# Are objects in a gravitational well shortened?

Bob is in a gravitational potential well, he moves a long vertical stick up and down a distance of 1 meters. Alice observes the upper end of the stick, at upper location.

There is the phenomenon of redshift of energy, so either Alice measures a shorter motion than 1 meters, or alternatively Alice measures a weakened force.

So I'm asking: is Bob weak, or is Bob short?

( EDIT: It interests me what REALLY happens. It seems that when you fall into a black hole, you maybe become contracted: your lower part is already slowed down, while your upper part is still moving a little bit faster.

I'm certain that when you are being lowered into a gravity well, you REALLY lose energy. Therefore there REALLY is no redshift of energy when energy travels back up from a gravity well.

Now I try to make a thought experiment out of these two things: a shortened person in a gravity well, sends his reduced energy up, by poking with a stick )

EDIT 2: Question 2: Bob moves the vertical stick horizontally a 10 meters distance, what distance and what force does Alice observe at the upper end of the stick?

• +1: It's actually a very good question--- I thought it was a bit silly at first. – Ron Maimon Feb 2 '12 at 14:06
• You don't become contracted at the horizon, there is no tidal force. You are pushed down by gravity, so you could be elastically compressed, but that's not what you mean. All these questions are best answered in Minkowski space using accelerated coordinates. You don't shrink, but you do find that lower down in the well, you have to push harder to raise a stick than further up in the well. – Ron Maimon Feb 3 '12 at 0:58
• Plus when you say REALLY, you mean "relative to the notion of energy defined by the static tau coordinate". There are different notions of energy when there are different notions of time where the geometry is time independent. So in Minkowski space, there is the "t-energy" or ordinary energy, and the "tau energy" or the gravitational well notion of energy that includes a time-shift. – Ron Maimon Feb 3 '12 at 1:00
• Well maybe it's possible that thing are not shortened by gravity. Photon in a box that is being lowered redshifts and slows down, wavelength is constant. Free falling photon slows down, no red or blue shift, wavelength shortens. Only free falling things are shortened by gravity. – kartsa Feb 4 '12 at 8:18
• I'll rewrite the previous comment: A photon experiences a redshift and a slowing down of propagation, when it's being lowered, so the wavelength stays the same. The only thing that a freely falling photon experiences is the slowing down of propagation, so the wavelength shortens. – kartsa Feb 5 '12 at 3:54

Bob is weak. Bob will find it harder to do the work of raising the meterstick, because Bob is slowed down by gravitational time dilation.

The proper way to answer this is to consider Rindler coordinates for an accelerating frame, where the metric is

$$r^2 d\tau^2 + dr^2$$

The "dr" coordinate is radial distance, and integrating this gives the meterstick length. The r^2 factor in front of $d\tau$ is the square of the redshift factor. The length of the meterstick is unchanged for all values of r, assuming it is neglibibly elastic in the range of gravitational fields considered, so that r is varying between two large values. The Rindler approximation is a valid local description of the gravitational field, in a region small enough so that the curvature is not important.

If you place two mirrors at two values of r, and let a photon bounce back and forth between the two mirrors, the answer becomes obvious. When the photon hits the lower mirror, it is more energetic, and it pushes the meterstick down by a large amount, then when it hits the upper mirror, it is less energetic, and it pushes the meterstick up by a smaller amount, but the center of mass of the stick over one photon cycle doesn't move.

Further, the back and forth bounces of the photon give an equal amount of pushes per unit tau at both locations (this is clear from the fact that tau is a killing vector for the metric, so that the process is tau stationary when averaged over many cycles). But a unit of tau at Bob's position is shorter than a unit of $\tau$ at Alice's position, so there are more strong pushes per unit proper time at Bob's position, balancing out the fewer weak pushes at Alice's position.

Each of these effects goes as the square root of the time component of the metric, so that Bob is weaker by the ratio of his $r^2$ to Alice's value of $r^2$.

In the Minkowski point of view, in order to keep the meterstick accelerating, you have to dump a certain amount of momentum per unit proper time into the stick. But you don't need to push as hard at the top to do this, both because the proper time is longer there so you don't need to put as much energy per second (because your second counts for more), and because the energy you put in gets blueshifted when it gets to the center of mass, so that it is giving you more oomph per kick. The effect is as the square of the time-dilation factor, from two cooperating effects.

• I don't REALLY understand this. But how's the other question, the one that I added – kartsa Feb 5 '12 at 4:17
• You aren't shortened by going down a gravity well, if your body can withstand the field stresses (if you aren't mushed by the gravity). But you are weakened. You asked if Bob is weak or Bob is short. Bob is weak, and Bob is not short. That's the answer. I don't know why you keep talking about shrinking, there is no shrinking, only weakening. – Ron Maimon Feb 5 '12 at 11:45
• Well OK Bob is weak. The Weak Bob cranks a vertical drive shaft, Alice at the upper end measures a tiny torque. Now Bob aims a beam of polarized photons at the end of the drive shaft. Still Alice measures smaller torque than Bob. What has happened to the spin of photons in the gravity well? What happens to the spin of photons that climb up from the gravity well? – kartsa Feb 6 '12 at 7:24