Re: Quiz
- From: Dono <sa_ge@xxxxxxxxxxx>
- Date: Sun, 30 Dec 2007 10:08:33 -0800 (PST)
On Dec 30, 9:33 am, "rot...@xxxxxxxxx" <rot...@xxxxxxxxx> wrote:
Note the words used: "looks" and "measured". As pointed out in the
replies, it still "looks" round due to the Terrell effects although it
will me "measured" to be L.contracted.
But there can be another catch to the problem. To 'see' (photograph)
implies that a visual frequency is arriving from the ball to your
eyes. Where/how is this visual signal coming from?
For example, if the ball is a source and emits a proper visual
frequency, then you will actually not 'see' anything at relativistic
speeds (depending on the angle of travel of the ball). If the ball is
not a source ( as a typical ball), then the light comes from the
exterior ( a light source or x-cosmic rays ?) These are reflected
from the ball to your eyes. Some of those frequencies will become
visible ones and some original visible one will no longer be visible.
Due to the range of sensitivity of the eye and the proportion of the
frequencies of the exterior sources, and the angle of travel of the
ball, you can see the ball no longer spherical (and of different
colors) and may even not 'see' it coming ... !
Let's assume v=.9c, the Doppler shift on approach will be:
fo=fe*sqrt ((1+v/c)/(1-v/c))=fe*sqrt(19) i.e. there is an upshift in
frequencies of about 4.2x
Since the visible spectrum is 400-790THz, a 4.2x shift will render to
object totally invisible, indeed.
.
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