# Local flatness for non-freely falling observers?

1. May 29, 2013

### center o bass

The equivalence principle implies that it's always possible, by an appropriate coordinate transformation, to locally transform away the effects of gravity by going into a freely falling frame.

Furthermore the local flatness theorem implies that it's always possible to but the metric $g_{\mu \nu}$ into it's canonical form at any event $P$, i.e.
$g_{\mu \nu}(P) = \eta_{\mu \nu}$ such that the first derivatives of the metric components vanish. This is thus the mathematical expression of the equivalence principle and by the (stronger) Einstein equivalence principle, one can apply the laws of special relativity in these coordinates.

These considerations are often applied when considering redshift effects in general relativity. Consider an observer; at the even A he sees a photon fly by. By going into inertial coordinates for which the observer is at rest his four-velocity will be $U = (1,0,0,0)$ and the energy-momentum of the photon will be $P=(E,p_x, p_y, p_z)$ and thus the energy the observer would measure is given by

$$E = - U \cdot P$$

and from this formula one can derive redshifts effects etc.

That is all nice, but doesn't the argument above require that the observer must be freely falling? If he is not freely falling, i.e. not on a geodesic, is it then also possible to construct local inertial coordinates for which U = (1,0,0,0)?

Often the formula above is applied to observers which are stationary in, say, Schwarzschild spacetime. But I would not suppose that these observers are freely falling, and if so is it then okay to use the above result for these observers?

If it is not possible to construct locally inertial coordinates for observers which is not freely falling, it seems like the result of such a computation would give the result of what a freely falling observer at the event A would measure and not necessarily what the observer of interest would measure.

2. May 29, 2013

### Bill_K

Locally Minkowskian coordinates based on an arbitrary timelike curve are called Fermi coordinates.

3. May 29, 2013

### center o bass

Is it always possible to transform the metric into Fermi coordinates? And is it physically allowed to apply SR in these coordinates when they are not freely falling?

If not, wouldn't it then be a problem that textbooks apply the formula above for non-freely falling observers?

4. May 29, 2013

### WannabeNewton

Hi center o bass! I'm not understanding your question. The fact that you can set up normal coordinates on some neighborhood of any point on a pseudo-Riemannian manifold $M$ has nothing at all to do with observers, it's just a property of the manifold (Normal coordinates just capture the intuitive notion that $T_p M$ is the "best linear approximation" of $M$ at $p$ by using the exponential map). I'm not seeing any link to observers.

5. May 29, 2013

### pervect

Staff Emeritus
You can always transform the metric so that it's diag(-1,1,1,1) locally. This is just a variable substitution - a linear one, at that.

That's all you really need to do what you want. In these local coordinates, you can interpret changes of coordinates as changes of distance.

The Christoffel symbols will be non-zero as the time derivatives of the metric will be non-zero, but this doesn't affect doppler shifts. I'm not going to try to make a list of what is and isn't affected by Christoffel symbols, Im hoping you'll be able to figure it out with some thought.

The cannonical example for understanding the effects of non-zero Christoffel symbols is Einstein's elevator. It's worth convincing yourself that the Christoffel symbols in the elevator must be the acceleration of the elevator if you're not familiar with the idea.

Fermi normal coordinates are slightly more involved, but they are applicable to both curved space-times as well as flat, so they'll include GR and SR.

Do you understand tangent spaces, also known as "local Lorentz frames"? I"m guessing you might not. Unfortunately, it's getting hard to continue to talk about Fermi normal coordinates without making some assumptions about what you do and don't know about regarding tangent spaces.

6. May 30, 2013

### BruceW

wikipedia seem to be saying that Fermi coordinates are only allowed on a geodesic. (i.e. only a freely-falling observer).

7. May 30, 2013

### Bill_K

You're right, they do say that!

Here's a better reference, which defines them for an arbitrary timelike curve, and also goes into more detail.

8. May 30, 2013

### BruceW

I think the idea is that if we define normal coordinates around the point, then if we move along a geodesic through that point, the Christoffel symbols vanish (as long as we are close to the point, and when we get further away, then generally Christoffel symbols become nonzero again).

So to answer the original question, yes the observer needs to be free-falling to experience local 'flatness'. And yes, in your textbook example, it seems they are really showing you what is observed by a free-falling observer in the Schwarzchild spacetime, not one which is stationary.

Last edited: May 30, 2013
9. May 30, 2013

### BruceW

ah, I see. thanks, its a good link. it defines Fermi coordinates in a better way than wikipedia. They define it for any timelike curve, and note that the Christoffel symbols vanish when that curve is a geodesic.

edit: Also, your link says that we can define Fermi normal coordinates along an entire timelike curve, and if the curve is a geodesic, then these 'Fermi normal coordinates' enforce local flatness along the entire geodesic. That's pretty cool. But I guess it makes sense, since at any point along the geodesic, we can define a coord system that has vanishing Christoffel symbols, therefore we can just let our coordinates evolve in such a way that at every point along the geodesic, the coordinates are exactly those which cause the Christoffel symbols to vanish.

Last edited: May 30, 2013
10. May 30, 2013

### WannabeNewton

Again, I'm not seeing what local flatness has to do with observers. It is simply a local properly of $C^{\infty}$ curved manifolds. The OP seems to be referring to the equivalence principle when asking about local flatness applied to freely falling vs. accelerating observers but the confusion is largely due to the horrible phrasing that some sources use with words like "freely falling frame" etc. when stating the equivalence principle. If you have a tensor equation, say $\nabla^{a}T_{ab} = 0$, and $p\in M$, then you can always find a neighborhood $U$ of $p$ with coordinate functions $x^{\mu}: U \rightarrow \mathbb{R}$ such that $\nabla^{\mu}T_{\mu\nu}(p) = \partial^{\mu}T_{\mu\nu}(p) = 0$ when represented in these coordinates i.e. the non-gravitational laws of general relativity reduce to the laws of special relativity locally in space-time; I never understood why some sources use terms like "freely falling frame" when stating this part of the equivalence principle (this doesn't even have anything to do with reference frames-it is simply a matter of employing normal coordinates). It is again just the intuitive statement that $T_p M$, which is isomorphic to $\mathbb{R}^{4}$, approximates $M$ at $p$.

There is no contradiction with regards to freely falling or accelerating observers: if an observer is freely falling and sets up normal coordinates in a neighborhood of an event on his/her worldline then he/she will find that at that event, he/she is traveling on a straight line (moving uniformly) i.e. $\frac{\mathrm{d} ^{2}x^{\mu}}{\mathrm{d} \tau^{2}}(p) = 0$. If an accelerating observer sets up normal coordinates in a neighborhood of an event on his/her worldline then he/she will still measure proper acceleration at that event, given in the normal coordinates by $\frac{\mathrm{d} ^{2}x^{\mu}}{\mathrm{d} \tau^{2}}(p) = a^{\mu}(p)$.

Last edited: May 30, 2013
11. May 30, 2013

### center o bass

You're probably right as to where my confusion lies. I've always recognized the local flatness of being a property of the manifold itself, but as you stated some authors conclude with sentences like: "since a freely falling frame and an inertial frame far away from gravitational fields are equivalent the laws of SR can be applied locally in the freely falling frame", giving the misconception that you have to be in a freely falling frame to use SR locally.

However the equivalence principle also expresses full equivalence with gravitation and acceleration locally, and since the laws of SR hold instantaneously in an accelerated frame they should also hold locally for any frame regardless of the motion of an observer stationary in that frame. Thus an accelerated observer (with 4-acceleration) could also write an equation like

$$P = (E, p_x, p_y, p_z).$$

Agree?

12. May 30, 2013

### WannabeNewton

I'm not sure what your question is. Consider a particle with 4-momentum $p$ at an event $q$ on its worldline. Let $O$ be an observer with 4-velocity $u$ who is present at $q$ ($O$ may or may not be accelerating). Choose a local Lorentz frame $\{e_{(a)}\}$ for $O$ at $q$. In this frame, $u = e_{(0)}$ and $p = p^{a}e_{(a)}$ so that $-g_{q}(p,u) = -p^{a}g_{q}(e_{{(a)}},e_{0}) = -p^{a}\eta_{a0} = -p^{0} = E$. Why do we need normal coordinates at all?

13. May 30, 2013

### BruceW

isn't a "freely falling frame" just the coordinate system you have described? (I always thought that 'coordinate system' and 'reference frame' were synonyms). But from what you are saying here, it seems that they are not?

You are correct that the manifold is always locally 'flat' in that we can choose the metric to be diagonal at any instant. (As someone already said). But we can't generally make the Christoffel symbols vanish, unless we have a freely falling frame. So we need a freely falling frame to be able to use SR locally.

If you are not travelling along a geodesic, then locally you will have fictitious forces that you cannot get rid of. When you have fictitious forces that you cannot get rid of, you do not have local SR. And so to be able to generally say that you have local SR, you need to be travelling along a geodesic.

p.s. I am relatively ;) new to GR, so I'm not 100% sure about what I'm saying.

14. May 30, 2013

### WannabeNewton

A coordinate system and a frame are not the same thing, although some sources use them synonymously which does lead to confusions. At any event on an observer's worldline, we can find an orthonormal basis for the tangent space at that event. This orthonormal basis represents a system of measuring rods and a clock held by the observer-this is what we call the frame; there is no apriori reason to think this should have anything to do with any kind of coordinate chart. We can use such a frame to define a coordinate system on some neighborhood of the event, but this is not the same thing as saying the frame is a coordinate system by itself.

See the following quote:

This relates to what I said above. First let me define what a normal neighborhood is. Let $M$ be a Riemannian manifold and let $p\in M$. There always exists a neighborhood $U\subseteq M$ of $p$ and a neighborhood $V\subseteq T_pM$ of $0\in T_pM$ such that $\exp_p:V\rightarrow U$ is a diffeomorphism. We call $U$ a normal neighborhood of $p$. Now, let $\{e_{(a)}\}$ be an orthonormal basis for $T_p M$. This defines an isomorphism $E:\mathbb{R}^{n}\rightarrow T_pM$ given by $E(x^{1},...,x^{n}) = x^{a}e_{(a)}$. Using $E$ and the exponential map on a normal neighborhood $U$ of $p$, we can define a coordinate map $\varphi = E^{-1}\circ \exp_p^{-1}:U\rightarrow \mathbb{R}^{n}$. It is possible to show that in these coordinates, $\Gamma ^{i}_{jk}(p) = 0$ identically. These coordinates are called Riemannian normal coordinates, or just normal coordinates.

So all I needed was an orthonormal basis for $T_p M$ and the fact that I can always find a normal neighborhood of $p$. If we have an accelerating observer and an event on his/her worldline, I can always find a momentarily comoving locally inertial frame (MCLIF) at that event i.e. find a freely falling observer whose worldline intersects the accelerating observer at the given event, and whose 4-velocity at that event coincides with that of the accelerating observer, and adopt the inertial frame of the freely falling observer at that event for the accelerating observer's use. What's stopping me from finding a normal neighborhood of this event and using the orthonormal basis of the MCLIF to employ Riemann normal coordinates as defined above? I'm personally failing to see why we cannot do this.

15. May 30, 2013

### BruceW

interesting stuff. I hope to learn more about those 'morphisms in the future. It seems a much nicer way to talk about GR. about the topic: I see what you're saying. If we choose 1 event on the worldline, then we can always define a geodesic through it, who sees the world as locally SR. But we can't relate this (in a useful way) to the person who is not travelling along the geodesic. This is because the person and the geodesic only intersect at exactly that 1 event. In the neighbourhood around that event, they will be doing different things. Therefore, the person does not see the world as locally SR. For him to see the world as locally SR, he would have to stay on the geodesic in some small neighbourhood around the event. But he does not.

Alright, to be more specific, his velocity is the same as that defined on the geodesic at that event, right? But his acceleration will not be. So even though his velocity carries him in the same direction as the geodesic, his acceleration will pull him away from this path. So I guess that to second order, he does not see the world as locally SR. He only sees the world as locally SR to first order.

for example, if he is measuring the path of a beam of light. If he measures the velocity of the beam of light, he gets the usual SR result. But if he measures the acceleration of the beam of light, he gets a result which does not match what he would expect from local SR.

And conversely, if a person travelling along a geodesic measures the acceleration of a beam of light, then he does get the result that agrees with local SR. This is because he is travelling along the geodesic, so he can use SR locally to get correct predictions.

16. May 30, 2013

### George Jones

Staff Emeritus
There two different types of normal coordinates being discussed in this thread, Fermi (by Bill) and Riemann (by WannabeNewton), and this might be causing some confusion.

Reimann normal coordinates. Consider an p as origin and some "nearby" point q. Pick an orthonormal frame of (tangent) vectors at p. Find the unique geodesic form p to q that has curve parameter 0 at p and 1 at q. At p, find the tangent vector to this curve. The Riemann normal coordinates of q are the components of this vector with respect to the chosen orthonormal frame at p. This establishes a coordinate system in an appropriate neighbourhood of q.

Fermi normal coordinates. Pick an event p on an observer's (accelerated or not) worldline, and choose an orthonormal frame at p for which e_0 is the observer's 4-velocity at p. Fermi transport the triad of spatial vectors all along the worldline. Consider a point q that is "nearby" the worldline, but not necessarily "nearby" p. From q, find the unique geodesic that orthogonally intersects the worldline (at p' say), and that has curve parameter 1 at q. At p', find the tangent vector to this curve. The Fermi normal spatial coordinates of q are the components of this vector with respect to the spatial triad at p' that was Fermi-transported from p. The Fermi normal time coordinate of q is difference in proper time between p and p'. This establishes a coordinate system in an appropriate neighbourhood tube along the *entire* worldline of the observer. The observer is *always* at the spatial origin of Fermi normal coordinates.

17. May 30, 2013

### WannabeNewton

Hi George. I scrapped my earlier post because I have a direct question. Now, if we happen to use Fermi Normal coordinates along the worldline of an accelerating observer, the Christoffel symbols don't vanish identically along the entire worldline in these coordinates so at any given event on the worldline, the accelerating observer would not be able to, in these coordinates, write down the non gravitational laws of GR in SR form.

However, what's confusing me is, why can't we choose some arbitrary event on the accelerating worldline and employ Riemann normal coordinates in a neighborhood of this event so that the Christoffel symbols do vanish identically at this event in these coordinates, making non-gravitational tensor laws reduce to ones in SR at this event? Is there something preventing us from doing this? It makes intuitive sense to me that even an accelerating observer should be able to use SR locally, via this coordinate system, since the fact that space-time is locally Minkowskian is a property of the manifold and not dependent on whether the observer in the local region is accelerating or not.

Why then, if my intuition is not horribly flawed, do some books (e.g. Stephani, MTW) say that a "locally flat coordinate system" (i.e. Riemann normal coordinates) about an event is the coordinate system set up exclusively by a freely falling observer at that event, as opposed to any time-like observer at that event? I'm just not getting why only a freely falling observer can setup these locally flat coordinates about events on his/her worldline. I don't see where in the construction of Riemann normal coordinates this requirement is imposed.

As far as I can see, the only difference between this and Fermi Normal coordinates (other than the fact that in Fermi Normal coordinates the Christoffel symbols won't vanish in general) is that the latter is defined all along the accelerating worldline and we have a preset transport law for the local Lorentz frame erected at some initial event (the Fermi Walker transport) whereas in the former case we would need to keep (in principle) reconstructing Riemann normal coordinates at each consecutive event. Could you clarify this? Thanks George (and sorry for all the questions, it's my fault for learning exclusively from Schutz and Wald :p)!

Last edited: May 30, 2013
18. May 30, 2013

### pervect

Staff Emeritus
I agree
I disagree. This is a common misconception. You can handle accelerating observers, and even accelerating frames just fine in SR, by which I mean without using Einstein's field equations.

The methods used typically aren't taught in introductory SR courses, but you don't need any new theoretical assumptions to handle fictitious forces.

See for instance http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html "Can special relativity handle accelerations."

OK, hope this helps

19. May 31, 2013

### Staff: Mentor

You can; but the coordinate chart you construct this way will be *different* than Fermi normal coordinates for that accelerating worldline.

Do they really say that? Can you give some specific quotes that give you this impression? I'm not familiar with Stephani, but in MTW, I don't get the sense that they are saying that Riemann normal coordinates can *only* be used by freely falling observers.

However, there is a sense in which Riemann normal coordinates give freely falling observers a "special" place: the worldlines of freely falling observers are straight lines in Riemann normal coordinates. See below.

This is true, but there's another difference, which I just alluded to. Let's take a concrete example: the worldline of an observer "hovering" at constant r in Schwarzschild coordinates. This observer is accelerated, so we can set up Fermi normal coordinates centered on his worldline. In these coordinates, the observer's worldline is a straight line.

Now pick an event E on that observer's worldline, and consider the worldline of a freely falling observer who just comes to rest for an instant at radial coordinate r at that event E. In the above Fermi normal coordinates (let's suppose we pick event E as the origin of those coordinates, to make it simple), the free-faller's worldline will be curved (I think it will look something like the way an inertial observer's worldline in Minkowski spacetime looks in Rindler coordinates). But if we set up Riemann normal coordinates centered on event E, in those coordinates the free-faller's worldline is a straight line, and the "hovering" observer's worldline will be curved (I think it will be a hyperbola, as a Rindler observer's worldline is in Minkowski coordinates in flat spacetime).

20. May 31, 2013

### Staff: Mentor

This reference also appears to indicate that, if the timelike curve is not a geodesic, the Fermi normal coordinates on that curve will *not* look like Minkowski coordinates. That may be the confusion underyling the Wikipedia article: what Wikipedia should have said is that Fermi normal coordinates can be set up for any timelike curve, but the metric will only take the Minkowski form at events on that curve if the curve is a geodesic.