Question about expansion and photon fatigue

[Grinkle]

@phinds noted in a recent thread that energy is not conserved on a cosmological scale, which hadn't occurred to me before, so I did some reading and happily the concepts were easy to digest and things I already understood but just hadn't connected the dots myself on regarding the implications for energy conservation.

I think a photon sees itself losing energy, the red shift is real as opposed to apparent. Is this correct?

Maybe a better phrasing is to ask if a freely moving observer traveling along with a photon sees the wavelength of the photon increasing over time. I think the answer is yes. Is that correct, at least according to GR?

There can also be an apparent red shift between successive photons caused by the source moving relative to the observer, I am not asking about that and at least I hope I am correct in thinking this is completely different than a red shift of an individual photon caused directly by expansion.

 

[Orodruin]

I think a photon sees itself losing energy, the red shift is real as opposed to apparent. Is this correct?

No. A photon does not have an inherent frequency or wavelength. The frequency and wavelength makes sense only in reference to a given observer. Since photons travel at the speed of light (locally), there is no local inertial frame where they are at rest.

Maybe a better phrasing is to ask if a freely moving observer traveling along with a photon sees the wavelength of the photon increasing over time. I think the answer is yes. Is that correct, at least according to GR?

An observer cannot move along with a photon for the reasons stated above.

There can also be an apparent red shift between successive photons caused by the source moving relative to the observer, I am not asking about that and at least I hope I am correct in thinking this is completely different than a red shift of an individual photon caused directly by expansion.

In fact, these effects are the same - just described in different ways. What matters is the projection of the photon's 4-momentum on the 4-velocity of the observer. In a general space-time, it is not possible to disentangle what is relative motion and what is cosmological or gravitational red shift.

 

[Grinkle]

No. A photon does not have an inherent frequency or wavelength.

Does that imply that a photon does not have an inherent energy?

If that is the case, my question does not make much sense.

 

[Orodruin]

Does that imply that a photon does not have an inherent energy?

Yes. No object has an inherent total energy. Energy is frame dependent. This is true in relativity as well as in classical mechanics.

 

[Grinkle]

Since photons travel at the speed of light (locally), there is no local inertial frame where they are at rest.

I see my blunder now - thanks!

 

[phinds]

Does that imply that a photon does not have an inherent energy?

If that is the case, my question does not make much sense.

To simply repeat what orodruin said, because it is so important, and to expand on it a bit, energy is frame dependent. The concept of a Frame Of Reference is SO important in cosmology that I want to make sure you get it. Of course, it isn't just light, which is special, that has frame dependent characteristics. Just think about the simple case of momentum. If someone throws a heavy rock at me while I'm going past them on a flatcar and the rock is traveling just a bit faster than the speed of the flatcar relative to the thrower, what is the rock's momentum? Well, in the frame of reference of the thrower, it has a fair amount of momentum. When it reaches me, it has almost no momentum.

Also very important is timing being frame dependent. The occurrence of a space-time event is frame independent but the TIMING of the event is frame defendant. So much so in fact that different observers can see two spatially separated events happen in reverse order. Look up "Relativity of Simultaneity" for more discussion.

 

[Grinkle]

I frustrate myself with how often I need to come back to the basics of relativity. I really appreciate the patience of the knowledgeable folks here.

 

[Chalnoth]

Does that imply that a photon does not have an inherent energy?

If that is the case, my question does not make much sense.

"Inherent energy" isn't a scientific term, but perhaps the closest thing to "inherent energy" that exists in physics would be mass: mass is the energy in the internal degrees of freedom of the object. A photon has zero mass. Its only energy is kinetic energy.

 

[nikkkom]

mass is the energy in the internal degrees of freedom of the object.

Wrong. Consider electron.

A photon has zero mass. Its only energy is kinetic energy.

Wrong.
In GR, mass is the T00 component of the stress-energy tensor. It is not zero for photons.

 

[Chalnoth]

Wrong. Consider electron.

I believe the current prevailing view is that the electron's mass is due to its interactions with the Higgs field. While my statement was slightly vague, I definitely intended to include this kind of interaction in "internal degrees of freedom." The reason why I used that wording specifically is that most of the mass in matter does not come from the masses of individual particles (which in turn likely comes from the Higgs interaction), but instead comes from the strong force binding energy between quarks.

Wrong.
In GR, mass is the T00 component of the stress-energy tensor. It is not zero for photons.

The T00 component is the energy density, and is frame-dependent. If I remember correctly, the mass (or at least the mass density) is the trace of the stress-energy tensor, which is frame-independent and identically zero for photons.

 

[nikkkom]

The T00 component is the energy density, and is frame-dependent. If I remember correctly, the mass (or at least the mass density) is the trace of the stress-energy tensor, which is frame-independent and identically zero for photons.

Everything is frame-dependent, the entire tensor is frame-dependent too. This does not invalidate anything.

We do not need to invoke GR here, though. For light, SR will suffice. In SR, energy and momentum is combined into a simpler object (than a tensor), a 4-vector. Four-momentum.

Basically, ordinary 3d momentum has three spatial components ("momentum into x-,y-,z-direction"). Four-momentum's zeroth component, E/c, is "momentum into time direction". Since everything, including photons, moves (in 4d space-time) into future, "momentum into time" can't be zero for them.

 

[Chalnoth]

Everything is frame-dependent, the entire tensor is frame-dependent too. This does not invalidate anything.

The trace of tensors in General Relativity is independent of choice of coordinates.

We do not need to invoke GR here, though. For light, SR will suffice. In SR, energy and momentum is combined into a simpler object (than a tensor), a 4-vector. Four-momentum.

Yes. Which can be written as:
$$v_\mu = (E, p_x, p_y, p_z)$$

The mass is $m = v_\mu v^\mu = E^2 - |\vec{p}|^2$ (I'm omitting the factors of c for clarity), and this mass will be unchanged no matter your frame of reference.

Basically, ordinary 3d momentum has three spatial components ("momentum into x-,y-,z-direction"). Four-momentum's zeroth component, E/c, is "momentum into time direction". Since everything, including photons, moves (in 4d space-time) into future, "momentum into time" can't be zero for them.

"Momentum into time direction" doesn't make sense. It's energy, which is a conserved current in a system that is time-independent in a specific way. Momentum is the conserved current in a system that is independent of spatial position in a specific way. The two are similar, but behave rather differently in some instances (e.g. in quantum mechanics there's a momentum-position uncertainty principle, as well as a similar energy-time uncertainty principle, but the two have rather different conceptual interpretations).

 

[Orodruin]

the entire tensor is frame-dependent too.

The entire point with tensors is that they are not frame dependent. Their components are frame dependent, but a tensor is a linear map from a number of copies of the tangent vector space and its dual to real numbers. Saying that tensors are frame-dependent is like saying that a vector points in different physical directions depending on the frame used to describe it.

In SR, energy and momentum is combined into a simpler object (than a tensor), a 4-vector. Four-momentum.

The implication of this, that you cannot deal with the 4-momentum in GR and do not need the stress-energy tensor in SR, is incorrect. You can deal with continua (e.g., the EM-field) in SR and GR alike and those require the stress-energy tensor. Looking at the 4-momentum instead of the stress-energy tensor is a matter of describing an ideal point-like object versus describing a continuum.