Notices
Results 1 to 47 of 47

Thread: Deflection of light

  1. #1 Deflection of light 
    Forum Junior whizkid's Avatar
    Join Date
    Jan 2013
    Posts
    282
    When light is deflected by gravity, (as when passing near the Sun) does frequency influence the value of the angle of deflection ?
    Any link is welcome.
    Thanks


    Reply With Quote  
     

  2.  
     

  3. #2  
    Suspended
    Join Date
    Feb 2013
    Posts
    1,774
    Quote Originally Posted by whizkid View Post
    When light is deflected by gravity, (as when passing near the Sun) does frequency influence the value of the angle of deflection ?
    Any link is welcome.
    Thanks
    No, it does not.


    Reply With Quote  
     

  4. #3  
    Forum Cosmic Wizard
    Join Date
    Dec 2013
    Posts
    2,408
    ...and yet I think it should because of the einstein frequency shift.
    Gravitational redshift - Wikipedia, the free encyclopedia

    https://www.google.ca/url?sa=t&rct=j...71198958,d.aWw

    Since the field strength of a gravitional lense changes as the distance from the gravitational source changes there should be some chromatic distortion.
    This should be observable as the lense moves in relation to the source and observer or as the distance between the observer and the lense changes.
    However in most cases the distances involved and the rate of motion would tend to minimize chromatic aberations so you would need strong gravitional fields at a fairly close distance to observe the effect.

    So far most attempts to observe gravitional lensing have relied on narrow frequency sensors (usually gamma ray) to record the transit so they are unable to observe any frquency shift since they simply see only a narrow frequency band.
    Reply With Quote  
     

  5. #4  
    Suspended
    Join Date
    Feb 2013
    Posts
    1,774
    Quote Originally Posted by dan hunter View Post
    ...and yet I think it should because of the einsten frequency shift.
    You are mixing up two DIFFERENT effects.

    Nothing to do with the light deflection.
    Reply With Quote  
     

  6. #5  
    Forum Junior whizkid's Avatar
    Join Date
    Jan 2013
    Posts
    282
    I read the original paper by Soldner : On the Deflection of a Light Ray from its Rectilinear Motion - Wikisource, the free online library
    He talks of a hyperbola, shouldn't the angle be calculated by an integration of the normal force?
    Last edited by whizkid; July 20th, 2014 at 02:21 AM.
    Reply With Quote  
     

  7. #6  
    Forum Cosmic Wizard
    Join Date
    Dec 2013
    Posts
    2,408
    Quote Originally Posted by xyzt View Post
    Quote Originally Posted by dan hunter View Post
    ...and yet I think it should because of the einsten frequency shift.
    You are mixing up two DIFFERENT effects.

    Nothing to do with the light deflection.


    OK. I read a bit more.
    So instead of chromatic aberrations the Einstein Redshift produces Temporal aberrations?
    Reply With Quote  
     

  8. #7  
    Suspended
    Join Date
    Feb 2013
    Posts
    1,774
    Quote Originally Posted by whizkid View Post
    I read the original paper by Soldner : On the Deflection of a Light Ray from its Rectilinear Motion - Wikisource, the free online library
    If I got it right he talks of a hyperbola, isn't it rather a parabola, since the two branches from infinity to the tangent point are symmetric?

    Shouldn't the angle be calculated by an integration rather by a tangent?
    Soldner calculation is known to be wrong. You can read the correct calculation in any modern textbook, I recommend that you get Rindler's "Relativitty: Special , General and Cosmological".
    Reply With Quote  
     

  9. #8  
    Moderator Moderator Markus Hanke's Avatar
    Join Date
    Nov 2011
    Location
    Ireland
    Posts
    7,302
    Quote Originally Posted by whizkid View Post
    Shouldn't the angle be calculated by an integration rather by a tangent?
    I have some years ago done the calculation for gravitational light deflection from start to finish - here it is :

    https://www.dropbox.com/s/50rswg7ft7...ction%20SM.pdf

    Gravitational red shift is an effect along the radial direction, whereas gravitational light deflection is an effect along the azimuthal angle. They are not the same things.
    Last edited by Markus Hanke; July 19th, 2014 at 01:27 AM.
    Reply With Quote  
     

  10. #9  
    KJW
    KJW is offline
    Forum Professor
    Join Date
    Jun 2013
    Posts
    1,431
    Quote Originally Posted by whizkid View Post
    When light is deflected by gravity, (as when passing near the Sun) does frequency influence the value of the angle of deflection ?
    No. A consequence of the equivalence principle is that the effect of gravity on light is the same for all frequencies of the light.
    There are no paradoxes in relativity, just people's misunderstandings of it.
    Reply With Quote  
     

  11. #10  
    Suspended
    Join Date
    Feb 2013
    Posts
    1,774
    Quote Originally Posted by Markus Hanke View Post
    Quote Originally Posted by whizkid View Post
    Shouldn't the angle be calculated by an integration rather by a tangent?
    I have some years ago done the calculation for gravitational light deflection from start to finish - here it is :

    https://www.dropbox.com/s/50rswg7ft7...ction%20SM.pdf

    Gravitational light shift is an effect along the radial direction, whereas gravitational light deflection is an effect along the azimuthal angle. They are not the same things.
    Beautiful!
    You can add to your paper the calculation of the Shapiro delay, this would be, in my opinion the most elegant calculation.
    Start with the fact that the trajectory is a parabola of equation that extends from to , makes an angle with the radial direction (you have already calculated the deflection angle in your paper) and has the value (the "closest impat parameter).
    Use the conditions to get the parameters:





    Now, calculate the length of the arc of the parabola .

    Divide by c in order to get the Shapiro delay.
    Last edited by Howard Roark; July 18th, 2014 at 05:32 PM.
    Reply With Quote  
     

  12. #11  
    Suspended
    Join Date
    Feb 2013
    Posts
    1,774
    Quote Originally Posted by KJW View Post
    Quote Originally Posted by whizkid View Post
    When light is deflected by gravity, (as when passing near the Sun) does frequency influence the value of the angle of deflection ?
    No. A consequence of the equivalence principle is that the effect of gravity on light is the same for all frequencies of the light.
    Otherwise the gravitational light deflection would look like plain vanilla diffraction. Which it doesn't.
    Reply With Quote  
     

  13. #12  
    Forum Cosmic Wizard
    Join Date
    Dec 2013
    Posts
    2,408
    So does the Shapiro Delay have anything to do with temporal distortions through moving gravitational lenses?
    Reply With Quote  
     

  14. #13  
    Suspended
    Join Date
    Feb 2013
    Posts
    1,774
    Quote Originally Posted by dan hunter View Post
    So does the Shapiro Delay have anything to do with temporal distortions through moving gravitational lenses?
    The sentence makes no sense, try rephrasing your question.
    Reply With Quote  
     

  15. #14  
    Forum Cosmic Wizard
    Join Date
    Dec 2013
    Posts
    2,408
    Quote Originally Posted by xyzt View Post
    Quote Originally Posted by dan hunter View Post
    So does the Shapiro Delay have anything to do with temporal distortions through moving gravitational lenses?
    The sentence makes no sense, try rephrasing your question.
    For clarification see:
    Aberration by gravitational lenses in motion
    Simonetta Frittelli
    Department of Physics, Duquesne University, Pittsburgh, PA 15282
    19 December 2013

    It appears the answer should be yes.
    Reply With Quote  
     

  16. #15  
    Suspended
    Join Date
    Feb 2013
    Posts
    1,774
    Quote Originally Posted by dan hunter View Post
    Quote Originally Posted by xyzt View Post
    Quote Originally Posted by dan hunter View Post
    So does the Shapiro Delay have anything to do with temporal distortions through moving gravitational lenses?
    The sentence makes no sense, try rephrasing your question.
    For clarification see:
    Aberration by gravitational lenses in motion
    Simonetta Frittelli
    Department of Physics, Duquesne University, Pittsburgh, PA 15282
    19 December 2013

    It appears the answer should be yes.
    The paper simply shows that there is an additional effect of the order of in the Shapiro delay when the gravitational object inducing the lensing is moving.
    Reply With Quote  
     

  17. #16  
    Forum Freshman
    Join Date
    May 2014
    Posts
    86
    Quote Originally Posted by dan hunter View Post
    So does the Shapiro Delay have anything to do with temporal distortions through moving gravitational lenses?
    The Shapiro delay is a GR effect based the same principle as gravitational lensing and frame dragging. It is essentially frame dragging in reverse. With frame dragging, a rapidly rotating massive body will accelerate the motion of a smaller body within its gravitational field and the acceleration is in the same direction as the rotation of the more massive body. The gravitational field of the massive body 'drags' the smaller body along with it. The orbit of Mercury about the sun is a familiar example of frame dragging. The Shapiro delay works in the opposite direction where a rapidly moving body enters the gravitational field of a slow or non-rotating body and the motion of the fast moving object is slowed as it passes through a static gravitational field. A non-rotating body is one that is not rotating relative to the distant galaxies.
    There is also a cosmological Shapiro delay where any rapidly moving object is slowed by its passage through the static gravitational field of the universe as a whole. This is a bit contrary to Newton's law where moving objects remain in motion. The cosmological Shapiro delay has yet to be observed. Possibly because it is too slight to be observed by present day technology but some think it may be a contributing factor to the redshifting of distant galaxies.
    Reply With Quote  
     

  18. #17  
    Forum Cosmic Wizard
    Join Date
    Dec 2013
    Posts
    2,408
    Quote Originally Posted by bangstrom View Post
    Quote Originally Posted by dan hunter View Post
    So does the Shapiro Delay have anything to do with temporal distortions through moving gravitational lenses?
    The Shapiro delay is a GR effect based the same principle as gravitational lensing and frame dragging. It is essentially frame dragging in reverse. With frame dragging, a rapidly rotating massive body will accelerate the motion of a smaller body within its gravitational field and the acceleration is in the same direction as the rotation of the more massive body. The gravitational field of the massive body 'drags' the smaller body along with it. The orbit of Mercury about the sun is a familiar example of frame dragging. The Shapiro delay works in the opposite direction where a rapidly moving body enters the gravitational field of a slow or non-rotating body and the motion of the fast moving object is slowed as it passes through a static gravitational field. A non-rotating body is one that is not rotating relative to the distant galaxies.
    There is also a cosmological Shapiro delay where any rapidly moving object is slowed by its passage through the static gravitational field of the universe as a whole. This is a bit contrary to Newton's law where moving objects remain in motion. The cosmological Shapiro delay has yet to be observed. Possibly because it is too slight to be observed by present day technology but some think it may be a contributing factor to the redshifting of distant galaxies.
    That is interesting. Can you tell me anything about how it works with moving lenses?

    I also wonder if the cosmological Shapiro delay would not just average out? Since the Einstein shift has to be observed from different gravitational wells with the difference between the deppth of the wells determining the degree and colour of the frequency shift shouldn't they simply cancel out on average?
    Reply With Quote  
     

  19. #18  
    Forum Junior whizkid's Avatar
    Join Date
    Jan 2013
    Posts
    282
    Thanks Markus, the discussion is very interesting, but I wanted to discuss the classical method.
    If someone can understand the rationale of the cited paper, please explain it to me, and tell me if with the current parameters 0.84 is still valid ,following that method.

    I would like to consider light as any other orbiting body, and the rationale should be:

    R = 7.5 *10^10 = 2.5 C, C/R = .4
    In order to bend the path by 1 radian F_r must be 0.4 C = 1.2*10^10 by 1'' .4 C/206265 F_1'' = 58 177
    the pull at the tangent (cos alpha = 1) is 23600
    the correct way to find the angle is deflection is to integrate the force normal to the direction of the light.
    Is this correct?
    Is this what Soldner did ? I saw no integration
    Reply With Quote  
     

  20. #19  
    Suspended
    Join Date
    Feb 2013
    Posts
    1,774
    Quote Originally Posted by whizkid View Post
    Thanks Markus, the discussion is very interesting, but I wanted to discuss the classical method.
    If someone can understand the rationale of the cited paper, please explain it to me, and tell me if with the current parameters 0.84 is still valid ,following that method.

    I would like to consider light as any other orbiting body, and the rationale should be:

    R = 7.5 *10^10 = 2.5 C, C/R = .4
    In order to bend the path by 1 radian F_r must be 0.4 C = 1.2*10^10 by 1'' .4 C/206265 F_1'' = 58 177
    the pull at the tangent (cos alpha = 1) is 23600
    the correct way to find the angle is deflection is to integrate the force normal to the direction of the light.
    Is this correct?
    Is this what Soldner did ? I saw no integration
    The fact that you can't recognize a simple integration shouldn't stop you from spamming. Contrary to your nonsense, Soldner integrates an ODE. Homework: can you find where? Hint: look for:

    "and if we again integrate, we will obtain:

    "
    Reply With Quote  
     

  21. #20  
    Forum Junior whizkid's Avatar
    Join Date
    Jan 2013
    Posts
    282
    Quote Originally Posted by xyzt View Post
    Soldner integrates an ODE. "
    I do not know calculus, let alone differentials, is it the integration of normal force?
    What I asked is if the rationale of Soldner is the one I described and if with current parameters we get same value of .84''

    Integrating force the way I showed, if the logic is correct, we get a much greater value , greater than GR

    P.S. It would be advisable that you stopped insulting or showing off, it doesn't do justice to your intelligence.
    Reply With Quote  
     

  22. #21  
    Suspended
    Join Date
    Feb 2013
    Posts
    1,774
    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by xyzt View Post
    Soldner integrates an ODE. "
    I do not know calculus, let alone differentials.
    Yes, that much is pretty obvious. You should try learning them. Much better use of your time than spamming threads on subjects that are way out of your level of understanding.





    Integrating force the way I showed, if the logic is correct, we get a much greater value , greater than GR
    There is no "integrating of force" in the method. You do not understand the physics because you cannot follow the simple calculus in the paper.
    Reply With Quote  
     

  23. #22  
    Forum Junior whizkid's Avatar
    Join Date
    Jan 2013
    Posts
    282
    Quote Originally Posted by whizkid View Post
    I would like to consider light as any other orbiting body, and the rationale should be:

    R = 7.5 *10^10 = 2.5 C, C/R = .4
    In order to bend the path by 1 radian F_r must be 0.4 C = 1.2*10^10 by 1'' .4 C/206265 F_1'' = 58 177
    the pull at the tangent (cos alpha = 1) is F_t = 23600

    the way to find the angle is deflection is to integrate the force normal to the direction of the light.
    Is this correct?
    Is this approach accepatable in classical physics? What is wrong in it?
    In order to get the correct GR value we must get F = 102 000, is this right?

    I do not understand why Soldner considers both sin /cos phi
    What is relevant here is F, which is cos phi or , better, CA/CM which multiplied by F_x (CM being the x-axis) gives the the value of the force at each point x

    If this approach is correct we need only one simple formula instead of the log derivation by Soldner
    Last edited by whizkid; July 20th, 2014 at 03:03 AM.
    Reply With Quote  
     

  24. #23  
    Moderator Moderator Markus Hanke's Avatar
    Join Date
    Nov 2011
    Location
    Ireland
    Posts
    7,302
    I do not know calculus, let alone differentials
    If you do not know calculus then you will find it very difficult to discuss this subject. I would advise you to first get a very solid grounding in multivariable calculus and linear algebra before getting involved with differential geometry and GR.

    Integrating force the way I showed, if the logic is correct, we get a much greater value , greater than GR
    I am not sure what you are trying to say, but there are no forces in GR. The derivation of gravitational light deflection is done by calculating the null geodesic along which light propagates, and then find the angle between their asymptotes. It's purely geometrical.
    Reply With Quote  
     

  25. #24  
    Forum Junior whizkid's Avatar
    Join Date
    Jan 2013
    Posts
    282
    Quote Originally Posted by Markus Hanke View Post
    a before getting involved with differential geometry and GR.
    ..I am not sure what you are trying to say, but there are no forces in GR. .
    Thanks Markus, but I would not dare to attempt to meddle with GR.

    I said that I want to discuss if the Soldner method is correct, I 'd like to understand what he did and if that value is correct with current parameters.

    Besides that, if nobody can tell me that, I'd like to know if my approach is valid in classical physics, applying to light the same laws we use for massive bodies : F = v^2/r
    The GR-Einstein value of 1.75'' corresponds in classical physics to F = 102000,
    I think we can get a similar value by classical physics, I do not know where Soldner went wrong, but only in the 5 seconds light takes to cover two R (1.5*10^11cm) while at the tangent it gets a force in the excess of 83000

    If you or anyone else is prepared to do that, I 'll explain my approach again, but I suppose it is standard physics.
    I do integrations at Wolfram, and so make up for my ignorance.

    Any idea is welcome
    Last edited by whizkid; July 20th, 2014 at 03:44 AM.
    Reply With Quote  
     

  26. #25  
    KJW
    KJW is offline
    Forum Professor
    Join Date
    Jun 2013
    Posts
    1,431
    Quote Originally Posted by bangstrom View Post
    The orbit of Mercury about the sun is a familiar example of frame dragging.
    This is incorrect. The calculation of the perihelion precession of Mercury's orbit does not use the rotation of the sun as a parameter. Indeed, the perihelion precession of Mercury's orbit is the result of the general relativistic deviation from the perfect inverse square law, this deviation causing non-conservation of the Laplace-Runge-Lenz vector. Frame dragging is an extremely tiny effect that has only recently been tested by Gravity Probe B.
    There are no paradoxes in relativity, just people's misunderstandings of it.
    Reply With Quote  
     

  27. #26  
    Forum Freshman
    Join Date
    May 2014
    Posts
    86
    Quote Originally Posted by dan hunter View Post
    Can you tell me anything about how it works with moving lenses?

    I also wonder if the cosmological Shapiro delay would not just average out? Since the Einstein shift has to be observed from different gravitational wells with the difference between the deppth of the wells determining the degree and colour of the frequency shift shouldn't they simply cancel out on average?
    The Shapiro delay only applies to slow or non-moving gravitational fields but even in this case it would have a negligible effect on gravitational lensing.

    As you mentioned, gravitational effects should cancel as objects or even light move in and out of a gravitational field but the cosmological Shapiro delay is a different animal because it involves the entire gravitational field of the universe where there is only an in but no out. We can't escape from the gravity of some 200 billion galaxies. There may appear to be zero gravity in deep space because gravity is equal in all directions but there is no place in space where there is no gravity. Gravity is curved spacetime and the entire universe is a spacetime curved upon itself. It is this static gravitational field that is suspected to have a Shapiro delaying effect on all moving bodies.
    Reply With Quote  
     

  28. #27  
    Forum Cosmic Wizard
    Join Date
    Dec 2013
    Posts
    2,408
    In some of the stuff I was reading there seemed to be a suggestion the Shapiro delay could account for the Hubble redshift.
    (Some of the claims seemed to focus on the idea that there is about six times as much dark matter as visible matter in the universe.)
    Is this possible?

    Also, if the universe is spherical (or nearly spherical) shouldn't most of the gravitational field inside the universe cancel naturally just as it would inside a thin shell?
    Reply With Quote  
     

  29. #28  
    Suspended
    Join Date
    Feb 2013
    Posts
    1,774
    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by Markus Hanke View Post
    a before getting involved with differential geometry and GR.
    ..I am not sure what you are trying to say, but there are no forces in GR. .
    Thanks Markus, but I would not dare to attempt to meddle with GR.

    I said that I want to discuss if the Soldner method is correct, I 'd like to understand what he did and if that value is correct with current parameters.

    Besides that, if nobody can tell me that, I'd like to know if my approach is valid in classical physics, applying to light the same laws we use for massive bodies : F = v^2/r
    The GR-Einstein value of 1.75'' corresponds in classical physics to F = 102000,
    I think we can get a similar value by classical physics, I do not know where Soldner went wrong, but only in the 5 seconds light takes to cover two R (1.5*10^11cm) while at the tangent it gets a force in the excess of 83000

    If you or anyone else is prepared to do that, I 'll explain my approach again, but I suppose it is standard physics.
    I do integrations at Wolfram, and so make up for my ignorance.

    Any idea is welcome
    Soldner method, using Newtonian mechanics produces an incorrect answer.
    GR predicts TWICE the angular deviation compared to Soldner method. Experiment confirms GR and contradicts Soldner.
    Even if you want to reproduce Soldner method you need to LEARN calculus, you cannot do it by using 9-th grade algebra as you keep trying. You cannot reduce a problem to your level of understanding.
    Last edited by Howard Roark; July 20th, 2014 at 09:35 AM.
    Reply With Quote  
     

  30. #29  
    Forum Freshman
    Join Date
    May 2014
    Posts
    86
    Quote Originally Posted by dan hunter View Post
    In some of the stuff I was reading there seemed to be a suggestion the Shapiro delay could account for the Hubble redshift.
    (Some of the claims seemed to focus on the idea that there is about six times as much dark matter as visible matter in the universe.)
    Is this possible?

    Also, if the universe is spherical (or nearly spherical) shouldn't most of the gravitational field inside the universe cancel naturally just as it would inside a thin shell?
    If you were in a capsule at the center of the Earth you could float weightlessly because gravity is acting equally in all directions but this does not mean that there is no gravity at the center of the Earth. Gravitational fields add rather than cancel in much the same way that air or water pressures increase with depth. We don't notice the air pressure around us because it is equal in all directions but the air pressure is an essential part of our environment and gravitational effects exists even when they are acting in all directions.

    Clocks run slower in a strong gravitational field than in one that is weak so a clock should run slower at the center of the Earth than at the surface and a clock at the surface should run slower than a clock in space. Likewise, the Shapiro delay should increase with the total strength of the gravitational field even when it is equal in all directions.

    Some speculation has come from neo-Machian physics that the Shapiro delay may be a part of the Hubble redshift. Amitabha Ghosh has even suggested that the universe is not expanding and the Hubble redshift is entirely Shapiro effect even without considering the possibility of exotic dark matter. The presence of exotic dark matter would amplify the effect but all this is hypothetical at this time. The part of this that I find incredible is the idea that there is six times as much dark matter as visible matter.
    Reply With Quote  
     

  31. #30  
    Forum Freshman Laurieag's Avatar
    Join Date
    Apr 2013
    Posts
    93
    Quote Originally Posted by bangstrom View Post
    The part of this that I find incredible is the idea that there is six times as much dark matter as visible matter.
    I also find it interesting that the ratio of (DM + VM)/VM seems to be converging on 2 * Pi as more accurate results come in.

    It's like what you would expect to see if rest mass and relative mass were being conflated on a universal scale (i.e. 2 * Pi is the difference between the reduced Compton wavelength and the reduced Planck constant and their standard counterparts).
    Reply With Quote  
     

  32. #31  
    Forum Cosmic Wizard
    Join Date
    Dec 2013
    Posts
    2,408
    Quote Originally Posted by bangstrom View Post
    If you were in a capsule at the center of the Earth you could float weightlessly because gravity is acting equally in all directions but this does not mean that there is no gravity at the center of the Earth. Gravitational fields add rather than cancel in much the same way that air or water pressures increase with depth. We don't notice the air pressure around us because it is equal in all directions but the air pressure is an essential part of our environment and gravitational effects exists even when they are acting in all directions....
    Actually I do not think this part of what you are saying is quite right.
    Shell theorem - Wikipedia, the free encyclopedia
    Reply With Quote  
     

  33. #32  
    Forum Junior whizkid's Avatar
    Join Date
    Jan 2013
    Posts
    282
    I'll start a new thread. please reply if you answer directly the questions
    Last edited by whizkid; July 21st, 2014 at 03:07 AM.
    Reply With Quote  
     

  34. #33  
    Moderator Moderator Janus's Avatar
    Join Date
    Jun 2007
    Posts
    2,174
    Quote Originally Posted by bangstrom View Post
    The presence of exotic dark matter would amplify the effect but all this is hypothetical at this time. The part of this that I find incredible is the idea that there is six times as much dark matter as visible matter.
    Since there does appear to be many times more DM than baryonic matter, Maybe it us that is made of the "exotic" matter and DM is the "norm".
    "Men are apt to mistake the strength of their feelings for the strength of their argument.
    The heated mind resents the chill touch & relentless scrutiny of logic"-W.E. Gladstone


    Edit/Delete Message
    Reply With Quote  
     

  35. #34  
    Forum Freshman
    Join Date
    May 2014
    Posts
    86
    Quote Originally Posted by dan hunter View Post
    Quote Originally Posted by bangstrom View Post
    If you were in a capsule at the center of the Earth you could float weightlessly because gravity is acting equally in all directions but this does not mean that there is no gravity at the center of the Earth. Gravitational fields add rather than cancel in much the same way that air or water pressures increase with depth. We don't notice the air pressure around us because it is equal in all directions but the air pressure is an essential part of our environment and gravitational effects exists even when they are acting in all directions....
    Actually I do not think this part of what you are saying is quite right.
    Shell theorem - Wikipedia, the free encyclopedia
    I don't see how the Shell theorem applies. The gravitational vectors may all go to zero at the center of a spherical body but this does not mean that gravity does not exist at that point. Where did it go? And when the spherical body is the universe itself, the universe is a 4-D hypersphere and every point within that sphere is essentially "at the center" where the gravitational forces are equal in all directions so we don't notice their effect. We only only notice the local effects of gravity from objects like the Earth, sun, and moon. Mach demonstrated the overwhelming effect of the total gravitational mass of the universe with a gyroscope. A freely rotating gyroscope orients itself with what Mach called the "combined masses of the distant stars" and a gyroscope's orientation rotates in sidereal time as the Earth rotates beneath it. In other words, a gyroscope always points to the same position in the heavens and it is not influenced by the rotation of the Earth or by the rotation of its surroundings if it is on a ship or an airplane. The local gravitational effects of the Earth and sun are not strong enough to effect the orientation of a gyroscope because a gyroscope aligns with the total mass of the universe which is far, far greater.
    Reply With Quote  
     

  36. #35  
    Forum Cosmic Wizard
    Join Date
    Dec 2013
    Posts
    2,408
    Quote Originally Posted by bangstrom View Post
    Quote Originally Posted by dan hunter View Post
    Quote Originally Posted by bangstrom View Post
    If you were in a capsule at the center of the Earth you could float weightlessly because gravity is acting equally in all directions but this does not mean that there is no gravity at the center of the Earth. Gravitational fields add rather than cancel in much the same way that air or water pressures increase with depth. We don't notice the air pressure around us because it is equal in all directions but the air pressure is an essential part of our environment and gravitational effects exists even when they are acting in all directions....
    Actually I do not think this part of what you are saying is quite right.
    Shell theorem - Wikipedia, the free encyclopedia
    I don't see how the Shell theorem applies. The gravitational vectors may all go to zero at the center of a spherical body but this does not mean that gravity does not exist at that point. Where did it go? And when the spherical body is the universe itself, the universe is a 4-D hypersphere and every point within that sphere is essentially "at the center" where the gravitational forces are equal in all directions so we don't notice their effect. We only only notice the local effects of gravity from objects like the Earth, sun, and moon. Mach demonstrated the overwhelming effect of the total gravitational mass of the universe with a gyroscope. A freely rotating gyroscope orients itself with what Mach called the "combined masses of the distant stars" and a gyroscope's orientation rotates in sidereal time as the Earth rotates beneath it. In other words, a gyroscope always points to the same position in the heavens and it is not influenced by the rotation of the Earth or by the rotation of its surroundings if it is on a ship or an airplane. The local gravitational effects of the Earth and sun are not strong enough to effect the orientation of a gyroscope because a gyroscope aligns with the total mass of the universe which is far, far greater.
    It is a common mistake to think that gravity only goes to zero at the centerpoint.
    Inside of a shell gravity is canceled at everypoint inside the shell, even if the point is against the inside wall of the shell.
    Reply With Quote  
     

  37. #36  
    Forum Freshman
    Join Date
    May 2014
    Posts
    86
    Quote Originally Posted by Janus View Post
    Since there does appear to be many times more DM than baryonic matter, Maybe it us that is made of the "exotic" matter and DM is the "norm".
    I have a bridge in New York made of EDM if anyone interested in buying it?
    Reply With Quote  
     

  38. #37  
    Forum Freshman
    Join Date
    May 2014
    Posts
    86
    Quote Originally Posted by dan hunter View Post
    It is a common mistake to think that gravity only goes to zero at the centerpoint.
    Inside of a shell gravity is canceled at everypoint inside the shell, even if the point is against the inside wall of the shell.
    This is true if your massive body is a hollow shell like a Van de Graff generator where the electrical charges all go to zero on the inside but the universe is not a hollow shell.
    Reply With Quote  
     

  39. #38  
    Forum Cosmic Wizard
    Join Date
    Dec 2013
    Posts
    2,408
    Quote Originally Posted by bangstrom View Post
    Quote Originally Posted by dan hunter View Post
    It is a common mistake to think that gravity only goes to zero at the centerpoint.
    Inside of a shell gravity is canceled at everypoint inside the shell, even if the point is against the inside wall of the shell.
    This is true if your massive body is a hollow shell like a Van de Graff generator where the electrical charges all go to zero on the inside but the universe is not a hollow shell.
    It also holds true for solid bodies because they can be considered as a series of hollow shells.
    Reply With Quote  
     

  40. #39  
    Forum Freshman
    Join Date
    May 2014
    Posts
    86
    Quote Originally Posted by dan hunter View Post
    It also holds true for solid bodies because they can be considered as a series of hollow shells.
    A series of hollow shells is a solid. The instant you pass through the first shell, gravity does not go to zero because you are still on the outside of a myriad of other shells. In a solid sphere, gravity is not equal in all directions until you reach the center of the center shell.
    Reply With Quote  
     

  41. #40  
    Suspended
    Join Date
    Feb 2013
    Posts
    1,774
    Quote Originally Posted by bangstrom View Post
    Quote Originally Posted by dan hunter View Post
    It also holds true for solid bodies because they can be considered as a series of hollow shells.
    A series of hollow shells is a solid. The instant you pass through the first shell, gravity does not go to zero because you are still on the outside of a myriad of other shells. In a solid sphere, gravity is not equal in all directions until you reach the center of the center shell.
    Depends what you mean by "gravity". If you mean "gravitational force", that is zero anywhere inside the shell, not only at the center. See here for exact proof.
    If, by "gravity" you mean "gravitational potential", that is not zero, indeed.
    So, what do you mean?
    Reply With Quote  
     

  42. #41  
    Forum Freshman
    Join Date
    May 2014
    Posts
    86
    The gravity I have in mind is both the net gravitational force of the total mass of the universe with its two billion some galaxies and the total gravitational potential of all the masses in the universe. I am not including anything smaller than the Whole Shebang. That is the "Big M" in calculations where M represents the total mass of the universe.

    The Shell theorem applies to smaller 3-D spheres that have a measurable radius and center unlike the 4-D universe that surrounds us. The Shell theorem can be used for calculating things like the weight of a miner deep inside a mine but it does not apply in the same way when the sphere under consideration is the entire universe.

    The net gravitational attraction of all the distant galaxies is essentially zero because it is uniform in all directions. We are at the apparent center of the universe and inside all the shells where the net gravity goes to zero but a remote observer in a far corner corner of the universe could make exactly the same observation. We are not within a 3-D sphere where one can calculate his distance from the center.

    The net gravitational attraction from the distant galaxies may be zero but we are still under the influence of their total gravitational potential which is considerable. It influences the rate at which time passes and our measurements of distance and all the other GR effects that we associate with a gravitational potential including the Shapiro delay.
    Reply With Quote  
     

  43. #42  
    Forum Cosmic Wizard
    Join Date
    Dec 2013
    Posts
    2,408
    Quote Originally Posted by bangstrom View Post
    Quote Originally Posted by dan hunter View Post
    It also holds true for solid bodies because they can be considered as a series of hollow shells.
    A series of hollow shells is a solid. The instant you pass through the first shell, gravity does not go to zero because you are still on the outside of a myriad of other shells. In a solid sphere, gravity is not equal in all directions until you reach the center of the center shell.
    So? As you said yourself " And when the spherical body is the universe itself, the universe is a 4-D hypersphere and every point within that sphere is essentially "at the center" where..." and since the universe is infinite we are at the center of an infinite series of shells no matter where we are so I think the point becomes a bit moot since the effect is the same.

    Xyzt's point about gravitational potential is worth noting too because I have been thinking in terms of force which relates to acceleration (or curvature) instead of gravitational potential which relates to work.
    Gravitational potential is negative almost everywhere and increasingly negative the deeper down a gravity well you are.



    Reply With Quote  
     

  44. #43  
    Moderator Moderator Markus Hanke's Avatar
    Join Date
    Nov 2011
    Location
    Ireland
    Posts
    7,302
    Quote Originally Posted by dan hunter View Post

    Xyzt's point about gravitational potential is worth noting too because I have been thinking in terms of force which relates to acceleration (or curvature) instead of gravitational potential which relates to work.

    Just to add to what has already been well explained by xyzt and others - the gravitational net force in the interior of a thin shell is everywhere zero, so a massive object placed anywhere inside the shell remains at rest. The gravitational potential is the same everywhere in the interior region of such a shell ( obviously, or else the net force couldn't vanish ). In terms of GR, space-time in the interior of a thin shell is completely flat and Minkowskian ( which follows directly from Birkhoff's theorem ).

    However, globally speaking, the uniform gravitational potential in the interior region of the shell has a value that is different from that of a reference observer at infinity; this means that a clock placed into the interior of a shell is gravitationally time dilated with respect to a far-away observer, even though space-time in that interior region is completely flat. This can be the source of immense confusion if not properly understood, hence I thought I'd point it out here.
    Reply With Quote  
     

  45. #44  
    Suspended
    Join Date
    Feb 2013
    Posts
    1,774
    Thank you, Markus



    Here
    is an excellent presentation that explains the variation of the gravitational potential on the Earth surface.
    Reply With Quote  
     

  46. #45  
    Forum Freshman
    Join Date
    May 2014
    Posts
    86
    Quote Originally Posted by Markus Hanke View Post
    Just to add to what has already been well explained by xyzt and others - the gravitational net force in the interior of a thin shell is everywhere zero, so a massive object placed anywhere inside the shell remains at rest. The gravitational potential is the same everywhere in the interior region of such a shell ( obviously, or else the net force couldn't vanish ). In terms of GR, space-time in the interior of a thin shell is completely flat and Minkowskian ( which follows directly from Birkhoff's theorem ).

    However, globally speaking, the uniform gravitational potential in the interior region of the shell has a value that is different from that of a reference observer at infinity; this means that a clock placed into the interior of a shell is gravitationally time dilated with respect to a far-away observer, even though space-time in that interior region is completely flat. This can be the source of immense confusion if not properly understood, hence I thought I'd point it out here.
    The physics of this appears clear as stated but could you clarify some of your terms? The term "thin shell" appears strange when speaking globally (universe wide) because the visible portion of our "thin shell" is the entire several billion ly of galaxies that surrounds us. Every observer at every point in the universe should find himself the apparent center of the universe so there is no part of our universe that is not a part of an observer's "thin shell." The term "thin" appears to be a bit of an understatement.

    And can you call it a "shell" when it has no outer surface?

    And, by "reference observer at infinity" would that be a god like observer ( outside ?) our universe?
    Reply With Quote  
     

  47. #46  
    Moderator Moderator Markus Hanke's Avatar
    Join Date
    Nov 2011
    Location
    Ireland
    Posts
    7,302
    Quote Originally Posted by bangstrom View Post
    The term "thin shell" appears strange when speaking globally (universe wide) because the visible portion of our "thin shell" is the entire several billion ly of galaxies that surrounds us.
    No, there seems to be a misunderstanding here. By "thin shell" I mean a hollow and perfectly spherical shell of ordinary matter and finite size; for example what you would get if you were to hollow out the Earth and just leave an empty shell a few kilometres thick. I wasn't referring to the universe as a whole, since Birkhoff's theorem does not apply to FLRW space-times. By "global" I just meant the totality of the shell, its interior, and a far-away observer, not the universe as it is modelled in cosmology. What I have said does not apply to the universe as a whole.
    Reply With Quote  
     

  48. #47  
    Forum Freshman
    Join Date
    May 2014
    Posts
    86
    Quote Originally Posted by Markus Hanke View Post
    No, there seems to be a misunderstanding here. By "thin shell" I mean a hollow and perfectly spherical shell of ordinary matter and finite size; for example what you would get if you were to hollow out the Earth and just leave an empty shell a few kilometres thick. I wasn't referring to the universe as a whole, since Birkhoff's theorem does not apply to FLRW space-times. By "global" I just meant the totality of the shell, its interior, and a far-away observer, not the universe as it is modelled in cosmology. What I have said does not apply to the universe as a whole.
    That makes sense and thanks for the clarification. You were NOT discussing the universe as a whole but I WAS discussing the universe as a whole and the misunderstanding predates your post. I was discussing the cosmological Shapiro delay and its possible effect on the Hubble redshift when Don Hunter introduced the Thin Shell Theorem in Post #31. I tried to explain in Post #35 that the Shell Theorem does not apply to the universe as a whole but the confusion continued. I thought you were also trying to apply the Shell Theorem as a cosmological effect which is why your phrasing sounded so strange to me.

    From Post #35
    Quote Originally Posted by bangstrom View Post
    I don't see how the Shell theorem applies. The gravitational vectors may all go to zero at the center of a spherical body but this does not mean that gravity does not exist at that point. Where did it go? And when the spherical body is the universe itself, the universe is a 4-D hypersphere and every point within that sphere is essentially "at the center" where the gravitational forces are equal in all directions so we don't notice their effect. We only only notice the local effects of gravity from objects like the Earth, sun, and moon.
    Reply With Quote  
     

Similar Threads

  1. What is trajectory Deflection Angle
    By Bjarne in forum Astronomy & Cosmology
    Replies: 3
    Last Post: January 27th, 2014, 01:16 AM
  2. Replies: 61
    Last Post: January 6th, 2014, 03:19 PM
  3. neglecting axial deformation is slope deflection method(Structural Engineering)
    By Sadeq in forum Mechanical, Structural and Chemical Engineering
    Replies: 0
    Last Post: August 9th, 2012, 06:19 AM
  4. Calculating the value of Deflection
    By Eagle9 in forum Mechanical, Structural and Chemical Engineering
    Replies: 1
    Last Post: March 26th, 2011, 06:02 AM
  5. Replies: 3
    Last Post: July 9th, 2008, 01:15 AM
Bookmarks
Bookmarks
Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •