When light waves are stretched and "red-shifted", is the amplitude of the light wave stretched as well, affecting the intensity/brightness of the light wave?

  • $\begingroup$ If you think about it in terms of energy arriving at the detector per unit time (power) you should be able to deduce the answer with confidence. $\endgroup$ – dmckee --- ex-moderator kitten Jul 28 '16 at 22:48
  • $\begingroup$ You need to think in terms of photons because a wave cannot account for this. Red-shift lowers the energy but the intensity relies only on the number of photons. $\endgroup$ – Bill Alsept Jul 29 '16 at 0:19
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    $\begingroup$ Bill's wrong here. Waves can account for it just fine, though it does takes rather more bookkeeping. $\endgroup$ – dmckee --- ex-moderator kitten Jul 29 '16 at 1:53
  • $\begingroup$ The photo electric effect showed it was photons or quanta. You can even count the impacts. The more there are the more intense. One blue photon may have more energy than a red photon but they have the same intensity. $\endgroup$ – Bill Alsept Jul 29 '16 at 2:39
  • $\begingroup$ @BillAlsept Yes, but your original comment said "a wave cannot account for this". That is incorrect. $\endgroup$ – garyp Jul 29 '16 at 10:58

The amplitude you may be thinking of is some sort of wave plot that shows the amplitude as the amplitude vector of the wave in one axis, with the wave advancing perpendicularly. That is just a visualization of a wave. For electromagnetism (EM, eg, light), that is just the amplitude of the electric field, or magnetic field, changing as the wave advances.

But it is simply the amplitude of the electric or magnetic field. It's square is power spectral density. It does not have dimensions of distance or time, but energy per unit area per unit time per unit freq. It does not stretch because it is not a distance. Nor a time. It's related to energy. In the classical wave description from classical EM, it is the intensity of the light, and related to a particle view as the number of photons. It does not change. Number of photons times the freq (or average it out right for multiple frequencies) is the intensity also.

What is easy to see as changing is the wavelength if space is stretching. Or freq. if an observer has velocity wrt to the emitter. Redshift can happen due to relative velocities, or space stretching (or actually a gravitational field). Either way you have to calculate it right, the two (or 3) cases are slightly different. But in either case both wavelength and freq change, such that c stays the same.

As for energy, the easiest way to calculate it, is from E = h X frequency, once you calculate the freq shift correctly. In gravity you can calculate the energy change as the change in the zeroth components of the energy momentum vector of a test zero mass particle going through its lightlike geodesic.

  • $\begingroup$ I know the question was badly worded, and worse, the question in essence seems to equate amplitude to some spatial $\endgroup$ – Bob Bee Jul 29 '16 at 18:27

The speed of light is constant in all reference frames. This is a principle of relativity derived from Maxwell's Equations.

The energy of a photon is given by $E = hf$ where $h$ is Planck's constant and $f$ is the frequency at which the photon propagates.

Now picture this: Kinetic energy is the energy of motion. When you throw a ball with a certain force, you give it a certain amount of energy. If you're running forward and you throw the ball with the same force, it has more kinetic energy. If you run backwards and throw the ball with the same force, it has less kinetic energy.

Now, for a photon, kinetic energy doesn't really apply, since it's massless (of course, it has energy associated with its 4-momentum, but that's more complicated than necessary). Instead, when a light ray is emitted, the motion of the emitter imparts it with more or less total energy, depending on its motion. Motion toward the observer imparts more energy, motion away imparts less.

By the above equation, more energy corresponds with higher frequency (blue) and less energy corresponds with lower frequency (red).


Light waves are not necessarily stretched. The formula

(frequency measured by the observer) = (speed of the light relative to the observer)/(wavelength)

suggests that the frequency shift (blueshift or redshift) can be due to the variation of the speed of the light relative to the observer, not to wavelength change. When the initially stationary observer starts moving towards the light source, this is obviously the case:

http://www.einstein-online.info/images/spotlights/doppler/doppler_static.gif (stationary observer)

http://www.einstein-online.info/images/spotlights/doppler/doppler_detector_blue.gif (moving observer)

I believe that the frequency shift is ALWAYS due to the variation of the speed of the light relative to the observer (the wavelength of the traveling light never changes).

  • $\begingroup$ "I believe that the frequency shift is ALWAYS due to the variation of the speed of the light relative to the observer " .....ahem. Cough...it sounds like you're saying the speed of light changes with frame of reference. "the wavelength of the traveling light never changes)." There would be no red shift if this were true. This is just plain wrong. The speed's the same, frequency and wavelength change. $\endgroup$ – R. Rankin Jul 29 '16 at 0:29
  • $\begingroup$ Yes, it is wrong. That's the principle of relativity (backed up by a zillion, more or less, experiments and observations) $\endgroup$ – Bob Bee Jul 29 '16 at 1:38
  • $\begingroup$ "The speed's the same, frequency and wavelength change." No. The motion of the observer cannot change the wavelength of the incoming light (that is, frequency and speed of light relative to the observer change, wavelength remains unchanged). This is obvious: farside.ph.utexas.edu/teaching/315/Waveshtml/node41.html University of Texas: "Thus, the moving observer sees a wave possessing the same wavelength [...] but a different frequency [...] to that seen by the stationary observer. This phenomenon is known as the Doppler effect." $\endgroup$ – Pentcho Valev Jul 29 '16 at 6:23
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    $\begingroup$ @PentchoValev That analysis is for mechanical wave in a medium (sound, water waves, etc). It does not apply to light. Light behaves very differently. The speed is the same for all observers. I suggest you delete this post because the analysis does not apply to the situation posed in the question. $\endgroup$ – garyp Jul 29 '16 at 11:10

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