Mentor

## Is a "uniform gravitational field" a gravitational field?

 Quote by Jorrie 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.

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 Quote by Boustrophedon 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.

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 Quote by George Jones 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....

Mentor
 Quote by pervect 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.
This looks right.

 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....
Consider a hovering pillar. Thinking in a Newtonian way, gravity tries to compress the pillar. For example, the cross-sectional slab in my attached diagram has it acting onthe wieght W of all the stuff above the slac, and also a normal force applied to the slab by the stuff below.

Still thinking in a Newtonian way, if the accleration due to gravity changes over the length of the pillar, the wieght of all the stuff in the portion of the pillar above the slab is found by "adding up" (integrating) the weights (dW = a(d) dm) of a "bunch" of infinitesimal slabs above the illustrated slabs. Here, if appropriate dimensions and units are used, the infinitesimal mass can taken to be your dd.

This leads to an infinite squashing force as the pillar approaches the event horizon.

I think both you and JesseM have talked about stuff like this in other threads.
Attached Thumbnails

 Recognitions: Science Advisor Staff Emeritus If one has a uniform bar in a Newtonian field, the stress in the bar will be non-linear (I believe it's quadratic). GR will have a different distribution of stress for a uniform bar than the Newtonian case, but I think it winds up the same for a short enough bar. But having a uniform bar is not the way to get a simple measure of tidal force. The way to get a linear relationship and thus a simple read on the tidal force is to have a "lightweight" bar of negligible mass supporting a test mass, i.e. (O)---------(m) where O is the observer, the ----- is the lightweight bar with a "small" mass, and m is the test mass. This arrangement makes the force in the bar essentially constant. But the above arrangement will work only if the observer is not accelerating. For an accelerating observer, one way to fix this, and the way I'm proposing, is this: (A1)-------------(A2) A1 is an accelerometer mounted on one end of the bar (some test mass and a spring, for example), and A2 is an accelerometer mounted on the other end of the bar. What I'm calling the "tidal force" is $$\lim_{L \rightarrow 0}(A2-A1)/L$$ where L is the length of the bar.

Mentor
 Quote by pervect If one has a uniform bar in a Newtonian field, the stress in the bar will be non-linear (I believe it's quadratic).
Quadatic in the length of the bar? I don't suppose you know a reference?

I think the effect in my previous post is real, and that this is why scenarios for human space travel have a ~ 1 g. Notice that I didn't actually call the effect a tidal force.

 But the above arrangement will work only if the observer is not accelerating.
Lately I have been working on wormholes (I get a big from Chris.), and the original Morris-Thorne paper calculates tidal forces for accelerated observers the same as they are calculated for freely falling observers. The paper says "The fact that the observer is accelerated (does not fall freely) has no influence on the relative acceleration that she feels."

Visser's book refers to box 37.1 of MTW for this argument, but I won't be able to look at MTW until tomorrow or Monday.

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 Quote by George Jones Quadatic in the length of the bar? I don't suppose you know a reference?
Gah. I should have said that pressure should be proportional to depth, displacement (assuming Hooke's law) should be quadratic.

 I think the effect in my previous post is real, and that this is why scenarios for human space travel have a ~ 1 g. Notice that I didn't actually call the effect a tidal force.
Yes, in casual conversation I'd just call that effect gravity, not tidal force. I'd interpret what you said as "when you stand up, the soles of your feet have to support all your weight".

IIRC the equations for pressure vs depth in GR can be derived from
$$\nabla_a T^{ab} = 0$$

and the appropriate metric of interest.

 Lately I have been working on wormholes (I get a big from Chris.), and the original Morris-Thorne paper calculates tidal forces for accelerated observers the same as they are calculated for freely falling observers. The paper says "The fact that the observer is accelerated (does not fall freely) has no influence on the relative acceleration that she feels." Visser's book refers to box 37.1 of MTW for this argument, but I won't be able to look at MTW until tomorrow or Monday.
Box 37.1 (pg 1007) looks VERY interesting. It's not quite as clearcut as Visser's summary makes it sound, there are some acceleration effects mentioned of the order of (1+ax) and (1+2ax) applied at various places in the formula. I think I need to study this section a lot more carefully.

Checking the index, pg 860 \$32.6 has a brief comment on tidal force as well, but it looks like the real meet is in box 37.1.

 Quote by JesseM 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.
You've missed the crux of the issue. Considering a single point A is no good - it will feel no force regardless of whether it's free falling in a "uniform" field or a "normal" non-uniform one. A uniform field is defined as one in which the equivalence principle holds exactly over an extended frame, so that an extended body feels no tidal force, or separated particles remain at constant distance, when in free fall.

If one adds a separated colinear observer B to your A then of course they'll stay at fixed separation when stationary in your scenario, but if one actually substitutes a gravitational field the only way that A and B can stay at constant distance is if the field is constant with height.

If you try and argue that the particle higher in the field needs to fall with greater acceleration to stay at the same distance you need a field "the wrong way round" - i.e. greater gravitational pull higher up and smaller gravitational pull lower down, which would be opposite to and in no way resemble the supposed "uniform acceleration".

Thus by elimination a "uniform gravitational field" is proven to be one where the g-force is the same at all heights, and by the same token "uniform acceleration" must involve the same g-force at all points along the frame.

 Quote by pervect 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).
I doubt this. Can you provide a reference please? What about for a freely falling observer who is just above the horizon and escaping to r=infinity (i.e. in free fall and moving away from the black hole)? Is the tidal force for such an observer independent of her velocity?

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MTW's textbook, "Gravitation", carries out the calculation of the tidal forces on an infalling observer in pg 821-822. I'll quote some of the relevant sections:

 The tidal forces felt by the explorer as he passes a given radius r are measured by the components of the Riemann curvature tensor with respect to his orthonormal frame there. To calculate these curvature components, proceed in two steps. (1) Calculate the components, not in the travellers frame, but rather in the "static" orthonormal frame.
and step (2) is to convert these to the traveler's frame.

 The payoff of this calculation, according to equation (31.6), none of the components of the Riemann (ed note:components of Riemann == tidal forces) in the explorers orthonormal frame become infinite at the gravitational radius. The tidal forces the traveler feels as he approaches r=2M are finite; they do not tear him apart.
The independence of the tidal forces on velocity is also discussed in these pages - the traveller's frame and the static orthonormal frame are related by a Lorentz boost if the object is outside the event horizon, and by a boost-like transformation (but with a velocity parameter greater than 'c') if the falling object is inside the event horizon.

In either case (with the ordinary boost or with the hyper-boost) the components of the Riemann (i.e. the tidal forces) are totally unaffected by the boost, and are equal to the components of the Riemann in the static orthonormal frame. In geometric units, the stretching tidal force is of magnitude 2M/r^3, where M is the mass of the body, and r is the radius.

To give the exact quote:

 The amazing result (a consequence of special algebraic properties of the Schwarzschild geometry, and somewhat analogous to what happens - or rather, does not happen - to the components of the electromagnetic field, E and B when they are both parallel to a boost) is this: all the components of Riemann are left completely unaffected by the boost.
This would also work for an outgoing traveler, though the textbook doesn't discuss this case explicitly.

One thing I should probably clarify. The results above apply only for a boost directly towards or directly away from the black hole. Though it is not directly discussed in MTW, one can see an effect on the Riemann / tidal forces due to a boost if the boost is in a non-radial direction. The point of the above remarks is that for a radial boost, there is no effect on the Riemann /tidal forces.

 Thanks so much for that great info pervect! But if MTW are right, then how is GR valid? Let a freely falling rod be escaping to r=infinity radially as it travels through a group of particles falling toward one another to eventually form a black hole, when a horizon forms along the rod so that the rod spans the horizon with the part of the rod that is above the horizon still escaping to r=infinity radially. If the rod can be small enough (or the black hole large enough) that the tidal force on the rod can be too weak to break the rod, and GR features no other force that can break it, then how does it break as GR predicts it must so that it does not pass outward through the horizon? To be a self-consistent theory it seems that GR must feature a force other than the tidal force to break the rod, but I don’t see one mentioned in my texts. The tidal force isn’t an attractive force, so it can’t pull the rod inward across the horizon. Do you know what breaks the rod in this case?
 Recognitions: Science Advisor Staff Emeritus There are a couple of problems with your argument, I think. The first problem is that one normally expects, in a well-behaved coordinate system, that two points with almost identical values of the 'r' coordinate will be "close together". Schwarzschild coordinates are not well-behaved at the horizon, however. In fact, g_rr goes to infinity. This means that two points with nearly the same r coordinate can be "far apart". Therefore it is not clear that a particle at the horizon, falling in, will ever be "close" to a particle that is escaping. Sure, they may have nearly the same r coordinate, but this does not mean that they are "close", because the metric goes to infinity. The second problem relates to the fact that your problem statement is different than the one worked out in the textbook. For any particle in the exterior region of the forming black hole, we can use the Schwarzschild metric, and the calculations from the textbook will apply. In this case there will be no difference between the forming black hole, and a static black hole, and the tidal forces will be finite. Therefore we can eliminate the forming black hole from the problem, and replace it with a static black hole, for any particle in the exterior of the forming black hole. We can do this both in the calculation, and in the problem statement. However, we can't use the Schwarzschild metric for any particle in the interior of your forming black hole. If we assume that your forming black hole is a presureless collapse, we'd have to use the interior metric for such a presureless collapse. I seem to recall that this is just the FRW metric in the interior, smoothly joined to a Schwarzschild metric in the interior. This needs a new calculation of the Riemann components.
 Recognitions: Science Advisor Staff Emeritus The simplest case to analyze appears to be when the black hole is formed by dust falling in from infinity - i.e. a FRW cosmology with a_max=infinity, i.e. a FRW cosmology with the critical density. In this case, for an observer moving with the dust, all tidal forces are compressive. So the radial component of the tidal force for someone comoving with the cloud will change sign depending on whether one is outside or inside the cloud. This surprised me, but it is similar to the Newtonian case. Radial tidal forces are of the stretching type outside a massive body, because gravitational acceleration (in the Newtonian sense) drops off with increasing radius. Newtonian radial forces are of the opposite type (i.e. compressive) inside a massive body, because gravitational acceleration increases with increasing radius, i.e. the acceleration is zero at the center and for a body of uniform density follows a linear "Hooke's law". As far as the effects of a boost on the tidal force goes: For the FRW cosmology, the boost doesn't seem to affect the components of tidal force in the direction of the boost, but it does affect (increase) the tidal forces in the transverse directions. So this suggests that the first explanation I offered is the correct one, the issue is that two objects with nearly identical 'r' coordinates may not actually be close near the horizon, because the metric coefficient multiplying dr^2 becomes infinite.

 Quote by pervect Therefore it is not clear that a particle at the horizon, falling in, will ever be "close" to a particle that is escaping. Sure, they may have nearly the same r coordinate, but this does not mean that they are "close", because the metric goes to infinity.
For any part of the rod to fall inward through the horizon, the rod must first break. But that’s the question, whether or not the rod breaks. If you’re talking about one of the cloud of particles that forms the black hole, I don’t see why it would need to be close to a particle of the rod that is escaping.

GR must predict that the tidal force on the rod is infinite, so the rod will always break regardless of its tensile strength, which can be arbitrarily high in principle. Otherwise the rod could pass outward through the horizon, violating GR’s prediction that nothing can do that.

 ... However, we can't use the Schwarzschild metric for any particle in the interior of your forming black hole. If we assume that your forming black hole is a presureless collapse, we'd have to use the interior metric for such a presureless collapse. I seem to recall that this is just the FRW metric in the interior, smoothly joined to a Schwarzschild metric in the interior. This needs a new calculation of the Riemann components.
I translate what you’re saying to this: Although the tidal force is independent of velocity, GR may predict that the tidal force on the rod is infinite, because part of the rod is below the horizon. Do I have that right?

 In this case, for an observer moving with the dust, all tidal forces are compressive. So the radial component of the tidal force for someone comoving with the cloud will change sign depending on whether one is outside or inside the cloud.
The rod isn’t comoving with the cloud of particles that formed the black hole. It was escaping to r=infinity from the cloud when the horizon formed along the rod.

It seems that GR would be self-inconsistent even if it predicted that the tidal force on the rod is infinite. GR postulates that SR holds locally. But SR cannot hold locally where the tidal force is infinite locally.

How does GR avoid self-inconsistency? I don’t see how it can.

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 Quote by Zanket GR must predict that the tidal force on the rod is infinite, so the rod will always break regardless of its tensile strength, which can be arbitrarily high in principle. Otherwise the rod could pass outward through the horizon, violating GR’s prediction that nothing can do that.
Wouldn't your argument also imply that if an accelerating observer in flat spacetime was dragging along a rod undergoing born rigid acceleration (so its length remains constant in the Rindler coordinate system where the accelerating observer is at rest), then by making the rod's tensile strength high enough the back end of the rod could pass through the Rindler horizon? But in fact this is impossible--as explained here, the rod must always break if it extends past the Rindler horizon. And the breaking has nothing to do with tidal forces, since this takes place in flat spacetime, it has to do with the fact that in order to pass the through the horizon in Rindler coordinates, the back end of the rod would have to move FTL as measured in an inertial coordinate system. It seems quite possible that the breaking of a relatively short rod moving outward that is cut in half by the formation of an event horizon would also have nothing to do with tidal forces (since after all tidal forces always disappear in the limit as you zoom in on a smaller and smaller local region of spacetime, and you're free to make the rod as short as you like, all that matters is that it gets cut in half by the event horizon). I believe observers at different ends of a rod undergoing Born rigid motion in flat spacetime do experience different G-forces (different proper acceleration), so perhaps the breaking of the rod could be explained in terms of some sort of "fictitious" tidal-like force that's a consequence of adopting a non-inertial coordinate system. But as I understand it a real tidal force always implies spacetime curvature.

 Quote by JesseM Wouldn't your argument also imply that if an accelerating observer in flat spacetime was dragging along a rod undergoing born rigid acceleration (so its length remains constant in the Rindler coordinate system where the accelerating observer is at rest), then by making the rod's tensile strength high enough the back end of the rod could pass through the Rindler horizon?
Assuming for the sake of argument it implied that, how does it answer the key question: how can GR be self-consistent? More on this below.

 It seems quite possible that the breaking of a relatively short rod moving outward that is cut in half by the formation of an event horizon would also have nothing to do with tidal forces (since after all tidal forces always disappear in the limit as you zoom in on a smaller and smaller local region of spacetime, and you're free to make the rod as short as you like, all that matters is that it gets cut in half by the event horizon).
The rod I’m talking about is freely falling. By the definition of “freely falling object”, no forces except gravity are acting on the rod. Then if the rod breaks, it must be the tidal force that broke it, because the only force of gravity in GR is the tidal force. Yes, GR says that the tidal force always disappears in the limit locally, but GR also predicts that the tidal force must be what breaks the rod if it breaks, by disallowing any other possibility.

 I believe observers at different ends of a rod undergoing Born rigid motion in flat spacetime do experience different G-forces (different proper acceleration), so perhaps the breaking of the rod could be explained in terms of some sort of "fictitious" tidal-like force that's a consequence of adopting a non-inertial coordinate system.
I don’t see how. According to GR, the tidal force is the only force on the rod. A fictitious force cannot break the rod; only a real force can do that.

The rod must break, or else GR is self-inconsistent (because then the rod would pass outward through the horizon in contradiction to GR’s prediction). Only an infinite tidal force is guaranteed to break the rod, since the tidal force is the only force on the rod (according to GR) and the rod’s tensile strength can be arbitrarily high in principle. So if the rod breaks, GR is self-inconsistent (because then SR would not hold locally as GR postulates). If GR is self-consistent then something must be wrong with my argument. I’d like to know in no uncertain terms what is wrong with it.