4 ms·

Antennas do not emit or absorb electrons at all. Radio waves propagate at most at the speed of light.

by fsh 4y ago

eurasiantiger 4y ago
What? I never claimed that. Radio waves are electromagnetic radiation just like visible light. Radio amplifiers pump electrons into the antenna, which transmits them as photons, and the receiving antenna absorbs them, turning them into electrons, the flow of which then becomes the audible radio signal which gets (demodulated and) amplified by the receiver. The photons do not travel any distance and they do so in zero time. This can be mathematically proven. See e.g. https://phys.org/news/2014-05-does-light-experience-time.amp https://phys.org/news/2014-05-does-light-experience-time.amp - the relevant parts: ”The closer you get to light speed, the less time you experience and the shorter a distance you experience. You may recall that these numbers begin to approach zero. According to relativity, mass can never move through the Universe at light speed. Mass will increase to infinity, and the amount of energy required to move it any faster will also be infinite. But for light itself, which is already moving at light speed… You guessed it, the photons reach zero distance and zero time. Photons can take hundreds of thousands of years to travel from the core of the Sun until they reach the surface and fly off into space. And yet, that final journey, that could take it billions of light years across space, was no different from jumping from atom to atom.”
- raattgift 4y ago
  Under your conception, how does RADAR work? Consider bouncing a RADAR signal off two objects at different distances (1 m, 10 m; or the Moon and Venus [1] when the two are very close to each other in the early evening or early morning sky). Do the RADAR reflections return instantly, or at the same time, or at different times, according to the wristwatch of the RADAR operator, or the RADAR apparatus's computer clock? How does that work with your comment, "the photons do not travel any distance and they do so in zero time. This can be mathematically proven."? Can we agree that each of the RADAR and the two RADAR-reflecting objects definitely occupy different points in space? - -- RADAR images (taken at e.g. Arecibo, here on Earth) of Venus, the Moon, and several other solar system bodies <https://www.naic.edu/~pradar/radarpage.html#venus https://www.naic.edu/~pradar/radarpage.html#venus>
  - tines 4y ago
    I believe the gp is correct, the closer you get to the speed of light the slower time passes. If a particle travels at the speed of light, no time passes for it.
    - raattgift 4y ago
      You're part-way there. First I'll explain how it works in standard textbook relativity, then I'll reword your first sentence a bit. p != p' are two points on a 3+1 dimensional Lorentzian manifold (M, g), modelling our spacetime. There is a minimizing geodesic from p to p', the "shortest straight-line path". In a Lorentzian spacetime, geodesics are classified into three groups: spacelike, null, and timelike. A null geodesic describes the freefall of the massless relativistic wave equation, which applies to e.g. photons (as the Maxwell equations using the Lorenz gauge); and a timelike geodesic describes the massive relativistic wave equation, which applies to e.g. neutrinos (as the Dirac equation) or the Higgs boson (as the Klein-Gordon equation). In a particle paradigm, photons move at c freely falling on null geodesics; neutrinos, electrons and so forth move at less than c freely falling (experiencing no accelerating force) on timelike geodesics. If p and p' are timelike-separated, they can be connected by a timelike geodesic (e.g., a path that could be taken by a massive particle), then we can calculate a nonzero spacetime interval S in some coordinate basis, and divide by c, giving us the proper time \tau such that d\tau = dS / c in a patch of flat spacetime (the metric g enters into the definition through general curved spacetime: on path P, \Delta \tau = \int_{P} 1/c \sqrt{g_{\mu\nu} dx^\mu dx^\nu (= \int_{P} d\tau), which is invariant under changes in coordinates, and has the same problem with dividing by c. However, if spacetime points p and p'are lightlike-separated, they cannot be connected by a timelike geodesic, but rather by a null one. The spacetime interval of a lightlike geodesic is defined such that dS = 0 (thus the name "null"). Any path through spacetime can be parametrized by applying arbitrary labels to its points. A geodesic is a species of path. We can label any geodesic from A to B any way we like: Apple Zebra Xray 3.1415 1 86 Xray-again Manitoba Xray Xray-again ... 911. Arbitrary labellings are not widely useful, though, so we will probably want to label every point with a unique value in such a way that there is a total ordering from A to B, and even more ideally so that there is some smooth monotonically non-decreasing function f on the geodesic, mapping all its points to unique values of the parameter. Proper time is just such a useful function on a timelike geodesic. It is not the only such function. On a null geodesic, proper time would give us a labelling 0 0 0 0 0 ... 0, which is obviously not useful, as the 0s aren't unique or ordered. However, we can use any monotonic function, setting out labels at every point along the path from p to p' where we would (if we measured) detect the photon. A good choice for a null geodesic is the affine parameter, as that preserves the photon's tangent vector under parallel transport along the path. There is a unique affine parametrization satisfying the geodesic equation for every geodesic, including every null geodesic. One of the properties of the affine parameter on a null geodesic is that we can define a momentum k^{\mu} = \dot{X}^{\mu}, the dot representing the derivative with respect to the affine parameter -- the momentum k can be calculated at every point on the null geodesic, and in a curved spacetime k will differ from point to point. This difference in momentum is exactly the gravitational redshift. So, it might be better to say: proper time is an unsuitable parameter for a photon in free fall, or for labelling points on its path from point A to point B. Instead we can use affine time (== the affine parameter) for a photon and for its path from point A to point B, and see that there is some definite timelike parameter for a photon on its journey that can [a] show where in spacetime it can be found during that journey and [b] what its momentum is at [a] and thus what its frequency, wavelength or energy is at [a] via the E = h{\nu} = \frac{hc}{\lambda} relation, and how those quantities change along the path in the presence of spacetime curvature. I would not say "no time passes for it", because we can always define such a time. Affine time is one such (useful) definition, but we can use any others we like. Although a proper time can be calculated for such a massless particle, the proper time is 0 everywhere, so not very useful. However that does not strictly speaking mean that there is no passage of time from the point of view of the photon. It just needs to "wear a different sort of wristwatch" than an electron might choose to. However there is an affine parameterization for an electron moving between points I and J, so electrons (and neutrinos and other non-massless waves, particles, or objects) can always "wear the same sort of wristwatch" as a photon. (Some mathematical details in Wald's graduate-level textbook _General Relativity_ §3.3 and see ch. 3, problem 5).
- scatters 4y ago
  Electrons are not transformed into photons; that would violate conservation of lepton number. Rather, an electron is accelerated by the antenna and as it accelerates it emits a photon. The electron still exists with the antenna afterward, with a lower energy.
akomtu 4y ago
I've read somewhere that photons, while travelling, casually turn into electron-positron pairs and back to photons. It's quite possible that both are just distortions in the magnetic field.

Antennas do not emit or absorb electrons at all. Radio waves propagate at most at the speed of light.

by fsh 4y ago

Antennas do not emit or absorb electrons at all. Radio waves propagate at most at the speed of light.

What? I never claimed that. Radio waves are electromagnetic radiation just like visible light. Radio amplifiers pump electrons into the antenna, which transmits them as photons, and the receiving antenna absorbs them, turning them into electrons, the flow of which then becomes the audible radio signal which gets (demodulated and) amplified by the receiver. The photons do not travel any distance and they do so in zero time. This can be mathematically proven. See e.g. https://phys.org/news/2014-05-does-light-experience-time.amp https://phys.org/news/2014-05-does-light-experience-time.amp - the relevant parts: ”The closer you get to light speed, the less time you experience and the shorter a distance you experience. You may recall that these numbers begin to approach zero. According to relativity, mass can never move through the Universe at light speed. Mass will increase to infinity, and the amount of energy required to move it any faster will also be infinite. But for light itself, which is already moving at light speed… You guessed it, the photons reach zero distance and zero time. Photons can take hundreds of thousands of years to travel from the core of the Sun until they reach the surface and fly off into space. And yet, that final journey, that could take it billions of light years across space, was no different from jumping from atom to atom.”

raattgift 4y ago
Under your conception, how does RADAR work? Consider bouncing a RADAR signal off two objects at different distances (1 m, 10 m; or the Moon and Venus [1] when the two are very close to each other in the early evening or early morning sky). Do the RADAR reflections return instantly, or at the same time, or at different times, according to the wristwatch of the RADAR operator, or the RADAR apparatus's computer clock? How does that work with your comment, "the photons do not travel any distance and they do so in zero time. This can be mathematically proven."? Can we agree that each of the RADAR and the two RADAR-reflecting objects definitely occupy different points in space? - -- RADAR images (taken at e.g. Arecibo, here on Earth) of Venus, the Moon, and several other solar system bodies <https://www.naic.edu/~pradar/radarpage.html#venus https://www.naic.edu/~pradar/radarpage.html#venus>
- tines 4y ago
  I believe the gp is correct, the closer you get to the speed of light the slower time passes. If a particle travels at the speed of light, no time passes for it.
  - raattgift 4y ago
    You're part-way there. First I'll explain how it works in standard textbook relativity, then I'll reword your first sentence a bit. p != p' are two points on a 3+1 dimensional Lorentzian manifold (M, g), modelling our spacetime. There is a minimizing geodesic from p to p', the "shortest straight-line path". In a Lorentzian spacetime, geodesics are classified into three groups: spacelike, null, and timelike. A null geodesic describes the freefall of the massless relativistic wave equation, which applies to e.g. photons (as the Maxwell equations using the Lorenz gauge); and a timelike geodesic describes the massive relativistic wave equation, which applies to e.g. neutrinos (as the Dirac equation) or the Higgs boson (as the Klein-Gordon equation). In a particle paradigm, photons move at c freely falling on null geodesics; neutrinos, electrons and so forth move at less than c freely falling (experiencing no accelerating force) on timelike geodesics. If p and p' are timelike-separated, they can be connected by a timelike geodesic (e.g., a path that could be taken by a massive particle), then we can calculate a nonzero spacetime interval S in some coordinate basis, and divide by c, giving us the proper time \tau such that d\tau = dS / c in a patch of flat spacetime (the metric g enters into the definition through general curved spacetime: on path P, \Delta \tau = \int_{P} 1/c \sqrt{g_{\mu\nu} dx^\mu dx^\nu (= \int_{P} d\tau), which is invariant under changes in coordinates, and has the same problem with dividing by c. However, if spacetime points p and p'are lightlike-separated, they cannot be connected by a timelike geodesic, but rather by a null one. The spacetime interval of a lightlike geodesic is defined such that dS = 0 (thus the name "null"). Any path through spacetime can be parametrized by applying arbitrary labels to its points. A geodesic is a species of path. We can label any geodesic from A to B any way we like: Apple Zebra Xray 3.1415 1 86 Xray-again Manitoba Xray Xray-again ... 911. Arbitrary labellings are not widely useful, though, so we will probably want to label every point with a unique value in such a way that there is a total ordering from A to B, and even more ideally so that there is some smooth monotonically non-decreasing function f on the geodesic, mapping all its points to unique values of the parameter. Proper time is just such a useful function on a timelike geodesic. It is not the only such function. On a null geodesic, proper time would give us a labelling 0 0 0 0 0 ... 0, which is obviously not useful, as the 0s aren't unique or ordered. However, we can use any monotonic function, setting out labels at every point along the path from p to p' where we would (if we measured) detect the photon. A good choice for a null geodesic is the affine parameter, as that preserves the photon's tangent vector under parallel transport along the path. There is a unique affine parametrization satisfying the geodesic equation for every geodesic, including every null geodesic. One of the properties of the affine parameter on a null geodesic is that we can define a momentum k^{\mu} = \dot{X}^{\mu}, the dot representing the derivative with respect to the affine parameter -- the momentum k can be calculated at every point on the null geodesic, and in a curved spacetime k will differ from point to point. This difference in momentum is exactly the gravitational redshift. So, it might be better to say: proper time is an unsuitable parameter for a photon in free fall, or for labelling points on its path from point A to point B. Instead we can use affine time (== the affine parameter) for a photon and for its path from point A to point B, and see that there is some definite timelike parameter for a photon on its journey that can [a] show where in spacetime it can be found during that journey and [b] what its momentum is at [a] and thus what its frequency, wavelength or energy is at [a] via the E = h{\nu} = \frac{hc}{\lambda} relation, and how those quantities change along the path in the presence of spacetime curvature. I would not say "no time passes for it", because we can always define such a time. Affine time is one such (useful) definition, but we can use any others we like. Although a proper time can be calculated for such a massless particle, the proper time is 0 everywhere, so not very useful. However that does not strictly speaking mean that there is no passage of time from the point of view of the photon. It just needs to "wear a different sort of wristwatch" than an electron might choose to. However there is an affine parameterization for an electron moving between points I and J, so electrons (and neutrinos and other non-massless waves, particles, or objects) can always "wear the same sort of wristwatch" as a photon. (Some mathematical details in Wald's graduate-level textbook _General Relativity_ §3.3 and see ch. 3, problem 5).
scatters 4y ago
Electrons are not transformed into photons; that would violate conservation of lepton number. Rather, an electron is accelerated by the antenna and as it accelerates it emits a photon. The electron still exists with the antenna afterward, with a lower energy.

akomtu 4y ago

I've read somewhere that photons, while travelling, casually turn into electron-positron pairs and back to photons. It's quite possible that both are just distortions in the magnetic field.