# 3.4: Local Inertial Reference Frames

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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?

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?

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.

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 $$r$$.

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.