On Feb 4, 4:38 pm, wrote:
On Feb 4, 4:35 pm, JanPB wrote:



On Feb 4, 4:01 pm, wrote:

On Feb 4, 3:51 pm, PD wrote:

On Feb 4, 5:23 pm, wrote:

On Feb 4, 8:11 am, PD wrote:

On Feb 3, 7:24 pm, wrote:

Bosons would be emitted simultaneously from every direcftion
continuously if they are to create force. They cannot aim as in a
Feynman diagram.

Mitch Raemsch Twice Nobel Laureate 2008

They don't aim. What on earth gives you the idea that they aim?
If you explode a shrapnel grenade, and one of the bits of shrapnel
kills a squirrel in a tree, and you draw a line showing the path of
the shrapnel from the grenade to the squirrel that accounts for why
the squirrel suddenly fell from the tree, why would you conclude that
you aimed that bit of shrapnel at the squirrel?

PD

Feynman diagrams aim.

Uh, no, they don't. Sorry. Repeating the same bonehead statement twice
in a row does not make it more true, especially coming from someone
who doesn't know the first thing about Feynman diagrams or what they
mean.

PD- Hide quoted text -

- Show quoted text -

Yes they do.

Cite reference.

--
Jan Bielawski- Hide quoted text -

- Show quoted text -

The reference is any Feynman diagram.

Mitch Raemsch Twice Nobel Laureate 2008

Circular references don't count.

--
Jan Bielawski

On Feb 4, 9:32*pm, PD wrote:
On Feb 4, 11:44*am, "Y.Porat" wrote:





On Feb 4, 6:11*pm, PD wrote:

On Feb 3, 7:24*pm, wrote:

Bosons would be emitted simultaneously from every direcftion
continuously if they are to create force. They cannot aim as in a
Feynman diagram.

Mitch Raemsch Twice Nobel Laureate 2008

They don't aim. What on earth gives you the idea that they aim?
If you explode a shrapnel grenade, and one of the bits of shrapnel
kills a squirrel in a tree, and you draw a line showing the path of
the shrapnel from the grenade to the squirrel that accounts for why
the squirrel suddenly fell from the tree, why would you conclude that
you aimed that bit of shrapnel at the squirrel?

PD

----------------------
idiot crook!!

no particle of physical entity that moves in a * straight line
can produce any attraction force

First of all, no one says the virtual particles move in straight
lines. It might help for you to read a short and friendly book by
Richard Feynman called "QED:
----------------
thanks for the advice
but i have seen Fenmans deal with it

it i s not a physical dealing
it is jsut a partial mathematical model!!
very partial and crippled in a way
that it cant give us more advance
you cant do advance if you dont get deeper into
the real process that happense in reality
you say
no one claimes that bosons or any other attraction
agent is moving in straight lines..
even just that say is some advance twards better understanding of
reality !!
but only vert 'some'
while making a real calculation you **cant ignore*
the real process ie the real path that those agents do !!

you miscalculate it and decieve others and youself
because say
it movesin a straight line it makes another distance
on its way untill hiting the target--compared another posibility
that it moves a big curved path tha tmakes the distance
of actions much bigger
as known if the distance is very short it needs' or hints about
a big mass

if OTHOA the distance of cation is much bigger
(due to a curved path) the agent needs to be
much less in mass.
so
the real geomertic components involved are criticslly
important and cannot be neglected
(in order of coming closer to reality)
not to mension fo r instance that the difference
between moving in a straight line or a curved line
is a PRINCIPAL factor
that by neglecting it will prevent more deeper understanding
and advance in other attraction forcess
and i dont have to remind you the deal lock
of QM in solving sat Gravitation .
i am not surprised about it
because i know what is rotten in the exiting situation
2
no force agent can be bigger than its 'mother'
that gave him 'birth' !!
(it doe s not make sense in biology as well as in physics
iow
a clear immediate 'red light' for something rotten!!
and over speculative

3
the photon can not the cause of say EM force
it is a result of it not a cause
once moving only in straight lines
they should be escaping form our material world
and lost in the infinity of the univers
iow
it should be deluted quickly by not hitting any traget
and getting lost
(no reason why they should move to targets
that 'you ' what them to move to !!)
and many others ereasoning that
you could do provided that you wouls start
to open your mind and (i refrain at this point from being
insultive )
and start to be a bit more skeptic than
a not religous beliver !!

The Strange Theory of Light and Matter"

----------------
nature (the inorganic one ) is not starnge it ha s ound a simple
principles
all the strangness is due only to ignorance !!


so that you have a little better idea what the theory actually says.

Secondly, another poster has already given a sizable hint which you no
doubt missed: What is the behavior of time-reversed repulsion?
------------------
had you realy undestood that TIME IS
you would not deal a second with

'time reversal' it is stupid and ignorant
time as is does not do reverse -- only forwards !!
nothing can stop it from moving forwwards

and as for amthematical junglarings and foolong around
just leave it for mathematicians to play with ......
save me the word (:-)

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

wrote:
On Feb 4, 4:35 pm, JanPB wrote:
On Feb 4, 4:01 pm, wrote:
Feynman diagrams aim.
Uh, no, they don't.
Yes they do.
Cite reference.

The reference is any Feynman diagram.

Nonsense. You CLEARLY do not know what a Feynman diagram represents.

Every Feynman diagram represents a specific INTEGRAL. In the QFT to
which the diagram belongs there is a 1-to-1 mapping between components
of the diagram and terms in the integral. For instance, in QED the
squiggly line corresponding to a photon corresponds to a term
p_slash/p^2; an electron-electron-photon vertex corresponds to
\integral d^3p/E delta^4(p1+p2+p3) \alpha; etc.

As I said before, it is implicit in the definition of the diagram that
one integrates every vertex over all of momentum space. This directly
implies that the directions of the lines connecting vertices are of no
consequence whatsoever, because they will "point" in all possible
directions during the integrations. As I said before, ALL that matters
is the topology of the diagram (that determines how the various terms of
the integral are inter-connected).

And if you think that because a given line connects to a given vertex in
a specific diagram that the object represented by the line "is aimed" at
that vertex, again you would be wrong -- one must SUM over all possible
diagrams, and (ANTI-)SYMMETRIZE over all external legs. This completely
"jumbles" any correspondence you might imagine between legs and vertices.


Tom Roberts

On Feb 5, 5:54*pm, PD wrote:
On Feb 4, 7:01*pm, Phil Cartwright
Putting things more simply: you're both wrong.

I think you read a little too much into the rather broad hint that I
mentioned.

You are right about virtual particles being able to carry momentum
that "points backward".

It is also true that in a tree-level diagram between two electrons,
there is a vertex term contributing to the Green's function for one of
the vertices that looks like -i(-e)[gammu_mu]. Changing that to a
positron vertex simply means changing (-e) to (e). Likewise the
Feynman propagator for the electron leg at that vertex is the
difference in two terms, both corresponding to a negative charge being
created at (x, t) and being annihilated at (x', t'), with the terms
corresponding to tt' and tt'. It's well-known that an electron
propagator proceeding forward in time is identical with a positron
propagator traveling backwards in time. Thus, by making an appropriate
transformation to time-reverse one of the electron legs in the lowest-
order tree diagram, you switch from an electron-electron (repulsive)
interaction to an electron-positron (attractive) interaction.


--
There's only four things you can be certain of: taxes, change, spam, and
death.- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

----------------------
you dont have to buy any stupid croocked manipolation
and play chess with Feynemen or youself

we are playing chess with NATURE!!!

not with ouselves !!
no stupid crooked diagram that takes the
whanter end results and manipolates them backwards
(from the WHANTED end result to the start 'reasonings'
)
can comvince people who are not parrots or suckers !!

iow
nature does not play your chess play with youself
(and Feynman)
it palyes its own play
and you are far away from natures play as far as earth is far
from galaxy No 10

and the fact is that this stupid self play
is stuck in the mudd
with idotic theories of virtual; bosons that are
90 times bigger than their mother
a schole boy will tell you that you are
a dumb parrot

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

On Feb 5, 12:09*pm, Tom Roberts wrote:
wrote:
On Feb 4, 4:35 pm, JanPB wrote:
On Feb 4, 4:01 pm, wrote:
Feynman diagrams aim.
Uh, no, they don't.
Yes they do.
Cite reference.

The reference is any Feynman diagram.

Nonsense. You CLEARLY do not know what a Feynman diagram represents.

Every Feynman diagram represents a specific INTEGRAL. In the QFT to
which the diagram belongs there is a 1-to-1 mapping between components
of the diagram and terms in the integral. For instance, in QED the
squiggly line corresponding to a photon corresponds to a term
p_slash/p^2; an electron-electron-photon vertex corresponds to
\integral d^3p/E delta^4(p1+p2+p3) \alpha; etc.

As I said before, it is implicit in the definition of the diagram that
one integrates every vertex over all of momentum space. This directly
implies that the directions of the lines connecting vertices are of no
consequence whatsoever, because they will "point" in all possible
directions during the integrations. As I said before, ALL that matters
is the topology of the diagram (that determines how the various terms of
the integral are inter-connected).

And if you think that because a given line connects to a given vertex in
a specific diagram that the object represented by the line "is aimed" at
that vertex, again you would be wrong -- one must SUM over all possible
diagrams, and (ANTI-)SYMMETRIZE over all external legs. This completely
"jumbles" any correspondence you might imagine between legs and vertices.

Tom Roberts

----------------
you can have all diagrams and integrals in the world
but if all thatis based on wrong assuptions ??

you can stuff all your 'sophystication:

you have a basic assumption that you deal with PHOTONS
AS force mediators
but allas ??
what would you do if one day you will find that
PHOTONS CANNOT BE ANY ATTRACTION FORCE ???!! MEDIATORS ??!!
Y.Porat
----------------------

On Feb 6, 8:23*am, "Y.Porat" wrote:
On Feb 5, 12:09*pm, Tom Roberts wrote:





wrote:
On Feb 4, 4:35 pm, JanPB wrote:
On Feb 4, 4:01 pm, wrote:
Feynman diagrams aim.
Uh, no, they don't.
Yes they do.
Cite reference.

The reference is any Feynman diagram.

Nonsense. You CLEARLY do not know what a Feynman diagram represents.

Every Feynman diagram represents a specific INTEGRAL. In the QFT to
which the diagram belongs there is a 1-to-1 mapping between components
of the diagram and terms in the integral. For instance, in QED the
squiggly line corresponding to a photon corresponds to a term
p_slash/p^2; an electron-electron-photon vertex corresponds to
\integral d^3p/E delta^4(p1+p2+p3) \alpha; etc.

As I said before, it is implicit in the definition of the diagram that
one integrates every vertex over all of momentum space. This directly
implies that the directions of the lines connecting vertices are of no
consequence whatsoever, because they will "point" in all possible
directions during the integrations. As I said before, ALL that matters
is the topology of the diagram (that determines how the various terms of
the integral are inter-connected).

And if you think that because a given line connects to a given vertex in
a specific diagram that the object represented by the line "is aimed" at
that vertex, again you would be wrong -- one must SUM over all possible
diagrams, and (ANTI-)SYMMETRIZE over all external legs. This completely
"jumbles" any correspondence you might imagine between legs and vertices..

Tom Roberts

----------------
you can have all diagrams and integrals in the world
but if all thatis based on wrong assuptions ??

you can stuff all your * 'sophystication:

you have a basic assumption that you deal with PHOTONS
*AS force mediators
but allas ??
what would you do if * *one day you will find that
PHOTONS CANNOT BE ANY *ATTRACTION *FORCE ???!! * MEDIATORS *??!!
Y.Porat
----------------------- Hide quoted text -

- Show quoted text -

The bosons in a Feynman diagram must aim. That is stupid.

MItch Raemsch Twice Nobel Laureate 2008

On Feb 6, 3:48 pm, wrote:
On Feb 6, 8:23 am, "Y.Porat" wrote:



On Feb 5, 12:09 pm, Tom Roberts wrote:

wrote:
On Feb 4, 4:35 pm, JanPB wrote:
On Feb 4, 4:01 pm, wrote:
Feynman diagrams aim.
Uh, no, they don't.
Yes they do.
Cite reference.

The reference is any Feynman diagram.

Nonsense. You CLEARLY do not know what a Feynman diagram represents.

Every Feynman diagram represents a specific INTEGRAL. In the QFT to
which the diagram belongs there is a 1-to-1 mapping between components
of the diagram and terms in the integral. For instance, in QED the
squiggly line corresponding to a photon corresponds to a term
p_slash/p^2; an electron-electron-photon vertex corresponds to
\integral d^3p/E delta^4(p1+p2+p3) \alpha; etc.

As I said before, it is implicit in the definition of the diagram that
one integrates every vertex over all of momentum space. This directly
implies that the directions of the lines connecting vertices are of no
consequence whatsoever, because they will "point" in all possible
directions during the integrations. As I said before, ALL that matters
is the topology of the diagram (that determines how the various terms of
the integral are inter-connected).

And if you think that because a given line connects to a given vertex in
a specific diagram that the object represented by the line "is aimed" at
that vertex, again you would be wrong -- one must SUM over all possible
diagrams, and (ANTI-)SYMMETRIZE over all external legs. This completely
"jumbles" any correspondence you might imagine between legs and vertices.

Tom Roberts

----------------
you can have all diagrams and integrals in the world
but if all thatis based on wrong assuptions ??

you can stuff all your 'sophystication:

you have a basic assumption that you deal with PHOTONS
AS force mediators
but allas ??
what would you do if one day you will find that
PHOTONS CANNOT BE ANY ATTRACTION FORCE ???!! MEDIATORS ??!!
Y.Porat
----------------------- Hide quoted text -

- Show quoted text -

The bosons in a Feynman diagram must aim.

Why?

--
Jan Bielawski

Motes eat. Shut up, Porat.

Deuterium's and americium's nucleòns are heavier than iron's and
nickel's nucleòns.

Silicon's and carbon's charges are heavier than silica's and carba's
charges.

Protòn's and neutròn's quarks are heavier than piòn's and kaòn's
quarks--if one gave them antiprotòn and antineutròn, they will happily
decay intom the latter pairs, and then leptòns and neutrinos which are
even leihter.

-Aut

Y.Porat wrote:
what would you do if one day you will find that
PHOTONS CANNOT BE ANY ATTRACTION FORCE ???!!

Cannot possibly happen. Photons are MATHEMATICAL OBJECTS that model
electromagnetic interactions. Mathematically they are quite clearly
capable of transmitting both attractive and repulsive forces (which it
is depends on the relative phases).

Now it could happen that the theories we have which include photons turn
out to not be the ultimate theory, but only an approximation (albeit an
extremely good one for all experiments so far). Indeed current progress
toward quantum gravity makes it seem likely that all current QFTs are
merely effective field theories -- approximations to a deeper theory.
But that is quite different from what you said.

You should actually LEARN something about the relevant theories before
being so quick to dispute them.


Tom Roberts

On Feb 8, 11:11 pm, "Y.Porat" wrote:
On Feb 7, 9:23 am, "Autymn D. C." wrote:

Motes eat. Shut up, Porat.

Deuterium's and americium's nucleòns are heavier than iron's and
nickel's nucleòns.

Silicon's and carbon's charges are heavier than silica's and carba's
charges.

Protòn's and neutròn's quarks are heavier than piòn's and kaòn's
quarks--if one gave them antiprotòn and antineutròn, they will happily
decay intom the latter pairs, and then leptòns and neutrinos which are
even leihter.

-Aut

-------------------
it is eather me or you that are idiots

There is a third option.

[...]