Cosmological Redshift Question

[Drakkith]

Let's say that an event releases a 1 second burst of EM radiation that propagates outward from the source in a spherical wavefront. If the EM wave travels long enough so that expansion doubles its wavelength, is that 1 second burst received as a 2 second burst? Has the total energy of the EM wave as a whole been reduced by redshift, or has it simply been spread out over time/space?

I assume that expansion has not only increased the wavelength, but it has also increased the diameter of the wavefront as well, correct? (Diameter meaning distance between two points of the wave traveling in opposite directions)

 

[Simon Bridge]

That's how I've seen it.
Each individual photon in the pulse increases in wavelength and the the distribution of photons gets more diffuse, by the same mechanism.

 

[timmdeeg]

Let's say that an event releases a 1 second burst of EM radiation that propagates outward from the source in a spherical wavefront. If the EM wave travels long enough so that expansion doubles its wavelength, is that 1 second burst received as a 2 second burst?

Yes, this effect is called 'cosmological time dilation', see here. Any clock in cosmological distance sending light pulses ticks slower, depending on the relative increase of the scale-factor between emission and absorption.

Has the total energy of the EM wave as a whole been reduced by redshift, or has it simply been spread out over time/space?

Energy isn't conserved in an expanding universe. Please search threads which deal with that.

I assume that expansion has not only increased the wavelength, but it has also increased the diameter of the wavefront as well, correct? (Diameter meaning distance between two points of the wave traveling in opposite directions)

Expansion means increasing distances.

 

[Chalnoth]

Let's say that an event releases a 1 second burst of EM radiation that propagates outward from the source in a spherical wavefront. If the EM wave travels long enough so that expansion doubles its wavelength, is that 1 second burst received as a 2 second burst?

Timmdeeg already answered this, but I thought I'd expand a little bit by explaining why this has to be the case.

The wavelength of this 1-second burst has doubled, so its frequency has been cut in half. Imagine that is a very long-wavelength bit of radiation, such that there's only one oscillation of the wave emitted over that initial second.

Since the frequency has been cut in half, it now takes two seconds for a detector to see that entire oscillation of the wave.

Has the total energy of the EM wave as a whole been reduced by redshift, or has it simply been spread out over time/space?

The total energy is reduced. The total energy of the wavefront is hf * number of photons. As the wavefront expands, it's still the same number of photons, but the frequency f is reduced (in this case, cut in half). So the total energy is halved.

I assume that expansion has not only increased the wavelength, but it has also increased the diameter of the wavefront as well, correct? (Diameter meaning distance between two points of the wave traveling in opposite directions)

Yep. See back to the explanation of why the time dilation occurs. The same thing happens here.

 

[Drakkith]

The total energy is reduced. The total energy of the wavefront is hf * number of photons. As the wavefront expands, it's still the same number of photons, but the frequency f is reduced (in this case, cut in half). So the total energy is halved.

Does classical EM theory also say that the total energy of the wavefront would be cut in half?

 

[Tanelorn]

Interesting that the energy is cut in half. Where would we say it went?

 

[Naty1]

Doppler and redshift can be thought of as the same pgenomena.

Chronos recently posted the following which is a slick way to think about photon energy:

...photons are lengthened precisely by the amount of expansion. That is, the wavelength is simply multiplied by the expansion factor; (z+1) is the factor which multiplies the redshift. You can use the stress-energy tensor for a photon gas to show the energy density drops as 1/a⁴. Since the number density of photons falls as 1/a³, it follows that the energy of each individual photon falls as 1/a.

The observed redshift will be equal to the total ‘expansion’ between the emission and absorption of the photon, regardless of what the rate of that expansion was at different times.

https://www.physicsforums.com/showthread.php?t=617506
Derivation of the formula for cosmological redshift

Interesting that the energy is cut ... Where would we say it went?

Is the Universe leaking energy:...http://astronomy.case.edu/heather/151/davis.pdf
Taara Davis...from another discussion in these forums...

In the end, therefore, there is no mystery to the energy loss of photons: the energies are being measured by galaxies that are receding from each other, and the drop in energy is just a matter of perspective and relative motion... the universe does not violate the conservation of energy; rather it lies outside that laws jurisdiction.”

 

[Naty1]

Let's say that an event releases a 1 second burst of EM...

You might as well just ask about CMBR...isn't it an analogous situation...

What I had trouble understanding for a while and never really confirmed via a discussion was the perspective illustrated by Tamara Davis:

...the drop in energy is just a matter of perspective and relative motion.

This can be reconciled relative to CMBR observations of the big bang or other source:... when the big bang started emitting CMBR photons, local temperatures were almost 3,000K...now it seems CMBR temperatures are under 3K...Did the photons shift frequency...not really...once emitted photons don't do that...space did of course expand.

This can be explained, I guess, by the fact the CMBR sphere is moving away really,really, fast now...shifting those high frequency energetic photons towards the weak infrared we now observe...otherwise we'd be fried. [In fact we could get fried if we took a high speed rocket ship off in any direction so as to cause blueshift. Seems like the energy is 'still there' !]

 

[Tanelorn]

"Is the Universe leaking energy:...http://astronomy.case.edu/heather/151/davis.pdf
Taara Davis...from another discussion in these forums...

In the end, therefore, there is no mystery to the energy loss of photons: the energies are being measured by galaxies that are receding from each other, and the drop in energy is just a matter of perspective and relative motion... the universe does not violate the conservation of energy; rather it lies outside that laws jurisdiction.” It is not in some how reappearing in the form of D.E?
I recall Light pressure was considered as a drive for a spacecraft but the contribution must be tiny..

 

[Drakkith]

I guess what I'm not getting is how energy loss is happening if the frequency is cut in half but the time it takes for the wavefront to pass is doubled. Of course, I know nothing of EM theory or the math.

To use a simpler example, if a 550 nm laser is pointed at an observer moving away from the source such that the measured wavelength is 1100 nm, has the energy delivered by the laser per second fallen to half, or to one quarter the amount a stationary observer would measure?

Also, I'd like to keep this strictly classical if possible.

 

[Naty1]

I'm not getting is how energy loss is happening...

Chalnoth first addressed the energy of the total wavefront:

The total energy of the wavefront is hf * number of photons. As the wavefront expands, it's still the same number of photons, but the frequency f is reduced (in this case, cut in half). So the total energy is halved.

If you are asking questions about a LASER, and assume there is no dispersion of the signal, that is no increase in area as the wavefront moves, the answer happens to be the same...
there is no 'loss of photons' at the detector.

However, as a typical wavefront expands, as in your original post, the number of photons per unit area [and of any detector] decreases ...as the square of the distance...plus your wavelength is doubled... further loss.

illustration here:
http://en.wikipedia.org/wiki/Inverse-square_law

 

[Drakkith]

I get that Naty. I guess what I'm asking is how the classical version looks. How does a classical EM wave work when it comes to transporting energy and being redshifted?

 

[Chronos]

The universe does not 'leak' energy. At worst, it is all eventually recovered as gravitational potential energy. Unfortunately, we have no convenient way to account for gravitational potential energy.

 

[jartsa]

Does classical EM theory also say that the total energy of the wavefront would be cut in half?

Here is Maxwell’s expression for the momentum of an electromagnetic wave:

p=E/c

I got it from here: http://www.adamauton.com/warp/emc2.html


Let's consider a collision between light that has momentum p and velocity c, and a large black static object. The object is heated by energy E, the energy of the wave.


Then we consider collision between light that has momentum p, velocity c, and a small black object with momentum -p and velocity -c. The object is heated by energy of the wave plus the energy of the object, because the object stops. We hope that from symmetry it follows that the kinetic energy of the object equals the energy of the light, so the heating energy should be twice the energy of the light.


Now we consider the frequency of the light, as observed by the object moving at velocity c. It's twice the frequency observed by a static object. (This is the same thing as when moving towards a sound source at speed of sound, the sound frequency is doubled)

So we have: when frequency is doubled by the Doppler shift, then the energy is also doubled, which is the same as what relativity says.


(We are being classical here, so it's not an error to assume an object with velocity c)

 

[Drakkith]

(We are being classical here, so it's not an error to assume an object with velocity c)

Relativity is a classical theory, and it itself says objects cannot travel at c, so you've lost me.

 

[jartsa]

Relativity is a classical theory, and it itself says objects cannot travel at c, so you've lost me.

I always try to avoid calculating anything. But the idea was this:

If light has energy E, we know its momentum. (p=E/c)

This light collides with some object which has some momentum, some velocity, and some kinetic energy.

The heat energy released in this inelastic collision is the same energy that the object observes the light having, because the object thinks all of the collision energy comes from the light.

Now who will calculate the heat energy relased in said collision?

 

[timmdeeg]

The universe does not 'leak' energy. At worst, it is all eventually recovered as gravitational potential energy. Unfortunately, we have no convenient way to account for gravitational potential energy.

Yes, in the case the universe expands forever.

Whereas in the Big Crunch case, then being increasingly blueshifted, the once redshifted photons would "pay back" their 'lost' energy, as λ goes with a.

People living in the Big Crunch epoch would ask the same question, however vice versa.

 

[timmdeeg]

I guess what I'm not getting is how energy loss is happening if the frequency is cut in half but the time it takes for the wavefront to pass is doubled.

Let's check what happens, if the universe doubles it's size.

Then during this period distances including the wavelength of photons are doubled. As the energy of photons is inverse to their wavelength, double wavelength means half energy. This is consistent with the earlier statement in this thread, E = h*f, because the photons's frequency f is inverse to it's wavelength, half frequency means double wavelength.

 

[Lino]

Drakkith, I hope that you don't mind me interjecting, but I am having difficulty following the discussion, because I don't see the problem ...

... How does a classical EM wave work when it comes to transporting energy and being redshifted?

Is it not simply a case that the wavefront looses energy for two reasons (i) ignoring redshift / universal expansion, in proportion to the enlargement of the surface area of the sphere, and (ii) considering only the universal expansion, in proportion to the redshift of the individual photons? Is there another factor that I am not considering?

(Apologies for the simplicity of the question.)

Regards,

Noel.

[Apologies, I realize the error in point (i) where the energy is spread over a greater area but not actually loosing energy.]

 

[Chalnoth]

Does classical EM theory also say that the total energy of the wavefront would be cut in half?

Only combined with General Relativity. Classical EM in flat space-time always conserves energy. It's the curvature of space-time that makes it so that energy is no longer conserved.

Or, if you want to get really pedantic:
http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

 

[jartsa]

I always try to avoid calculating anything. But the idea was this:

If light has energy E, we know its momentum. (p=E/c)

This light collides with some object which has some momentum, some velocity, and some kinetic energy.

The heat energy released in this inelastic collision is the same energy that the object observes the light having, because the object thinks all of the collision energy comes from the light.

Now who will calculate the heat energy relased in said collision?

So, here's my calculation:Let's say an observer has momentum p and velocity 0.1 c.

And a light beam has momentum -p and velocity -c.

The light beam and the observer collide, for both colliders the duration of collision is the same, and the force of the collision is constant, and the magnitude of the force is the same for both colliders.

Work done is F*s. Force times distance.

The object's s is one tenth of the light's s.

Correcton, it's 1/20 because object's average speed during the collision is 0.05 c.

No, that's wrong, we must assume light slows down from -c to 0. So the average speed of the light during the collision is 0.5 c.

So the object's s is one tenth of the light's s.

So collision energy is increased by 10% at velocity 0.1 c .(That was a calculation of change of energy of light due to observer motion, calculated using one law probably invented by Maxwell, p=E/c. I remember it was asked what Maxwell's laws say about the change of energy when the observer moves. )

 

[Chalnoth]

So, here's my calculation:


Let's say an observer has momentum p and velocity 0.1 c.

And a light beam has momentum -p and velocity -c.

The light beam and the observer collide, for both colliders the duration of collision is the same, and the force of the collision is constant, and the magnitude of the force is the same for both colliders.

Work done is F*s. Force times distance.

The object's s is one tenth of the light's s.

Correcton, it's 1/20 because object's average speed during the collision is 0.05 c.

No, that's wrong, we must assume light slows down from -c to 0. So the average speed of the light during the collision is 0.5 c.

So the object's s is one tenth of the light's s.

So collision energy is increased by 10% at velocity 0.1 c .


(That was a calculation of change of energy of light due to observer motion, calculated using one law probably invented by Maxwell, p=E/c. I remember it was asked what Maxwell's laws say about the change of energy when the observer moves. )

The photon doesn't "slow down". It either passes through, bounces off, or gets absorbed.

In practice this means that for any macroscopic object, the photon will have so much energy that it will either pass straight through the object (potentially sending a few electrons flying), or it will shatter an atomic nucleus.

But let's imagine that you built a sort of bucket to capture the energy of the light ray. This would involve making something that makes it very likely that the photon will shatter an atomic nucleus, and will also capture all of the particles that result. This would mean an initial nucleus being shattered, those particles going off to shatter a bunch more nuclei, with the energy of each piece reducing every time. Eventually, after the energy of the particles of the next reaction gets low enough, those particles won't be able to shatter more nuclei, and will come to a stop within the material of the bucket.

What you would feel, from outside the bucket, is each individual particle coming to a stop. That means that in the end, it would "feel" like your bucket, instead of being hit by a single photon, got hit by a cloud of subatomic particles with the same momentum.