Local flatness for non-freely falling observers?

[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 :)