tag:blogger.com,1999:blog-2202092988208583550.post5931817491849542973..comments2024-03-04T15:09:00.479-08:00Comments on The Scientific Worldview: General Relativity Theory “confirmed” by cosmogonists once againGlenn Borchardthttp://www.blogger.com/profile/09394474754821945146noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-2202092988208583550.post-84991061206707695582019-08-22T09:10:46.117-07:002019-08-22T09:10:46.117-07:00Bligh:
Reread this from IUT p. 207:
"The re...Bligh:<br /><br />Reread this from IUT p. 207:<br /><br />"The refractive index (n) for water is (300,000,000 m/s)/(225,000,000 m/s)=1.333. The wavelength of light is shortened to 75% of what it is in air. Thus, red light with a wavelength of 650 nm in air has a wavelength of 488 nm in water. As in the NOAA example, the number of cycles per second (frequency), however, remains unchanged."<br /><br />That is why red laser light at 650 nm in air is still red at 488 nm in water. Maybe you are thinking of the supposed 1:1 wavelength/frequency relationship in "vacuum." Because there is no perfectly empty space, that relationship is never perfect either. Because Pound-Rebka assumed a perfect vacuum, they had to use the time-worn "time dilation" for explaining Cosmological Redshift instead of distally increasing aether pressure.Glenn Borchardthttps://www.blogger.com/profile/09394474754821945146noreply@blogger.comtag:blogger.com,1999:blog-2202092988208583550.post-4250936548219266442019-08-21T12:15:06.548-07:002019-08-21T12:15:06.548-07:00It is impossible for light to have its wavelength ...It is impossible for light to have its wavelength altered without the reciprocal change in frequency.<br />I forget the equation. Is it C=lamda/frequency???Blighhttps://www.blogger.com/profile/10160829900151513063noreply@blogger.com