Question about the rotating disk


by dude222
Tags: disk, general, relativity, rotating
dude222
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#1
Sep9-11, 03:31 PM
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Hey, does anyone know where I can find the correct metric for a rotating disk? Is the article at this link correct? If not, where is the error?

[edit]Link to non-mainstream journal removed.[/edit]
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PeterDonis
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Sep9-11, 04:06 PM
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It would help to first decide what is meant by "the correct metric for a rotating disk". I'm not sure the paper you linked to is entirely clear about this. I would recommend checking out this Wikipedia page for a start:

http://en.wikipedia.org/wiki/Born_coordinates

I haven't tried to follow the paper's logic in detail, but on a quick skim, it looks like he's made assumptions that are only valid at a single point (for example, he sets [itex]t_{0} = 0[/itex]), and then acted as though the equations he derives are general, applying to the entire spacetime. However, even if the equations he derives are correct, I don't see how they justify his conclusion that time dilation and length contraction are not present. His equations for dy and dt both contain [itex]\sqrt{1 - v^{2}}[/itex] in the denominator, which is exactly the time dilation/length contraction factor.
PAllen
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Sep9-11, 04:17 PM
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Agreeing with Peter, the phrase "the correct metric" is meaningless. You have to say "a correct metric". Obviously, if the disk has minimal mass, one correct metric is the inertial one flat metric. Presumably you mean a metric in coordinates where the disk appears stationary. However, there are obviously an infinite number of such choices, even many very reasonable ones. In some sense, they are all the same metric: once you have properly defined your coordinates, you can simply transform the flat Minkowski metric.

PeterDonis
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Sep9-11, 04:36 PM
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Question about the rotating disk


Quote Quote by PAllen View Post
Agreeing with Peter, the phrase "the correct metric" is meaningless. You have to say "a correct metric". Obviously, if the disk has minimal mass, one correct metric is the inertial one flat metric.
Just to be clear, the fact that the disk is rotating in itself rules out a flat metric for the disk as viewed by observers rotating with it. A flat metric would only apply for non-rotating observers (who don't see the disk as stationary).

Also, as the Wikipedia page I linked to notes (this is also discussed in the Usenet Physics FAQ page I link to below), since observers at different points on the disk can't agree on a notion of simultaneity, there is in fact *no* single frame that can be called "the" frame of the disk. That means that whatever "metric" you adopt will violate some common intuitions about how metrics "ought" to behave. For example, not only is the "spatial geometry" of the disk not flat as seen by the rotating observers, it's not even clear that that term has any well-defined meaning at all for the rotating observers (because infinitesimal pieces of spatial hyperslices for rotating observers at various locations can't be "fit together" into a single spatial hyperslice for the whole disk).

Another good page to read, from the Usenet Physics FAQ:

http://math.ucr.edu/home/baez/physic...igid_disk.html
PAllen
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Sep9-11, 04:52 PM
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Quote Quote by PeterDonis View Post
Just to be clear, the fact that the disk is rotating in itself rules out a flat metric for the disk as viewed by observers rotating with it. A flat metric would only apply for non-rotating observers (who don't see the disk as stationary).
Just to be clear, as well, the metric characterizes the geometry of the manifold. Viewed independent of coordinates, the metric for a vanishing mass rotating disk is the flat space-time metric. Analyzing the disk in inertial coordinates is also clearly the easiest way to compute any possible measurement. Pick any coordinates in which the disk is stationary, you have a different expression of the metric. However the curvature tensor still must vanish.
DaleSpam
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Sep9-11, 05:36 PM
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Quote Quote by PeterDonis View Post
Just to be clear, the fact that the disk is rotating in itself rules out a flat metric for the disk as viewed by observers rotating with it. A flat metric would only apply for non-rotating observers (who don't see the disk as stationary).
Careful here. If the spacetime is flat for inertial observers then is is flat for rotating observers. What you are probably refering to is the spatial metric for some definition of simultaneity appropriate to the rotating observers.
PeterDonis
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Sep9-11, 06:10 PM
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Quote Quote by PAllen View Post
Just to be clear, as well, the metric characterizes the geometry of the manifold. Viewed independent of coordinates, the metric for a vanishing mass rotating disk is the flat space-time metric. Analyzing the disk in inertial coordinates is also clearly the easiest way to compute any possible measurement. Pick any coordinates in which the disk is stationary, you have a different expression of the metric. However the curvature tensor still must vanish.
I agree; the manifold as a whole doesn't change for observers in different states of motion.
PeterDonis
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Sep9-11, 06:12 PM
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Quote Quote by DaleSpam View Post
Careful here. If the spacetime is flat for inertial observers then is is flat for rotating observers. What you are probably refering to is the spatial metric for some definition of simultaneity appropriate to the rotating observers.
Yes, I should have said a flat *spatial* metric applies for non-rotating observers. For rotating observers, there isn't really a well-defined spatial metric at all (or at least, not one that uses a definition of simultaneity matching the natural one for any of the rotating observers). The spacetime as a whole is flat for all observers.
dude222
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#9
Sep30-11, 01:29 PM
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PeterDonis,
do you really think there is no unique spatial metric? Doesn't the article here

[edit]Link to non-mainstream journal removed.[/edit]

clear that up, so that the only consistent coordinate values are those for the inertial observer. The non-inertial, disk-riding observers must use their measured acceleration to calculate the proper, non-paradoxical coordinates.
harrylin
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#10
Oct2-11, 04:01 PM
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Quote Quote by dude222 View Post
PeterDonis,
do you really think there is no unique spatial metric? [..] so that the only consistent coordinate values are those for the inertial observer. [..]
First of all, he said that there isn't really a well-defined spatial metric at all for rotating observers, or at least, not one that uses a definition of simultaneity matching the natural one for any of the rotating observers. Indeed, one has to use an inertial reference system to avoid a complicated mess.

And of course (not sure if that is what you mean), the description will be the simplest with the origin fixed at the centre, because of the rotation of the disk around that centre. However, the coordinate values of any inertial reference system are consistent.

Note: I think that we should only discuss peer-reviewed papers here.
dude222
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#11
Oct2-11, 05:13 PM
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harrylin,

thank you so much for the profound advice...and don't tell me what I can or cannot discuss.
Doc Al
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#12
Oct2-11, 05:33 PM
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Quote Quote by dude222 View Post
...and don't tell me what I can or cannot discuss.
Allow me then. Our posting rules, which are linked at the top of every page, prohibit linking to non-mainstream (crackpot) sites.

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