# Finite tidal forces at black hole event horizon redux

1. Aug 18, 2015

### pervect

Staff Emeritus
What's the best way to explain why tidal forces for an observer free-falling through an event horizon are finite?

My first thought was to say that "gravity isn't a force, it's a curved space-time". On further thought, however, it seems to me that consideration of the Rindler horizon shows that the essential features of the issue don't really require curvature to resolve, and that it wouldn't really be a good idea to claim otherwise.

My best take on using the Rindler horizon approach so far is to consider a rigid rod, with absolutely no forces on it of any kind, it's just floating in free space.

Then we switch to an accelerating frame (the Rindler frame), and consider what happens to said rigid rod. It still has no forces on it, even when it passes through the Rindler horizon. It may not even "know" that it passed through the horizon, as the horizon is a property of the accelerating observer, and there could be several different observers with different horizon. As far as the rod is concerned, though, it's happily floating in empty space and doesn't care at all that someone else thinks it is at, near, passing through, or already past a horizon.

Perhaps we can mention a few other things at the same time about how things appear in the Rindler frame, such as the Rindler frame assigning an infinite time coordinate to the rod's penetration, and the Rindler frames notion of "gravity", i.e. the proper acceleration required to hold oneself stationary in the Rindler frame, increasing towards infinity as one approaches the horizon.

This is the best approach I can think of, but I suspect it's still too advanced if one is not already familiar with the ins and outs of the Rindler horizon :(.

This was inspired by a recent thread, I thought my question was sufficiently different in focus, narrowness, and tone that I'd start a new thread rather than derail the existing one.

2. Aug 18, 2015

### PAllen

I think the Rindler horizon is the best approach, supplemented with a coherent argument based on e.g. the principle of equivalence, that a BH horizon over a small region must be equivalent to a Rindler horizon. Of course, nowadays someone will always raise firewalls, which leaves open whether the classical description of horizons matches our universe. You can bypass that by simply stating the context is classical, not quantum.

3. Aug 18, 2015

### Staff: Mentor

The other thing that you can do is make a Newtonian gravity based argument that the gravitational force falls off as 1/r^2 while the tidal force falls off as 1/r^3. GR of course replicates that in the weak field, but even in the strong field as M increases the tidal forces decrease at a fixed proper acceleration.

4. Aug 18, 2015

### bcrowell

Staff Emeritus
I don't see why one would expect infinite tidal forces at the horizon. If someone asked me whether Sandy Koufax was a woman, I would just say no, he wasn't. I wouldn't feel the need to offer arguments as to why he couldn't possibly have been a woman.

5. Aug 18, 2015

### WannabeNewton

This can be answered using just simple dimensional analysis. Why make it so complicated?

6. Aug 18, 2015

### Staff: Mentor

I'm not sure I agree with this. The Rindler horizon occurs in flat spacetime, and the whole point of the question is what happens in curved spacetime. Any question about "tidal forces" presupposes curved spacetime; if you try to answer it using an argument based in flat spacetime, you're dodging the question, IMO.

To put this another way: if I make the mass of the hole smaller, the tidal forces at the horizon increase; so given any object of finite size, I can find a hole small enough that tidal forces at the horizon will break the object. But the analogy with the Rindler horizon is independent of the mass of the hole; given a hole of any mass, you can always find a small enough patch of spacetime around the horizon that looks flat, and within that patch, the hole's horizon is equivalent to a Rindler horizon. It's just that, for a small enough hole, that patch will be smaller than the size of an object that can be broken apart by the tidal forces at the horizon, so any argument making use of the Rindler horizon will simply fail for that object. That doesn't look to me like a good strategy for a general argument to show that tidal forces at the horizon must be finite.

I also don't agree with this. Tidal forces affect free-falling objects; bringing in accelerated objects adds complication with no real gain that I can see.

To me the answer to this question is simple: compute the Riemann tensor at the horizon and show that all its nonzero components are finite. MTW has this as an exercise.

The only other angle I can see is that a newbie might have the (mistaken) idea that "tidal forces" is the same as "differences in acceleration due to gravity". That's true in the weak field approximation, but not in general (obviously, since "acceleration due to gravity" diverges at the horizon). So the answer to that newbie is simply what I just said, that his belief is only true in the weak field approximation, not in general.

7. Aug 18, 2015

### pervect

Staff Emeritus
How's that? Is this equivalent to the m/r^3 vs m/r^2 argument? (Which seems like a reasonable and simple approach, one I didn't think of).

8. Aug 18, 2015

### pervect

Staff Emeritus
I agree that this is a good approach for an advanced student/poster. It may also be either necessary or just useful to "appeal to authority" (i.e. textbooks) to point out that the answer is what it is. I believe direct computation is the approach most textbooks would take to the problem.

For educational purposes, though, it leaves a bit to be desired, as some understanding of "why" beyond "because the textbooks say so" is useful - when it can be obtained.

I would pretty much guess that this is not only true, but the main origin of the (false) argument that the tidal forces "should be" infinite.

Because your approach has more textbook support, it might be better in many cases.

9. Aug 18, 2015

### Staff: Mentor

I think one can certainly give some explanation of what the Riemann tensor means in terms of tidal gravity. The usual illustration I give is two rocks dropped from rest above the Earth: first, drop them along the same radial line from slightly different heights, and watch them diverge; second, drop them from the same height at slightly different horizontal locations and watch them converge.

Of course this illustration still makes use of the student's intuition about how "acceleration due to gravity" changes; but then you can say something like: in many cases (such as stronger fields like at the horizon of a black hole), there will no longer be a direct correspondence between the convergence/divergence of free-falling worldlines and changes in the "acceleration due to gravity"; but that's because, while convergence/divergence is well-defined everywhere in any spacetime, "acceleration due to gravity" often either diverges (like at a black hole horizon), or fails to be well-defined at all (like in any non-stationary spacetime). Thus, convergence/divergence of free-falling worldlines is more general and more fundamental, so that's what we use in GR (and what the Riemann tensor describes).

10. Aug 19, 2015

### Markus Hanke

Is it not sufficient to calculate any one of the curvature invariants from the Riemann tensor ( such as Kretschmann, Euler, Chern-Pontryagin ), and show they are not singular at the event horizon, to demonstrate this ?