Would the one accelerating please stand up?

[Dale]

The geometric concept is this: in Euclidean space there is a single straight line which connects two points and it is the shortest path connecting those two points.

In curved space there can be multiple straight lines (geodesics) which connect two points, and they may be different lengths. Consider a sphere, if you pick two nearby points you can go "straight there" as you usually think of it, but you can also walk straight in the exact opposite direction all the way around the globe to get there.

 

[collinsmark]

I'm not sure I understand what you mean by this. Changing your own proper acceleration does not change the geometry of spacetime. It only changes your particular worldline. So on the most obvious interpretation I can find of your statement above (that changing the "landscape" means changing the geometry of spacetime), it is false.

Right; perhaps it was a poor choice of words on my part. Of course spacetime itself is not significantly altered by a single observer's acceleration, but it can alter the observer's observed spatial components and time components separating distinct spacetime events. Given sufficient timelike separation of two spacetime events, and given sufficient curvature between and around the events, there might exist worldlines that ultimately result in greater proper time than a purely freefall worldline between the same events (assuming such a purely freefall worldline exists).

 

[PeterDonis]

it can alter the observer's observed spatial components and time components separating distinct spacetime events

The word "observed" is somewhat misleading here; the "spacial and time components" are coordinates, not observables. In flat spacetime, there is no unique way to construct coordinates for an accelerated observer, but there is, at least arguably, a unique way to do it for a free-falling observer, namely, the standard Minkowski chart. So the temptation is there to view the space and time coordinates in this chart as being "directly observable".

However, in curved spacetime, there is no longer a unique way to construct coordinates even for a free-falling observer, if they are to cover more than a small patch of spacetime. (In a small enough patch of spacetime, you can always construct local inertial coordinates in which a given free-falling observer is at rest.) So there are no unique "space and time components" for a given observer between events outside a single local inertial frame, even if that observer is free-falling. This is an important reason why the geometric viewpoint--viewing everything in terms of geometric invariants instead of coordinates--is desirable in general relativity.

Given the above, viewing proper acceleration as "altering the space and time components" is not a good way to view what's going on; those components aren't uniquely determined even without the acceleration. I'm also not sure that it's a good idea to single out proper acceleration as what "changes your view of spacetime" so that worldlines with more proper time than a given free-fall worldline are possible. See further comments below.

Given sufficient timelike separation of two spacetime events, and given sufficient curvature between and around the events, there might exist worldlines that ultimately result in greater proper time than a purely freefall worldline between the same events (assuming such a purely freefall worldline exists).

I'm not sure this is the best way to state what spacetime curvature can do. Here's how I would state it: in curved spacetime, there can be multiple free-falling worldlines between the same two events (given sufficient timelike separation between the events and sufficient curvature between and around the events). These different free-falling worldlines can have different lengths (proper times); at least one of these must be the globally maximal length for all worldlines between those two events, but other lengths can be shorter.

Adding accelerated worldlines to the mix does not change anything about the above; it just draws attention to the fact that, if some of the multiple free-falling worldlines between two given events have lengths shorter than the globally maximal length, there will, in general, be accelerated worldlines between those two events that are longer than some of the free-falling worldlines between those events. But that will only be the case if there are multiple free-falling worldlines between those events, and if some of them are shorter than the globally maximal length. So the effects of spacetime curvature can be described entirely in terms of free-falling worldlines; you don't need to consider accelerated worldlines.

In the example under discussion, there are three free-falling worldlines between the same pair of events: the worldline of the clock falling through the hole, the worldline of the clock in the "skimming the surface" orbit, and the worldline of the clock thrown upward that comes back after exactly one orbit of the other clocks (i.e., after 84 min). The last of these is the one whose length is globally maximal (for all worldlines between that pair of events). The others are shorter. (The fourth worldline, that of the clock at rest at one end of the hole, is accelerated, and is longer than the two free-falling worldlines whose lengths are not globally maximal.)

 

[Orodruin]

Did you find that, for the idealized case of a constant density spherical mass, the two proper times are actually the same?

To first order in the gravitational potential, yes. I did not have time to pursue it further than that.

I'm not sure this is the best way to state what spacetime curvature can do. Here's how I would state it: in curved spacetime, there can be multiple free-falling worldlines between the same two events (given sufficient timelike separation between the events and sufficient curvature between and around the events). These different free-falling worldlines can have different lengths (proper times); at least one of these must be the globally maximal length for all worldlines between those two events, but other lengths can be shorter.

Perhaps a comparison to a Riemannian case can help a bit here. If one looks at a sphere, the shortest distance between two points which are not antipodal is along a great circle. However, if you follow the same great circle in the other direction, you will still be following a geodesic curve, but it will not be a global minimum of the length between the points (it is a saddle point).

 

[PeterDonis]

To first order in the gravitational potential, yes.

I wonder about that. The integral I get for the clock falling through the hole is

$$
\tau = \int_{0}^{T} \sqrt{1 - \frac{3M}{R} - \frac{M}{R} \left[ \sin^2 ( 2 \pi t / T ) - \cos^2 ( 2 \pi t / T ) \right]} dt
$$

The average value of the expression in brackets above is zero over the interval of integration. I thought I remembered some theorem or other that says the contribution of that term to the integral vanishes in that case (which would mean we wouldn't have to expand out the square root to show that $\tau$ is the same here as for the clock in the circular orbit), but I haven't been able to find a statement of the theorem.

 

[Gary Feierbach]

A practical problem is 2 rocket probes, clocks synchronized, take off from high Earth orbit. One will loop venus for a gravitational velocity assist to go to Mars while the other takes a more direct approach to Mars with the same initial acceleration. They both arrive at Mars at the same Earth time. How do their internal clocks differ.

 

[PeterDonis]

One will loop venus for a gravitational velocity assist to go to Mars while the other takes a more direct approach to Mars with the same initial acceleration. They both arrive at Mars at the same Earth time. How do their internal clocks differ.

I strongly doubt this scenario is even possible in our solar system (at least not without one or both probes having to accelerate for a large part of their trip, not just at the start and end).

However, if we assume that we're in some solar system with planets arranged just right to make it possible, the general answer is that you would have to do the detailed computation; there's no simple rule of thumb that will give you the right answer for a scenario like this.

 

[Gary Feierbach]

We've been going to Mars this way for quite some time. The exception being ion rockets and mid course corrections of conventional rockets. Of course you have to decellerate when you get to Mars unless it is okay to crash into Mars. let's assume both probes crash into Mars but both have hardened blue boxes containing the clocks. The point of this is to determine if acceleration due to gravity alters the clock relative to the direct probe. I'm also assuming that the planets are aligned in such a way that this is possible.

 

[PeterDonis]

We've been going to Mars this way for quite some time.

I didn't mean that the general idea of sending spaceships to other planets this way is impossible; I meant that the specific scenario you describe, where spaceships traveling on the two routes you describe leave Earth orbit at the same time and arrive at Mars at the same time, is most likely impossible.

The point of this is to determine if acceleration due to gravity alters the clock relative to the direct probe.

Attributing any difference in elapsed time (on the assumption that the two trips take the same amount of Earth time) to "acceleration due to gravity" during the Venus flyby is much too simplistic. If elapsed proper time along the two routes is different, it's not any particular point in either journey that makes the difference; it's the entire journey, as compared with the other.

I'm also assuming that the planets are aligned in such a way that this is possible.

And that's the assumption that I think is extremely unlikely in our actual solar system; I don't think there is any possible alignment of the three planets, given their actual orbits, that would make the scenario possible.

 

[Gary Feierbach]

Well I can visualize it but can't prove it without a NASA solar navigation model.

 

[A.T.]

Have you tried to calculate the clock in the hole vs. the clock in the "skimming the surface" orbit? That to me is the most interesting comparison.

Also interesting: Clock oscillating within the hole vs. clock at the center of the sphere.

Old thread on this:
https://www.physicsforums.com/threads/tossing-a-clock.249722

 

[Dale]

A practical problem is 2 rocket probes, clocks synchronized, take off from high Earth orbit. One will loop venus for a gravitational velocity assist to go to Mars while the other takes a more direct approach to Mars with the same initial acceleration. They both arrive at Mars at the same Earth time. How do their internal clocks differ.

I strongly doubt this scenario is even possible in our solar system (at least not without one or both probes having to accelerate for a large part of their trip, not just at the start and end).

I agree that, as stated, it is probably not possible. However, if you relax the "same initial acceleration" part and just have them launch from the same event at different initial velocities, then I think it would be possible. You would calculate the time required for the gravitational assist path for the first rocket and then choose the initial velocity of the other rocket to arrive at Mars at that time. I am quite certain that the initial accelerations will be different, but for any arrival time there should be some initial velocity which will get you there.

 

[PeterDonis]

if you relax the "same initial acceleration" part and just have them launch from the same event at different initial velocities, then I think it would be possible. You would calculate the time required for the gravitational assist path for the first rocket and then choose the initial velocity of the other rocket to arrive at Mars at that time.

Hm, yes, you're right, if the initial velocities are unconstrained (except for the obvious speed of light limit), then there should always be some geodesic on the direct path that matches the endpoints of the gravity assist geodesic.

 

[PeterDonis]

Also interesting: Clock oscillating within the hole vs. clock at the center of the sphere.

Yes, that case is interesting too. Since I need to correct the equations I posted previously anyway , I'll go ahead and post one for that case too.

First, the general expression for $g_{tt}$ in the interior of a spherically symmetric mass of constant density is:

$$
g_{tt} = \left( \frac{3}{2} \sqrt{1 - \frac{2M}{R}} - \frac{1}{2} \sqrt{1 - \frac{2M r^2}{R^3}} \right)^2
$$

where $R$ is the radius at the surface and $r$ is the radial coordinate. I had misremembered this before, which led to the incorrect formula I gave in post #35 for the clock falling through the hole (though, as we'll see, that formula would still lead to the same approximation for small $M / R$). The general formula for time dilation (relative to a clock at rest at infinity) is

$$
\tau = \int \sqrt{g_{tt} dt^2 - g_{rr} dr^2 - g_{\phi \phi} d\phi^2} = \int \sqrt{g_{tt} - g_{rr} v_r^2 - r^2 \omega^2} dt
$$

where I have used the fact that $g_{\phi \phi} = r^2$ in the "equatorial plane" ($\theta = \pi / 2$) of Schwarzschild coordinates, and $\omega$ is the angular velocity (again, relative to an observer at rest at infinity).

Now we can specialize these expressions for each particular case: clocks A (clock tossed up and falling back), B (at rest on the surface), C (in circular orbit skimming the surface), D (falling through the hole), and E (at rest at the center of the Earth). The simplest is clock B, which gives:

$$
\frac{d\tau_B}{dt} = \sqrt{1 - \frac{2M}{R}} \approx 1 - \frac{M}{R}
$$

where for each case the last expression will be what is obtained to first order in $M / R$ in the approxmation that $M / R << 1$.

The next simplest is clock C. We know that $r^2 \omega^2 = M / R$ for this clock, and $g_{tt}$ is the same as for the clock at rest on the surface, so we have

$$
\frac{d\tau_C}{dt} = \sqrt{1 - \frac{3M}{R}} \approx 1 - \frac{3}{2} \frac{M}{R}
$$

Clock A is a bit more complicated because both $g_{tt}$ and $g_{rr}$ vary over its trajectory. We have:

$$
\frac{d\tau_A}{dt} = \sqrt{1 - \frac{2M}{r} - \frac{v_r^2}{1 - 2M / r}} \approx 1 - \frac{M}{r} - \frac{1}{2} v_r^2
$$

where the factor $1/ \left( 1 - 2M / r \right)$ in the last term drops out at this order.

Clock E is next, since it is simpler than clock D because it's at rest:

$$
\frac{d\tau_E}{dt} = \frac{3}{2} \sqrt{1 - \frac{2M}{R}} - \frac{1}{2} \approx 1 - \frac{3}{2} \frac{M}{R}
$$

Finally, we have clock D:

$$
\frac{d\tau_D}{dt} = \sqrt{ \left( \frac{3}{2} \sqrt{1 - \frac{2M}{R}} - \frac{1}{2} \sqrt{1 - \frac{2Mr^2}{R^3}} \right)^2 - \frac{v_r^2}{1 - 2Mr^2 / R^3} } \approx 1 - \frac{3}{2} \frac{M}{R} + \frac{1}{2} \frac{M r^2}{R^3} - \frac{1}{2} v_r^2
$$
$$\approx 1 - \frac{3}{2} \frac{M}{R} - \frac{1}{2} \frac{M}{R} \left[ \sin^2 \left( 2 \pi t / T \right) - \cos^2 \left( 2 \pi t / T \right) \right]
$$

where the second approximate expression is obtained by plugging in the sinusoidal solution for $r$ and $v$. As I noted above, this last expression is the same as what would be obtained from the formula I gave in post #35 by expanding the square root to first order in $M / R$.

So to first order in $M / R$, we have $\tau_C < \tau_B < \tau_A$, as expected; but we also have $\tau_E = \tau_C$, and if we assume that the sinusoidal contributions drop out of the integral for $\tau_D$ since they average to zero over the range of integration, we have $\tau_E = \tau_D = \tau_C$. The differences between these three are in the next order; I'll defer comment on that to a separate post.

 

[PeterDonis]

Ok, so looking at the formulas for $\tau_C$, $\tau_D$, and $\tau_E$ from my previous posts, the exact relationship between them is interesting.

It is evident that, contrary to intuition (at least mine), $\tau_E > \tau_C$, i.e., the clock at rest at the center of the spherical mass runs faster than the clock in the circular orbit skimming the surface. This is easy to show; we just have to verify that

$$
\frac{3}{2} \sqrt{1 - \frac{2M}{R}} - \frac{1}{2} > \sqrt{1 - \frac{3M}{R}}
$$

which can be done by noting that both expressions go to 1 as $M / R \rightarrow 0$, and showing that the derivative with respect to $M / R$ of the left-hand expression is always less negative than the derivative with respect to $M / R$ of the right-hand expression. This works out to

$$
- \frac{3}{2} \frac{1}{\sqrt{1 - 2M / R}} > - \frac{3}{2} \frac{1}{\sqrt{1 - 3M / R}}
$$

which confirms that, as $M / R$ gets larger (i.e., as the object gets more compact), $\tau_E$ decreases more slowly than $\tau_C$, so we will always have $\tau_E > \tau_C$.

The formula for $\tau_D$ is more complicated, but we can gain insight by looking at its behavior at some useful "test" values: $R = 3M$ and $R = 6M$. I will also define the dimensionless radial coordinate $\rho = r / R$ to simplify the formulas.

For $R = 6M$, we have $\tau_C = 1 / \sqrt{2} \approx 0.7071$ and $\tau_E = \left( \sqrt{6} - 1 \right)/2 \approx 0.7247$. The formula for $\tau_D$ works out to

$$
\tau_D = \sqrt{\frac{7}{4} - \sqrt{\frac{3 - \rho^2}{2}} - \frac{1}{12} \rho^2 - \frac{3 v^2}{3 - \rho^2}}
$$

This is easy to evaluate at the endpoints, $\rho = 1$ (where $v = 0$) and $\rho = 0$ (where $v$ has its maximum value, which we'll leave as $v$ in the formula). For $\rho = 1$, we end up with $\tau_D = \sqrt{2/3} \approx 0.8165$ (note that this is equal to $\tau_B$, the clock rate for the observer at rest on the surface at this radius), and for $\rho = 0$, we end up with $\tau_D = \sqrt{7/4 - \sqrt{3/2} - v^2} \approx \sqrt{0.5253 - v^2}$. If we assume that the maximum value of $v$ is approximately the same as the circular orbit velocity for $R = 6M$, which is $1/2$, then for $\rho = 0$ we obtain $\tau_D \approx 0.5246$. This strongly suggests that a full numerical integration would show $\tau_D < \tau_C < \tau_E$ (but see further comments below on the possible behavior of the maximum value of $v$).

For $R = 3M$, it is even more interesting. We have $\tau_C = 0$ at this radius, corresponding to the fact that this radius is the photon sphere. We have $\tau_E = \left( \sqrt{3} - 1 \right) / 2 \approx 0.3660$. The formula for $\tau_D$ becomes

$$
\tau_D = \sqrt{1 - \frac{1}{2} \sqrt{3 - 2 \rho^2} - \frac{1}{6} \rho^2 - \frac{3 v^2}{3 - 2 \rho^2}}
$$

This gives $\tau_D = \sqrt{1/3} \approx 0.5774$ for $\rho = 1$ (again, the same as $\tau_B$ for this radius). For $\rho = 0$ it gives $\tau_D = \sqrt{1 - \sqrt{3} / 2 - v^2} \approx \sqrt{0.1340 - v^2}$. If $v \rightarrow 1$ for an object this compact, this obviously gives a negative quantity under the square root; but even if $v$ is anything larger than $\sqrt{1 - \sqrt{3} / 2} \approx 0.3660$, the quantity under the square root will be negative (since $v^2$ is subtracted there). This indicates that, for an object this compact, either the "dropping the clock through the hole" experiment can't actually be done (unlikely since the value of $\tau_E$ shows us that a clock can sit at rest at the center of the object perfectly well), or the maximum velocity achieved in such an experiment, for objects this compact, is much less than the corresponding circular orbit velocity.

If the latter is the case, then we have to re-evaluate the $R = 6M$ case. If the maximum velocity achieved for that case is anything less than $\sqrt{5/4 - \sqrt{3/2}} \approx 0.1589$, then the minimum value of $\tau_D$ (at $\rho = 0$) will be greater than $\tau_C$. Even a somewhat larger maximum velocity will still give an average value of $\tau_D$ greater than $\tau_C$. This would be nice since we know we must have $\tau_C < \tau_D$ for the $R = 3M$ case (since $\tau_C = 0$ for this case), and we would expect the same inequality between the magnitudes to hold for all values of $M / R$ (at least, it seems to me that that ought to be the case).

That still leaves open the question of the relationship between $\tau_D$ and $\tau_E$. Given the values above, I am inclined to think that $\tau_D > \tau_E$ holds; however, checking this would involve doing the full numerical integration, which is more than I want to tackle.

So, bottom line, it looks to me like the full set of inequalities is:

$$
\tau_C < \tau_E < \tau_D < \tau_B < \tau_A
$$

But numerical integration would be required to confirm whether $\tau_D$ belongs where it is, or between $\tau_C$ and $\tau_E$ (and to confirm that the same inequalities hold for all allowed values of $M / R$).

 

[MikeGomez]

... There is no accelerating elevator in flat spacetime where a dropped clock will fall down and then fall back up.

Are you sure?

 

[A.T.]

There is no accelerating elevator in flat spacetime where a dropped clock will fall down and then fall back up.

Are you sure?

I think he means a linearly uniformly accelerating elevator.

 

[MikeGomez]

I think he means a linearly uniformly accelerating elevator.

I see. My thinking was that for the "hole in the earth" scenario, there might be an equivalent situation with a clock in free-fall relative to non-linearly accelerating elevator. Of course the "elevator" wouldn't look much like your everyday elevator.

[Edit] Changed "non-uniformly" to "non-linearly"

 

[Dale]

Are you sure?

Yes, I am sure, but I guess I wasn't clear. I meant an elevator with constant proper acceleration.

This was intended to show that the equivalence principle cannot be applied here. By the equivalence principle an elevator at rest in a gravitational field is locally equivalent to one with constant proper acceleration in flat spacetime.

There is no "equivalence principle" elevator for the hole-in-the-earth scenario.

 

[MikeGomez]

Yes, I am sure, but I guess I wasn't clear. I meant an elevator with constant proper acceleration.

This was intended to show that the equivalence principle cannot be applied here. By the equivalence principle an elevator at rest in a gravitational field is locally equivalent to one with constant proper acceleration in flat spacetime.

There is no "equivalence principle" elevator for the hole-in-the-earth scenario.

Ok. I think you are taking the equivalence principle to mean strictly a linearly accelerating elevator and the static gravitational field at the surface of the earth. I am thinking of it in a more general sense, in that any gravitational effect has an equivalent inertial effect, not just the comparison with gravitation at the surface of the Earth and an elevator accelerating at 1g.

To me it's just semantics to say that there is no "equivalence principle" elevator for the hole-in-the-earth scenario, and therefore the equivalence principle does not apply. The clock falling in the hole and then falling back up has a equivalent inertial counterpart.

 

[A.T.]

The clock falling in the hole and then falling back up has a equivalent inertial counterpart.

To qualify as a "principle" it cannot just apply to special cases like a single dropped object, but to all experiments of the same scope.

For example, when you drop two clocks into the hole though the massive sphere, so they oscillate with a phase shift of half a period, there is no way you can reproduce their kinematics by free fall in an accelerated frame in flat space-time. So the equivalence principle doesn't apply to the gravitational effect, that the massive sphere produces across it's entire size. It applies only locally, for small regions.

 

[PeterDonis]

I am thinking of it in a more general sense, in that any gravitational effect has an equivalent inertial effect, not just the comparison with gravitation at the surface of the Earth and an elevator accelerating at 1g.

The equivalence principle is not about dropping a single clock, but about any experiment done in a gravitational filed, versus accelerated frame in flat space-time.

The scope of the EP is actually between the two extremes described in these quotes. It covers any proper acceleration, not just 1 g; but it only covers constant proper acceleration for a short time and over a short distance, as compared with being at rest in the equivalent gravitational field for a short time and over a short distance ("short" meaning "short enough that tidal gravity is negligible"). So it doesn't cover "any" gravitational effect or "any" experiment done in a gravitational field.

 

[A.T.]

The scope of the EP is actually between the two extremes described in these quotes.

I actually meant to explain "principle" vs. a special case there. Corrected it now.

it only covers constant proper acceleration

Right, but even if you generalize it to variable proper acceleration, to emulate the single object in the hole (as Mike seems to propose), you still cannot emulate things like two objects in counter phase in the hole. That was my point.

 

[MikeGomez]

Right, but even if you generalize it to variable proper acceleration, to emulate the single object in the hole (as Mike seems to propose), you still cannot emulate things like two objects in counter phase in the hole. That was my point.

I was thinking that variable proper acceleration could be emulated, not just proper constant proper acceleration, but the argument I have in regards to the clock in the hole has to do with free-fall, so wouldn't that be coordinate acceleration and not proper acceleration?

I agree it would be problematic with more than one body.

As for the single body case, I've gone back and forth a few times, but my current thinking is along the lines of accelerating a spheroid shaped body (perhaps in the shape of a football or balloon) which has a hole drilled through the major axis. I'm not sure though if it needs to be a spheroid, maybe a perfectly round sphere would work also.

 

[PeterDonis]

even if you generalize it to variable proper acceleration

You can't. Variable proper acceleration is outside the scope of the EP.

to emulate the single object in the hole

That object's path does not fit in a single local inertial frame, so it is outside the scope of the EP.

you still cannot emulate things like two objects in counter phase in the hole

There is a true statement that can be made along these lines: you can set up Riemann normal coordinates along the worldline of one of the objects, but those coordinates won't cover the worldline of the other object (except in the small patches where the worldlines cross). But Riemann normal coordinates are outside the scope of a single local inertial frame, hence outside the scope of the EP.

 

[A.T.]

outside the scope of the EP

Yes, we get that. "Generalize" meant here modify to widen that scope. I agree that it cannot be done, for the reasons I stated.

 

[A.T.]

I have in regards to the clock in the hole has to do with free-fall, so wouldn't that be coordinate acceleration and not proper acceleration?

Yes, their proper acceleration is zero all the time.

I agree it would be problematic with more than one body.

I don't see how you could reproduce their trajectories while they have zero proper acceleration all the time. That would be a requirement for an equivalence to the above.

 

[PeterDonis]

I was thinking that variable proper acceleration could be emulated, not just proper constant proper acceleration

You can certainly have an object with variable proper acceleration in flat spacetime. But using that to try to "emulate" variable proper acceleration in curved spacetime is outside the scope of the EP.

In fact, thiking of the EP in terms of using flat spacetime to "emulate" curved spacetime is not really correct. The EP only talks about what happens within a single local inertial frame. Within a single local inertial frame, spacetime curvature is negligible; that's the whole point. As soon as you are considering effects that depend on spacetime curvature (like the path of the clock in the hole), you're outside a single local inertial frame, and therefore outside the scope of the EP.

the argument I have in regards to the clock in the hole has to do with free-fall, so wouldn't that be coordinate acceleration and not proper acceleration?

Yes. But as far as setting up a local inertial frame is concerned, the clock in the hole will be at rest in such a frame, so it won't have any coordinate acceleration either.

 

[MikeGomez]

BTW, just in case there might have been some confusion, when I said post #50 when I said that that any gravitational effect has an equivalent inertial effect, I meant inertial as in the inertial properties of a body, not as in an inertial reference frame. The equivalence of gravitational mass and inertial mass have to do with the inertial properties of matter, and the relative relationships of (the inertia of) bodies with respect to each other. Inertial and non-inertial frames of reference are used merely as a tool for discussions concerning that.

You can't. Variable proper acceleration is outside the scope of the EP.

That object's path does not fit in a single local inertial frame, so it is outside the scope of the EP.

I disagree that the equivalence principle should be interpreted as requiring an object’s path to fit into a single local inertial frame. That is only the case where appropriate. In EP the comparison between a body on the surface of the Earth is compared to a body in a uniformly accelerated elevator because those two situations are equivalent, not because of any specific requirement for uniform acceleration. For cases where a (single) body is accelerated non-uniformly by gravity, there is an equivalent non-gravitational situation, at least in principle.

 

[MikeGomez]

Hang on. I sent post #59 before I received and read post #58. Give me a minute to check that out.