Is a uniform gravitational field a gravitational field?

[pmb_phy]

So would this mean the G-forces experienced in any accelerating coordinate system in flat spacetime could be treated as the result of a uniform gravitational field, even if the coordinate system was accelerating in a non-uniform way? The G-forces measured by an accelerometer in a uniform gravitational field can be changing with time as well as space, in other words?

There can still be a g-force present but the field won't be uniform. A rotating frame is a good example of this.

Pete

 

[JesseM]

There can still be a g-force present but the field won't be uniform. A rotating frame is a good example of this.

Pete

But you said "uniform gravitational field" in GR just meant "no tidal forces"--I thought that tidal forces could only occur in curved spacetime by definition (see the comment here for example), while I was just talking about the "gravitational field" experienced by an non-uniformly accelerating observer in flat spacetime. Am I mistaken about the definition of tidal force, or did I misunderstand your comment?

 

[Boustrophedon]

From what I gather, a uniform gravitational field is actually not uniform in terms of the G-forces experienced at different heights

How about whether it increases or decreases with height ? I don't see how you can have it both ways - and the snag is: it needs to increase with height for a free-falling frame to be inertial and yet it needs to decrease with height to resemble 'uniform acceleration'

 

[pmb_phy]

But you said "uniform gravitational field" in GR just meant "no tidal forces"--I thought that tidal forces could only occur in curved spacetime by definition (see the comment here for example), while I was just talking about the "gravitational field" experienced by an non-uniformly accelerating observer in flat spacetime. Am I mistaken about the definition of tidal force, or did I misunderstand your comment?

A uniform g-field requires a zero Riemann tensor. But this is not the only requirement. Just a necessary one.

Pete

 

[JesseM]

A uniform g-field requires a zero Riemann tensor. But this is not the only requirement. Just a necessary one.

OK, so was I wrong in inferring from pervect's post that an additional requirement is that a uniform g-field is equivalent to what would be measured in a coordinate system whose measuring-rods are undergoing Born rigid acceleration?

 

[pmb_phy]

OK, so was I wrong in inferring from pervect's post that an additional requirement is that a uniform g-field is equivalent to what would be measured in a coordinate system whose measuring-rods are undergoing Born rigid acceleration?

Hmmm! I never heard of Born rigid acceleration. That's what I just love about forums like these. Things come up where they've never come up before.

What is Born rigid acceleration?

Pete

 

[coalquay404]

Hmmm! I never heard of Born rigid acceleration. That's what I just love about forums like these. Things come up where they've never come up before.

What is Born rigid acceleration?

Pete

Born rigidity is a well-known concept, particularly in the context of the rigid rotating disk in special relativity. See, for example, Vol. 1 of Held.

I haven't read the thread since it seems to be an exercise in pedantry, but presumably that's what JesseM is talking about.

 

[JesseM]

Hmmm! I never heard of Born rigid acceleration. That's what I just love about forums like these. Things come up where they've never come up before.

What is Born rigid acceleration?

Pete

There are some links in pervect's post which I quoted in my post #57 earlier. This one looks useful, for example. In the case of linear acceleration, my understanding is that Born rigidity means the different parts of the object are accelerating in such a way that the length of the object in the instantaneous inertial reference frame of any part of it will remain constant from one moment to another.

 

[pmb_phy]

There are some links in pervect's post which I quoted in my post #57 earlier. This one looks useful, for example. In the case of linear acceleration, my understanding is that Born rigidity means the different parts of the object are accelerating in such a way that the length of the object in the instantaneous inertial reference frame of any part of it will remain constant from one moment to another.

As I recall, Born rigid acceleration is a requirement for either a uniformly acceleration of a uniform gravitational field. If the metrics you end up with are not identical then something has gone wrong with the derivation. Only experimentation can tell the two apart and to do such an experiment would require the use of a uniform gravitational field.

I'll try to scan in Mould's derivation for a uniformly accerating frame of reference but I'm extremely busy now. Is there anyone who would want to read it? Otherwise I see no sense in posting the section on my website in PDF format.

Kind Regards

Pete

ps - When I get some time later I'll read pervects post in more detail to see exactly what he was saying.

 

[Boustrophedon]

my understanding is that Born rigidity means the different parts of the object are accelerating in such a way that the length of the object in the instantaneous inertial reference frame of any part of it will remain constant from one moment to another.

This is an exact and precisely correct definition of "Born rigid (accelerated) motion". What cannot be relied upon, however, are the various other 'interpretations' of it and 'constructions' put upon it - such as that the acceleration 'g-force' should diminish towards the forward end.

Earlier in this thread it was stated that "uniform acceleration" is the same thing as what has just been defined as "Born rigid acceleration" and this is quite correct.
However, one property that a uniformly accelerating frame (or Born rigid frame) absolutely must have is that it should be perfectly inertial to co-moving observers when freely falling in a "uniform gravitational field". Any version of the equivalence principle makes this mandatory.

Unfortunately, if the g-force of a uniform field diminishes with increasing height (by "indistinguishability" equivalence), the free falling frame by the same token will have to experience diminishing g-forces towards its 'forward' end, which will be 'lower in the field'.
So the g-forces existing and the g-forces required (for uniform acceleration) decrease in opposite directions and cannot possibly cancel across any finite sized falling frame.

Thus the equivalence principle itself forces the conclusion that the only tenable definition of either a 'uniform gravitational field' or of 'uniform (or Born rigid) acceleration' is such that the g-force and the acceleration is precisely constant across and in all parts of a given frame.

 

[pervect]

In the first place nobody in the GR community defines tidal force as the difference in proper accelerations between two ends of a rigid bar. If the spacetime is Minkowskian then there will be no stress in the rod and that is indicative of the presence of a 4-vector.

I told you we'd have a big argument over what a tidal force is :-).

The point I was trying to make is that by at least some definitions of the term "tidal force", the Riemann tensor isn't always the same as the tidal force, specifically for observers not following geodesics.

The geodesic deviation equation only tells us that the Riemann tensor gives us the tidal force for an observer following a geodesic - it says nothing about accelerated observers.

The Riemann tensor is independent of coordinates. (More specifically, the Riemann transforms in a certain known manner under changes of coordinates that imply that it can be interpreted as a coordinate-independent geometric entity.)

This means that in a frame-field, the Riemann tensor as an abstract entity must be independent of the motion of the frame field. Or in coordinate dependent language, it means that for any two observer in relative motion at the same point, that the components of their Riemann are related by the appropriate Lorentz transform.

But consider a static observer near a black hole - we know that the tidal force for such an observer approaches infinity as the observer approaches the event horizon, for it requires infinite acceleration to hold station at the horizon and a finite acceleration to hold station just above the horizon.

Now consider an observer free-falling through the event horizon of a black hole. The tidal force for such an observer is finite and independent of his velocity. (I can provide a reference if needed).

One can achieve similar results by considering an observer on rigid bar. If the bar is not accelerating, there is no tidal force. If the bar accelerates, there is a (usually very small) tidal force. But there is no Riemann tensor.

This demonstrates that the concept of the tidal force is not quite the same as the coordinate-independent concept of the Riemann - for the tidal force for the stationary observer is not the same as the tidal force for the free-falling observer.

Usually this difference between the Riemann and the tidal force doesn't matter - unfortunately, sometimes it does, and this particular issue (the accelerated rod) is one of those cases where it does matter.

 

[Jorrie]

Free-falling through a black hole?

Now consider an observer free-falling through a black hole. The tidal force for such an observer is finite and independent of his velocity. (I can provide a reference if needed).

Great post, as usual. I'm sure that you meant "free-falling through the event horizon"

Jorrie

 

[pervect]

Great post, as usual. I'm sure that you meant "free-falling through the event horizon"

Jorrie

Oops, yep - I fixed it.

 

[pmb_phy]

I told you we'd have a big argument over what a tidal force is :-).

That was understood as soon as I read read your comment predicting this. Howewever I have no intention to include myself in such a discussion. I guess you would say that I've already recognized it as an arguements in semantics and ha ve decided to steer myself clear away from it.

Pete

 

[MeJennifer]

Unfortunately, if the g-force of a uniform field diminishes with increasing height (by "indistinguishability" equivalence), the free falling frame by the same token will have to experience diminishing g-forces towards its 'forward' end, which will be 'lower in the field'.

Looks like you have it upside down Boustrophedon.
The free falling frame will encounter an increasing g-force when it approaches the center of mass.

 

[MeJennifer]

But consider a static observer near a black hole - we know that the tidal force for such an observer approaches infinity as the observer approaches the event horizon, for it requires infinite acceleration to hold station at the horizon and a finite acceleration to hold station just above the horizon.

Now consider an observer free-falling through the event horizon of a black hole. The tidal force for such an observer is finite and independent of his velocity. (I can provide a reference if needed).

One can achieve similar results by considering an observer on rigid bar. If the bar is not accelerating, there is no tidal force. If the bar accelerates, there is a (usually very small) tidal force. But there is no Riemann tensor.

I am not sure what you are trying to demonstrate here.
What has your example with the accelerating bar to do with the curvature in a gravitational field?
Furthermore the suggestion that the tidal forces are infinite at the event horizon is simply a coordinate effect.
A coordinate independent interpretation is that all future paths are curved towards the singularity at the event horizon, but that does not mean that the curvature is infinite there.

This demonstrates that the concept of the tidal force is not quite the same as the coordinate-independent concept of the Riemann - for the tidal force for the stationary observer is not the same as the tidal force for the free-falling observer.

Sorry but mentioning acceleration and space-time curvature together in one posting does not demonstrate anything.

I have the impression that for some reason or another you like people to think that accelerated movement is in some way the same as curved space-time.
Einstein never suggested that!
And it is not what the equivalence principle is saying.

Is it just a personal claim you make that "a tidal force is the difference in proper accelerations between two ends of a rigid bar" or do you claim that this is globally accepted as such? If so, would you care to provide at least one reference to a published book on relativity?

 

[Boustrophedon]

Looks like you have it upside down Boustrophedon.
The free falling frame will encounter an increasing g-force when it approaches the center of mass.

I haven't got it 'upside down'. I meant "...experience diminishing inertial g-force towards its forward end..." (ie. oppositely directed to the 'pull' of the g-field) I am pointing out a contradiction in the "decreasing g-force" definition of a 'uniform' field' or 'uniform acceleration'.
If you care to check what I have written it should now be clearer that I take the incorrect definition of 'uniform acceleration' as having diminishing 'inertial' g-force towards its forward-moving end, and then...

(a) Draw the conclusion that an 'equivalent' uniform 'gravitational' field must have g-force diminish with height since it should be 'indistinguishable' to inhabitants/travellers.

(b) Apply the principle that a free-falling frame (accelerating 'downwards') in a uniform gravitational field should be perfectly inertial.

(c) Show immediate contradiction that: the free-falling frame inertial g-force will have to be smaller at the 'forward' end - which is the end lower in the field since it's falling - while the gravitational pull is greater lower in the field from (a).

Thus the gravitational pull and the inertial g-force (which point in opposite directions) can never cancel over any vertical distance greater than zero and the finite frame can never be inertial ! [Try drawing a diagram if it's still not clear.]

It follows easily that the only 'uniform g-field' that satisfies both the indistinguishability condition and the inertial free-fall condition is a field having constant g-force with height.
It also follows that a 'uniformly accelerating frame' must have a constant g-force (accelerometer reading) along the direction of travel.

 

[Jorrie]

Furthermore the suggestion that the tidal forces are infinite at the event horizon is simply a coordinate effect.
A coordinate independent interpretation is that all future paths are curved towards the singularity at the event horizon, but that does not mean that the curvature is infinite there.

My understanding of what pervect was illustrating is that since the static acceleration at the event horizon of a hole diverges, while the free-fall acceleration remains finite there, the tidal gravity for static and free-fall frames cannot be the same everywhere. The Riemann tensor is however coordinate independent.

This may have implications for the definition of a 'uniform gravitational field', which I think pervect is best at explaining.

 

[MeJennifer]

It follows easily that the only 'uniform g-field' that satisfies both the indistinguishability condition and the inertial free-fall condition is a field having constant g-force with height.

Looks like you are changing tunes now Boustrophedon.
And, you are confused about what you call "the indistinguishability condition".

Einstein never claimed that the equivalence principle works "at a distance", it only applies locally.

 

[MeJennifer]

My understanding of what pervect was illustrating is that since the static acceleration at the event horizon of a hole diverges, while the free-fall acceleration remains finite there, the tidal gravity for static and free-fall frames cannot be the same everywhere.

All those "conclusions" are coordinate dependent observations. Tidal effects are variations of the curvature of space-time in a given region. See for instance Schutz: "A First Course in General Relativity". If I recall correctly (I am currently in Thailand and have no access to books) his definition is quite literal using the Earth and moon as examples.

 

[Boustrophedon]

I don't quite see in what way I've 'changed tunes' - perhaps you might amplify ? What I have abbreviated to the 'indistinguishabilty condition' is that the equivalence of a uniformly accelerated reference frame and a 'uniform gravitational field' is such that a laboratory in either situation should be unable by any experiment to determine which condition is the source of the g-forces that are experienced. This means that if uniform acceleration involves a g-force that is measured higher as one checks over a distance in the direction the g-force is pointing (as proposed) then the 'indistinguishable' uniform gravitational field should be identical, ie. have (downward) g-force diminishing with height.
The reason that the equivalence principle only applies 'at a point' is that 'real' gravitational fields are all non-uniform and thus have 'tidal effects'. A hypothetical perfectly uniform gravitational field would be one where such tidal effects do not occur so that the equivalence principle would hold over significant distances. From the arguments given it appears that the only sense of 'uniformity' that satisfies these criteria is where the g-force (gravitational or inertial) is constant along the direction of its action.

 

[Jorrie]

Really?

All those "conclusions" are coordinate dependent observations. Tidal effects are variations of the curvature of space-time in a given region.

Do you mean that the 'g-force' felt by a static observer just outside of a Schwarzschild event horizon (wasting an enormous amount of fuel to stay static) is coordinate dependent?

Is the fact that an observer attempting to remain static at such an horizon needs an infinite force (-> infinite acceleration required) coordinate dependent?

I somehow don't think so, but I would like to know the real answer.

 

[pmb_phy]

Do you mean that the 'g-force' felt by a static observer just outside of a Schwarzschild event horizon (wasting an enormous amount of fuel to stay static) is coordinate dependent?

That is one way to stay static but not the only way. I'm sitting in a chair in my living room and I'm at rest in the gravitatiomal field.

Pete

 

[pervect]

Do you mean that the 'g-force' felt by a static observer just outside of a Schwarzschild event horizon (wasting an enormous amount of fuel to stay static) is coordinate dependent?

Is the fact that an observer attempting to remain static at such an horizon needs an infinite force (-> infinite acceleration required) coordinate dependent?

I somehow don't think so, but I would like to know the real answer.

Given a particular path through space-time the magnitude of the 4-acceleration along that path is a geometric (i.e. coordinate independent) quantity - i.e. the 4-acceleration is a 4-vector, and like other 4-vectors, its length is an invariant.

This scalar quantity has a physical interpretation as the magnitude of the felt g-force.

Treating the acceleration 4-vector as a 3-vector does involve some coordinate dependence - basically, one is defining the time $\partial / \partial t$to point along the path, and furthermore one is requiring space to be orthogonal to time.

 

[MeJennifer]

Do you mean that the 'g-force' felt by a static observer just outside of a Schwarzschild event horizon (wasting an enormous amount of fuel to stay static) is coordinate dependent?

See Pervect's comments.

Is the fact that an observer attempting to remain static at such an horizon needs an infinite force (-> infinite acceleration required) coordinate dependent?

See Pervect's comments.

I somehow don't think so, but I would like to know the real answer.

Well we agree, so you are fighting the straw man.

But the notion, as some present here, that the tidal force is infinite at the event horizon for an accelerating observer is not correct IMHO.

Tidal forces appear when a gravitational field is inhomogeneous, or in other terms, in a region where the Riemann curvature tensor is not constant. Tidal forces are curvature differentials in space-time not some property of the state of an observer's acceleration.

If you wish to believe that tidal forces depend on the state of acceleration instead of being a property of space-time, or even, talking about confusion, a combination of that, then that is of course entirely up to you. And then you, consequently, can state that, when approaching the event horizon, the tidal forces are infinite for an accelerating observer while they are not infinite for an observer in free fall.
But I frankly fail to see how one could develop a better understanding of general relativity by such an idiosyncratic view.

My advice in developing a better understanding of general relativity is to learn how the properties and shape of space-time determine how observers see their world, as opposed to understanding the observer's world through fancy metrics with cross terms and coordinate dependent views or in attempt to somewhat equate the hyperbolic properties of flat Minkowski space-time with space-time curvature.

Anyhow, don't be surprised that many will disagree with such a notion of what tidal forces are.

 

[Boustrophedon]

Tidal forces appear when a gravitational field is inhomogeneous, or in other terms, in a region where the Riemann curvature tensor is not constant.

You mean - ...in a region where the Riemann curvature tensor is not zero, surely !?

 

[Jorrie]

Anyhow, don't be surprised that many will disagree with such a notion of what tidal forces are.

Yep, it seems that tidal gravity is as difficult to define as "uniform gravitational field"!

 

[Jorrie]

That is one way to stay static but not the only way. I'm sitting in a chair in my living room and I'm at rest in the gravitatiomal field.

Pete

Your favorite armchair would not work so nicely just outside of an event horizon - well, maybe with a BIG rocket attached.

I believe one can actually enjoy a 'static 1g' just outside of a Schwarzschild hole's horizon if it (the hole) is BIG, really BIG.

 

[George Jones]

Do you mean that the 'g-force' felt by a static observer just outside of a Schwarzschild event horizon (wasting an enormous amount of fuel to stay static) is coordinate dependent?

Is the fact that an observer attempting to remain static at such an horizon needs an infinite force (-> infinite acceleration required) coordinate dependent?

No.

I somehow don't think so, but I would like to know the real answer.

An observer static at $r$ experiences a 4-acceleration (and thus a 3-accleration, since 4-acceleration is orhtogonal to 4-velocity) given by

$$\left( 1 - \frac{2M}{r} \right)^{-\frac{1}{2}} \frac{M}{r^2},$$

which blows up at the event horizon of a black hole of any mass.

Tidal stresses at the event horizon decrease as the mass of the black hole increases.

 

[Jorrie]

No.
Tidal stresses at the event horizon decrease as the mass of the black hole increases.

Thanks George, I'm comfortable with that (and with the equation you gave).

What's not clear to me is how the 'tidal forces' experienced by 'static' and 'free-falling' observers differ in the vicinity of the event horizon of a hole, irrespective of its mass. It appears that pervect and MeJennifer have different views.

However, I also understand that there may be different definitions of "static observers" and 'tidal forces' at play here...

Jorrie

 

[pervect]

This thread is going around in circles a bit. I have not researched the usage of "tidal force" in the literature, unfortunately, but I've always understood a tidal force to basically be a map from a displacement to a force.

I'm sure we could spend a lot of time wrangling over whether or not this definition is correct or not. But I unfortunately I don't really have the time to do the research or at this point, the inclination. So I'll talk about the consequences of using this definition.

Physically, the components of the Riemann tensor at some point can be described as a the tidal force experienced by a geodesic and non-rotating observer at that point using this definition of tidal force. The caveats that the observer be geodesic and non-rotating are often omitted, but under some circumstances they can be important as some of the (rather extreme) examples illustrate.

The example of rotation is IMO interesting and hasn't been mentioned.

Suppose we have an inertial frame, where the tidal force (by this definition) is zero everywhere. The interesting point is that there is a non-zero tidal force in a rotating frame of reference even at the origin.

This shows that (at least using the above definition of tidal force) that the tidal force is not a tensor quantity. It can be shown that the value of a tensor quantity at a point must be independent of rotation about that specific point. For instance, the electric field at some point in the origin of a rotating frame is the same as the electric field at that same point in a non-rotating frame. This can be formally derived from the transformation properties of a tensor, but it is easier to think of it (IMO) as following from the fact that a tensor is a geometric object independent of coordinates.

If there is a documented definition of "tidal force" which avoids this issue and makes it a true tensor quantity, great, but I'm not aware of it. I expect that any such definition will wind up being equivalent to the definition of the Riemann tensor. But the main point of this exercise (in my opinion) was to provide some more approachable, physically based, definition of the Riemann while still being accurate enough to be useful.

 

[Jorrie]

This thread is going around in circles a bit. I have not researched the usage of "tidal force" in the literature, unfortunately, but I've always understood a tidal force to basically be a map from a displacement to a force.

Looks like that's as good a definition as we will get. I'm not quite sure that MeJennifer's original question 'Is a "uniform gravitational field" a gravitational field?' has been answered yet. (Apologies if I missed the answer or just plainly did not understand that it was the answer).

 

[George Jones]

Thanks George, I'm comfortable with that (and with the equation you gave).

What's not clear to me is how the 'tidal forces' experienced by 'static' and 'free-falling' observers differ in the vicinity of the event horizon of a hole, irrespective of its mass. It appears that pervect and MeJennifer have different views.

I meant the tidal force on a freely falling observer. I haven't been following this thread in detail, so, before posting, I was not aware of all the views expressed, and I am not perpared to talk about any case other than the freely falling one.

However, I also understand that there may be different definitions of "static observers" and 'tidal forces' at play here...

I think that in a static spacetime like Schwarzschild there is a standard definition of static observer.

The (image of) the worldline of a static observer is (the image of) an integral curve (flow line) of the Killing vector $\partial_t$, and the static observer's 4-velocity is proportional to $\partial_t$ along the worldline. Such an observer has constant r, theta, and phi coordinates.

 

[JesseM]

This is an exact and precisely correct definition of "Born rigid (accelerated) motion". What cannot be relied upon, however, are the various other 'interpretations' of it and 'constructions' put upon it - such as that the acceleration 'g-force' should diminish towards the forward end.

Earlier in this thread it was stated that "uniform acceleration" is the same thing as what has just been defined as "Born rigid acceleration" and this is quite correct.
However, one property that a uniformly accelerating frame (or Born rigid frame) absolutely must have is that it should be perfectly inertial to co-moving observers when freely falling in a "uniform gravitational field". Any version of the equivalence principle makes this mandatory.

Unfortunately, if the g-force of a uniform field diminishes with increasing height (by "indistinguishability" equivalence), the free falling frame by the same token will have to experience diminishing g-forces towards its 'forward' end, which will be 'lower in the field'.

Why do you say that? Is this just an intuition, or have you worked out the math?

It seems to me that the primary question here is just whether or not two accelerometers situated on opposite ends of a rod undergoing Born rigid motion will measure the same G-force or different G-forces. In pervect's post #9 on this thread, I thought he was saying the G-forces would be different:

If two observers are undergoing Born rigid acceleration each observer (the front and back observer) will measure a different acceleration with his or her local accelerometer.

See for instance http://www.mathpages.com/home/kmath422/kmath422.htm

Quote:
Trailing sections of the rod must undergo a greater acceleration in order to maintain Born rigidity with the leading end, and the required acceleration is inversely proportional to the distance from the pivot event

http://arxiv.org/abs/physics/9810017 also discusses this. This is the simplest peer reviewed English language reference I've been able to find on the topic. (The mathpages article is also pretty good IMO and may be easier to follow though of course it is not peer reviewed).

What the literature calls a "uniform gravitational field" isn't actually uniform as measured by local accelerometers.

See for instance http://arxiv.org/PS_cache/physics/pdf/0604/0604025.pdf for an example of this usage.

If the G-forces are different for different points on a rod undergoing Born rigid acceleration, and if a "uniform gravitational field" is just what must be postulated by a coordinate system undergoing Born rigid acceleration in order to view itself as being at rest, then there should be no problem with freely-falling observers--if the accelerating system sees an observer A as freely-falling, then A will see herself as being at rest in flat spacetime with no gravitational field while the coordinate system accelerates past her, so of course she will feel no G-forces.

 

[pervect]

No.
An observer static at $r$ experiences a 4-acceleration (and thus a 3-accleration, since 4-acceleration is orhtogonal to 4-velocity) given by

$$\left( 1 - \frac{2M}{r} \right)^{-\frac{1}{2}} \frac{M}{r^2},$$

which blows up at the event horizon of a black hole of any mass.

Tidal stresses at the event horizon decrease as the mass of the black hole increases.

I could use a sanity check, if you have the time.

The rate of change of proper acceleration with respect to the r coordinate should be

df/dr, where
$$f=\left( 1 - \frac{2M}{r} \right)^{-\frac{1}{2}} \frac{M}{r^2},$$

The rate of change of proper acceleration with respect to distance d should then be, by the chain rule

$$\frac{df}{dr} \frac{dr}{dd}$$

From the metric coefficients, we can say that
$$dd = \frac{1}{\sqrt{1-2m/r}} dr$$

where dd is the differential change in distance. dd is always greater than dr, so dr/dd is less than 1.

Thus we can say that the rate of change of acceleration with distance is

$${\sqrt{1-2m/r}} \frac{df}{dr}$$

The term on the left goes to zero at r=2m, but the term on the right blows up, so it's not clear without a detailed calculation what happens.

Plugging this into Maple, I get

$$-{\frac {m \left( 2\,r-3\,m \right) }{{r}^{3} \left( r-2\,m \right) }}$$

assuming I haven't messed anything up. This blows up at r=2m.

Of course, we still haven't really settled whether or not calculating the rate of change of proper acceleration with respect to distance is the correct definition of tidal force...