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Does Newtonian gravity bend light?

  1. Jul 10, 2006 #1
    Looking for help interpreting a proof of G.R. please. . .

    One of the earliest proofs of G.R. was the deflection of light by a gravitational field, first shown in 1919 during a solar eclipse.

    I think Einstein predicted approximately 1.7 seconds of arc deflection during that eclipse. But in his book, Relativity (written for the general reader), Einstein says that "half of this deflection is produced by the Newtonian field of attraction of the Sun, and the other half by the geometrical curvature of space"

    1) Does that mean that classical Newtonian gravitational theory would have predicted roughly 0.8 seconds of arc deflection of light, and G.R. predicts 1.7?

    But wasn't light considered massless to Newton?

    If so, how does Newton's gravity theory predict the attraction of light?

    2) If the answer is 'no' to #1, then what does Einstein mean by the quote above?

    thanks !

  2. jcsd
  3. Jul 10, 2006 #2


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  4. Jul 11, 2006 #3
    Hi smithpa9,

    In Classical Mechanics, the formula for gravitational acceleration is :

    F = Gmm'/r²

    According to Newton's second law, F = ma, we find :

    a = Gm'/r²

    This equation does not imply the attracted object's mass. So, even if a photon is massless, according to this last equation, light should be curved by an astronomical object. After calculation, we find that light should be curved.

    GR introduces some corrections (relativistic effects). Compare GR light curvature to classical light curvature and you find a quotient equal at 2.
  5. Jul 11, 2006 #4
    Hi Zeit,

    I think there is a problem.

    In Classical Mechanics, the formula for coulomb acceleration is :

    F = m'qq'/r²

    According to Newton's second law, F = ma, we find :

    a = m'q'/r²

    This equation does not imply the attracted object's mass. So, even if a neutron is chargeless, according to this last equation, neutrons should be curved by an electrically charged object.

    ----------> I don't think a massless photon would be affected by Newtonian gravitation under your argument
  6. Jul 11, 2006 #5


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    Wasn't it Laplace who calculated that (using Newton's gravitation and his corpuscular definition of light) a sufficiently massive body would have an escape speed greater than c?
  7. Jul 11, 2006 #6


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  8. Jul 11, 2006 #7

    Coulomb's formula is :

    F = (1/4πε)(qq'/r²)

    Where q and q' are charges (coulomb :C) of two motionless particles , ε is the electric constant ε≈8.854 C/Vm (coulomb per volt-meter).

    Because Newton's second law, we found :

    a = F/m = (1/4πε)(qq'/r²)(1/m)

    I have never seen this equation somewhere, but the dimension r/t² is good.

    However, gravitational field is Gm'/r² and electric field is (1/4πε)(q'/r²). So, the electric field is not, at the contrary of the gravitational field, an acceleration. But the effect is the same : it describes the force applied in any point in space.

    Yes, I think so (robphy alredy gave us a link). But, the Dark Star concept was different than the concept of black hole of course.


    EDIT : You can look at this, especially at "Parallels between electrostatics and gravity" : http://en.wikipedia.org/wiki/Electric_field" [Broken]
    Last edited by a moderator: May 2, 2017
  9. Jul 11, 2006 #8
    Zeit - In the Newtonian formula for Gravitational force you gave,

    F = Gmm'/r²

    So, for my question about if light bends when close to the Sun under Newtonian gravity theory, I think m = mass of Sun (or photon of light) and m' = mass of photon of light (or Sun).

    If either one of these masses is zero, then the right hand size of the equation = 0, hence F=0, hence no Force of gravity between the two, hence no bending of light.

    What's wrong with my logic?
    Last edited: Jul 11, 2006
  10. Jul 11, 2006 #9
    OK, I've thought about this a little more. . . let's see if this is correct:

    Prior to SPECIAL relativity, photons were considered massless, always. So everyone before Einstein would have believed that light does NOT bend near a massive object, like the Sun.

    Nowdays, photons are thought to have a rest-mass of zero, but of course you never find one "resting." At the speed of light, photons have some mass due to their speed, hence energy and the equivalence of mass and energy (E=mc^2). (True???) so, they would bend due to gravity near the Sun, and would do so approximately 0.8 arc-seconds.

    Then, as part of Einstein's discussion of GENERAL relativity, he postulated that light passing near the Sun would bend EVEN FURTHER due to the additional effect of the curvature of space, which added another 0.9 arc-seconds of bending, for a total of 1.7. ???

    In sum:

    1) under purely pre-Relativity mechanics, light would NOT be thought to bend near the Sun. Photons are massless, space-time is flat, hence no bending of light for any reason.

    2) considering only SPECIAL RELATIVITY, photons have energy and hence mass-equivalence, and are therefore affected by gravity. Photons have mass, spacetime is still flat however, hence, SOME bending of light near the Sun.

    3) considering SPECIAL and GENERAL RELATIVITY, photons have mass, spacetime is curved, hence MORE bending of light near the Sun, for both reasons.

    Does this sound right?
  11. Jul 12, 2006 #10


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    No. E = mc^2 relates the rest energy of an object to its rest mass, i.e., it relates energy in the reference frame in which the object is at rest to the invariant mass of the object. There is no such thing as a rest frame for a photon - it has no rest energy, and no rest mass.

    See https://www.physicsforums.com/showthread.php?t=125613"

    Additionally, things with mass cannot travel at c. In the same vein, anything massless must travel at c.
    Last edited by a moderator: Apr 22, 2017
  12. Jul 12, 2006 #11


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    With a=F/m, a stays constant, regardless of m'. In the limit m'->0, a is still the same. At m'=0, a is mathematically defined by L'Hôpital's rule, if you prefer not to cancel it earlier.
  13. Jul 12, 2006 #12
    Hi everybody,

    No, it means that you haven't to exert force to bend light. As Ich noted, a = F/m and it's a "constant". Because F = Gmm'/r² (m' is Sun's mass here), g = Gm'/r² and is independent from the attracted object's mass, even if it's a photon, you or Earth. Because when m increases, F does too ; when m decreases, F does too. However, it suggests that we don't need the force concept to describe gravitation, only the gravitational field concept. It's closer to space-time curvature than to Newton's force.

    Here a link (in French... I haven't found one in English about our subject) :
    http://home.tiscali.be/jp.delavallee/Gravitation/DeviationLumiere/deviation_lumiere.htm" [Broken]

    The entire equation for energy in special relativity is :

    E² = (pc)² + (mc²)²

    where, usually, p = ymv (y is Lorentz factor).

    If the object is motionless, (pc)² = 0, so we have E = mc². If the object is massless, (mc²)² = 0, so E = pc.

    But, using the "normal" formula for momentum, E = 0. We used de Broglie formula for momentum : p = h/λ , where h is Planck constant and λ is wavelenght.

    You can't use m = E/c²


    PS : Here another link about deflection of light : http://www.mathpages.com/rr/s6-03/6-03.htm" [Broken]
    Last edited by a moderator: May 2, 2017
  14. Jul 12, 2006 #13


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    The easiest way to understand GRs bending of light is that half of it comes from the equivalence principle (essentially 'time curvature') and the other half from space curvature. Newton's gravity only bends light if the 'corpuscle theory' of light is used. In the 'wave theory' there is no bending under Newton's gravity.
  15. Jul 12, 2006 #14
    Does the corpuscle theory of light assign a mass to these corpuscles?
  16. Jul 12, 2006 #15


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    I agree with Zeit here: in a Newtonian corpuscular light theory with massless light particles, light would be bend by gravity. Of course it's a different issue with light described by classical EM: there, one has in fact the choice (and it would lead, in any case, to some kind of inconsistency).

    Let us consider the Newtonian case where a body A exerts a gravitational force on body B. We denote the (active) gravitational mass of body A by m2, the (passive) gravitational mass of body B by m3, then the force acting upon body B by body A is given by:

    F = - G m2 m3 / R^2 with R the distance between both objects and the force F in the radial direction.

    Now, the force F acting upon body B, accelerates body B, by Newton's equation: F = m1 a
    Here, m1 is the INERTIAL MASS of body B.

    Now, it is an experimental fact that for a given body, the inertial mass equals the passive gravitational mass, so we have that m1 = m3. In Newton's theory, this didn't need to be the case, because m3 is the "gravitational charge" which defines the gravity force, while m1 is the "inertial resistance". A priori, those two concepts have nothing to do with one another, but they turn out to be equal.

    It also turns out that the gravitational force F acting on body B is equal and opposite to the gravitational force F' acting upon body A. In the last case, however, the "active" role is now played by B, and the passive role by A. So this also means that the active and passive gravitational mass for a body are equal. This, however, IS a consequence of the action=reaction principle.

    So everything together:

    for a given body, it turns out that active gravitational mass = passive gravitational mass = intertial mass. hence we call this quantity simply the mass of the body, without distinguishing between the 3 different functions of mass.

    This is (a version of) the *equivalence principle*.

    From the equivalence principle follows now an important concept:

    A body B, with mass m, will undergo a gravitational force F which will be the sum of different contributions of different other bodies A1, A2, ...:

    F = G m m1 / r1^2 1_r1 + G m m2 / r2^2 1_r2 + ....

    and we can factor out G m:

    F = G m (m1 / r1^2 1_r1 + m2 / r2^2 1_r2 +....)

    here, the factor in () corresponds to the geometry and all the (active) contributions of all other bodies, but does not depend upon anything else but the place of body B (does not depend on its mass).

    Applying Newton's law:

    m . a = F = G m (m1 / r1^2 1_r1 + m2 / r2^2 1_r2 +....)

    allows us to cancel m if m is non-zero (or, as was pointed out, to use l'Hopital's rule for the ratio m.a/F in the limit of m -> 0):

    a = G (m1 / r1^2 1_r1 + m2 / r2^2 1_r2 +....)

    So we see that the acceleration of body B is independent of any of its properties, but only of its LOCATION. So we can just as well say that this is a property of the LOCATION of B, and not of B itself.

    ANY body that is at the said location, will undergo an acceleration a.

    So also a massless particle. If we know the velocity of the massless particle, we can calculate its trajectory. It will be the same trajectory as a particle with mass, and the same velocity.

    This reduces gravity already to a geometrical effect, even in Newtonian physics. I think it was Cartan who worked out a "Newtonian general relativity" by reformulating the above concept in entirely geometrical terms of curved spacetime. However, the theory is more complicated (and less accurate) than general relativity.
  17. Jul 12, 2006 #16


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  18. Jul 12, 2006 #17


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    No, light-bending exercises simply assumes that the trajectory will be a hyperbola - like "tiny, high speed comets", e.g. Faber 1983: Differential Geometry and Relativity.
  19. Jul 12, 2006 #18
    Wow! Thanks to everyone for some very complete and thorough answers. This was very enlightening (no pun intended).

  20. Jul 13, 2006 #19
    It may be of interest to some of you posting here that the full (1.7 sec.of arc) relativistic light deflection can be (and was actually derived) without the calculations of general relativity by L.I. Schiff in 1959.

    Simply using equivalence principle and special relativity considerations Schiff provided a very simple derivation whereby the full angular solar deflection (theta) was shown to be :

    Theta = (4GM/Rc^2), where R is the radius of the sun. (Put in the figures to get the 1.7 ").

    This was important because solar/starlight deflection was suppose to be one of 3 'crucial' tests of Gen Rel. , (gravitational doppler shift and the perihelion precession of Mercury's orbit being the other two).

    Since grav. doppler shift and starlight deflection could both be derived outside of Gen Relativity (that is, without the goedesic or field eqns. of GR), Schff concluded that only the Mercurial orbital precession provided any real 'crucial' test of GR and that seemed very insufficient.

    (See: "On Experimental Tests of General Relativity", L. I. Schiff, Amer. J. of Phys., 28, 340,(1960).

    The point is: This lack of 'crucial' substantiation of GR became the motivation for Schiff to look for possible NEW experimental tests of GR, and was the precurser of his arriving (in very next year) of calculations showing that Gen. Relativistic precessional changes of a gyroscope in earth orbit could fit the bill.
    Thus was born the beginnings of an experiment later to be called Gravity Probe B.

    Creator :smile:
    Last edited: Jul 13, 2006
  21. Jul 13, 2006 #20
    I please for additional explanation.
    (As first, special relativity enter in the game, because it limiting speed of light to c. Newton gravity gave it unlimited velocity.)
    I calculated angle at sun classically and it is 2mG/(Rc2) = 0.729’’. It is calculated by supposition that light can be a little accelerated above speed of light.
    But, what is (simply) additional reason in general relativity, that angle is duplicated? How equivalence principle is included in calculation?
    It was also written that this factor is because of curvature of space. I please for more explanation.

    I think also that all formulae in high physics should be explained graphically and simpler. For instance factor 2 above. Or principle of uncertainty is well explained by by two Gaussian curves, where one rises the other becomes thinner.
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