It is said that we can introduce local inertial coordinates/Fermi normal coordinates for any timelike geodesic. But why only for timelike geodesics? What about null geodesics? Perhaps it has to do with invertibility or something?

# [Physics] Local inertial coordinates/Fermi normal coordinates

coordinate systemsdifferential-geometrygeneral-relativitygeodesicsreference frames

#### Related Solutions

Even in curved spacetime, you can perform a coordinate transformation at any location ("move to a freely falling frame") such that your metric is locally flat , and takes the form \begin{equation} ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2\end{equation}

If you consider a null trajectory where $ds^2 = 0$, then the above equation takes the form

\begin{equation} cdt = \sqrt{dx^2 + dy^2 + dz^2}. \end{equation}

This is the statement that "the speed of light times the differential time interval, as measured by an observer in a freely falling frame at the location in consideration, is equal to the differential physical distance traveled along the trajectory, measured by that same observer." From Einstein's equivalence principle, this is precisely the way that light must behave.

In general, coordinates mean almost nothing. And there are many conformally related metrics that are topologically compact that can contain a subset that has a metric (and domain, the subset) conformally related to a given spacetime. In other words you can put lots of different boundaries on a spacetime.

waves can reach the boundary in finite time

Reaching or not reaching something in finite coordinate is physically meaningless. For instance you can take Minkowski space $\{(t,x,y,z,): x,y,z,t \in \mathbb R\}=\mathbb R^4.$ And give it the Minkowski metric $ds^2=c^2dt^2-dx^2-dy^2-dz^2$ and you can't reach a border in finite time. But change to the coordinate $T=\arctan t$ and now you can and the border there is spacelike. Or you can instead let $X=\arctan x$ and now there is a timelike boundary a finite coordinate away. Coordinates by themselves have almost zero physical meaning. All you need is to have coordinate patches where you can have an open set and to have a locally one to one map between the coordinates and a neighborhood of events and to have a metric defined on your local coordinate system.

The relationship between the metric and the coordinates is where the true physics is. And there are lots of different spacetimes with boundary that contain Minkowski space as a subset. And it doesn't mean anything physically if one of them has a timelike boundary and one of them have a lightlike boundary. They both contain Minkowski space and Minkowski space is geodesically complete.

and, since said boundary is timelike, they can be reflected back into the bulk.

Again, I can embedded Minkowski space into a spacetime with a boundary that has a timelike boundary, but it means nothing physically. So just saying you have a timelike boundary means nothing by itself.

null rays take finite COORDINATE TIME to reach the AdS boundary but that this corresponds to infinite affine parameter.

And coordinate differences mean nothing physically.

Q1: Does this mean an observer staying inside the bulk and measuring coordinate time could see the light ray go to infinity and back in finite time?

You don't measure coordinate time. A well designed clock (that is also a test object) measures proper time along a timelike curve. It doesn't know what coordinate system different people may or may not choose to use.

But, if the light ray itself needs infinite affine parameter to reach r=infinity, how could it ever get back into the bulk? How are these two ideas consistent?

There aren't two ideas. Coordinate times aren't physical.

Q2: Secondly, assuming our wave is travelling on a null geodesic, what does the boundary being timelike have to do with the ability to reflect the wave back into the bulk?

How does a boundary that can't affect your experiments have any ability whatsoever. It's a drawing. When you use the coordinate system with $T=\arctan t$ then you can make a surface of $T=+\pi/2$ but it is basically a surface of $t=+\infty$ and can't be reached or affect any experiments or ever come up in a prediction. It could be helpful for discussing or drawing things and for classifying or grouping things that are real curves in the actual space, but it is just wrong to talk as if things reach it.

What would have happened if it was a null boundary or a spacelike boundary?

By itself, it doesn't mean anything.

I have thought about this from the perspective of Penrose diagrams

You can draw multiple inequivalent Penrose diagrams that have the spacetime embedded inside. So don't read anything into it at all. Think of it as a tool that can help you understand not as an actual larger spacetime.

if I drew the boundary as null (at 45 degrees), then any null wave that hit it (coming in at 45 degrees), would be reflected and start travelling along the r=infinity surface, right?

Do you have a physical basis to say it would ever reach a boundary? For instance if you are using AdS for duality purposes, have you taken the correct notions of what is physical on the dual theory and mapped them over? If you are concerned about AdS as its own spacetime and a solution to Einstein's Equation have you showed that the paths taken by geodesics are the limiting case of a family of physical metrics with an actual metric there. Because when you take that limit the things in the limit are not asymptotically vacuum AdS in when the curves are at the boundary. So any analysis based on that will fail. Plus in the paper you cite the waves eventually have significant back reaction on the geometry, so the whole idea of curves on a background spacetime becomes physicslly meaningless.

Q3: How does one show the boundary is timelike?

Again, the larger spacetime can have boundaries, but it is possible to have the smaller (physical) spacetime not have boundaries and be complete. You are adding something to a spacetime that is already complete and you can add a single event or a timelike boundary or a spacelike boundary or add whole other spacetimes right next to it. Its just your choice of what larger spacetime to pretend tour already complete spacetime is sitting inside. Since you can put different boundaries on the space that AdS is a subset of, then AdS itself can have different boundaries.

Q4: If the above argument is correct, then why do I get a weird result when I apply it to Minkowski space?

You aren't making an argument. You are treating arbitrarily choices as if they were physical. And haven't even stated your criteria for physicality (regular, dual, test particle, etc.)

But we know from the Penrose diagram that the boundary of Minkowski is null with a null normal vector so what has gone wrong?

You know of **a** Penrose diagram with a lightlike boundary. It's totally wrong to say that diagram is **the** diagram because that wrongly implies there is just one.

And indeed, assuming my reasoning is wrong, how does one correctly show AdS has a timelike boundary and Minkowski has a null boundary?

Are you sure that's a question? Since we can embed AdS as a subset of a larger dimensional $\mathbb R^{2,3}$ space and that can have different boundaries I think the burden is on you to argue that it has to have only certain boundaries. The paper you cited wants boundaries it calls energy momentum conserving.

Q5: In Minkowski, null waves take an infinite coordinate time to arrive at the boundary.

In some coordinate systems. Not in others. So there is no physical content in that statement.

## Best Answer

We assume OP's question (v2) is the following:

The answer is

Yes,see. e.g. Ref. 1. (As OP correctly notes, most textbooks deal only with Fermi normal coordinates fortimelikegeodesics, cf. e.g. Ref. 2 and Ref. 3.)References:

M. Blau, D. Frank, and S. Weiss,

Fermi Coordinates and Penrose Limits,Class. Quant. Grav 23 (2006) 3993, http://arxiv.org/abs/hep-th/0603109MTW.

E. Poisson,

The Motion of Point Particles in Curved Spacetime,(2004), http://www.livingreviews.org/lrr-2004-6