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B Confusion about closed timelike curves

  1. Dec 23, 2017 #21

    RUTA

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    I’ve had students solve some problems using the Schwarzschild metric in my GR course over the years, but I’ve never done this one — simply throwing a ball up into the air from the surface of Earth. Turns out it’s tougher than the others, but let me outline it here, since the OP asked about this. Please point out any errors, so I can fix them before making this an Insight.
    Here is the Schwarzschild metric in units of ##G = c = 1##:
    $$ ds^2 = -\left(1 - \frac{2M}{r}\right)dt^2 + \left(1 - \frac{2M}{r}\right)^{-1}dr^2 + r^2d\Omega^2 \label{metric} $$
    ##d\Omega## is the solid angle, we won’t be using the angular coordinates since our motion is strictly radial. Here are the two geodesic equations of relevance:
    $$ \frac{d^2r}{dp^2} + \Gamma^{r}_{tt}\left(\frac{dt}{dp}\right)^2 + \Gamma^{r}_{rr}\left(\frac{dr}{dp}\right)^2 = 0 \label{RgdscEq} $$
    and
    $$ \frac{d^2t}{dp^2} + 2\Gamma^{t}_{rt}\left(\frac{dt}{dp}\right) \left(\frac{dr}{dp}\right) = 0 \label{TgdscEq} $$
    where ##p## is proper time along the geodesic. We have:
    $$ \Gamma^{r}_{tt} = \frac{M}{r^2} \left(1 - \frac{2M}{r}\right) \label{Christoffelrtt} $$
    $$ \Gamma^{r}_{rr} = -\frac{M}{r^2} \left(1 - \frac{2M}{r}\right)^{-1} \label{Christoffelrrr} $$
    $$ \Gamma^{t}_{rt} = \frac{M}{r^2} \left(1 - \frac{2M}{r}\right)^{-1} \label{Christoffelttr} $$
    This gives:
    $$ \frac{d}{dp}\left(\left(1 - \frac{2M}{r}\right)\frac{dt}{dp}\right) = 0 \label{TgdscEq2} $$
    for our second geodesic equation, which means:
    $$ \left(1 - \frac{2M}{r}\right)\frac{dt}{dp} = B \label{TgdscEq3} $$
    with ##B## a constant. At ##r = \infty## and ##v = 0 ## the metric tells us that ##dp = dt ##, since ## ds^2 = -dp^2 ## along the geodesic, so the second geodesic equations tells us ##B = 1 ## when our object is launched with escape velocity ##v = v_e ##. If we get to ##r = \infty## with ## v > 0 ##, then our metric tells us
    $$ \frac{dp}{\sqrt{1-\frac{v^2}{c^2}}} = dt \label{SReqn} $$
    (I restored ##c## here) which tells us ##\frac{dt}{dp} > 1 ## so our second geodesic equation says ##B > 1 ## when the object is launched with ##v > v_e ##. Therefore, we assume ##B < 1## when the object is launched with ##v < v_e ##. We’ll need this info on ##B## when we get to our result.
    Using this result and putting our other two Christoffel symbols into our first geodesic equation gives
    $$ \left(1 - \frac{2M}{r}\right)\frac{d^2r}{dp^2} + \frac{M}{r^2}\left(B^2 - \left(\frac{dr}{dp}\right)^2 \right) = 0 \label{RgdscEq2} $$
    Now I will start making approximations to show this leads to the Newtonian conservation of energy equation:
    $$ \frac{1}{2}v^2 - \frac{GM}{r} = \frac{E}{m} \label{Newton} $$
    where ##\frac{E}{m}## is the (conserved) total energy per unit mass of the launched object (I restored ##G##). Start by assuming ##\frac{d^2r}{dp^2} = -g = -\frac{GM}{R^2}## where ##R## is the radius of Earth and M is the mass of Earth, i.e., ##g = 9.8 m/s^2## per usual. Next let ##r = R+y## so that ##\frac{dr}{dp} = \frac{dy}{dp} = v##, i.e., ##y## is the height above Earth’s surface of our projectile; we will assume ##y## is small compared to ##R##. Now our first geodesic equation gives us:
    $$ -\left(1 - \frac{2GMR}{c^2R^2\left(1+y/R\right)}\right)g + \frac{GM}{R^2\left(1+y/R\right)^2}\left(B^2 - \frac{v^2}{c^2} \right) = 0 \label{RgdscEq3} $$
    where I have restored ##G## and ##c##. Expanding:
    $$\left(1 + y/R \right)^{-1} \approx 1 – y/R \label{approx1} $$
    and
    $$ \left(1 + y/R \right)^{-2} \approx 1 – 2y/R \label{approx2} $$
    and putting these into our first geodesic equation and rearranging we obtain:
    $$\left(\frac{1}{2} - \frac{y}{R}\right)v^2 + gy = \left(\frac{1}{2} - \frac{y}{R}\right)B^2c^2 - \frac{c^2}{2} + gR \label{RgdscEq4} $$
    Since ##\frac{y}{R} \ll 1 ##, ##\left(\frac{1}{2} - \frac{y}{R}\right) \approx \frac{1}{2}## and our first geodesic equation is now:
    $$\frac{1}{2}v^2 + gy = \frac{\left(B^2 - 1\right)c^2}{2} + gR \label{RgdscEq5} $$
    The ##gy## on the LHS and ##gR## on the RHS come from ##-\frac{GM}{R\left(1+y/R\right)}## expanded on the LHS of the Newtonian conservation of energy equation then using ## g = \frac{GM}{R^2}##. That means ## \frac{\left(B^2 - 1\right)c^2}{2}## is our total conserved energy per unit mass ##\frac{E}{m}## from the RHS. Notice that for ##v = v_e## at launch we have ##B = 1## and the RHS of our first geodesic equation is just ##gR## as expected (total energy equals zero). If ##v > v_e## at launch, ##B > 1## and the RHS of our first geodesic equation is a little larger than ##gR## (total energy is positive). If ##v < v_e## at launch, ##B < 1## and the RHS is a little smaller than ##gR## (total energy is negative).
    That concludes my outline of how the geodesic equations for the Schwarzschild metric give rise to standard Newtonian mechanics for an object launched slowly upwards near the surface of Earth. I’ll provide the details in an Insight along with other problems one can solve with the Schwarzschild metric at a later date.
     
    Last edited: Dec 26, 2017
  2. Dec 23, 2017 #22
    Above my head right now, but I’m sure others can benefit from your considerable effort. Thanks again.
     
  3. Dec 23, 2017 #23

    PeterDonis

    Staff: Mentor

    As @laymanB implies, this is probably too heavyweight for a "B" level thread. But it does look like excellent material for an Insight (or a series of them). :wink:
     
  4. Dec 24, 2017 #24

    RUTA

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    There are several Insights on the Schwarzschild metric and problems it can solve (a series written by you, for example). I’ll look through them carefully and add the problems I’ve solved that don’t appear :-)
     
  5. Feb 6, 2018 #25
    What kinds of physical phenomena would we expect to see in a spacetime containing closed timelike curves?
     
  6. Mar 9, 2018 #26

    JMz

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    I think a better phrasing of this conclusion is that CTCs are inconsistent with inflation.
     
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