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Repulsing objects by photons

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  • abelov0927
    Message 1 of 2 , Mar 17, 2011
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      http://knol.google.com/k/alex-belov/repulsing-objects-by-photons/1xmqm1l0s4ys/23#view 

      This system shows repulsing two objects by photons.



      Base on radiation pressure or light pressure phenomenon, the the particle photon without stationary mass has a momentum which can be transfered to an object. 
      The light radiation and objects repulsion must be follow by law of momentum conservation. The figure1 shows repulsing two identical objects with mass m. After a period of time the objects are utilizing photons momentums and start conduct a translational motion. If laser system has identical ray intensity in both directions then objects will have translational momentums P with same value. 
      -vec P=vec P
      Other words, the objects with same mass m will have same velocities v1.
      m(-vec v_1)=mvec v_1
      The laser system with identical ray intensity in both directions has zero net momentum.

      However, the objects will utilize less photon momentums when these objects increase it's own velocities relatively to laser. The speed of light is constant. However, base on Doppler effect the photons will increase wavelength which reduce photons momentum
      p=frac{h}{lambda}
      Where: - photon momentum, - Planck constant, lambda - photons wavelength

      Other words, the objects linear velocities will gain in non-linear mode.

      How photons will repulse objects if one of them is rotating?

      The figure2 shows repulsing objects by photons where one of them conducts rotation around it's own center of mass.


      In this case, this objects will utilize photon momentum differently. After some period of time, the non-rotated object will utilize photons momentums on velocity v. However, the rotated object will utilize photons momentums on velocity v+wR. Base on Doppler effect the rotated object will utilize photons momentums less than non-rotated object. Therefore, after some period of time the rotated and non-rotated objects will have different translational velocities v1 and v2.
      The figure 2_2 shows objects repulsing with mirrors. 

      Here's objects reflect photon back to light source. The photons wavelengths will be different on receiver i.e. photons have a different momentums. 



      A few calculations.


      Will it happen relativistic Doppler effect between source and receivers? How big is it? Let's calculate it.

      The figure 3 shows parts what will described in details.


      The part A shows left non-rotated objects R and source S.

      Let's define frames of referces of these objects R ans S.

      frame of reference in physics, may refer to a coordinate system or set of axes within which to measure the position, orientation, and other properties of objects in it, or it may refer to an observational reference frame tied to the state of motion of an observer. It may also refer to both an observational reference frame and an attached coordinate system, as a unit.

      The objects R and S have owns frames of references. Are these frames of references are equal?
      The object R has translational velocity –V1 relatively to observer on object S. However, the object Shas translational velocity V1 relatively to observer on object R. These objects have translational velocity |V1| relatively to each other. Therefore, these objects and S frames of references are not identical and objects R and S are registering wavelength of light of source S differently.
      Base on Doppler effect:
      gamma=frac{1}{sqrt 1-(frac{v_1}{c})^2}
      f_r=frac{f_s}{gamma}\f=frac{c}{lamb}\frac{c}{lamb_r}=frac{c}{lamb_sgam}\lam_r=lam_sfrac{1}{sqrt 1-(frac{v_1}{c})^2}
      Where: Y – Lorentz factor, c – speed of light, V1 – velocity of source S relatively receiver Rfr – light frequency registered on receiver Rfs – light frequency registered on source Slr – light wavelength registered on receiver Rfs – light wavelength registered on source S.

      The part B shows right rotated objects R and source S.

      As it was shown for part A, the receiver R and source S frames of references are not equal. These objects have translational velocity |V2+wR| relatively to each other. Therefore, base on Doppler effect, the objects and S are registering wavelength of light of source S differently.
      gamma=frac{1}{sqrt 1-(frac{v_2+ome R}{c})^2}
      f_r=frac{f_s}{gamma}\f=frac{c}{lamb}\frac{c}{lamb_r}=frac{c}{lamb_sgam}\lam_r=lam_sfrac{1}{sqrt 1-(frac{v_2+ome R}{c})^2}
      Where: Y – Lorentz factor, c – speed of light, V2 – velocity of source S relatively receiver Rfr – light frequency registered on receiver Rfs – light frequency registered on source Slr – light wavelength registered on receiver Rls – light wavelength registered on source S, w - angular velocity of rotated object, R – distance between center of mass of rotated object and point of light receiving of rotated object.

      Since source has two identical rays with opposite direction it's net momentum is zero
      -vec P_l_1=vec P_l_2
      Where: Pl1,Pl2 - momentums of rays of source

      If calculate for one photon then:
      P=frac{h}{lam}\lam_s_1=lam_s_1\frac{h}{lam_s_1}= frac{h}{lam_s_2}
      Where: lamda_s1,lambda_s2 - wavelength of light of source, h - constant Planck, P - momentum

      The left non-rotated object has momentum for one photon equal to:
      P_l=frac{h}{lam_r}\P_l=frac{hsqrt(1-(frac{v_1}{c})^2)}{lam_s}
      Where: Pl - photon momentum, c – speed of light, V1 – velocity of source S relatively receiver R,  lr – light wavelength registered on receiver Rls – light wavelength registered on source S

      The right rotated object has momentum for one photon equal to:
      P_r=frac{h}{lam_r}\P_r=frac{hsqrt(1-(frac{v_2+ome R}{c})^2)}{lam_s}
      Where: Pr - photon momentum, c – speed of light, V2 – velocity of source S relatively receiver R,  lr – light wavelength registered on receiver Rls – light wavelength registered on source S, w - angular velocity of rotated object, R – distance between center of mass of rotated object and point of light receiving of rotated object.

      The photons momentums of left and right receivers are no equal to each other.
      The difference of these momentums will be equal to:
      P_r-P_l=frac{hsqrt(1-(frac{v_1}{c})^2)}{lam_s}-frac{hsqrt(1-(frac{v_2+ome R}{c})^2)}{lam_s}\
      Del P=frac{h}{lam_s}(sqrt(1-(frac{v_1}{c})^2)}-sqrt(1-(frac{v_2+ome R}{c})^2))
      Where: Pr - photon momentum for right receiver, Pl - photon momentum for left receiver, c – speed of light, V1 – velocity of source S relatively to left receiver RV2 – velocity of source S relatively to right receiver R ls – light wavelength registered on source S, w - angular velocity of rotated object, R – distance between center of mass of rotated object and point of light receiving of rotated object.



      The light propulsion system


      As it was shown before, the receivers may register light with different wavelengths relatively to light source. The photon is particle without mass rest. However this particle has a momentum. This propulsion system use this Doppler effect.
      This figure 1_1 show this light propulsion system.
      The receivers have particles which conduct round trip. The Ray of source rich these particle where they have velocity V relatively to light source.
      As it was shown before, these particles frames of references are not equal to light source frame of reference. Therefore, these particles will receive the source light with difference wavelength for each receiver. The particles of left receiver are moving forward to light source. The particles of this receiver are getting light wavelength difference as blueshift. 
      The momentum for one photon of source light for this receiver is:
      P_r=frac{h}{lam_r}\P_r=frac{h}{lam_ssqrt(1-(frac{v}{c})^2)}
      Where: lambda_r,lambda_s - wavelength of light of source on receiver and source, c - speed of light,h - constant Planck, Pr - photon momentum registered on right receiver.

      Otherwise, the particles of left receiver are moving forward to light source. The particles of left receiver are moving away from light source. The particles of this receiver are getting light wavelength difference as redshift.
      The momentum for one photon of source light for this receiver is:
      P_l=frac{h}{lam_r}\P_l=frac{hsqrt(1-(frac{v}{c})^2)}{lam_s}
      Where: lambda_r,lambda_s - wavelength of light of source on receiver and source, c - speed of light,h - constant Planck, Pl - photon momentum registered on right receiver.

      The difference photon momentum for left and right receivers equal to:
      P_r-P_l=frac{h}{lam_ssqrt(1-(frac{v}{c})^2)}-frac{hsqrt(1-(frac{v}{c})^2)}{lam_s}


      Delta P=frac{h}{lam_s}(frac{v^2}{c^2sqrt(1-(frac{v}{c})^2)})
      Where: lambda_s - wavelength of light of source on source, c - speed of light, h - constant Planck, Pr - photon momentum registered on right receiver, Pl - photon momentum registered on left receiver, dP- difference between photon momentums,V - velocity of particles relatively to source of light


      Conclusion

      Base on Doppler effect and particles(photons) with zero rest mass, the repulsed objects may have different translational velocities by value inside isolated system.


      Reference:

    • Jim S
      Hello Abelov927, Forgive me for replying directly to your post here in the forum as I know that much of the time the Yahoo autoposting system makes errors in
      Message 2 of 2 , Mar 28, 2011
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        Hello Abelov927,

             Forgive me for replying directly to your post here in the forum as I know that much of the time the Yahoo autoposting system makes errors in reposting posts that contain illustrations. However I felt that some of these illustrations are vital for an understanding of what you are speaking of and wanted others to have a chance to review them along with the comments I make. If any of the images do fail or if there are some odd code errors anyone can go back to the original post to which I am replying on the main forum, which should be the prior post.

              I too have studied photo optrics and photo optics, there is a distinction. The amazing properties of photons and other particles at various frequencies is astounding, I will try to keep this in layman's terms, but as any involved in particle physics know: there is a relationship between frequency and wavelength, and a vector scale is quite easy to demonstrate such behavior. I am speaking here of a defined particles behavior in a vacuum, as other alterations do happen when particles pass through other mediums, although those other mediums too may have an important "inertial effect" as science progresses, in relationship to the passing of say a photon through the medium, do note however that a photon after having passed through a medium may have a slightly different frequency but none the less when going back into a vacuum will alter it's speed back up to that of the normal speed of light for a photon in a vacuum. Amazing little waving balls what?

           Lol. Forgive me for a little levity, but imagine what immense power is enclosed within each "packet" of light, whether a wave or sphere in interpretation the incredible energy required for a photon to escape from the deep gravity well of an incredibly distant sun and travel perhaps billions of light years and still have enough energy left after such a long journey to make an energy impact upon some sensitive instrument on Earth, whether mechanical or the glassy organic eye of an owl at night. A photon contains it seems almost enough energy as to be virtually perpetual in nature, on and on it propagates. Certainly has a much longer life than I. This propagation of light holds a mystery key I believe, an answer to many perplexing puzzles, if we could grasp such astounding congealed energy. And Abelov927 I think one or more of your illustrations does tickle on these supersecrets.

           I have long been intrigued by the "Optical Magnus Effect" and do think it holds the key to the actual propulsive potential of the photon. Here is one link to a paper written on that in 1992;

         

        http://pra.aps.org/abstract/PRA/v45/i11/p8204_1

        Where they say this;

        "It is predicted theoretically and registered experimentally that the speckle pattern of a laser beam transmitted through a multimode fiber undergoes an angular shift from the switching of the chirality of the polarization. The effect may be considered as the result of the spin-orbit interaction for the photon in the inhomogeneous medium."

        The actual paper goes on to convey much more data including hints at what the authors think is possible, or at least as far as they dare go within the limits of their tenure and accepted physics papers, (not to rock the boat as it were). Shame it is hosted on one of those sites that want too much of your personal data and maybe even some cash to read what was free for all to read just a few years ago, I was lucky enough to have read it then.

        More recent was this release through physorg on similar and slightly more advanced research on photon behavior;

        http://www.physorg.com/news148140149.html

        Where they say,

        "The researchers also believe their ongoing work can provide results that are useful to other fields of physics. According to Prof. Hasman, "There are a number of systems where the spin of a particle couples with its trajectory in high-energy and condensed matter physics. The math is the same in all cases, but experimentally it's very hard to understand what's going on. Our experimental system offers a new way to get at some of these fundamental questions clearly and precisely.""

              Then there is the opposite side of particles, when they are stopped and when they release their stored up energy. It is said that the rest Mass of a photon is Zero. True enough perhaps since at rest, most photons lose all of their identity as a photon and become released energy, so they cease to exist. Nothing is nothing. But the energy they were can be used to many purposes as anyone with a solar panel can attest. As recalled from Relativity E=MC\2 and so the photon has given up it's energy when no longer moving at speed, but just because at rest it has no mass, does not mean that it had no mass when at speed. In fact as told in the relativity equation in order for both sides of the equation to be true, if the photon on one side of the equation had Energy, then on the other side of the equation, it had to have something akin to Mass, and moving at considered velocity as well. This was also foretold by Einstein when he projected that light would be effected by gravity, since proven true and even used by some astronomy equipment to obtain images through an effect called gravitational lensing. Some 5 years ago I answered a question here on the yahoo answers about photons at rest and at speed. I used different words and again layman's terms but fit to that specific question as asked, gave a fair answer. 5 years, my how time flys .

           http://answers.yahoo.com/question/index?qid=20060720074926AAOhDYb

        I seem to be beating about the bush a bit perhaps but wanted to let you know that I agree you are on the right track, especially if we can sometime admit that light can have a "motional mass" attributed to itself. After all mass does have direct connection with gravity and gravity interacting with light that is at immense speed may be said to be having a fairly powerful motional reaction, especially if we arrange factors to take greatest possible advantage of said interaction. I am speaking of far more than a solar sail here, although that too prove the ability of light to cause mass to become in motion. But where you are headed with your preliminary designs shows potential of amplification of propulsive potentials. Keep up with this, who knows how fast, we may someday go...

          Years ago I built a light coil, very similar to an analogy of an electric coil. It was used for temporal purposes, and of course some energy was lost in the system, but I think it may be possible to build a light transformer as well, and see some potential in that direction with one of your designs below, although will watch what I say here where all public eyes can review but since you are already on right track, thought I'd bring it up. You see I am building something which I call an "Inter Dimensional Imaging Chamber", have been working on it for over 20 years, a temporal device for viewing more than is apparent in our normal 3D experiences, and have interjected some live images from distant stars. Gravity becomes very apparent using this device, although it still needs considerable work to have it finished and truly well and finely tuned. But I do have enough results already to say your ideas have some merit, do continue.

           Thank You for your time and considerations,

             Mystery B-)

        Jim

         

        ________________________________________________________________________________

         


        --- In gravitationalpropulsionstevenson@yahoogroups.com, "abelov0927" <abelov0927@...> wrote:

        http://knol.google.com/k/alex-belov/repulsing-objects-by-photons/1xmqm1\ l0s4ys/23#view>
         http://knol.google.com/k/alex-belov/repulsing-objects-by-photons/1xmqm1l\
         0s4ys/23#view
        http://knol.google.com/k/alex-belov/repulsing-objects-by-photons/1xmqm1\l0s4ys/23#view> This system shows repulsing two objects by photons.
        http://knol.google.com/k/-/-/1xmqm1l0s4ys/qpjoee/lightrepulsion1.jpg
         Base on radiation pressure
        http://en.wikipedia.org/wiki/Radiation_pressure or light pressurephenomenon 

         http://en.wikipedia.org/wiki/Light#Light_pressure ,  the the particle photon without stationary mass has a momentum which can be
        transfered to an object. The light radiation and objects repulsion must
        be follow by law of momentum conservation
        http://en.wikipedia.org/wiki/Momentum#Conservation_of_linear_momentum

         . The figure1 shows repulsing two identical objects with mass m. After
        > period of time the objects are utilizing photons momentums and start
        > conduct a translational motion. If laser system has identical ray
        > intensity in both directions then objects will have translational
        > momentums P with same value. [-\vec P=\vec P]
        > Other words, the objects with same mass m will have same velocities v1.
        > [m(-\vec v_1)=m\vec v_1] The laser system with identical ray intensity
        > in both directions has zero net momentum.
        > However, the objects will utilize less photon momentums when these
        > objects increase it's own velocities relatively to laser. The speed of
        > light is constant. However, base on Doppler effect
        http://en.wikipedia.org/wiki/Doppler_effect the photons will increase
         wavelength which reduce photons momentum
        http://en.wikipedia.org/wiki/Photon#Physical_properties
        [p=\frac{h}{\lambda}]
        Where: p - photon momentum, h - Planck constant, lambda - photons
         wavelength
        Other words, the objects linear velocities will gain in non-linear mode.
        How photons will repulse objects if one of them is rotating?The figure2
        shows repulsing objects by photons where one of them conducts rotation
        around it's own center of mass.
        http://knol.google.com/k/-/-/1xmqm1l0s4ys/qpjoee/lightrepulsion2.jpg>
        In this case, this objects will utilize photon momentum differently. After some period of time, the non-rotated object will utilize photons
         momentums on velocity v. However, the rotated object will utilize
        photons momentums on velocity v+wR. Base on Doppler effect the rotated
         object will utilize photons momentums less than non-rotated object.
        Therefore, after some period of time the rotated and non-rotated objects
        will have different translational velocities v1 and v2.The figure 2_2
        shows objects repulsing with mirrors.
        http://knol.google.com/k/-/-/1xmqm1l0s4ys/qpjoee/lightrepulsionm.jpg>
        Here's objects reflect photon back to light source. The photons
        wavelengths will be different on receiver i.e. photons have a different
         momentums.
         

        A few calculations.
        Will it happen relativistic Doppler effect between source and receivers?
        How big is it? Let's calculate it.The figure 3 shows parts what will
        described in details.

        http://knol.google.com/k/-/-/1xmqm1l0s4ys/qpjoee/lightrepulsionpartsab. jpg
        The part A shows left non-rotated objects R and source S.
        http://knol.google.com/k/-/-/1xmqm1l0s4ys/qpjoee/lightparta.jpg>
        Let's define frames of referces of these objects R ans S.
        A frame of reference <http://en.wikipedia.org/wiki/Frame_of_reference>  in physics <http://en.wikipedia.org/wiki/Physics> , may refer to a
         coordinate system  http://en.wikipedia.org/wiki/Coordinate_system or
        set of axes  http://en.wikipedia.org/wiki/Cartesian_coordinate_system
        within which to measure the position, orientation
        http://en.wikipedia.org/wiki/Orientation_(geometry) , and other
         properties of objects in it, or it may refer to an observational
         reference frame tied to the state of motion of an observer
        http://en.wikipedia.org/wiki/Observer_(special_relativity) . It may
        also refer to both an observational reference frame and an attached
         coordinate system, as a unit.
        The objects R and S have owns frames of references. Are these frames of
        > references are equal?The object R has translational velocity –V1
        > relatively to observer on object S. However, the object Shas
        > translational velocity V1 relatively to observer on object R. These
        > objects have translational velocity |V1| relatively to each other.
        > Therefore, these objects R and S frames of references are not identical
        > and objects R and S are registering wavelength of light of source S
        > differently.Base on Doppler effect: [\gamma=\frac{1}{\sqrt
        > 1-(\frac{v_1}{c})^2}]
        >
        > [f_r=\frac{f_s}{\gamma}\\f=\frac{c}{\lamb}\\\frac{c}{\lamb_r}=\frac{c}{\\
        > lamb_s\gam}\\\lam_r=\lam_s\frac{1}{\sqrt 1-(\frac{v_1}{c})^2}]
        > Where: Y – Lorentz factor, c – speed of light, V1 – velocity
        > of source S relatively receiver R, fr – light frequency registered
        > on receiver R, fs – light frequency registered on source S, lr –
        > light wavelength registered on receiver R, fs – light wavelength
        > registered on source S.
        > The part B shows right rotated objects R and source S.
        http://knol.google.com/k/-/-/1xmqm1l0s4ys/qpjoee/lightpartb.jpg As it
        > was shown for part A, the receiver R and source S frames of references
        > are not equal. These objects have translational velocity |V2+wR|
        > relatively to each other. Therefore, base on Doppler effect, the objects
        > R and S are registering wavelength of light of source S differently.
        > [\gamma=\frac{1}{\sqrt 1-(\frac{v_2+\ome R}{c})^2}]
        >
        > [f_r=\frac{f_s}{\gamma}\\f=\frac{c}{\lamb}\\\frac{c}{\lamb_r}=\frac{c}{\\
        > lamb_s\gam}\\\lam_r=\lam_s\frac{1}{\sqrt 1-(\frac{v_2+\ome R}{c})^2}]
        > Where: Y – Lorentz factor, c – speed of light, V2 – velocity
        > of source S relatively receiver R, fr – light frequency registered
        > on receiver R, fs – light frequency registered on source S, lr –
        > light wavelength registered on receiver R, ls – light wavelength
        > registered on source S, w - angular velocity of rotated object, R
        > – distance between center of mass of rotated object and point of
        > light receiving of rotated object.
        > Since source has two identical rays with opposite direction it's net
        > momentum is zero [-\vec P_l_1=\vec P_l_2]
        > Where: Pl1,Pl2 - momentums of rays of source
        > If calculate for one photon then:
        > [P=\frac{h}{\lam}\\\lam_s_1=\lam_s_1\\\frac{h}{\lam_s_1}=
        > \frac{h}{\lam_s_2} ]
        > Where: lamda_s1,lambda_s2 - wavelength of light of source, h - constant
        > Planck, P - momentum
        > The left non-rotated object has momentum for one photon equal to:
        > [P_l=\frac{h}{\lam_r}\\P_l=\frac{h\sqrt(1-(\frac{v_1}{c})^2)}{\lam_s}]
        > Where: Pl - photon momentum, c – speed of light, V1 – velocity
        > of source S relatively receiver R, lr – light wavelength registered
        > on receiver R, ls – light wavelength registered on source S
        > The right rotated object has momentum for one photon equal to:
        > [P_r=\frac{h}{\lam_r}\\P_r=\frac{h\sqrt(1-(\frac{v_2+\ome
        > R}{c})^2)}{\lam_s}]
        > Where: Pr - photon momentum, c – speed of light, V2 – velocity
        > of source S relatively receiver R, lr – light wavelength registered
        > on receiver R, ls – light wavelength registered on source S,
        > w - angular velocity of rotated object, R – distance between center
        > of mass of rotated object and point of light receiving of rotated
        > object.
        > The photons momentums of left and right receivers are no equal to each
        > other.The difference of these momentums will be equal to:
        > [P_r-P_l=\frac{h\sqrt(1-(\frac{v_1}{c})^2)}{\lam_s}-\frac{h\sqrt(1-(\fra\
        > c{v_2+\ome R}{c})^2)}{\lam_s}\\ ]
        > [\Del
        > P=\frac{h}{\lam_s}(\sqrt(1-(\frac{v_1}{c})^2)}-\sqrt(1-(\frac{v_2+\ome
        > R}{c})^2))]
        > Where: Pr - photon momentum for right receiver, Pl - photon momentum for
        > left receiver, c – speed of light, V1 – velocity of source S
        > relatively to left receiver R, V2 – velocity of source S relatively
        > to right receiver R, ls – light wavelength registered on source
        > S, w - angular velocity of rotated object, R – distance
        > between center of mass of rotated object and point of light receiving of
        > rotated object.
        >
        >
        > The light propulsion system
        > As it was shown before, the receivers may register light with different
        > wavelengths relatively to light source. The photon is particle without
        > mass rest. However this particle has a momentum. This propulsion system
        > use this Doppler effect.This figure 1_1 show this light propulsion
        > system.
        http://knol.google.com/k/-/-/1xmqm1l0s4ys/qpjoee/lasersystem11.jpg  The
        > receivers have particles which conduct round trip. The Ray of source
        > rich these particle where they have velocity V relatively to light
        > source.As it was shown before, these particles frames of references are
        > not equal to light source frame of reference. Therefore, these particles
        > will receive the source light with difference wavelength for each
        > receiver. The particles of left receiver are moving forward to light
        > source. The particles of this receiver are getting light wavelength
        > difference as blueshift. The momentum for one photon of source light for
        > this receiver is:
        > [P_r=\frac{h}{\lam_r}\\P_r=\frac{h}{\lam_s\sqrt(1-(\frac{v}{c})^2)}]
        > Where: lambda_r,lambda_s - wavelength of light of source on receiver and
        > source, c - speed of light,h - constant Planck, Pr - photon momentum
        > registered on right receiver.
        > Otherwise, the particles of left receiver are moving forward to light
        > source. The particles of left receiver are moving away from light
        > source. The particles of this receiver are getting light wavelength
        > difference as redshift.The momentum for one photon of source light for
        > this receiver is:
        > [P_l=\frac{h}{\lam_r}\\P_l=\frac{h\sqrt(1-(\frac{v}{c})^2)}{\lam_s}]
        > Where: lambda_r,lambda_s - wavelength of light of source on receiver and
        > source, c - speed of light,h - constant Planck, Pl - photon momentum
        > registered on right receiver.
        > The difference photon momentum for left and right receivers equal to:
        > [P_r-P_l=\frac{h}{\lam_s\sqrt(1-(\frac{v}{c})^2)}-\frac{h\sqrt(1-(\frac{\
        > v}{c})^2)}{\lam_s}]
        >
        >
        > [\Delta P=\frac{h}{\lam_s}(\frac{v^2}{c^2\sqrt(1-(\frac{v}{c})^2)})]
        > Where: lambda_s - wavelength of light of source on source, c - speed of
        > light, h - constant Planck, Pr - photon momentum registered on right
        > receiver, Pl - photon momentum registered on left receiver, dP-
        > difference between photon momentums,V - velocity of particles relatively
        > to source of light
        >
        > ConclusionBase on Doppler effect and particles(photons) with zero rest
        > mass, the repulsed objects may have different translational velocities
        > by value inside isolated system.
        >
        > Reference:1. Frame of reference
        http://en.wikipedia.org/wiki/Frame_of_reference> 2. Relativistic
        Doppler effect

        http://en.wikipedia.org/wiki/Relativistic_Doppler_effect

         3. Lorentz factor  http://en.wikipedia.org/wiki/Lorentz_factor

         4. Redshift
        http://en.wikipedia.org/wiki/Redshift 

         5. Blueshift
        http://en.wikipedia.org/wiki/Blueshift

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