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3.4: Local Inertial Reference Frames

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    43247
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    The constants of motion \(E\) and \(L\) from a previous section are expressed in global coordinates, which means that they don't correspond to anything directly measurable. To turn them into something measurable, we have to use a global-to-local coordinate transformation. We did this back in Section 2.6. As a reminder the results were

    \[\begin{align*}dt_\text{LIRF}&=\sqrt{1-\frac{2M}{r}}dt\\ dx_\text{LIRF}&=r\ d\phi\\ dy_\text{LIRF}&=\frac{1}{\sqrt{1-\frac{2M}{r}}}dr\end{align*}.\]

    After applying the transformation, we get that the energy in the LIRF is

    \[\frac{E_\text{LIRF}}{m}=\gamma=\frac{1}{\sqrt{1-\frac{2M}{r}}}\frac{E}{m},\label{eq:LIRFEnergy}\]

    where \(\gamma=\frac{1}{\sqrt{1-v_\text{LIRF}^2}}\) (recall that Special Relativity rules in the LIRF, so we can use the Special Relativity definition of energy). What this allows us to do is take a known value of \(E\) and use it to determine the speed at any given r-coordinate.

    Exercise \(\PageIndex{1}\)

    Is \(E_\text{LIRF}\) a constant of motion? Is it an invariant?

    Answer

    Given that \(E_\text{LIRF}\) depends on speed, you may think that it is not a constant of motion. Recall, however, that an LIRF exists only for an infinitesimal amount of time, which means that the speed is effectively constant and therefore that \(E_\text{LIRF}\) is a constant of motion. It is not an invariant because it has different values in different reference frames.

    Exercise \(\PageIndex{2}\)

    According to Equation \ref{eq:LIRFEnergy}, does speed increase, decrease, or remain the same as an object approaches the event horizon?

    Answer

    Let's start by solving Equation \ref{eq:LIRFEnergy} for the speed.

    \[\begin{align*}\frac{1}{\sqrt{1-v_\text{LIRF}^2}}&=\frac{1}{\sqrt{1-\frac{2M}{r}}}\frac{E}{m}&&\\ \sqrt{1-v_\text{LIRF}^2}&=\sqrt{1-\frac{2M}{r}}\frac{m}{E}&&\text{take reciprocal of both sides}\\ 1-v_\text{LIRF}^2&=\left(1-\frac{2M}{r}\right)\left(\frac{m}{E}\right)^2&&\text{square both sides}\\ v_\text{LIRF}^2&=1-\left(1-\frac{2M}{r}\right)\left(\frac{m}{E}\right)^2&&\text{isolate }v_\text{LIRF}^2\\ v_\text{LIRF}&=\sqrt{1-\left(1-\frac{2M}{r}\right)\left(\frac{m}{E}\right)^2}&&\text{final answer}\end{align*}\]

    As \(r\to 2M\), \(v_\text{LIRF}\to 1\). In other words, the object approaches the speed of light! This might seem like a contradiction since in global coordinates we saw objects seemed to slow down at the event horizon. But there is no reason that someone hovering just outside the event horizon of a black hole needs to make the same observations as someone who is very far away. As weird as it seems, there is no contradiction. Welcome to General Relativity!

    Exercise \(\PageIndex{3}\)

    While nobody actually measures \(E\), anyone in an LIRF can determine it. Explain how.

    Answer

    Equation \ref{eq:LIRFEnergy} provides a prescription for how to do it. Anyone in a LIRF can take local measurements of speed. Given that they know the r-value of their reference frame, they can simply plug in to determine \(E\). Even though observers at different r-coordinates will have different measurements of the energy in their LIRF, they will all agree on the global energy \(E\).

    Exercise \(\PageIndex{4}\)

    Suppose a small stone of mass \(m\) is released from rest very far from a black hole of mass \(M\). What is the global energy \(E\) of the stone? Determine \(v_\text{LIRF}\) for the stone as a function of \(r\).

    Answer

    The global energy is given by \(E=\left(1-\frac{2M}{r}\right)\frac{dt}{d\tau}m\). Very far from the black hole, \(dt=d\tau\) and \(\frac{2M}{r}\to 0\), which means that \(E=m\). We can plug this into Equation \ref{eq:LIRFEnergy} to determine \(v_\text{LIRF}\). In fact, we already derived \(v_\text{LIRF}\) as a function of \(r\) in a previous exercise, so all we have to do is substitute \(E=m\). As a reminder, the result was

    \[v_\text{LIRF}=\sqrt{1-\left(1-\frac{2M}{r}\right)\left(\frac{m}{E}\right)^2}.\nonumber\]

    Substituting \(E=m\), we get

    \[v_\text{LIRF}=\sqrt{\frac{2M}{r}}.\]

    We see that the stone accelerates as it approaches the event horizon.


    Box \(\PageIndex{1}\)

    Use the definitions of the global-to-local coordinate transformations along with \(\frac{E}{m}=\left(1-\frac{2M}{r}\right)\frac{dt}{d\tau}\) and \(\frac{E_\text{LIRF}}{m}=\frac{dt_\text{LIRF}}{d\tau}\) to derive Equation \ref{eq:LIRFEnergy}.

    Box \(\PageIndex{2}\)

    In one of the exercises, we showed how to determine \(v_\text{LIRF}\) as a function of \(r\) for a small stone released from rest infinitely far away from a non-spinning black hole of mass \(M\). Let's repeat the process for a stone that is released from rest from a position that is not infinitely far away. That is, the stone is released from some radius \(r=r_0\), where \(r_0\) is a finite number.

    a) Write an expression for the global map energy per unit mass \(\frac{E}{m}\) in terms of \(r_0\).
    b) Use your previous answer as well as the fact that \(\frac{E}{m}\) is a constant of motion to determine an expression for \(v_\text{LIRF}\)in terms of \(r\). That is, write a function that could be used to determine the LIRF velocity at any \(r\) for a stone that is released from rest at \(r_0\). Your answer should only contain the variables \(r\), \(r_0\), and \(M\). (You can check your answer by taking the limit \(r_0\to\infty\).)

    Box \(\PageIndex{3}\)

    Now let's repeat the process from the previous Box, except this time for a stone that is thrown inward at a speed \(v_0\) from infinitely far away.

    a) Write an expression for the global map energy per unit mass \(\frac{E}{m}\) in terms of \(v_0\).
    b) Use your previous answer as well as the fact that \(\frac{E}{m}\) is a constant of motion to determine an expression for \(v_\text{LIRF}\) in terms of \(r\). Your answer should only contain the variables \(r\), \(v_0\), and \(M\). (You can check your answer by taking the limit \(v_0\to 0\).)

    Box \(\PageIndex{4}\)

    We can use the LIRF transformations to determine the speed of circular orbits in General Relativity. It can be shown (though we won't do so here) that

    \[v_\text{LIRF}=\left[1-\left(\frac{mr}{L}\right)^2\right]^{-1/2}.\label{eq:vLIRF_circular}\]

    We also saw in a previous section that, for a circular orbit,

    \[\left(\frac{L}{m}\right)^2=\frac{Mr^2}{r-3M}.\nonumber\]

    Substitute this into Equation \ref{eq:vLIRF_circular} to show that

    \[v_\text{LIRF}=\sqrt{\frac{\frac{M}{r}}{1-\frac{2M}{r}}}\]

    then use it to determine \(v_\text{LIRF}\) at the minimum stable orbit (\(r=6M\)) and the minimum unstable circular orbit (\(r=3M\)).


    This page titled 3.4: Local Inertial Reference Frames is shared under a not declared license and was authored, remixed, and/or curated by Evan Halstead.

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