I have an equation for the mass of the photon!
Peter Christensen
Undefined is the same as undefined.
The equation is EASY to remember. -Use it iff (if and only if) you feel,
that you NEED an equation for the 'mass of the photon'...
PC
(Ps. Don't confuse the second term with C^C :>)
_________________________
Computer Science: 10 = 2D
Math: 0^1 = 0 and 1^0 = 1
PC-Joke: 5* on Google.com
Why 25 characters 5 times,
Did this take 't' to make?
Peter Christensen
>
> Undefined is the same as undefined.
PC-Joke: 5* on Google.com
(http://groups.google.com/group/sci.physics.relativity?hl=en)
Maybe just my own rating, but a 'photon-mass' IS a good joke...
PC
________________
Pe...@MailAPS.org
April 1st today
mme...@cars3.uchicago.edu
>I say: m_photon '=' 0^0
>
>Undefined is the same as undefined.
>
are strictly correct) it evaluates to zero.
Mati Meron | "When you argue with a fool,
me...@cars.uchicago.edu | chances are he is doing just the same"
Peter Christensen
<mme...@cars3.uchicago.edu> skrev i en meddelelse
news:krsXf.9$45....@news.uchicago.edu...
> In article <442e50bf$0$15792$1472...@news.sunsite.dk>, "Peter
> Christensen" <Pe...@MailAPS.org> writes:
>>I say: m_photon '=' 0^0
>>
>>Undefined is the same as undefined.
>>
> No, it is perfectly well defined. And (assuming Maxwell's equations
> are strictly correct) it evaluates to zero.
How does that work?
PC
________________
Pe...@MailAPS.org
mme...@cars3.uchicago.edu
4-vector. Meaning
m^2*c^4 = E^2 - p^2*c^2
Assuming that Maxwell's equations are rigorously correct, for planar
monochromatic wave (that's what photon represents you've E = pc
Substitute to the above and you get m = 0.
Peter Christensen
> In article <442e55e3$0$15795$1472...@news.sunsite.dk>, "Peter
> Christensen" <Pe...@MailAPS.org> writes:
snip
> Mass is defined (up to scaling) as the length of the energy-momentum
> 4-vector. Meaning
>
> m^2*c^4 = E^2 - p^2*c^2
>
> Assuming that Maxwell's equations are rigorously correct, for planar
> monochromatic wave (that's what photon represents you've E = pc
>
> Substitute to the above and you get m = 0.
I get the same result, if I use this first equation. But, with all respect,
I don't think that it applies for photons. I agree with you, that E = pc for
photons. It is known, from quantum mechanics, that for photons E = h*f and p
= h/lambda. It is also known, that c = lambda*f, so we do get E = pc. (f is
the frequency and lambda is the wavelength.)
PC
________________
Pe...@MailAPS.org
Y.Porat
th e mass of the photon has been aleady defined: it is
(you have just to forget about the Lorentz
factor while you deal with the photon
it does not apply to the photon
once you realize it
every thing becomes simple as cvould be.)
mass photon =hf/C^2
Y.Porat
------------------------
Peter Christensen
"Y.Porat" <map...@012.net.il> skrev i en meddelelse
news:1143889209....@j33g2000cwa.googlegroups.com...
> you are too late and too slow!!
>
> th e mass of the photon has been aleady defined: it is
> (you have just to forget about the Lorentz
> factor while you deal with the photon
> it does not apply to the photon
> once you realize it
> every thing becomes simple as cvould be.)
>
>
> mass photon =hf/C^2
I agree, that the energy is E = h*f for a photon, but E=m*c^2 does not apply
for photons.
m_photon can't be defined.
PC
________________
Pe...@MailAPS.org
Peter Webb
<mme...@cars3.uchicago.edu> wrote in message
news:mMsXf.10$45....@news.uchicago.edu...
**Rest mass** is zero. A somewhat abstract measure, given a photon cannot be
at rest.
The OP was in a sense not that far off. The photon's relativistic mass if
you naively plug it into Special Relativisties mass equation is
m = 0/sqrt(1-c^2/c^2) = 0/0.
Of course, its correct relativistic mass is hf/c^2, where h is Plancks
constant and f is its frequency.
I do agree that the OP's discovery of the mass of the photon is 101 years
late, and wrong.
Sue...
mme...@cars3.uchicago.edu wrote:
> In article <442e55e3$0$15795$1472...@news.sunsite.dk>, "Peter Christensen" <Pe...@MailAPS.org> writes:
> >
> ><mme...@cars3.uchicago.edu> skrev i en meddelelse
> >news:krsXf.9$45....@news.uchicago.edu...
> >> In article <442e50bf$0$15792$1472...@news.sunsite.dk>, "Peter
> >> Christensen" <Pe...@MailAPS.org> writes:
> >>>I say: m_photon '=' 0^0
> >>>
> >>>Undefined is the same as undefined.
> >>>
> >> No, it is perfectly well defined. And (assuming Maxwell's equations
> >> are strictly correct) it evaluates to zero.
> >
> >How does that work?
> >
> Mass is defined (up to scaling) as the length of the energy-momentum
> 4-vector. Meaning
>
> m^2*c^4 = E^2 - p^2*c^2
>
> Assuming that Maxwell's equations are rigorously correct, for planar
> monochromatic wave (that's what photon represents you've E = pc
That would be a false assumption. A photon represents a probabilistic
connection between an atomic emission and absorbtion.
Maxwell's equation's can descrbe standing waves where there
is no energy and thus no equivalent mass transfered over a path.
Sue...
Peter Webb
"Sue..." <suzyse...@yahoo.com.au> wrote in message
news:1143890806.6...@i40g2000cwc.googlegroups.com...
I agree that what you and Peter Christensen say is correct, but I don't
think that is required to show the photon is massless. Einstein's papers of
1905 on the photoelectric effect and Special Relativity between them produce
a (rest) massless photon as an immediate corollary. All you need from
quantum theory is the value of Planck's constant, the machinery of complex
number probability spaces is unneccessary.
Sue...
Do you compute the petrol consumtion of a horse and carriage
when you fill your automoble for a long trip? :o)
"Einstein, Albert. 1920. Relativity: The Special and General Theory."
http://www.bartleby.com/173/
<< The Nobel Committee avoids committing itself to the particle
concept. Light-quanta or with modern terminology, photons, were
explicitly mentioned in the reports on which the prize decision
rested only in connection with emission and absorption processes.
The Committee says that the most important application of
Einstein's photoelectric law and also its most convincing confirmation
has come from the use Bohr made of it in his theory of atoms, which
explains a vast amount of spectroscopic data.>>
http://nobelprize.org/physics/articles/ekspong/index.html
Sue...
Peter Christensen
I would also say undefined, as for example 0/0 or 0^0, rather than just
use m_photon = 0.
I still don't think that E=m*c^2 (with E=h*f) can be used for photons.
If the photon had a mass, then it would be affected by gravitational
fields, which it doesn't.
PC
________________
Pe...@MailAPS.org
Sue...
In a sense it affected by gravity because it is defined by *either*
emission or absorbtion. If the nuclear resonance of the absorber shifts
with altitude, then a more accurate statement might be that a photon
shifts with gravity but its light does not.
Pound Rebka's Doppler modulator shifts the light to match that
of an absorbed photon.
'On the Interpretation of the Redshift in a Static Gravitational Field'
http://arxiv.org/abs/physics/9907017
Sue...
Sue...
>
> PC
> ________________
> Pe...@MailAPS.org
Y.Porat
doe s not apply to the photon??
that is the orriginal formula of Einstien
did he ever said what you say ?
if not the burden of prove id on you!!!
ATB
Y.Porat
----------------------
Daniel Pitts
The point was that "E^2 = m^2*c^4 + p^2*c^2" and all the energy of a
photon comes from the "p^2*c^2" part of the equation. So, the energy
density of a photon is represented by E=p*c. Any energy density DOES
bend space time. It has been proven that light is affected by and does
effect gravity (e.g. the warping of space-time).
As a side note:
There is a big difference between the undefined values of 0/0 and 0^0,
especially if you arrive at those values through a limit equation. x/x
(x lim->0) is "almost 0" where as x^0 (x lim->0) is "almost 1" since
you can get arbitrarilty close to x=0 and the answer will always be 1.
I'm not saying that 0/0 is always 0, or that 0^0 is always 1, just that
it depends on the context, and that 0/0 is not the same as 0^0
Peter Christensen
>> I still don't think that E=m*c^2 (with E=h*f) can be used for photons.
>> If the photon had a mass, then it would be affected by gravitational
>> fields, which it doesn't.
>
> In a sense it affected by gravity because it is defined by *either*
> emission or absorbtion. If the nuclear resonance of the absorber shifts
>
> with altitude, then a more accurate statement might be that a photon
> shifts with gravity but its light does not.
> Pound Rebka's Doppler modulator shifts the light to match that
> of an absorbed photon.
> 'On the Interpretation of the Redshift in a Static Gravitational Field'
> http://arxiv.org/abs/physics/9907017
I would say, that the photon will loose energy and momentum, because of the
way that the space-time effects are changing f (frequency) and lambda
(wavelength) of the photon. This will change both E=h*f and p=h/lambda.
I don't think about "gravitational mass" or the "potential energy" of the
photon.
PC
________________
Pe...@MailAPS.org
Y.Porat
is ..........
mc*c
in simpler words mC^2
how could all the 'smarties mathematicians ' screw such a simple thing
so far away ??!!!
and in addition please tell us what are the dimensions of p
and what of Energy!!
ATB
Y.Porat
-------------
Hexenmeister
"Peter Christensen" <Pe...@MailAPS.org> wrote in message
news:442e50bf$0$15792$1472...@news.sunsite.dk...
|I say: m_photon '=' 0^0
Ok. :-)
Androcles.
Y.Porat
btw i think you are a bit not updated
the relativistic mass as already abandoned
ATB
Y.Porat
Hexenmeister
<mme...@cars3.uchicago.edu> wrote in message
news:krsXf.9$45....@news.uchicago.edu...
0^3 = 0*0*0
0^2 = 0*0
0^1 = 0
0^0 = "Meron is an idiot."
Androcles | "When you agree with a fool,
andr...@hotmail.com | the guarantee is he is doing just the same"
Tom Roberts
> m_photon can't be defined.
Sure it can! The current experimental upper bound on the mass of the
photon is 6*10^-17 eV/c^2. That is, the experiments cannot distinguish
between QED and an alternative theory in which m_photon!=0, as long as
m_photon is less than this limit.
Those experiments are, at base, why we believe QED is a good theory of
electromagnetism. In QED of course, m_photon=0 rigorously.
Note, please, that "the mass of the photon" cannot be understood outside
of a specific theoretical context. In the usual context of modern
physics, m_photon=0. Period. The other _guesses_ in this thread are
INconsistent with the theoretical context those posters appear to be
trying to use (e.g. they use the equation E_photon=hf, which comes from
this context).
In particular, the above limit is _wildly_ INconsistent with the claim
m_photon=hf/c^2.
> I would also say undefined, as for example 0/0 or 00, rather than just
> use m_photon = 0.
That is just plain wrong in modern physics. The mass of the photon is
well defined, and consistent with zero, as explained above.
Tom Roberts tjroberts2lucent.com
Peter Webb
> I still don't think that E=m*c^2 (with E=h*f) can be used for photons.
> If the photon had a mass, then it would be affected by gravitational
> fields, which it doesn't.
>
Which it is. Hence gravitational lensing.
dlzc1 D:cox T:net@nospam.com N:dlzc D:aol T:com (dlzc)
<mme...@cars3.uchicago.edu> wrote in message
news:krsXf.9$45....@news.uchicago.edu...
> In article <442e50bf$0$15792$1472...@news.sunsite.dk>, "Peter
> Christensen" <Pe...@MailAPS.org> writes:
>>I say: m_photon '=' 0^0
>>
>>Undefined is the same as undefined.
>>
> No, it is perfectly well defined. And (assuming Maxwell's
> equations
> are strictly correct) it evaluates to zero.
Note the date of the post, Mati...
... think "All Fools Day".
David A. Smith
Hexenmeister
"Peter Webb" <webbfamily...@optusnet.com.au> wrote in message
news:442eb288$0$7533$afc3...@news.optusnet.com.au...
Androcles.
LeoK
somewhere here around:
h = [Eemf] - mPH + h
If one makes a broader view one perhaps also can express
X-ray and gamma "particles" with this formula.
Just add a multiply factor to each part of the formula, for example
an X-ray "particle":
b*h = b*[Eemf] - b*mPH + b*h
Pros and cons?
LeoK
Sue...
Peter Christensen wrote:
> snip
>
> >> I still don't think that E=m*c^2 (with E=h*f) can be used for photons.
> >> If the photon had a mass, then it would be affected by gravitational
> >> fields, which it doesn't.
> >
> > In a sense it affected by gravity because it is defined by *either*
> > emission or absorbtion. If the nuclear resonance of the absorber shifts
> >
> > with altitude, then a more accurate statement might be that a photon
> > shifts with gravity but its light does not.
> > Pound Rebka's Doppler modulator shifts the light to match that
> > of an absorbed photon.
> > 'On the Interpretation of the Redshift in a Static Gravitational Field'
> > http://arxiv.org/abs/physics/9907017
>
> I would say, that the photon will loose energy and momentum, because of the
> way that the space-time effects are changing f (frequency) and lambda
> (wavelength) of the photon. This will change both E=h*f and p=h/lambda.
Hmmm..Maybe there is a picture of that here:
http://web.mit.edu/8.02t/www/802TEAL3D/teal_tour.htm
or a discription here:
http://farside.ph.utexas.edu/teaching/em/lectures/lectures.html
Sue...
LeoK
somewhere here around:
h = [Eemf] - mPH + h
If one makes a broader view one perhaps also can express
X-ray and gamma "particles" with this formula.
Just add a common multiply factor to each part of the formula,
Sue...
X-ray particles?
<< The ways that neutrons interact with matter are very
different from the way x rays interact with matter.
X rays interact with the electron cloud surrounding the
nucleus of an atom. Neutrons interact with the nucleus itself >>
http://www.spie.org/web/oer/june/jun00/cover1.html
Sue...
>
> LeoK
mme...@cars3.uchicago.edu
>
><mme...@cars3.uchicago.edu> skrev i en meddelelse
>news:mMsXf.10$45....@news.uchicago.edu...
>> In article <442e55e3$0$15795$1472...@news.sunsite.dk>, "Peter
>> Christensen" <Pe...@MailAPS.org> writes:
>
>snip
>
>> Mass is defined (up to scaling) as the length of the energy-momentum
>> 4-vector. Meaning
>>
>> m^2*c^4 = E^2 - p^2*c^2
>>
>> Assuming that Maxwell's equations are rigorously correct, for planar
>> monochromatic wave (that's what photon represents you've E = pc
>>
>> Substitute to the above and you get m = 0.
>
>I get the same result, if I use this first equation. But, with all respect,
>I don't think that it applies for photons.
It certainly does. It is a definition, not a derived result. It
applies to everything.
> I agree with you, that E = pc for
>photons. It is known, from quantum mechanics, that for photons E = h*f and p
>= h/lambda. It is also known, that c = lambda*f, so we do get E = pc. (f is
>the frequency and lambda is the wavelength.)
>
You've to be careful not to confuse to things. That E = pc for EM
radiation, that follows from Maxwell's equations *regardless* of
quantum mechanics. That EM radiation is quantized, *that* follows
from quantum mechanics and if it is quantized then its particles, the
photons, must follow E = pc as well. What you do in the above is
taking results which were *derived* using E = pc and using them to
show that E = pc. That's wrong practice.
mme...@cars3.uchicago.edu
>
><mme...@cars3.uchicago.edu> wrote in message
>news:mMsXf.10$45....@news.uchicago.edu...
>> In article <442e55e3$0$15795$1472...@news.sunsite.dk>, "Peter
>> Christensen" <Pe...@MailAPS.org> writes:
>>>
>>><mme...@cars3.uchicago.edu> skrev i en meddelelse
>>>news:krsXf.9$45....@news.uchicago.edu...
>>>> In article <442e50bf$0$15792$1472...@news.sunsite.dk>, "Peter
>>>> Christensen" <Pe...@MailAPS.org> writes:
>>>>>I say: m_photon '=' 0^0
>>>>>
>>>>>Undefined is the same as undefined.
>>>>>
>>>> No, it is perfectly well defined. And (assuming Maxwell's equations
>>>> are strictly correct) it evaluates to zero.
>>>
>>>How does that work?
>>>
>> Mass is defined (up to scaling) as the length of the energy-momentum
>> 4-vector. Meaning
>>
>> m^2*c^4 = E^2 - p^2*c^2
>>
>> Assuming that Maxwell's equations are rigorously correct, for planar
>> monochromatic wave (that's what photon represents you've E = pc
>>
>> Substitute to the above and you get m = 0.
>>
>> Mati Meron | "When you argue with a fool,
>> me...@cars.uchicago.edu | chances are he is doing just the same"
>
>**Rest mass** is zero.
We don't call it "rest mass" any more, just "mass".
> A somewhat abstract measure, given a photon cannot be at rest.
Indeed, but the deintion stands regardless.
>
>The OP was in a sense not that far off. The photon's relativistic mass if
>you naively plug it into Special Relativisties mass equation is
>m = 0/sqrt(1-c^2/c^2) = 0/0.
>
>Of course, its correct relativistic mass is hf/c^2, where h is Plancks
>constant and f is its frequency.
The term "relativistic mass" is no longer in use (except for
popularizations and beginning textbooks which are always at least a
generation behind the times, as it serves no useful purpose. All it
is, is another name for energy in scaled units.
mme...@cars3.uchicago.edu
to show that a given quantity (mass, in this case) has a given value
for a given entity, the quantity needs to be clearly defined first, in
terms of measurable quantities. The equation above provides the
modern definition of mass. This is the starting point.
mme...@cars3.uchicago.edu
mass. You can't use relativity with Newtonian gravitation (except as
an approximation), you've to use GR. And in GR there is no longer a
direct connection between the mass and the way trajectories are
affected (in truth, with Newtonian mechanincs the trajectory is
independent of the mass too, it only took a while to grasp the
significance of this).
mme...@cars3.uchicago.edu
all the year around.
T Wake
"Y.Porat" <map...@012.net.il> wrote in message
news:1143905479....@z34g2000cwc.googlegroups.com...
> who told you that E=mC^2
<pedanty>
C is centigrade.
c is (I suspect) what you meant.
The formula you have shown is an approximation of the correct one, so in
theory Einstein never said *it* did apply to the photon.
</pedantry>
> doe s not apply to the photon??
>
> that is the orriginal formula of Einstien
Not quite.
Bill Hobba
<mme...@cars3.uchicago.edu> wrote in message
news:mxBXf.16$45....@news.uchicago.edu...
> In article <sNyXf.2391$kT4.2313@fed1read02>, "N:dlzc D:aol T:com \(dlzc\)"
> <N: dlzc1 D:cox T:n...@nospam.com> writes:
>>Dear mmeron:
>>
>><mme...@cars3.uchicago.edu> wrote in message
>>news:krsXf.9$45....@news.uchicago.edu...
>>> In article <442e50bf$0$15792$1472...@news.sunsite.dk>, "Peter
>>> Christensen" <Pe...@MailAPS.org> writes:
>>>>I say: m_photon '=' 0^0
>>>>
>>>>Undefined is the same as undefined.
>>>>
>>> No, it is perfectly well defined. And (assuming Maxwell's
>>> equations
>>> are strictly correct) it evaluates to zero.
>>
>>Note the date of the post, Mati...
>>
>>... think "All Fools Day".
>>
>>David A. Smith
>>
> True, got a point:-) But then, on sci.physics it is "All Fools Day"
> all the year around.
On sci.physics.relativity as well. That is one thing you need never worry
about - is it an April fools joke - you simply can not tell the difference.
Thanks
Bill
Gregory L. Hansen
Peter Christensen <Pe...@MailAPS.org> wrote:
>
>"Y.Porat" <map...@012.net.il> skrev i en meddelelse
>news:1143889209....@j33g2000cwa.googlegroups.com...
>> you are too late and too slow!!
>>
>> th e mass of the photon has been aleady defined: it is
>> (you have just to forget about the Lorentz
>> factor while you deal with the photon
>> it does not apply to the photon
>> once you realize it
>> every thing becomes simple as cvould be.)
>>
>>
>> mass photon =hf/C^2
>
>I agree, that the energy is E = h*f for a photon, but E=m*c^2 does not apply
>for photons.
Some people would rather assume that it does and then find a new
definition of mass, or leave mass itself not rigorously defined.
--
"He who only sees business in business is a fool."
Eric Gisse
Y.Porat wrote:
> who told you that E=mC^2
> doe s not apply to the photon??
>
> that is the orriginal formula of Einstien
> did he ever said what you say ?
>
> if not the burden of prove id on you!!!
The burden of proof is never on the crank, eh Porat?
>
> ATB
> Y.Porat
> ----------------------
Y.Porat
T Wake wrote:
> "Y.Porat" <map...@012.net.il> wrote in message
> news:1143905479....@z34g2000cwc.googlegroups.com...
> > who told you that E=mC^2
>
> <pedanty>
>
> C is centigrade.
>
> c is (I suspect) what you meant.
>
> The formula you have shown is an approximation of the correct one, so in
> theory Einstein never said *it* did apply to the photon.
yet did he said that* it does** not **apply to the photon**??
>
> </pedantry>
>
> > doe s not apply to the photon??
> >
> > that is the orriginal formula of Einstien
>
> Not quite.
not quite ???
so please show us the orriginal formula created by Einstein
-------------------
TIA
Y.Porat
-------------------------------------
Y.Porat
please tell us what are the dimensions of momentum of the photon
and tell us what are those kilograms there
and if it is relativistic mass
please tell us what is the dimensions of your 'reletivistic mss'
is it relativisic kilograms ??
sowe have a new system of dimewnsions ie
instead of meter kilograms seconds
we have :
meter , relativisticMassrelativistic
klidimensionsIErelativistickilogramsgrams,, seconds (M RK S ) !!!
and the *relativistic kilograms* new dimension system is a copyright
of whom ??
(may be Eric Shaise can give a hand ??anyway Eric shaise
dont answer before Wake he is a big boy may be even bigger than you ..)
TIA
Y.Porat
-----------------------------
Y.Porat
no need for a new definition of mass
there is an urgent need to find a new definition of modern scientists
(:-)
the new understanding of *mathematrical formulas IN PHYSICS
AND THEIR LIMITS OF VALIDATIONS )
IS MUCH LESS PROBLEMETIC
than new definitions of one of the most basic physical entity
that is called mass
btw i heared a strage argument that since th ephoton is always
in motion its mass 'cannt be rest mass' !!
our earth is in a constant motion- therefore 'it does not have rest
mass'!!
hint No 2
the main problem of modern 'physicists' (imho)is that
the y dont reallise that even th e Lorentz factor has its limits
of validation.
ATB
Y.Porat
--------------------------
Peter Christensen
If you say, that physicists have decided to define m_photon to be 0, then
you are probably very sure about it, so I will take your word for it. But
there are some very special problems with THIS particle, because it can be
said, that it simply can't exist at rest, because it always must have the
speed c. The other thing is, that it is already known from the relations
E=h*f and p=h/lambda (lambda*f=c), that E=pc always applies for photons. We
do not have to be able to derive E=pc from Einsteins relation with m=0. So I
thought, that it wouldn't be necessary to have a definition of m_photon = 0
in physics. -And that E=pc was simply used
instead of E(p,m=0), m^2*c^4 = E^2 - p^2*c^2, for photons...
Today masses can be defined from accelerator experiments from the relation
m=E_0/c^2. In the old days Newtons law F=m*a was used. And of course, these
two relations will stil give the same results for the rest-masses of
particles. There are now some problems with the mass of the photon. It
appears, that it can't be defined from experiments like for other particles:
(a) Has m = 0 ever been measured on a 'photon at rest'?
(b) Can F=m*a be used to determine m for a photon, when the speed of the
light is always c?
That's the reason, that I said that m_photon was 'somehow undefined' in
physics.
jOkE: When I wrote something about '0^0', or later '0/0', then I was just
using it in the sense of something that's not defined. I obviously don't
associate these expressions with physics, it was ONLY meant as a part of a
joke in some of my postings from APRIL 1st. A part of the text was intended
as a joke, but I've probably caused a lot of confusion, so maybe the joke
wasn't very good. (Or was it? ;>)
I didn't intend to cause confusion on the subject, and I didn't 'know' that
physicists have decided to define m_photon = 0 (theoretically). I actually
'thought', that it was decided NOT to define it, because: (a) It can't be
measured at rest, and (b) the photon can't be accelerated.
PC
* m_photon = 0^0 (undefined) was meant as a joke for April 1st. And so was
some comments about '5 stars' (toprating of postings on Groups.Google.com /
http://groups.google.com/group/sci.physics.relativity?hl=en). I also rated
the posting to ***** myself (from another account), only meant as a part of
the joke.
* I wrote something about April 1st today in some of my first postings, but
suddenly I got a lot of replies, so I couldn't really get things
explained. -On the other hand, maybe this shouldn't be done before later,
isn't this what the traditions says..? ;>)
* If I had known, how much time I will have to use writing replies today,
then I wouldn't have written my posting about photon-masses yesterday. This
is also true.
Ps. Another 'new law' of mine: a_photon = 0, because the speed of light, c,
is always the same. Maybe I will write a long and complicated posting about
this important 'acceleration problem' next year (April 1st). This only
depend on the feedback on "I have an equation for the MESS of the
photon". -Please feel free to rate this posting (on Google.Com) as you like.
If I get only one star (not worth to read), then I promise, that I will NOT
post anything like this next year!
April 2nd
(Sorry, if my post from yesterday wasn't funny ;>)
_______________________________
Pe...@MailAPS.org
m_photon = 0^0 (undefined), but only on April st!
Peter Christensen
explanation, why I wrote these (lies) yesterday April 1st. I was meant as a
joke. Here is some text from my posting from today:
"When I wrote something about '0^0', or later '0/0', then I was just
using it in the sense of something that's not defined. I obviously don't
associate these expressions with physics, it was ONLY meant as a part of a
joke in some of my postings from April 1st. A part of the text was intended
as a joke, but I've probably caused a lot of confusion, so maybe the joke
wasn't very good. (Or was it? ;>) "
"I didn't intend to cause confusion on the subject, and I didn't 'know' that
physicists have decided to define m_photon = 0 (theoretically). I actually
'thought', that it was decided NOT to define it, because: (a) It can't be
measured at rest, and (b) the photon can't be accelerated."
I'm sorry if it wasn't funny, or if people spended to much effort replying
on it.
PC
"Tom Roberts" <tjro...@lucent.com> skrev i en meddelelse
news:1qxXf.54532$F_3....@newssvr29.news.prodigy.net...
mme...@cars3.uchicago.edu
>Important nOtE:
>
>If you say, that physicists have decided to define m_photon to be 0, then
>you are probably very sure about it, so I will take your word for it.
It is not a matter of "decided", just that by the definition of mass
we're using m_photon *comes out* to be zero.
> But
>there are some very special problems with THIS particle, because it can be
>said, that it simply can't exist at rest, because it always must have the
>speed c.
Indeed.
> The other thing is, that it is already known from the relations
>E=h*f and p=h/lambda (lambda*f=c), that E=pc always applies for photons.
Again, no. The second of the ralationships you quote is *result* of
the first one and of E = pc. E = pc, by itself, results from the
Maxwell's equations. You should take care to distinguish what results
from what, else you get trapped in circularity.
> We do not have to be able to derive E=pc from Einsteins relation with m=0.
Again, E = pc is derived frojm Maxwell's equations. That preceeds
relativity.
>So I thought, that it wouldn't be necessary to have a definition of
>m_photon = 0 in physics.
It is not a definition but a derived result.
> -And that E=pc was simply used
>instead of E(p,m=0), m^2*c^4 = E^2 - p^2*c^2, for photons...
>
You're confused. There is no "instead" here. We use *both*
m^2*c^4 = E^2 - p^2*c^2
which is the definition of mass, for *everything* and
E = pc
Which *results* from Maxwell's equation, for EM radiation. From both
equations together you get m_photon = 0. As I said, this is a derived
result.
>Today masses can be defined from accelerator experiments from the relation
>m=E_0/c^2. In the old days Newtons law F=m*a was used. And of course, these
>two relations will stil give the same results for the rest-masses of
>particles.
As I mentioned elsewhere, we no longer use "rest mass". Only "mass".
> There are now some problems with the mass of the photon.
There are absolutely no problems with the mass of the photon.
> It appears, that it can't be defined from experiments like for other
> particles:
>
It certainly can, you just have to use the right experiments.
>(a) Has m = 0 ever been measured on a 'photon at rest'?
You cannot measure photon at rest.
>
>(b) Can F=m*a be used to determine m for a photon, when the speed of the
>light is always c?
F = ma no longer serves as the definition of mass.
>
>That's the reason, that I said that m_photon was 'somehow undefined' in
>physics.
No, there is nothing "undefined", only the definitions have changed.
Learn the new ones.
>I didn't intend to cause confusion on the subject, and I didn't 'know' that
>physicists have decided to define m_photon = 0 (theoretically).
Wrong, see above.
> I actually
>'thought', that it was decided NOT to define it, because: (a) It can't be
>measured at rest, and (b) the photon can't be accelerated.
>
Not relevant, since the definition of mass has changed. n It involves
neither a measurement at rest, nor acceleration.
Peter Christensen
explanation, why I wrote these (lies) yesterday April 1st. I was meant as a
joke. Here is some text from my posting from today:
(Actually the one, that you just replied on)
"When I wrote something about '0^0', or later '0/0', then I was just
using it in the sense of something that's not defined. I obviously don't
associate these expressions with physics, it was ONLY meant as a part of a
joke in some of my postings from April 1st. A part of the text was intended
as a joke, but I've probably caused a lot of confusion, so maybe the joke
wasn't very good. (Or was it? ;>) "
"I didn't intend to cause confusion on the subject, and I didn't 'know' that
physicists have decided to define m_photon = 0 (theoretically). I actually
'thought', that it was decided NOT to define it, because: (a) It can't be
measured at rest, and (b) the photon can't be accelerated."
I'm sorry if it wasn't funny, or if people spended to much effort replying
on it.
PC
<mme...@cars3.uchicago.edu> skrev i en meddelelse
news:SrMXf.18$45....@news.uchicago.edu...
mme...@cars3.uchicago.edu
>I will not start an argument on this 'mass-problem', just give the
>explanation, why I wrote these (lies) yesterday April 1st. I was meant as a
>joke. Here is some text from my posting from today:
>
>(Actually the one, that you just replied on)
>
>"When I wrote something about '0^0', or later '0/0', then I was just
>using it in the sense of something that's not defined. I obviously don't
>associate these expressions with physics, it was ONLY meant as a part of a
>joke in some of my postings from April 1st. A part of the text was intended
>as a joke, but I've probably caused a lot of confusion, so maybe the joke
>wasn't very good. (Or was it? ;>) "
>
>"I didn't intend to cause confusion on the subject, and I didn't 'know' that
>physicists have decided to define m_photon = 0 (theoretically). I actually
>'thought', that it was decided NOT to define it, because: (a) It can't be
>measured at rest, and (b) the photon can't be accelerated."
I understand the attempt at a joke. What you should understand is
that the definitions of mass currently used relies neither on
measurement at rest nor on acceleration and that m_photon is not
*defined* to be zero, rather it being zero is a result of the current
definition of mass and of the Maxwell's equations.
Peter Christensen
> In article <442e50bf$0$15792$1472...@news.sunsite.dk>, "Peter
>>I say: m_photon '=' 0^0
>>
>>Undefined is the same as undefined.
>>
> No, it is perfectly well defined. And (assuming Maxwell's equations
> are strictly correct) it evaluates to zero.
>
> me...@cars.uchicago.edu | chances are he is doing just the same"
April 1st is Fool's Day
I accept m_photon = 0 today, April 2nd.
I'm sorry, if this joke wasn't funny...
PC
_________________________________________________
T Wake
"Peter Christensen" <Pe...@MailAPS.org> wrote in message
news:442f97e3$0$15783$1472...@news.sunsite.dk...
>
> <mme...@cars3.uchicago.edu> skrev i en meddelelse
> news:krsXf.9$45....@news.uchicago.edu...
>> In article <442e50bf$0$15792$1472...@news.sunsite.dk>, "Peter
>> Christensen" <Pe...@MailAPS.org> writes:
>>>I say: m_photon '=' 0^0
>>>
>>>Undefined is the same as undefined.
>>>
>> No, it is perfectly well defined. And (assuming Maxwell's equations
>> are strictly correct) it evaluates to zero.
>>
>> Mati Meron | "When you argue with a fool,
>> me...@cars.uchicago.edu | chances are he is doing just the same"
>
> April 1st is Fool's Day
>
> I accept m_photon = 0 today, April 2nd.
>
> I'm sorry, if this joke wasn't funny...
It's ok. I found it mildly amusing (mainly in how it brings the cranks out
of the woodwork).
T Wake
>
> T Wake wrote:
>> "Y.Porat" <map...@012.net.il> wrote in message
>> news:1143905479....@z34g2000cwc.googlegroups.com...
>> > who told you that E=mC^2
>>
>> <pedanty>
>>
>> C is centigrade.
>>
>> c is (I suspect) what you meant.
>>
>> The formula you have shown is an approximation of the correct one, so in
>> theory Einstein never said *it* did apply to the photon.
>
>
> yet did he said that* it does** not **apply to the photon**??
I dont know. Did he say that? I never spoke to him.
Do you demand that unless Einstein said "x" did not apply to "y" then it
must apply? Interesting perversion of logic there.
>> </pedantry>
>>
>> > doe s not apply to the photon??
>> >
>> > that is the orriginal formula of Einstien
>>
>> Not quite.
>
> not quite ???
> so please show us the orriginal formula created by Einstein
This has been done to death. You have had it posted to you multiple times.
As a side line, the copy of the 1905 paper I have in PDF format shows it as
m=E/c^2
Peter Christensen
"T Wake" <Usenet...@gishpuppy.com> skrev i en meddelelse
news:AfCdnTipqOj...@pipex.net...
Thanks, I'm glad that somebody found it (at least a bit) amusing.
(Actually, it also took some time to make up :>)
PC
__________________________________________________
m_photon = 0^0 (undefined), but only on April 1st!
Does intelligent people have more humour? §:>) NO
PC, Fool's Day 2006 (01-04-2006)
T Wake
"Peter Christensen" <Pe...@MailAPS.org> wrote in message
>
> "T Wake" <Usenet...@gishpuppy.com> skrev i en meddelelse
> news:AfCdnTipqOj...@pipex.net...
>>
>> "Peter Christensen" <Pe...@MailAPS.org> wrote in message
>> news:442f97e3$0$15783$1472...@news.sunsite.dk...
>>>
>>> <mme...@cars3.uchicago.edu> skrev i en meddelelse
>>> news:krsXf.9$45....@news.uchicago.edu...
>>>> In article <442e50bf$0$15792$1472...@news.sunsite.dk>, "Peter
>>>> Christensen" <Pe...@MailAPS.org> writes:
>>>>>I say: m_photon '=' 0^0
>>>>>
>>>>>Undefined is the same as undefined.
>>>>>
>>>> No, it is perfectly well defined. And (assuming Maxwell's equations
>>>> are strictly correct) it evaluates to zero.
>>>>
>>>> Mati Meron | "When you argue with a fool,
>>>> me...@cars.uchicago.edu | chances are he is doing just the
>>>> same"
>>>
>>> April 1st is Fool's Day
>>>
>>> I accept m_photon = 0 today, April 2nd.
>>>
>>> I'm sorry, if this joke wasn't funny...
>>
>> It's ok. I found it mildly amusing (mainly in how it brings the cranks
>> out of the woodwork).
>
> Thanks, I'm glad that somebody found it (at least a bit) amusing.
> (Actually, it also took some time to make up :>)
>
Problem on sci.physics* is that it is hard to tell what is a crank and what
is humour. Every day of the year head cases like Spaceman, traveller,
porat, Relf et al., come up with farcical ideas about how the universe
*should* be.
Still, getting them frothing at the mouth is amusing enough for me :-)
Peter Christensen
I just gave both your postings five stars on Google.com
(http://groups.google.com/group/sci.physics.relativity?hl=en)
:-)
T Wake
"Peter Christensen" <Pe...@MailAPS.org> wrote in message
Thanks. :-)
Peter Christensen
> that the definitions of mass currently used relies neither on
> measurement at rest nor on acceleration and that m_photon is not
> *defined* to be zero, rather it being zero is a result of the current
> definition of mass and of the Maxwell's equations.
Ok, I understand that.
Rgds,
PC
:>)
tony fleming
thread
http://www.unifiedphysics.com/SFT_Mathematics.pdf
you'll find out HOW the mass of the photon is crucial to the
development of mathematical physics.
Tony Fleming PhD
visit www.unifiedphysics.com
Hexenmeister
"tony fleming" <tfle...@hotkey.net.au> wrote in message
news:1143984269.3...@i39g2000cwa.googlegroups.com...
Hey Fleming, why don't you learn the basics before you
pretend to know mathematics?
http://mathworld.wolfram.com/VectorSpace.html
http://mathworld.wolfram.com/SingularMatrix.html
http://mathworld.wolfram.com/Determinant.html
You have a typo in 1.11 (c subscript y).
Androcles PhD
Tom Roberts
>> I still don't think that E=m*c^2 (with E=h*f) can be used for photons.
>> If the photon had a mass, then it would be affected by gravitational
>> fields, which it doesn't.
>
Photons have a mass that is consistent with zero. In GR, EM waves follow
null geodesics which are affected by gravitation, in spite of being
modeled are EM waves which are completely massless. In GR the source of
gravitation is the energy-momentum tensor, not just mass, and Em fields
possess energy and momentum.
For photons, the supposed m=hf/c^2 is completely and utterly refuted by
experiments, by a factor of 10^13 at least.
Tom Roberts tjro...@lucent.com
Timo Nieminen
> E = pc, by itself, results from the
> Maxwell's equations.
E = pv is general for waves, classically. From Maxwell, v=c for photons.
One can obtain E=pc directly from Maxwell bypassing the general result,
but I think that's a tad inelegant. Umov did some good work on this. (Is
Umov's paper available in English? I only have a Russian reprint, which
makes reading it slower and more painful.)
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
LeoK
> experiments, by a factor of 10^13 at least.
> Tom Roberts
Can anyone please specify why E=mc^2 applies to the photon mass?
That is classified as a quantum object during a century.
Saw someone arguing about all matter being quantum objects.
But is for example an electron a quantum object? - it is not.
So all matter is not quantum objects.
The mass of the photon cannot be defined with E=mc^2.
LeoK
brian a m stuckless
> Photons have a [Coup GR] mass that is consistent with zero.
>
> In [Coup] GR, EM waves -=-
$$ Tom is STiLL using the Coup GR "definition" ..of "REST mass".
$$
$$ -- The CRANK is hopeless. --
$$
$$ The NEWTONiAN part is the ONLY verifiable part of "predictions".
$$ The NEWTONiAN part is the ONLY part of "predictions", that work.
> -=- follow null geodesics which are affected by gravitation, in
> spite of being modeled are EM waves which are completely [Coup GR
> REST] massless.
>
> In [Coup] GR the source of gravitation is the energy-momentum
> tensor, not just [the Coup GR REST] mass, and Em fields possess
> energy and momentum.
>
> For photons, the supposed m=hf/c^2 is completely and utterly
> refuted by experiments, by a factor of 10^13 at least.
$$ Tom is CONFUSED between ABSORBED photon (REST) mass h*fL/c^2
$$ and a theoretical iN-TRANSiT photon, which nobody CAN measure.
$$
$$ Photon DETECTOR REST mass ..moving at v.
$$ For example, if a body of REST mass m_o, moving with velocity
$$ v, ABSORBs a photon of energy E withOUT change in velocity, it
$$ will have it's REST mass iNCREASED by the QUANTiTY, m = E/c^2.
$$
$$ The EQUiVALENT mass (m_o + E/c^2) is the RESULTANT REST mass m_v.
$$
$$ [ Hamiltonian ENTHALPY E = pL*c ..ONLY where m=0 AND pA*fA=0 ].
$$
$$ -- Hamiltonian ENTHALPY E = m*c^2 + pL*c + pA*fA --
$$ -- [..the PROPER GR Equation of State.] --
> Tom [ He between his error-bars ] Roberts tjro...@lucent.com
$$ Sincerely, ```Brian A M Stuckless, Ph.T (Tivity).
Re: Google < http://groups.google.ca/grphp?hl=en&tab=wg&ie=UTF-8 >.!!
Re: < GUESS nomenclature >< My BiGGER bang.!! >< My SACRED Meter >.!!
Re: I have an equation for the [Coup GR REST] mass of the photon!
Re: Chris Hillman ..the GR coup, after Einstein died.
mme...@cars3.uchicago.edu
>On Sun, 2 Apr 2006 mme...@cars3.uchicago.edu wrote:
>
>> E = pc, by itself, results from the
>> Maxwell's equations.
>
>E = pv is general for waves, classically. From Maxwell, v=c for photons.
>One can obtain E=pc directly from Maxwell bypassing the general result,
>but I think that's a tad inelegant.
Yes, true. And you're right, no need to reinvent wheels.
> Umov did some good work on this. (Is
>Umov's paper available in English? I only have a Russian reprint, which
>makes reading it slower and more painful.)
>
I've no idea whether it is available. I vaguely recall seeing some
stuff quoted from it, but no more than this.
Peter Webb
"Tom Roberts" <tjro...@lucent.com> wrote in message
news:tMUXf.7795$4L1....@newssvr11.news.prodigy.com...
Yeah, yeah, all true, but beside the point.
I have a big space ship. In the middle are a mass of matter and an equal
mass of antimatter inside a perfect gamma wave reflector. Lets say the mass
of the ship is M+2m, where M is the ship and 2m are the lumps of matter and
antimatter.
The ship has momentum of (M+2m)v, and kinetic energy of 0.5*(M+2m)*v^2. It
attracts other bodies gravitationally according to (M+2m).
Lets now let the antimatter and matter react to form gamma rays. The
reaction is entirely contained within the perfect gamma ray detector inside
the space ship.
From the outside, the ship is unchanged. It still has momentum of (M+2m)v,
and kinetic energy of 0.5*(M+2m)*v^2. It attracts other bodies
gravitationally according to (M+2m). In every sense, it behaves as if the
photons that were created have mass equal to 2m. In what sense do the
photons that are created (which are not at rest), not have a mass of 2m?
What experiment could you do from the outside which could tell the
difference in mass between two lumps of matter of mass 2m and the equivalent
"mass" of gamma rays? If the gamma rays act in all experiments affecting the
ship as a whole as if they have this mass, in what sense don't they have
this mass?
mme...@cars3.uchicago.edu
>
>"Tom Roberts" <tjro...@lucent.com> wrote in message
>news:tMUXf.7795$4L1....@newssvr11.news.prodigy.com...
>
>I have a big space ship. In the middle are a mass of matter and an equal
>mass of antimatter inside a perfect gamma wave reflector. Lets say the mass
>of the ship is M+2m, where M is the ship and 2m are the lumps of matter and
>antimatter.
>
>The ship has momentum of (M+2m)v, and kinetic energy of 0.5*(M+2m)*v^2. It
>attracts other bodies gravitationally according to (M+2m).
>
>Lets now let the antimatter and matter react to form gamma rays. The
>reaction is entirely contained within the perfect gamma ray detector inside
>the space ship.
>
>From the outside, the ship is unchanged. It still has momentum of (M+2m)v,
>and kinetic energy of 0.5*(M+2m)*v^2. It attracts other bodies
>gravitationally according to (M+2m).
That's all non relativistic, but no matter.
> In every sense, it behaves as if the
>photons that were created have mass equal to 2m.
Yes.
> In what sense do the
>photons that are created (which are not at rest), not have a mass of 2m?
They do.
>What experiment could you do from the outside which could tell the
>difference in mass between two lumps of matter of mass 2m and the equivalent
>"mass" of gamma rays? If the gamma rays act in all experiments affecting the
>ship as a whole as if they have this mass, in what sense don't they have
>this mass?
>
The photons do have this mass. The mass of a photon is zero. There
is no contradiction between the two once you realize that in
relativity mass is *not* an additive property. The mass of a system
doesn't, in general, equal the sum of the masses of its components.
So, in the case of an annihilation, the mass of the final system
(comprising of all the photons relesed in the reaction) equals the
mass of the initial system (the particles about to annihilate). But
the mass of each photon is zero. All this follows in a
straightforward manner once you apply the definition of mass, i.e.
m^2*c^4 = e^2 - p^2*c^2.
Try it.
dlzc1 D:cox T:net@nospam.com N:dlzc D:aol T:com (dlzc)
"Peter Webb" <webbfamily...@optusnet.com.au> wrote in
message news:44305cd8$0$7533$afc3...@news.optusnet.com.au...
...
> I have a big space ship. In the middle are a mass of matter
> and an equal mass of antimatter inside a perfect gamma
> wave reflector. Lets say the mass of the ship is M+2m,
> where M is the ship and 2m are the lumps of matter and
> antimatter.
>
> The ship has momentum of (M+2m)v, and kinetic energy
> of 0.5*(M+2m)*v^2.
Not a good choice for high speeds.
> It attracts other bodies gravitationally according to (M+2m).
>
> Lets now let the antimatter and matter react to form
> gamma rays. The reaction is entirely contained within
> the perfect gamma ray detector
... reflector ...
> inside the space ship.
>
> From the outside, the ship is unchanged. It still has
> momentum of (M+2m)v, and kinetic energy of
> 0.5*(M+2m)*v^2. It attracts other bodies
> gravitationally according to (M+2m). In every sense,
> it behaves as if the photons that were created have
> mass equal to 2m. In what sense do the photons
> that are created (which are not at rest), not have a
> mass of 2m?
The rest mass of pairs of particles in their "center of momentum"
frame does not have to be zero. It is the "assembly" that has
mass, and not the individual members.
> What experiment could you do from the outside
> which could tell the difference in mass between two
> lumps of matter of mass 2m and the equivalent "mass" of gamma
> rays?
You could detect a difference of the two lumps of mass were
separated...
> If the gamma rays act in all experiments affecting
> the ship as a whole as if they have this mass, in
> what sense don't they have this mass?
In the sense that mass is not a fundamental property, but a
"composite" property also.
David A. Smith
brian a m stuckless
> Photons have a [Coup GR] mass that is consistent with zero.
>
> In [Coup] GR, EM waves -=-
$$ Tom is STiLL using the Coup GR "definition" ..of "REST mass".
$$
$$ -- The CRANK is hopeless. --
$$
$$ The NEWTONiAN part is the ONLY verifiable part of "predictions".
$$ The NEWTONiAN part is the ONLY part of "predictions", that work.
> -=- follow null geodesics which are affected by gravitation, in
> spite of being modeled are EM waves which are completely [Coup GR
> REST] massless.
>
> In [Coup] GR the source of gravitation is the energy-momentum
> tensor, not just [the Coup GR REST] mass, and Em fields possess
> energy and momentum.
>
> For photons, the supposed m=hf/c^2 is completely and utterly
> refuted by experiments, by a factor of 10^13 at least.
$$ Tom is CONFUSED between ABSORBED photon (REST) mass h*fL/c^2
Y.Porat
> >
> > T Wake wrote:
> > >>
> >> c is (I suspect) what you meant.
> >>
> >> The formula you have shown is an approximation of the correct one, so in
> >> theory Einstein never said *it* did apply to the photon.
> >
> >
> > yet did he said that* it does** not **apply to the photon**??
>
> I dont know. Did he say that? I never spoke to him.
>
> Do you demand that unless Einstein said "x" did not apply to "y" then it
> must apply? Interesting perversion of logic there.
it was you that claimed that Einstein claimed
that this formula does nt apply to the photon
*while he never said that !!
if he never said that how gave you the permision to
say waht he didnt say??
if he never said that no reason that anyone else interpret it
in a way he never did!!
iow it is only *your interpretation* not his
so
waht is your interpretation better than my while
any bling man can see that in the
mC^2 THERE IS MASS !! incorporated in the letetr m
it is black on whilte there !!!!
----------------------------
>
> >> </pedantry>
> >>
> >> > doe s not apply to the photon??
> >> >
> >> > that is the orriginal formula of Einstien
> >>
> >> Not quite.
> >
> > not quite ???
> > so please show us the orriginal formula created by Einstein
>
> This has been done to death. You have had it posted to you multiple times.
and if it has been done by idiot parrots a thousand times
doe s it maked it more valid??
500 years ago millions of peoiple said that the sun orbits the earth
they said :you can even see with your own eyes that the sun orbits the
earth
and waht is more eevident than eye sight ??
and only one said the opposite so ???!!
in our case it is not even 'eyc sight' nor exoerimental sight
there is no experiment that can say : the photon is massless
whikle at least the most basic formula E=mC^2
says it as clear as possible
morover
that is indisputable in macrososm that energy is
mass in motion and so momentum
if one claimes different in microcosm
thje burden of prove is on him !!!
because it is him who changesb masic principles of physics
not me !!!
--------------------
>
> As a side line, the copy of the 1905 paper I have in PDF format shows it as
> m=E/c^2
that is exactly as E=mC^2 !! (algebaicly)
and actually you strengthen my view
because m is here the 'main hero' of the formula
and not a zero
that is a universal foirmula
and universal meand photons as well !!
and at the same time
E=hf
so
m(photon) = hf/C^2
nothing simpler than that with non of all the junglarings .
ATB
Y.Porat
---------------------------------
Eric Gisse
Y.Porat wrote:
[snip]
Yeesh. When are you going to give up this pointless pursuit? Nobody
accepts what you say, and those who are trying to can barely get
through your indecipherable English.
I think there is a strong correlation between your inability to learn
English and your inability to learn physics. You babble on and on in
your typical incoherant manner, never showing any sign of correcting
either the many misspelled words nor actually learning the details
behind the topic you have been ranting about for months now.
The "fertz" fixation was amusing, this isn't. Time to move on.
Y.Porat
*any physics formula has its limits of validations*???
including the Maxells equations
while in addition the phton is A LIMIT CASE EVEN MATHEMATICALLY
i onc eheared anidiot like you
who claimes based of those eqautions that
the re is a photon with a ferquency of (sit still on your chair
because there is nolomit to stopidity of some mathematiciance)
so he claimed tha there is a photon with a frequency of
ONE CYCLE PER BILLION YEARS!!
do you jion to that claim??
if yes i will suggest that you will be fired from your university
as a teacger of physics !!
(you could apparently be a betetr teaher of crook lawyers !!)and just
stickit to your stiff corrupted - mind- what old crackpot Porat said:
NO MASS - NO REAL PHYSICS .(copyright Y.Porat )
keep well
Y.Porat
----------------------
Bill Hobba
"Peter Webb" <webbfamily...@optusnet.com.au> wrote in message
news:44305cd8$0$7533$afc3...@news.optusnet.com.au...
>
> news:tMUXf.7795$4L1....@newssvr11.news.prodigy.com...
>> Peter Webb wrote:
>>>> I still don't think that E=m*c^2 (with E=h*f) can be used for photons.
>>>> If the photon had a mass, then it would be affected by gravitational
>>>> fields, which it doesn't.
>>>
>>> Which it is. Hence gravitational lensing.
>>
>> Photons have a mass that is consistent with zero. In GR, EM waves follow
>> null geodesics which are affected by gravitation, in spite of being
>> modeled are EM waves which are completely massless. In GR the source of
>> gravitation is the energy-momentum tensor, not just mass, and Em fields
>> possess energy and momentum.
>>
>> For photons, the supposed m=hf/c^2 is completely and utterly refuted by
>> experiments, by a factor of 10^13 at least.
>>
>>
>> Tom Roberts tjro...@lucent.com
>
> Yeah, yeah, all true, but beside the point.
>
> I have a big space ship. In the middle are a mass of matter and an equal
> mass of antimatter inside a perfect gamma wave reflector. Lets say the
> mass of the ship is M+2m, where M is the ship and 2m are the lumps of
> matter and antimatter.
>
> The ship has momentum of (M+2m)v, and kinetic energy of 0.5*(M+2m)*v^2. It
> attracts other bodies gravitationally according to (M+2m).
Nope - according to GR it has a stress energy tensor so is a source of
gravity. One must not use Newtonian gravitation when dealing with energy.
>
> Lets now let the antimatter and matter react to form gamma rays. The
> reaction is entirely contained within the perfect gamma ray detector
> inside the space ship.
>
> From the outside, the ship is unchanged. It still has momentum of (M+2m)v,
> and kinetic energy of 0.5*(M+2m)*v^2. It attracts other bodies
> gravitationally according to (M+2m).
Well since it did not attract it according to that equation it still does
not.
> In every sense, it behaves as if the photons that were created have mass
> equal to 2m.
Just because photons have a rest mass of zero does not mean they do not have
a mass when bound to a system. One way to looking at it (not necessarily
the best because it enshrines perturbation theory) is that during the time
the photons are reflected and the photon is absorbed the energy of the
reflector is increased a bit as per how Feynman explains it in QED - The
Strange Theory of Light and Matter. This increase in energy is reflected in
an increase in mass. However there is no need to do that - one simply needs
to look at the relativistic definition of mass M^2 = E^2 - P^2 (units C=1).
> In what sense do the photons that are created (which are not at rest), not
> have a mass of 2m? What experiment could you do from the outside which
> could tell the difference in mass between two lumps of matter of mass 2m
> and the equivalent "mass" of gamma rays? If the gamma rays act in all
> experiments affecting the ship as a whole as if they have this mass, in
> what sense don't they have this mass?
It is obvious - while not interacting with other things they have a mass of
zero - when interacting they can contribute to the mass of what they are
interacting with.
Bill
>
>
>
>
>
>
>
Hexenmeister
"Peter Webb" <webbfamily...@optusnet.com.au> wrote in message
news:44305cd8$0$7533$afc3...@news.optusnet.com.au...
|
| news:tMUXf.7795$4L1....@newssvr11.news.prodigy.com...
| > Peter Webb wrote:
| >>> I still don't think that E=m*c^2 (with E=h*f) can be used for photons.
| >>> If the photon had a mass, then it would be affected by gravitational
| >>> fields, which it doesn't.
| >>
| >> Which it is. Hence gravitational lensing.
| >
| > Photons have a mass that is consistent with zero. In GR, EM waves follow
| > null geodesics which are affected by gravitation, in spite of being
| > modeled are EM waves which are completely massless. In GR the source of
| > gravitation is the energy-momentum tensor, not just mass, and Em fields
| > possess energy and momentum.
| >
| > For photons, the supposed m=hf/c^2 is completely and utterly refuted by
| > experiments, by a factor of 10^13 at least.
| >
| >
| > Tom Roberts tjro...@lucent.com
|
| Yeah, yeah, all true, but beside the point.
"Real" has nothing to do with it. -- Tom Roberts.
| I have a big space ship.
Oh, ok... as long as we are not discussing reality, you two kooks carry on.
I have big balls.
Androcles.
Hexenmeister
<mme...@cars3.uchicago.edu> wrote in message
news:FeZXf.5$45....@news.uchicago.edu...
Another shithead who doesn't know what a square is.
http://www.androcles01.pwp.blueyonder.co.uk/E^2/EnergySquare.htm
Androcles | "When you agree with a fool,
andr...@hotmail.com | the guarantee is he is doing just the same"
Y.Porat
N:dlzc D:aol T:com (dlzc) wrote:
> Dear Peter Webb:
>
> "Peter Webb" <webbfamily...@optusnet.com.au> wrote in
> message news:44305cd8$0$7533$afc3...@news.optusnet.com.au...
> ...
> > >
> > of 0.5*(M+2m)*v^2.
>
> Not a good choice for high speeds.
>
> > It attracts other bodies gravitationally according to (M+2m).
> >
> > Lets now let the antimatter and matter react to form
> > gamma rays. The reaction is entirely contained within
> > the perfect gamma ray detector
>
> ... reflector ...
>
> > >
'center of momentum' can you show us anywhere else except your
fruteful immagination
how center of momentum is defined?
and who else execpt you use that new concept in physics ??
> frame does not have to be zero. It is the "assembly" that has
> mass, and not the individual members.
very interesting !!
ie before ;assembly' it (that system or physical entity ) didnt
have mass
after assembly it got mass!!
are you of the Hurry Potter association of magicians ??
because what you did above is creating mass from nothing !!
TIA
Y.Porat
-----------------------------
CrankHater
Hexenmeister wrote:
>
> Oh, ok... as long as we are not discussing reality, you two kooks carry on.
> I have big balls.
> Androcles.
>
yare a big testicle.
CrankHater
Hexenmeister wrote:
>
> Another shithead who doesn't know what a square is.
>
> http://www.androcles01.pwp.blueyonder.co.uk/E^2/EnergySquare.htm
>
> Androcles | "When you agree with a fool,
> andr...@hotmail.com | the guarantee is he is doing just the same"
you website sucks. your science is non existent.
you are just one big swollen penis.
Y.Porat
please show us the *experimental data* that is 10^13 difference
2 you forgot even to indicate if 10^13 more than the hf/C^2
or less than that
3 how do you measure the mass of a photon??
and if you can measure it it means ........... you can measue photon*
mass*!!
(does you ears hear what your mouth is talking ??)
TIA
Y.Porat
-----------------------
>
>
> Tom Roberts tjro...@lucent.com
brian a m stuckless
> Photons have a [Coup GR] mass that is consistent with zero.
>
> In [Coup] GR, EM waves -=-
$$ Tom is STiLL using the Coup GR "definition" ..of "REST mass".
$$
$$ -- The CRANK is hopeless. --
$$
$$ The NEWTONiAN part is the ONLY verifiable part of "predictions".
$$ The NEWTONiAN part is the ONLY part of "predictions", that work.
> -=- follow null geodesics which are affected by gravitation, in
> spite of being modeled are EM waves which are completely [Coup GR
> REST] massless.
>
> In [Coup] GR the source of gravitation is the energy-momentum
> tensor, not just [the Coup GR REST] mass, and Em fields possess
> energy and momentum.
>
> For photons, the supposed m=hf/c^2 is completely and utterly
> refuted by experiments, by a factor of 10^13 at least.
$$ Tom is CONFUSED between ABSORBED photon (REST) mass h*fL/c^2
T Wake
>
> T Wake wrote:
>> "Y.Porat" <map...@012.net.il> wrote in message
>> news:1143963696....@v46g2000cwv.googlegroups.com...
>> >
>> > T Wake wrote:
>> > >>
>> >> c is (I suspect) what you meant.
>> >>
>> >> The formula you have shown is an approximation of the correct one, so
>> >> in
>> >> theory Einstein never said *it* did apply to the photon.
>> >
>> >
>> > yet did he said that* it does** not **apply to the photon**??
>>
>> I dont know. Did he say that? I never spoke to him.
>>
>> Do you demand that unless Einstein said "x" did not apply to "y" then it
>> must apply? Interesting perversion of logic there.
> no Sir
>
> it was you that claimed that Einstein claimed
> that this formula does nt apply to the photon
> *while he never said that !!
No Sir,
I never said that. I said "so in theory Einstein never said *it* did apply
to the photon." That is very different from saying "Einstein said it didnt
apply to the photon."
Feel free to disagree with me all you like.
> if he never said that how gave you the permision to
> say waht he didnt say??
No idea what you are talking about here.
Who gave you the permission to say what you think I said and didnt say? Or
what Einstein said or didnt say?
> if he never said that no reason that anyone else interpret it
> in a way he never did!!
Nope, lost me.
I dont mean to be rude (well any more than normal) but I have no idea what
this is supposed to mean.
> iow it is only *your interpretation* not his
We only have interpretations.
> so
> waht is your interpretation better than my while
> any bling man can see that in the
> mC^2 THERE IS MASS !! incorporated in the letetr m
The formula you are grasping to show is E_0 = mc^2. Please note the c is
lower case. C is centigrade and used as an "old fashioned" measurement of
temperature. If you want to talk temperature use K instead.
Critcally this formula is based on rest energy and mass.
> it is black on whilte there !!!!
So use the proper, expanded formula then.
>>
>> This has been done to death. You have had it posted to you multiple
>> times.
>
> and if it has been done by idiot parrots a thousand times
> doe s it maked it more valid??
Well, the fact you refuse to accept it certainly doesnt make it any less
valid.
Calling people a "parrot" as if it is an insult is interesting. You make up
nonsense and repeat yourself constantly then call others "parrots."
I do hope the irony isn't lost upon you.
> 500 years ago millions of peoiple said that the sun orbits the earth
> they said :you can even see with your own eyes that the sun orbits the
> earth
> and waht is more eevident than eye sight ??
Well, the wonders of scientific experimentation have done us proud.
You are demanding your logical arument supports the mass of a photon not
being zero, in face of all experimental and theoretical evidence - yet you
bring in this example, when it clearly shows how the immediately logical
approach sometimes produces the wrong answer.
Amazing.
Did you mean it to come out this way?
> and only one said the opposite so ???!!
> in our case it is not even 'eyc sight' nor exoerimental sight
> there is no experiment that can say : the photon is massless
> whikle at least the most basic formula E=mC^2
> says it as clear as possible
> morover
> that is indisputable in macrososm that energy is
> mass in motion and so momentum
> if one claimes different in microcosm
> thje burden of prove is on him !!!
No it isnt.
> because it is him who changesb masic principles of physics
> not me !!!
Stop crying about "basic" principles. Makes you read like spaceman and we
all know what a loon he is.
You are trying to determine the rest mass of a particle that is not at rest.
Can you see the problem with this?
> --------------------
>
>
>
>>
>> As a side line, the copy of the 1905 paper I have in PDF format shows it
>> as
>> m=E/c^2
>
> that is exactly as E=mC^2 !! (algebaicly)
Yes.
It is talking about rest mass though.
> and actually you strengthen my view
> because m is here the 'main hero' of the formula
> and not a zero
> that is a universal foirmula
> and universal meand photons as well !!
Says who?
> and at the same time
> E=hf
> so
> m(photon) = hf/C^2
>
> nothing simpler than that with non of all the junglarings .
Well then genius, what is the mass of a Gamma ray photon with a frequency of
3x10^19Hz?
You seem to imply that it would have a mass in the region of 2x10^-31 kg -
is this what you intend to state?
Peter Christensen
Honestly, I was surpriced, that my posting made some people quite upset, and
that some spend quite some effort to write replies with corrections. Sorry
about that. I just expected a few people to reply with maybe a few lines
about such a joke. I thought, that it would be just something like, what I
would have replied on a joke-posting myself. Could have been something like
this:
"It's obvious, that your posting is nothing but a JOKE. Tomorrow we will be
using m_photon = 0 again! This 'physics' is NOT to 5 stars. Here it is 1/4
!" (Just 1/4 star, or 1/4 like 01-04 for April 1st)
-And not more than that, in the same style.
Here in Denmark, we have a great tradition with jokes on April 1st. All
major newspapers always has one, and there are also a lot in TV. Thought,
that it was like that everywhere; we are always looking for these jokes on
this day. -And we usually catch most of them, when we se them. I thought,
that it would be easily seen as a joke-posting. For example, I wrote "April
1st today" in the first reply in the thread. Personally, I think, that it
was at least a little bit funny, with some of those discussions and the
'mess' of the photon. But maybe my Danish homour doesn't fit into the
English newsgroups.
-It is NOT my fault, that so many people still make the error m_photon =
h*f/c^2. I don't, and I never did. The mass of the photon IS determined to
be 0.
:-)
PC
____________________
m_photon = a_photon = 0
Hexenmeister
"Peter Christensen" <Pe...@MailAPS.org> wrote in message
news:443127b8$0$15791$1472...@news.sunsite.dk...
The problem is the same joke is heard 365 days a year, it's called
"relativity". Nobody can distinguish your joke from it.
Androcles.
Peter Webb
"Peter Christensen" <Pe...@MailAPS.org> wrote in message
news:443127b8$0$15791$1472...@news.sunsite.dk...
>
> Honestly, I was surpriced, that my posting made some people quite upset,
> and that some spend quite some effort to write replies with corrections.
> Sorry about that. I just expected a few people to reply with maybe a few
> lines about such a joke. I thought, that it would be just something like,
> what I would have replied on a joke-posting myself. Could have been
> something like this:
>
> "It's obvious, that your posting is nothing but a JOKE. Tomorrow we will
> be using m_photon = 0 again! This 'physics' is NOT to 5 stars. Here it
> is 1/4 !" (Just 1/4 star, or 1/4 like 01-04 for April 1st)
I understand you (now), but I am from a part of the world where dates are
written that way. In the US it would be 4/1.
>
> -And not more than that, in the same style.
>
> Here in Denmark, we have a great tradition with jokes on April 1st. All
> major newspapers always has one, and there are also a lot in TV. Thought,
> that it was like that everywhere; we are always looking for these jokes on
> this day. -And we usually catch most of them, when we se them. I thought,
> that it would be easily seen as a joke-posting. For example, I wrote
> "April 1st today" in the first reply in the thread. Personally, I think,
> that it was at least a little bit funny, with some of those discussions
> and the 'mess' of the photon. But maybe my Danish homour doesn't fit into
> the English newsgroups.
>
The only thing that wasn't 100% pure kook was the absence of spelling
mistakes, and that is probably why you were taken seriously.
> -It is NOT my fault, that so many people still make the error m_photon =
> h*f/c^2. I don't, and I never did. The mass of the photon IS determined to
> be 0.
>
> :-)
> PC
It all depends upon how you look at it ...
Y.Porat
he was 'half serious' i will not add
his next name
BTW are you a new incarnation of
Feuerbacher??
i have a strong hunch you are ......
where has he disappeared ??
or may be change you name again ??
---------
2 not A SINGLE WORD OF PHYSICS
and how do you want anyone will take you seriously??
you post is a 100 percent personal
attack
so how do you want to call yourself
a serious responsible scientists
3 what from are you making your living
while having so much time for
personal fightings here
while obviously you are not a pensioner.??
4 you r 'argument about my English
without any ability to defy
my new claim
is another prove you are not a physicist but a personal politician .
5
it is the last time i am going to write more than two lines discussion
with you
though you make considerable efforts to drag me to a heavy fight.
you will not succeed to wat my energy
on zeros like you
keep well
Y.Porat
----------------
Peter Christensen
"Tom Roberts" <tjro...@lucent.com> skrev i en meddelelse
news:tMUXf.7795$4L1....@newssvr11.news.prodigy.com...
> Peter Webb wrote:
>>> I still don't think that E=m*c^2 (with E=h*f) can be used for photons.
>>> If the photon had a mass, then it would be affected by gravitational
>>> fields, which it doesn't.
>>
>> Which it is. Hence gravitational lensing.
I don't think so.
For example, when a star is affecting space-time, the geodesics for the
photons are also affected. This can cause a lens-effect. It's not caused
directly by gravitational effects on the massless photon, but by the
curvature of space-time caused by the star. I hope, this is the right way to
phrase it. The lens-effect is not difficult to imagine, I think. If the
space had been absolute, in the Newtonian way, there couldn't be such an
effect on the massless photon. That the photon is affected by the star, must
mean that the space is relative like in GR, and not like in the old
Newtonian mechanics.
On lenses: The effect is very different from the way optical lenses (glass
or plastic) is working. Here the lens effect is caused by the slower
propagation of light in the media: c / (refractive index n). Also here
Maxwell's Equations apply for EM. But it is a completely different effect,
so one can not compare a gravitational lens with just a normal lens made of
glass. The optical lens has nothing to do with relativity. But the way
space-time curvature can 'focus' light gives a result similar to the effect
from refractive lenses in optics.
The lens-effect from stars is an interesting effect from relativity. I
think, that it was seen as an important confirmation of Einsteins GR, when
it was first observed in astronomy in practice. I can't remember when this
was done, but GR was confirmed by such experiments almost 100 years ago.
It's not like in SR where Einstein explained some experiments; this time
Einstein was first with the theory. And then, later, the experimental
results, that confirmed the theory, came. -An impressive scientist, who made
both SR and GR...
> Photons have a mass that is consistent with zero. In GR, EM waves follow
> null geodesics which are affected by gravitation, in spite of being
> modeled are EM waves which are completely massless. In GR the source of
> gravitation is the energy-momentum tensor, not just mass, and Em fields
> possess energy and momentum.
Related simply by E = pc.
E=m*c^2 = 'E(p=0,m)' doesn't apply here. It's the other way around: m is 0
and p is nonzero. Both E = 0 and m = h*f/c^2 are wrong. Here people could
make two errors with just one equation, so to speak. Einsteins equation
E(p,m=0) from SR gives the right result E=pc. Not surpricing. But also
E=h*f, p=h/lambda and c=lambda/f from QM gives E=pc.
I'm not sure if relativity or QM should be used to derive E=pc. One should
probably start with relativity and then derive results from QM. If we know p
and can measure f, then h can be determined. But it also appear to me, that
h = p*lambda can be used to determine h, because it should also be possible
to measure lambda.
Also c can be determined, if lambda and f are known. I guess that it's like
this: We know both E and p from relativity, and we can measure lambda and f
in space-time. Then one can determine the two physical constants c and h.
But I'm a little bit confused right here...
> For photons, the supposed m=hf/c^2 is completely and utterly refuted by
> experiments, by a factor of 10^13 at least.
It's an implicit assumption in starting with E=m*c^2, that p=0. This is
simply wrong for a photon.
Using E=m*c^2 = 'E(p=0,m)' is wrong. For the photon m=0 and p is nonzero, so
using E(p,m=0) = p*c is correct. It's easy just to remember that we have E
as a function of m and p: 'E(p,m)'. For a photon E(p,0) = pc is used, and
for a massive particle at rest E(0,m) = m*c^2 is used.
If people just think of the equation, m^2*c^4 = E^2 - p^2*c^2, as an
expression for the energy E(p,m), then things are easy to remember.
-E(p,m) also looks better than m^2*c^4 = E^2 - p^2*c^2 in a txt-file...
"The mass of the photon IS determined to be 0". I agree today.
(news:443127b8$0$15791$1472...@news.sunsite.dk)
Please just continue the discussion on photons, even though this thread
itself was just meant as a joke (1/4), when I started it last saturday. I
just change the headline, that's easy enough...
:-)
PC
________________
Pe...@MailAPS.org
Y.Porat
i will put it again simply:
my claim As that Einstein (AFAIK)
created only the formula:
E=mc^2 (now a little c to make you happy
and to block your demagogies ....)
that was a universal formula
here attributed to rest mass or not rest mass
the 'rest mass or relativistic mass' was invented only later by
mathematics physicists .
so no one can say it occludes or not includes the photon
because the original inventor didnt address that question
at the first place it was invented
(until now right ??)
so you interpretation that it applies or not app lyes
to the photon is not better than my interpretation
iow
no one has a monopoly (interpretation)
on that universal formula right ??
----------------
2 you say that it does not apply to the photon
why ?? because you say ??
you have to reason your claim !!
so what is your reasoning for that??
because once you have a formula it apples
to all cases *unless you prove otherwise*
and proving is by experiment
so
what is your experimental tool with which you can
measure
THE MASS OF A PHOTON ??(DIRECTLY
IE not through energy or momentum
because if it is through energy we are back
to the start question:
what is energy made of
i suggest a simple test:
energy dimensions !!
it is
meter kilogram second
your dimension is
meter 'relativistic kilogram' seconds
did you ever hard about a physics system
of meter relativistic kilograms seconds ??
now i thought you might be sophisticated and claim that
the m there in E=mc^2 is actually
LF times m (LF is the Lorentz factor 1/ (1- v^2/c^2)
is that your base for 'relativistic mass ??
instiller words: as velocity of mass comes close to c
m becomes close to infinity ??
is That your claim ??
TIA
Y.Porat
--------------------
Peter Christensen
that's true.
No more (stupid) jokes from me now!
PC
mme...@cars3.uchicago.edu
>> April 1st today
>
>Honestly, I was surpriced, that my posting made some people quite upset,
I didn't notice "upset" (but then I don't read most posts).
> and that some spend quite some effort to write replies with corrections.
Stick around sci.physics for a while and things will become much
clearer to you. On second thought, well, you sound like a nice guy, I
wouldn't want to recommend to you to waste your time.
Mati Meron | "When you argue with a fool,
me...@cars.uchicago.edu | chances are he is doing just the same"
Peter Christensen
<mme...@cars3.uchicago.edu> skrev i en meddelelse
news:laXXf.3$45....@news.uchicago.edu...
> In article <2006040305...@emu.uq.edu.au>, Timo Nieminen
> <uqtn...@mailbox.uq.edu.au> writes:
>>On Sun, 2 Apr 2006 mme...@cars3.uchicago.edu wrote:
>>
>>> E = pc, by itself, results from the
>>> Maxwell's equations.
>>
>>E = pv is general for waves, classically. From Maxwell, v=c for photons.
>>One can obtain E=pc directly from Maxwell bypassing the general result,
>>but I think that's a tad inelegant.
>
> Yes, true. And you're right, no need to reinvent wheels.
E=pc
As the result is from Maxwells equations, I now realize, why it sometimes
appeared to me, that the result could be derived from both QM and SR. It's
obviously coming from Maxwells equations in both cases. When E=pv for waves
in general, then it's just to use v=c for photons. That's easy enough.
PC
________________
Pe...@MailAPS.org
Peter Christensen
> probably start with relativity and then derive results from QM. If we know
> p and can measure f, then h can be determined. But it also appear to me,
> that h = p*lambda can be used to determine h, because it should also be
> possible to measure lambda.
I just found out. It's basically from Maxwells equations, so it appears both
places.
PC
________________
Pe...@MailAPS.org
Timo Nieminen
> In article <443127b8$0$15791$1472...@news.sunsite.dk>, "Peter Christensen" <Pe...@MailAPS.org> writes:
>>> April 1st today
>>
>> Honestly, I was surpriced, that my posting made some people quite upset,
>
> I didn't notice "upset" (but then I don't read most posts).
>
>> and that some spend quite some effort to write replies with corrections.
>
> Stick around sci.physics for a while and things will become much
> clearer to you. On second thought, well, you sound like a nice guy, I
> wouldn't want to recommend to you to waste your time.
OTOH, that's about the whole point of usenet: wasting time. If you're
looking for a way to waste time that beats out much TV, stick around. (I'd
rather watch Iron Chef than post here, but it's only on once a week!)
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
LeoK
I've seen you, and others, arguing for that the mass of the
photon can be derived from E=mc^2.
Let me make an example that not even the mass of the electron
can be derived from E=mc^2.
1. We know the mass of the electron, let's call it m_e.
2. We know the speed of light, c.
3. We know the energy to split an electron to quarks, and let's
call this energy for E_e.
This should give the following equation:
E_e = m_e*c^2
But the energies on both side of the equation are not the same
we know from CERN and other particle collision studies.
So not even the electron, with a KNOWN mass, applies to the formula
of : E=mc^2. Why should the photon do so then? It doesn't apply either.
The energy for splitting any electron E_e is far more greater the
mathematical (E=mc^2) for the electron by exponential magnitudes.
This can be a proof that also the photon with a unknown smaller
mass than the electron also cannot be derived from E=mc^2.
LeoK
Spaceman
"LeoK" <rof...@yahoo.com> wrote in message
news:1144091864.3...@j33g2000cwa.googlegroups.com...
>> NO MASS - NO REAL PHYSICS .(copyright Y.Porat )
>> Y.Porat
>
> I've seen you, and others, arguing for that the mass of the
> photon can be derived from E=mc^2.
>
> Let me make an example that not even the mass of the electron
> can be derived from E=mc^2.
>
> 1. We know the mass of the electron, let's call it m_e.
What did you use to find such a mass of an electron?
:)
Peter Christensen
"Timo Nieminen" <uqtn...@mailbox.uq.edu.au> skrev i en meddelelse
news:2006040405...@emu.uq.edu.au...
> On Mon, 3 Apr 2006 mme...@cars3.uchicago.edu wrote:
>
>> In article <443127b8$0$15791$1472...@news.sunsite.dk>, "Peter
>> Christensen" <Pe...@MailAPS.org> writes:
>>>> April 1st today
>>>
>>> Honestly, I was surpriced, that my posting made some people quite upset,
>>
>> I didn't notice "upset" (but then I don't read most posts).
>>
>>> and that some spend quite some effort to write replies with corrections.
>>
>> Stick around sci.physics for a while and things will become much
>> clearer to you. On second thought, well, you sound like a nice guy, I
>> wouldn't want to recommend to you to waste your time.
>
> OTOH, that's about the whole point of usenet: wasting time. If you're
> looking for a way to waste time that beats out much TV, stick around. (I'd
> rather watch Iron Chef than post here, but it's only on once a week!)
Maybe you are right, but I think, that studying GR is better than just
watching TV.
-By the way, may I guess that you are from Finland? Some time ago I had a
friend from Finland, whos name was also Timo, and I understood from him,
that it was a name almost only used in Finland. This language is quite
different from both Scandinavian (west) and Russian (east), and so are the
names.
PC
________________
Pe...@MailAPS.org
Peter Christensen
> clearer to you. On second thought, well, you sound like a nice guy, I
> wouldn't want to recommend to you to waste your time.
Thanks, at the moment I study GR. I didn't have this opportunity, when I
studied technical physics at a technical university. They didn't have
courses on GR at all, only on SR. So I will have to study GR myself, and
that's what I'm doing at the moment. I've always liked both mathematics and
physics.
It's always nice to have a place to discuss things, as for example the
physics groups.
You appear to be one of the more professional in SPR, so I will read your
posts with interest...
PC
________________
Pe...@MailAPS.org
T Wake
> since you lowyers tactics was to complicate the discussion
> i will put it again simply:
I assume you are talking to me here. Can I suggest you get a decent news
reader or learn to use Google Groups in a manner that allows the people you
are talking to, to work out you are talking to them.
When you accuse me of using lawyers tactics, do you mean I simply responded
to your ranting in the same manner by which you ranted?
> my claim As that Einstein (AFAIK)
> created only the formula:
>
> E=mc^2 (now a little c to make you happy
> and to block your demagogies ....)
The sad thing is you dont realise why it is important to be accurate.
> that was a universal formula
Well, no.
> here attributed to rest mass or not rest mass
> the 'rest mass or relativistic mass' was invented only later by
> mathematics physicists .
You are using plays on words which arent really appropriate. Remember your
complaint about my use of "lowyers tactics" - well you are doing the same,
only your tacitcs disagree with experimental data.
> so no one can say it occludes or not includes the photon
> because the original inventor didnt address that question
> at the first place it was invented
> (until now right ??)
Ok, I accept that Einstein never addressed the issue of photons in the
original draft of the formula. However, Einstien is also credited with
stating nothing with mass can reach the speed of light.
> so you interpretation that it applies or not app lyes
> to the photon is not better than my interpretation
> iow
> no one has a monopoly (interpretation)
> on that universal formula right ??
Well, not true. In science theories are refined and redesigned on an almost
constant basis. Theories are often deveopled and, where required, made more
accurate and more specific.
OK, word salad aside and without mentioning your blatant ignorance and
determination to argue over non-topics - if you think I have claimed
something then refute that. Do not spend 30 odd lines stating things which I
have never said. (cf. Strawman Fallacy).
I asked you:
"Well then genius, what is the mass of a Gamma ray photon with a frequency
of
3x10^19Hz?
You seem to imply that it would have a mass in the region of 2x10^-31 kg -
is this what you intend to state? "
Do you have an answer for that?
Robert Low
> "Tom Roberts" <tjro...@lucent.com> skrev i en meddelelse
> news:tMUXf.7795$4L1....@newssvr11.news.prodigy.com...
>
>>Peter Webb wrote:
>>
>>>>I still don't think that E=m*c^2 (with E=h*f) can be used for photons.
>>>>If the photon had a mass, then it would be affected by gravitational
>>>>fields, which it doesn't.
>>>Which it is. Hence gravitational lensing.
> I don't think so.
> For example, when a star is affecting space-time, the geodesics for the
> photons are also affected. This can cause a lens-effect. It's not caused
> directly by gravitational effects on the massless photon, but by the
> curvature of space-time caused by the star.
The curvature of space-time *is* the gravitational field, inasmuchas
it makes sense to talk about a gravitational field in GR.
mme...@cars3.uchicago.edu
either relativity or QM is that due to the way physics is taught (the
"gap" issue), that's where they first encounter this relationship.
mme...@cars3.uchicago.edu
>On Mon, 3 Apr 2006 mme...@cars3.uchicago.edu wrote:
>
>> In article <443127b8$0$15791$1472...@news.sunsite.dk>, "Peter Christensen" <Pe...@MailAPS.org> writes:
>>>> April 1st today
>>>
>>> Honestly, I was surpriced, that my posting made some people quite upset,
>>
>> I didn't notice "upset" (but then I don't read most posts).
>>
>>> and that some spend quite some effort to write replies with corrections.
>>
>> Stick around sci.physics for a while and things will become much
>> clearer to you. On second thought, well, you sound like a nice guy, I
>> wouldn't want to recommend to you to waste your time.
>
>OTOH, that's about the whole point of usenet: wasting time. If you're
>looking for a way to waste time that beats out much TV, stick around.
Aye, no argument about this:-) As a procrastination device, it is
marvellous.
mme...@cars3.uchicago.edu
>> Stick around sci.physics for a while and things will become much
>> clearer to you. On second thought, well, you sound like a nice guy, I
>> wouldn't want to recommend to you to waste your time.
>
>Thanks, at the moment I study GR. I didn't have this opportunity, when I
>studied technical physics at a technical university. They didn't have
>courses on GR at all, only on SR. So I will have to study GR myself, and
>that's what I'm doing at the moment.
Well, I wish you much success with this.
> I've always liked both mathematics and physics.
Same here.
>
>It's always nice to have a place to discuss things, as for example the
>physics groups.
As long as you take proper care to filter away the noise, yes.
>
>You appear to be one of the more professional in SPR, so I will read your
>posts with interest...
>
Well, I hope not to disappoint you. And I don't usually post to SPR
unless in response to something that was posted to sci.physics. I
mean, SP is bad enough, SPR is worse. But, it is not for me to
prejudge thinks for you.
Peter Christensen
"Peter Webb" <webbfamily...@optusnet.com.au> skrev i en meddelelse
news:44313bfa$0$10673$afc3...@news.optusnet.com.au...
>
> "Peter Christensen" <Pe...@MailAPS.org> wrote in message
> news:443127b8$0$15791$1472...@news.sunsite.dk...
> I understand you (now), but I am from a part of the world where dates are
> written that way. In the US it would be 4/1.
It's actually quite irritating, that somewhere it's called 1/4 and somewhere
else it's called 4/1. There should be an international standard. -A lot of
confusion.
> The only thing that wasn't 100% pure kook was the absence of spelling
> mistakes, and that is probably why you were taken seriously.
Because of TV and Internet, my English has become better than my native
Danish, I think. Another thing is to use the spell check, in for example MS
Outlook Express.
>> -It is NOT my fault, that so many people still make the error m_photon =
>> h*f/c^2. I don't, and I never did. The mass of the photon IS determined
>> to be 0.
>>
>> :-)
>> PC
>
> It all depends upon how you look at it ...
Yes, probably...
PC
_______________________________________
m_photon = 0^0 (undefined), but only on April 1st!
Peter Christensen
"Hexenmeister" <vanq...@broom.Mickey_a> skrev i en meddelelse
news:BDaYf.166051$zk4.1...@fe3.news.blueyonder.co.uk...
> The problem is the same joke is heard 365 days a year, it's called
> "relativity". Nobody can distinguish your joke from it.
> Androcles.
It's more than just a joke, I think.
T Wake
"Peter Christensen" <Pe...@MailAPS.org> wrote in message
>
> "Peter Webb" <webbfamily...@optusnet.com.au> skrev i en meddelelse
> news:44313bfa$0$10673$afc3...@news.optusnet.com.au...
>>
>> "Peter Christensen" <Pe...@MailAPS.org> wrote in message
>> news:443127b8$0$15791$1472...@news.sunsite.dk...
>
>> I understand you (now), but I am from a part of the world where dates are
>> written that way. In the US it would be 4/1.
>
> It's actually quite irritating, that somewhere it's called 1/4 and
> somewhere else it's called 4/1. There should be an international
> standard. -A lot of confusion.
>
IIRC the international standard is YYYY/MM/DD so it would read as
2006/04/01.
The "colonials" seem to like doing it MM/DD/YYYY which seems odd to me
personally but I suspect they view the DD/MM/YYYY I think is normal as a bit
strange.
As an example, for me, 9/11 is a totally different date to when the WTC got
hit....