# Disk rotating in a gravitational field, like a clock

1. Nov 20, 2014

### Karl Coryat

Suppose we have a large circular disk with pegs around the perimeter, and we set the disk rotating so that one pegs passes us every second. Then, we move this disk to a gravitational field, such that one side of the disk is closer to the gravitational mass than the other.

From our perspective at the far side of the disc, we continue to observe one peg passing per second. Meanwhile, one assumes, an observer placed at the near end (closer to the gravitational mass) similarly observes one peg passing per second on his side. My questions:

1. Does the far observer, looking toward the side of the disk closer to the mass, see one peg per second passing the other observer? (Seems like he couldn't, since he would observe a clock at that location running slow.)

2. In order to accommodate that, does the far observer notice some aspect of the disc changing (spatial or otherwise)? In other words, what must happen in the observation of the disc to prevent the "faster" pegs from appearing to catch up with the "slower" ones"?

I am trying to imagine how this scenario would appear to the far observer as the disc rotates. Thank you.

2. Nov 20, 2014

### Simon Bridge

The object is only a disk in flat space. In the gravitational field it is no longer and exact circle, nor is it entirely rigid.
You want to do the math for a rotating disk in the Swarzchilde metric.

3. Nov 21, 2014

### A.T.

Why would you assume that?

4. Nov 21, 2014

### Jonathan Scott

A second at one observer is not the same as a second at the other because of gravitational time dilation of the observer's clocks. Both observers find that the number of pegs passing per second is the same both for the near and far side of the disk (as otherwise pegs would not be conserved) but the exact rate would depend on the gravitational potential (and hence time dilation) of the observer.

5. Nov 21, 2014

### Karl Coryat

>A second at one observer is not the same as a second at the other because of gravitational time dilation of the observer's clocks.

That is precisely why I'm asking this question. It seems to me that the disk would appear spatially distorted from the far observer's perspective (and also from other perspectives) -- that is the only way, I figure, that that the disk could still rotate. My question was how this distortion would appear. In the time since I asked the question, I've imagined that perhaps for the "far" observer, the other side of the disk (the side closer to the mass) would appear spread out, with the pegs "stretched apart," but the angular velocity the same. For the "near" observer meanwhile, the far end would appear pinched together, so that more than one peg per second is seen to pass the distant observer. Is that correct? I'm only looking for a qualitative description.

>Why would you assume that [an observer placed closer to the gravitational mass similarly observes one peg passing per second on his side]?

Because locally, the passing of the pegs and the observer's own clock would be synchronized. I continue to assume that.

6. Nov 21, 2014

### A.T.

You assume that you can keep the passing of the pegs synchronized with local clocks on both ends? Maybe that is the problem?

7. Nov 21, 2014

### Karl Coryat

Perhaps it is.

So we set up the rotating disk in flat spacetime, with an observer at each end counting one peg per second. Then we place the observers and disk in the field. Are you saying that the observers, particularly the one closer to the mass, would no longer count one peg per second at his location? My assumption was that the passing pegs would serve as a valid clock, and that this clock would slow down the same as any other local clock, thus preserving the (locally measured) one-peg-per-second rate at any point around the disk. Is that not the case?

8. Nov 21, 2014

### A.T.

A valid clock for the local time at the circumference? Why would that be? The wheel is obviously not a local object, and the number of pegs is globally constrained.

9. Nov 21, 2014

### pervect

Staff Emeritus
If the disk were rigid, the velocity of the top of the disk would be equal and opposite to the velocity of the bottom of the disk. Without having a mathematically detailed answer, which would probably get very messy, probably messy enough to obscure the answer, one might ask the following questions:

1) Do rigid objects exist in special relativity?
(This was supposed to be semi-rhetorical, I'm not sure if it comes out that way!)

2) If we assume an ideal of total non-rigidity for ease of calculation - a marble (or a series of them) going round and round a circular track, with no forces between the marbles, only the forces exerted by the track upon the marbles - what does the conservation of "energy" say about the velocity of the marbles at the top of the track and at the bottom of the track?

10. Nov 21, 2014

### Karl Coryat

A.T.: I take your response to mean that when we take the disk into the field, the pegs appear to speed up -- and the closer we are to the mass, the faster the pegs appear to go, measured by a local clock.

This would make sense, since if an observer at the bottom sends a click for each passing peg to the observer at the top, the gravitational redshifting of that signal would then synchronize with the passing of pegs at the top (far end). Do I have that right?

And, to answer my original question, an observer at any point along the circumference would measure the same peg rate both locally and through a telescope aimed at any other point -- although since he is in the field, this rate would always be faster than the one-peg-per-second measured in flat spacetime. Do I have that right as well?

11. Nov 21, 2014

### A.T.

Yes, I think so.

12. Nov 21, 2014

### Karl Coryat

This is why I love Physics Forums! Thank you all.

13. Nov 21, 2014

### Jonathan Scott

Mostly right apart from "faster". Time dilation depends on the potential, not the field, and is relative. If the middle of the wheel still appears to be rotating at the same angular rate, then an observer at the high side of the wheel would have a faster clock and hence measure the pegs to be passing more slowly, and an observer at the low side of the wheel would have a slower clock so would measure the pegs to be passing faster.

14. Nov 21, 2014

### Karl Coryat

Jonathan -- Thank you for that important clarification.