Let r be the axis of motion, and t be the axis of time. Supposing for a moment we express velocity, as observed from a sufficiently distant observer, as v(r)=c * sin(θ), and rate-of-time as v(t)=c * sin(ω) - such that, in terms of the Lorentz factor γ, v(t) = c*γ, or sin(ω) = γ. That is, supposing we express velocity as a rotation of the plane of the axes of time and the direction of travel relative to some observer, such that θ+ω=pi/2. θ (and thus ω) is defined as rotation relative to the observer’s orientation.
Let the curvature at a given point in spacetime be expressed as κ. I think the equation for acceleration might take something like the form dθ/dr = κ(r) * cos(θ).
Struggling with the math for this. Curvature expresses the radius of a circle; what I’m looking for is something more like torque. Rotational acceleration.
Why is there rotational acceleration? Because the near side and far side of a particle are instantaneously moving at different velocities. Why is the rotation in time? Because the disparity in velocity exists with respect to the axes corresponding to time (future light-cone) and distance (to the gravitational body).
This is going to be proportional to the difference in velocity. κ(r) represents the instantaneous curvature. cos(θ) is expected because as θ approaches pi/2 (as spacial velocity approaches the speed of light), the acceleration approaches 0. We need to multiply by c here as well, since that is the rate of change; dθ/dr = c * κ(r) * cos(θ).
For acceleration in terms of m/s, I get the following equation:
Let M be the mass of the gravitational body, let G be the gravitational constant, let c be the speed of light.
Let rs=2*M*G/c^2 (Schwarzschild radius)
Let κ(r) = c / (2*r^2 * (1-rs/r)^1/2) ← I don’t understand tensors well enough to use the Ricci curvature, so I invented my own curvature that I know represents the values I’m interested in, which is the derivative of Schwartzschild time dilation as t0/tf.
Let θ = asin(v(r) / c) ← We need to convert the velocity into a form we can work with. I am assuming all velocity is either directly towards the source of gravity; I don’t know if, or how, the equation will change if this assumption is invalid.
Then the acceleration experienced by a particle in a gravitational field, as measured by an independent observer, should be:
acceleration (m/s^2) = c * k(r) * cos(θ) * rs
I’ve tested the equation; it appears to work, except for one problem: That last factor, the Schwarzschild radius, makes no sense to me here; the actual acceleration the particle experienced using my original equation was incorrect. I expected -a- value there, because otherwise the units wouldn’t make sense. This may be a product of the way I reinvented curvature.
However, none of this is actually what I set out to do, since we’re using m/s^2, instead of radians/s^2; that is, I was originally setting out to figure out either dθ/dt^2 or dθ/dr. What I ended up with doesn’t actually look like what I wanted.
Notes to myself:
Let r be the axis of motion, and t be the axis of time. Supposing for a moment we express velocity, as observed from a sufficiently distant observer, as v(r)=c * sin(θ), and rate-of-time as v(t)=c * sin(ω) - such that, in terms of the Lorentz factor γ, v(t) = c*γ, or sin(ω) = γ. That is, supposing we express velocity as a rotation of the plane of the axes of time and the direction of travel relative to some observer, such that θ+ω=pi/2. θ (and thus ω) is defined as rotation relative to the observer’s orientation.
Let the curvature at a given point in spacetime be expressed as κ. I think the equation for acceleration might take something like the form dθ/dr = κ(r) * cos(θ).
Struggling with the math for this. Curvature expresses the radius of a circle; what I’m looking for is something more like torque. Rotational acceleration.
Why is there rotational acceleration? Because the near side and far side of a particle are instantaneously moving at different velocities. Why is the rotation in time? Because the disparity in velocity exists with respect to the axes corresponding to time (future light-cone) and distance (to the gravitational body).
This is going to be proportional to the difference in velocity. κ(r) represents the instantaneous curvature. cos(θ) is expected because as θ approaches pi/2 (as spacial velocity approaches the speed of light), the acceleration approaches 0. We need to multiply by c here as well, since that is the rate of change; dθ/dr = c * κ(r) * cos(θ).
For acceleration in terms of m/s, I get the following equation:
Let M be the mass of the gravitational body, let G be the gravitational constant, let c be the speed of light.
Let rs=2*M*G/c^2 (Schwarzschild radius)
Let κ(r) = c / (2*r^2 * (1-rs/r)^1/2) ← I don’t understand tensors well enough to use the Ricci curvature, so I invented my own curvature that I know represents the values I’m interested in, which is the derivative of Schwartzschild time dilation as t0/tf.
Let θ = asin(v(r) / c) ← We need to convert the velocity into a form we can work with. I am assuming all velocity is either directly towards the source of gravity; I don’t know if, or how, the equation will change if this assumption is invalid.
Then the acceleration experienced by a particle in a gravitational field, as measured by an independent observer, should be:
acceleration (m/s^2) = c * k(r) * cos(θ) * rs
I’ve tested the equation; it appears to work, except for one problem: That last factor, the Schwarzschild radius, makes no sense to me here; the actual acceleration the particle experienced using my original equation was incorrect. I expected -a- value there, because otherwise the units wouldn’t make sense. This may be a product of the way I reinvented curvature.
However, none of this is actually what I set out to do, since we’re using m/s^2, instead of radians/s^2; that is, I was originally setting out to figure out either dθ/dt^2 or dθ/dr. What I ended up with doesn’t actually look like what I wanted.