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Gravitational lensing derivation using equivalence principle

  1. Jul 15, 2012 #1
    I have been trying to work this out for the last couple of weeks, but I just keep getting the Newtonian deviation in angle for a path of a photon travelling from x=-∞ to x=∞. At first I tried putting the actual path into a computer simulation, transforming back and forth between the hovering observer's frame locally with the photon and that of a distant observer, then again using the coordinate radial and tangent accelerations that the distant observer measures for the photon. My last attempts were to simply find the change in angle mathematically, where asin(vy / v) ≈ vy / v for a small change in angle with small M and large point of closest approach R (= y), per dx = v dt for a photon that otherwise travels a straight line path along x with an instantaneous coordinate speed v according to the distant observer, then integrating that along x to find the total change in angle.

    Locally I am applying the equivalence principle, using the relativistic acceleration formulas to find the change in radial speed for a freefaller initially falling at the same coordinate speed vr1 as the photon, radially only with no tangent speed. Since that freefaller is inertial and falling at the same radial speed as the photon, from the freefaller's point of view, then, the photon just travels directly away tangentially at c, and as the freefaller continues to travel inertially, this should continue to be true. So whatever final coordinate radial speed vr2 is achieved by the freefaller over time dt, this will also be achieved by the photon in the radial direction. The tangent speed according to the hovering observer, then, since the freefaller and hovering observer both measure the total speed c of the photon, is just vt2 = sqrt(c^2 - vr2^2). The change in speed vy is then found from the radial and tangent changes in speed, giving the change in angle along y according to the distant observer, then integrated for the entire path of the photon along x to find the total change in angle.

    But this hasn't been working out so far, giving only the Newtonian change in angle instead of twice that value which GR predicts. Could someone please show me how to derive the GR value using the equivalence principle in this way? Or if it is usually found in some different way, that would be fine too so that I can compare the difference to what I am attempting. I only understand algebra and basic calculus, though, so please present your derivations that way. I also want to see it visually, worked out according to what each observer actually measures, bit by bit over small portions of space, then integrated to find the whole, so no quicky math or matrix solutions please. Thanks.
     
    Last edited: Jul 15, 2012
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  3. Jul 15, 2012 #2

    Bill_K

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    grav-universe, I'm not aware of any way to derive the light deflection from just the equivalence principle. It sounds like you're trying to use it nonlocally, not just in a local neighborhood.

    The standard way to derive light deflection is to write down and solve the geodesic equation. This is fairly easy to do, and gives you a result of the form Δφ = ∫ f(r) dr. The only tricky part is then evaluating (approximately) the integral!
     
  4. Jul 15, 2012 #3
    Thanks Bill K. For reference, I'll show how I worked through the integration. Maybe someone can find where I went wrong. With arbitrarily small mass M of the gravitating body and a large distance to the point of closest approach R, the change in angle is extremely small, so I am integrating for the deviation in the y direction along a straight line path of the photon along x. We will integrate for a photon starting at x=-∞ and travelling in the +x direction with the gravitating mass at the origin and the photon travelling at a constant y = R. We'll use the absolute value for the distance x in the calculations.

    Okay, so a distant observer measures the speed of the photon to be v, travelling along the x axis, distances R and x from the gravitating mass M, whereby d = sqrt(x^2 + R^2). The distant observer measures the speed in the radial direction to be vr1 = v (x / d) and in the tangent direction to be vt1 = v (R / d). A hovering observer at the same place as the current position as the photon, then, will measure vr1' = v (x / d) / (1 - g / d) and vt2' = v (R / d) / sqrt(1 - g / d), where g = 2 G M / c^2. The hovering observer must measure the speed of the photon to be c, so we have

    vr1'^2 + vt1'^2 = c^2

    v^2 (x / d)^2 / (1 - g / d) + v^2 (R / d)^2 / (1 - g / d) = c^2

    v^2 = c^2 (1 - g / d)^2 / [(x / d)^2 + (R / d)^2 (1 - g / d)]

    v^2 = c^2 (1 - g / d)^2 / [(x / d)^2 + (R / d)^2 - (g / d) (R / d)^2]

    v = c (1 - g / d) / sqrt[1 - (g / d) (R / d)^2]

    So we have the speed of the photon v that the distant observer measures, whereby

    vr1' = v (x / d) / (1 - g / d)

    = c (x / d) / sqrt[1 - (g / d) (R / d)^2]

    So this is the radial speed of the photon according to the hovering observer. Now let's say that another observer is freefalling only radially with no tangent speed with a current radial speed of vr1', at the same place and matching the speed of the photon. In that case, the freefaller will measure the photon to be travelling directly tangent to him. Since the freefaller is inertial, then this should continue to be true, so after a time dt' in the hovering observer's frame, the freefaller will have accelerated radially to vr2' and the radial speed of the photon will match that, so the hovering observer will measure the tangent speed of the photon to be vt2' = sqrt(c^2 - vr2'^2).

    Okay, so now we need to find the speed vr2' that the freefaller accelerates to after a time dt' in the hovering observer's frame. The relativistic formula for the speed attained by an object undergoing constant proper acceleration is

    (v / c) = (a t / c) / sqrt(1 + (a t / c)^2)

    which can also be re-arranged to get

    (a t / c) = (v / c) / sqrt(1 - (v / c)) and

    sqrt(1 - (v / c)) = 1 / sqrt(1 + (a t / c)^2)

    all of which we will use. In this case, however, since the freefaller is the one that is inertial and the hovering observer is the one accelerating, the result will just be the coordinate speed attained by applying a coordinate acceleration according to the hovering observer, which must reduce to a coordinate acceleration of just a = G M / d^2 locally for an object falling from rest, so giving a speed for the freefaller of v = (a t) / sqrt(1 + (a t / c^2)) = (G M t' / d^2) / sqrt[1 + (G M t / (c^2 d^2))] after some short time t.

    If the freefaller were to constantly accelerate from rest according to the hovering observer, the freefaller would attain a speed of vr1' after a time t1 = (c / a) (vr1' / c) / sqrt(1 - (vr1' / c)^2) and a speed vr2' after a time t2 = (c / a) (vr2' / c) / sqrt(1 - (vr2' / c)^2). Now, it doesn't matter whether or not the freefaller actually accelerated at a constant rate during the time t1 since we are only taking the difference in speed during the time dt' = t2 - t1. This can be seen more directly through calculus, of course, it is the same thing, but I am just taking the long way. So we have

    Δvr' = vr2' - vr1'

    = (a t2) / sqrt(1 + (a t2 /c)^2) - (a t1) / sqrt(1 + (a t1 / c)^2)

    = (a dt') / sqrt(1 + (a t1 / c)^2)

    = (a dt') sqrt(1 - (vr1' / c)^2)

    The change in tangent speed, then, is

    Δvt' = vt2' - vt1'

    = c sqrt(1 - (vr2' / c)^2) - c sqrt(1 - (vr1' / c)^2)

    = c / sqrt(1 + (a t2 / c)^2) - c / sqrt(1 + (a t1 / c)^2)

    = c [sqrt(1 + (a t2 / c)^2) - sqrt(1 + (a t1 / c)^2)] / (1 + (a t1 / c)^2)

    = - c [(a dt' / c) (a t1 / c) / sqrt(1 + (a t1 / c)^2)] / (1 + (a t1 / c)^2)

    = - c [(a dt' / c) (vr1' / c)] (1 - (vr1' / c)^2)

    = - (a dt') (vr1' / c) (1 - (vr1' / c)^2)

    Again, all of this can also be shown more directly through calculus, I'm just presenting it in the way I know best, long and drawn out. :) Okay, so reverting back to the change in speeds that the distant observer measures, we have for the change in radial speed,

    Δvr = Δvr' (1 - g / d)

    = [(a dt') sqrt(1 - (vr1' / c)^2)] (1 - g / d)

    = (a dt / sqrt(1 - g / d)) sqrt(1 - (vr1 / c)^2 / (1 - g / d)^2) (1 - g / d)

    where (vr1 / c) / (1 - g / d) = vr1 ' / c = (x / d) / sqrt(1 - (g / d) (R / d)^2) from before, so

    Δvr = a dt sqrt(1 - g / d) sqrt(1 - (x / d)^2 / (1 - (g / d) (R / d)^2))

    and for the change in tangent speed,

    Δvt = Δvt' sqrt(1 - g / d)

    = [- (a dt') (vr1' / c) (1 - (vr1' / c)^2) sqrt(1 - g / d)

    = (a dt / sqrt(1 - g / d)) ((vr1 / c) / (1 - g / d)) (1 - (vr1 / c)^2 / (1 - g / d)^2) sqrt(1 - g / d)

    = - a dt [(x / d) / sqrt(1 - (g / d) (R / d)^2)] (1 - (x / d)^2 / (1 - (g / d) (R / d)^2))

    where further

    1 - (x / d)^2 / (1 - (g / d) (R / d)^2)

    = [(1 - (g / d)(R / d)^2) - (x / d)^2] / (1 - (g / d) (R / d)^2)

    = [1 - (g / d) (R / d)^2 - (1 - (R / d)^2)] / (1 - (g / d) (R / d)^2)

    = (R / d)^2 (1 - g / d) / (1 - (g / d) (R / d)^2)

    giving

    Δvr = a dt (1 - g / d) (R / d) / sqrt(1 - (g / d) (R / d)^2)

    Δvt = - a dt (x / d) (1 - g / d) (R / d)^2 / (1 - (g / d) (R / d)^2)^(3/2)

    The change in angle is asin(vy / v) ≈ vy / v for small M and large R, and dx = dt / v. We only want the change in speed in the y direction, so we have

    Δvy = Δvr (R / d) - Δvt (x / d)

    = [a (dx / v) (1 - g / d) (R / d)^2 / sqrt(1 - (g / d) (R / d)^2)] [1 + (x / d)^2 / (1 - (g / d) (R / d)^2)]

    = [(G M / d^2) (dx sqrt(1 - (g / d) (R / d)^2) / (c (1 - g / d)) (1 - g / d) (R / d)^2 / sqrt(1 - (g / d) (R / d)^2)][1 + (x / d)^2 / (1 - (g / d) (R / d)^2)]

    = [(G M / d^2) dx (R / d)^2 / c] [1 + (x / d)^2 / (1 - (g / d) (R / d)^2)]

    Δangle ≈ Δvy / v

    = (G M / c^2) dx (R / d)^2 [1 + (x / d)^2 / (1 - (g / d)(R / d)^2)] sqrt(1 - (g / d) (R / d)^2) / (d^2 (1 - g / d))

    = (G M / c^2) dx (R / d)^2 [(1 - (g / d) (R / d)^2) + (x / d)^2] / (d^2 sqrt(1 - (g / d) (R / d)^2) (1 - g / d))

    For arbitrarily small g, that becomes

    Δangle ≈ Δvy / v = (G M / c^2) (R / d)^2 (1 + (x / d)^2) dx / d^2

    and integrating along x, we finally get

    angle = (G M / c^2) R^2 [∫dx / (x^2 + R^2)^2 + ∫x^2 dx / (x^2 + R^2)^3

    = (G M / c^2) 2 [((∏ / 2) / (2 R^3)) + ((∏ / 2) / (8 R^3))]

    = (5 ∏ / 8) (G M / (R c^2))

    = 1.9635 (G M / (R c^2))

    which is just the Newtonian deflection. I'm assuming the small difference here from 2 G M / (R c^2) would be from taking a straight line path instead of the slightly curved one. So where did I go wrong?
     
  5. Jul 15, 2012 #4

    Bill_K

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    I have no idea. I do know that no matter how you do it, by dimensional analysis the deflection angle must come out N M/b, where N is some numerical constant. The only trick is getting N right! Would you like to see the correct derivation? I think it's considerably simpler.

    A null geodesic in Schwarzschild is given by

    0 = (1 - 2m/r)-1(dr/ds)2 + r2(dφ/ds)2 - (1 - 2m/r)(dt/ds)2

    where s is an affine parameter. There are two immediate first integrals,

    L ≡ r2(dφ/ds)
    E ≡ (1 - 2m/r)(dt/ds)

    We choose to scale s so that s → t as r → ∞. This determines the value of E = 1. The orbit equation then reduces to

    (dr/ds)2 = 1 - (L2/r2)(1 - 2m/r)
    (dr/dφ)2 = (dr/ds)2/(dφ/ds)2 = r4/L2 - r2(1 - 2m/r)

    Let r = b be the perihelion, the place where dr/ds = 0. This determines L:

    L2 = b2/(1 - 2m/b)

    and the final form of the orbit equation:

    (dr/dφ)2 = (r4/b2)(1 - 2m/b) - r2(1 - 2m/r) ≡ f(r)

    This is exact. The only thing that remains is to approximate m/b << 1 and integrate:

    Δφ = ∫dr/√f(r) ≈ ∫dr/(r√(r2/b2 - 1)) + m ∫dr(r3/b3 - 1)/r2(r2/b2 - 1)3/2)

    The first integral is easy, while the second one takes an integral table or computer. But the result is the correct one:

    Δφ = π + 4m/b
     
  6. Jul 16, 2012 #5
    Great post, Bill K, thanks. I don't know what those parameters are or how they are to be applied, but I will compare the result for the change in angle per dr to my own results and work through examples of each with some actual numbers to see what the difference is. I also notice the angle you found gives 180 degrees plus 4 r_s/r, so I'm assuming it gives the angle the positioning of the photon travels around the gravitating body rather than just the deviation from the original path of the photon, so I will re-work what I was doing using that angle to see what it gives.
     
  7. Jul 19, 2012 #6
    Well, I have compared the results, that between using the metric as you gave and using the equivalence principle as I was trying to do, and the difference is astounding. I certainly didn't expect to see the results I am seeing when using the metric, since it doesn't appear to leave room for the principle of equivalence to apply at all. Of course, please verify these results and make sure that I have performed it correctly. It is always possible and even likely in this case that I have missed something.

    Okay, so I wasn't sure at first at what point I could compare the curvature, speed, and direction of travel between the two methods, especially if each give completely different results, but there is one place where everyone can agree what is taking place. That is at the point of closest approach. At that point, the photon is travelling perfectly perpendicular to the gravitating body with a speed of c as measured by the local hovering observer and a coordinate speed of sqrt(1 - 2 m / b) c as measured by the distant observer. The latter is according to the metric, and we will apply the metric all the way here to see what it gives.

    So using the metric and starting at the point of closest approach b, the distant observer will measure the photon to be travelling perpendicular to the body with a speed of sqrt(1 - 2 m / b) c. After some infinitesimal time, the photon will be at a distance r = b + dr from the body. Using the integration you gave, the change in angle of its position will be

    ψ = (∏ / 2) - atan(b / sqrt(r^2 - b^2)) + m sqrt(r^2 / b^2 - 1) (1 / (r + b) + 1 / r)

    Its original coordinates were x1 = 0, y1 = b, while its new coordinates are x2 = sin(ψ) r, y2 = cos(ψ) r. With infinitesimal distance dr, the tangent speed won't change much, so the time of travel can be approximately found with dt = x2 / (c sqrt(1 - 2 m / b)). For what I want to find for, we will also need the change in position along y, where dy = b - y2.

    Okay, so for a photon travelling tangent to the body for an infinitesimal amount of time, with zero radial speed initially, its coordinate acceleration along the y axis is a_y = 2 dy / dt^2. Running the actual numbers for this in a computer program, given m = G M / c^2, b, and dr close to that for the sun but rounded off so we can find the related equations better, and finding for dt and dy as described above, we get

    Code (Text):

      m          b            dr               dt                    dy
    10^3       10^10       10^(-20)      4.71731*10^(-14)     3.000000225*10^(-27)
    10^3       10^10       2*10^(-20)    6.67128*10^(-14)     6.00000045*10^(-27)
    10^3       2*10^10     10^(-20)      6.67128*10^(-14)     1.500000563*10^(-27)
    2*10^3     10^10       10^(-20)      4.71731*10^(-14)     6.0000009*10^(-27)
     
    To first order, that is just

    dt = sqrt(2 dr b) / c and dy = 3 dr m / b, giving

    a_y = 2 dy / dt^2

    = 2 (3 dr m / b) / (2 dr b / c^2)

    = 3 m c^2 / b^2

    Second order can be found by applying (1 - 2 m / b) ^ n, which works out to n = 17 / 8 using the dt = x / (c sqrt(1 - 2 m / b)) approximation, which drops it from a margin of error of 1 part in a million to 1 part in a million million for dt and dy using

    dt = (sqrt(2 dr b) / c) / (1 - 2 m / b)^(5 / 4) and dy = (3 dr m / b) / (1 - 2 m / b)^(3 / 8)

    although again, only for that particular approximation. I also tried it a couple of other ways, transforming to the perspective of the hovering observer, for instance, and the exponents will change somewhat, but remains the same to first order as measured by both the hovering observer and the distant observer for small m / b, with a coordinate acceleration along the y axis of a_y = 3 m c^2 / b^2, 3 times the Newtonian value.

    That of course is the surprising result. It must reduce to Newtonian for non-relativistic speeds, so the speed dependent equation might be something like (1 + 2 (v/c)^2) m c^2 / b^2 applying the local scalar speed or whatever, but there is still a problem when it comes to the equivalence principle. If we were to drop an observer in an elevator from rest at b, then the elevator will coordinately accelerate according to a hovering observer there at about m c^2 / b^2. If the freefalling elevator observer emits a photon precisely in the tangent direction at the moment he begins to fall, then since the elevator is inertial, according to the equivalence principle the photon should just travel straight across the elevator and strike a point on the opposite wall that is the same height above the floor as when it was emitted (ignoring the gravity gradient when considering it to be insignificant over a small distance). But according to the metric, at least what I have worked out if it is correct, the elevator will coordinately accelerate at m c^2 / b^2 according to a hovering observer while the photon accelerates in the direction of freefall of the elevator at 3 times the rate, so the freefalling elevator observer should instead measure a coordinate acceleration of the photon towards the floor of the elevator of about 2 m c^2 / b^2. So have I done this correctly, and if so, what does it mean?
     
    Last edited: Jul 19, 2012
  8. Jul 21, 2012 #7
    Anybody? Here is a post that I made in another forum.

    However, as also mentioned there, even if the tube is immensely strong so that it resists bending due to general stresses of gravity, it may be possible that the metric affects it in some way that it is still forced to bend such that the ends droop down 2/3 the distance the light travels, about 1.39252604358*10^(-15) m, so that the photon and the center of end B of the tube will coincide at the same place upon freefalling, which would preserve the equivalence principle if that is the case. What are your thoughts?
     
    Last edited: Jul 21, 2012
  9. Jul 21, 2012 #8
    If it is the case that the tube bends, however, in order to preserve the equivalence principle, then if instead of a photon, we fire a massive particle that travels at a low non-relativistic speed, the center of end B should still coincide with the particle when the tube is inertial upon freefalling. What is dψ/dr for a particle travelling from the tangent point b at a speed v?
     
    Last edited: Jul 21, 2012
  10. Jul 21, 2012 #9
    I made a computer program a while back that was sensitive enough to measure the precession of Mercury and other planets. From the precession program, I found the radial acceleration of a particle to be something between a = (G M / r^2) (1 + 6 m / r) and a = (G M / r^2) (1 + 3 (v/c)^2), both of which give the accurate precession. I didn't know until now, though, what the proportion of (v/c)^2 to m / r should be. It might also be angle dependent, but we know now that for a photon travelling perfectly tangent to the gravitating body at least, its radial acceleration is rather precisely 3 (G M / r^2) / (1 - 2 m / r) according to a distant observer. The precession program is also observed from the perspective of a distant observer. In order to line up the equations to match both precession and gravitational lensing, then, it would have to be a = (G M / r^2) (1 + 2 (v/c)^2) / (1 - 2 m / r) to second order for a particle travelling in the tangent direction. Of course, it must also reduce to Newton. For freefall from rest with initial tangent speed v = 0, that gives a = (G M / r^2) / (1 - 2 m / r).

    So now let's try those 3 accelerations, a_c = 3 (G M / r^2) / (1 - 2 m / r) for a photon, a_p = (G M / r^2) (1 + 2 (v/c)^2) / (1 - 2 m / r) for a particle, and a_f = (G M / r^2) / (1 - 2 m / r) for the tube freefalling from rest to see what we get in terms of the bend of the tube. If the length of the tube is d, then the photon travels the length of the tube in a time of about t_c = d / c. During that time, it accelerates radially while the tube in freefall accelerates radially also at 1/3 the rate, and the equivalence principle says the two must coincide, so the distance of the radial bend at the end of the tube from the straight length must be

    d_bend = a_c t_c^2 / 2 - a_f t_c^2 / 2

    = [3 (G M / r^2) / (1 - 2 m / r) - (G M / r^2) / (1 - 2 m / r)] (d / c)^2 / 2

    = (G M / r^2) (d / c)^2 / (1 - 2 m / r)

    Now let's find out what the bend would have to be in order for a particle travelling at v to coincide with the end. The time for the particle to traverse the tube will be about t_p = d / v, so we have

    d_bend = a_p t_p^2 / 2 - a_f t_p^2 / 2

    = [(G M / r^2) (1 + 2 (v/c)^2) / (1 - 2 m / r) - (G M / r^2) / (1 - 2 m / r)] (d / v)^2 / 2

    = [(G M / r^2) (2 (v/c)^2) / (1 - 2 m / r)] (d / v)^2 / 2

    = (G M / r^2) (d / c)^2 / (1 - 2 m / r)

    which is exactly the same as what we found for the photon. So the equivalence principle definitely can work out if the tube is bent in this way, and so probably does, and we just found a formula that describes the rigidity of objects in a gravitational field. :)
     
    Last edited: Jul 21, 2012
  11. Jul 21, 2012 #10
    Oh whoops, the acceleration for the photon that was found by taking the integration to four orders was a = 3 (G M / r^2) (1 - 2 m / r), not 3 (G M / r^2) / (1 - 2 m / r), so we may still need more orders for the precession formula or something, and the time of travel was also an approximation, but the equations line up well enough to verify that the equivalence principle can apply with a bend at the end of the tube of about (G M / r^2) (d / c)^2 to first order.
     
  12. Jul 22, 2012 #11

    Jonathan Scott

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    I don't have time to address the complete thread, but here are some points which may be relevant:

    The principle of equivalence is not enough to determine the path of a fast-moving object, as this depends on the curvature of local space relative to the overall coordinate system.

    In the simple central case, for a weak field, the curvature of space relative to an isotropic coordinate system is the same as the Newtonian acceleration (given by the curvature of space-time with respect to time). This means for example that local horizontal rulers are curved relative to the coordinate system and vertical rulers which appear to be parallel locally are not parallel relative to the coordinate system.

    The results are still consistent with the principle of equivalence, in that relative to local rulers, horizontal photons only appear to accelerate with the usual Newtonian acceleration, the same as a brick or anything else. However, relative to the coordinate system, the rulers themselves are bent, so horizontal photons accelerate twice as much.

    In this case, if all quantities are expressed relative to an isotropic coordinate system (including the coordinate speed of light, which I'll still call c even though it isn't exactly the standard value), then the following equation holds for the rate of change of coordinate momentum in terms of the Newtonian acceleration, regardless of the direction of travel:

    [tex]
    \frac{d\mathbf{p}}{dt} = \frac{E}{c^2} \mathbf{g} \left ( 1 + \frac{v^2}{c^2} \right )
    [/tex]

    Since [itex]p = E\mathbf{v}/c^2[/itex], we can divide this by [itex]E[/itex] this to show that it does not depend on the energy of the particle:

    [tex]
    \frac{d}{dt} \left( \frac{\mathbf{v}}{c^2} \right ) = \frac{1}{c^2} \mathbf{g} \left ( 1 + \frac{v^2}{c^2} \right )
    [/tex]
     
  13. Jul 22, 2012 #12
    Hmm. I can see that if the ruler is bent to conserve the equivalence principle, the locally measured acceleration of the photon must necessarily be Newtonian, the same as for the freefalling tube or elevator, but using the metric to gain the coordinate acceleration of the photon in the y direction, the direction of freefall, the metric says that the distant observer should measure three times the Newtonian acceleration for a horizontally travelling photon, not twice. Is there something else that I'm missing?
     
    Last edited: Jul 22, 2012
  14. Jul 22, 2012 #13

    JDoolin

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    Thanks for the derivation Bill. I'm still working my way through it, but thought I'd add a couple little questions and comments.

    Is this a null geodesic? I think this is a more general case, here. The null geodesic is a much simpler problem; you can set c*dτ=0. Here, you are letting the ds vary, which means its applicable to any path.

    I was able to make some headway here by trying dt/ds → 1 as r → ∞.



    To derive this part, it helps greatly to apply the approximation 2m/r ≈ 0.

     
  15. Jul 22, 2012 #14

    Jonathan Scott

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    I don't know what it is, but yes, there must be, as the coordinate deflection of light is definitely only twice that for the Newtonian acceleration (assuming the weak field approximation), and the equations I gave above were determined from the Schwarzschild solution in isotropic coordinates.

    Note that the rate of change of coordinate momentum equation which I gave above applies to all freely falling test objects, not just light, and not just moving horizontally. In that sense, it helps show the relationship with the Newtonian point of view where gravity acts as a force, in that the main difference is simply the extra term in v2/c2, which arises from motion through space which is curved or changing in scale factor relative to the coordinate system.

    For vertically moving light, this may seem a little unexpected, but if you work it out, you will find that even though the coordinate value of c decreases as light falls, and the coordinate magnitude of v is equal to c, the momentum simplifies to E/c which increases downwards as the light falls (and by the right amount too) because of the decrease in c.
     
    Last edited: Jul 22, 2012
  16. Jul 22, 2012 #15
    Did you use Eddington's isotropic coordinates?
     
  17. Jul 22, 2012 #16
    I don't understand this part of the derivation. If we make E = 1 with large r, then we are making 1 - 2 m / r equal 1 for that part also. Why are we allowed to do that and if we can do it there, why not for (1 - 2m/r)-1(dr/ds)2, reducing it to just (dr/ds)2?
     
  18. Jul 22, 2012 #17
    Yep, sure enough. :) I just took the program I already had, ran it using Schwarzschild coordinates the same way as before, then kept the same angle and time, but had it go back and change the radiuses from b to b' = b / 2 - m / 2 + sqrt((b / 4) (b - 2 m)) and the same thing for r to r', transforming to Eddington isotropic coordinates. Then it found for dy' = b' - (cos θ) r', which ends up being 2/3 the distance, giving a coordinate acceleration of about 2 (G M / r^2) (1 - 2 m / r)^2 in Eddington's isotropic coordinates instead of the 3 (G M / r^2) (1 - 2 m / r) we get from Schwarzschild. Cool.
     
  19. Jul 23, 2012 #18
    I hope to find time this evening to check it in a GR book* at home, which, I think, describes how to do it. Including length contraction and time dilation should suffice to obtain the light bending of GR directly from SR together with the equivalence principle - but I never actually calculated it myself (lazy me - I always postpone doing what you now are doing!).

    *Adler et al, Introduction to General Relativity
     
    Last edited: Jul 23, 2012
  20. Jul 23, 2012 #19

    JDoolin

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    On further reflection, I realized that because of the zero on the left-hand-side, this does have to be a null-geodesic after all.
     
  21. Jul 23, 2012 #20
    You guys don't know how much time I spent trying to work through the metric using the isotropic coordinates and then integrating it, only to get a bunch of complex results and elliptical integrals, before realizing I could just transform the radiuses in the original program lol. I've been working on some more stuff and it's starting to get involved, so for future reference, I want to go ahead and post what I've found so far. A while back in another forum, someone asked what the coordinate speed of light would be that a distant observer would measure depending upon the angle to the gravitating body, so I posted this:

    I'm glad I already had that worked out. :) Okay, so below is an image of the path a photon will travel in relation to the angle and radius. Over infinitesimal amounts of time and distance, the path can be considered straight for the derivation, but of course it will curve over a larger finite time and distance. I already used θ (now used for what the distant observer measures) for the angle of travel of the photon from the radial direction, so I will use dφ for the infinitesimal change in angle of the radius (although greatly exaggerated in the image). Following the image, one can see that that

    ((r1 + dy)^2 + dx^2 = r2^2

    (r1 + (cos θ) c' dt)^2 + ((sin θ) c' dt)^2 = r2^2

    r1^2 + 2 (cos θ) c' dt r1 + (cos θ)^2 c'^2 dt^2 + (sin θ)^2 c'^2 dt^2 = r2^2

    where (cos θ)^2 + (sin θ)^2 = 1 so

    r2 = r1 sqrt(1 + 2 (cos θ) c' dt / r1 + c'^2 dt^2 / r1^2)

    and expanding the square root and taking only first order infinitesimals,

    r2 = r1 (1 + (cos θ) c' dt / r1)

    dr = r2 - r1 = (cos θ) c' dt

    (dr / dt) = (cos θ) c'

    So we now have dr / dt. We can also find

    (sin dφ) r2 = dx = (sin θ) c' dt

    where (sin dφ) = dφ for first order infinitesimal dφ, so

    dφ r2 = (sin θ) c' dt

    (dφ / dt) = (sin θ) c' / r

    using r since we are not finding for the difference between r1 and r2 and the difference is infinitesimal, so r = r1 = r2 here. We now have dφ / dt also, but both values we have found for are expressed in terms of θ and c', so we will need to get rid of those.

    From those two values, we have

    (dr / dt)^2 = (cos θ)^2 c'^2

    (dr / dt)^2 = c'^2 - (sin θ)^2 c'^2

    (dr / dt)^2 = c'^2 - (dφ / dt)^2 r^2

    c'^2 = (dr / dt)^2 + (dφ / dt)^2 r^2

    From the quote, we get

    c'^2 = c^2 (1 - 2 m / r)^2 / (1 - (sin θ)^2 (2 m / r))

    c'^2 - (sin θ)^2 (2 m / r) c'^2 = c^2 (1 - 2 m / r)^2

    and substituting, we get

    [(dr / dt)^2 + (dφ / dt)^2 r^2] - [(dφ / dt)^2 r^2] (2 m / r) = c^2 (1 - 2 m / r)^2

    (dr / dt)^2 + (1 - 2 m / r) (dφ / dt)^2 r^2 = c^2 (1 - 2 m / r)^2

    c^2 dt^2 (1 - 2 m / r) - dr^2 / (1 - 2 m / r) - dφ^2 r^2 = 0

    which gives us the metric.

    Here's something more. If we replace the speed of light with the speed of a massive particle, the coordinate speed v' will be found in exactly the same way as c' was, just replacing c with v_loc. The same thing goes for the rest of the post. But of course c and v_loc are the locally measured scalar speeds. So the metric then becomes

    v_loc^2 dt^2 (1 - 2 m / r) - dr^2 / (1 - 2 m / r) - dφ^2 r^2 = 0

    Used in this way, the metric is always null, even for speeds less than c, at least according to the distant observer. If we substitute the function for v_loc into the metric, then we can solve the metric using that. Does anybody know what that would be, if v_loc is given for a particular radius? Does it depend upon the direction of travel also?

    I'm trying to find another way to solve the metric. I still don't understand what BillK posted about s approaching t as r approaches infinity making E = 1, which if that is meant to be ds approaching c dt, then since ds = 0 always, not even infinitesimal, I don't see how it could approach c dt or how we could divide the other terms by it in the first place, but if ds were non-zero, then that would also make the metric equal unity instead of zero when divided by ds as far as I can tell. Anyway, I'll keep working on it.
     

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    Last edited: Jul 23, 2012
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