On Mar 4, 9:32 pm, Koobee Wublee wrote:
On Mar 4, 6:03 pm, wrote:

In sci.physics Koobee Wublee wrote:
On Mar 4, 9:59 am, wrote:
In sci.physics Koobee Wublee wrote:
2 O ~= pi + 2 u (1 + 1 / B^2)

This is correct.
But according to you, Lightman gave a result below.
2 O = pi + 2 u (1 + B^2)

I did not say that. I said that the result was proportional to
1+B^2. It is -- the GR prediction is 2u/B^2 (1+B^2), that is,
1+B^2 times the Newtonian prediction for a given speed and impact
parameter.

You were very misleading then. You left out a very important factor,
1 / B^2.

I suggest that you try for a graceful retreat.

I did say the result was not what I initially expected. What else do
you want me to do?

However, I need no graceful retreat from the comment on the absurdity
of the geodesics following the path with the least accumulated
spacetime. When Christoffel derived the geodesic equations, there was
nothing wrong with his assumption that the shortest distance between
two points in 3-dimensional space is the actual distance through each
local point even if that space is curved. It is not necessarily a
straight line as an observer observes it using his choice of
coordinate system. However, when the Goettingen group of
mathematicians including Hilbert, Klein, Schwarzschild, and Minkowski
worked on this problem, they simply extended to the 4-dimensional
spacetime. In doing so, no one really thought out except 100 years
later by yours truly that this model does not allow photons to
propagate with a coherent path. Are you still standing by your
conjecture that this model of geodesic is a valid one?

Oh look, another stunning display of your typical arrogant stupidity.
Learn what the parameterizations you use mean, then you won't make
cosmic ****ups like that over and over and over.

On Mar 4, 6:03 pm, wrote:

I suggest that you try for a graceful retreat.

Well, I found a mistake in the boundary condition. As you have
suggested, I will execute a graceful retreat this time. In doing so,
my instinct might still be correct about any high-speed particle
having a discontinuity as its speed goes to the speed of light.

On Mar 5, 9:55 pm, Koobee Wublee wrote:
On Mar 4, 6:03 pm, wrote:

I suggest that you try for a graceful retreat.

Well, I found a mistake in the boundary condition. As you have
suggested, I will execute a graceful retreat this time. In doing so,
my instinct might still be correct about any high-speed particle
having a discontinuity as its speed goes to the speed of light.

My god your arrogance is astounding. DO THE COMPUTATION.

In sci.physics.relativity, Eric Gisse

wrote
on Thu, 6 Mar 2008 06:50:41 -0800 (PST)
:
On Mar 5, 9:55 pm, Koobee Wublee wrote:
On Mar 4, 6:03 pm, wrote:

I suggest that you try for a graceful retreat.

Well, I found a mistake in the boundary condition. As you have
suggested, I will execute a graceful retreat this time. In doing so,
my instinct might still be correct about any high-speed particle
having a discontinuity as its speed goes to the speed of light.

My god your arrogance is astounding. DO THE COMPUTATION.

What math would you have him do? ;-) There is indeed a
discontinuity in the SR energy equation

E = m c^2 / sqrt(1-v^2/c^2)

Going through infinity to become imaginary energy sounds like
a pretty big discontinuity to me....

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Supertroll Eric Gisse trolled:

Yes, more things to confuse poor Eric.

http://www.helinium.nl/trolltech.gif

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"The Ghost In The Machine" wrote in message
...
| In sci.physics.relativity, Eric Gisse
|
| wrote
| on Thu, 6 Mar 2008 06:50:41 -0800 (PST)
| :
| On Mar 5, 9:55 pm, Koobee Wublee wrote:
| On Mar 4, 6:03 pm, wrote:
|
| I suggest that you try for a graceful retreat.
|
| Well, I found a mistake in the boundary condition. As you have
| suggested, I will execute a graceful retreat this time. In doing so,
| my instinct might still be correct about any high-speed particle
| having a discontinuity as its speed goes to the speed of light.
|
| My god your arrogance is astounding. DO THE COMPUTATION.
|
| What math would you have him do? ;-) There is indeed a
| discontinuity in the SR energy equation
|
| E = m c^2 / sqrt(1-v^2/c^2)
|
| Going through infinity to become imaginary energy sounds like
| a pretty big discontinuity to me....

v = 2c here,
http://hands-on-cern.physto.se/ani/a.../lhc_atlas.swf
so iE = mc^2

v is not a velocity, though, it's a closing rate; just ask Professor Poe.

Actually the discontinuity is between the real world and
theoretical (aka ****head) physics.

On Mar 6, 9:38*am, The Ghost In The Machine
wrote:
In sci.physics.relativity, Eric Gisse

*wrote
on Thu, 6 Mar 2008 06:50:41 -0800 (PST)
:

On Mar 5, 9:55 pm, Koobee Wublee wrote:
On Mar 4, 6:03 pm, wrote:

I suggest that you try for a graceful retreat.

Well, I found a mistake in the boundary condition. *As you have
suggested, I will execute a graceful retreat this time. *In doing so,
my instinct might still be correct about any high-speed particle
having a discontinuity as its speed goes to the speed of light.

My god your arrogance is astounding. DO THE COMPUTATION.

What math would you have him do? *;-) *There is indeed a
discontinuity in the SR energy equation

E = m c^2 / sqrt(1-v^2/c^2)

Going through infinity to become imaginary energy sounds like
a pretty big discontinuity to me....

There's no going *through* infinity. There is an *approach* to
infinity. A function that has an infinite asymptote is not
discontinuous.


--
#191,
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On Mar 6, 6:38 am, The Ghost In The Machine
wrote:
In sci.physics.relativity, Eric Gisse

wrote
on Thu, 6 Mar 2008 06:50:41 -0800 (PST)
:

On Mar 5, 9:55 pm, Koobee Wublee wrote:
On Mar 4, 6:03 pm, wrote:

I suggest that you try for a graceful retreat.

Well, I found a mistake in the boundary condition. As you have
suggested, I will execute a graceful retreat this time. In doing so,
my instinct might still be correct about any high-speed particle
having a discontinuity as its speed goes to the speed of light.

My god your arrogance is astounding. DO THE COMPUTATION.

What math would you have him do? ;-) There is indeed a
discontinuity in the SR energy equation

E = m c^2 / sqrt(1-v^2/c^2)

Going through infinity to become imaginary energy sounds like
a pretty big discontinuity to me....

GR, not SR.


--
#191,
/dev/signatu Not a text file

--
Posted via a free Usenet account fromhttp://www.teranews.com

In sci.physics.relativity, Androcles

wrote
on Thu, 06 Mar 2008 16:37:11 GMT
:

"The Ghost In The Machine" wrote in message
...
| In sci.physics.relativity, Eric Gisse
|
| wrote
| on Thu, 6 Mar 2008 06:50:41 -0800 (PST)
| :
| On Mar 5, 9:55 pm, Koobee Wublee wrote:
| On Mar 4, 6:03 pm, wrote:
|
| I suggest that you try for a graceful retreat.
|
| Well, I found a mistake in the boundary condition. As you have
| suggested, I will execute a graceful retreat this time. In doing so,
| my instinct might still be correct about any high-speed particle
| having a discontinuity as its speed goes to the speed of light.
|
| My god your arrogance is astounding. DO THE COMPUTATION.
|
| What math would you have him do? ;-) There is indeed a
| discontinuity in the SR energy equation
|
| E = m c^2 / sqrt(1-v^2/c^2)
|
| Going through infinity to become imaginary energy sounds like
| a pretty big discontinuity to me....

v = 2c here,
http://hands-on-cern.physto.se/ani/a.../lhc_atlas.swf
so iE = mc^2

v is not a velocity, though, it's a closing rate; just ask Professor Poe.

Actually the discontinuity is between the real world and
theoretical (aka ****head) physics.


Very true. The collision energy of two photons, however,
is 1/2 mc^2 + 1/2 mc^2 = mc^2 in Newtonian space,
not 1/2 m (2c)^2 = 2mc^2.

In any event, there's no discontinuity at v = c.

(BTW, if E = 1/2 m v^2, which way is the vector E pointing?)

--
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elderberries!" - Monty Python and the Holy Grail

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"PD" wrote in message
...
On Mar 6, 9:38 am, The Ghost In The Machine
wrote:
In sci.physics.relativity, Eric Gisse

wrote
on Thu, 6 Mar 2008 06:50:41 -0800 (PST)
:

On Mar 5, 9:55 pm, Koobee Wublee wrote:
On Mar 4, 6:03 pm, wrote:

I suggest that you try for a graceful retreat.

Well, I found a mistake in the boundary condition. As you have
suggested, I will execute a graceful retreat this time. In doing so,
my instinct might still be correct about any high-speed particle
having a discontinuity as its speed goes to the speed of light.

My god your arrogance is astounding. DO THE COMPUTATION.

What math would you have him do? ;-) There is indeed a
discontinuity in the SR energy equation

E = m c^2 / sqrt(1-v^2/c^2)

Going through infinity to become imaginary energy sounds like
a pretty big discontinuity to me....

| There's no going *through* infinity. There is an *approach* to
| infinity. A function that has an infinite asymptote is not
| discontinuous.


y = tan(x) for x = 0 to pi has no discontinuity at pi/2?
Why yes, yes it does.

****ing idiot!
HAHAHAHAHAHA!