michel123456 Posted June 17, 2011 Share Posted June 17, 2011 This below is from a site where declared non mainstream physics is presented: •Either the universe is expanding, atoms are getting smaller, or both perspectives could be considered valid. Meaning that 2 options are open: 1. the Universe is expanding and atoms (all elementary particles) have a defined dimension :that is mainstream physics. 2. the atoms (all elementary particles) are getting smaller and the Universe has a defined dimension. My question is: is point 2 acceptable through mainstream physics, or do we have elements to debunk point 2? Link to comment Share on other sites More sharing options...
ajb Posted June 17, 2011 Share Posted June 17, 2011 The expansion of the universe is really a large scale phenomena that applies to clusters of galaxies. Locally we do not feel this expansion, the gravitational attraction between individual galaxies in a cluster is more than sufficient to counteract and tendency to get pushed apart. Thus I cannot see 2. as being a viable explanation. Link to comment Share on other sites More sharing options...
swansont Posted June 17, 2011 Share Posted June 17, 2011 There are properties that scale with different powers of r. if atoms were changing size, these properties would be varying with respect to each other. Anyone wanting to propose this needs to present evidence for it, since it should be observable, or present arguments why it isn't observable. Link to comment Share on other sites More sharing options...
michel123456 Posted June 17, 2011 Author Share Posted June 17, 2011 O.K. so the short answer is: No, it is not mainstream physics. Link to comment Share on other sites More sharing options...
michel123456 Posted June 18, 2011 Author Share Posted June 18, 2011 There are properties that scale with different powers of r. Could you give an example? Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted June 18, 2011 Share Posted June 18, 2011 (edited) This below is from a site where declared non mainstream physics is presented: Meaning that 2 options are open: 1. the Universe is expanding and atoms (all elementary particles) have a defined dimension :that is mainstream physics. 2. the atoms (all elementary particles) are getting smaller and the Universe has a defined dimension. My question is: is point 2 acceptable through mainstream physics, or do we have elements to debunk point 2? This is possible, though extra assumptions are required, and Occam's beard starts looking pretty wild and unkempt. His wife would get after him. The speed of light would probably be changing or the rate of time or some combination to match up with the shrinking microscopic vs expanding macroscopic and I think there would be other "adjustments" required as well. That said, I have mused on some (interesting but half baked) steady state theories with a kind of balance of expansion and contraction in extra dimensions, but with no success. Edited June 18, 2011 by J.C.MacSwell Link to comment Share on other sites More sharing options...
swansont Posted June 18, 2011 Share Posted June 18, 2011 Could you give an example? The time it takes for a material to dissolve depends on the surface/volume ratio. If the dimensions if atoms were changing, this would vary. Link to comment Share on other sites More sharing options...
michel123456 Posted June 19, 2011 Author Share Posted June 19, 2011 The surface/volume ratio of a sphere is always the same no matter the r of the sphere. If the scaling of atoms is radial (along the radius), I don't see any reason why r should have an influence in your example. Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted June 19, 2011 Share Posted June 19, 2011 Volume of a sphere: [math]V = \frac{4}{3} \pi r^3[/math] Surface area of a sphere: [math]S = 4\pi r^2[/math] So the ratio of volume to surface area is [imath]\frac{V}{S} = \frac{r}{3}[/imath], which certainly does depend on r. Link to comment Share on other sites More sharing options...
michel123456 Posted June 19, 2011 Author Share Posted June 19, 2011 (edited) Cap'n right. Michel wrong. Sorry. ------------------ But If you multiply r by a scale factor F, the left side must be multiplied by the same F factor. Edited June 19, 2011 by michel123456 Link to comment Share on other sites More sharing options...
michel123456 Posted June 20, 2011 Author Share Posted June 20, 2011 My edit din't reach the audience. Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted June 20, 2011 Share Posted June 20, 2011 Cap'n right. Michel wrong. Sorry. ------------------ But If you multiply r by a scale factor F, the left side must be multiplied by the same F factor. The F factor will be cubed in one case and squared in the other so the ratio does change, correct? Link to comment Share on other sites More sharing options...
michel123456 Posted June 21, 2011 Author Share Posted June 21, 2011 The F factor will be cubed in one case and squared in the other so the ratio does change, correct? I don't understand your point [imath]F\frac{V}{S} =F \frac{r}{3}[/imath] Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted June 21, 2011 Share Posted June 21, 2011 I don't understand your point [imath]F\frac{V}{S} =F \frac{r}{3}[/imath] The new ratio is F (Vo/So) where Vo and So are the original volume and surface areas. The ratio changes by a factor of F. It does not remain the same. Do you agree with that? Link to comment Share on other sites More sharing options...
michel123456 Posted June 23, 2011 Author Share Posted June 23, 2011 yes. Link to comment Share on other sites More sharing options...
michel123456 Posted June 25, 2011 Author Share Posted June 25, 2011 R changes only for an external observator and there is not. Both vitruvian men will measure the same ratio. Isn't it? Link to comment Share on other sites More sharing options...
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