# Acceleration and gravity, conceptual problems.

1. Jun 27, 2013

### tolove

The book "Black Hole Wars" used the following to explain the concept of a difference in time near black holes.

Step 1: Find a massive black hole, large enough so that tidal forces will be harmless to a human near the Schwarzschild radius.
Step 2: Lower a person down via a cable.
Result: The person lowered will experience significantly lower time than the people at the other end of a rope.

I have a couple questions with this thought. Could someone explain a general idea about the numbers involved in such a situation?
The person lowered will be experiencing an enormous gravity, right? Enough to crush a person? Enough to flatten steel?
Wouldn't the pull on the cord and lowered object be absurdly large? If the cable was attached to a distant planet, wouldn't the entire planet be pulled in at a rapid pace.

Could someone check my understanding? I am going to type a small paragraph from my understanding of acceleration/gravity. I assume most of it is a bit off:

"There is no difference in time variance resulting from an accelerating object or an object inside a gravitational field. If a trampoline is being used as an analogy, then space-time is curved around an object placed on the surface. Acceleration may be viewed as the trampoline being shifted up and down. Eg, an object placed on the surface causes a constant indentation. The object is then given acceleration, resulting in the trampoline being shifted. The trampoline's surface in now perfectly flat, or can be, with the correct acceleration." A shifting trampoline may be a bad analogy, but I'm trying to find a way to visualize how acceleration causes a variance in space-time without creating a gravitational field. Gravity is a field, but acceleration is one object's property?

Further, if my statement "There is no difference in time variance resulting from an accelerating object or an object inside a gravitational field" is true, then how else can this be worded? Can this same message be written without the word 'time?' I ask, because a new wording may help me understand the principle better, or help me see physical equivalencies between space, time, and whatever else is present in this situation.

Thank you very much for your time. Any comments on the subject will be appreciated.

2. Jun 27, 2013

### Staff: Mentor

Not necessarily. The more massive the hole is, the closer you have to get to its horizon to experience a large acceleration. If the hole is large enough that tidal gravity is negligible near the horizon, it is also large enough that you can get quite close to the horizon without experiencing crushing acceleration.

This is really the same question as above: the acceleration experienced by an object being held steady at a constant altitude is what is resisting the "pull" you are talking about.

Not locally. But an accelerating object, in and of itself, does not curve spacetime; so as soon as you make observations over a large enough range of space or time that tidal gravity can be detected, you can detect the difference between being in an accelerating rocket in flat spacetime, and being at rest in the curved spacetime produced by the gravitational field of a planet, star, or other massive object.

The trampoline analogy is a bad one for at least two reasons: it only includes space curvature, not spacetime curvature; and the space curvature it describes is frame-dependent--it is only valid for observers who are hovering at rest above the gravitating object.

No, this is not correct. The trampoline analogy can't really deal with the gravitating object being accelerated, since, as I just noted, it's only vaild for observers who hover at rest above the object, meaning the object itself must also be at rest.

You can't, because acceleration doesn't cause a change in spacetime. See above.

3. Jun 27, 2013

### WannabeNewton

It essentially comes down to the equivalence principle. A static observer in a sufficiently small region of a stationary space-time (small enough so that the gravitational field is uniform) can always transform away the gravitational field and instead describe him/herself as a uniformly accelerating observer. However as Peter pointed out, space-time can be curved globally so you can, on a global scale, detect tidal forces whereas the metric of a uniformly accelerating observer (Rindler metric) describes a portion of flat space-time (called the Rindler wedge) so there won't be tidal forces in the above sense.

4. Jun 27, 2013

### PAllen

Well, it depends. Expressed as e.g. 1.01 times horizon radius, yes. The bigger the BH mass, the lower the proper acceleration for hovering here. However, if you ask what is the proper acceleration 1 meter from a BH, it goes to infinity as BH mass increases; and to zero as BH mass decreases. For discussions of lowering ropes, I think the second meaning is more appropriate. [Of course, you have to be careful of what you mean by 1 meter from the horizon, but for any reasonable definition, what I say is true].

[edit: Also, to get a large difference in rate of lower and higher (hovering) clocks over a stated distance (e.g. 10 meters), you need a large magnitude of proper acceleration (for both - else tidal gravity would be extreme)].

Last edited: Jun 27, 2013
5. Jun 27, 2013

### WannabeNewton

Even when an observer is at rest in the uniform gravitational field within a planet in the solar system e.g. standing on the ground of the planet (if the planet is terrestrial and not gas), he/she feels a proper acceleration (justified via equivalence principle) due to the normal reaction force from the ground on him/her (we imagine the ground as fixed to the terrestrial planet itself) that increases the more massive the planet is. More explicitly, since this is a weak field region, we may choose a fixed background global inertial frame and global inertial coordinates $(t,x,y,z)$ and note that the observer follows an orbit of $(\frac{\partial }{\partial t})^{a}$ in which case the proper acceleration is $a^{\mu} = \delta^{\nu}_{t}\nabla_{\nu}\delta^{\mu}_{t} = \Gamma ^{\mu}_{tt} = g\delta^{\mu}_{z}$ (I have oriented the coordinates so that $z$ is aligned with the direction of the gravitational field) i.e. $a = g$ as we would expect. $g$ of course scales proportionally with the mass of the source.

6. Jun 27, 2013

### Staff: Mentor

Yes, this is what I meant, but I was thinking of it slightly differently: suppose I want to experience a given proper acceleration $a$ while hovering at some radius $r$ = $2M \bar{r}$ (so $\bar{r}$ is 1.01 in the example in your quote above). The larger $M$ (the mass of the hole), the smaller $\bar{r}$ is for a given fixed $a$.

7. Jun 27, 2013

### Naty1

tolove:

You have experts already posting, but maybe an 'amateur' can offer some insights ....Somebody will critique, I am sure, if I say something incorrect. I have been struggling on and off for some months with a question similar to yours:

tolove:
PeterDonis replies:
ok,let's start from there....

[None of what I express here is intended to contradict the above posts. Merely offer a different perspective, hopefully accurate.]

[1]Only two things 'affect' spacetime: velocity and gravitational potential. Acceleration does not in and of itself affect spacetime, but different velocities do....that's special relativity in which we say a faster moving observer experiences 'time dilation' and 'length contraction'. Space and time morph together in such as way as to enable the speed of light 'c' to be the same for all observers, fast or slow. [Mathematically, the Lorentz Transform describes these relationships.] So for observers in relative motion, one's observed space may be partially the other's time, and vice versa. This is a type of spacetime 'distortion' but is NOT gravitational spacetime curvature.

[Someone posted elsewhere the difference is that gravity has tidal forces....squeezing and stretching...a particular type of curvature. ]

This REALLY bothered me for a time until several years ago I got an explanation that went like this: [My edited version of a DrGreg post]

[2]
So blending these two descriptions, [1 & 2] together, it turns out special relativity where space and time' change places' is represented by the non square grid lines of FLAT graph paper...and is observer dependent. The gravitational spacetime curvature is still represented by the 'curved' graph paper itself....and is observer INdependent.

Just keep in mind the term 'gravitational field' and 'curvature' are complex subjects. Misner, Thorne and Wheeler, a reliable source textbook puts it this way...