronAlexandria-Dateien/image002.jpg
Is the guitar the only instrument you know, Mickey?
-- This message is brought to you from the keyboard of
Dirk Van de moortel
Apr 9, 2013, 2:39:47 AM4/9/13
to
Indeed, but unless At^2 + Bt + C is indeed a perfect square,
the original equations don't produce a constant angle p(t), but
probably some function involving an arctangent, which is not
what MTW was looking for.
Dirk Vdm
Dirk Van de moortel
Apr 9, 2013, 2:57:33 AM4/9/13
to
"Hetware" <hat...@speakyeasy.net> schreef in bericht news:Duidna9WrZBIBv7M...@megapath.net...
> On 4/8/2013 9:27 AM, Dirk Van de moortel wrote:
>> Hetware <hat...@speakyeasy.net> wrote:
>>> This is the geodesic equation under discussion:
>>>
>>> d^2(r)/dt^2 = r(dp/dt)^2
>>>
>>> d^2(p)/dt^2 = -(2/r)(dp/dt)(dr/dt).
>>>
>>> r is radius in polar coordinates, p is the angle, and t is a path
>>> parameter.
>>> The authors ask me to "[S]olve the geodesic equation for r(t) and
>>> p(t), and show that the solution is a uniformly parametrized straight
>>> line(x===r cos(p) = at+p for some a and b; y===r sin(p) = jt+k for
>>> some j and k).
>>
>> Normally we'd write dotted variables, but with quotes it's easier.
>> So write
>
> Like this?
>
> <math xmlns='http://www.w3.org/1998/Math/MathML'
[snip]
>> Hetware <hat...@speakyeasy.net> wrote:
>>> This is the geodesic equation under discussion:
>>>
>>> d^2(r)/dt^2 = r(dp/dt)^2
>>>
>>> d^2(p)/dt^2 = -(2/r)(dp/dt)(dr/dt).
>>>
>>> r is radius in polar coordinates, p is the angle, and t is a path
>>> parameter.
>>> The authors ask me to "[S]olve the geodesic equation for r(t) and
>>> p(t), and show that the solution is a uniformly parametrized straight
>>> line(x===r cos(p) = at+p for some a and b; y===r sin(p) = jt+k for
>>> some j and k).
>>
>> Normally we'd write dotted variables, but with quotes it's easier.
>> So write
>
> Like this?
>
> <math xmlns='http://www.w3.org/1998/Math/MathML'
This is a text group :-)
Dirk Vdm
Lord Androcles, Zeroth Earl of Medway
Apr 9, 2013, 3:05:53 AM4/9/13
to
"Dirk Van de moortel" wrote in message
news:kk0e6p$7g2$1...@speranza.aioe.org...
==============================================
I knew you would snip, you are such a predictable imbecile.
-- "Dork Van de faggot" <dirkvand...@hotspam.not> wrote in message
news:ke1gcs$f2d$1...@speranza.aioe.org...
> No, imbecile, v' = 0.
"Dork Van de faggot" wrote in message news:kfp3ba$ku0$1...@speranza.aioe.org...
How hard is it to listen to the definitions and stick with the rules?
"Did you ever had algebra?" - Dork Van de faggot
"the transformation equations are valid only for speeds below or up to c" --
Dork Van de faggot
-- So if T = 5 years and v = 0.8c, then the stay at home twin will
have aged 10 years (2T) while his travelling twin sister will have
aged 6 years (2T/g). <no silly grin>
-- Psychodork Van de improper faggot.
news:kk0e6p$7g2$1...@speranza.aioe.org...
I knew you would snip, you are such a predictable imbecile.
-- "Dork Van de faggot" <dirkvand...@hotspam.not> wrote in message
news:ke1gcs$f2d$1...@speranza.aioe.org...
> Indeed, writing LT and inverse:
> x' = g ( x - v t ) [1]
> t' = g ( t - v x ) [2]
> x = g ( x' + v t' ) [3]
> t = g ( t' + v x' ) [4]
[ but v' = x'/t', the inverse velocity ]
> x' = g ( x - v t ) [1]
> t' = g ( t - v x ) [2]
> x = g ( x' + v t' ) [3]
> t = g ( t' + v x' ) [4]
> No, imbecile, v' = 0.
"Dork Van de faggot" wrote in message news:kfp3ba$ku0$1...@speranza.aioe.org...
How hard is it to listen to the definitions and stick with the rules?
"Did you ever had algebra?" - Dork Van de faggot
"the transformation equations are valid only for speeds below or up to c" --
Dork Van de faggot
-- So if T = 5 years and v = 0.8c, then the stay at home twin will
have aged 10 years (2T) while his travelling twin sister will have
aged 6 years (2T/g). <no silly grin>
-- Psychodork Van de improper faggot.
Hetware
Apr 9, 2013, 9:47:48 PM4/9/13
to
On 4/9/2013 1:37 AM, Lord Androcles, Zeroth Earl of Medway wrote:
> "Hetware" wrote in message
> news:spGdnRP1stmXOf7M...@megapath.net...
>
> On 4/9/2013 12:44 AM, Lord Androcles, Zeroth Earl of Medway wrote:
>> "Hetware" wrote in message
>
>> I don't have time to follow up on this, but this is from a guy who
>> doesn't spend his time trying to prove how smart he is. He just shows
>> you what he thinks in the clearest manner available to him. The mark of
>> a true genius.
>>
>> http://people.oregonstate.edu/~drayt/Courses/MTH437/2007/hw/geodesic.pdf
>> =======================================
>> Fine, but it has no more connection with physics than politics or music.
>>
>
> This is most apropos and irrelevant:
> http://www.youtube.com/watch?v=dh0woT7NkKI
>
> ===========================================
> I don't have time for horse drawn seed drill inventors, it was only a
> variation on the odometer anyway.
> http://en.wikipedia.org/wiki/Jethro_Tull_(agriculturist)
> http://www.mlahanas.de/Greeks/HeronAlexandria-Dateien/image002.jpg
>
> Is the guitar the only instrument you know, Mickey?
>
http://www.youtube.com/watch?v=qWabhnt91Uc
> "Hetware" wrote in message
> news:spGdnRP1stmXOf7M...@megapath.net...
>
> On 4/9/2013 12:44 AM, Lord Androcles, Zeroth Earl of Medway wrote:
>> "Hetware" wrote in message
>
>> I don't have time to follow up on this, but this is from a guy who
>> doesn't spend his time trying to prove how smart he is. He just shows
>> you what he thinks in the clearest manner available to him. The mark of
>> a true genius.
>>
>> http://people.oregonstate.edu/~drayt/Courses/MTH437/2007/hw/geodesic.pdf
>> =======================================
>> Fine, but it has no more connection with physics than politics or music.
>>
>
> This is most apropos and irrelevant:
> http://www.youtube.com/watch?v=dh0woT7NkKI
>
> ===========================================
> I don't have time for horse drawn seed drill inventors, it was only a
> variation on the odometer anyway.
> http://en.wikipedia.org/wiki/Jethro_Tull_(agriculturist)
> http://www.mlahanas.de/Greeks/HeronAlexandria-Dateien/image002.jpg
>
> Is the guitar the only instrument you know, Mickey?
>
Hetware
Apr 9, 2013, 10:43:57 PM4/9/13
to
<?xml version='1.0' encoding='UTF-8'?>
<!DOCTYPE html PUBLIC '-//W3C//DTD XHTML 1.1 plus MathML 2.0//EN'
'http://www.w3.org/Math/DTD/mathml2/xhtml-math11-f.dtd'>
<html xmlns='http://www.w3.org/1999/xhtml'>
<head>
<title>Or something like that</title>
</head>
<body>
<p>[MathML goes here.]</p>
</body>
</html>
Hetware
Apr 10, 2013, 12:04:25 AM4/10/13
to
Dirk Van de moortel
Apr 10, 2013, 4:24:22 AM4/10/13
to
Hetware <hat...@speakyeasy.net> wrote:
> On 4/9/2013 10:43 PM, Hetware wrote:
>> On 4/9/2013 2:57 AM, Dirk Van de moortel wrote:
>>>
>>> "Hetware" <hat...@speakyeasy.net> schreef in bericht
>>> news:Duidna9WrZBIBv7M...@megapath.net...
>>>> On 4/8/2013 9:27 AM, Dirk Van de moortel wrote:
>>>>> Hetware <hat...@speakyeasy.net> wrote:
>>>>>> This is the geodesic equation under discussion:
>>>>>>
>>>>>> d^2(r)/dt^2 = r(dp/dt)^2
>>>>>>
>>>>>> d^2(p)/dt^2 = -(2/r)(dp/dt)(dr/dt).
>>>>>>
>>>>>> r is radius in polar coordinates, p is the angle, and t is a path
>>>>>> parameter.
>>>>>> The authors ask me to "[S]olve the geodesic equation for r(t) and
>>>>>> p(t), and show that the solution is a uniformly parametrized
>>>>>> straight line(x===r cos(p) = at+p for some a and b; y===r sin(p)
>>>>>> = jt+k for some j and k).
>>>>>
>>>>> Normally we'd write dotted variables, but with quotes it's easier.
>>>>> So write
>>>>
>>>> Like this?
>>>>
>>>> <math xmlns='http://www.w3.org/1998/Math/MathML'
>>>
>>>
>>> [snip]
>>>
>>> This is a text group :-)
>>
>> Everything I posted was ASCII text.
Sure, ASCII text representing HTML.
> On 4/9/2013 10:43 PM, Hetware wrote:
>> On 4/9/2013 2:57 AM, Dirk Van de moortel wrote:
>>>
>>> "Hetware" <hat...@speakyeasy.net> schreef in bericht
>>> news:Duidna9WrZBIBv7M...@megapath.net...
>>>> On 4/8/2013 9:27 AM, Dirk Van de moortel wrote:
>>>>> Hetware <hat...@speakyeasy.net> wrote:
>>>>>> This is the geodesic equation under discussion:
>>>>>>
>>>>>> d^2(r)/dt^2 = r(dp/dt)^2
>>>>>>
>>>>>> d^2(p)/dt^2 = -(2/r)(dp/dt)(dr/dt).
>>>>>>
>>>>>> r is radius in polar coordinates, p is the angle, and t is a path
>>>>>> parameter.
>>>>>> The authors ask me to "[S]olve the geodesic equation for r(t) and
>>>>>> p(t), and show that the solution is a uniformly parametrized
>>>>>> straight line(x===r cos(p) = at+p for some a and b; y===r sin(p)
>>>>>> = jt+k for some j and k).
>>>>>
>>>>> Normally we'd write dotted variables, but with quotes it's easier.
>>>>> So write
>>>>
>>>> Like this?
>>>>
>>>> <math xmlns='http://www.w3.org/1998/Math/MathML'
>>>
>>>
>>> [snip]
>>>
>>> This is a text group :-)
>>
>> Everything I posted was ASCII text.
This is a text group.
>>
>> <?xml version='1.0' encoding='UTF-8'?>
>> <!DOCTYPE html PUBLIC '-//W3C//DTD XHTML 1.1 plus MathML 2.0//EN'
>> 'http://www.w3.org/Math/DTD/mathml2/xhtml-math11-f.dtd'>
>> <html xmlns='http://www.w3.org/1999/xhtml'>
>> <head>
>> <title>Or something like that</title>
>> </head>
>> <body>
>>
>> <p>[MathML goes here.]</p>
>>
>> </body>
>> </html>
>
> Ach! I mixed adjacent meta-levels. Scratch the '<p>[' and ']</p>'
your ASCII with <html> and <body> tags.
I saw nothing new from your OP.
Dirk Vdm
Wesley Burns
Apr 10, 2013, 12:40:42 PM4/10/13
to
Dirk Van de moortel wrote:
> Normally we'd write dotted variables, but with quotes it's easier.
> So write
> r' = dr/dt r'' = d^2(r)/dt^2
> p = dp/dt
> p'' = d^2(p)/dt^2
If your
> So write
> r' = dr/dt r'' = d^2(r)/dt^2
> p = dp/dt
> p'' = d^2(p)/dt^2
dr = (r1-r0)
dt = (t1-t0)
r' = (r1-r0) / (t1-t0)
how would you express your
r'' = d^2(r)/dt^2
Let me see your tensors, maybe you are wrong.
Dirk Van de moortel
Apr 11, 2013, 4:28:46 AM4/11/13
to
Wesley Burns <wb...@gmail.com> wrote:
> Dirk Van de moortel wrote:
>
>> Normally we'd write dotted variables, but with quotes it's easier.
>> So write
>> r' = dr/dt r'' = d^2(r)/dt^2
>> p = dp/dt
>> p'' = d^2(p)/dt^2
>
> If your
> dr = (r1-r0)
It isn't.
> Dirk Van de moortel wrote:
>
>> Normally we'd write dotted variables, but with quotes it's easier.
>> So write
>> r' = dr/dt r'' = d^2(r)/dt^2
>> p = dp/dt
>> p'' = d^2(p)/dt^2
>
> If your
> dr = (r1-r0)
Stop pretending to be an imbecile -- it's counterproductive.
Dirk Vdm
Wesley Burns
Apr 11, 2013, 11:19:25 AM4/11/13
to
Dirk Van de moortel wrote:
>> If your
>> dr = (r1-r0)
>
> It isn't.
> Stop pretending to be an imbecile -- it's counterproductive.
I wonder what it is then, is not d*t right?
>> If your
>> dr = (r1-r0)
>
> It isn't.
> Stop pretending to be an imbecile -- it's counterproductive.
Dont take it as an offence, I try to figure out whether you do your
models by yourself, or abstractizise at a higher level. How would you
implement those equation numerically, i mean evaluate, get a number out
of it. I do not want transformations from tables. Thanks
How would you interpret this one algorithmically?
r'' = d^2(r)/dt^2
rotchm
Apr 11, 2013, 11:25:04 AM4/11/13
to
> >> If your
> I wonder what it is then, is not d*t right?
> Dont take it as an offence, I try to figure out whether you do your
> models by yourself, or abstractizise at a higher level. How would you
> implement those equation numerically, i mean evaluate, get a number out
> of it. I do not want transformations from tables. Thanks
>
> How would you interpret this one algorithmically?
>
> r'' = d^2(r)/dt^2
You wont get smart that way.
Admitting to your failings is the first step to recovery.
Dono.
Apr 11, 2013, 11:40:45 AM4/11/13
to
reality, you are the same imbecile that was peddling LET over SR.
Dono.
Apr 11, 2013, 11:41:43 AM4/11/13
to
On Apr 11, 8:25 am, rotchm <rot...@gmail.com> wrote:
You are pretending to being smart when you are actually an imbecile.
Wesley Burns
Apr 11, 2013, 2:09:35 PM4/11/13
to
rotchm wrote:
>> How would you interpret this one algorithmically?
>>
>> r'' = d^2(r)/dt^2
>
> Stop pretending to be an imbecile to hide the fact that you are one.
Since you tell you ARE a scientists, how would you interpret that
>> How would you interpret this one algorithmically?
>>
>> r'' = d^2(r)/dt^2
>
> Stop pretending to be an imbecile to hide the fact that you are one.
equation algorithmically? I guess you own a PhD and do many models
in physics. What do you use, high-level, low-level?
Hetware
Apr 13, 2013, 12:55:27 AM4/13/13
to
Dono.
Apr 13, 2013, 1:47:06 AM4/13/13
to
On Apr 11, 8:25 am, rotchm <rot...@gmail.com> wrote:
You, Crotchm, pretend to be smart while you're actually an imbecile.
Dono.
Apr 13, 2013, 1:54:45 AM4/13/13
to
On Apr 8, 6:08 pm, rotchm <rot...@gmail.com> wrote:
> > which again with [1] gives
> > r r''' + 3 r'' r' = 0
>
> > So if you find some r(t) that has r''(t) = r'''(t) = 0 for all t,
> > you're done. Easy. Take
> > r(t) = A t + B [3]
>
> Yes, thats a "trivial" solution and as you hinted, there are others.
> The general solution to this is
> r = s(At2 + Bt + C), which can be ~ at + b if its a perfect square.
>
No, pretentious imbecile, what you put down is not a solution since r"
> > which again with [1] gives
> > r r''' + 3 r'' r' = 0
>
> > So if you find some r(t) that has r''(t) = r'''(t) = 0 for all t,
> > you're done. Easy. Take
> > r(t) = A t + B [3]
>
> Yes, thats a "trivial" solution and as you hinted, there are others.
> The general solution to this is
> r = s(At2 + Bt + C), which can be ~ at + b if its a perfect square.
>
isn't null for all t (while r"'=0). You are not only a pretender, you
are an outright idiot. And to think that the other french idiot, YBM,
rushes to suck your cock and lick your ass every time you post one of
your patented idiocies.
Lord Androcles, Zeroth Earl of Medway
Apr 13, 2013, 2:15:42 AM4/13/13
to
"Hetware" wrote in message
news:TsOdnb5lEIHZePXM...@megapath.net...
==========================================================
<yawn>
news:TsOdnb5lEIHZePXM...@megapath.net...
<yawn>
Dono.
Apr 13, 2013, 10:03:58 AM4/13/13
to
rotchm
Apr 13, 2013, 10:55:37 AM4/13/13
to
On Apr 13, 1:54 am, "Dono." <sa...@comcast.net> wrote:
> On Apr 8, 6:08 pm, rotchm <rot...@gmail.com> wrote:> > which again with [1] gives
> > > r r''' + 3 r'' r' = 0
>
> > > So if you find some r(t) that has r''(t) = r'''(t) = 0 for all t,
> > > you're done. Easy. Take
> > > r(t) = A t + B [3]
>
> > Yes, thats a "trivial" solution and as you hinted, there are others.
> > The general solution to this is
> > r = s(At2 + Bt + C), which can be ~ at + b if its a perfect square.
>
> No, pretentious imbecile, what you put down is not a solution since r"
> isn't null for all t (while r"'=0).
1) ?? Didnt you see DvDM's reply to me? He agreed that what I put down
> On Apr 8, 6:08 pm, rotchm <rot...@gmail.com> wrote:> > which again with [1] gives
> > > r r''' + 3 r'' r' = 0
>
> > > So if you find some r(t) that has r''(t) = r'''(t) = 0 for all t,
> > > you're done. Easy. Take
> > > r(t) = A t + B [3]
>
> > Yes, thats a "trivial" solution and as you hinted, there are others.
> > The general solution to this is
> > r = s(At2 + Bt + C), which can be ~ at + b if its a perfect square.
>
> No, pretentious imbecile, what you put down is not a solution since r"
> isn't null for all t (while r"'=0).
is a solution
to r r''' + 3 r'' r' = 0.
Try it. You can do it by direct substitution!! TRY IT !!
2) You seem to impose that r'' must equal zero. Why? It need not be.
3) Now, you seem to have also misinterpreted (your incapacity to
deduce) the notation of s(...). DvDM did understand it. You didnt read
those who replied to me? That is quite sloppy on your part.
4) Try again.
Dono.
Apr 13, 2013, 11:05:04 AM4/13/13
to
On Apr 13, 7:55 am, rotchm <rot...@gmail.com> wrote:
> On Apr 13, 1:54 am, "Dono." <sa...@comcast.net> wrote:
>
> > On Apr 8, 6:08 pm, rotchm <rot...@gmail.com> wrote:> > which again with [1] gives
> > > > r r''' + 3 r'' r' = 0
>
> > > > So if you find some r(t) that has r''(t) = r'''(t) = 0 for all t,
> > > > you're done. Easy. Take
> > > > r(t) = A t + B [3]
>
> > > Yes, thats a "trivial" solution and as you hinted, there are others.
> > > The general solution to this is
> > > r = s(At2 + Bt + C), which can be ~ at + b if its a perfect square.
>
> > No, pretentious imbecile, what you put down is not a solution since r"
> > isn't null for all t (while r"'=0).
>
> 1) ?? Didnt you see DvDM's reply to me? He agreed that what I put down
> is a solution
> to r r''' + 3 r'' r' = 0.
>
> Try it. You can do it by direct substitution!! TRY IT !!
>
No imbecile, he didn't agree, he simply did not notice your utter
> On Apr 13, 1:54 am, "Dono." <sa...@comcast.net> wrote:
>
> > On Apr 8, 6:08 pm, rotchm <rot...@gmail.com> wrote:> > which again with [1] gives
> > > > r r''' + 3 r'' r' = 0
>
> > > > So if you find some r(t) that has r''(t) = r'''(t) = 0 for all t,
> > > > you're done. Easy. Take
> > > > r(t) = A t + B [3]
>
> > > Yes, thats a "trivial" solution and as you hinted, there are others.
> > > The general solution to this is
> > > r = s(At2 + Bt + C), which can be ~ at + b if its a perfect square.
>
> > No, pretentious imbecile, what you put down is not a solution since r"
> > isn't null for all t (while r"'=0).
>
> 1) ?? Didnt you see DvDM's reply to me? He agreed that what I put down
> is a solution
> to r r''' + 3 r'' r' = 0.
>
> Try it. You can do it by direct substitution!! TRY IT !!
>
cretinism.
> 2) You seem to impose that r'' must equal zero. Why? It need not be.
>
the equation reduces to 3 r'' r' = 0
Crotchm,
You are a patented imbecile.
Dono.
Apr 13, 2013, 7:12:19 PM4/13/13
to
On Apr 8, 6:08 pm, rotchm <rot...@gmail.com> wrote:
>
> Yes, thats a "trivial" solution and as you hinted, there are others.
> The general solution to this is
> r = s(At2 + Bt + C), which can be ~ at + b if its a perfect square.
>
You mean r=\sqrt(At^2+Bt+C), arithmetic ignoramus.
>
> Yes, thats a "trivial" solution and as you hinted, there are others.
> The general solution to this is
> r = s(At2 + Bt + C), which can be ~ at + b if its a perfect square.
>
r = s(At2 + Bt + C) is nothing but a polynomial degree 2 multiplied
by "s", whatever that may be. And, as such, this isn't a solution for
the ODE.
You can't even write the solution correctly. I hate to break it to you
but even as you copied this solution from a free software package, it
isn't the correct solution. You were too lazy in your copying, A, B
and C aren't arbitrary, they are tied to each other through some
constants of integration, c1=r(0), c2=r'(0) and c3=r"(0), as the final
solution needs to depend on the initial conditions. But you are too
big of an idiot to know this.
rotchm
Apr 13, 2013, 7:59:31 PM4/13/13
to
On Apr 13, 7:12 pm, "Dono." <sa...@comcast.net> wrote:
> > r = s(At2 + Bt + C), which can be ~ at + b if its a perfect square.
>
> You mean r=\sqrt(At^2+Bt+C), arithmetic ignoramus.
Smart people knew what it meant. DvDM knew it. You read his post and
> > r = s(At2 + Bt + C), which can be ~ at + b if its a perfect square.
>
> You mean r=\sqrt(At^2+Bt+C), arithmetic ignoramus.
still failed to understand the notation. The I explicited it to you
and yo STILL failed to understand it.
> A, B and C aren't arbitrary, they are tied to each other through some
> constants of integration, c1=r(0), c2=r'(0) and c3=r"(0),
You are again trying to hide your failures....
Dono.
Apr 13, 2013, 8:03:47 PM4/13/13
to
On Apr 13, 4:59 pm, rotchm <rot...@gmail.com> wrote:
> On Apr 13, 7:12 pm, "Dono." <sa...@comcast.net> wrote:
>
> > > r = s(At2 + Bt + C), which can be ~ at + b if its a perfect square.
>
> > You mean r=\sqrt(At^2+Bt+C), arithmetic ignoramus.
>
> Smart people knew what it meant.
Only an imbecile like you would write s(At2 + Bt + C). You are
> On Apr 13, 7:12 pm, "Dono." <sa...@comcast.net> wrote:
>
> > > r = s(At2 + Bt + C), which can be ~ at + b if its a perfect square.
>
> > You mean r=\sqrt(At^2+Bt+C), arithmetic ignoramus.
>
> Smart people knew what it meant.
obviously ignorant of the very basics of mathematical notation,
pretender.
> The I explicited it to you
> and yo STILL failed to understand it.
>
> > A, B and C aren't arbitrary, they are tied to each other through some
> > constants of integration, c1=r(0), c2=r'(0) and c3=r"(0),
>
> Well DuH... tell me something I dont know!
>
Dono.
Apr 13, 2013, 8:08:07 PM4/13/13
to
On Apr 13, 4:59 pm, rotchm <rot...@gmail.com> wrote:
> On Apr 13, 7:12 pm, "Dono." <sa...@comcast.net> wrote:
>
> > > r = s(At2 + Bt + C), which can be ~ at + b if its a perfect square.
>
> > You mean r=\sqrt(At^2+Bt+C), arithmetic ignoramus.
>
>
> > > r = s(At2 + Bt + C), which can be ~ at + b if its a perfect square.
>
> > You mean r=\sqrt(At^2+Bt+C), arithmetic ignoramus.
>
> The I explicited it to you and yo STILL failed to understand it.
You are not only an imbecile, you are also a liar, here is what you
wrote:
https://groups.google.com/group/sci.physics.relativity/msg/5c060547daa3491d?hl=en&dmode=source
>
> > A, B and C aren't arbitrary, they are tied to each other through some
> > constants of integration, c1=r(0), c2=r'(0) and c3=r"(0),
>
> Well DuH... tell me something I dont know!
>
rotchm
Apr 13, 2013, 8:20:22 PM4/13/13
to
On Apr 13, 8:03 pm, "Dono." <sa...@comcast.net> wrote:
> You obviously don't, since you wrote A,B and C.
Obviously those constant represent some constants depending on the
> You obviously don't, since you wrote A,B and C.
constants of integration. Its is so obvious to smart people like me
and DvDM that we dont need to specify it. This is what DvDM wrote too.
"At + B" w/o ever invoking initial conditions (as r(0) etc). But
you... you always fail to understand.
Dono.
Apr 13, 2013, 8:22:37 PM4/13/13
to
On Apr 13, 5:20 pm, rotchm <rot...@gmail.com> wrote:
> On Apr 13, 8:03 pm, "Dono." <sa...@comcast.net> wrote:
>
> > You obviously don't, since you wrote A,B and C.
>
> Obviously those constant represent some constants depending on the
> constants of integration.
Ok, pretender, A has a very specific expression, what is the
> On Apr 13, 8:03 pm, "Dono." <sa...@comcast.net> wrote:
>
> > You obviously don't, since you wrote A,B and C.
>
> Obviously those constant represent some constants depending on the
> constants of integration.
expression for A?
> Its is so obvious to smart people like me
Dono.
Apr 13, 2013, 8:24:22 PM4/13/13
to
On Apr 7, 9:05 pm, Hetware <hatt...@speakyeasy.net> wrote:
>
> I am banned for life for mentioning the name Germar Rudolf. I don't like
> zionazis. Never have, and never will.
What do we have here? A neo-nazi, it seems.
>
> I am banned for life for mentioning the name Germar Rudolf. I don't like
> zionazis. Never have, and never will.
What do we have here? A neo-nazi, it seems.
rotchm
Apr 13, 2013, 9:59:52 PM4/13/13
to
On Apr 13, 8:08 pm, "Dono." <sa...@comcast.net> wrote:
> > The I explicited it to you and yo STILL failed to understand it.
>
> You are not only an imbecile, you are also a liar, here is what you
> wrote:
>
> https://groups.google.com/group/sci.physics.relativity/msg/5c060547da...
> > The I explicited it to you and yo STILL failed to understand it.
>
> You are not only an imbecile, you are also a liar, here is what you
> wrote:
>
Ha! that support my claim idiot.
DvDM replied to me on April 9th, where he emphasizes s(...) =
sqrt(...) and agrees with my answer.
You posted you disagreement on April 13. You had a few days to read
all of the posts but failed to or failed to understand them.
Then AFTER your post of april 13 I replied to you pointing out to
you what was to be understood by s(...).
Then AFTER that, after you read what Dirk said, you posted back
falsely claiming that Dirk didnt agree with me and that you still
disagreed with me.
That was the timeline, all kept by Google for all to see.
You were informed what s(...) meant and after all that you still
disagreed.
Google kept a record of your inability to deduce s(...) and of your
inability to read (what DvDM wrote).
Eventually you did get it.
> OK, let's try a simple test, what is the expression for A?
and now you want us to hold your pencil !? Grow up!
Tell you what, you give A and I will give you C.
Dono.
Apr 13, 2013, 10:17:15 PM4/13/13
to
On Apr 13, 6:59 pm, rotchm <rot...@gmail.com> wrote:
> On Apr 13, 8:08 pm, "Dono." <sa...@comcast.net> wrote:
>
> > > The I explicited it to you and yo STILL failed to understand it.
>
> > You are not only an imbecile, you are also a liar, here is what you
> > wrote:
>
> >https://groups.google.com/group/sci.physics.relativity/msg/5c060547da...
>
> Ha! that support my claim
>
No, it simply points out that you are lying and that you continue to
> On Apr 13, 8:08 pm, "Dono." <sa...@comcast.net> wrote:
>
> > > The I explicited it to you and yo STILL failed to understand it.
>
> > You are not only an imbecile, you are also a liar, here is what you
> > wrote:
>
> >https://groups.google.com/group/sci.physics.relativity/msg/5c060547da...
>
> Ha! that support my claim
>
lie.
https://groups.google.com/group/sci.physics.relativity/msg/5c060547daa3491d?hl=en&dmode=source
> Then AFTER your post of april 13 I replied to you pointing out to
> you what was to be understood by s(...).
>
> Google kept a record of your inability to deduce s(...) and of your
> inability to read (what DvDM wrote).
>
you.
> > OK, let's try a simple test, what is the expression for A?
>
> Ha! We solved the ODE for you,
> we explained it to you a zillion times
> and now you want us to hold your pencil !? Grow up!
>
> Tell you what, you give A and I will give you C.
value for A, B and C. I'll call your bluff:
C=c_1*c_3^2-c_2/c_1
A is much simpler, so what is A, Crotchm?
rotchm
Apr 13, 2013, 11:01:07 PM4/13/13
to
On Apr 13, 10:17 pm, "Dono." <sa...@comcast.net> wrote:
> > Tell you what, you give A and I will give you C.
>
> You have no clue. You are still trolling the internet to find out the
> value for A, B and C. I'll call your bluff:
>
> C=c_1*c_3^2-c_2/c_1
>
> A is much simpler, so what is A, Crotchm?
I asked you to give A. You failed; you gave an answer for C, not A.
> > Tell you what, you give A and I will give you C.
>
> You have no clue. You are still trolling the internet to find out the
> value for A, B and C. I'll call your bluff:
>
> C=c_1*c_3^2-c_2/c_1
>
> A is much simpler, so what is A, Crotchm?
Now lets see what you answer is.
Recap the problem: r = sqr(At^2 + Bt + C)
Then YOU wanted to include c1=r(0), c2=r'(0) and c3=r"(0)
Ok...
r(0) = sqr(A*0^2+B*0 + C) = sqr(C). Thus
c1 = sqr(C) thus C = c1^2. This is much different from your answer of
> C=c_1*c_3^2-c_2/c_1
You failed again.
Its useless to teach you math and physics. You need to start over at
the kindergarten level, something we dont do here in this NG.
Dono.
Apr 13, 2013, 11:57:31 PM4/13/13
to
On Apr 13, 8:01 pm, rotchm <rot...@gmail.com> wrote:
> On Apr 13, 10:17 pm, "Dono." <sa...@comcast.net> wrote:
>
> > > Tell you what, you give A and I will give you C.
>
> > You have no clue. You are still trolling the internet to find out the
> > value for A, B and C. I'll call your bluff:
>
> > C=c_1*c_3^2-c_2/c_1
>
> > A is much simpler, so what is A, Crotchm?
>
> I asked you to give A. You failed; you gave an answer for C, not A.
>
> Now lets see what you answer is.
>
> Recap the problem: r = sqr(At^2 + Bt + C)
> Then YOU wanted to include c1=r(0), c2=r'(0) and c3=r"(0)
>
> Ok...
>
> r(0) = sqr(A*0^2+B*0 + C) = sqr(C). Thus
> c1 = sqr(C) thus C = c1^2. This is much different from your answer of
>
> > C=c_1*c_3^2-c_2/c_1
>
> You failed again.
>
Err, you have no idea what the value for A, B and C is.
> On Apr 13, 10:17 pm, "Dono." <sa...@comcast.net> wrote:
>
> > > Tell you what, you give A and I will give you C.
>
> > You have no clue. You are still trolling the internet to find out the
> > value for A, B and C. I'll call your bluff:
>
> > C=c_1*c_3^2-c_2/c_1
>
> > A is much simpler, so what is A, Crotchm?
>
> I asked you to give A. You failed; you gave an answer for C, not A.
>
> Now lets see what you answer is.
>
> Recap the problem: r = sqr(At^2 + Bt + C)
> Then YOU wanted to include c1=r(0), c2=r'(0) and c3=r"(0)
>
> Ok...
>
> r(0) = sqr(A*0^2+B*0 + C) = sqr(C). Thus
> c1 = sqr(C) thus C = c1^2. This is much different from your answer of
>
> > C=c_1*c_3^2-c_2/c_1
>
> You failed again.
>
C=c_1*c_3^2-c_2/c_1
and
A=c1
Last chance, fraudster, what is the value for B?
You aren't only an imbecile, you are a fraud as well.
Hetware
Apr 14, 2013, 5:12:54 PM4/14/13
to
Dono.
Apr 14, 2013, 5:54:13 PM4/14/13
to
0 new messages