Amplitude of wave stretched as well as red shift?

[SteveinLondon]

The expansion of spacetime stretches and red shifts the wavelength of light. Is the amplitude of the wave stretched as well? So that very distant stars appear brighter, and therefor nearer?

 

[bcrowell]

Electromagnetic waves are not transverse vibrations of a medium. There is no actual sideways motion. It's not like photographically enlarging an image of a water wave, where the amplitude grows by the same scale factor as the wavelength.

Although this question can be answered purely classically, I think it's easier to analyze if you think in terms of photons. The photon's frequency is reduced by a factor z, so E=hf is also reduced by that factor. The volume occupied by the wave is increased by a factor of z^3, so the energy density is reduced by a factor of z^4. Since the energy density is reduced by z^4, the electric and magnetic fields are also reduced, by factors of z^2.

[EDIT] Marcus pointed out that z should be replaced with 1+z everywhere above in order to be consistent with standard notation.

 

[bcrowell]

SteveinLondon, you've posted this in two different places: here and https://www.physicsforums.com/showthread.php?t=484705 Please don't do that. It's inconsiderate.

 

[Naty1]

So that very distant stars appear brighter, and therefor nearer?

Are you suggesting a distant star is "brighter" than a nearer star?? Is that what you think you observe? no.

Also, cosmic redshift does not occur over relatively small galactic distances but rather much larger interstellar distances...in the smaller regions gravity keeps everything pretty much in the same relative position...and nearby galaxies can be moving towards one another as well as away from each other...this could result in some redshift or blue shift...

 

[Naty1]

Crowell:

Electromagnetic waves are not transverse vibrations of a medium. There is no actual sideways motion.

No medium, I get that..what does the second part mean..."actual sideways motion" ??
The field oscillates that way, right...?

Wiki has a (correct) concise explanation and associated diagram here:

http://en.wikipedia.org/wiki/Electromagnetic_wave

Electromagnetic radiation (often abbreviated E-M radiation or EMR) is a form of energy exhibiting wave-like behavior as it travels through space. EMR has both electric and magnetic field components, which oscillate in phase perpendicular to each other and perpendicular to the direction of energy propagation.

 

[marcus]

...

Although this question can be answered purely classically, I think it's easier to analyze if you think in terms of photons. The photon's frequency is reduced by a factor z, so E=hf is also reduced by that factor. The volume occupied by the wave is increased by a factor of z^3, so the energy density is reduced by a factor of z^4. Since the energy density is reduced by z^4, the electric and magnetic fields are also reduced, by factors of z^2.

This is a clear complete response to Steve, so there's really nothing to add, but it illustrates an ambiguity in the English language---how we talk about fractional increase and increase "by a factor".

Or, in this case the reverse: "decrease by a factor".

The energy density of the photons is actually decreased by a factor of (1+z)⁴
in the sense that that is what you divide the old density by to get the new density.

Beginners can get confused by this ambiguity in English.

For example if the redshift of some light is z = 0.5, then the energy of each photon has been reduced by a factor of 1.5 (you divide by that to get the present energy).
And the volume has been increased by a factor of 1.5³
so the number of photons per unit volume is reduced by 1.5³ (they are spread out in a larger volume).

So the whole effect on the energy density is to reduce it by a factor of 1.5⁴ = a bit over 5.
So you divide the old energy density by a factor of 5 or so.

This does not have any simple intuitive relation to 0.5⁴ = 1/16 so a beginner might get confused by saying "reduced by a factor of z⁴ = 1/16

 

[bcrowell]

The energy density of the photons is actually decreased by a factor of (1+z)⁴

Hi, marcus -- Thanks for the correction. I just didn't realize I wasn't using the standard definition of z.

No medium, I get that..what does the second part mean..."actual sideways motion" ??
The field oscillates that way, right...?

Wiki has a (correct) concise explanation and associated diagram here:

http://en.wikipedia.org/wiki/Electromagnetic_wave

That type of diagram is misleading if you don't interpret it correctly. The transverse stuff is the field vectors, which don't have units of meters.

 

[Bill_K]

The energy density of the photons is actually decreased by a factor of (1+z)4

Sorry, but the original question was not about the energy density, was it. It says 'does the amplitude change'? The amplitude of an electromagnetic wave is just |E|. The issue is simply a matter of how the Lorentz transformation acts on E and B. So I say it should be just one factor of 1+z.

 

[bcrowell]

Sorry, but the original question was not about the energy density, was it. It says 'does the amplitude change'?

Please take another look at the final sentence of #2.

The amplitude of an electromagnetic wave is just |E|. The issue is simply a matter of how the Lorentz transformation acts on E and B. So I say it should be just one factor of 1+z.

It's not a Lorentz transformation. This isn't a special-relativistic kinematic Doppler shift. There are no global Lorentz transformations in GR.

 

[Phrak]

so the number of photons per unit volume is reduced by 1.5³ (they are spread out in a larger volume).

I'm sure you meant 1.5² here for the inverse square law, didn't you?

So the whole effect on the energy density is to reduce it by a factor of 1.5⁴ = a bit over 5.

 

[Naty1]

"I'm sure you meant 1.5² here for the inverse square law, didn't you?

That's for the unit area...

Marcus was addressing the unit volume...

V = 4/3(pi)r³

 

[bcrowell]

"I'm sure you meant 1.5² here for the inverse square law, didn't you?

That's for the unit area...

Marcus was addressing the unit volume...

V = 4/3(pi)r³

There's no sphere involved. For example, a photon from a laser would experience the same increase in its volume, by a factor of (1+z)^3. The increase in volume is because of cosmological expansion, not expansion of a spherical wavefront.