FrediFizzx wrote:

"Jesse Mazer" wrote in message
...
|
|
| Jesse Mazer wrote:
|
|
|
| FrediFizzx wrote:
|
| "Jesse Mazer" wrote in message
| ...

[...]
| | You think the sentence "emu is the unit of magnetic moment in cgs
| units"
| | is garbage? Have a look at this page:
| |
| | http://www.unc.edu/~rowlett/units/cgsmks.html
| |
| | It says that emu is the unit of the magnetic dipole moment in CGS
| units,
| | and that the conversion from CGS to SI is 1 emu = 0.001
| Ampere*meter^2.
|
| It is wrong. An emu is a unit of current. It is equal to 10 ampere
SI.
|
| FrediFizzx
|
|
|
| Well, it's possible you're right since I don't claim to be an expert
| in unit systems, but do you have a source for that? The website above
| comes from a professor of mathematics at the University of North
| Carolina at Chapel Hill, so it seems pretty credible.
|
| Jesse
|
|
| This page also says that emu is the unit of magnetic moment in CGS
| units, and that 1 Ampere*meter^2 = 1000 emu:
|
| http://www.geo.umn.edu/orgs/irm/hg2m/hg2m_a/hg2m_a.html

Well, then there are two usages of the term emu but you can plainly see at
the link below that in the emu unit system the main emu unit is for current.


You are misreading that page. The table says that within the e.m.u.
*system*, the unit of current is the "abampere", and that it is equal to
10 Ampere.

Jesse

"Jesse Mazer" wrote in message
...


Androcles wrote:

"Jesse Mazer" wrote in message
...


Androcles wrote:



"Jesse Mazer" wrote in message
...



In any reference frame which Maxwell's laws hold, it can be proved
mathematically that electromagnetic waves must travel with a
velocity
of 1/squareroot(permittivity*permeability).



Not by you I cannot, but you are welcome to try.

The day you can prove
c = 1/squareroot(permittivity*permeability) = c+v
without assuming
1/2 [AB/(c+v) + BA/(c-v)] = AB/(c+v)

publish it in a mathematical journal (not a scientific journal,
they'll believe
anything).



You really don't know what you're talking about--the proof has
nothing
to do with any relativistic assumptions, Maxwell would certainly have
believed that velocities in different reference frames work as they
do
in Newtonian physics. He would only have thought that Maxwell's laws
hold in one frame, and that in this frame light waves always move at
c, but in other frames he'd have said light waves move at c+v or c-v.
So, he'd have thought that if you were moving with respect to the
frame where Maxwell's laws held, then the time for light to go from
points A to B would be *different* than the time to for light to go
from point B to A.



It is. I can go halfway to B at 0.5c and be back at A again to meet
the
returning light at A, and claiming my watch has slowed as I did so
fails, the speed of light says so, or I can continue on and meet
the returning light head on.

If you change directions, you are changing your rest frame, the
Lorentz
transformations only transform between coordinate systems of observers
who move at constant velocity.

Bull****. As with any dumb relativist, you don't know your subject.

"From this there ensues the following peculiar consequence. If at the
points A and B of K there are stationary clocks which, viewed in the
stationary system, are synchronous; and if the clock at A is moved with
the velocity v along the line AB to B, then on its arrival at B the two
clocks no longer synchronize, but the clock moved from A to B lags
behind the other which has remained at B by (up to magnitudes of fourth
and higher order), t being the time occupied in the journey from A to B.

It is at once apparent that this result still holds good if the clock
moves from A to B in any polygonal line, and also when the points A and
B coincide.

If we assume that the result proved for a polygonal line is also valid
for a continuously curved line, we arrive at this result: If one of two
synchronous clocks at A is moved in a closed curve with constant
velocity until it returns to A, the journey lasting t seconds, then by
the clock which has remained at rest the travelled clock on its arrival
at A will be tv^2/(2^c^2) second slow."

http://www.fourmilab.ch/etexts/einstein/specrel/www/ Section 4.





If you continue and meet the returning
light head on, then you will measure the light going at c in both
directions in your frame, and an observer who is in the rest frame of
A
and B will *also* measure the light going at c, and both you and this
other observer will agree in their predictions about what time your
clock will read when you meet the returning light head on. As always,
I
can demonstrate this with some actual numbers for the velocities and
the
rate that your clock is ticking if you like.

I'm discussing **Einstein's** relatvity, and you are disagreeing with
him.
I have no interest in YOUR relativity at all. You have no idea what
rigour is.

[Reminder snipped unread, you are wasting my time]

Androcles.

"Jesse Mazer" wrote in message
...


Androcles wrote:

"Jesse Mazer" wrote in message
...

Androcles wrote:


"Jesse Mazer" wrote in message
...


No, but the earth isn't very dense, and thus is not very massive.
The earth has a density of about 5 gm/cm^3, while estimates of the
density of white dwarf stars range from 100,000 through
1,000,000,000 gm/cm^3. If the earth was as dense as a white dwarf,
and if the moon had an atmosphere, then yes, the earth might pull
some of that atmosphere away.


So the moon would get pretty darn small, quickly.
Poof! It's gone.


If the earth was a white dwarf star, and the moon was a companion
star, and the white dwarf star's pull was strong enough to pull away
some of the gas from the companion star's photosphere, then I have no
idea what the rate it would pull this gas away would be. Maybe it
would only pull away 1% of the mass of the companion star every 10
million years, for example. Without actually doing some calculations
or building a computer simulation, I don't know, and neither do you.


Then it would only go 'pop' every 10,000,000 years.
Gotta be fast enough to keep popping. Look up Roche limit, too.


Why would you assume that 1% of the mass of the companion star is
exactly the mass needed to cause a new explosion?

I'm not like you, I don't make assumptions.
Science is observation, investigation and explanation, not a guessing
game.
My explanation is consistent with the observation, consistent
mathematically,
consistent with the PoR and shaven by Occam's Razor as the simplest
solution.
You are wasting my time.
Androcles.

"Jesse Mazer" wrote in message
...
|
|
| FrediFizzx wrote:
|
| "Jesse Mazer" wrote in message
| ...
| |
| |
| | Jesse Mazer wrote:
| |
| |
| |
| | FrediFizzx wrote:
| |
| | "Jesse Mazer" wrote in message
| | ...
|
| [...]
| | | You think the sentence "emu is the unit of magnetic moment in cgs
| | units"
| | | is garbage? Have a look at this page:
| | |
| | | http://www.unc.edu/~rowlett/units/cgsmks.html
| | |
| | | It says that emu is the unit of the magnetic dipole moment in CGS
| | units,
| | | and that the conversion from CGS to SI is 1 emu = 0.001
| | Ampere*meter^2.
| |
| | It is wrong. An emu is a unit of current. It is equal to 10 ampere
| SI.
| |
| | FrediFizzx
| |
| |
| |
| | Well, it's possible you're right since I don't claim to be an expert
| | in unit systems, but do you have a source for that? The website above
| | comes from a professor of mathematics at the University of North
| | Carolina at Chapel Hill, so it seems pretty credible.
| |
| | Jesse
| |
| |
| | This page also says that emu is the unit of magnetic moment in CGS
| | units, and that 1 Ampere*meter^2 = 1000 emu:
| |
| | http://www.geo.umn.edu/orgs/irm/hg2m/hg2m_a/hg2m_a.html
|
| Well, then there are two usages of the term emu but you can plainly see
at
| the link below that in the emu unit system the main emu unit is for
current.
|
|
| You are misreading that page. The table says that within the e.m.u.
| *system*, the unit of current is the "abampere", and that it is equal to
| 10 Ampere.

http://www.ee.surrey.ac.uk/Workshop/...s/unit_systems

Look right above Equation USL. He says, "...1 e.m.u. of current experiences
a force of two dynes..." What do you think that means? It means there is
also two names; emu and abampere. I think the common usage was emu and not
abampere. I never even heard of the name abampere until I saw that web
site. But I had heard of emu before.

FrediFizzx

http://www.vacuum-physics.com/QVC/qu...uum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/qu...cuum_charge.ps

Androcles wrote:

"Jesse Mazer" wrote in message
...


Androcles wrote:



"Jesse Mazer" wrote in message
...




Androcles wrote:





"Jesse Mazer" wrote in message
...





In any reference frame which Maxwell's laws hold, it can be proved
mathematically that electromagnetic waves must travel with a
velocity
of 1/squareroot(permittivity*permeability).





Not by you I cannot, but you are welcome to try.

The day you can prove
c = 1/squareroot(permittivity*permeability) = c+v
without assuming
1/2 [AB/(c+v) + BA/(c-v)] = AB/(c+v)

publish it in a mathematical journal (not a scientific journal,
they'll believe
anything).





You really don't know what you're talking about--the proof has
nothing
to do with any relativistic assumptions, Maxwell would certainly have
believed that velocities in different reference frames work as they
do
in Newtonian physics. He would only have thought that Maxwell's laws
hold in one frame, and that in this frame light waves always move at
c, but in other frames he'd have said light waves move at c+v or c-v.
So, he'd have thought that if you were moving with respect to the
frame where Maxwell's laws held, then the time for light to go from
points A to B would be *different* than the time to for light to go


from point B to A.




It is. I can go halfway to B at 0.5c and be back at A again to meet
the
returning light at A, and claiming my watch has slowed as I did so
fails, the speed of light says so, or I can continue on and meet
the returning light head on.



If you change directions, you are changing your rest frame, the
Lorentz
transformations only transform between coordinate systems of observers
who move at constant velocity.



Bull****. As with any dumb relativist, you don't know your subject.


If you think the rules of special relativity work in non-inertial
reference frames, you are making a very basic error. The fact that the
rules only work in inertial frames is one of the *most basic* ideas of
relativity, like the idea that the speed of light goes the same in all
frames. Read the two fundamental postulates of relativity in section 2
again:

1. The laws by which the states of physical systems undergo change
are not affected, whether these changes of state be referred to
the one or the other of two systems of co-ordinates in uniform
translatory motion.
2. Any ray of light moves in the ``stationary'' system of
co-ordinates with the determined velocity c, whether the ray be
emitted by a stationary or by a moving body. Hence

velocity = light path/time interval

where time interval is to be taken in the sense of the
definition in 1.


When Einstein says "two systems of co-ordinates in uniform translatory
motion", the "uniform" part means a frame that is not accelerating, ie
an inertial reference frame. Just about any website on relativity will
explain that you must consider things from the point of view of inertial
reference frames, I can come up with as many links as you like to
support this; likewise, every single textbook on relativity will tell
you this from the start, I can provide quotes from some of these too.

Instead of just reading Einstein's paper, which is aimed at fellow
physicists, I would strongly suggest supplementing your study of
relativity by reading texts designed more for pedagogical purposes,
since they would devote more explanation to core concepts like this
instead of assuming that a few words like "uniform translatory motion"
are sufficient.


"From this there ensues the following peculiar consequence. If at the
points A and B of K there are stationary clocks which, viewed in the
stationary system, are synchronous; and if the clock at A is moved with
the velocity v along the line AB to B, then on its arrival at B the two
clocks no longer synchronize, but the clock moved from A to B lags
behind the other which has remained at B by (up to magnitudes of fourth
and higher order), t being the time occupied in the journey from A to B.

It is at once apparent that this result still holds good if the clock
moves from A to B in any polygonal line, and also when the points A and
B coincide.

If we assume that the result proved for a polygonal line is also valid
for a continuously curved line, we arrive at this result: If one of two
synchronous clocks at A is moved in a closed curve with constant
velocity until it returns to A, the journey lasting t seconds, then by
the clock which has remained at rest the travelled clock on its arrival
at A will be tv^2/(2^c^2) second slow."

http://www.fourmilab.ch/etexts/einstein/specrel/www/ Section 4.


I already explained what he meant in this quote, but I guess you just
snipped it without reading. Here it is again:

"he's talking about considering things from the perspective of an
inertial observer, not from the perspective of the clock itself. In
fact, I touched on the same idea in my last post, when I said:


The Lorentz transformations only work for observers moving at constant
velocity, you can't use them to transform into the frame of an
observer who changes velocity (like one who turns around). You can
figure out what the clock of a person who turns around, but you do it
from within the frame of an observer moving at constant velocity--for
example, if I see you move at velocity v1 for t1 seconds in my frame,
then during that time I know your clock will only tick sqrt(1 -
v1^2/c^2)*t1 seconds; then if you switch to velocity v2 for t2
seconds, your clock will tick sqrt(1 - v2^2/c^2)*t2, so I know that at
the end of t1 + t2 seconds in my frame your clock will have ticked
forward by sqrt(1 - v1^2/c^2)*t1 + sqrt(1 - v2^2/c^2)*t2 seconds.


You can see the method if your path is made up of two line segments, so
you could imagine extending this to a path made up of more line
segments. A curved path can be approximated by a path involving a lot of
tiny line segments--that's what he was talking about when he talked
about the "polygonal line" above. Using calculus, we can imagine the
limit as the line segments get smaller and smaller, which gives us a
path integral which we can use the time ticked by a clock moving on a
curved path. But again, this is all done from the perspective of an
inertial reference frame, not from the frame of the clock moving on a
curved path."

Jesse

"Jesse Mazer" wrote in message
...


Androcles wrote:

"Jesse Mazer" wrote in message
...

Androcles wrote:


"Jesse Mazer" wrote in message
...



Androcles wrote:




"Jesse Mazer" wrote in message
...




In any reference frame which Maxwell's laws hold, it can be
proved
mathematically that electromagnetic waves must travel with a
velocity
of 1/squareroot(permittivity*permeability).




Not by you I cannot, but you are welcome to try.

The day you can prove
c = 1/squareroot(permittivity*permeability) = c+v
without assuming
1/2 [AB/(c+v) + BA/(c-v)] = AB/(c+v)

publish it in a mathematical journal (not a scientific journal,
they'll believe
anything).




You really don't know what you're talking about--the proof has
nothing
to do with any relativistic assumptions, Maxwell would certainly
have
believed that velocities in different reference frames work as they
do
in Newtonian physics. He would only have thought that Maxwell's
laws
hold in one frame, and that in this frame light waves always move
at
c, but in other frames he'd have said light waves move at c+v or
c-v.
So, he'd have thought that if you were moving with respect to the
frame where Maxwell's laws held, then the time for light to go from
points A to B would be *different* than the time to for light to go

from point B to A.


It is. I can go halfway to B at 0.5c and be back at A again to meet
the
returning light at A, and claiming my watch has slowed as I did so
fails, the speed of light says so, or I can continue on and meet
the returning light head on.


If you change directions, you are changing your rest frame, the
Lorentz
transformations only transform between coordinate systems of
observers
who move at constant velocity.


Bull****. As with any dumb relativist, you don't know your subject.


If you think the rules of special relativity work in non-inertial
reference frames, you are making a very basic error.


Bull****. As with any dumb relativist, you don't know your subject,
too lazy to read a paragraph, and leap too instant comclusions.


The fact that the rules only work in inertial frames is one of the
*most basic* ideas of relativity, like the idea that the speed of
light goes the same in all frames.

Bull****. As with any dumb relativist, you don't know your subject,
too lazy to read a paragraph, and leap too instant comclusions.


Read the two fundamental postulates of relativity in section 2
again:

Read section 4 again. I've even copied it to you.

[snip crap]

"From this there ensues the following peculiar consequence. If at the
points A and B of K there are stationary clocks which, viewed in the
stationary system, are synchronous; and if the clock at A is moved
with the velocity v along the line AB to B, then on its arrival at B
the two clocks no longer synchronize, but the clock moved from A to B
lags behind the other which has remained at B by (up to magnitudes of
fourth and higher order), t being the time occupied in the journey
from A to B.

It is at once apparent that this result still holds good if the clock
moves from A to B in any polygonal line, and also when the points A
and B coincide.

If we assume that the result proved for a polygonal line is also valid
for a continuously curved line, we arrive at this result: If one of
two synchronous clocks at A is moved in a closed curve with constant
velocity until it returns to A, the journey lasting t seconds, then by
the clock which has remained at rest the travelled clock on its
arrival at A will be tv^2/(2^c^2) second slow."

http://www.fourmilab.ch/etexts/einstein/specrel/www/ Section 4.


I already explained what he meant in this quote, but I guess you just
snipped it without reading. Here it is again:

"he's talking about considering things from the perspective of an
inertial observer,

No such animal exists. Therefore SR is fiction.
You are wasting my time. I'm done with you and your religion. Go away.
*plonk*.
Androcles

Androcles wrote:

"Jesse Mazer" wrote in message
...


Androcles wrote:



"Jesse Mazer" wrote in message
...



Androcles wrote:




"Jesse Mazer" wrote in message
...





Androcles wrote:






"Jesse Mazer" wrote in message
...






In any reference frame which Maxwell's laws hold, it can be
proved
mathematically that electromagnetic waves must travel with a
velocity
of 1/squareroot(permittivity*permeability).






Not by you I cannot, but you are welcome to try.

The day you can prove
c = 1/squareroot(permittivity*permeability) = c+v
without assuming
1/2 [AB/(c+v) + BA/(c-v)] = AB/(c+v)

publish it in a mathematical journal (not a scientific journal,
they'll believe
anything).






You really don't know what you're talking about--the proof has
nothing
to do with any relativistic assumptions, Maxwell would certainly
have
believed that velocities in different reference frames work as they
do
in Newtonian physics. He would only have thought that Maxwell's
laws
hold in one frame, and that in this frame light waves always move
at
c, but in other frames he'd have said light waves move at c+v or
c-v.
So, he'd have thought that if you were moving with respect to the
frame where Maxwell's laws held, then the time for light to go from
points A to B would be *different* than the time to for light to go



from point B to A.



It is. I can go halfway to B at 0.5c and be back at A again to meet
the
returning light at A, and claiming my watch has slowed as I did so
fails, the speed of light says so, or I can continue on and meet
the returning light head on.




If you change directions, you are changing your rest frame, the
Lorentz
transformations only transform between coordinate systems of
observers
who move at constant velocity.



Bull****. As with any dumb relativist, you don't know your subject.



If you think the rules of special relativity work in non-inertial
reference frames, you are making a very basic error.




Bull****. As with any dumb relativist, you don't know your subject,
too lazy to read a paragraph, and leap too instant comclusions.


Sigh. Do a google search for "special relativity inertial", you will
find thousands of websites confirming this. Crack open *any* relativity
textbook, you will also see confirmation of this. Learn what Einstein
meant when he said "uniform translational motion", and you will understand.

Here are just a few of those websites:

http://www.sparknotes.com/physics/sp.../section1.html

There are many possible places to start in Special Relativity. That
is, it is a matter of choice what is to be called a 'postulate' and
what a 'theorem.' However, it is best to start with the most
believable and simple. With this in mind, our first postulate states:
All inertial reference frames are equivalent.
An inertial reference frame (see the introduction to this topic for an
explanation of reference frames) is simply in which Newton's First Law
holds (that is, the law of inertia: all bodies remain at rest or in
constant motion unless a force acts upon them). This means any
non-accelerating reference frame is an inertial one. In an
accelerating (including rotating) frame we have to invent 'imaginary
forces,' such as the (coriolis or centrifugal forces) in order for the
law of inertia to be valid. Our postulate then says that any reference
frame at rest or in constant motion is as good as any other--there is
no absolute frame.

http://www.upscale.utoronto.ca/Gener...ey/relspec.htm


Introduction.

The Principle of Relativity states that:

The laws of mechanics are the same in all inertial reference frames

or, in different words,

All inertial frames are equivalent; there is no preferred inertial
frame ...

(An early form of this principle was first enunciated by Galileo in
"The Dialogue on the Great World Systems". )

What's an Inertial Reference Frame? One in which the Law of Inertia
holds (circular definition??).

Meaning that ----A non-spinning frame, isolated in deep space, with no
local sources of gravitation or other forces is an inertial frame. All
other frames of reference which are moving uniformly (i.e. with
constant velocity) with respect to this one are inertial frames. Any
frame that is accelerating in any way with respect to such inertial
frames, is not an inertial frame.

All inertial frames are equivalent ---- means that there is no way to
determine which frame is "moving" or which is "at rest". Which, in
turn, implies that the laws of physics are the same for observers in
all inertial frames.

Where can I find an Inertial Frame? Rather surprisingly, in spite of
the fact that we are more or less fixed on the surface of a spinning
planet, which is orbiting a star which is itself orbiting the centre
of our galaxy, which itself is moving through the universe, frames
fixed in our earth-bound laboratories are good approximations to
inertial frames.



http://galileoandeinstein.physics.vi.../spec_rel.html

We now come to Einstein's major insight: the Theory of Special
Relativity. It is deceptively simple. Einstein first dusted off
Galileo's discussion of experiments below decks on a uniformly moving
ship, and restated it as :

The Laws of Physics are the same in all Inertial Frames.

Einstein then simply brought this up to date, by pointing out that the
Laws of Physics must now include Maxwell's equations describing
electric and magnetic fields as well as Newton's laws describing
motion of masses under gravity and other forces.


http://www.usd.edu/phys/courses/phys...es/notes2.html

* Postulates of Special Relativity
o The Laws of Physics should be the same to all observers in
inertial frames of reference.
o The speed of light is the same for all inertial observers


http://en.wikipedia.org/wiki/Special_relativity

1. First postulate (principle of relativity)

Observation of physical phenomena by more than one inertial
observer must result in agreement between the observers as to the
nature of reality. Or, the nature of the universe must not change
for an observer if their inertial state changes.

Every physical theory should look the same mathematically to every
inertial observer.

To state that simply, no property of the universe will change if
the observer is in motion. The laws if the universe are the same
regardless of inertial frame of referance.

2. Second postulate (invariance of c)

The speed of light in vacuum, commonly denoted c, is the same to
all inertial observers, is the same in all directions, and does
not depend on the velocity of the object emitting the light. When
combined with the First Postulate, this Second Postulate is
equivalent to stating that light does not require any medium (such
as "aether") in

which to propagate.

http://ffden-2.phys.uaf.edu/212_fall...ssumptions.htm

Special Relativity begins with two basic assumptions which are
fundamental to all the conclusions that can be drawn from it.

1) The laws of physics are identical in all inertial frames, or
equivalently, the outcome of any physical experiment is the same when
performed with identical initial conditions relative to any inertial
frame.


http://www.physics.nyu.edu/courses/V85.0020/node38.html

The first postulate of the special theory of relativity is that there
are no preferred inertial frames. This postulate is nothing more than
a statement of the principle of relativity that we associated with
Galileo. Thus the first postulate is ``old hat.'' The second postulate
can be stated in two, equivalent ways. In can be stated as ``light (or
any electromagnetic radiation) does not require a medium in which to
propagate'' or ``the speed of light is the same in all inertial
frames.'' These statements are equivalent. For example, if the speed
of light is the same in all inertial frames, then light must propagate
without a medium; otherwise the medium would affect the speed of light
as viewed from different inertial frames. This second postulate is the
``nonintuitive'' one and leads to all the ``weird'' effects of special
relativity. For mechanical waves, such as sound or water waves, a
medium is essential for wave propagation. The medium carries along the
waves and there are no surprising results. For electromagnetic waves,
no medium is needed and we are led to new conclusions about the nature
of space and time. Of course, the postulates can be tested only by
experiment. The experimental confirmations of the special theory of
relativity are overwhelming.


http://www.cec.mtu.edu/csa/courses/p...ons/rosten.htm

There are two postulates to the special theory of relativity:

1. The laws of physics are the same in all inertial (non accelerating)
reference frames

2. The speed of light in a vacuum, c, is a universal constant with a
value of about 3.00*108 m/s, in all inertial frames. It is constant
regardless of the speed of the source or the observer.


....and so on. I can provide many more if you like, and also quotes from
textbooks.





The fact that the rules only work in inertial frames is one of the
*most basic* ideas of relativity, like the idea that the speed of
light goes the same in all frames.



Bull****. As with any dumb relativist, you don't know your subject,
too lazy to read a paragraph, and leap too instant comclusions.


Read the two fundamental postulates of relativity in section 2


again:



Read section 4 again. I've even copied it to you.

[snip crap]


"From this there ensues the following peculiar consequence. If at the
points A and B of K there are stationary clocks which, viewed in the
stationary system, are synchronous; and if the clock at A is moved
with the velocity v along the line AB to B, then on its arrival at B
the two clocks no longer synchronize, but the clock moved from A to B
lags behind the other which has remained at B by (up to magnitudes of
fourth and higher order), t being the time occupied in the journey


from A to B.


It is at once apparent that this result still holds good if the clock
moves from A to B in any polygonal line, and also when the points A
and B coincide.

If we assume that the result proved for a polygonal line is also valid
for a continuously curved line, we arrive at this result: If one of
two synchronous clocks at A is moved in a closed curve with constant
velocity until it returns to A, the journey lasting t seconds, then by
the clock which has remained at rest the travelled clock on its
arrival at A will be tv^2/(2^c^2) second slow."

http://www.fourmilab.ch/etexts/einstein/specrel/www/ Section 4.



I already explained what he meant in this quote, but I guess you just
snipped it without reading. Here it is again:

"he's talking about considering things from the perspective of an
inertial observer,



No such animal exists. Therefore SR is fiction.


Uh, do you even know what an "inertial observer" is? Hint: Newtonian
laws also only work in the reference frame of inertial observers.

Jesse

"Androcles" wrote in message . uk...

"Jesse Mazer" wrote in message
...

[snip]

I already explained what he meant in this quote, but I guess you just
snipped it without reading. Here it is again:

"he's talking about considering things from the perspective of an
inertial observer,

No such animal exists. Therefore SR is fiction.
You are wasting my time. I'm done with you and your religion. Go away.
*plonk*.
Androcles

Translation: "I win by kicking the table":
http://users.pandora.be/vdmoortel/dirk/Stuff/Beaten.gif
http://groups-beta.google.com/groups...ndrocles+plonk

Dirk Vdm

"Jesse Mazer" wrote in message
...


Franz Heymann wrote:

"Jesse Mazer" wrote in message
...


Franz Heymann wrote:



"Jesse Mazer" wrote in message
...




I wrote:





Well, you're thinking about it wrong. It would be possible to




choose




units of mass such that the gravitational constant G was equal
to




1,




but that wouldn't make it any less real as a physical
constant--similarly, in cgs units, the basic units of charge


(emu)




Sorry, that should be esu...emu is the unit of magnetic charge
in




cgs units.

You might know what you were trying to say both in the original
statement and the correction.
I doubt if anybody else would.
Comment on this garbage, for instance
"emu is the unit of magnetic charge in cgs units"



OK, magnetic moment, I confused my terms, Jeez. I think you know


what I


meant.



No. I had not the foggiest notion. I still do not know what you
are
trying to say.
How about being effective by giving a complete sentence with the
phrase "magnetic moment" in its apropriate place.
If I were to just replace "magnetic charge" by "magnetic moment",
the
sentence would still be garbage, so I feel sure you must be meaning
something else.

Franz




You think the sentence "emu is the unit of magnetic moment in cgs
units"
is garbage?

Yes, assuredly.
The emu, is a generic term which is the name for a complete set of
units encompassing all magnetic and all electrostatic quantities.
What emu is *not* is what you refer to as "the unit of magnetic moment
in cgs units. Moreover there are two differently sized units for
magnetic moment in cgs, namely the cgs emu of magnetic moment and the
cgs esu of magnetic moment.

Have a look at this page:

http://www.unc.edu/~rowlett/units/cgsmks.html

It says that emu is the unit of the magnetic dipole moment in CGS
units,
and that the conversion from CGS to SI is 1 emu = 0.001
Ampere*meter^2.

I don't need to.
What it should say is that 1 emu *of magnetic moment* is the unit of
magnetic dipole moment.
And yes, the conversion factor is correct.

Franz

"FrediFizzx" wrote in message
...
"Jesse Mazer" wrote in message
...
|
|
| Franz Heymann wrote:
|
| "Jesse Mazer" wrote in message
| ...
|
|
| Franz Heymann wrote:
|
|
|
| "Jesse Mazer" wrote in message
| ...
|
|
|
|
| I wrote:
|
|
|
|
|
| Well, you're thinking about it wrong. It would be possible to
|
|
|
|
| choose
|
|
|
|
| units of mass such that the gravitational constant G was
equal to
|
|
|
|
| 1,
|
|
|
|
| but that wouldn't make it any less real as a physical
| constant--similarly, in cgs units, the basic units of charge
|
|
| (emu)
|
|
|
|
| Sorry, that should be esu...emu is the unit of magnetic charge
in
|
|
|
|
| cgs units.
|
| You might know what you were trying to say both in the original
| statement and the correction.
| I doubt if anybody else would.
| Comment on this garbage, for instance
| "emu is the unit of magnetic charge in cgs units"
|
|
|
| OK, magnetic moment, I confused my terms, Jeez. I think you know
|
|
| what I
|
|
| meant.
|
|
|
| No. I had not the foggiest notion. I still do not know what you
are
| trying to say.
| How about being effective by giving a complete sentence with the
| phrase "magnetic moment" in its apropriate place.
| If I were to just replace "magnetic charge" by "magnetic moment",
the
| sentence would still be garbage, so I feel sure you must be
meaning
| something else.
|
| Franz
|
|
|
|
| You think the sentence "emu is the unit of magnetic moment in cgs
units"
| is garbage? Have a look at this page:
|
| http://www.unc.edu/~rowlett/units/cgsmks.html
|
| It says that emu is the unit of the magnetic dipole moment in CGS
units,
| and that the conversion from CGS to SI is 1 emu = 0.001
Ampere*meter^2.

It is wrong. An emu is a unit of current. It is equal to 10 ampere
SI.

Balls. The emu is a complete system of electrical a=nd magnetic units.
The emu of current is sometimes called the Abamp. (An abbreviation
for "absolute Ampere")
1 Abamp = 10 Amperes

Franz