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OK.Let’s have an example of this.I would like to calculate what speed a ball is travelling at when it hits the ground after I drop it from a height h. How long does it take to hit the ground?I will solve it both ways: once using Newtonian physics, and once using Einstein’s General Relativity. I will use minimal assumptions — just purely using the equations of motion.Let’s see which method is better, right?This answer has a fair amount of mathematics (that’s sort of the point I’m trying to make), so stick with it. You don’t need to understand exactly what I’m doing — just the general gist.The Classical SolutionThe really simple solution to this would be to say that we know that [math]g \approx 9.8[/math]m/s[math]^2[/math], and then use “suvat” to solve it — but that is only an approximation when the gravitational field is constant, which it isn’t. Let’s do the full treatment with a varying gravitational field. It adds a few lines, but you’ll see why we do it later.Newton’s theory (along with Gauss’ theorem) says that we can say that the gravitational attraction due to the Earth (mass [math]M[/math]) on an object of mass [math]m[/math], which is a distance [math]r[/math][math][/math] away from the centre of the Earth, is given by:[math]\displaystyle \mathbf{F} = - \frac{GMm}{r^2} \hat{r} \tag*{}[/math]Since the object in freefall is of constant mass, we can write:[math]\displaystyle \mathbf{F} = m \mathbf{a} = m \frac{\mathrm d \mathbf{v}}{\mathrm d t} \tag*{}[/math]We now make the simplifying assertion that since our ball starts out at rest ([math]\mathbf{v}(0) = 0[/math]), and the only force present is radial in direction, that [math]\mathbf{v}[/math] is only ever radial in direction — i.e. we ignore everything apart from the radial coordinate.This means:[math]\displaystyle \frac{\mathrm d v_r}{\mathrm d t} = - \frac{GM}{r^2} \tag*{}[/math]We now use the following fact:[math]\displaystyle \frac{\mathrm d v_r}{\mathrm d t} = \frac{\mathrm d v_r}{\mathrm d [/math][math]r[/math][math]} \frac{\mathrm d r}{\mathrm d t} \tag*{}[/math]But [math]\frac{\mathrm d r}{\mathrm d t}[/math]is just the radial velocity:[math]\displaystyle v dv = -GM \frac{dr}{r^2} \tag*{}[/math]Integrate, starting from [math]r[/math][math] = R_p + h[/math] at [math]v=0[/math] to [math]r[/math][math] = R_p + x[/math] at [math]v = v(t)[/math].Here “[math]x[/math]" is the height above the ground, and [math]R_p[/math] is the radius of the planet.Hence:[math]\displaystyle v(x) = \sqrt{ 2 GM \left(\frac{1}{R_p + x} - \frac{1}{R_p + h} \right)} \tag*{}[/math]We can now answer the “speed when it hits the ground” question.Setting [math]x = 0[/math], gives:[math]\displaystyle v(0) = \sqrt{ 2GM \left(\frac{1}{R_p } - \frac{1}{R_p + h} \right)} \tag*{}[/math]For a height of 100m, this is:[math]\displaystyle v(0) \approx 44.3 \text{m/s} \tag*{}[/math]The time taken to fall is actually somewhat more difficult — it’s not a particularly nice closed-form expression, but it is the answer to the following integral:[math]\displaystyle t_{fall} = \int_0^h \frac{dx}{\sqrt{ 2GM \left( \frac{1}{R_p + x} - \frac{1}{R_p + h} \right) }} \tag*{}[/math]For [math]h = 100[/math], this comes out as…[math]\displaystyle t_{fall} = 4.5 \text{seconds} \tag*{}[/math]There. That wasn’t too bad, was it?Of course — it’s Newtonian physics, so we know that it must be totally wrong, right?General Relativity SolutionHere’s the General Relativistic approach to the same question — the same height, and the same planet.The equations of motion in GR are the geodesic equations, which take the form[math]\displaystyle \ddot{x}^\mu = \Gamma^\mu_{\nu \lambda} \dot{x}^\nu \dot{x}^\lambda \tag*{}[/math]Where[math]\dot{x}^\mu = \frac{\mathrm d x^\mu}{d p}[/math] where [math]p[/math] is some affine parameter.[math]\Gamma^\alpha_{\beta \gamma}[/math] are the Christoffel symbols derived from the metric.We will model the Earth as being spherically symmetric, which means (by Birkhoff's theorem), it is described by the Schwarzschild metric:[math]\displaystyle -c^2 \mathrm d\tau^2 = g_{\mu \nu} \mathrm d x^\mu \mathrm d x^\nu \tag*{}[/math][math]\displaystyle -c^2 \mathrm d \tau^2 = -c^2 \left( 1 - \frac{R_s}{r}\right) \mathrm d t^2 + \left( 1 - \frac{R_s}{r}\right)^{-1} \mathrm dr^2 + r^2 \mathrm d \Omega^2 \tag*{}[/math]We make our same approximation as before, that there will only be radial motion, so all [math]\theta[/math] and [math]\phi[/math] components drop out.We are therefore left with a need to calculate the following components:[math]\Gamma^r_{r r}[/math][math]\Gamma^r_{r t} = \Gamma^r_{t r}[/math][math]\Gamma^r_{t t}[/math][math]\Gamma^t_{r r}[/math][math]\Gamma^t_{t r} = \Gamma^t_{r t}[/math][math]\Gamma^t_{t t}[/math]Where:[math]\displaystyle \Gamma^a_{b c} = \frac{1}{2} g^{a \sigma} \left( \partial_c g_{\sigma b} + \partial_b g_{\sigma c} - \partial_\sigma g_{bc} \right) \tag*{}[/math]Normally, this would be a whole exercise in itself, but I’m feeling kind today, so I’ll just look them up:[math]\Gamma^r_{r r} = -\frac{GM}{c^2r^2} \left( 1- \frac{R_s}{r}\right)^{-1}[/math][math]\Gamma^r_{r t} = \Gamma^r_{t r} = 0[/math][math]\Gamma^r_{t t} = \frac{GM}{c^2 r^2} \left( 1- \frac{R_s}{r}\right)[/math][math]\Gamma^t_{r r} = 0[/math][math]\Gamma^t_{t r} = \Gamma^t_{r t} = \frac{GM}{c^2 r^2} \left( 1- \frac{R_s}{r}\right)^{-1}[/math][math]\Gamma^t_{t t} = 0[/math]We now perform a transform from our affine parameter [math]p[/math], to the coordinate time [math]t[/math] to give our geodesic equations in the form:[math]\displaystyle \ddot{x}^\mu = -\Gamma^\mu_{\alpha \beta} \dot{x}^\alpha \dot{x}^\beta + \Gamma^t_{\alpha \beta} \dot{x}^\alpha \dot{x}^\beta \dot{x}^\mu \tag*{}[/math]If [math]\mu = t, \theta, \phi[/math], this equation is equal to zero.Hence, we are left with:[math]\displaystyle \ddot{r} = -\Gamma^r_{\alpha \beta} \dot{x}^\alpha \dot{x}^\beta + \Gamma^t_{\alpha \beta} \dot{x}^\alpha \dot{x}^\beta \dot{r} \tag*{}[/math][math]\displaystyle \ddot{r} = -\Gamma^r_{r [/math][math]r[/math][math]} \dot{r}^2 - \Gamma^r_{t t} c^2 \dot{t}^2 + 2 \Gamma^t_{r t} \dot{r} \dot{t} \dot{r} \tag*{}[/math]Plugging in all of the values of [math]\Gamma[/math], and noting that [math]\dot{t} = 1[/math], gives:[math]\displaystyle \ddot{r} = \frac{GM}{c^2 r^2} \left( 1- \frac{R_s}{r} \right)^{-1} \dot{r}^2 - \frac{GM}{r^2} \left( 1- \frac{R_s}{r} \right) + \frac{2GM}{c^2 r^2} \left( 1- \frac{R_s}{r} \right)^{-1} \dot{r}^2 \tag*{}[/math]Which simplifies to:[math]\displaystyle \ddot{r} = \frac{3GM}{c^2 r^2} \left( 1- \frac{R_s}{r} \right)^{-1} \dot{r}^2 - \frac{GM}{r^2} \left( 1- \frac{R_s}{r} \right) \tag*{}[/math]We need to integrate this.Unfortunately, this is not a very nice thing to integrate — you can perfectly well plug it into a computer, and it will do the calculations for you.But we notice that we can plug in some numbers, and for example:[math]\displaystyle \frac{3 GM}{c^2 r^2}\left( 1- \frac{R_s}{r} \right)^{-1} \approx 3 \times 10^{-16} \tag*{}[/math]This is true for all values of [math]r[/math][math][/math] above the planet’s surface.In other words — unless the velocity of our ball is around 20% the speed of light, this term is tiny — absolutely tiny.If the ball is moving at “normal” speeds, then it is of order [math]10^{-13}[/math]In comparison, the second term is approximately [math]10^{16}[/math] times larger — this term is definitely going to be the one that matters.So, we make the approximation that:[math]\displaystyle \ddot{r} \approx - \frac{GM}{r^2} \left( 1- \frac{R_s}{r} \right) + \approx 10^{-13} \tag*{}[/math]But then, we ask — what is [math]R_s[/math]?[math]R_s[/math] is small for a planet the size of Earth — around 9mm.Therefore, [math]R_s/r[/math] is around [math]10^{-9}[/math].Do you think we can ignore that as well? I think we probably can….[math]\displaystyle \ddot{r} \approx - \frac{GM}{r^2} + \approx 10^{-9} \tag*{}[/math]Ah! Now we can solve this!But….But….But….That equation is the almost exact same one that we solved in the classical section of this answer!Classical equation:[math]\displaystyle \ddot{r} = \frac{\mathrm d v_r}{\mathrm d t} = - \frac{GM}{r^2} \tag*{}[/math]GR equation:[math]\displaystyle \ddot{r} \approx - \frac{GM}{r^2} + \approx 10^{-9} \tag*{}[/math]So we spent all that time and effort messing around with GR stuff and metrics….And what does it get us?For a planet the size of Earth, a correction on the order of [math]10^{-9}[/math].Honestly, if you exhaled on the ball on the way down it would have a bigger effect than [math]10^{-9}[/math].We spent sodding ages on that mathematics, and it got us a correction smaller than farting on the goddamn ball.I’m honestly not joking. Accidental nearby flatulence is a more serious correction than General Relativity.So why do we still learn Newtonian mechanics?Because it is so much quicker!Even the method I used in this answer was unnecessarily long and laborious.You can solve the “how fast is it going” question in a single line, using conservation of energy.In general relativity, however?The maths is much, much, much harder — and we weren’t even dealing with any of the conceptually difficult stuff, this was just raw number crunching.And how much more accurate was it?It was no more accurate.To all intents and purposes, both results gave the exact same answer.Relativity is necessary in some regimes — but in the “everyday” regime….using General Relativity is like googling how to spell your own name.Sure, you’ll get the right answer, but there are quicker ways to do it….So that is why we still learn Newtonian mechanics — because 99.99% of the time there is no realistic difference between the two answers.Using General Relativity is a pain.You want to avoid it and use approximations as often as possible.And Newtonian mechanics is a bloody good approximation.

Nice![math]r\cdot r\cdot y=\ln\chi+\ln m^{sa-1}[/math][math]r\cdot r\cdot y=\ln\chi m^{sa-1}[/math][math]e^{r\cdot r\cdot y}=\chi m^{as}\cdot m^{-1}[/math][math]me^{rry}=\chi m^{as}[/math]

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