Expansion of space affecting photon's energy

[stuart100]

TL;DR

Light is travelling as a photon, what extent is change in journey distance affecting wavelength
given it's only at one place at a time? Also, a reduction in frequency is a reduction in energy, so how and where is that energy gone?

On a long trip the photon goes, but it occupies a wavelength of space at any particular time. If the space between start and finish (inspection) is expanding all the way all the journey time, then most of the expansion has no effect on the photon. Like eg second tenth is section currently passed thru, but the effect takes place on the other say ninety percent all the total time period. So, did I get that right? Also, the photon lost energy by reducing it's frequency but where went the energy? And, in the vicinity of that proton within a wavelength, does space expand? Last, why does light travel below c in a medium? Is it curving next to atoms because space is not empty?

 

[PeroK]

Summary:: Light is traveling as a photon, what extent is change in journey distance affecting wavelength
given it's only at one place at a time? Also, a reduction in frequency is a reduction in energy, so how and where is that energy gone?

On a long trip the photon goes, but it occupies a wavelength of space at any particular time. If the space between start and finish (inspection) is expanding all the way all the journey time, then most of the expansion has no effect on the photon. Like eg second tenth is section currently passed thru, but the effect takes place on the other say ninety percent all the total time period. So, did I get that right? Also, the photon lost energy by reducing it's frequency but where went the energy? And, in the vicinity of that proton within a wavelength, does space expand? Last, why does light travel below c in a medium? Is it curving next to atoms because space is not empty?

Energy is, and always has been, dependent on the frame of reference. A photon does not have a definitive energy (or a definitive frequency or wavelength). Its energy, frequency and wavelength depend on the frame of reference of the detection device. It's true, of course, that a photon has a specific energy relative to its source - but its energy if measured depends on the relationship between the source and the detector. E.g. if the source and detector are in relative motion, then the energy of the photon may be greater or less than the energy if measured at the source.

The expansion of space also creates a relationship between the source of light and where it is detected. In an expanding universe, this results in a reduction of the measured energy the further from the source a photon is detected. This is called redshift. This can be equated to a "recession" velocity, so in some sense it is not fundamentally different from the redshift associated with simple motion between source and receiver.

In that sense, the energy does not "go anywhere". That said, in an expanding universe, energy is not necessarily conserved. The conservation of energy fundamentally is a consequence of time symmetry - which is lost in an expanding universe.

On your final question, the best answer is to consider light as an electromagnetic (EM) wave. It travels through vacuum at $c$. In a medium, however, if it can travel at all, the propagation of the wave is affecetd by the electromagnetic field within the medium. Ultimately, EM fields must satisfy Maxwell's equations and this results in a reduced speed for propagation of an EM wave through a medium.

 

[phinds]

Energy isn't conserved in General Relativity. Here's a good, detailed look if you're interested:
http://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/

 

[Drakkith]

On a long trip the photon goes, but it occupies a wavelength of space at any particular time.

Not true. The position of a photon cannot be defined until it interacts with something. Until then, it is better thought of as being a large, spread out wave, possibly of more than one wavelength.

Last, why does light travel below c in a medium? Is it curving next to atoms because space is not empty?

If I remember correctly, a very simplified answer is that the incoming EM wave interacts with electric charges in the medium and these charges create another EM wave that interferes with the original in such a way as to slow it down.

 

[stuart100]

Energy is, and always has been, dependent on the frame of reference. A photon does not have a definitive energy (or a definitive frequency or wavelength). Its energy, frequency and wavelength depend on the frame of reference of the detection device. It's true, of course, that a photon has a specific energy relative to its source - but its energy if measured depends on the relationship between the source and the detector. E.g. if the source and detector are in relative motion, then the energy of the photon may be greater or less than the energy if measured at the source.

The expansion of space also creates a relationship between the source of light and where it is detected. In an expanding universe, this results in a reduction of the measured energy the further from the source a photon is detected. This is called redshift. This can be equated to a "recession" velocity, so in some sense it is not fundamentally different from the redshift associated with simple motion between source and receiver.

In that sense, the energy does not "go anywhere". That said, in an expanding universe, energy is not necessarily conserved. The conservation of energy fundamentally is a consequence of time symmetry - which is lost in an expanding universe.

On your final question, the best answer is to consider light as an electromagnetic (EM) wave. It travels through vacuum at $c$. In a medium, however, if it can travel at all, the propagation of the wave is affecetd by the electromagnetic field within the medium. Ultimately, EM fields must satisfy Maxwell's equations and this results in a reduced speed for propagation of an EM wave through a medium.

Thanks for the info, especially; "This can be equated to a 'recession' velocity, so in some sense it is not fundamentally different from the redshift associated with simple motion..." I keep forgetting how fuzzy things get in the micro. So, even though the distance ahead of the photon expands in the journeytime, the whole effect self-determines based on source-detector relationship? On the last point, I suspected your explanation.

 

[stuart100]

Not true. The position of a photon cannot be defined until it interacts with something. Until then, it is better thought of as being a large, spread out wave, possibly of more than one wavelength.

If I remember correctly, a very simplified answer is that the incoming EM wave interacts with electric charges in the medium and these charges create another EM wave that interferes with the original in such a way as to slow it down.

Thanks. I did suspect something like that given the photon is an EM wave. With the universe running on probabilities it's doing very well. I guess space over billions of light years is not a sufficiently dense medium to produce an apparent slowdown.

 

[PeroK]

The whole effect self-determines based on source-detector relationship?

The redshift calculation in an expanding universe is remarkably simple. It boils down to:
$$\frac{\lambda_r}{\lambda_s} = \frac{a(t_s)}{a(t_r)}$$where $\lambda_r, \lambda_s$ are the measured wavelength at the receiver and the wavelength emitted at source, and $a(t_r), a(t_s)$ are the universal scale factor at the times received $t_r$ and emitted $t_s$.

In an expanding universe, $a(t)$ increases over time, so the measured wavelength of light gets longer the further it travels.

PS $t$ is so-called comoving time. And we assume the light is emitted and received by comoving objects. The Earth is almost comoving; as are most distant objects.

 

[mbond]

Is there really a consensus to say that energy is not conserved in GR ?
For example, according to A. Guth, during inflation, the density is constant while the universe is exponentially expanding, and this does not violate energy conservation because negative gravitational energy is produced at the same time. For a flat universe, the total energy is constant and 0.

Considering the photons, may their energy loss due to the expansion be considered as going into the gravitational potential, slowing down the expansion? Indeed the expansion is slower for a radiation-dominated universe than for a matter-dominated one (time power 1/2 versus 2/3 respectively).

 

[PeterDonis]

Is there really a consensus to say that energy is not conserved in GR ?

The short answer is "no". The slightly longer answer is "it's complicated".

@pervect made a good post on this in another recent thread:

https://www.physicsforums.com/threads/is-gravity-a-force.1008409/post-6557489

Both of the references he gives there are worth reading in full.

according to A. Guth, during inflation, the density is constant while the universe is exponentially expanding, and this does not violate energy conservation because negative gravitational energy is produced at the same time.

The problem with this (for those physicists who think it's a problem) is that there is no invariant way to define the "negative gravitational energy" Guth is talking about. (In more technical language that goes beyond a "B" level discussion, but you'll see it if you start digging into more advanced sources, there is no way to define a tensor for "gravitational energy". You can define various pseudo-tensors that capture aspects of "gravitational energy", but all of them depend on particular coordinate choices and different physicists who like the "gravitational energy" concept prefer different ones.)

Considering the photons, may their energy loss due to the expansion be considered as going into the gravitational potential, slowing down the expansion?

Some physicists (of whom Guth might be one) might describe it this way. But again, there is a problem (for those physicists who think it's a problem): the concept of "gravitational potential", if you try to make it rigorous, is only well-defined in a stationary spacetime, and the spacetime of an expanding universe is not stationary. So using the concept of "gravitational potential" this way in an expanding universe is really just hand-waving that falls apart on closer inspection.

the expansion is slower for a radiation-dominated universe than for a matter-dominated one

More precisely, the deceleration is larger for a radiation dominated universe. The reason for this, if you want a reason that will stand up to a more rigorous treatment (unlike the "energy loss" reason I discussed above), is that pressure gravitates, and a radiation dominated universe has pressure equal to 1/3 the energy density, while a matter dominated universe has zero pressure. (More precisely, negligible pressure compared to the energy density.) So, given the same energy density, a radiation dominated universe will "gravitate more" and hence decelerate the expansion more.

 

[archamja]

Energy is, and always has been, dependent on the frame of reference. A photon does not have a definitive energy (or a definitive frequency or wavelength). Its energy, frequency and wavelength depend on the frame of reference of the detection device. It's true, of course, that a photon has a specific energy relative to its source - but its energy if measured depends on the relationship between the source and the detector. E.g. if the source and detector are in relative motion, then the energy of the photon may be greater or less than the energy if measured at the source.

The expansion of space also creates a relationship between the source of light and where it is detected. In an expanding universe, this results in a reduction of the measured energy the further from the source a photon is detected. This is called redshift. This can be equated to a "recession" velocity, so in some sense it is not fundamentally different from the redshift associated with simple motion between source and receiver.

In that sense, the energy does not "go anywhere". That said, in an expanding universe, energy is not necessarily conserved. The conservation of energy fundamentally is a consequence of time symmetry - which is lost in an expanding universe.

On your final question, the best answer is to consider light as an electromagnetic (EM) wave. It travels through vacuum at $c$. In a medium, however, if it can travel at all, the propagation of the wave is affecetd by the electromagnetic field within the medium. Ultimately, EM fields must satisfy Maxwell's equations and this results in a reduced speed for propagation of an EM wave through a medium.

The photon emitted for say a remote hydrogen atom can be calculated, and is the same as one emitted locally. This does not change with the frame of reference, both are local to the event and give the same result. I I measured it at both ends, I will measure a different energy, and there is a loss. How does this happen, and how much it challenges how we define energy in relation to wavelength other than locally?

 

[Ibix]

How does this happen, and how much it challenges how we define energy in relation to wavelength other than locally?

The curvature of space means that there's a changing inner product between the worldlines of local co-moving observers and that of a travelling light pulse, which is not the case in flat spacetime. This is no challenge to how we define energy, since that's frame dependent anyway as already noted, and you can only measure it locally.

 

[vanhees71]

The socalled "photon" is described by it's four-momentum vector $q^{\mu}$, fullfilling $g_{\mu \nu}(x) q^{\mu} q^{\nu}=0$. To get the measured energy you need the four-velocity $u^{\mu}$ of the detector, normalized such that $g_{\mu \nu}(x) u^{\mu} u^{\nu}=c^2$. Then the energy of the photon measured by this detector is
$$E_{\gamma}=\hbar \omega_{\gamma}=g_{\mu \nu}(x) q^{\mu} u^{\nu}.$$
Also this is, of course, independent of the choice of coordinates/reference frames as it must be for a measured quantity.

This formula implements both the gravitational frequency shift as well as the Doppler effect for moving detectors.

For a detailed treatment of electromagnetic waves in GR, particularly in FLRW spacetime, see

https://itp.uni-frankfurt.de/~hees/pf-faq/gr-edyn.pdf

This also clarifies, who to understand the "naive photon picture", which is unfortunately used without clear definitions in most textbooks. It's to be understood as the geomtric-optic approximation (leading-order eikonal approximation) of Maxwell's equations for (free) em. fields.