Calculating the force exerted by the photon on the elementary particle

Transcription

1 Calculating the force exerted b the hoton on the elementar article R. MANJUNATH #16,8 TH Main road, Shivanagar, Rajajinagar Bangalore: , INDIA, manjunath5496@gmail.com Abstract: - Photons, like all elementar articles, exhibit both wave like and article -like roerties.the dual nature of hotons can be visualized through their interaction with other objects. The hotons ossess momentum; carr mass during its motion. Hence the also exert force and ressure on other objects. The recent astronomical observations stated that the radiation ressure is strong enough to ush an asteroid. This statement shocked the scientific comit and made them to gae in an astonished wa, thinking that do radiation ressure reall ushes an asteroid. The inherent goal of roosal of this mathematical model is to calculate the force exerted b the hoton on the other elementar articles, in a simle and consistent wa. The basic concets of quantum mechanics and classical mechanics are incororated together in a single frame work to la the foundation for the emergence of the equation for the calculation of the force imarted b the hoton on other elementar articles like roton, electron etc. [Nature and Science 2010;8(10):82-87]. (ISSN: ). Ke words: - seed of light, scattering angle, hoton, article, energ, force and time. Quantum mechanics is a well defined mathematical model that grew out of Planck s quantum theor, considered as one of the major revolutions of 12 th centur. It takes into account the dual nature of articles, Heisenberg uncertaint rincile and quantum mechanical behavior of the microscoic articles. Let us consider a article at rest with resect to the observer frame of reference. Total energ associated with the article will be equal to [m o c 2 ].Suose a hoton of quantum energ [hf] is incident on that article. A art of the energ [hf ] is absorbed b the article and rest of the energ [hf ] is scattered b the article. The energ of the article {after the absortion of energ [hf ] from the hoton} will be equal to m o c 2 +hf = mc 2.Let [θ] be the scattering angle formed between the incident hoton and scattered hoton resectivel. Energ 1/stabilit Higher the energ, lesser is the stabilit.in order to attain stabilit, absorbed energ is converted into motion of the article. Hence the article travels a distance [X] in time [t]. Figure: 1 Schematic diagram of scattering of energ of the hoton b the article X= Linear dislacement of the article hf = Energ of incident hoton hf = Energ of scattered hoton θ = scattering angle Consider a arallelogram ABCD constructed as shown in the FIGURE:-1. Let AB=CD=X, AD=BC=hf, AC=hf [oosite sides in a arallelogram are alwas equal] Law of cosine is mathematicall reresented b: - a 2 =b 2 +c 2-2bc cosθ. Let a =X, b=hf, c=hf, cos A = cosθ. htt:// 82 marslandress@gmail.com

2 B aling the law of cosine to the triangle ADC We get:-x 2 = (hf) 2 + (hf ) 2-2(hf) (hf ) cosθ [1] B the law of conservation of momentum of the hoton We get = + ; where " ',, is the " ' momentum of the incident hoton, absorbed momentum of the hoton b the article, momentum of the scattered hoton. When the article-hoton collision takes lace.a art of the motion of the hoton is transferred to the article, resulting in the motion of the article. The Absorbed momentum of the hoton equals the momentum of the article. = " = + Where = momentum of the article = Squaring on the both sides; we get: 2 = ' 2 ' ' (a-b) 2 =a 2 +b 2-2ab Thus the above equation becomes 2 = * According to dot roduct rule: - * = cos θ a b = a b cosθ Then we get 2 = cos θ Let us multil the above equation b c 2 We get: - 2 c 2 = 2 c 2 + c 2-2 c 2 cos θ Where c = seed of light in vacuum [ m /s] The hotons carr momentum roortional to their frequenc f hf = c 2 c 2 = (hf) 2 + (hf ) 2-2(hf) (hf ) cos θ [2] B comarison of [1] and [2] We get: X 2 = 2 c 2 X= c [3] Here X=linear dislacement of the article = amount of motion associated with the article. The linear dislacement of the article is defined as the function of the amount of motion associated with the article. The dislacement of the article equals the momentum ossessed b the article times the seed of light in vacuum. The value of [X]is too large comared to that of []. X> Although X, dislacement of the article [X] is alwas greater than the momentum[] ossessed b the article. The mass of the with the hoton= m o The mass of the article after collision with the hoton= m m=m o + hf Here hf =loss of energ from the hoton htt:// 83 marslandress@gmail.com

3 Loss of energ from the hoton =energ absorbed b the article after collision with the hoton The absorbed energ stored in the article decreases with the distance travel of the article. If [hf ] = 0, then m=m o The amount of motion associated with the article is given b: - = m v Here m= mass of the article after collision v= velocit of the article X= (m v) c Velocit of the article is defined as the dislacement of the article er unit time. [v=x/t] work against gravitational field of the black hole. Total energ content of the hoton is a measure of mass known as relativistic mass. We also observe that the relativistic mass of the hoton has been decreased followed b the decrease in the frequenc of hoton. Hence frequenc[f] of the hoton is directl roortional to its mass[m]. f [Mass of the article after collision mass of the ]= loss of energ from the hoton When a article hoton collision takes lace, mass of the article will be:- m=m o + m * Here m= mass of the article after collision m t=mc [4] Here t=time of travel of the article m= mass of the article after collision The time of travel of the article is defined as the function of the mass of the article after collision. The time of travel of the article equals the mass of the article after collision times the seed of light in vacuum. The value of [t]is too large comared to that of [m]. t>m Although t m, time of travel of the article [t] is alwas greater than the mass of the article after collision[m]. m o = rest mass of the article or mass of the m * = mass of absorbed energ of the hoton or loss of mass from the hoton after collision with the article Multiling b c 2, we get: mc 2 =m o c 2 +m * c 2 m * c 2 =hf m * f Conversion of energ into mass and its vice versa is beautifull exlained b Einstein s famous equation E=mc 2,here c is not just the velocit of a certain henomenon namel the roagation of electromagnetic radiation but rather a fundamental feature of the wa sace and time are unified as sace- time continuum. The equation imlies conversion of energ into mass and it s vice versa accounts the unification of sace and time. In other words in resence of mass there is unification of sace and time.in absence of mass, sace and time behave as two searate factors. The sace, time, mass are different concets in hsics and these concets are incororated in a single mathematical equation. Moreover the question arises in human mind, the need of unification of sace and time in conversion of The rincile of relativit in conjunction with Maxwell equations exlained, in a consistent wa that mass is a direct measure of the energ content of the bod. The hotons carr mass roortional to their frequenc.let us imagine a traed hoton escaes from the gravitational field of the black hole, we observe difference in the frequenc of hoton i.e. frequenc [f] of hoton has been decreased. A Part of the energ of the traed hoton is utilized to do energ into mass and it s vice versa. htt:// 84 marslandress@gmail.com

4 mc 2 =m o c 2 +hf From (3) we know: X=c Absorbed momentum of the hoton equals the momentum of the article = X= c The absorbed frequenc of the hoton is directl roortional to its momentum f c= hf X= hf [5] Absorbed energ is comletel utilized b the article to travel a distance [X] from oint [A] to oint [B] in a time [t]. When a article hoton collision takes lace.a art of the energ of the hoton is transferred to the article, resulting in the decrease of energ of the hoton and simultaneous increase of energ of the article ;energ of the article increases from [m o c 2 ] to [mc 2 ]. mc 2 =m o c 2 +X m=m o + X /c 2 Here m=mass of the article after collision m o =rest mass of the article or mass of the From (4) we know t=mc m=m o + X /c 2 Time of travel of the article [t] exressed in terms of the mass of the [m o ] and linear dislacement of the article [X]:- t=m o c + X/c [6] The theor of general relativit oints out that hotons have no mass.exerimental value of mass of hoton was found to be grams. However if the mass of the hoton is not considered to zero, then quantum mechanics would be in trouble.it also an uhill task to conduct a exeriment which roves the hoton mass to be exactl zero. If we al the above equation to hotons:-m o = 0 [since hotons can never be at rest in an frame of reference]. We get X=c t c=x / t The Particle onl with a zero rest mass can travel with a velocit equal to that of light. From (6), we know: - t=m o c + X/c X=c X/=c t= m o c + [7] The momentum ossessed b the article is given b: =mv X=linear dislacement of the article The osition and time are undefined arameters, most fundamental assumtions in hsics. For our convincible urose of understanding, we assume [X] =zero, when the article is at rest.since we can measure no value for [X]. If [X] =0, then [m=m o ] The [m o ] aroaches[m] and [m]aroaches [m o ] b the factor X/c 2. Thus [7] becomes: t= m o c + mv [8] Ever article continues in its state of rest with resect to observer frame of reference, unless comelled b an external force to change that state. When a article hoton collision takes lace, the hoton exerts force [F] on the mass [m o ] and roduces acceleration [a] in it. F=m o a Here m o =rest mass of the article or mass of the a=acceleration roduced in the article htt:// 85 marslandress@gmail.com

5 Acceleration is defined as velocit of the article er unit time. a=v/t Here v= velocit of the article t= time of travel of the article Ft=m o v Imulse is the effect of force acting for short interval of time.it is measured as the roduct of the force and time. I =Ft I= m o v [9] Here m o =rest mass of the article or mass of the v=velocit of the article I= imulse v=i/m o Thus [8] becomes: - t= m o c + (mi /m o ) [10] From [4], we know: - t=mc t /c= m Here t=time of travel of the article m= mass of the article after collision Thus [10] becomes: - t= m o c + (ti /m o c) t - (ti /m o c) = m o c t [1- (I /m o c)] = m o c [1- (I /m o c)] = m o c /t [1- (m o c /t)] = I /m o c I= [m o c- (m o 2 c 2 /t)] [11] Thus [11] becomes: - F= [m o c/t - (m o 2 c 2 /t 2 )] [12] Here F=force exerted b the hoton m o = rest mass of the article t= time of travel of the article Thus, we have the formula [12] for the calculation of the force exerted b the hoton on the other elementar articles, which takes into consideration the time of travel of the article and rest mass of the article. The discover of laws of motion b Sir Isaac Newton is one of the most remarkable events in the histor of hsics. The Newton laws of motion are being constantl tested b man hsicists from ast several centuries. So far these laws have with stood ever test of exerimentation. These laws of motion holds good for all ractical uroses in an inertial frame of reference. The statement: force alied on the bod varies inversel with the time of travel of the bod or time after which the bod again comes to its initial state, redicted b Newton laws of motion. The above statement has been exerimentall verified b several scientists across the world. F [1/ t] More the force alied on the article, less will be time of travel of the article. If the equation [12] obes the above cited statement then the equation holds good for all ractical uroses of exerimentation. If the equation fails to satisf the above condition then the equation brings about its own itfall. We can t just accet the mathematicall derived equations without testing them. The allowed orbits are those for which the orbital angular momentum of the electron about the nucleus is an integral multile of h/2π.that is, angular momentum of the electron about the nucleus is mvr = n [h/2π] Here m= mass of electron v= orbital velocit of the article r=radius of electron orbit Imulse is measured as the roduct of the force and time: - I=Ft h=planck s constant [ Js] htt:// 86 marslandress@gmail.com

6 For the electron at a infinite distance from the nucleus v r = v=0 1/ r The electron in the ver distant orbit is assumed to be at rest since their orbital velocit is ver negligible [aroximatel equal to zero]. Newl derived mathematical formula: - F= [m o c /t - (m o 2 c 2 /t 2 )] [12] For electron: - m o = kg The value of [F] is calculated for different values of [t] using the equation [12].The values are noted in the tabular column as shown below: SLNO Time of travel of the article Force exerted b the hoton 1] 1s N 2] 2s N From the above observations, we come to the final conclusion that the value of [F] decreases with increase in the value of [t]. Hence the newl framed mathematical equation [12] holds good for all ractical uroses of examination. 5. en.wikiedia.org/wiki/momentum - Cached Similar 6. en.wikiedia.org/wiki/comton scattering - Cached Similar 7. en.wikiedia.org/wiki/force - Cached Similar 8. en.wikiedia.org/wiki/imulse_(hsics) Cached 9. en.wikiedia.org/wiki/newton's_laws_of_m otion - Cached Similar 10. mathworld.wolfram.com Algebra Vector Algebra - Cached Similar 11. en.wikiedia.org/wiki/bohr_model - Cached Similar 12. en.wikiedia.org/wiki/photon - Cached Similar df - Similar Acknowledgments: - I would like to exress m dee gratitude to all those who gave me oortunit to comlete this thesis. Date of submission: - 15/6/2010 References 1. en.wikiedia.org/wiki/parallelogram - Cached Similar 2. en.wikiedia.org/wiki/law of cosines - Cached Similar 3. en.wikiedia.org/wiki/planck's law - Cached Similar 4. en.wikiedia.org/wiki/mass energ equivalence - Cached Similar htt:// 87 marslandress@gmail.com