
#1
Sep911, 03:31 PM

P: 20

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 nonmainstream journal removed.[/edit] 



#2
Sep911, 04:06 PM

Physics
Sci Advisor
PF Gold
P: 5,507

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. 



#3
Sep911, 04:17 PM

Sci Advisor
PF Gold
P: 4,862

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.




#4
Sep911, 04:36 PM

Physics
Sci Advisor
PF Gold
P: 5,507

Question about the rotating diskAlso, 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 welldefined 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 



#5
Sep911, 04:52 PM

Sci Advisor
PF Gold
P: 4,862





#6
Sep911, 05:36 PM

Mentor
P: 16,477





#7
Sep911, 06:10 PM

Physics
Sci Advisor
PF Gold
P: 5,507





#8
Sep911, 06:12 PM

Physics
Sci Advisor
PF Gold
P: 5,507





#9
Sep3011, 01:29 PM

P: 20

PeterDonis,
do you really think there is no unique spatial metric? Doesn't the article here [edit]Link to nonmainstream journal removed.[/edit] clear that up, so that the only consistent coordinate values are those for the inertial observer. The noninertial, diskriding observers must use their measured acceleration to calculate the proper, nonparadoxical coordinates. 



#10
Oct211, 04:01 PM

P: 3,178

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 peerreviewed papers here. 



#11
Oct211, 05:13 PM

P: 20

harrylin,
thank you so much for the profound advice...and don't tell me what I can or cannot discuss. 


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