3.4: Local Inertial Reference Frames

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.

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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 ($r=3M$).