Gravitational lensing derivation using equivalence principle

[grav-universe]

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 traveling 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.

 

[Bill_K]

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!

 

[grav-universe]

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 traveling in the +x direction with the gravitating mass at the origin and the photon traveling 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, traveling 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 traveling 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?

 

[Bill_K]

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

 

[grav-universe]

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.

 

[grav-universe]

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 traveling 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 traveling 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 traveling 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

  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?

 

[grav-universe]

Anybody? Here is a post that I made in another forum.

We can perform an experiment like this right here on Earth. Let's say a photon is emitted directly along the surface of the Earth at b = 6.371*10^6 m. First let's find the angle of curvature for the photon to travel from a radius of b to b + 1 micrometer. Integrating to four orders to find the angle, I get

θ = [-atan(B / sqrt(R^2 - B^2)) + pi/2]
+ [M sqrt(R^2/B^2-1) (1/(B+R) + 1/R) - 0]
+ [1.5 M^2 ((sqrt(R^2/B^2-1)/6) (3/R^2 - 2 (5 B + 4 R) / (B (B+R)^2)) - (5 / (2 B^2)) atan(B / sqrt(R^2-B^2)))
+ 1.5 M^2 (5 / (2 B^2)) (Pi/2)]
+ [(2.5 M^3 / (15 B^3)) (45 atan(B / sqrt(R^2-B^2)) + (B sqrt(R^2/B^2-1) / (R^3 (B+R)^3)) (5 B^5 + 15 B^4 R + 70 B^3 X^2 + 247 B^2 R^3 + 306 B R^4 + 122 R^5))
- (2.5 M^3 / (15 B^3)) 45 (Pi/2)]

Using UBasic (prog"lense170" for my own notes) and plugging in b = 6.371*10^6 and r = b + 1 / 10^6 with m = G M / c^2 = 4.43589170257*10^(-3), I get an angle of θ = 5.60287838354*10^(-7) radians. So if the original coordinates of the photon were x1 = 0, y1 = 6.371*10^6, they are now

x2 = (sin θ) r = 3.56959381816 m

y2 = (cos θ) r = 6370999.99999999999999791121093462 m

So let's say that we design an immensely strong metal tube, out in free space away from any gravitation, so that we can use a light beam while inertial to be sure that it is perfectly straight. Its dimensions will be 7.1391876363 m long with a radius of 2.0887890653751*10^(-15) m. Then we will bring it to the surface of the Earth and align it perfectly level with its center at b, lying perfectly along y = b. We will then use a laser to cut it in half carefully and remove one half, so that one end of the tube will now be centered on xA = 0 and yA = b while the other end is centered at xB = 3.5659381816 m and yB = b. From end A we will fire a laser perfectly tangent down the center of the tube. From earlier, we see that the beam will bend so that it strikes right at the edge of the tube at the other end, having fallen a distance in the y direction of dy = 2.0887890653751*10^(-15) m, the radius of the tube.

Now let's say we only fire a single photon down the tube. The time it will take to traverse the tube is approximately dt = dx / c = 1.19068833218*10^(-8) sec. Let's also say that at the same time that we fire the photon at end A, we drop a massive particle from rest at the center of end B. The particle will just fall due to Earth's gravity at a rate of g = G M / b^2 = 9.82216285185 m/sec^2 (for the values used). In the time dt, then, the particle will fall a distance dh = g dt^2 / 2 = 6.96263021792*10^(-16) m. This is only about 1/2.999999999999607595510104 the distance the photon falls in the same direction over the same time. So then, if we were to release the tube at the moment we fire the photon, allowing it to freefall, the tube will fall only about a third of the distance the photon will, while if it were to be considered inertial, the tube should fall at the same rate so that the photon would continue to travel along the center of the tube, but that is not what occurs according to the metric.

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?

 

[grav-universe]

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 traveling from the tangent point b at a speed v?

 

[grav-universe]

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 traveling 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 traveling 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 traveling 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. :)

 

[grav-universe]

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.

 

[Jonathan Scott]

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:

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

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

$$\frac{d}{dt} \left( \frac{\mathbf{v}}{c^2} \right ) = \frac{1}{c^2} \mathbf{g} \left ( 1 + \frac{v^2}{c^2} \right )$$

 

[grav-universe]

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:

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

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

$$\frac{d}{dt} \left( \frac{\mathbf{v}}{c^2} \right ) = \frac{1}{c^2} \mathbf{g} \left ( 1 + \frac{v^2}{c^2} \right )$$

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 traveling photon, not twice. Is there something else that I'm missing?

 

[JDoolin]

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

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

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.

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

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

(dr/ds)² = 1 - (L²/r²)(1 - 2m/r)

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

(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

 

[Jonathan Scott]

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 traveling photon, not twice. Is there something else that I'm missing?

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.

 

[grav-universe]

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.

Did you use Eddington's isotropic coordinates?

 

[grav-universe]

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)

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?

 

[grav-universe]

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.

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.

 

[harrylin]

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 traveling from x=-∞ to x=∞. [..]

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. [..]

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? [..]

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

 

[JDoolin]

A null geodesic in Schwarzschild is given by

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

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 $c d\tau=0$. Here, you are letting the ds vary, which means its applicable to any path.

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.

 

[grav-universe]

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:

Well, let's see. Let's say the x-axis lies along the radial direction and a hovering observer emits a pulse of light locally at c along the angle θ from the x-axis toward the mass. The pulse travels a very short distance so that we can ignore the gravity gradient and our measurements remain local. According to the local observer, then, the pulse travels a distance d in a time t, ending up at coordinates x = (cos θ) d = (cos θ) c t, y = (sin θ) d = (sin θ) c t, y being revolved to include the z axis. To the distant observer using regular Schwarzschild coordinates, the distance the pulse travels along y is the same, the distance traveled along x is contracted to x' = sqrt(1 - r_s / r) x, the time that passes is t' = t / sqrt(1 - r_s / r), and there is no simultaneity difference, where r is the distance of the hovering observer from the mass according to the distant observer. So the distant observer measures a speed for the light pulse at r in the gravitational field using the locally measured angle θ of

c' = d' / t' = sqrt[(1 - r_s / r) x^2 + y^2] / [t / sqrt(1 - r_s / r)]

= sqrt[(1 - r_s / r) (cos θ)^2 c t + (sin θ)^2 c t] / [t / sqrt(1 - r_s / r)]

= c sqrt[(1 - r_s / r) (cos θ)^2 + (sin θ)^2] sqrt(1 - r_s / r)

= c sqrt[(1 - r_s / r) (cos θ)^2 + 1 - (cos θ)^2] sqrt(1 - r_s / r)

= c sqrt[1 - (cos θ)^2 (r_s / r)] sqrt(1 - r_s / r)

The angle that the pulse travels according to the distant observer is

cos θ' = x' / d' = sqrt(1 - r_s / r) (cos θ) / sqrt[(1 - r_s / r) (cos θ)^2 + (sin θ)^2]

= sqrt(1 - r_s / r) (cos θ) / sqrt[1 - (cos θ)^2 (r_s / r)]

Solving for θ, we get

[1 - (cos θ)^2 (r_s / r)] (cos θ')^2 = (1 - r_s / r) (cos θ)^2

(cos θ)^2 [1 - r_s / r + (cos θ')^2 (r_s / r)] = (cos θ')^2

(cos θ) = (cos θ') / sqrt[1 - r_s / r + (cos θ')^2 (r_s / r)]

And substituting that into the equation to gain the coordinate speed c' in terms of the angle θ' the distant observer measures, we have

c' = c sqrt(1 - (cos θ')^2 (r_s / r) / [1 - r_s / r + (cos θ')^2 (r_s / r)]) sqrt(1 - r_s / r)

= c sqrt([1 - r_s / r + (cos θ')^2 (r_s / r)] - (cos θ')^2 (r_s / r)) sqrt(1 - r_s / r) / sqrt[1 - r_s / r + (cos θ')^2 (r_s / r)]

= c (1 - r_s / r) / sqrt[1 - r_s / r + (cos θ')^2 (r_s / r)]

= c (1 - r_s / r) / sqrt[1 - (sin θ')^2 (r_s / r)]

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.

 

[grav-universe]

After all of that, I visited JDoolin's site and from what he had on a page about black holes, realized that since it was already given that the coordinate speed of light that the distant observer measures is just v_r' = v_r (1 - 2 m / r) and v_t' = v_r sqrt(1 - 2 m / r) by simply comparing the temporal and distance transforms in those directions, one can derive

vt^2 + vr^2 = c^2 locally

(vt' / sqrt(1 - 2 m / r))^2 + (vr' / (1 - 2 m / r))^2 = c^2

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

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

although I didn't see that it could be worked out so directly until having worked through it, um, more rigorously. Hah. :)

 

[harrylin]

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

OK, it's not clear if you already found the solution but indeed, in a nutshell (sorry not enough time now):

- from SR + equivalence principle:
time dilation and vertical length contraction, approx. correction factor 1 +/- gh/c² = 1 +/- Δψ/c²
They obtained from that and the equation for a Lorentz line element an approximate Schwartzschild equation.

Then apply the Huygens principle just like Einstein did in 1916: p.198, 199 of the English version of
- http://web.archive.org/web/20060829045130/http://www.Alberteinstein.info/gallery/gtext3.html

However, those few lines correspond to about a page of derivation in the textbook...
Interestingly, the time dilation and the length contraction contribute equally.

 

[grav-universe]

Thanks harrylin. I found what I needed for the most part, but now I'm backtracking and trying to find a way to solve the metric that makes sense to me, solving for dθ/dr, and then find what I was looking for more precisely from that. I also still need to know the relation for a massive particle between the local speed v<c and r.

 

[JDoolin]

It appears that others have noted the same difficulty that I have:

From Wikipedia:
http://en.wikipedia.org/wiki/Schwar...hematical_derivations_of_the_orbital_equation

"For lightlike (or null) orbits (which are traveled by massless particles such as the photon), the proper time is zero and, strictly speaking, cannot be used as the variable q. Nevertheless, lightlike orbits can be derived as the ultrarelativistic limit of timelike orbits, that is, the limit as the particle mass m goes to zero while holding its total energy fixed."

Now, when I tried it (last summer), I took no such "limit as the particle mass m goes to zero" but instead, just plugged a zero into the left-hand-side of the Schwarzschild metric directly. (I believe this may be what grav-universe was referring to in post #21)

However, at the time, I was unable to make progress on what seemed to be a much more difficult problem of a non-null geodesic.

 

[grav-universe]

The last few days I have been trying to transform between the coordinate systems of the local observer and the distant observer. If I figure that many local observers are at a distant r on the surface of a planet, then the distant observer will measure the planet as spherical, and the local observers, all observing the same thing from any position and since the contraction is radial only, might determine their planet as spherical also. At what radius though, I'm not sure, since the inferred radial distance is greater but the circumference is the same as the distant observer measures. If the tube is perfectly straight in free space and is brought down to the planet, local observers should still measure the tube as straight, right? If they then lay the tube level on the ground, each end will rise a slight distance from the ground. After finding that distance that the local observers measure in the radial direction for the end of the tube from the ground of a spherical planet, I then contracted that radial distance for what the distant observer measures between the ends of the tube and the ground, bringing the ends closer to the ground, while the center of the tube remained on the ground at the same place on the surface. The distant observer, then, will measure a slight bend in the tube this way that is the same as what was found earlier that it would need to be, but only half the necessary value, so there's still something missing.

If I simply extend a single local observer's local coordinate system globally, so that if the local observer is located at some point upon the y-axis in the radial direction, say, then all distances in the y direction are 1 / sqrt(1 - r_s / r) greater than the distant observer measures and all distances along the x-axis are the same. This would make the planet look like an ellipse to that local observer with him at the peak. If I then place the straight tube with the center at the peak, then transform in the same way back to the distant observer's coordinate system, then there is no bend. But then, if we're dealing with single local observers at this point, then there should be another at the end of the tube to measure that distance, but would then observe the planet and tube differently than the first local observer that was at the center of the tube, so might not say the tube is straight, hard to say. My latest attempts were to start with the distant observer's coordinate system, already knowing what the bend should be, then trying to transform that back to what the local observer would measure that would make the tube appear straight, but the results are so far indefinite. So I'm stuck, help please.

 

[JDoolin]

I've also been trying to work this problem out (again); I was working on the same problem about a year ago. Grav-universe, your approach seems unfamiliar; I don't know whether it might be ultimately equivalent, but I can show you the direction someone else took, and whose footsteps I tried to follow in.

Kevin Brown appears to have the entire derivation online here...
http://www.mathpages.com/rr/s8-09/8-09.htmI was working through the derivation on physics-forums here, but I did not get all the way through it. (In fact, I could not even follow the derivation to Kevin Brown's equation (8.9.2), at the time)
https://www.physicsforums.com/showthread.php?t=510985&page=2

I spent some time trying to remember what I was doing, retracing my steps from last year, and recording it with screen-recording software. Maybe this might give you some ideas.

http://www.spoonfedrelativity.com/misc/2012-07-28_1642-Deriving_-Scwarzschild-Metric.swf

http://www.spoonfedrelativity.com/misc/2012-07-29_1017-Deriving-Schwarzschild.swf

http://www.spoonfedrelativity.com/misc/2012-07-29_1039-Schwarzschild-Derivation.swf

http://www.spoonfedrelativity.com/misc/2012-07-29_1141-DerivingSchwarzschild.swf

 

[grav-universe]

Yay, I finally got it. The tangent speed from which the metric was derived in an earlier post was

$$v_t = (sin \ d\theta) r / dt$$

The numerator is the tangent distance which the distant observer and local observer will measure the same, so

$$(sin \ d\theta') r' = (sin \ d\theta)r$$ (primed for the local observer's measurements)

where $$r' = r / \sqrt{1 - r_s/r}$$, so

$$sin \ d\theta' = (sin \ d\theta) \sqrt{1 - r_s/r}$$

The angle we're finding for is that between the radial direction and along the line from the center of the planet to the end of the tube, with some very small x'/r' and x/r, which the local observer will measure as

$$sin \ \theta' = x' / \sqrt{r'^2 + x'^2}$$

with the radial distance r' and the length of the straight tube x' in the tangent direction. For the distant observer, we have

$$sin \ \theta = x / \sqrt{(r - y)^2 + x^2}$$

where the radial distance to the end of the tube is r-y and the tangent distance x, where the tube bends some distance y in the radial direction.

Since both agree upon the tangent distance, then x' = x, and we have

$$x' / \sqrt{r'^2 + x'^2} = (x \sqrt{1 - r_s / r}) / \sqrt{(r-y)^2 + x^2}$$

$$\sqrt{(r-y)^2 + x^2} = \sqrt{r'^2 + x^2} \sqrt{1 - r_s/r}$$

$$r^2 - 2 r y + y^2 + x^2 = r^2 + x^2 (1 - r_s/r)$$

$$y^2 - 2 r y + x^2 (r_s / r) = 0$$

Applying the quadratic formula, we get

$$y = \left(2 r _-^+ \sqrt{4 r^2 - 4 x^2 (r_s/r)}\right)/2$$

and taking the negative value since the positive would give us 2 r, we have

$$y = r - r \sqrt{1 - (x / r)^2 (r_s / r)}$$

which to first order gives us

$$y = r \left(1 - \left(1 - \left(\frac{x}{r}\right)^2 \left(\frac{r_s}{r}\right) / 2\right)\right)$$

$$y = \left(\frac{G M}{c^2}\right) \left(\frac{x}{r}\right)^2$$

precisely as it would need to be for the equivalence principle to work out. Cool. $$\ddot{\smile}$$

 

[grav-universe]

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

Here also is a Wiki link to some other methods for solving the metric. I don't understand half of them and the other half don't make sense to me. For instance, as I stated earlier in the thread, I don't see that it is mathematically rigorous to divide by ds or dτ when either is zero for a photon. Even if we were to consider a particle that travels just under c, we are not actually using the particle as the origin as we would for SR, where the distance of the particle from the particle's own origin would be zero, leaving just ds^2 = c^2 dτ^2 in terms of the proper time of the particle, but rather we are using the distance of the particle from the gravitating body, so even in the particle's own frame, this distance would be non-zero and so should be included in the metric. Anyway, I think I have found a very simple way to solve the metric. If L is the locally measured angular momentum, and if this quantity is conserved, then we have

L = m v_t' r' = constant

Since m is also a constant, we can drop that, leaving

L = v_t' r' = (v_t / sqrt(1 - 2m/r)) (r / sqrt(1 - 2m/r)) = v_t r / (1 - 2m/r)

where the primed values are locally measured and unprimed is measured by a distant observer. Since L is constant, then at the point of closest approach b, where the photon travels perfectly tangent to the gravitating body, we have

L = v_t' b' = c (b / sqrt(1 - 2m/b))

L_b = L_r, c b / sqrt(1 - 2m/b) = v_t r / (1 - 2m/r)

c b / sqrt(1 - 2m/b) = (dφ r / dt) r / (1 - 2m/r)

dt = dφ r^2 sqrt(1 - 2m/b) / (c b (1 - 2m/r))

So applying that to the metric, we get

v_r'^2 + v_t'^2 = c^2

(v_r / (1 - 2m/r))^2 + (v_t / sqrt(1 - 2m/r))^2 = c^2

where v_r = dr / dt and v_t = (sin dφ) r / dt = dφ r / dt

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

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

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

dr^2 / (1 - 2m/r) = dφ^2 [r^4 (1 - 2m/b) / (b^2 (1 - 2m/r)) - r^2]

dr^2 / dφ^2 = r^4 (1 - 2m/b) / b^2 - r^2 (1 - 2m/r)

This seems much simpler and somewhat more mathematically rigorous than any of the other methods, don't you think? Now I'm only left with trying to find a way to prove that the angular momentum is conserved as measured locally. Does anybody know a way to show that?

 

[grav-universe]

Here's something very interesting. If we just start with the already known solution to the metric and reverse engineer it, we'll end up with

(dr / dθ)^2 = r^4 (1 - 2m/b) / b^2 - r^2 (1 - 2m/r)

dr^2 = dθ^2 r^4 (1 - 2m/b) / b^2 - dθ^2 r^2 (1 - 2m/r)

dr^2 + dθ^2 r^2 (1 - 2m/r) = dθ^2 r^4 (1 - 2m/b) / b^2

dr^2 / (1 - 2m/r) + dθ^2 r^2 = dθ^2 r^4 (1 - 2m/b) / (b^2 (1 - 2m/r))

and substituting according to the metric,

c^2 dt^2 (1 - 2m/r) = dθ^2 r^4 (1 - 2m/b) / (b^2 (1 - 2m/r))

and re-arranging gives

c^2 b^2 / (1 - 2m/b) = (dθ r / dt)^2 r^2 / (1 - 2m/r)^2

c^2 b^2 / (1 - 2m/b) = v_t(r)^2 r^2 / (1 - 2m/r)^2

c b / sqrt(1 - 2m/b) = v_t(r) r / (1 - 2m/r)

and transforming c to the tangent speed measured by the distant observer so that all values will be as measured by the distant observer, we have

c_dist b / (1 - 2m/b) = v_t(r) r / (1 - 2m/r)

so according to the metric and its solution, this relation must be conserved regardless of whether we consider it angular momentum or whatever, but that's definitely what I would call it.

That's not the interesting part though. The angular momentum was found at the point where the photon travels tangently at a distance b. For a photon that approaches the body and then escapes, that's fine. But in the case of a photon orbitting the body, there will be two tangent points, at the aphelion and perihelion, with the photon traveling tangent to the body with a locally measured speed of c at both points. If the angular momentum is conserved at each of those points, then we have

c_a a / (1 - 2m/a) = c_p p / (1 - 2m/p)

and transforming to the locally measured speeds,

(c_a' sqrt(1 - 2m/a)) a / (1 - 2m/a) = (c_p' sqrt(1 - 2m/p) p / (1 - 2m/p)

c_a' a / sqrt(1 - 2m/a) = c_p' p / sqrt(1 - 2m/p)

But the locally measured tangent speeds at each of these points is just

c_a' = c_p' = c, giving

c a / sqrt(1 - 2m/a) = c p / sqrt(1 - 2m/p)

a = p

And so it would appear that a photon cannot have an elliptical orbit. A circular orbit is still allowed, but only if it is perfectly circular, so it looks like photons do not orbit a body at all, only approach and escape or fall in.

 

[harrylin]

[..] And so it would appear that a photon cannot have an elliptical orbit. A circular orbit is still allowed, but only if it is perfectly circular, so it looks like photons do not orbit a body at all, only approach and escape or fall in.

That's interesting! I never thought of that, but it makes perfect sense to me as a photon cannot store potential energy. Thanks.

 

[grav-universe]

Here is the conservation of energy. The locally measured energy of a photon just depends upon the local time t, since the observed frequency of a photon depends only upon the local gravitational time dilation. If a hovering observer emits a photon, then another photon a time t later, the first photon will follow some path, curved or otherwise, to some other point in the gravitational field, and the second will follow exactly the same path and arrive at the same point a time t later, so the frequency that the photons pass any point remains the same to the observer, but a observer at the point where they arrived will measure a different frequency that depends only upon his own rate of time. So we'll have

$$\frac{f_q}{f_p} = \frac{dt_p}{dt_q} = \frac{\sqrt{1 - 2m/p}}{\sqrt{1-2m/q}}$$

for a photon traveling from a radial distance p to a radial distance q in the field, although the path can be curved and non-radial. So there is also a dependence upon r. Since $$f_p$$ and $$f_q$$ are directly proportional to the locally measured energy of the photon, we have

$$\frac{E_q}{E_p} = \frac{\sqrt{1 - 2m/p}}{\sqrt{1 - 2m/q}}$$

Since this ratio of energy holds for light, we can try it for massive particles as well and see how that holds up. If we just take this ratio of energies directly for massive particles, we get

$$\frac{E_q}{E_p} = \frac{\sqrt{1 - 2m/p}}{\sqrt{1 - 2m/q}}$$

$$\frac{(m \ c^2 / \sqrt{1 - (v_q'/c)^2})}{(m \ c^2 / \sqrt{1 - (v_p'/c)^2})} = \frac{\sqrt{1 - 2m/p}}{\sqrt{1 - 2m/q}}$$

$$\frac{(1 - (v_p'/c)^2)}{(1 - 2m/p)} = \frac{(1 - (v_q'/c)^2)}{(1 - 2m/q)} = K (constant)$$

K = 0 for a photon. Of course, that must also reduce to Newtonian gravity, so let's find that for a massive body traveling from p to q = p - dp. The scalar speeds for the energy are locally measured while the distances are measured by a distance observer, whereas $$dp = p - q$$ and $$dp' = (p - q) / \sqrt{1 - 2m/p}$$.

$$(1-(v_q'/c)^2) = \frac{(1-2m/q) (1-(v_p'/c)^2)}{(1-2m/p)}$$

$$(v_q'/c)^2 = 1 - \frac{(1-2m/q) (1-(v_p'/c)^2)}{(1-2m/p)}$$

$$(v_q'/c)^2 - (v_p'/c)^2 = 1 - (v_p'/c)^2 - \frac{(1-2m/q) (1-(v_p'/c)^2)}{(1-2m/p)}$$

$$(v_q'/c)^2 - (v_p'/c)^2 = \frac{(1 - (v_p'/c)^2)}{(1-2m/p)} [(1 - 2m/p) - (1 - 2m/q)]$$

$$(v_q'/c)^2 - (v_p'/c)^2 = \frac{(1 - (v_p'/c)^2)}{(1-2m/p)} \left[\frac{2 m (p - q)}{p \ q}\right]$$

$$a' = \frac{d(v'^2)}{2 dp'} = \frac{[(v_q'/c)^2 - (v_p'/c)^2] \sqrt{1 - 2m/p}}{2 \ (p-q)} = \frac{m \ (1 - (v_p'/c)^2)}{p \ q \ \sqrt{1 - 2 m/p}}$$

and dropping infinitesimals at this point, the locally measured acceleration at r = p with instantaneous speed v' is

$$a' = \frac{(1 - (v'/c)^2) (G M / r^2)}{\sqrt{1 - 2m/r}}$$

We can see that for a body falling from rest at r with v' = 0, the acceleration reduces to Newtonian to first order. Okay, but this is an acceleration meant for a scalar speed. For instance, according to the equation, the acceleration of a photon with v' = c is zero, so only its direction changes, but not its scalar speed. So we'll want to divide that up into its radial and tangent components. We can use the equivalence principle for that. If a rod freefalls at r with original radial speed $$v_r'$$, matching the radial component of a particle that also freefalls with some tangent speed, then in order for the equivalence principle to hold, the radial acceleration of both must match over an infinitesimal time of freefall. With no tangent component, the radial speed of the rod is its scalar speed, so for the rod we have just

$$a'_{rod} = \frac{(1 - (v'/c)^2) (G M / r^2)}{\sqrt{1 - 2m/r}} = \frac{(1 - (v_r'/c)^2) (G M / r^2)}{\sqrt{1 - 2m/r}}$$

The radial component of the particle must match that, so we have

$$a'_r = \frac{(1 - (v_r'/c)^2) (G M / r^2)}{\sqrt{1 - 2m/r}}$$

There should be no tangent acceleration, so the apparent tangent component of the acceleration comes from the tidal gradient I think. For instance, for a circular orbit, as a particle passes the x-axis at a vertical distance y, its tangent speed begins to drop while its radial speed increases, although the scalar speed stays the same. Technically, then, we would really be extending the accelerations over some distances in the x and y directions instead of radially and tangent, but the apparent acceleration along infinitesimal y should still match the instantaneous radial acceleration if gravity only acts in that direction. So then, from that equation, a photon traveling radially will not accelerate at all, always measured at c locally in the radial direction, while a photon traveling in a circular orbit will have zero radial speed, so reduces to just

$$a'_r = \frac{(G M / r^2)}{\sqrt{1 - 2m/r}} = \frac{c^2}{r'}$$

$$\frac{(m / r^2)}{\sqrt{1 - 2 m / r}} = \frac{\sqrt{1 - 2 m / r}}{r}$$

$$(m / r) = 1 - 2 m / r$$

$$3 m / r = 1$$

$$r = 3 m$$

giving r = 3m for a photon traveling in a perfectly circular orbit. As far as photons go, as a photon falls toward a body, it spirals inward. If it doesn't pass a tangent point, it will fall all the way in. If it passes a tangent point, however, then it will spiral back out and eventually escape. It cannot turn around to spiral back in again because that would require that it passes another tangent point at a greater radius than the first, so angular momentum wouldn't be conserved. So other than a perfectly circular orbit at r = 3m, which would require that it be "born" there, and so precise with no disturbances as to make it virtually impossible, these are really the only two options for photons.

 

[grav-universe]

$$\frac{(1 - (v_p'/c)^2)}{(1 - 2m/p)} = \frac{(1 - (v_q'/c)^2)}{(1 - 2m/q)} = K (constant)$$

I introduced the constant K for energy conservation because I want to use it to show that ds = c dτ for the proper time of a particle or freefaller, as I wasn't sure about that earlier in the thread. For a freefaller passing a static observer at r with a scalar speed of v' in any direction, the locally measured time dilation of the freefaller's clock is sqrt(1 - (v'/c)^2). Since there are no simultaneity differences between static observers, only a gravitational time dilation between them, the time dilation that the distant observer measures of the freefaller's clock becomes dτ / dt' = sqrt(1 - (v'/c)^2) slower than for the local static observer that coincides in the same place, with the static observer's clock ticking dt' / dt = sqrt(1 - 2 m / r) slower than the distant observer's own clock, so the distant observer measure's a time dilation for the freefaller of (dτ / dt) = (dτ / dt') (dt' / dt) = sqrt(1 - (v'/c)^2) sqrt(1 - 2 m / r). Okay, so we have

K = (1 - (v'/c)^2) / (1 - 2 m / r)

1 - (v'/c)^2 = K (1 - 2 m / r)

v'^2 = c^2 [1 - K (1 - 2 m / r)]

and the metric becomes

v'^2 = v_r'^2 + v_t'^2

v'^2 = (dr' / dt')^2 + (dθ' r'/ dt')^2

v'^2 dt'^2 = dr'^2 + (dθ' r')^2

v'^2 dt^2 (1 - 2 m / r) - dr^2 / (1 - 2 m / r) - dθ^2 r^2 = 0

and substituting for v'^2,

c^2 [1 - K (1 - 2 m / r)] dt^2 (1 - 2 m / r) - dr^2 / (1 - 2 m / r) - dθ^2 r^2 = 0

c^2 dt^2 (1 - 2 m / r) - dr^2 / (1 - 2 m / r) - dθ^2 r^2 = c^2 dt^2 K (1 - 2 m / r)^2

ds^2 = c^2 dt^2 K (1 - 2 m / r)^2

ds^2 = c^2 dt^2 [(1 - (v'/c)^2) / (1 - 2 m / r)] (1 - 2 m / r)^2

ds^2 = c^2 dt^2 (1 - (v'/c)^2) (1 - 2 m / r)

ds = c [dt sqrt(1 - (v'/c)^2 sqrt(1 - 2 m / r)]

ds = c dτ

 

[grav-universe]

Okay, now I want to work out what the proper time of the freefaller will be upon reaching a radius h when falling from rest from a radius b with an original clock reading of zero. So we have

$$v'^2 = v_r'^2 + v_t'^2$$

where $$v'^2 = c^2 [1 - K (1 - 2 m / r)]$$ and $$v_t' = 0$$ for a particle falling radially, giving

$$c^2 [1 - K (1 - 2 m / r)] = (dr' / dt')^2$$

$$c^2 [1 - K (1 - 2 m / r)] \left(\frac{d\tau}{\sqrt{1 - (v'/c)^2}}\right)^2 = \left(\frac{dr}{\sqrt{1 - 2 m / r}}\right)^2$$

$$c^2 [1 - K (1 - 2 m / r)] \ d\tau^2 = K \ dr^2$$

$$d\tau^2 = \frac{dr^2}{c^2 [\frac{1}{K} - (1 - 2 m / r)]}$$

where

$$K = \frac{(1 - (v_b/c)^2)}{(1 - 2 m / b)}$$ for a particle falling from rest at b, where also $$v_b = 0$$ in that case, so

$$K = \frac{1}{(1 - 2 m / b)}$$

$$d\tau^2 = \frac{dr^2}{c^2 [(1 - 2 m / b) - (1 - 2 m / r)]}$$

$$\tau = \ _h\int^b \frac{dr \sqrt{r}}{c \ \sqrt{2 m} \sqrt{1 - \frac{r}{b}}}$$

$$\tau = \left(\frac{b}{c}\right)\left[\sqrt{\frac{h}{2 m}} \sqrt{1 - h / b} + \sqrt{\frac{b}{2 m}} \tan^{-1} (\sqrt{b / h - 1}) \right]$$

When falling to h = 2 m where b >> 2m, the atan approaches π / 2 and the first term in brackets helps to make up the difference, so we can approximate (the same as using h = 0) with

$$\tau \approx \left(\frac{b}{c}\right) \sqrt{\frac{b}{2 m}} \left(\frac{\pi}{2}\right)$$

 

[grav-universe]

Geesh. All I really needed to show that ds = c dτ, without any reference to energy, is

v'^2 = v_r'^2 + v_r'^2

for the locally measured radial and tangent speed of a freefaller with scalar speed v', which transforms to

v'^2 = (dr' / dt')^2 + (dθ' r' / dt')^2

v'^2 dt'^2 = dr'^2 + dθ'^2 r'^2

c^2 dt'^2 - [v'^2 dt'^2] = c^2 dt'^2 - [dr'^2 + dθ'^2 r'^2]

c^2 (1 - (v'/c)^2) dt'^2 = c^2 dt'^2 - dr'^2 - dθ'^2 r'^2

where dτ = sqrt(1 - (v'/c)^2) dt' as the clock of the freefaller is observed locally, so

c^2 dτ^2 = c^2 dt'^2 - dr'^2 - dθ'^2 r'^2

ds^2 = c^2 dτ^2 = c^2 dt'^2 - dr'^2 - dθ'^2 r'^2 = c^2 dt^2 (1 - 2 m / r) - dr^2 / (1 - 2 m / r) - dθ^2 r^2