"The Ghost In The Machine" wrote in message
...
| 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,

Oh, the LHC is a photon collider, is it? Why not call
it the Large Photon Collider, I wonder?

On Mar 4, 3:40*am, "Y.Porat" wrote:
On Mar 4, 2:53*am, wrote:





Apologies if this is a duplicate -- I'm having some news problems.

In sci.physics Koobee Wublee wrote:

On Feb 25, 7:47 am, Tom Roberts wrote:
The best model we have for the propagation of light near a massive
object like the sun is GR, in which the curvature of spacetime is the
important aspect in determining the path light follows. And it agrees
with measurements to part-per-million accuracy over an enormous range..
First, derive a set of geodesic equations a massed particle traveling
at high speed near the sun. *Then, gradually reducing the mass to zero
and increasing the speed to c, do you see a discontinuity at mass = 0
and speed = c?

This is definitely a worthwhile exercise. *I recommend that you do it.
If you get stuck, you can find the details in Lightman et al., _Problem
book in relativity and gravitation_, problem 15.9.

As you know, the geodesic equations are independent of mass. *What
does that tell you when the model predicts a 1x deflection traveling
at speed just a hair below c and suddenly jumps to 2x deflection at
speed = c?

It doesn't. *The model predicts a deflection proportional to 1+v^2/c^2,
which varies smoothly from the "Newtonian" value of 1 for small velocities
to 2 as v approaches c.

The moral is that before you decide that a model doesn't make sense,
you should check what the model actually predicts.

Steve Carlip

------------------
lie!!
the model ddint predict
it was **fiddling** the data to the model

Nitwit. If you could actually do the analysis then you might not be so
inclined to make such a gargantuan fool of yourself. Calling somebody
a liar, based on ignorance, just proves you're a blusterinng
ignoramous.

Y.Porat
------------------------------ Hide quoted text -

- Show quoted text -

On Mar 6, 9:19*am, "Androcles" wrote:
"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.

Graph it nitwit.

tan pi = .054886...
tan pi/2 = .027422...
tan pi/100 = .0005483...
tan pi/1000 = .00005483...


****ing idiot!
HAHAHAHAHAHA!- Hide quoted text -

- Show quoted text -

wrote in message
...
On Mar 4, 3:40 am, "Y.Porat" wrote:
On Mar 4, 2:53 am, wrote:





Apologies if this is a duplicate -- I'm having some news problems.

In sci.physics Koobee Wublee wrote:

On Feb 25, 7:47 am, Tom Roberts wrote:
The best model we have for the propagation of light near a massive
object like the sun is GR, in which the curvature of spacetime is the
important aspect in determining the path light follows. And it agrees
with measurements to part-per-million accuracy over an enormous
range.
First, derive a set of geodesic equations a massed particle traveling
at high speed near the sun. Then, gradually reducing the mass to zero
and increasing the speed to c, do you see a discontinuity at mass = 0
and speed = c?

This is definitely a worthwhile exercise. I recommend that you do it.
If you get stuck, you can find the details in Lightman et al., _Problem
book in relativity and gravitation_, problem 15.9.

As you know, the geodesic equations are independent of mass. What
does that tell you when the model predicts a 1x deflection traveling
at speed just a hair below c and suddenly jumps to 2x deflection at
speed = c?

It doesn't. The model predicts a deflection proportional to 1+v^2/c^2,
which varies smoothly from the "Newtonian" value of 1 for small
velocities
to 2 as v approaches c.

The moral is that before you decide that a model doesn't make sense,
you should check what the model actually predicts.

Steve Carlip

------------------
lie!!
the model ddint predict
it was **fiddling** the data to the model

| Nitwit. If you could actually do the analysis then you might not be so
| inclined to make such a gargantuan fool of yourself. Calling somebody
| a liar, based on ignorance, just proves you're a blusterinng
| ignoramous.

Ignorant ****head. You've just proven you're a "blusterinng ignoramous"
who has made a gargantuan fool of himself by being unable to spell
"blustering ignoramus".

On Mar 6, 5:13*pm, "Androcles" wrote:
wrote in message

...
On Mar 4, 3:40 am, "Y.Porat" wrote:





On Mar 4, 2:53 am, wrote:

Apologies if this is a duplicate -- I'm having some news problems.

In sci.physics Koobee Wublee wrote:

On Feb 25, 7:47 am, Tom Roberts wrote:
The best model we have for the propagation of light near a massive
object like the sun is GR, in which the curvature of spacetime is the
important aspect in determining the path light follows. And it agrees
with measurements to part-per-million accuracy over an enormous
range.
First, derive a set of geodesic equations a massed particle traveling
at high speed near the sun. Then, gradually reducing the mass to zero
and increasing the speed to c, do you see a discontinuity at mass = 0
and speed = c?

This is definitely a worthwhile exercise. I recommend that you do it.
If you get stuck, you can find the details in Lightman et al., _Problem
book in relativity and gravitation_, problem 15.9.

As you know, the geodesic equations are independent of mass. What
does that tell you when the model predicts a 1x deflection traveling
at speed just a hair below c and suddenly jumps to 2x deflection at
speed = c?

It doesn't. The model predicts a deflection proportional to 1+v^2/c^2,
which varies smoothly from the "Newtonian" value of 1 for small
velocities
to 2 as v approaches c.

The moral is that before you decide that a model doesn't make sense,
you should check what the model actually predicts.

Steve Carlip

------------------
lie!!
the model ddint predict
it was **fiddling** the data to the model

| Nitwit. If you could actually do the analysis then you might not be so
| inclined to make such a gargantuan fool of yourself. Calling somebody
| a liar, based on ignorance, just proves you're a blusterinng
| ignoramous.

Ignorant ****head. You've just proven you're a "blusterinng ignoramous"
who has made a gargantuan fool of himself by being unable to spell
"blustering ignoramus".- Hide quoted text -

Goofball, at least I know there is no discontinuity @ tan pi/2. My
guess is, in your drunken stupor, you punched in tan pi 2 in your hand
help calc.. Dummy.

- Show quoted text -

On Mar 6, 4:53*pm, wrote:
On Mar 6, 9:19*am, "Androcles" wrote:

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

*Graph it nitwit.

tan pi = .054886...
tan pi/2 = .027422...
tan pi/100 = .0005483...
tan pi/1000 = .00005483...

Are you really that ignorant? Try set your calculator to 'radian'
mode instead of the default 'degree' mode. Duh!

Is this one of Dono's many different identities? Sort of like multi-
personality disorder. However, each one is as stupid as the next.
Unbelievable!

wrote in message
...
On Mar 6, 9:19 am, "Androcles" wrote:
"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.

| Graph it nitwit.

| tan pi = .054886...
| tan pi/2 = .027422...
| tan pi/100 = .0005483...
| tan pi/1000 = .00005483...

HAHAHAHA!

Try pressing the radians button on your calculator, ****wit,
or entering 180 degrees for pi radians.

tan 180 degrees = 0
tan pi radians = 0
tan 90 degrees = ????
tan pi/2 radians = ????

What a dumb****, calling me nitwit!

wrote in message
...
On Mar 6, 5:13 pm, "Androcles" wrote:
wrote in message

...
On Mar 4, 3:40 am, "Y.Porat" wrote:





On Mar 4, 2:53 am, wrote:

Apologies if this is a duplicate -- I'm having some news problems.

In sci.physics Koobee Wublee wrote:

On Feb 25, 7:47 am, Tom Roberts wrote:
The best model we have for the propagation of light near a massive
object like the sun is GR, in which the curvature of spacetime is
the
important aspect in determining the path light follows. And it
agrees
with measurements to part-per-million accuracy over an enormous
range.
First, derive a set of geodesic equations a massed particle
traveling
at high speed near the sun. Then, gradually reducing the mass to
zero
and increasing the speed to c, do you see a discontinuity at mass =
0
and speed = c?

This is definitely a worthwhile exercise. I recommend that you do it.
If you get stuck, you can find the details in Lightman et al.,
_Problem
book in relativity and gravitation_, problem 15.9.

As you know, the geodesic equations are independent of mass. What
does that tell you when the model predicts a 1x deflection traveling
at speed just a hair below c and suddenly jumps to 2x deflection at
speed = c?

It doesn't. The model predicts a deflection proportional to 1+v^2/c^2,
which varies smoothly from the "Newtonian" value of 1 for small
velocities
to 2 as v approaches c.

The moral is that before you decide that a model doesn't make sense,
you should check what the model actually predicts.

Steve Carlip

------------------
lie!!
the model ddint predict
it was **fiddling** the data to the model

| Nitwit. If you could actually do the analysis then you might not be so
| inclined to make such a gargantuan fool of yourself. Calling somebody
| a liar, based on ignorance, just proves you're a blusterinng
| ignoramous.

Ignorant ****head. You've just proven you're a "blusterinng ignoramous"
who has made a gargantuan fool of himself by being unable to spell
"blustering ignoramus".- Hide quoted text -

| Goofball, at least I know there is no discontinuity @ tan pi/2. My
| guess is, in your drunken stupor, you punched in tan pi 2 in your hand
| help calc.. Dummy.

Not only can you not spell, you are so stupid you don't know how
to operate the "radians" button on your hand "help" calculator and
calculated tan 3.14159 degrees in your drunken stupor, you ****in'
ignoramus.

On Mar 4, 2:40 am, "Y.Porat" wrote:
On Mar 4, 2:53 am, wrote:



Apologies if this is a duplicate -- I'm having some news problems.

In sci.physics Koobee Wublee wrote:

On Feb 25, 7:47 am, Tom Roberts wrote:
The best model we have for the propagation of light near a massive
object like the sun is GR, in which the curvature of spacetime is the
important aspect in determining the path light follows. And it agrees
with measurements to part-per-million accuracy over an enormous range.
First, derive a set of geodesic equations a massed particle traveling
at high speed near the sun. Then, gradually reducing the mass to zero
and increasing the speed to c, do you see a discontinuity at mass = 0
and speed = c?

This is definitely a worthwhile exercise. I recommend that you do it.
If you get stuck, you can find the details in Lightman et al., _Problem
book in relativity and gravitation_, problem 15.9.

As you know, the geodesic equations are independent of mass. What
does that tell you when the model predicts a 1x deflection traveling
at speed just a hair below c and suddenly jumps to 2x deflection at
speed = c?

It doesn't. The model predicts a deflection proportional to 1+v^2/c^2,
which varies smoothly from the "Newtonian" value of 1 for small velocities
to 2 as v approaches c.

The moral is that before you decide that a model doesn't make sense,
you should check what the model actually predicts.

Steve Carlip

------------------
lie!!
the model ddint predict
it was **fiddling** the data to the model

Y.Porat
-----------------------------

Oh shut the **** up.

On Mar 6, 3:53 pm, wrote:
On Mar 6, 9:19 am, "Androcles" wrote:



"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.

Graph it nitwit.

tan pi = .054886...
tan pi/2 = .027422...
tan pi/100 = .0005483...
tan pi/1000 = .00005483...

Dude.

tan(x) = sin(x) / cos(x)

sin(pi/2) = 1
cos(pi/2) = 0

Plot it on a calculator.




****ing idiot!
HAHAHAHAHAHA!- Hide quoted text -

- Show quoted text -