Gauge:
(Bilge) wrote in message
e-al.net...

The quantity r_o is not the wavelenth of the pion. It's the range of
the Yukawa potential,

E^2 = p^2 + (mc^2)^2

(\hbar w) = (\hbar k)^2 + (\hbar k_0)^2

You forgot to square the first part sparky

I'm impressed. You found a typo.

[...]
Do you simply copy words from textbooks without reading the material
or trying to understand what it means?

This is a good example of when you should think twice about posting a
flame like this sparky - And you never did state why you have this
need to flame - Why is that?

Because you are an idiot who posts nothing but misinformation
under the guise of relativity and then "flames" anyone who corrects
you.

Anyway - I posted warnings here about this directly to you and you
ignored them.

If you are referring to your posts as some sort of warning by example,
then I haven't ignored them. Everytime I post I think to myself, "have
I failed to check this any better than pmb would have done"? If the answer
is no, then I check it again for mistakes and physics errots before
posting it. So far, I've never answered that question with a yes.

And here you are making the mistake I warned you about.

A typo? Try running one of your own posts (any one will do) through a
spell checker and then tell me about typos.

E is usually used to represent total energy. In this case E = hbar*w.
But the correct relation for the 4-momentum magnitude is

(E - V)^2 = p^2 + m^2

Where E = T + V
where T = time component of 4-momentum

I knew this would only be a matter of time before you made this
mistake.

You mean the typo?

(Bilge) wrote


Because you are an idiot who posts nothing but misinformation
under the guise of relativity and then "flames" anyone who corrects
you.

Bald faced lie. I've never flamed anyone for having a different
opinion. You've never corrected me or pointed out a valid error.

However all you have ever done is to flame.

Anyway - I posted warnings here about this directly to you and you
ignored them.

If you are referring to your posts as some sort of warning by example,
then I haven't ignored them. Everytime I post I think to myself, "have
I failed to check this any better than pmb would have done"? If the answer
is no, then I check it again for mistakes and physics errots before
posting it. So far, I've never answered that question with a yes.

And here you are making the mistake I warned you about.

A typo? Try running one of your own posts (any one will do) through a
spell checker and then tell me about typos.




E is usually used to represent total energy. In this case E = hbar*w.
But the correct relation for the 4-momentum magnitude is

(E - V)^2 = p^2 + m^2

Where E = T + V
where T = time component of 4-momentum

I knew this would only be a matter of time before you made this
mistake.

You mean the typo?


Nope - another example of bige making an error and is unable to admit
it.

What you posted, i.e.

(\hbar w) = (\hbar k)^2 + (\hbar k_0)^2

Is not a valid relation. For a particle moving in a potential the
correct equation is

(E - V)^2 = p^2 + m^2

where E = "total energy" = "kinetic energy" + "potential energy" +
"rest energy"

The eigenvalues of E are hbar*w, i.e. H*psi = E*psi where E = hbar*w.
In which case the correct relation is

(hbar - V)^2 = p^2 + m^2


Even when you see the correction you still don't undertsand and an
instead post flames? Get a grip.

pmb

Gauge:

I know what it means flamer. r_o is defined in terms of mass - not the
otherway around. I.e. The Yukawa potential is (a, V_o = constants)


You are a complete dunce who argues against anything based upon your
extremely limited exposure to physics and the (blatantly incorrect)
inferences you make about things which you've read anything about.


V(r) = V_o*exp(-ar)/r

r_o is defined to be that point where the exponential in the potential
drops to 1/e.

And to what do you think `a' corresponds?

Once the form of the potential is found then a is determined.

Obviously, that is not true, since you have written the form of the
potential and `a' is arbitrary. You need to actually measure something
to determine `a'.

However that does not mean that a single measurement defines the mass
of a particle.

For experimental reasons, of course not. By the same token, a
a single measurement of any quantity doesn't determine its value.

Once one has statistical data on experiments then such
constants can be adjusted so that the potential fits the data.

Do you really believe you can measure any experimental quantity
without statistics?

However that doesn't mean that such a distance is the definition of
mass.

Why don't you go buy a nuclear physics book or particle physics
book?

waite is claiming that in all concievable cases that something he
calls "length" is suppose to define mass. But that kind of "length" is
statistical in nature.

Don't be an imbecile. That's the way masses are dealt with in all
solid state physics. In fact, in solid state physics one even defines
masses via:

(1/m) = (1/hbar)^2 (d2^E/dk^2)

Do you really think that is unrelated to other physics?

Its just plain bogus for a definition.

Why? In relativity, the natural unit for mass is a length. The mass of
the sun, for example, is 1.6 km. Since I haven't examined his particular
way of defining things, I can't attest to it's validity, but there is
certain;y nothing wrong with the concept, a priori. I can define k just
as easily as p.

I'll post why in more detail later.

Don't bother.

If waite is not speaking in general and is refering strictly to a free
particle then E^2 - p^2 = m^2 has the meaning he thinks. If not and
he's being general, i.e. the particle may or may not be free, then,
like you, he's confusing the time component of the 4-momentum with
total energy. And it's the total energy that is related to omega.

Don't be a dunce. Has it never occurred to you that the physics from
different disciplines is all interrelated? I'm guessing not, since you
can't seem to connect quantities together.

(Bilge) wrote in message ...
Gauge:


the whole idea waite is trying to put forth is the there is a length
associated with a particle - anyparticle anywhere at any time under
any situation - by definition.

It's called the "compton wavelength".

My opponents write about mass all time.
I asked my opponents, Joe Fischer, Jeff Kimmel, Waite David,
Vetwannabel, Tom Roberts, Bilge, to define the mass in the spirit of
operationism.
I asked for an operation to measure the mass.
I have no answer.
It seems to be lack of seriousness

Jeff Krimmel wrote in message . ..
On Fri, 01 Aug 2003 01:50:59 -0700, Gauge wrote:

Apparently more than ever have from what I see you posting all the
time.

Honestly, is that English?

Jeff

My opponents write about mass all time.
I asked my opponents, Joe Fischer, Jeff Kimmel, Waite David,
Vetwannabel, Tom Roberts, Bilge, to define the mass in the spirit of
operationism.
I asked for an operation to measure the mass.
I have no answer.
It seems to be lack of seriousness

Radi Khrapko:
(Bilge) wrote in message ...
Gauge:


the whole idea waite is trying to put forth is the there is a length
associated with a particle - anyparticle anywhere at any time under
any situation - by definition.

It's called the "compton wavelength".

My opponents write about mass all time.
I asked my opponents, Joe Fischer, Jeff Kimmel, Waite David,
Vetwannabel, Tom Roberts, Bilge, to define the mass in the spirit of
operationism.

It's a poincare invariant.

I asked for an operation to measure the mass.
I have no answer.

I answered this. Use a balance.

It seems to be lack of seriousness

To an extent, yes.

Gauge: crackpot
(Bilge) wrote in message

And to what do you think `a' corresponds?

Yep.

It does not correspond to "Yep".

You're not reading again

Yes, I am. I'm just laughing as I read, since I can do both at the same
time.

- I explained to you that r_o is DEFINED
as the place where ar_o = 1 - Sheesh! Pay attention huh?


The relation, ar_0 = 1, doesn't make any sense. `a' is 1/r_0. In
case you weren't aware of it, the expression is often written:

\exp(-r/r_0)

You obviously also don't know that `a' is the 1/compton wavelength:

\exp(-ar) = \exp(-mcr/hbar)

Why don't you try deriving the yukawa potential? Here's a hint.
Start with the low-energy limit to the first order scattering
amplitude using the klein-gordon propagator:

M = (g_0)^2/[|p|^2 + m^2]

and fourier transform it.


By the way. Why don't you tell me how to determine `a' for the
weak interaction. Is it perhaps: m_W c/hbar =~ 2.5 x 10^-18 m,
the compton wavelength of the W (or Z)?

Obviously, that is not true, since you have written the form of the
potential and `a' is arbitrary. You need to actually measure something
to determine `a'.

By "form" I mean the exact value of a - pleas4e pay closer attention

"Form of" does not mean "exact value". Are you stoned or just stupid?

Do you really believe you can measure any experimental quantity
without statistics?

Of course - and the fact that you claim different shows your
ignorance.

How is that? You just agreed with what I said. Or are you trying to
say that you don't understand that the statistics associated with
experimental error and quantum statistics are different?

Take spin. I can measure it to be up or down with exactness
- doesn't mean its in an eigen state.

That is a classic and is certainly one for dirk's fumbles page.
By definition, if you measure the spin to be up or down, you've
measured an eigenstate. Where do you think the relation:

J_z\Psi = m\hbar\Psi

(with m either integral or half integral), comes from? Measuring J_z
gives the eigenvalue for J_z. Thumbed through a quantum mechanics book
lately?

However waite's so-called "definition" is meaningless.

So, is the compton wavelength meaningless?

[...]
Why don't you go buy a nuclear physics book or particle physics
book?

Why? Do you nee more help with your homework?

No, so you'll have enough paper to keep from freezing to death in
the winter. I'm sure your other books have been used up.

Don't be an imbecile. That's the way masses are dealt with in all
solid state physics. In fact, in solid state physics one even defines
masses via:

There's a huge difference between an equality and a definition. The
defintion comes before you can write down the Hamiltonian - Sheeesh

Is that sentence supposed to make sense? To whom?

[...]

Before

Before what?

wchogg:
On Mon, 4 Aug 2003, Bilge wrote:

M = (g_0)^2/[|p|^2 + m^2]

and fourier transform it.

Do you mean take the limit |p| m in the propagator first and then take
the fourier transform? If you do, then I think I might have to try to do
that.

No, just fourier transform it directly:


\integral \exp(ip.r)/(p^2 + m^2) d3p

= \integral d\phi \integral dp p^2/(p^2+m^2)

x \integral \exp(ipr cos(\theta)) sin(\theta)d\theta


= 2\pi \integral dp sin(pr) p/(p^2 + m^2)

Which you can evaluate by either looking it up or by re-writing
p sin(pr) as -(1/r)(d/dr) cos(pr) to get the integral:


(1/r)(d/dr) \integral cos(pr)dp/(p^2 + m^2)

or
(1/r)(d/dr) Re \integral dp \exp(ipr)/(p^2 + m^2)

then find the residues and take the derivative, d/dr.


Oh, btw, so you're just using the klein-gordon propagator because the
yukawa potential assumes a scalar interaction?

Yes and sort of. Yes, since a pion is a pseudoscalar, that definitly
works for the one-pion nuclear potential. Sort of for the weak interaction,
where in reality you would have to consider the spins. However, I just
ignored the spins.


What's the process for recovering the macro-scopic potential for other
forces? Is it just converting the low energy limit of the momentum space
propagator into co-ordinate space with a fourier?

Well... For the most part. I'm not sure quite how you go about it for
something like gluons in qcd.

(Bilge) wrote in message ...
Gauge: crackpot
(Bilge) wrote in message

And to what do you think `a' corresponds?

Yep.

It does not correspond to "Yep".

You're not reading again

Yes, I am. I'm just laughing as I read, since I can do both at the same
time.

- I explained to you that r_o is DEFINED
as the place where ar_o = 1 - Sheesh! Pay attention huh?


The relation, ar_0 = 1, doesn't make any sense.

What that means is that you don't understand it - as usual.

`a' is 1/r_0. In
case you weren't aware of it, the expression is often written:

\exp(-r/r_0)


Duh! That's just what I wrote with the value a = 1/r_o substituted
into it. Sheesh! As I've said, and what you don't understand, is
that's how r_o is defined


[schoolboy insults snipped]

When you start acting like an adult then let me know. Otherwise you've
reduced this to your usual flaming when someone critisizes your posts
- get a life bilge. And grow up.

Pmb

On Tue, 5 Aug 2003, Bilge wrote:

wchogg:
On Mon, 4 Aug 2003, Bilge wrote:

M = (g_0)^2/[|p|^2 + m^2]

and fourier transform it.

Do you mean take the limit |p| m in the propagator first and then take
the fourier transform? If you do, then I think I might have to try to do
that.

No, just fourier transform it directly:


\integral \exp(ip.r)/(p^2 + m^2) d3p

= \integral d\phi \integral dp p^2/(p^2+m^2)

x \integral \exp(ipr cos(\theta)) sin(\theta)d\theta


= 2\pi \integral dp sin(pr) p/(p^2 + m^2)

Okay then where does the low energy limit you mentioned enter in to it?
Is it because you're only doing a 3-space momentum integral instead of the
4-d fourier transform that would give you the co-ordinate space green's
function? Sorry if I'm just sounding confused.


--
William C. Hogg