From: Y.Porat on
On Jun 25, 5:50 am, "Y.Porat" <y.y.po...(a)gmail.com> wrote:
> On Jun 24, 11:17 pm, PD <thedraperfam...(a)gmail.com> wrote:
>
>
>
> > On Jun 23, 3:59 am, "Y.Porat" <y.y.po...(a)gmail.com> wrote:
>
> > > the photon momentum can be presented as
>
> > > P =  hf/c
> > > right
> > > it is the full   comprehensive presentation
> > > of the photon momentum
> > > nothing missing - nothing excessive   right ?? !!
>
> > > now lets take it as is
> > > (without changing anything in it as the formula
> > > presenting the **photon momentum *
> > > momentum ie not energy   .....!! )
>
> > > and present it by its dimensions and
> > > dimensionless  figures
>
> > > h is
> > > 6.6 exp -34
>
> > > f is
> > > fs/second
>
> > > while   fs is*** the dimensionless
> > > figure that is attached to the  1/second ****
>
> > > c is say (aprox )
> > > 3  exp10   meter/second
> > > now if we combine all of it
> > > we get
>
> > > P  6.6 exp -Kg    meter ^2 /second  times fs/Sec
> > > divided by  meter/Second times   3 exp10
>
> > >  ie if we present it without the
> > > dimensions that are canceling  themselves
> > > in    nominator and denominator
>
> > > we get
> > > ====================================
> > > 6.6exp-34   Kg     MET /SEC times fs/3 exp10
> > > ====================================
> > > now my question is
> > > where do  you   see anything relativistic in it  ??!!
>
> > > TIA
> > > Y.Porat
> > > ----------------------
>
> > Just a suggestion, Porat.
>
> > Don't title a post that implies that you're going to do dimensional
> > analysis, if you don't know how to do dimensional analysis, if you
> > don't know the dimensions or the units of the quantities you're going
> > to look at, if you think that relativistic quantities have different
> > dimensions than nonrelativistic quantities, or if you think you can
> > learn something about the fundamental nature of a thing by looking at
> > the dimensions of one of its properties.
>
> > This would be like someone proposing to do the plumbing for a kitchen
> > sink while not knowing what a pipe wrench is, what the hole in the
> > bottom of the sink is for, how to tell which spigot is hot and which
> > one is cold, or why a sink should not be made out of rye bread.
>
> > PD
>
> --------------------
> and Mr PD
> i have a suggestion for you: !!!
> instead of being a   CHEAP   demagogue !!!
>
> just tell   us what is ***YOUR**
> AGAIN   ***YOUR***
> DIMENSION ANALYSIS** OF THE PHOTON MOMENTUM
> (momentum not energy
> again momentum and not energy   !!!)
>
> P = hf/Lambda
> AND TEL US WHAT IS RELATIVISTIC THERE !!!
>
> 2
> as far as i know you
> a crook like you  -- will not do it  !!!
> for your   'tactical '   cheating  reasons !!!
> because you know i am right   !!!
>
>  if  you want to   prove that i am wrong with  with my analysis  ---
>
> JUST BRING YOUR ALTERNATIVE ANALYSIS!.
> THAT IS WHAT AN HONEST PERSON
> AND AN HONEST SCIENTIST --  WOULD DO !!!
> and btw
> do   it by the MKS system unless you claim that
> the MKS is '''not good enough  or not legitimate  for it '''  !!!
> and dont  just hand wave demagogically
> adding relevant  obfuscating philosophy   stories   !!!
>
> JUST A SIMPLE DIMENSION ANALYSIS
> INCLUDING THE DIMENSION LESS FIGURES !!!
> OF YOURS   !!!
>
> DON T FORGET OR IGNORE THE DIMENSIONLESS   FIGURES   !!!
>
> TIA
> Y.Porat
> ---------------------------------

sorry Typo
he formula of photon momentum is

h/Lambda !!! or hf/c!!
yet please note
hf /c
IS A MOMENTUM FORMULA
SOPLEASE DONT BOGGLE OUR MIDS PRESENTING IT AS AN SORT OF ENERGY
FORMULA!!
TH EMOMENT IT IS DIVIDED BY c
it isnot anymore an energy formula
but MOMENTUM FORMULA
and it makes a principaldifference as you will see later because
momentum is not energy !!)

so sorry my typo mistake
it is not
not
hf/Lambda
but
h/lambda or hf/lambda (momentum !!!)

(i respond quickly and have in my mind just the main arguments
so such mistakes can occure
in my **quick responds **
***unlike** my main analysts in which i spend more time checking
it !!
and i noticed it before anyone to correct me (:-))

so please MR PD ---as above
just give us *your analysis **of it

TIA
Y.Porat
--------------------
From: artful on
On Jun 25, 2:25 pm, "Y.Porat" <y.y.po...(a)gmail.com> wrote:
[snip]
> sorry Typo
> he formula of photon momentum is
>
> h/Lambda !!!  or hf/c!!

Yes .. we know

> yet please note
> hf /c
> IS A MOMENTUM FORMULA

Yes .. we know. You just said that

> SOPLEASE DONT BOGGLE OUR MIDS PRESENTING IT AS AN SORT OF ENERGY
> FORMULA!!

I didn't. I gave a number of different equations for photon momentum,
all equally valid

P = Mc = E/c = hf/c = h/lambda

> TH EMOMENT IT IS DIVIDED BY c

Yes .. we know .. you divide the energy of the photon by c. E = hf,
so E/c = hf/c = momentum of the photon

> it isnot anymore an energy formula
> but MOMENTUM FORMULA

Yes .. we know

> and it makes a principaldifference as you will  see later because
> momentum is not energy !!)

Yes .. we know

> so sorry my typo mistake
> it is not
> not
> hf/Lambda
> but
> h/lambda  or    hf/lambda   (momentum  !!!)

I think you mean "h/lamda or hf/c"

> (i respond quickly  and have in my mind just  the main arguments
>  so such mistakes can occure
> in my **quick responds **
> ***unlike** my main analysts in which i spend more time checking
> it !!
> and i noticed it before anyone to correct me (:-))
>
> so please    MR    PD ---as above
> just give    us *your analysis **of it

I've done this before for you .. its trivial

Dimensions of momentum are

[Mass] x [Length] x [Time]^-1

Dimensions of h are

[Mass] x [Length]^2 x [Time]^-1

Dimensions of lambda are

[Length]

Dimensions of f are

[Time]^-1

Dimensions of c are

[Length] x [time]^-1

So h/lambda has dimensions

[Mass] x [Length]^2 x [Time]^-1 / [Length] = [Mass] x [Length] x
[Time]^-1

So it is a valid momentum formula

Similarly hf/c has dimensions

( [Mass] x [Length]^2 x [Time]^-1 ) x ( [Time]^-1 ) / ( [Length] x
[Time]-1 ) = [Mass] x [Length] x [Time]^-1

So it is a valid momentum formula

(hopefully no typos there)

That is a complete dimensional analysis. Do you understand it?
From: Y.Porat on
On Jun 25, 7:02 am, artful <artful...(a)hotmail.com> wrote:
> On Jun 25, 2:25 pm, "Y.Porat" <y.y.po...(a)gmail.com> wrote:
> [snip]
>
> > sorry Typo
> > he formula of photon momentum is
>
> > h/Lambda !!!  or hf/c!!
>
> Yes .. we know
>
> > yet please note
> > hf /c
> > IS A MOMENTUM FORMULA
>
> Yes .. we know.  You just said that
>
> > SOPLEASE DONT BOGGLE OUR MIDS PRESENTING IT AS AN SORT OF ENERGY
> > FORMULA!!
>
> I didn't.  I gave a number of different equations for photon momentum,
> all equally valid
>
> P = Mc = E/c = hf/c = h/lambda
>
> > TH EMOMENT IT IS DIVIDED BY c
>
> Yes .. we know .. you divide the energy of the photon by c.  E = hf,
> so E/c = hf/c = momentum of the photon
>
> > it isnot anymore an energy formula
> > but MOMENTUM FORMULA
>
> Yes .. we know
>
> > and it makes a principaldifference as you will  see later because
> > momentum is not energy !!)
>
> Yes .. we know
>
> > so sorry my typo mistake
> > it is not
> > not
> > hf/Lambda
> > but
> > h/lambda  or    hf/lambda   (momentum  !!!)
>
> I think you mean "h/lamda or hf/c"
>
> > (i respond quickly  and have in my mind just  the main arguments
> >  so such mistakes can occure
> > in my **quick responds **
> > ***unlike** my main analysts in which i spend more time checking
> > it !!
> > and i noticed it before anyone to correct me (:-))
>
> > so please    MR    PD ---as above
> > just give    us *your analysis **of it
>
> I've done this before for you .. its trivial
>
> Dimensions of momentum are
>
> [Mass] x [Length] x [Time]^-1
>
> Dimensions of h are
>
> [Mass] x [Length]^2 x [Time]^-1
>
> Dimensions of lambda are
>
> [Length]
>
> Dimensions of f are
>
> [Time]^-1
>
> Dimensions of c are
>
> [Length] x [time]^-1
>
> So h/lambda has dimensions
>
> [Mass] x [Length]^2 x [Time]^-1 / [Length] = [Mass] x [Length] x
> [Time]^-1
>
> So it is a valid momentum formula
>
> Similarly hf/c has dimensions
>
> ( [Mass] x [Length]^2 x [Time]^-1 ) x ( [Time]^-1 ) / ( [Length] x
> [Time]-1 ) = [Mass] x [Length] x [Time]^-1
>
> So it is a valid momentum formula
>
> (hopefully no typos there)
>
> That is a complete dimensional analysis.  Do you understand it?

------------
psychopath crook
with 3 different anonymous names
DO YOU THINK THAT PD NEEDS YOUR FUCKEN crooked HELP?!!
2
NASTY PIGGY
YOU 'FORGOT'' THE DIMENSION LESS
FIGURES !!! (not accidentally (:-)

so little Josef Goebbels

jut let PD do it ** properly and honestly **
and completely to the last detail !! !!

'THE DEVIL IS IN THE DETAILS' !!!)

TIA to PD

Y.Porat
-------------------------------
From: artful on
On Jun 25, 3:51 pm, "Y.Porat" <y.y.po...(a)gmail.com> wrote:
> On Jun 25, 7:02 am, artful <artful...(a)hotmail.com> wrote:
>
>
>
>
>
> > On Jun 25, 2:25 pm, "Y.Porat" <y.y.po...(a)gmail.com> wrote:
> > [snip]
>
> > > sorry Typo
> > > he formula of photon momentum is
>
> > > h/Lambda !!!  or hf/c!!
>
> > Yes .. we know
>
> > > yet please note
> > > hf /c
> > > IS A MOMENTUM FORMULA
>
> > Yes .. we know.  You just said that
>
> > > SOPLEASE DONT BOGGLE OUR MIDS PRESENTING IT AS AN SORT OF ENERGY
> > > FORMULA!!
>
> > I didn't.  I gave a number of different equations for photon momentum,
> > all equally valid
>
> > P = Mc = E/c = hf/c = h/lambda
>
> > > TH EMOMENT IT IS DIVIDED BY c
>
> > Yes .. we know .. you divide the energy of the photon by c.  E = hf,
> > so E/c = hf/c = momentum of the photon
>
> > > it isnot anymore an energy formula
> > > but MOMENTUM FORMULA
>
> > Yes .. we know
>
> > > and it makes a principaldifference as you will  see later because
> > > momentum is not energy !!)
>
> > Yes .. we know
>
> > > so sorry my typo mistake
> > > it is not
> > > not
> > > hf/Lambda
> > > but
> > > h/lambda  or    hf/lambda   (momentum  !!!)
>
> > I think you mean "h/lamda or hf/c"
>
> > > (i respond quickly  and have in my mind just  the main arguments
> > >  so such mistakes can occure
> > > in my **quick responds **
> > > ***unlike** my main analysts in which i spend more time checking
> > > it !!
> > > and i noticed it before anyone to correct me (:-))
>
> > > so please    MR    PD ---as above
> > > just give    us *your analysis **of it
>
> > I've done this before for you .. its trivial
>
> > Dimensions of momentum are
>
> > [Mass] x [Length] x [Time]^-1
>
> > Dimensions of h are
>
> > [Mass] x [Length]^2 x [Time]^-1
>
> > Dimensions of lambda are
>
> > [Length]
>
> > Dimensions of f are
>
> > [Time]^-1
>
> > Dimensions of c are
>
> > [Length] x [time]^-1
>
> > So h/lambda has dimensions
>
> > [Mass] x [Length]^2 x [Time]^-1 / [Length] = [Mass] x [Length] x
> > [Time]^-1
>
> > So it is a valid momentum formula
>
> > Similarly hf/c has dimensions
>
> > ( [Mass] x [Length]^2 x [Time]^-1 ) x ( [Time]^-1 ) / ( [Length] x
> > [Time]-1 ) = [Mass] x [Length] x [Time]^-1
>
> > So it is a valid momentum formula
>
> > (hopefully no typos there)
>
> > That is a complete dimensional analysis.  Do you understand it?
>
> ------------
> psychopath crook

Nope

> with 3 different anonymous names

As I've explained the reason for many time .. get over it.

> DO YOU THINK THAT PD NEEDS YOUR FUCKEN    crooked      HELP?!!

Yes .. you most certainly do

> 2
> NASTY PIGGY
> YOU 'FORGOT'' THE DIMENSION LESS
> FIGURES  !!! (not accidentally   (:-)

No .. I didn't FORGET them .. and it wasn't accidentally .. you asked
for a DIMENSIONAL ANALYSIS which does NOT include the dimensioless
values .. only the dimensions. THAT is why it is called DIMENSIONAL
analysis. Get it? Probably not.

> so little       Josef Goebbels

Not me

> jut let PD do   it ** properly and honestly **

No need .. I already have done it properly and honestly .. But if he
wants to do it again, he can.

> and completely to  the last detail  !!  !!

Like I did. Complete and detailed.

> 'THE DEVIL IS   IN THE DETAILS'      !!!)

Yes it is .. you seem to ignore details like: what is a dimensional
analysis.

> TIA to PD

Note that even including a dimensionless number (so its no longer
dimensional anlysis) does not change anything .. the value is
arbitrary (depending on the units you use and the values of the
variables involved). All you'll end up showing is what we already
know .. photon momentum is proportional to frequency and inversely
proportional to wavelength (with some unit-dependent ratio).
From: Y.Porat on
On Jun 25, 8:04 am, artful <artful...(a)hotmail.com> wrote:
> On Jun 25, 3:51 pm, "Y.Porat" <y.y.po...(a)gmail.com> wrote:
>
>
>
> > On Jun 25, 7:02 am, artful <artful...(a)hotmail.com> wrote:
>
> > > On Jun 25, 2:25 pm, "Y.Porat" <y.y.po...(a)gmail.com> wrote:
> > > [snip]
>
> > > > sorry Typo
> > > > he formula of photon momentum is
>
> > > > h/Lambda !!!  or hf/c!!
>
> > > Yes .. we know
>
> > > > yet please note
> > > > hf /c
> > > > IS A MOMENTUM FORMULA
>
> > > Yes .. we know.  You just said that
>
> > > > SOPLEASE DONT BOGGLE OUR MIDS PRESENTING IT AS AN SORT OF ENERGY
> > > > FORMULA!!
>
> > > I didn't.  I gave a number of different equations for photon momentum,
> > > all equally valid
>
> > > P = Mc = E/c = hf/c = h/lambda
>
> > > > TH EMOMENT IT IS DIVIDED BY c
>
> > > Yes .. we know .. you divide the energy of the photon by c.  E = hf,
> > > so E/c = hf/c = momentum of the photon
>
> > > > it isnot anymore an energy formula
> > > > but MOMENTUM FORMULA
>
> > > Yes .. we know
>
> > > > and it makes a principaldifference as you will  see later because
> > > > momentum is not energy !!)
>
> > > Yes .. we know
>
> > > > so sorry my typo mistake
> > > > it is not
> > > > not
> > > > hf/Lambda
> > > > but
> > > > h/lambda  or    hf/lambda   (momentum  !!!)
>
> > > I think you mean "h/lamda or hf/c"
>
> > > > (i respond quickly  and have in my mind just  the main arguments
> > > >  so such mistakes can occure
> > > > in my **quick responds **
> > > > ***unlike** my main analysts in which i spend more time checking
> > > > it !!
> > > > and i noticed it before anyone to correct me (:-))
>
> > > > so please    MR    PD ---as above
> > > > just give    us *your analysis **of it
>
> > > I've done this before for you .. its trivial
>
> > > Dimensions of momentum are
>
> > > [Mass] x [Length] x [Time]^-1
>
> > > Dimensions of h are
>
> > > [Mass] x [Length]^2 x [Time]^-1
>
> > > Dimensions of lambda are
>
> > > [Length]
>
> > > Dimensions of f are
>
> > > [Time]^-1
>
> > > Dimensions of c are
>
> > > [Length] x [time]^-1
>
> > > So h/lambda has dimensions
>
> > > [Mass] x [Length]^2 x [Time]^-1 / [Length] = [Mass] x [Length] x
> > > [Time]^-1
>
> > > So it is a valid momentum formula
>
> > > Similarly hf/c has dimensions
>
> > > ( [Mass] x [Length]^2 x [Time]^-1 ) x ( [Time]^-1 ) / ( [Length] x
> > > [Time]-1 ) = [Mass] x [Length] x [Time]^-1
>
> > > So it is a valid momentum formula
>
> > > (hopefully no typos there)
>
> > > That is a complete dimensional analysis.  Do you understand it?
>
> > ------------
> > psychopath crook
>
> Nope
>
> > with 3 different anonymous names
>
> As I've explained the reason for many time .. get over it.
>
> > DO YOU THINK THAT PD NEEDS YOUR FUCKEN    crooked      HELP?!!
>
> Yes .. you most certainly do
>
> > 2
> > NASTY PIGGY
> > YOU 'FORGOT'' THE DIMENSION LESS
> > FIGURES  !!! (not accidentally   (:-)
>
> No .. I didn't FORGET them .. and it wasn't accidentally .. you asked
> for a DIMENSIONAL ANALYSIS which does NOT include the dimensioless
> values .. only the dimensions.  THAT is why it is called DIMENSIONAL
> analysis.  Get it?  Probably not.
>
> > so little       Josef Goebbels
>
> Not me
>
> > jut let PD do   it ** properly and honestly **
>
> No need .. I already have done it properly and honestly .. But if he
> wants to do it again, he can.
>
> > and completely to  the last detail  !!  !!
>
> Like I did.  Complete and detailed.
>
> > 'THE DEVIL IS   IN THE DETAILS'      !!!)
>
> Yes it is .. you seem to ignore details like: what is a dimensional
> analysis.
>
> > TIA to PD
>
> Note that even including a dimensionless number (so its no longer
> dimensional anlysis) does not change anything .. the value is
> arbitrary (depending on the units you use and the values of the
> variables involved).  All you'll end up showing is what we already
> know .. photon momentum is proportional to frequency and inversely
> proportional to wavelength (with some unit-dependent ratio).

-------------------
i said that i have no intention todsicuss with
little Josef Goebbels

neat please not for the psycho pigshit anonymous
and please include the dimension less figures
that are indispensable for that analysis
iow
without them the --formula ***means nothing !!
as a real physicist should know !
****as its time for psycho pigshit
(that wants to teach me physics and to obfuscate
and to wast time ) ----to know it as well!!!****
as i said

THE DEVIL IS IN THE DE TALES!!

not for little pigshits !!!
so next please to PD
TIA
Y.Porat
--------------------------------------

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