[thedoc (OP)]

Magnus Carlsson  asked the Naked Scientists:
   
Hello Dr Chris.

Firstly thanks for excellent programmes. Have been following them for the last 2-3 years via podcast (thanks technology) and find them both informative and amusing.

I have understood that longer wavelength photons have lower energy than shorter. Photons get longer by expansion of universe. Two questions:

What is the process? What is interactiong with the photons so that they get longer wavelength? Must there be some density in universe that is "pulling"?

If the radiation energy is reduced, where is it going considering energy conservation?

Thanks,
Magnus Carlsson,
Sweden

What do you think?

[Phractality]

The principle of conservation of energy applies only within a given inertial reference frame, though that part of the principle is usually omitted in literature. A photon has different amounts of energy in different reference frames. The redshift is associated with a change of inertial reference frames. When the origin of the reference frame is accelerating, it is not an inertial reference frame. The photon's energy in the original inertial reference frame is constant, but in an accelerating reference frame, it is changing.

Cosmologists prefer "comoving coordinates", which confuse the issue of whether the reference frame is inertial or accelerating. The expansion of space causes distant galaxies to accelerate away from one another, but in comoving coordinates, both galaxies are stationary, even though the distance between them is growing at an accelerating rate.

[Pmb]

In a static gravitional field a photon doesn't loose energy. I showed all this in a web page I created on the subject. Let me know if you can follow it

http://home.comcast.net/~peter.m.brown/gr/grav_red_shift.htm

[Geezer]

In a static gravitional field a photon doesn't loose energy. I showed all this in a web page I created on the subject. Let me know if you can follow it

Pete,

We'd appreciate it a lot if you actually participated in the discussion rather than simply posting links to your site. It's OK to provide a reference in support of your position, but we'd prefer that you not make an assertion then post a link.

That said, I was under the misguided impression that energy cannot be destroyed and that the frequency of photons determines their energy. If both of those statements are true, the energy is no longer in the photon, but it is somewhere else.

[Phractality]

That said, I was under the misguided impression that energy cannot be destroyed and that the frequency of photons determines their energy. If both of those statements are true, the energy is no longer in the photon, but it is somewhere else.

As I said earlier, energy is conserved in any inertial reference frame. When a photon is emitted in one reference frame and absorbed in a different reference frame the energy absorbed isn't necessarily the same as the energy emitted.

The same is true of kinetic energy where a mass moves from one reference frame to another. For example, a bullet fired at an approaching target will impact with greater energy than one fired at a receding target. In the reference frame of the gun, the bullet has the same energy in both cases, and the same amount of energy is imparted to the target in the gun's reference frame. It is only when you calculate the energy in the target's reference frame that you get a different amount of energy. There are transformation formulas for calculating the energy of a system in different reference frames.

The frequency and wavelength of the photon do not change in a non-accelerating and non-expanding reference frame. However, cosmologists do not use that sort of reference frame where great distances and times are involved. Instead, they use comoving coordinates, in which two comoving galaxies are considered to have zero relative velocity, even if they are moving away from one another the distance between them is increasing at nearly the speed of light.
Energy is not conserved in comoving coordinates.

[Geezer]

That said, I was under the misguided impression that energy cannot be destroyed and that the frequency of photons determines their energy. If both of those statements are true, the energy is no longer in the photon, but it is somewhere else.

As I said earlier, energy is conserved in any inertial reference frame. When a photon is emitted in one reference frame and absorbed in a different reference frame the energy absorbed isn't necessarily the same as the energy emitted.

The same is true of kinetic energy where a mass moves from one reference frame to another. For example, a bullet fired at an approaching target will impact with greater energy than one fired at a receding target. In the reference frame of the gun, the bullet has the same energy in both cases, and the same amount of energy is imparted to the target in the gun's reference frame. It is only when you calculate the energy in the target's reference frame that you get a different amount of energy. There are transformation formulas for calculating the energy of a system in different reference frames.

The frequency and wavelength of the photon do not change in a non-accelerating and non-expanding reference frame. However, cosmologists do not use that sort of reference frame where great distances and times are involved. Instead, they use comoving coordinates, in which two comoving galaxies are considered to have zero relative velocity, even if they are moving away from one another the distance between them is increasing at nearly the speed of light.
Energy is not conserved in comoving coordinates.

Reference frames are artifacts. While they are useful in helping to explain certain phenomena, they have no basis in physical reality.

Photons can red-shift (and, presumably, they can blue-shift too). Either way, their energy must have changed. Assuming you don't believe the energy was destroyed, where did the energy go?

[Pmb]

Pete,

We'd appreciate it a lot if you actually participated in the discussion rather than simply posting links to your site.

There's always a very good reason for everything I do. In this case its been many many years since I've put thought into the mechanism of gravitational redshift. I created that web page because it rerquires a lot of caution when answering that problem. I created that webpage when I was at my peak of understanding the subject matter. If I posted everything from recall then it wouldn't be as good as the answer I created all those years ago. It's not as if I'm not answering the question. It's simply that my memory is flimsy so I don't do it on the fly. Think of it as me quoting me, just as a supplement to my contribution. I asked of they could follow it so I was really preparing myself for more to come. Don't get the impression that my first post was my only post. There may be members who read it but who don't have the mathematical skills to understand it.

The purpose of those links are to provide a detailed derivation/explantion of the physics. I wrote them up as a way to give the best response that I can. That isn't the end behind frequency asked questions. I often use them as supplements to my posts for thos who enjoy the math and rigor.

I wrote each page at a time when my understanding of the relevant physics was at its peak. That was many years ago. I forget details over the years due to a memory problem that I'm cursed with.

If both of those statements are true, the energy is no longer in the photon, but it is somewhere else.

Yes. The energy is in the gravitational field in the form of gravitational potential energy.

[evan_au]

This discussion overlaps with: "Topic: where does the energy go from a redshift?" http://www.thenakedscientists.com/forum/index.php?topic=44583.new#new

[JP]

A lot of this argument comes down to the idea that if we measure the photon on the surface of the planet and then in space, it's energy has changed.  I'm not sure how valid that is.  Let's simplify things by looking at a Newtonian analogue:

I stand at the top of the flight of stairs an you stand at the bottom.  You throw a ball up to me and I catch it.  You measure the kinetic energy it has when it leaves your hand, and I measure the kinetic energy when it arrives at my hand.  I measure significantly less kinetic energy than you do.  Did the ball lose energy?  It certainly lost kinetic energy, but it gained potential energy. 
-
Now if I move up into orbit and you stay on the earth's surface and we replace the ball with a photon, did the photon lose energy? 

The big difference between these cases is that Newton would argue that both observers are in the same inertial reference frame, so energy conservation holds between their measurements and potential energy accounts for the difference.  (Potential energy being the integral of gravitational force over distance moved by the ball).

I don't know enough GR to get into the details, but from what I do know, Einstein says the two observers are in different reference frames and therefore they need not measure (locally) the same total energy.  Since gravity here isn't a force in the sense of Newton, you can't simply define gravitational potential energy to account for the discrepancy: you have to do a more detailed analysis to show how the reference frame change makes their measurements not agree.  Once you properly account for the reference frames, energy should be conserved.  Would that be correct?

And of course from a fixed external reference frame watching the entire process, the total energy has to be conserved.

[Phractality]

Reference frames are artifacts. While they are useful in helping to explain certain phenomena, they have no basis in physical reality.

If reference frames are artifacts, then so is energy. Energy is defined and quantified in terms of a reference frame. Change the reference frame and you change the quantity of energy. Of course, there is something real which we quantify as "energy"; but the quantity of that something is undefined except in a reference frame. It makes no sense to say that something is conserved unless you can quantify it and say that the quantity remains constant.

[Pmb]

Now if I move up into orbit and you stay on the earth's surface and we replace the ball with a photon, did the photon lose energy? 

Yes. The photon lost kinetic energy. The kinetic energy of a photon is K = E = hf so since h is a constant and there was a reduction in the amount of kinetic energy then the frequency will have dropped, i.e. gravitational redshift.

I don't know enough GR to get into the details, but from what I do know, Einstein says the two observers are in different reference frames and therefore they need not measure (locally) the same total energy.  Since gravity here isn't a force in the sense of Newton, ..

Sure it is. Where’d you get the idea that it wasn’t? In GR there are two kinds of forces. Ones represented by a 4-force and those which are classified as inertial forces. Gravity is an inertial force.

… you can't simply define gravitational potential energy to account for the discrepancy: you have to do a more detailed analysis to show how the reference frame change makes their measurements not agree.  Once you properly account for the reference frames, energy should be conserved.  Would that be correct?

I don’t understand.

Did you read the web page I wrote up on this subject? If not then I recommend looking it over. It might help you understand it better.

[Pmb]

Reference frames are artifacts. While they are useful in helping to explain certain phenomena, they have no basis in physical reality.

I don't understand what you mean by artifacts. Can you clarify for me?

A coordinate system is merely a way to quantitatively state what we experience in real  life. E.g. "The ceiling of this large room is tall and the couch is near the center of the room facing west and to the side wall of the room." That's very vauge. When we want precision then we use numbers. Then we can say things like "The ceiling is 15 feet above the floor. The width of the room is 20 feet and its length is 30 feet. The edge of the couch is 14 feet from the rear wall and is nest to the wall. The other end of he couch is seven feet from the wall."  Thats more precise. Whe we get to the precision what is required for experiment then we have established a system of measurements, i.e. a coordinate system. You can't paint a coordinate system green. You can't weigh it. You can't push it and you can't kick it. But it exists in that it is the description of something very real. I can't pick up "huge" but I know that Mt. Everest is huge.

Assuming you don't believe the energy was destroyed, where did the energy go?

You're speaking of energy as if it had a mechanical existance. Energy is merely a book keeping system where the sum of individual types of quantities associated with different things is quantized, the sum being a constasnt of motion. Potential energy is energy by virtue of position in a field of force. It went to potential energy, i.e. to energy by virtue of its position in the gravitational field.

[JP]

Now if I move up into orbit and you stay on the earth's surface and we replace the ball with a photon, did the photon lose energy? 

Yes. The photon lost kinetic energy. The kinetic energy of a photon is K = E = hf so since h is a constant and there was a reduction in the amount of kinetic energy then the frequency will have dropped, i.e. gravitational redshift.

I don't know enough GR to get into the details, but from what I do know, Einstein says the two observers are in different reference frames and therefore they need not measure (locally) the same total energy.  Since gravity here isn't a force in the sense of Newton, ..

Sure it is. Where’d you get the idea that it wasn’t? In GR there are two kinds of forces. Ones represented by a 4-force and those which are classified as inertial forces. Gravity is an inertial force.

… you can't simply define gravitational potential energy to account for the discrepancy: you have to do a more detailed analysis to show how the reference frame change makes their measurements not agree.  Once you properly account for the reference frames, energy should be conserved.  Would that be correct?

I don’t understand.

Did you read the web page I wrote up on this subject? If not then I recommend looking it over. It might help you understand it better.

Yes, I was essentially paraphrasing what I'd gained from other sources and from your site.  I think we essentially agree.  Yes, gravity is an inertial force.  I'm still stuck in the mindset of forces as non-inertial (because I haven't had to do any GR work since grad. school). 

My last question is about how you account for the discrepancy in measured kinetic energy?  Can it be naturally accounted for by a non-inertial potential energy term or do you have to do a reference frame change?

[Geezer]

Reference frames are artifacts. While they are useful in helping to explain certain phenomena, they have no basis in physical reality.

If reference frames are artifacts, then so is energy. Energy is defined and quantified in terms of a reference frame. Change the reference frame and you change the quantity of energy. Of course, there is something real which we quantify as "energy"; but the quantity of that something is undefined except in a reference frame. It makes no sense to say that something is conserved unless you can quantify it and say that the quantity remains constant.

You may be missing my point. If I use a particular frame of reference, I can convince myself that ficticious forces are real, even though they have no basis in physics. I think that means they are artifacts.

[pinballed]

Interesting discussion, but I guess I cannot really follow....sorry for asking again....

As I understand it a Photon is a quantifiable unit. This unit has different energy levels (by it's own) and this is manifested by wavelength.
If we have a Doppler effect, with us seeing a red shift (sign of lower energy), there should really not be lower energy, since the longer wavelength
is only created by the speed of the object that emits the photon - away from us. This should in this case mean that the photon has the same energy, but we
see it different depending on our frame of reference. Are we travelling toward the object (and photon) we would even see a blue shift I guess? So in this case
the observers motion relative the source of the photon decides if it is red or blue shift, but in "absolute terms" the photon has the same energy
as when it was emitted?

Is it the same when the photon get stretched by an expanding universe? If so, the photon should have the same energy, and we should have the same
effects as above?

If the photon looses "absolute" energy, where does that go?

Second - or first question - why does universe drag the photon "apart"? What is interacting with the photon so that it cannot keep the wavelength?
Other particles does not seem to get spread out due to universe expansion?

Or, is this just an accelerated Doppler effect?

Thanks,
Magnus Carlsson

[Geezer]

Let's try this:

First, get some sharks and some "lasers" ..........

OK - seriously:

We set up a battery powered laser on Earth. We know how much energy the laser is consuming, and we can quantify the amount of energy in the photons produced by the laser.

We now put the laser on something rather fast, and point the beam back at us so we can observe it.

I think we all agree that the frequency of the light observed from the laser when it is traveling away from us will be less. Even though it is still consuming energy from the battery at the same rate, it is beaming energy back to us at a reduced rate because of its motion. In other words, energy density pumped into spacetime by the laser has to be reduced from our perspective. If the energy transmission speed was reduced, the frequency would not have to change, but because the speed cannot change, the frequency must.

Does that explain it, or is it a red herring?

[bizerl]

I'm not one for formulas and like Magnus, have found the discussion difficult to follow, but in Geezer's example, it seems to me that we would be able to measure the same amount of energy as if it were stationary, it would just take more time to do it.

If the laser was pulsed for say, 10 seconds (according to a timer attached to the laser), and we measure the energy from the photons, we get a reference. Then we shoot it into space as per the example, wouldn't we measure the same amount of energy if we measured the entire pulse - which would take longer than 10 seconds? Similarly, if it were blue shifted towards us, it would presumably be a higher energy state but the pulse wouldn't last as long?

I hope you can follow my point.

Luke.

[pinballed]

Hi,

I have the same understanding as bizerl. If I am a photon, I will have the same energy independent the observers speed?
I have a certain energy and I cannot know if i meet  an observer that is closing into me (blueshift) or one leaving me (redshift).
The energy is there.....and if not; then it is even more interesting

Second question; Disregarding the redshift (or blue) seen due to Doppler, will the wavelength change due to the expansion
of universe during the travel pf the photon? I have heard that it is said that the radiation energy in universe is reduced due to
this fact, but if it is;

1. What "stretches out" the photon? It must then in my view interact with something?
2. If energy is lost, where does it go? Energy conservation?

Thanks,
Magnus

[CZARCAR]

imagine the moving light source having 2 lights shining in opposite directions= 1 goes red & the other goes blue. Turn off the blue light & then what?

[yor_on]

What is acting on the photons are the way the 'frames of reference' are acting between each other Magnus. If you think of it as photon-balls coming in a even stream from your garden hose, you and me standing still, one ball at a second. Then you take hold of it (the garden hose) and start running away from me, me still watching those 'photons' streaming out of it, will they have a larger distance between them as perceived by me? Between each 'photon' that comes out you will have moved, let's say a meter, away from me. If we now assume that 'each photon' instead represented a decided position of part of the 'wave', would that wave then become 'stretched out' for me.

And, would you, running with it, notice any difference?

So it goes back to 'relative motion', not to the energy intrinsic to those photons/waves. That doesn't state that the energy you measure doesn't change, but that's just the same as you see on earth, someone running towards, or from, you while throwing a ball at you.

When it comes to the cosmological redshift the idea is that 'space' somehow 'expands', and then if you think of that as a 'distance' growing in all points, your wave has to be elongated, and down shifted to red. but it is still a theory, and a very tricky one to me