Re: How can photons be massless
From: Y.Porat (maporat_at_012.net.il)
Date: 07/17/04
- Next message: John A. Bailo: "Re: Looking for the best way to browse threads."
- Previous message: DAH: "re:The Cost of Relativity"
- In reply to: Bjoern Feuerbacher: "Re: How can photons be massless"
- Next in thread: alistair: "Re: How can photons be massless"
- Messages sorted by: [ date ] [ thread ]
Date: 16 Jul 2004 20:48:29 -0700
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<cd8b2g$agb$1@news.urz.uni-heidelberg.de>...
> Y.Porat wrote:
> > Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<cd5t0f$cer$1@news.urz.uni-heidelberg.de>...
> >
>
> >>>
>
> >>his ideas.
> >
> > -----------------
> > yes i also recomend anyone to see the thread '1 FERTZ'
>
> Yes, I also recommend that. It will also give you a nice
> picture of how clueless Porat is, and how little he understands science.
>
>
> > The pompous ashole Feuerbacher invented the FERTZ
> > so you can get some idea about his intellectual abilitis.
>
> And here you can see that Porat thinks that insults are arguments.
> -----------------
as far as i remember it is you who always strt that stile
may be you even not notice it
that once you say 'you dont know what physics is' or alike
that is the beginning of that process
because you are not in a position to patronize me
--------------
>
>
> >>>ie is
> >>>Photon mass (the smallest) = H/C^2 times 1/ 1 time unit
> >>
> >>It was pointed out *lots* of times that this does not make sense,
> >>since if one chooses a different time unit, the result will be
> >>different.
> >>
> >>I'll show it again for you: if one uses the standard SI units
> >>(meters, kilograms and seconds), one has
> >> h = 6.626 * 10^(-34) J s = 6.626 * 10^(-34) kg m^2 / s
> >> c = 3 * 10^8 m / min
> >>and therefore:
> >> m_photon = 7.36 * 10^(-51) kg
> >>
> >>However, if one uses one minute as the time unit, one has:
> >> h = 3.976 * 10^(-32) kg m^2 / min
> >> c = 1.8 * 10^10 m / min
> >>and therefore:
> >> m_photon = 1.23 * 10^(-52) kg
> >
> > ----------------
> > no minute and no shminute
> > the formula is time free foermula
>
> A formula which has "time unit" in it is a "time free formula"?????
> --------------------
it has only mass units at the left side
therefore should have only mass units at the right side
if you get it time dependant on anyway
you have a mistake on your way!!
---------------
>
> > it was derived from the formula E=mC^2 =Hf
>
> You have no clue what a "derivation" is.
-------------
and that is not supoposed to be an insult .....
i got it from MC^2 = hf
if thse basic formulas are right the aritmetic
out come of it is also legitimate
now if you isolate m
it is a pure mass formula nothinjg like time in it
and it should not be dependant on any unit system you use
provided it is done correctly
if you get it time dependant
there is a mistake on your way.
f should become 1.0000 on any unit system you take
----------------------
>
> You simply say that one has to insert "1/ 1 time unit" for f.
>
> And conveniently ignore that the photon mass which comes out
> depends on the time unit one chooses.
because apparently you miss something may be the g factor
which is the difference between mks and cgs?
-------------
>
>
> > both are legitimate
>
> Yes. However, your application of them is not.
>
>
> > and while you extract m to the left side it is m nothing else
>
> I have no clue what this is supposed to mean.
-------------
if it is just m and nothing more
time should be irrelevant
-----------
>
>
> > you ave to chose the smallest time unit
> > no matter what system you use.
>
> Hint: as I showed above, different systems lead to different
> results.
-----------
so look for your mistake
i will try as well to find the mistake
other readers are invidet to solve the 'riddle'
---------------
>
>
> > a formula of mass is just mass nothing else
>
> I have no clue what this is supposed to mean.
>
>
> > and you can always chse the smallest integer number for your time system
>
> Or that.
>
>
> > you use the same time units as you did for h and for c
>
> I did do exactly that above. Result: for different unit systems,
> the photon mass is different.
but what is your answr to that strange phenomena
that a formula that has no time factoer at all
is sensitive to time input??
---------------
>
>
> > while you do that Time is nonexistant at your formula.
>
> I have no clue what this is supposed to mean.
at this moment i dont have an explanation unless
there is a hiden mistake in one of your results.
it cant be otherwise by definition of a time free formula.
--------------
>
>
> Look at the calculation above and tell me what is wrong with it, please.
>
> You are in plain denial now. It is *obvious* to *anyone* that you
> are wrong. LOPK AT THE CALCULATION!!!!!
--------------
there must be a hidden mistake dont you see that??
-------------
>
Y.Porat
-----------------
- Next message: John A. Bailo: "Re: Looking for the best way to browse threads."
- Previous message: DAH: "re:The Cost of Relativity"
- In reply to: Bjoern Feuerbacher: "Re: How can photons be massless"
- Next in thread: alistair: "Re: How can photons be massless"
- Messages sorted by: [ date ] [ thread ]
Relevant Pages
|
