Gravity ~ Acceleration ~ Centrifuge & GR

[gonegahgah]

Gravity & Acceleration are said to be fairly equivalent.
This includes their respective effects on time dilation doesn't it?

Is artificial gravity generated by centrifugal motion considered to be fairly equivalent to these?

 

[sweet springs]

Hi.
Artificial or natural doesn't matter with the time dilation effect of gravity.
Regards.

 

[gonegahgah]

Thanks sweet.
What happens if one of these effectively cancels out the same type or a different type?
Does that have any effect on the time dilation due to GR?

Examples of same type cancellation:
1. If you are at the gravitational balance point between two planets (or planet & moon)?
2. If you were at the centre of the Earth?

Example of different type cancellation:
2. If you were on a travelator (flat escalator) around the entire Earth diameter that was moving fast enough to make you weightless?

 

[A.T.]

Gravity & Acceleration are said to be fairly equivalent.
This includes their respective effects on time dilation doesn't it?

Is artificial gravity generated by centrifugal motion considered to be fairly equivalent to these?

Clocks at rest in a rotating frame run at different rates along the radial direction. This is equivalent to gravitational-time-dilation.

In the inertial non-rotating frame their desynchronization is blamed on their different speeds which cause different movement-time-dilation.

 

[yuiop]

What happens if one of these effectively cancels out the same type or a different type?
Does that have any effect on the time dilation due to GR?

Examples of same type cancellation:
1. If you are at the gravitational balance point between two planets (or planet & moon)?
2. If you were at the centre of the Earth?

Gravitational time dilation does not cancel out like gravitational force. The time dilation is more related to the gravitational potential than to the gravitational force. In the examples you gave:

1. The gravitational time dilation at the balance point between two planets is greater than an infinity.

2. The gravitational time dilation at the centre of the Earth is greater than an infinity and also greater than at the surface of the Earth. You can calculate this using the interior Schwarzschild solution.

In a very rough visualisation, you can use the rubber sheet model. The depth of the depression of the rubber sheet is roughly proportional to the gravitational time dilation. If you have two masses on the rubber sheet, the sheet is depressed more between the two masses than at the far edges of the sheet.

Note that I am considering the simplified case of a non rotating Earth. The rotation of the Earth effectively increases the time dilation factor at the surface. See comment below about combining gravitational time dilation and velocity time dilation.

Example of different type cancellation:
2. If you were on a travelator (flat escalator) around the entire Earth diameter that was moving fast enough to make you weightless?

This is essentially the same situation as an orbiting satellite. The time dilation on the satellite is the product of time dilation due to the orbital velocity of the satellite and the gravitational time dilation due to the height of the satellite. In the case of low velocities and non extreme gravity as in the case of an Earth satellite this approximates to the sum of the gravitational and velocity time dilation factors.

 

[gonegahgah]

Thanks Kev.

Gravitational time dilation does not cancel out like gravitational force. The time dilation is more related to the gravitational potential than to the gravitational force. In the examples you gave:

I thought the potential was represented by the slope and not the depth? Is that wrong?
Just wondering because the slope of the balance point between the two planets has a gradient of 0?

1. The gravitational time dilation at the balance point between two planets is greater than an infinity.

2. The gravitational time dilation at the centre of the Earth is greater than an infinity and also greater than at the surface of the Earth. You can calculate this using the interior Schwarzschild solution.

Infinity?

In a very rough visualisation, you can use the rubber sheet model. The depth of the depression of the rubber sheet is roughly proportional to the gravitational time dilation. If you have two masses on the rubber sheet, the sheet is depressed more between the two masses than at the far edges of the sheet.

Fair enough but I just need that clarification about if gravitational potential is represented by the slope or the depth?

Note that I am considering the simplified case of a non rotating Earth. The rotation of the Earth effectively increases the time dilation factor at the surface. See comment below about combining gravitational time dilation and velocity time dilation.

That's fine. I haven't included frame dragging or SR effects just to simplify it for me.

This is essentially the same situation as an orbiting satellite. The time dilation on the satellite is the product of time dilation due to the orbital velocity of the satellite and the gravitational time dilation due to the height of the satellite. In the case of low velocities and non extreme gravity as in the case of an Earth satellite this approximates to the sum of the gravitational and velocity time dilation factors.

I think I read that the GR time dilation is the major component of the time dilation with the SR time dilation only cancelling it a very tiny amount. That's cool. But I just want to understand the non-frame drag GR component a bit better first.

Just as another example.
If you had a centrifugal wheel spinning at 1G and you were in a gravity well near a planet at 1G also then at the 'bottom' of the rotation your net gravity would be 2G and at the top it would be 0G. At the bottom of rotation is your 'potential' going to be 2G or 1G still?

Or, if it is depth related instead of slope related then does this mean that your potential between the 2 planets is the sum of the two opposite potentials at that point? Does this mean that instead of 0 time dilation that it is instead proportional to the sum of the opposite potentials. Or is only the greater potential used as the multiplier?

 

[yuiop]

I thought the potential was represented by the slope and not the depth? Is that wrong?
Just wondering because the slope of the balance point between the two planets has a gradient of 0?

In my crude understanding, "potential" and "potential gradient" are two different things. The potential gradient between the two planets is indeed zero, but it is the potential or depth that is important rather than the slope as far as gravitational time dilation is concerned.

Infinity?

By infinity I was referring to a reference clock that is very far from any gravitational body. In the Schwarzschild metric the coordinate time of a clock "at infinity" is compared to the proper time of a local clock. The "infinity clock" runs at the maximum possible rate and is unaffected by any time dilation. Of course actually getting a clock to infinity is impractical and we can consider the limiting case of a clock very far from any gravitational sources, where the time dilation tends towards zero.

Just as another example.
If you had a centrifugal wheel spinning at 1G and you were in a gravity well near a planet at 1G also then at the 'bottom' of the rotation your net gravity would be 2G and at the top it would be 0G. At the bottom of rotation is your 'potential' going to be 2G or 1G still?

OK, let's clarify the situation a little. Imagine you are far above the North pole and the Earth appears to be rotating anticlockwise. The vertical centrifugal wheel is at the equator. Now if the wheel is rotating clockwise (from your viewpoint above the North pole), the velocity of the wheel at the bottom of the rotation is adding to the velocity of the Earth's rotation so the increased effective velocity is increasing the velocity time dilation factor. The gravitational time dilation is simply what it would normally be at the surface of the Earth and is unaffected by how many g's are experienced by the clock on the wheel. If the wheel is rotating anticlockwise, its velocity at the bottom of the rotation is subtracting from the velocity of the Earth, so there is less velocity time dilation in this case. Note that the total time dilation is dependent on the direction the wheel is rotating in and independent of the number of g's experienced.

To clarify further, consider a collection of different size wheels that all have the same rim velocity, but experience different g forces due to the different radii. All the clocks on the rims of all the wheels will run at the same rate as each other, independent of the g forces measured at the rims of the different wheels.

Or, if it is depth related instead of slope related then does this mean that your potential between the 2 planets is the sum of the two opposite potentials at that point? Does this mean that instead of 0 time dilation that it is instead proportional to the sum of the opposite potentials. Or is only the greater potential used as the multiplier?

If I recall correctly, the total gravitational potential between the two planets is simply the sum of the individual contributions from the two planets and these are not "opposite potentials" and do not cancel out.

 

[Janus]

Thanks Kev.


I thought the potential was represented by the slope and not the depth? Is that wrong?
Just wondering because the slope of the balance point between the two planets has a gradient of 0?

potential would be represented by the depth, local strength would be the slope.

Infinity?

I believe he means at at infinity. Gravitational time dilation is measured with respect to an observer an infinite distance from the gravity source.

I think I read that the GR time dilation is the major component of the time dilation with the SR time dilation only cancelling it a very tiny amount.

it depends on the orbit.

Just as another example.
If you had a centrifugal wheel spinning at 1G and you were in a gravity well near a planet at 1G also then at the 'bottom' of the rotation your net gravity would be 2G and at the top it would be 0G. At the bottom of rotation is your 'potential' going to be 2G or 1G still?

Neither, potential isn't measured in gs. Potential is related to how work it would take to lift the object to a different height in the field.
Let's two centrifuges, They have different length arms, and spun such that a clock at the end experiences each feels 1g. The clock on the centrifuge on the longer arm will run slower, even though it feels the same g-force as the other clock. Or you could arrange it so that the speed of the ends of the arms are the same for each centrifuge. In this case both clocks will run at the same rate, even though one will feel a greater g0force than the other.

Or, if it is depth related instead of slope related then does this mean that your potential between the 2 planets is the sum of the two opposite potentials at that point?

One way to look at it would be to use the rubber sheet analogy. The planets make "dips" int he rubber sheet. Between the two dips is a shallow "trough". The depth of the trough at any point represents the potential.

 

[gonegahgah]

Neither, potential isn't measured in gs. Potential is related to how work it would take to lift the object to a different height in the field.
Let's two centrifuges, They have different length arms, and spun such that a clock at the end experiences each feels 1g. The clock on the centrifuge on the longer arm will run slower, even though it feels the same g-force as the other clock. Or you could arrange it so that the speed of the ends of the arms are the same for each centrifuge. In this case both clocks will run at the same rate, even though one will feel a greater g0force than the other.

Thanks Kev & Janus.
Just a question about yours Janus first because I just need clarification on your answer sorry just because it surprised me.
If you were in 0g space (impossible I imagine but can approach hopefully) inside one of two spinning wheels and one wheel was bigger than the other but they spun with the same G force at the edge then would two equal masses - one on each wheel - weigh the same as each other on those different scales?

 

[Janus]

Thanks Kev & Janus.
Just a question about yours Janus first because I just need clarification on your answer sorry just because it surprised me.
If you were in 0g space (impossible I imagine but can approach hopefully) inside one of two spinning wheels and one wheel was bigger than the other but they spun with the same G force at the edge then would two equal masses - one on each wheel - weigh the same as each other on those different scales?

Yes.

 

[gonegahgah]

Let's two centrifuges, They have different length arms, and spun such that a clock at the end experiences each feels 1g. The clock on the centrifuge on the longer arm will run slower, even though it feels the same g-force as the other clock. Or you could arrange it so that the speed of the ends of the arms are the same for each centrifuge. In this case both clocks will run at the same rate, even though one will feel a greater g0force than the other.

Relating the above back to gravitational potential, are you saying with the above that:
In case 1 (same g different speed) that it will take different forces to move the same weight an initial tiny identical distance in the same time to observers inside the separate wheels?
In case 2 (different g same speed) that it will take the same force to move the same weight an initial tiny identical distance in the same time to observers inside the separate wheels?
Am I interpreting it right or wrongly?
Just to clarify, I'm talking about the internal wheel observers who don't see themselves as moving just that they are under gravity (albeit it is artificial).
That clarification may be the difference but I'll just check with you.

 

[gonegahgah]

Potential is related to how work it would take to lift the object to a different height in the field.
One way to look at it would be to use the rubber sheet analogy. The planets make "dips" int he rubber sheet. Between the two dips is a shallow "trough". The depth of the trough at any point represents the potential.

I'm also a little confused about this. For the example where you have two planets you are at the balance point between (or being in a bubble in the centre of the Earth) it takes no more effort to 'move' an object than it would if it were in 0g space.

This is in contrast to the rubber sheet example which would suggest it would be harder to move the object at these points?

Can someone help me with understanding the physical distinction?

 

[gonegahgah]

So is the amount of space time distortion at a point on the curve, that is due to GR, determined by the slope of the curve that its at or the depth of the point?

 

[Janus]

Relating the above back to gravitational potential, are you saying with the above that:
In case 1 (same g different speed) that it will take different forces to move the same weight an initial tiny identical distance in the same time to observers inside the separate wheels?

No

In case 2 (different g same speed) that it will take the same force to move the same weight an initial tiny identical distance in the same time to observers inside the separate wheels?

No

Am I interpreting it right or wrongly?
Just to clarify, I'm talking about the internal wheel observers who don't see themselves as moving just that they are under gravity (albeit it is artificial).
That clarification may be the difference but I'll just check with you.

With the the centrifuges, the potential is equal to the amount of total work needed to move the clock from the edge to the center of the centrifuge. This is what determines what the time dilation will be.

 

[Ich]

So is the amount of space time distortion at a point on the curve, that is due to GR, determined by the slope of the curve that its at or the depth of the point?

In the rubber sheet example:

The depth is the potential. Difference in potential = work per kg = difference in "clock rates".

The slope is gravitational acceleration = gravitational force per kg = work per kg and m lift height. Has no effect on time dilatation (see "clock hypothesis").

Spacetime distortion = curvature = tidal acceleration is the second derivation of depth. Has nothing to do with clock rates either.

 

[Janus]

I'm also a little confused about this. For the example where you have two planets you are at the balance point between (or being in a bubble in the centre of the Earth) it takes no more effort to 'move' an object than it would if it were in 0g space.

This is in contrast to the rubber sheet example which would suggest it would be harder to move the object at these points?

Can someone help me with understanding the physical distinction?

We are talking about the total effort needed to move the object from its point of equilibrium to a point far removed from either planet. Imagine you try and move an object from this position along a line perpendicular to the line joining the planets. The instant you start to move the object, the gravity of the planets will try to pull it back. The total amount of energy needed to lift the object from this "sweet spot" is the potential.

For the hollow of the center of the Earth, it takes work to lift the object from the center to the surface, and then additional energy to lift it to an far distance above the Earth's surface. The far removed clock runs faster than the surface clock, and the surface clock runs faster than the clock at the center.

 

[gonegahgah]

Now I'm starting to understand it a bit better I think.
For gravity wells it depends upon a point at infinity away because it is only at that point that the potential is 0 because it can be 'lifted' no further.
Then you start running backwards from there and determine the work required to get something from the new measure from point to this original infinity point which has the potential of 0?
Is that correct?

 

[Ich]

You don't need that "potential of 0"; there's no need to define an "absolute clock rate", and it's not always possible. We're always talking about relative clock rates.

 

[gonegahgah]

Thanks Ich. Sure, I get that. No highest point measurable because it is at infinity. So things are instead measured relative to each other in terms of how deep they are in a space-time well where they are from that points respective infinity. You only need to get the difference between the highest object and the lowest object in space-time for GR purposes.

I can't fathom the underlying physical mechanism that generates this 'instantaneous' relativity at a distance but I accept that it is the process by which time-space relativity is determined.
As everyone is saying it is not weight based in any way, only relative depth based but it is a curiosity to me, I admit, as to how they 'simultaneously' conform to relative depths.
The only physical medium we have is this 'space-time' which I have yet to come to grips with how it has this immaterial pervasive presence.
Space-time is not made of any strange particles is it?
It just is; isn't it?

 

[gonegahgah]

One way to look at it would be to use the rubber sheet analogy. The planets make "dips" int he rubber sheet. Between the two dips is a shallow "trough". The depth of the trough at any point represents the potential.

Someone on here a little while ago helped me to understand that instead of the 'trough' that you describe that you actually end up with something called a saddle ridge.
That is that the two planets in the example that we are between have their low points shifted towards each other (hence why they move towards each other) but the low points don't combine to create a common trough. Instead a ridge remains between the two low points.

If you add us we create our own little space time dimple. If we are at the balance point then we will sit at the bottom of our dimple and will cause the low gravity points of the two planets to shift a tiny little more towards the common centre; but still without combining.

If we are more to one side of the balance than the other then our little dimple will be a little bit more towards the stronger side than us so we will move towards the stronger side because of this.

But I have another question in relation to that.
I assume you would measure our relative potential as being our height relative to the height of the planets in the overall space-time topology. So that actually makes our potential less than that of the planets. So for us at our point in-between the planets does that mean our clocks due to GR should actually be traveling faster?

 

[Ich]

I can't fathom the underlying physical mechanism that generates this 'instantaneous' relativity at a distance but I accept that it is the process by which time-space relativity is determined.

There is no such mechanism. We were talking about potential; this is not a basic ingredient of GR, it's rather a familiar Newtonian term that can be defined only in a static spacetime in static coordinates. So there is no instantaneous relativity, the terms "potential" and "clock rate" only work with a static background.
Generally, it's all about path length, just as in SR. Two observers take different paths in spacetime, the paths have different length = different elapsed time when they meet again. As long as both observers are at different locations, there is generally no well-defined meaning of simultaneousity or "relative clock rate". It doesn't matter anyway.

 

[turin]

With the the centrifuges, the potential is equal to the amount of total work needed to move the clock from the edge to the center of the centrifuge. This is what determines what the time dilation will be.

Can you describe how you would calculate this? I figure that the amount of work depends on how you move the clock. The clock has some initial kinetic energy due to its circular motion, I'll call it KE. Then, I suppose that we place the clock at the center with zero KE. So, the work done on the clock is W = -KE? I must be missing something. Maybe I'm not allowed to stop the rotation first? Does the Coriolis effect come into play?

 

[yuiop]

Someone on here a little while ago helped me to understand that instead of the 'trough' that you describe that you actually end up with something called a saddle ridge.
That is that the two planets in the example that we are between have their low points shifted towards each other (hence why they move towards each other) but the low points don't combine to create a common trough. Instead a ridge remains between the two low points.

The balance point is the highest point on a transect connecting the centres of the two planets (in the rubber sheet model) but this point is still not as high as points inifinitely far away. In fact if you start at the saddle point at move on a line that is at right angles to the connecting line, you continue to go uphill to a higher (less negative) potential.

But I have another question in relation to that.
I assume you would measure our relative potential as being our height relative to the height of the planets in the overall space-time topology. So that actually makes our potential less than that of the planets.

Your potential is actually greater than that of the planets (but less than the potential at infinity). Potential refers to the potential (possibility) of recovering energy by falling. If you fall your potential energy is converted to kinetic energy which is your falling velocity and when you hit the ground the kinetic energy is converted to sound energy and heat energy in a very rough Newtonian sense.

So for us at our point in-between the planets does that mean our clocks due to GR should actually be traveling faster?

Yes, but it would probably be better to say the clocks are ticking faster than the clocks on the planets (but not as fast as a clock at infinity).

We have been banging on about potential in this thread, as that is probably the easiest to way to visualise things (but bear in mind that the rubber sheet model is a crude analogy) and because the Newtonian term for gravitational potential GM/r appears in the gravitation time dilation factor:

\sqrt{1-\frac{2GM}{rc^2}}

However as Ich as hinted at, that is not the whole story.

It is interesting to note that the Newtonian expression for escape velocity is:

v_e = \sqrt{\frac{2GM}{r}}

and when this expression is inserted into the factor above it, the gravitational time dilation factor becomes:

\sqrt{1-\frac{v_{e}^2}{c^2}}

which has a very obvious analogy with time dilation in Special Relativity.

Could it be that escape velocity is the essential ingredient for determining the time dilation at a given point?

All we have to do now, is figure out what the escape velocity on the rim of a rotating wheel is.

Can you describe how you would calculate this? I figure that the amount of work depends on how you move the clock. The clock has some initial kinetic energy due to its circular motion, I'll call it KE. Then, I suppose that we place the clock at the center with zero KE. So, the work done on the clock is W = -KE? I must be missing something. Maybe I'm not allowed to stop the rotation first? Does the Coriolis effect come into play?

This relates closely to my comments above about the escape velocity. One definition of escape velocity is the velocity a particle requires for its KE to be equal to the gravitational potential energy at its location. Escape velocity is also equal to the terminal velocity of a particle that falls from infinity to a given height. I imagine this "escape velocity" as it applies to a wheel is the velocity a test weight achieves when it slides frictionlessly along a spoke from the centre of the wheel to the rim of the wheel.

 

[yuiop]

...
Generally, it's all about path length, just as in SR. Two observers take different paths in spacetime, the paths have different length = different elapsed time when they meet again. As long as both observers are at different locations, there is generally no well-defined meaning of simultaneousity or "relative clock rate". It doesn't matter anyway.

How would you define the path lengths in this scenario? On twin descends at a controlled rate deep into a gravitational well. The second twin waits one hundred years and then descends down at the same controlled rate to meet his sibling that is biologically twenty years old. The relative time difference is very physically manifested here.

 

[Jonathan Scott]

Can you describe how you would calculate this? I figure that the amount of work depends on how you move the clock. The clock has some initial kinetic energy due to its circular motion, I'll call it KE. Then, I suppose that we place the clock at the center with zero KE. So, the work done on the clock is W = -KE? I must be missing something. Maybe I'm not allowed to stop the rotation first? Does the Coriolis effect come into play?

In Newtonian gravity for height h in a constant field g, the potential difference, giving the potential energy per mass, is gh. If you divide that by c², giving gh/c², you have potential energy per energy, which is the dimensionless fraction by which the time rate changes (assuming that the field is weak enough that the fraction is small compared with 1). That is, a clock which is higher by distance h runs (1+gh/c²) times faster than the lower one.

If you want to treat as rotating system as being like a gravitational field, then for non-relativistic speeds the acceleration g is replaced with v^2/r, which is equal to r \omega^2 where \omega is the angular velocity). This varies for different r values, so you have to integrate the equivalent of gh/c^2 from the centre to the desired radius to get the time dilation fraction.

<br /> <br /> \int_0^r \, \frac{R \omega^2}{c^2} \, dR = \frac{1}{2} \, \frac{r^2 \omega^2}{c^2} = \frac{1}{2} \, \frac{v^2}{c^2}<br /> <br />

That says your time dilation is approximately \tfrac{1}{2} (v^2/c^2) which is of course the same as the ratio of the classical kinetic energy to the total energy, which is the same as the fraction by which time is dilated according to special relativity for speed v, which is the easier way of doing it. In the relativistic case, the usual full special relativistic time dilation factor applies.

 

[Ich]

How would you define the path lengths in this scenario?

The length of the respective world lines, as caculated http://en.wikipedia.org/wiki/Proper_time#In_general_relativity". Their path in spacetime, not in space.

On twin descends at a controlled rate deep into a gravitational well. The second twin waits one hundred years and then descends down at the same controlled rate to meet his sibling that is biologically twenty years old. The relative time difference is very physically manifested here.

Just as a visualization, have a look at the drawing. It symbolizes how same synchronous coordinate time (the angle) still means different proper time (path length), depending on the potential.

 

[yuiop]

...
Just as a visualization, have a look at the drawing. It symbolizes how same synchronous coordinate time (the angle) still means different proper time (path length), depending on the potential.

Nice demonstration.
However we agree that when the twins get back together again in this gravitational version, they have aged differently and for most people that would suggest gravitational time dilation has a physical "reality". No?

 

[Ich]

However we agree that when the twins get back together again in this gravitational version, they have aged differently and for most people that would suggest gravitational time dilation has a physical "reality".

Did I dispute that?
I don't know exactly what part of my answers you interpreted as saying otherwise.
I'm just saying that, in a dynamic spacetime, the very notion of "gravitational time dilation" gets a bit fuzzy, so you can't use it to create 'instantaneous' relativity at a distance. That's all.

 

[yuiop]

Could it be that escape velocity is the essential ingredient for determining the time dilation at a given point?

All we have to do now, is figure out what the escape velocity on the rim of a rotating wheel is.

It cost me a warning to ask that question in a sub forum, but it turns out the "escape velocity" v_e on the rim is equal to the tangential velocity of the rim of the wheel and that time dilation in the gravitational context and in the rotational context can equally and equivalently be expressed by the single equation:

\frac{\tau}{t} = \sqrt{1-\frac{v_{e}^2}{c^2}}

In the context of a wheel the "escape velocity" is the minimum velocity that a particle would require when launched "upward" (inward) orthogonal to the rim up a spoke in order to arrive at the centre of the wheel, where the "potential" in the artificial gravity field is zero.

Using an analogy of how Newtonian potential is calculated in a gravitational field, the kinetic energy acquired by an object falling from zero potential (infinity in the gravity case and the centre of the wheel in the rotational case) to the point under consideration is equal to the potential energy at that point.

The terminal kinetic energy of a particle falling outward on a wheel is :

KE = \int_0^r \, {mr \omega^2} \, dr = \frac{mr^2 \omega^2}{2}= \frac{mv^2}{2}

Since this is the same as the kinetic energy of a particle that remains on the rim it can be concluded that the radial "escape velocity" (v_e) of a location on the wheel, is equal in magnitude to the tangential velocity (v_t) of that point relative to the non rotating frame and so in the case of a wheel these two expressions for the time dilation are equal:

\sqrt{1-\frac{v_{e}^2}{c^2}} = \sqrt{1-\frac{v_{t}^2}{c^2}}

 

[turin]

... up a spoke ...
...
... the kinetic energy acquired by an object falling from zero potential ... to the point under consideration is equal to the potential energy at that point.
...
KE = \int_0^r \, {mr \omega^2} \, dr
...

Thanks, kev. That seems straightforward and correct. Basically, your are using dT=dW=F.dr and F=-dV/dr, so that dT=-dV. (Note the inclusion of the minus sign, a minor detail.)

However, something still bothers me: What is the gravitational analogy of the constraint force due to the spoke?

Apparently, the mass on the wheel spoke only allows a 1-D gravitational equivalent. It doesn't seem to work if you allow, for instance, a tangential degree of freedom. So, this doesn't apply to, for instance, space stations that generate their artificial gravity from rotation, does it? To put it another way, there is no concept of artificial gravitational (scalar) potential energy if you allow for a tangential degree of freedom, because there would be a velocity-dependent (tangential) force (in the rotating frame).

I'm not saying that any of this is wrong, I just wonder if anyone else is bothered by this. Anyway, the end result must be true, since it agrees with the more fundamental calculation based on proper time.