General Response to #48, #49, #51, #52
Thanks for all the helpful comments, especially the detailed explanation in #51. I wanted to initially respond by trying to explain my general approach and then respond to specific points after I have had a chance to work through the details provided.
- First, apologises for the incorrect sign in equation [1] #47, as picked up by kev #51.
- Second, while appreciating Nereid’s comments, my reasons for wanting to compare the approximation of Newtonian physics with GR were mainly in response to the Jim-Jon exchanges. However, as a general comment, while the issues surrounding the perihelion of Mercury and bending of light support GR, the effects in absolute terms are quite small. Therefore, I will try to better explain my rationale below.
From a learning perspective, I find it useful to try to anchor some sort of physical interpretation on all the maths; otherwise concepts can quickly be lost in abstraction. Hence my correlation of GR theory to Newtonian physics, but please note I am not advocating a Newtonian preference. However, I did want to test Jim’s statement that `
Newtonian predictions are wrong` rather than the usual implication that they were still a good approximation under most circumstances. Therefore, the point of my original equations (1-4) in #47 was to try to narrow the overall complexity, as alluded to in Garth’s comments in #48, to a very specific, and possibly over-simplistic, example in the hope that some physical interpretation might be more obvious. For example, equation [2] is the free-fall velocity as perceived by both the onboard observer and any local observer at [r].
[2] v_R = c\sqrt {\frac {Rs}{r}}
It is possibly worth highlighting that [2] is also the perspective derived from Newtonian physics. As such, what are the real physical implications of equation [#47:1], if the velocity, as perceived by the distant observer, slows under free-fall gravitational acceleration to zero?
Based on what appears to be the most
‘real’ physical interpretation, and I realize that some may question the applicability of the word ‘
real’, I then tried to correlate the implications of equation #2 on equations [3,4]. In this respect, I was focusing on the onboard observer, while Kev in #51 appears to have focused more on the distant observer. Equation [3] being the Newtonian form, while [4] is derived from Schwarzschild’s metric and therefore assumed consistent with GR under the general caveats assumed by this solution. However, the implication on [3] and [4] for an onboard free-falling observer is that orbital velocity [vo=0] and, as such, both equations appear to collapse to the form in [5], i.e.
[3] Newtionian: E_T = 1/2 m\left(v_r^2 + v_o^2\right) +\left (-\frac{GMm}{r} \right)
[4] GR: E_T =1/2 m\left(v_r^2 + v_o^2\right) - \frac{GMm}{r} \left( 1+ \frac{v_o^2}{c^2}\right)
[5] E_T = 1/2 mv_R^2 - \frac{GMm}{r}
Substituting for [v_R] and Rs=2GM/c^2] leads to the form:
[6] E_T = \frac{GMm}{r} - \frac{GMm}{r} = 0
Note, this is only valid for the radially solution, not an orbiting one; also this equation appears not to account for any relativistic effects, i.e. velocity or gravitation as normally associated with [\gamma]. However, the form of equation [6] has transpose the kinetic energy associated with [v_R] into an equivalence amount of potential energy via equation [2]. In this form, the only quantity that would be subject to relativistic effects is [m], so any value of [\gamma] would cancel out in [6].
As such, energy would seem to be conserved based on potential energy being converted to kinetic energy, noting that classically, potential energy is negative ranging from a maximum of zero at an infinite [r] to a minimum, i.e. maximum negative, as the centre of mass [M] is approach. It is noted that free-falling into a black hole raises an anomaly when [v=c] at [r=Rs]. However, generally, the conclusion being forwarded seems to contradict the statement made in #51and so I would like to better understand if I have made a wrong assumption or misinterpreted what was being said in this specific case?
#51: It should also be clear that energy is not conserved according to measurements by local observers and Garth has also indicated it is not conserved from the point of view of the falling particle.
Referencing
`Exploring Black Holes` by Taylor and Wheeler p3-12/section-5:
The fact that E/m is constant for a free particle yields a great simplification in describing the motion of a radially plunging particle
Equation [10] in this reference also gives an equation of the form:
[7] \frac{E}{mc^2} = \left(1-Rs/r\right) \frac{dt}{d\tau} = 1
From which it might be assumed that:
[8] \frac{dt}{d\tau} = \left(1-Rs/r\right)^{-1}
This raised the question in my mind as to whether the relativistic factor [\gamma] could be physically interpreted as follows:
[9] \gamma = \frac{1}{\sqrt(1-v^2/c^2)}*\frac{1}{\sqrt(1-Rs/r)}
However, in the specific case of a free-falling observer, velocity is also proportional to radius such that:
[10] \frac{1}{\sqrt(1-v^2/c^2)} = \frac{1}{\sqrt(1-Rs/r)}
I believe this is consistent to Kev’s statement in #51. However, in #49, Jim presented the following equation and I would therefore like to clarify that the use of the metric tensor [g00] is essentially equivalent to equation [7] given the assumption of this specific case?
E = \frac{m}{\sqrt{1-\frac{v^2}{c^2}}}c^2\sqrt{g_{00}}
This post is already too long, and possibly drifting off the main thrust of this thread, so I will terminate at this point, but will continue to review the detailed points already raised. Many thanks
.