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Suppose a body is travelling at the speed of light minus 2.998 $m/s$ when it gets near the Sun where nominal $g = 270 m/s^2$, what is the actual increase of the speed in a second?

What is the formula to find the effective acceleration and increase of KE ?

edit: if speed is .99 99 99 99 c (2997945500.2 m/s) I find that considering g constant for a whole second before it crashes onto the Sun, speed/s increases roughly by 5 microns. is that anytwhere near the right value?

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    $\begingroup$ by "real" do you mean in the rest frame of the sun? $\endgroup$
    – JMLCarter
    Jan 27 '17 at 12:19
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    $\begingroup$ While it is not my downvote, (a) what you are asking about here is exactly equivalent to every other 'I'll add velocities and go faster than the speed of light' notion and is resolved with the velocity addition formula just like all the others, and (b) the word 'real' in this kind of concept is usually a marker for not really internalizing the relativity principle---coordinate acceleration is a relative context and all observers have the same claim to correctness. $\endgroup$ Jan 27 '17 at 20:12
  • $\begingroup$ @lambertwhite: I didn't downvote, but I can sympathize with those who did because you've shown zero effort in calculating anything and are asking us to do a somewhat complicated problem. Did you get that "1.85 micron" prior to John's answer or after? And how does a micron, a unit of distance, relate to a reduction in velocity here? $\endgroup$
    – Kyle Kanos
    Jan 28 '17 at 12:00
  • $\begingroup$ But to get to that formula involves solving a complicated problem. You also need to check your units. $c$ has units m/s so it makes little sense to "subtract" 1 m from it (or 1.85 microns). $\endgroup$
    – Kyle Kanos
    Jan 28 '17 at 12:11
  • $\begingroup$ @dmckee, what you are asking about here is exactly equivalent to every other 'I'll add velocities and go faster than the speed of light' notion and is resolved with the velocity addition formula just like all the others...could you please confirm that I can simply use the velocity addition calculator and forget the complicated way suggested by John Rennie? $\endgroup$
    – user137879
    Feb 9 '17 at 15:56
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The trajectory of a freely falling body, like a comet is given by the geodesic equation:

$$ \frac{d^2x^\alpha}{d\tau^2} = - \Gamma^\alpha{}_{\mu\nu}\frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} \tag{1} $$

Unless you're a GR head this probably looks appalling, but it isn't as complicated as it seems. The symbol $x^\alpha$ is just a position in spacetime. For example if we are using polar coordinates, $(t, r, \theta, \phi)$, $x^0$ is time, $x^1$ is radial distance $r$, $x^2$ is the equatorial angle $\theta$ and $x^3$ is the latitude angle $\phi$.

If we take the comet to be falling directly towards the Sun then the equatorial and latitude angles don't change so:

$$ \frac{d\theta}{d\tau} = \frac{d\phi}{d\tau} = 0$$

and the acceleration towards the Sun is just given by the second derivative of $x^1 = r$ i.e.

$$ \frac{d^2r}{d\tau^2} = - \Gamma^r{}_{\mu\nu}\frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} \tag{2} $$

And equation (2) is now starting to look more useful because the quantity on the left looks like an acceleration $d^2/dt^2$. However there is one more step to take. The variable $\tau$ is the proper time of the comet, i.e. the time recorded by a clock bolted to the comet, and this isn't the same as our time because for a fast moving comet time is dilated. What we want is to replace the comet time $\tau$ with our time $t$. I won't go into exactly how this is done because it's fiddly (though straightforward) and the result is:

$$ \frac{d^2r}{dt^2} = - \Gamma^r{}_{\mu\nu}\frac{dx^\mu}{dt} \frac{dx^\nu}{dt} + \Gamma^t{}_{\mu\nu}\frac{dx^\mu}{dt} \frac{dx^\nu}{dt} \frac{dr}{dt} \tag{3} $$

Now we have the radial acceleration on the left side, which is exactly what you asked for i.e. the acceleration we measure the comet to have. So we just need to calculate all the stuff on the right hand side. But this quickly gets tedious. The symbols $\Gamma^\alpha_{\,\,\mu\nu}$ are called Christoffel symbols and they describe the spacetime curvature caused by the Sun. These are complicated functions, for example:

$$ \Gamma^r{}_{tt} = \frac{GM}{r^2}\left(1 - \frac{2GM}{c^2r}\right) $$

so although it's straightforward algebra it quickly gets long and messy to write out in full.

I'm going to leave the explanation at this point (due to pressure of time). If you want to have a go at calculating it yourself the Christoffel symbols for a spherical body like the Sun are given here. For a radial trajectory, i.e. constant $\theta$ and $\phi$, the only relevant non-zero Christoffel symbols are:

$$ \Gamma^r{}_{tt} = \frac{GM}{r^2}\left(1 - \frac{2GM}{c^2r}\right) $$

$$ \Gamma^r{}_{rr} = - \frac{\frac{GM}{r^2}}{1 - \frac{2GM}{c^2r}} $$

$$ \Gamma^t{}_{tr} = \Gamma^t{}_{rt} = -\Gamma^r{}_{rr} = \frac{\frac{GM}{r^2}}{1 - \frac{2GM}{c^2r}} $$

And equation (3) becomes:

$$\begin{align} \frac{d^2r}{dt^2} &= - \Gamma^r{}_{tt} - \Gamma^r{}_{rr}\frac{dr}{dt} \frac{dr}{dt} + 2\Gamma^t{}_{tr}\frac{dr}{dt} \frac{dr}{dt} \\ &= - \Gamma^r{}_{tt} - 3\Gamma^r{}_{rr}\left(\frac{dr}{dt}\right)^2 \end{align}$$

Substitute in the expressions for the Christoffel symbols, and remember that $dr/dt$ is just the radial speed of the comet e.g. $dr/dt = c-1$ in your question.

If doing the full calculation seems intimidating you might be interested to calculate the acceleration of a stationary object i.e. an object with $dr/dt = 0$. In that case the expression simplifies to:

$$ \frac{d^2r}{dt^2} = - \Gamma^r{}_{tt} = -\frac{GM}{r^2}\left(1 - \frac{2GM}{c^2r}\right) = -\frac{GM}{r^2}\left(1 - \frac{r_s}{r}\right) $$

where $r_s$ is the event horizon radius:

$$ r_s = \frac{2GM}{c^2} $$

If the radial distance is far greater than the event horizon radius then the equation simplifies to:

$$ \frac{d^2r}{dt^2} \approx -\frac{GM}{r^2} $$

which is just the Newtonian expression for the gravitational acceleration. So reassuringly our equation gives the correct result in the non-relativistic limit. However something rather unexpected happens as we move inwards, i.e. as $r \to r_s$, but I'll leave that for you to explore.

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  • $\begingroup$ In a comment above dmckee said: "what you are asking about here is exactly equivalent to every other 'I'll add velocities and go faster than the speed of light' notion and is resolved with the velocity addition formula just like all the others", can I simply use a velocity addition calculator like this to spare all complex operations? Do I get an acceptable approximation? $\endgroup$
    – user137879
    Feb 9 '17 at 16:11