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originally posted by: JUhrman
Everything you post in this thread shows the same level of understand of physics. That of a child.
originally posted by: Bedlam
Meaning, I understand it and don't have to visualize it to do so. Math: it is your friend.
It is a small wooden ball, typically painted yellow.
The reason you have to have a "what is it like" is because you can only deal with concepts verbally. That's not easy when it's something you will have difficulty visualizing or describing in high-school English. As you have had with light from day 1.
Light is like light. It's like other EM radiation.
Wrong-o. When an electron makes an orbital transition that emits a photon, there is no 'vibrating source'.
Map/territory. This is not a pipe.
originally posted by: Bedlam
originally posted by: ImaFungi
a reply to: Arbitrageur
How does a photon lose energy?
Colorfully.
originally posted by: anonentity
originally posted by: Bedlam
originally posted by: ImaFungi
a reply to: Arbitrageur
How does a photon lose energy?
Colorfully.
Where does it loose its energy? if its not in the observers linear time frame. Then its the observers perception of it, which has changed. Perhaps its our perception of light that needs a makeover, and the maths, have been massaged to fit the perception. If it looses energy, then the wavelength lengthens, to the red . But that all depends on whether the source is moving to us or away from us . If the source, was moving toward us , we would perceive it as moving to the violet, because the observed wavelength would be shorter. But its only the perceived wavelength that we assume in our makeup of reality.
Energy is not always conserved in general relativity, between the reference frames of different observers.
originally posted by: anonentity
Where does it loose its energy?
I don't know what you mean by "changed". Relativity says that observations differ depending on the observer's reference frame. Our reference frame is on earth. If a photon came from a distant star with an earth-like planet around it, observers on that planet would have observed the photons having higher energy because they wouldn't see cosmological redshift effects from such a close distance.
if its not in the observers linear time frame. Then its the observers perception of it, which has changed.
Relativity is widely accepted because the math does fit most observations pretty well. However lacking a theory of quantum gravity we can only surmise that either relativity or quantum mechanics or both are at best incomplete.
Perhaps its our perception of light that needs a makeover, and the maths, have been massaged to fit the perception.
originally posted by: ImaFungi
Everything that is non nothing, can be compared to geometry. That is what math is. Geometry is the visualization of math (points, lines, arcs, waves, forms etc.)
I colloquially use the term light to refer to all EM radiation, like is acceptably done in physics conversations between people who know this is acceptably done, but now you will know that from now on.
Marking over time and space, the point of an electron prior to it making an orbital transition, through to it finishing its orbital transformation, is what I referred to as a vibration.
originally posted by: ImaFungi
originally posted by: Bedlam
Not even slightly. However, half of light is a magnetic field.
Em radiation is a magnetic field.
But ok, going along with your statement, what is the other half?
originally posted by: anonentity
originally posted by: Bedlam
originally posted by: ImaFungi
a reply to: Arbitrageur
How does a photon lose energy?
Colorfully.
Where does it loose its energy?
Take a monochromatic light source, measure the frequency (thus the energy) of the light, from a detector stationary with respect to the light. Move the detector toward the light source and the light appears blue-shifted (higher frequency thus higher energy) and move the detector away and the light appears red-shifted (thus lower energy).
originally posted by: ImaFungi
Would the same energy signature photon be measured differently because one would hit the detector as the detector was moving away from the photon
And one would hit the detector as the detector was moving toward the photon?
originally posted by: Arbitrageur
Take a monochromatic light source, measure the frequency (thus the energy) of the light, from a detector stationary with respect to the light. Move the detector toward the light source and the light appears blue-shifted (higher frequency thus higher energy) and move the detector away and the light appears red-shifted (thus lower energy).
originally posted by: ImaFungi
Would the same energy signature photon be measured differently because one would hit the detector as the detector was moving away from the photon
And one would hit the detector as the detector was moving toward the photon?
The redshifts of galaxies include both a component related to recessional velocity from expansion of the Universe, and a component related to peculiar motion (Doppler shift). The redshift due to expansion of the Universe depends upon the recessional velocity in a fashion determined by the cosmological model chosen to describe the expansion of the Universe, which is very different from how Doppler redshift depends upon local velocity. Describing the cosmological expansion origin of redshift, cosmologist Edward Robert Harrison said, "Light leaves a galaxy, which is stationary in its local region of space, and is eventually received by observers who are stationary in their own local region of space. Between the galaxy and the observer, light travels through vast regions of expanding space. As a result, all wavelengths of the light are stretched by the expansion of space. It is as simple as that..."