- #1
quincy harman
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If the speed of light is constant in any inertial frame then how do we measure a red shift or blue shift or why?
If the speed of light is constant in any inertial frame
- Thread starter quincy harman
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- #2
quincy harman
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Sorry I didn't think the question through before I posted. I just remembered the wavelength has nothing to do with the speed.
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Imagine you have a light source at rest between two detectors which starts emitting light waves.
It would look something like this
Each light wave spreads out in a circle at c, with each wave front hitting each detector at a rate equal to the emission frequency.
Now consider what happens if the the light source is moving towards one detector and away for the other as seen from the frame of the detectors.
The light waves still spread out in a circle at c, but the origin point of each wave front is closer to one detector than it is from the other and the center of each circular wave front is displaced from the previous one. The wave fronts hitting the blue dot are closer together and the one hitting the red dot are further apart. since they are still traveling at c with respect to the detectors, the Blue dot see a higher frequency of light and the red dot a lower frequency.
As far as the light source is concerned, the waves still are moving outward from it at c in circles, but the red dot is running away from the wave fronts traveling in its direction and the blue dot is rushing towards the wave fronts in its direction. So according to the Light source frame we get the same result, wave fronts hit the blue dot at a faster rate than they hit the red dot.
So no matter which frame you pick, you expect the detectors to register Doppler shifts.
It would look something like this
Each light wave spreads out in a circle at c, with each wave front hitting each detector at a rate equal to the emission frequency.
Now consider what happens if the the light source is moving towards one detector and away for the other as seen from the frame of the detectors.
The light waves still spread out in a circle at c, but the origin point of each wave front is closer to one detector than it is from the other and the center of each circular wave front is displaced from the previous one. The wave fronts hitting the blue dot are closer together and the one hitting the red dot are further apart. since they are still traveling at c with respect to the detectors, the Blue dot see a higher frequency of light and the red dot a lower frequency.
As far as the light source is concerned, the waves still are moving outward from it at c in circles, but the red dot is running away from the wave fronts traveling in its direction and the blue dot is rushing towards the wave fronts in its direction. So according to the Light source frame we get the same result, wave fronts hit the blue dot at a faster rate than they hit the red dot.
So no matter which frame you pick, you expect the detectors to register Doppler shifts.
1. The Big Idea:
According to Einstein’s relativity, all motion is relative. You can’t tell if you’re moving at a constant velocity without looking outside. But what if there is a universal “rest frame” (like the old idea of the “ether”)? This experiment tries to find out by looking for tiny, directional differences in how objects move inside a sealed box.
2. How It Works: The Two-Stage Process
Imagine a perfectly isolated spacecraft (our lab) moving through space at some unknown speed V...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles.
The Relativator was sold by (as printed) Atomic Laboratories, Inc. 3086 Claremont Ave, Berkeley 5, California , which seems to be a division of Cenco Instruments (Central Scientific Company)...
Source: https://www.physicsforums.com/insights/relativator-circular-slide-rule-simulated-with-desmos/
by @robphy
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this:
$$
\partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}.
$$
The integrability conditions for the existence of a global solution ##F_{lj}## is:
$$
R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0
$$
Then from the equation:
$$\nabla_b e_a= \Gamma^c_{ab} e_c$$
Using cartesian basis ## e_I...
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