Subject:
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Re: Relativity Question
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Newsgroups:
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lugnet.off-topic.geek
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Date:
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Wed, 8 Nov 2000 21:32:13 GMT
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Viewed:
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2404 times
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In lugnet.off-topic.geek, Dave Schuler writes:
> In lugnet.off-topic.geek, Jay Jacinto writes:
>
> > For example: an observer accelerating away from a stationary observer would
> > experience the sensation of acceleration. Conversely, the stationary • observer
> > would experience a gravitational field generated by the Universe accelerating
> > from the other observer. So neither could determine which is in the
> > accelerated inertia frame. Thus, all motion including acceleration is
> > relative.
>
> Well, thanks for throwing a monkey wrench into it for me. How do the
> famous de-synchronized clocks fit into this? Obviously one of them slowed
> down relative to the other. Do we infer that the ground-based clock
> decelerated sufficiently to speed up time's local passage? I thought I had
> a handle on this, but now I see I'm missing something.
Acceleration is bad ju-ju, since it isn't accounted for in special relativity.
It's only in General Relativity that we get gravity thrown into the mix, and
that's taught in grad school. However, from a strictly special relativistic
POV the observer that is actually moving measures time slower than a stationary
observer as determined by the Lorentz transform. In this example there are
only two inertia frames, the rest of the universe is ignored.
> > Also the doppler shift measured by each observer will be the same, all
> > acceleration does is add an additional term as determined by the Lorentz
> > transformation.
I think I goofed here concerning the additional term from acceleration. The
frequency of light emitted is determined only by *instantaneously* velocity and
the wavelength of light.
> Hmm. This is the part that always messes me up. So if there is a distant
> star whose light we see as red-shifted, and X is standing on Earth, while Y
> takes off at .9c toward the distant star, will both X and Y perceive the
> same red-shift?
>
> Dave!
Yep, if we consider at least one of the objects to be at "rest." The actual
factor that determines the doppler shift is the following:
sqrt((1 + ß)/(1 - ß))
which is the classical doppler shift with some simplication.
ß = v/c
v = velocity
I feel like I'm digging a very deep hole for myself with all this.
Thanks
Jay Jacinto
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Message is in Reply To:
 | | Re: Relativity Question
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| (...) Well, thanks for throwing a monkey wrench into it for me. How do the famous de-synchronized clocks fit into this? Obviously one of them slowed down relative to the other. Do we infer that the ground-based clock decelerated sufficiently to (...) (26 years ago, 8-Nov-00, to lugnet.off-topic.geek)
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