Just a dimension analysis of photon momentum!!
in [Physics]
|
Prev: 9-11 Truth makes HUGE showing at Los Angeles Peace March as bushkultie Iarndud Crackwhore Kook for War screeches in helpless rage
Next: FLUSHING THE TOILET
From: Mathal on 24 Jun 2010 01:56 On Jun 23, 1:59 am, "Y.Porat" <y.y.po...(a)gmail.com> wrote: > the photon momentum can be presented as > > P = hf/c > right > it is the full comprehensive presentation > of the photon momentum > nothing missing - nothing excessive right ?? !! > > now lets take it as is > (without changing anything in it as the formula > presenting the **photon momentum * > momentum ie not energy .....!! ) > > and present it by its dimensions and > dimensionless figures > > h is > 6.6 exp -34 > > f is > fs/second > > while fs is*** the dimensionless > figure that is attached to the 1/second **** > > c is say (aprox ) > 3 exp10 meter/second > now if we combine all of it > we get > > P 6.6 exp -Kg meter ^2 /second times fs/Sec > divided by meter/Second times 3 exp10 > > ie if we present it without the > dimensions that are canceling themselves > in nominator and denominator > > we get > ==================================== > 6.6exp-34 Kg MET /SEC times fs/3 exp10 > ==================================== > now my question is > where do you see anything relativistic in it ??!! > > TIA > Y.Porat > ---------------------- P = hf/c h is constant. c is constant. f is not constant. Since the measured frequency of a received photon is dependant on the initial frequency and the difference between the velocity of this sending frame and the receiving frame one has to take relativity (SR) into consideration if the two frames are not motionless WRT each other. This is where gamma enters the picture. How can the momentum be anything but relativistic. Mathal
From: Y.Porat on 24 Jun 2010 02:42 On Jun 24, 7:56 am, Mathal <mathmusi...(a)gmail.com> wrote: > On Jun 23, 1:59 am, "Y.Porat" <y.y.po...(a)gmail.com> wrote: > > > > > the photon momentum can be presented as > > > P = hf/c > > right > > it is the full comprehensive presentation > > of the photon momentum > > nothing missing - nothing excessive right ?? !! > > > now lets take it as is > > (without changing anything in it as the formula > > presenting the **photon momentum * > > momentum ie not energy .....!! ) > > > and present it by its dimensions and > > dimensionless figures > > > h is > > 6.6 exp -34 > > > f is > > fs/second > > > while fs is*** the dimensionless > > figure that is attached to the 1/second **** > > > c is say (aprox ) > > 3 exp10 meter/second > > now if we combine all of it > > we get > > > P 6.6 exp -Kg meter ^2 /second times fs/Sec > > divided by meter/Second times 3 exp10 > > > ie if we present it without the > > dimensions that are canceling themselves > > in nominator and denominator > > > we get > > ==================================== > > 6.6exp-34 Kg MET /SEC times fs/3 exp10 > > ==================================== > > now my question is > > where do you see anything relativistic in it ??!! > > > TIA > > Y.Porat > > ---------------------- > > P = hf/c > > h is constant. > c is constant. > f is not constant. > > Since the measured frequency of a received photon is dependant on the > initial frequency and the difference between the velocity of this > sending frame and the receiving frame one has to take relativity (SR) > into consideration if the two frames are not motionless WRT each -------------------------- 1 very nice!! if you take the Doppler effect yiou see that the** f **is changing!! AND YOU CAN SEE IT IN THE FORMULA AS I PRESENTED IT (PRESENTED by THAT fs so IT IS THE fs THAT IS CHANGING NOT **THE INITIAL*** MASS UNITS ** it is their number thaqt can change IOW MORE OR LESS **MASS UNITS **ARE COMING INTO THE SECONDARY FRAME BUT LISTEN CAREFULLY TH E MASS ** UNITS** ARE NOT CHANGING IT IS THE **NUMBER OF THOSE ** MASS UNITS !!! and it i s presented nicely in my above formula analysis presented in that ** fs ** the number of mas s units is linearity (nothing like the relativistic ***second order** )!!!!! and nicely presented by that fs no need to look for more formula or 'interpretations'' btw it is as well my new insight about how the Doppler effect is a prove that the hf is not the right definition of the REAL SINGLE PHOTON !!! 2 you forgot that TH E VELOCITY OF PHOTON ****IS ALWAYS c IN ALL FRAMES!! *** no mater if in relative motion or not so ??!! even relative to the secondary frame it remains c!!! as well as c -- in the original; frame !! 2 being in motion for itself ***is not enough to be relativistic !!!*** as long it is relative to***** itself **** !!! and that is why YOU CANT SEE ANYTHING RELATIVISTIC IN THAT PHOTON MOMENTUM FORMULA !!! except that that photons declared himself to be 'pope of Rome' !!! (:-) 3 WHIL WE WE DELL WITH PHOTON MOMENTUM WE DEAL JUST IN ** ONE FRAME** NOT IN MANY FRAMES !!! about many FRAMES see my above explanations about how it works in the Doppler case two frames but no need to obfuscate it we have to conclude (resume) first about just one frame !!! > other. This is where gamma enters the picture. How can the momentum > be anything but relativistic. see about the pope of Rome (:-) ie how can i not be the pope of Rome ??!! -------------------- TIA Y.Porat ------------------------------ > > Mathal
From: waldofj on 24 Jun 2010 09:39 On Jun 23, 4:59 am, "Y.Porat" <y.y.po...(a)gmail.com> wrote: > the photon momentum can be presented as > > P = hf/c > right > it is the full comprehensive presentation > of the photon momentum > nothing missing - nothing excessive right ?? !! > > now lets take it as is > (without changing anything in it as the formula > presenting the **photon momentum * > momentum ie not energy .....!! ) > > and present it by its dimensions and > dimensionless figures > > h is > 6.6 exp -34 > > f is > fs/second > > while fs is*** the dimensionless > figure that is attached to the 1/second **** > > c is say (aprox ) > 3 exp10 meter/second > now if we combine all of it > we get > > P 6.6 exp -Kg meter ^2 /second times fs/Sec > divided by meter/Second times 3 exp10 > > ie if we present it without the > dimensions that are canceling themselves > in nominator and denominator > > we get > ==================================== > 6.6exp-34 Kg MET /SEC times fs/3 exp10 > ==================================== > now my question is > where do you see anything relativistic in it ??!! > > TIA > Y.Porat > ---------------------- it depends on what you mean by "relativistic"
From: Y.Porat on 24 Jun 2010 11:04 On Jun 24, 3:54 pm, Robert > find the units of momentum (in SI) as "kg m s^-1", which are the > correct units. > > > > > --------------------- > > and according to Higgins correction: > > > ====================================== > > 6.6 exp -34 KILOGRAM MET/SEC times fs / > > 3/exp 8 > > ======================================== > > Thank you Higgins for saving me and my above analysis (:-) > > No one can save you. > > > > > TIA > > Y.Porat > > --------------------- > > > > > now my question is > > > > where do you see anything relativistic in it ??!! > > > > I have to ask - where did you get your "engineering" degree? > > Please answer this question, so young people everywhere know which > school to avoid. > > > > > > > TIA > > > > Y.Porat > > > > ---------------------- Dear readers ddi you hear above a single word of physics arguments ??~~ you dint answer a simple question: where do you see in my equation or in your equation as you will ike it SOMETHING RELATIVISTIC !!! ??? now big scientist (i will skip you cheap demagogism that cannot convince a secondary scol boy not tomention inteligent peopel as are here and HAS NOTHING TO DO WITH PHYSICS ARGUMENTS BUT RATHER TO A FISH MARKET)) so my question is: we know from binding energy analysis of particles that **the mass that was lost from the particle IS EXACTLY **QUANTITATIVELY** THE SAME AS THE ***RELATIVISTIC MASS** THAT WAS GAINED BY (into) THE ENERGY created by that process ??? SO WHAT IS (AT THE ABOVE CASE ) THE ***QUANTITATIVE DIFFERENCE *** BETWEEN THE 'RELATIVISTIC MASS** (of energy gain) AND THE REST MASS THAT WAS TAKEN FROM --- the above PARTICLE MASS ?? TIA Y.Porat -----------------------
From: Mathal on 24 Jun 2010 11:30
On Jun 23, 11:42 pm, "Y.Porat" <y.y.po...(a)gmail.com> wrote: > On Jun 24, 7:56 am, Mathal <mathmusi...(a)gmail.com> wrote: > > > > > On Jun 23, 1:59 am, "Y.Porat" <y.y.po...(a)gmail.com> wrote: > > > > the photon momentum can be presented as > > > > P = hf/c > > > right > > > it is the full comprehensive presentation > > > of the photon momentum > > > nothing missing - nothing excessive right ?? !! > > > > now lets take it as is > > > (without changing anything in it as the formula > > > presenting the **photon momentum * > > > momentum ie not energy .....!! ) > > > > and present it by its dimensions and > > > dimensionless figures > > > > h is > > > 6.6 exp -34 > > > > f is > > > fs/second > > > > while fs is*** the dimensionless > > > figure that is attached to the 1/second **** > > > > c is say (aprox ) > > > 3 exp10 meter/second > > > now if we combine all of it > > > we get > > > > P 6.6 exp -Kg meter ^2 /second times fs/Sec > > > divided by meter/Second times 3 exp10 > > > > ie if we present it without the > > > dimensions that are canceling themselves > > > in nominator and denominator > > > > we get > > > ==================================== > > > 6.6exp-34 Kg MET /SEC times fs/3 exp10 > > > ==================================== > > > now my question is > > > where do you see anything relativistic in it ??!! > > > > TIA > > > Y.Porat > > > ---------------------- > > > P = hf/c > > > h is constant. > > c is constant. > > f is not constant. > > > Since the measured frequency of a received photon is dependant on the > > initial frequency and the difference between the velocity of this > > sending frame and the receiving frame one has to take relativity (SR) > > into consideration if the two frames are not motionless WRT each > > -------------------------- > 1 > very nice!! > if you take the Doppler effect > yiou see that the** f **is changing!! > AND YOU CAN SEE IT IN THE FORMULA AS I PRESENTED IT (PRESENTED by THAT > fs > so IT IS THE fs THAT IS CHANGING NOT > **THE INITIAL*** MASS UNITS ** > it is their number thaqt can change > IOW > MORE OR LESS **MASS UNITS **ARE COMING INTO THE SECONDARY FRAME > BUT > LISTEN CAREFULLY > TH E MASS ** UNITS** ARE NOT CHANGING > IT IS THE **NUMBER OF THOSE ** MASS UNITS !!! > and it i s presented nicely in my above formula analysis presented > in that ** fs ** > the number of mas s units is linearity > (nothing like the relativistic ***second order** )!!!!! and nicely > presented by that fs > no need to look for more formula > or 'interpretations'' > > btw > it is as well my new insight > about how the Doppler effect is a prove > that the hf is not the right definition of the REAL SINGLE > PHOTON !!! > 2 > you forgot that > TH E VELOCITY OF PHOTON > ****IS ALWAYS c > IN ALL FRAMES!! *** > > no mater if in relative motion or not > so ??!! > even relative to the secondary frame > it remains c!!! > as well as c -- in the original; frame !! > 2 > being in motion for itself > ***is not enough to be relativistic !!!*** > as long it is relative to***** itself **** !!! > > and that is why > YOU CANT SEE ANYTHING RELATIVISTIC IN THAT PHOTON MOMENTUM > FORMULA !!! > > except that that photons > declared himself to be > 'pope of Rome' !!! (:-) > 3 > WHIL WE WE DELL WITH > PHOTON MOMENTUM > > WE DEAL JUST IN ** ONE FRAME** > NOT IN MANY FRAMES !!! > > about many FRAMES > see my above explanations > about how it works in the Doppler case > two frames > but no need to obfuscate it > we have to conclude (resume) first about > just one frame !!! > > > other. This is where gamma enters the picture. How can the momentum > > be anything but relativistic. > > see about the pope of Rome (:-) > ie > how can i not be the pope of Rome ??!! > -------------------- > > TIA > Y.Porat > ------------------------------ > > > > > > > Mathal I said the velocity difference in frames is where relativity enters into the picture. The notion that relativity has no bearing on the results of using the formula is silly. The Doppler effect has a more significant effect on the frequency than the relativistic frame effect. This notion that frequency of light is the number of 'mass units' for a specific period of time is absurd on many levels. 1. photons don't have mass. 2.Your hypothesis means necessarily that an individual photon does not have a frequency or a wavelength. It is just this 'mass unit'. How does your FM radio receiver decide which mass units it wants to acknowledge and which 'mass' units it chooses to ignore? Mathal |