Cosmic Redshift & The Redshifted Photon

[vinven7]

Hello all,

I have a question on cosmic redshift and I am really hoping that one of you can solve this for me, as you have many times in the past.

Consider a quasar that is emitting a steady stream of X-ray photons many billions of years into the Universe's past. When the X-rays finally reach the Earth now, they appear to be visible light radiation due to cosmic redshift.
If we were to consider just one photon that started out at the quasar and reached the earth, we clearly find that the frequency of this photon is vastly reduced by the redshift. However, Since the energy of the photon is proportional to the frequency, the photon has apparently lost energy due to cosmic redshift - without interacting with anything at all on the way. How can we explain the change in energy of the photon by cosmic redshift alone?

Thanks!

 

[Drakkith]

You just explained it. The EM wave is stretched out by the time it reaches us, which causes redshift and a loss of energy. Similar to normal Doppler shift, the wave doesn't need to interact with anything in order for redshift or blueshift to occur.

 

[vinven7]

Hi - my question was not what causes the change in frequency or wavelength. And I did specify that we would consider this as a photon and not a wave. To rephrase my question, considering light as a stream of photons (i.e., not a wave) how do we account for the change in frequency due to cosmic redshift?

 

[Bill_K]

the photon has apparently lost energy due to cosmic redshift - without interacting with anything at all on the way. How can we explain the change in energy of the photon by cosmic redshift alone?

I think what you're asking is, "Since the photon lost energy, and energy must be conserved, where did it go?" Well I hope that's what you're asking, because that's a good question! It's often stated in terms of the Cosmic Microwave Background. All those photons from the CMB started out energetic but by the time we receive them they have much less energy. Where did it go?

There are three answers...

1) For cosmological solutions in GR, total energy is not conserved. (!) Energy is the conserved quantity that results from time translation symmetry. It's conserved in gravitational fields that are time-independent, such as for particle trajectories in the field of a black hole. But a cosmological solution varies with time, and such a field has no time-translation invariance, and hence no conserved quantity.

2) Total energy is conserved, even when the gravitational field is time-varying. What happens is that the photon's energy is turned into gravitational potential energy, and when that is included you get a conserved quantity. Two troubles with this. One, the local value of the conserved quantity is not unique, only its global integral. Second, the globally integrated energy always comes out zero.

3) But, some still argue, it's not trivially zero, it's zero only by virtue of the Einstein field equations being obeyed.

The issue is discussed in various places. For example:
http://motls.blogspot.com/2010/08/why-and-how-energy-is-not-conserved-in.html
http://blogs.discovermagazine.com/cosmicvariance/2010/02/22/energy-is-not-conserved/#.UzS3kIWi398
http://arxiv.org/abs/gr-qc/9701028

 

[vinven7]

Thank you so much! This was exactly my question and the very answer I was looking for.

 

[Chronos]

A photon that is red shifted is also time dilated. I suspect its energy is merely delayed, not 'lost'.

 

[vinven7]

A photon that is red shifted is also time dilated. I suspect its energy is merely delayed, not 'lost'.

Could you please explain this ? If you are talking about the Lorenz time dilation of the photon from the reference frame of the earth, the dilation is infinite. If "energy is being delayed" - what does that mean?

 

[Nugatory]

Could you please explain this ? If you are talking about the Lorenz time dilation of the photon from the reference frame of the earth, the dilation is infinite. If "energy is being delayed" - what does that mean?

Red shifting a light wave means decreasing the frequency and increasing the wavelength, both classically and relativistically. The longer wavelength means that the time that passes on Earth between the arrival of one crest on Earth and the next is longer the time that passed at the source between the departure of the two crests. You can choose to interpret this as the amount of energy emitted in given time at the source being received over a longer time at the destination.

First footnote:

If you are talking about the Lorenz time dilation of the photon from the reference frame of the earth, the dilation is infinite.

That is not right; see the FAQ in the relativity section at https://www.physicsforums.com/showthread.php?t=511170

Second footnote:

To rephrase my question, considering light as a stream of photons (i.e., not a wave)

Don't consider the light that way . Seriously, kidding aside, thinking about photons as little particles that stream away from the light source is very misleading. There are many discussions in the quantum mechanics forum about the relationship between photons and light as electromagnetic waves - I'll see if I can dig up a pointer to one of the helpful threads.
[Edit: Here's one https://www.physicsforums.com/showthread.php?t=741790]

 

[Drakkith]

Don't consider the light that way . Seriously, kidding aside, thinking about photons as little particles that stream away from the light source is very misleading.

Yep. Which is why I chose to use EM wave instead of photon in my response.

 

[HeLiXe]

Vinven7 another useful way to think about a photon is a quanta of energy..like a wave packet

 

[Drakkith]

Vinven7 another useful way to think about a photon is a quanta of energy..like a wave packet

I don't believe a wave packet is related to energy quanta (photons) except that it determines their location.

 

[HeLiXe]

Can you explain that Drakkith? Not following you.

 

[Ken G]

Let's be clear that the energy the photon will be observed as having certainly does drop, so if one wishes to incorporate wave-packet treatments that include time dilation, then one will still find less energy once the wave packet has been fully absorbed. In other words, the time dilation effect must be included if you are looking at the time dependence of the photon-detector interaction, but you will still get that the photon energy depends on the redshift, the reasons for which having been explained by Bill_K.

If it bothers you that the energy of the photon is not staying the same, notice that this even happens for bullets, so depends neither on waves nor on quantum mechanics. If you shoot a gun straight up, the bullet has less energy at high altitude. In a Newtonian description, that's because you've generated gravitational potential energy, but in GR, there's no such potential energy, there's just a warping of spacetime, such that if you impose a Euclidean type of space-time coordinatization, then energy is not conserved in those coordinates. In other words, if I am allowed to choose any coordinates I like, I can always make the kinetic energy of a particle moving through empty space appear to drop. The easiest way to do that is to shoot a bullet in the reference frame of a stationary gun, but observe its energy in the frame of an observer moving away from the gun.

Nontechnically, we can say that the rules of conservation of energy are generally intended to apply within a given locally inertial reference frame, they don't let you use any coordinates you like or change reference frames any way you like, which is pretty much what is happening in the cosmological example.

 

[herbertweidner]

@Drakkith: Take the spin-flip in a hydrogen atom, which emits the energy 5.87433 μeV. Does the atom emit a photon or a wave packet? There are no energy-detectors for such a low energy. But Astronomers receive EM-waves at 1420 MHz. I think they related! Where and how is the wave packet generated? Near the hydrogen atom or near the receiving dipole?

 

[Ken G]

Where and how the wave packet is generated depends on what you think a quantum mechanical wave function is. But the added effects of lots of those wave packets becomes a measurable electromagnetic field, and that is pretty uncontroversial to say is generated at the source, given that it could in principle be measured there.