Would the one accelerating please stand up?

[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.

 

[A.T.]

For cases where a (single) body is accelerated non-uniformly by gravity, there is an equivalent non-gravitational situation, at least in principle.

A single point like object. As soon it's big enough to be deformed by the tidal effects, you cannot reproduce that non-gravitationally.

 

[MikeGomez]

If I understand what is meant by curvature, it has to do with tidal effect and those can be ignored with a body as small as a clock. If I do not understand curvature, further explanation would be greatly appreciated.
--------------------------------------------------------------------------------------

Clock dropped in a hole through the Earth experiment.

Situation A (Gravitational)

Clock frame:
-Clock has 0 proper acceleration.
-Clock has 0 coordinate acceleration.

Earth frame:
-Clock has 0 proper acceleration.
-Clock has non-linear coordinate acceleration.

Situation B (Non-gravitational) Not to be confused with the classic accelerated elevator analogy which implies uniform acceleration. This “elevator” is a non-linearly accelerated spheroid with a hole through the major axis, and it accelerates such that a clock in space (clock goes through the hole of the elevator) sees a relative coordinate acceleration with respect to a position on the elevator which is exactly the same as the Earth situation.

Clock frame:
-Clock has 0 proper acceleration.
-Clock has 0 coordinate acceleration.

Elevator frame:
-Clock has 0 proper acceleration.
-Clock has non-linear coordinate acceleration. The clock has the same (relative) non-linear coordinate acceleration as the case for the hole-in-the-earth situation.

 

[A.T.]

If I understand what is meant by curvature, it has to do with tidal effect and those can be ignored with a body as small as a clock.

Not only the bodies have to be small, but also their trajectory. When two objects oscillate in the hole, and accelerate (coordinate wise) towards each other, that's a tidal effect of curvature too. And it cannot be reproduced in flat space time.

 

[PeterDonis]

I disagree that the equivalence principle should be interpreted as requiring an object’s path to fit into a single local inertial frame.

It isn't a matter of agreement or disagreement. The fact that, in order to have the laws of physics take the same form as they do in flat spacetime, you have to work within a single local inertial frame, is just a fact; it's how the universe works. Whether you want to use the term "equivalence principle" to refer to that fact, or something else, is a matter of terminology, not physics; but since that's the accepted use of the term, you won't get very far trying to make it mean something else.

If I understand what is meant by curvature, it has to do with tidal effect and those can be ignored with a body as small as a clock.

No, they can't. Tidal effects don't just work in the spatial dimensions; they work in the time dimension too.

Here's a simple example: take two spaceships in free fall. One is at some altitude $R$ above the center of the Earth, and is moving at exactly the speed required for a circular orbit about the Earth at that altitude. The second is at a slightly higher altitude $R + \delta R$, and is moving (at some particular instant) at the same speed as the first, so they are at rest relative to each other at that instant. Both spaceships are in free fall.

As time goes on, the two spaceships will not stay at rest relative to each other. (In Newtonian terms, this is because the second spaceship is in an elliptical orbit whose perigee is $R + \delta R$, while the first is in a circular orbit with radius $R$, so the two will move apart as the second gains altitude.) You can make the spaceships as small as you like and this will still happen, because it doesn't depend on the size of the ships, it depends only on the curvature of spacetime, and the effect shows up over time, not over some distance in space at the same instant of time.

This “elevator” is a non-linearly accelerated spheroid with a hole through the major axis, and it accelerates such that a clock in space (clock goes through the hole of the elevator) sees a relative coordinate acceleration with respect to a position on the elevator which is exactly the same as the Earth situation.

Does the spheroid have gravity, or not? I'm assuming not, since spacetime is flat. If it doesn't, what's the point of having it there? Obviously I can make the clock's motion relative to the spheroid match the first clock's motion relative to the Earth, but that's not what the equivalence principle is about. The EP doesn't say that, given any scenario in curved spacetime, we can mock up some scenario in flat spacetime that has the same relative motions. It says that the laws of physics in a local inertial frame in curved spacetime are the same as they would be in a similarly sized patch of flat spacetime. If the spheroid in the second scenario doesn't produce any gravity, it's not affecting any physics (except for the relative motions, which as I've just said, aren't the point), so it's just superfluous.

(That statement about relative motions, btw, is still false. As A.T. said, if you try putting a second free-falling clock in your spheroid scenario, with the same motion relative to the first clock as in the hole in the Earth scenario, it won't work.)

 

[Dale]

There have been a lot of posts here today, so I may be re-hashing stuff that has already been covered.

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 taking the equivalence principle to have its standard meaning. Applying the standard meaning to this specific scenario requires a uniformly accelerating elevator since only such an elevator would have a uniform accelerometer reading of 1g as would be observed in this specific scenario for the clock at rest on the surface of the earth.

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.

No it doesn't. There is no way, in flat spacetime, to have one clock with proper acceleration of 1g and to have a second clock fall down and then up all while remaining inertial.

The equivalence principle specifically refers to local experiments. Local is defined to mean that tidal effects are negligible. So saying that there is no "equivalence principle" elevator for the hole-in-the-earth scenario is a way of saying that the hole-in-the-earth scenario is non-local. It cannot be replicated in flat spacetime.

 

[PeterDonis]

There is no way, in flat spacetime, to have one clock with proper acceleration of 1g and to have a second clock fall down and then up all while remaining inertial.

This is a good point that I didn't include in my previous post. The "accelerated spheroid" flat spacetime scenario works if all you want to do is reproduce the relative motion of the clock falling through the hole and the "Earth" (i.e., some large object with a tunnel in it); but it will not reproduce the relative motion of the clock falling through the hole and a clock sitting at rest on the surface of the Earth with a constant 1 g acceleration (the acceleration of a clock in flat spacetime with the same motion relative to the free-falling clock would have to be variable, just like that of the spheroid itself).

 

[MikeGomez]

From Post#63

...When two objects oscillate in the hole, and accelerate (coordinate wise) towards each other, that's a tidal effect of curvature too. And it cannot be reproduced in flat space time.

From Post #64

...two spaceships in free fall. One is at some altitude R above the center of the Earth, and is moving at exactly the speed required for a circular orbit about the Earth at that altitude. The second is at a slightly higher altitude R+δR...

From Post #66

...The "accelerated spheroid" flat spacetime scenario works if all you want to do is reproduce the relative motion of the clock falling through the hole and the "Earth" (i.e., some large object with a tunnel in it); but it will not reproduce the relative motion of the clock falling through the hole and a clock sitting at rest on the surface of the Earth...

I would like to point out that all of these remarks were made after Post# 54 where I indicated in regards to the equivalent hole in the Earth scenario that I was not envisioning a system composed of more than one body.

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

 

[MikeGomez]

It isn't a matter of agreement or disagreement.

It is a matter a agreement or disagreement.

...but since that's the accepted use of the term, you won't get very far trying to make it mean something else.

No, I am not trying to make it mean something else. How would you like it if I were to do to the same, and say that what you are saying is to make the equivalence principle something different than what it is? We do have a disagreement on that subject. Please, if you feel that I have an error in my logic or that I misunderstand something, then simply state your arguments without any hyperbole.

As my argument (and I believe your counter argument) hinges the meaning of the equivalence principle, let us begin there. The equivalence in the equivalence principle concerns the equivalence of gravitational and inertial mass. This has been known since the time of Galileo. What Einstein did was to provide the correct interpretation. Mass is mass. There is no distinction between gravitational mass and inertial mass because they are one and the same. At a fundamental level, the reason that gravitational mass and inertial mass seem to be that same is that they are the same, not because nature conspires to make them appear the same.

Let us forget for a moment the hole in the Earth scenario, and return to the comparison between a body in freefall above the Earth versus a body in freefall in an accelerating elevator. The body above the Earth experiences a gravitational field. What is surprising is that what appears to be a different situation in the case of the body in freefall in the accelerating elevator is actually no different at all. There is a real gravitational field experienced by the body in freefall in the elevator.

Now consider the body in the accelerating elevator on the floor of the elevator. The body experiences a 1g acceleration due to the acceleration of the elevator. Comparing this with situation with a body on the surface of the Earth under the influence of gravity, it might appear that the two situations are different, but that is not the case. The body on the surface of the Earth experiences a real inertial acceleration.

Not only is gravitational mass equivalent to inertial mass, but the gravitational field is equivalent to an inertial field. Einstein presents the comparison of the freefalling body above the Earth with a freefalling body in an elevator to show that the two situations are the same. The reason that he chose a uniformly accelerated elevator is to show that for this example of equivalence, it is an appropriate comparison.

These examples assume the size of the bodies under consideration are small enough that they have negligible internal stress (particle). That is to say that they experience no local tidal effects, but will experience time tidal effects if their trajectory takes them along a path that will do that as has been pointed out.

Einstein could have chosen another situation to describe, such as a particle in a varying gravitational field, in which case an elevator might not be the appropriate non-gravitational counter part, but the equivalence principle would still apply, and at the heart of the equivalence principle is the equivalence between inertial mass and gravitation mass, not any inertial or non-inertial reference frame. What is pertinent to the equivalence principle is that sufficiently small regions in spacetime (such that the effect of gravitation can be neglected) can be considered such that the invariant laws of physics of Special Relativity can be extended to include the non-inertial frames of reference in General Relativity.

Sigh, I'm ready. Tell me how wrong I am.

 

[HiggsBoson1]

Yeah, I realize about the frame of reference part. What I mean is: if someone flew down from outer space, placed a watch on the falling guy, stole it back during mid-fall and brought it to flying spaghetti monster or what have you and flying spaghetti monster looked at the watch, wouldn't he see that the watch read the same time as his (after correcting for the slight acceleration from and to the earth), whereas if he picked up a clock from the Dead Sea he would see the clock show an earlier time than his own (however slight a difference it would be)? Because when the falling guy was in free-fall, he was not effected by gravitational time/length/mass warps (according to spaghetti monster's frame of reference). Or am I way off?

That sounds right to me.

 

[HiggsBoson1]

It is a matter a agreement or disagreement.
No, I am not trying to make it mean something else. How would you like it if I were to do to the same, and say that what you are saying is to make the equivalence principle something different than what it is? We do have a disagreement on that subject. Please, if you feel that I have an error in my logic or that I misunderstand something, then simply state your arguments without any hyperbole.

As my argument (and I believe your counter argument) hinges the meaning of the equivalence principle, let us begin there. The equivalence in the equivalence principle concerns the equivalence of gravitational and inertial mass. This has been known since the time of Galileo. What Einstein did was to provide the correct interpretation. Mass is mass. There is no distinction between gravitational mass and inertial mass because they are one and the same. At a fundamental level, the reason that gravitational mass and inertial mass seem to be that same is that they are the same, not because nature conspires to make them appear the same.

Let us forget for a moment the hole in the Earth scenario, and return to the comparison between a body in freefall above the Earth versus a body in freefall in an accelerating elevator. The body above the Earth experiences a gravitational field. What is surprising is that what appears to be a different situation in the case of the body in freefall in the accelerating elevator is actually no different at all. There is a real gravitational field experienced by the body in freefall in the elevator.

Now consider the body in the accelerating elevator on the floor of the elevator. The body experiences a 1g acceleration due to the acceleration of the elevator. Comparing this with situation with a body on the surface of the Earth under the influence of gravity, it might appear that the two situations are different, but that is not the case. The body on the surface of the Earth experiences a real inertial acceleration.

Not only is gravitational mass equivalent to inertial mass, but the gravitational field is equivalent to an inertial field. Einstein presents the comparison of the freefalling body above the Earth with a freefalling body in an elevator to show that the two situations are the same. The reason that he chose a uniformly accelerated elevator is to show that for this example of equivalence, it is an appropriate comparison.

These examples assume the size of the bodies under consideration are small enough that they have negligible internal stress (particle). That is to say that they experience no local tidal effects, but will experience time tidal effects if their trajectory takes them along a path that will do that as has been pointed out.

Einstein could have chosen another situation to describe, such as a particle in a varying gravitational field, in which case an elevator might not be the appropriate non-gravitational counter part, but the equivalence principle would still apply, and at the heart of the equivalence principle is the equivalence between inertial mass and gravitation mass, not any inertial or non-inertial reference frame. What is pertinent to the equivalence principle is that sufficiently small regions in spacetime (such that the effect of gravitation can be neglected) can be considered such that the invariant laws of physics of Special Relativity can be extended to include the non-inertial frames of reference in General Relativity.

Sigh, I'm ready. Tell me how wrong I am.

We are all comparable to bodies in an elevator. The earth, orbiting is the elevator, although it's speed is constant. I agree with what you said.