Local flatness for non-freely falling observers?

[PeterDonis]

It makes it seem like even an accelerating observer who adopts a momentarily comoving freely falling frame at $E$ will get SR results identical to those of a freely falling observer at $E$ when in reality he just gets SR results identical to those of the same accelerating observer in SR. Correct?

Yes.

The two frame fields will be different in the small patch of space-time because for the accelerating observer we will need a different freely falling frame at each event along his worldline contained in the patch whereas for the freely falling observer we can trivially have the same freely falling frame at each point along his worldline within the patch since this frame would get parallel transported along the worldline right?

Yes, although I would say it's even simpler: the frames carried by each observer can only match up at a single event, because one is accelerated and the other is inertial.

if we are to think of the "freely falling frame" as really describing a frame field, we can only make sense of it for the freely falling observer since for the accelerating observer we won't have a single "freely falling frame" in the patch, which would make the concept of ascribing a "freely falling frame" to a frame field associated with the accelerating observer ill defined?

Yes.

In what way would the EEP then be too weak with regards to this? Would it then be unable to state the equivalence of local experiments performed by accelerating observers in space-time with identical experiments performed by the same accelerating observers in SR?

If you use the term "freely falling frame" in stating the EEP, then it seems obvious, given the above, that it can only apply to measurements made by a freely falling observer, yes?

(which we know the EEP does state, event by event)

We "know" this because we intuitively extend the EEP to accelerated observers, since that's obviously the intent. But stating the EEP using the term "freely falling frame" does not, strictly speaking, justify this extension of its scope (at least IMO).

 

[WannabeNewton]

If you use the term "freely falling frame" in stating the EEP, then it seems obvious, given the above, that it can only apply to measurements made by a freely falling observer, yes?
...
We "know" this because we intuitively extend the EEP to accelerated observers, since that's obviously the intent. But stating the EEP using the term "freely falling frame" does not, strictly speaking, justify this extension of its scope (at least IMO).

So in light of these two statements and everything discussed above, should we just be cautious with the way many textbooks phrase this part of the EEP, that is when they use the term "freely falling frame" (because I see it a lot)?

The full power of the EEP does allow an accelerating observer to use SR locally in the manner discussed amongst the previous posts i.e. via RN coordinates setup by the accelerating observer at an event, using some orthonormal frame carried by a momentarily comoving freely falling observer (and the orthonormal frame may or may not correspond to the rest frame of the comoving observers at that event). There is no contradiction because as you mentioned the accelerating observers will not get the same results from local experiments as purely freely falling observers, even if they are comoving at the event in question.

I think this is why the way in which many textbooks state the EEP, i.e. in such an ambiguous way using the term "freely falling frame", manages to confuse me and the OP. It makes it seem like the EEP / local flatness about an event is somehow restricted exclusively to freely falling observers even though it is a local property of space-time itself (honestly I think this is the easiest way to state it).

Interestingly, Wald never once uses these confusing terms in the text. He doesn't even mention the terms locally inertial coordinates and locally inertial / freely falling frames once in the text. This can be a good thing if you want to avoid the confusions you could potentially get due to the way such things are presented in other texts but it is also a bad thing simply because you will never gain experience with those concepts if you only use Wald (which was where I got screwed over ).

 

[PeterDonis]

should we just be cautious with the way many textbooks phrase this part of the EEP, that is when they use the term "freely falling frame" (because I see it a lot)?

I think that would help, yes.

 

[WannabeNewton]

I think that would help, yes.

Thank you so much for all the help Peter. This discussion has been really enlightening; it has cleared up many of the confusions I have acquired from the various texts mentioned (except Wald xD) and hopefully for the OP as well. Until next time

 

[BruceW]

so, specifically to the OP's question, he wanted to know if a non-free-falling observer could measure the energy of a photon (or beam of light really) and would this give the same answer as for an observer who was free-falling. (if I have interpreted it correctly). From what I've seen, I think the answer is yes. This is because even though there will be fictitious forces acting on the beam of light, we are not making any measurements that depend on its acceleration. Our measurement is of its frequency, so the acceleration of the beam of light does not matter. It will cause the frequency of the light beam to change as it passes the observer, but the observer can just choose the value of the light's frequency when it passes by closest to him.

And if the observer wanted to do an experiment that did involve accelerations, then he would get different results compared to the freely-falling observer. We just happen to be lucky that the experiment described by the OP'er doesn't depend on accelerations. Does this all sound about right?

 

[PAllen]

so, specifically to the OP's question, he wanted to know if a non-free-falling observer could measure the energy of a photon (or beam of light really) and would this give the same answer as for an observer who was free-falling. (if I have interpreted it correctly). From what I've seen, I think the answer is yes. This is because even though there will be fictitious forces acting on the beam of light, we are not making any measurements that depend on its acceleration. Our measurement is of its frequency, so the acceleration of the beam of light does not matter. It will cause the frequency of the light beam to change as it passes the observer, but the observer can just choose the value of the light's frequency when it passes by closest to him.

And if the observer wanted to do an experiment that did involve accelerations, then he would get different results compared to the freely-falling observer. We just happen to be lucky that the experiment described by the OP'er doesn't depend on accelerations. Does this all sound about right?

Yes.

 

[WannabeNewton]

Yes.

I have a question about this related to what Bruce said. Going back to the OP's original equation, it was $E = -g_q(p,u)$ where $p$ is the 4-momentum of a time-like particle at an event $q$ and $u$ is the 4-velocity of an observer $O$ at $q$ ($O$ may or may not be accelerating); subsequently $E$ is the energy as measured by $O$. In proving this equation, we can choose a local Lorentz frame $\{e_{a}\}$ at $q$ for $O$, which physically corresponds to finding a freely falling observer at $q$ whose 4-velocity equals that of $O$ at $q$. Since both their 4-velocities are the same, it doesn't matter if $O$ makes the measurement or if the momentarily comoving freely falling observer makes the measurement because regardless we will get the same result for the value of $g_q(p,u)$.

Now in this local Lorentz frame, $u = e_0, p = p^a e_a$ so $-g_q(p,u) = -p^{a}g_q(e_0,e_a) = -p^{a}\eta_{0a} = p^{0}$. Since the local Lorentz frame at $q$ spans Minkowski space-time at that event, we can apply SR at this event and say that $p^0 = E$ giving us $E = - g_q(p,u)$ which is a covariant equation so we can then claim it holds in full generality. Now this is the usual procedure for things like this.


What if we now had an equation that depended on the acceleration of the observer making the measurement? If we wanted to prove the equation point-wise by going to a local Lorentz frame at an event and using SR, there would be an ambiguity as to whether the momentarily comoving freely falling observer was making the measurement or the accelerating observer because, unlike before where we only cared about the 4-velocty which was equal for both at said event, the acceleration of the momentarily comoving freely falling observer at the event is not the same as that of the accelerating observer so a measurement conducted by one would give a different value from a measurement conducted by the other. In such a case, how do we physically interpret the possible use of a local Lorentz frame to prove the generality of an acceleration dependent equation, in the above manner? Is it even valid to do so for an equation that depends on acceleration?

Thank you in advance :)

 

[PeterDonis]

In such a case, how do we physically interpret the possible use of a local Lorentz frame to prove the generality of an acceleration dependent equation, in the above manner?

Following on from what I said before, I would say that specifying a "frame" at a single event does not specify any derivatives, so you can't even express equations that depend on derivatives (like acceleration) using a frame at a single event. You have to express them using a frame field on a patch of spacetime surrounding a given event, since a frame field allows derivatives to be defined. And, as I said before, the frame field of a free-falling observer will be different than the frame field of an accelerated observer.

Another way of looking at this is to take the view that, when properly defined, frames (and frame fields) do not describe spacetime; they describe observers (or families of observers). So the term "local Lorentz frame" is really a misnomer, since it's intended to refer to something that describes spacetime, or at least a small patch of it around some chosen event. What most people really mean when the say "local Lorentz frame" is "local R-N coordinate chart", which allows you to describe experiments done in the same small patch of spacetime by observers in different states of motion.

 

[WannabeNewton]

What most people really mean when the say "local Lorentz frame" is "local R-N coordinate chart", which allows you to describe experiments done in the same small patch of spacetime by observers in different states of motion.

Thanks Peter but I think I'm missing something here. Since the setup of the RN coordinates themselves in the neighborhood of an event depend on just choosing an orthonormal frame at that event, won't the RN coordinates setup in that neighborhood by an accelerating observer and a momentarily comoving freely falling observer be identical, as far as the coordinates themselves go? How then would we distinguish between a measurement of some observer acceleration dependent quantity made by the accelerating observer at that event and one made by a comoving freely falling observer at that event?

Is it because, even if the coordinates themselves are identical, the representation of the accelerating observer's velocity on the entire neighborhood won't be the same as the representation of the freely falling observer's velocity on that neighborhood? More precisely, the RN representation of the acceleration observer's velocity will be the solution to $\frac{\mathrm{d} u^{\mu}}{\mathrm{d} \tau} = a^{\mu}$ on that neighborhood whereas for the freely falling observer it will be the solution to $\frac{\mathrm{d} u^{\mu}}{\mathrm{d} \tau} = 0$.

As such, even if the velocities agree at the event in question where the acceleration dependent quantity is being measured, the coordinates will take into account the fact that the velocities on the entire neighborhood do not agree (and we need velocity on the entire neighborhood for an acceleration quantity) and this will be what distinguishes who measures what. The ambiguity only comes in if all we have is a momentarily comoving LIF for the accelerating observer at a single event, because simply knowing velocity at a point will not be enough to determine an acceleration -the comoving LIF will not be able to distinguish the accelerating observer from the comoving freely falling one when the experiment is performed since all it can do is check point-wise velocities which are the same for both at that event; this seems like a problem since the experimental results won't be the same. However an RN coordinate chart at that event can distinguish the two and then there is no ambiguity with regards to who performs the experiment when the acceleration dependent quantity is evaluated at whatever event in question.

Does all of this seem right to you or am I still missing something? Thanks again.

 

[PeterDonis]

won't the RN coordinates setup in that neighborhood by an accelerating observer and a momentarily comoving freely falling observer be identical, as far as the coordinates themselves go?

Yes. But the worldlines of the two observers, as described in those coordinates, will be different, except at the single event where they are momentarily comoving.

How then would we distinguish between a measurement of some observer acceleration dependent quantity made by the accelerating observer at that event and one made by a comoving freely falling observer at that event?

Because any acceleration-dependent quantity will have derivatives in its description, and the derivatives will be different for the two observers, because their worldlines are different, even though they happen to coincide at one particular event.

Is it because, even if the coordinates themselves are identical, the representation of the accelerating observer's velocity on the entire neighborhood won't be the same as the representation of the freely falling observer's velocity on that neighborhood?

Yes.

As such, even if the velocities agree at the event in question where the acceleration dependent quantity is being measured, the coordinates will take into account the fact that the velocities on the entire neighborhood do not agree (and we need velocity on the entire neighborhood for an acceleration quantity) and this will be what distinguishes who measures what.

Yes.

The ambiguity only comes in if all we have is a momentarily comoving LIF for the accelerating observer at a single event

But this will not be a coordinate chart on a local patch of spacetime surrounding the event; it will be a frame in the sense of 4 orthonormal basis vectors at a single event, nothing else, i.e., no information about derivatives. That's what creates the ambiguity: the difference between the two observers is only "visible" as derivatives at the event in question, and a frame at a single event can't deal with derivatives.

 

[WannabeNewton]

Perfect, thank you very much Peter. I am forever indebted :)