CSSlemaker said:
Thanks George. Please let me start over. First, I assume that apparent magnitudes of the SN light curve are corrected for redshift dimming and the time axis of the light curve is corrected for time dilation and that these are done before determining absolute magnitude of the SN. Correct?
Yes.
CSSlemaker said:
Then:
1. There are several "flavors" of distance: Luminosity, Radial Comoving, Light Travel, Angular Diameter.
Right, in cosmology, there are several operational definitions of distance, with no one definition being, in general, best.
CSSlemaker said:
Is the distance determined from the SN distance modulus (m-M) the Luminosity Distance, or the Light Travel Distance?
Distance modulus is defined in terms of luminosity distance.
CSSlemaker said:
2. All the literature I have seen state that the "Luminosity Distance" (DL) is not a true distance, that it serves mainly to indicate the extreme faintness of high-Z objects, and numerous charts of DL vs. Z show the DL for a Z of 10 to have the outrageous value of something like 350 billion LYs!
Yes, this is what I get when I do the calculation for our standard model of the universe.
CSSlemaker said:
There are conversion equations based upon the Standard Model of the universe, however, which permit converting a DL value to any of the other distance types.
3. Ordinarily I would think that Light Travel Distance would govern the apparent brightness of a SN because it would be the radius of the expanding spherical shell of light, and this is certainly true for relatively nearby SNs (i.e., the inverse-square law). However, if the distance calculated for a very remote SN is, in fact, the Luminosity Distance (my question #1) rather than the Light Travel Distance, I have no idea what physics would account for an apparent magnitude so faint as to imply (falsely) an outrageous "distance".
I wouldn't put it this way. Again, there is no "true" or "best" distance in cosmology.
Thanks for the expanded explanation; now, I think I can, at least partially, answer your questions.
Suppose we observe a Type Ia supernova that has apparent magnitude 23.8, and we know that its absolute magnitude is -19.2. We calculate (the model-dependent) luminosity distance to be about 4000 Mpc.
Now, imagine that we live in a universe governed by special relativity, not in an expanding universe. Assume also that we are stationary with respect to the supernova. How far away is the supernova if it has absolute magnitude -19.2 and an apparent magnitude of 23.8? 4000 Mpc. Consequently, luminosity distance is an effective distance. It is the special relativistic distance that a stationary object would have to be at in order for us to observe the same magnitude.
Why is this distance so large? Well, you have given the physical reasons for this, but let me elaborate.
The expansion of the universe has, in two ways, diminished the energy flux that we receive. The energy of light is inversely proportional to its wavelength. As the supernova's light travels to us, the expansion of the universe expands the wavelength of the light by a factor of 1+z. Also, the expansion of the universe decreases the rate at which we receive photons, as compared to the rate at which photons left the supernova, by another factor of 1+z. Thus, for z=10, expansion of the universe results in a reduction of energy flux by a factor of 121. These reduction effects are not present in an non-expanding universe, so, in a non-expanding universe, a stationary similar-magnitude object would be have to be very distant. Blame the expansion of the universe! Not, just the expansion, though, but also the way in which the universe expands (I might try and explain this better).
Note: the factor 1+z equals the ratio of the (linear) scale of the universe "now" to the scale of the universe "then". (1+z)^2 appears in the energy, but only 1+z appears in the distance.