# Luminosity Distance, Determination From Type Ia SN

1. Nov 14, 2012

### CSSlemaker

The light curve of a type Ia supernova in a cosmologically remote galaxy requires two particular corrections. The time axis must be corrected for time dilation, and the apparent-magnitude axis must be corrected for redshift dimming. It is my understanding that the "distance" determined directly from such a SN is the "luminosity distance".

Question #1: Is this determination performed before, or after making these corrections?

Question #2: If after (as I would expect), what is the physical explanation for the extreme faintness and for the extreme values of the "distance"?

2. Nov 15, 2012

### Lino

CSSIemaker, I’m just a starter on this stuff, but my understanding is that these are two different measurement methods and “corrections” are not made when using either. Luminosity Distance is used for objects at reasonably close to intermediate distances. Redshifts are used at greater distances, but do overlap with the use of Luminosity Distance - the overlap allows the calibration of the RedShift measures. Given that the overlap happens at low Redshilfts, they don’t need to be corrected to equate to Luminosity Distance measures, and at greater distances, there is no overlap so there is nothing to correct to! (To the best of my knowledge, there are also other distance measurement methods that can be used, that also overlap and provide further verifications.)

However, I am very interested if anyone has any information / pointers about measured time dilation for SN 1a light curves.

*

*

Regards,

Noel.

3. Nov 15, 2012

### CSSlemaker

Thanks Lino. But I doubt that redshift is ever used to determine distance to a type Ia supernova. Rather, it is these supernovae, serving as "standard candles", which are used to calibrate the redshift-distance relation. In fact, it was SN 1997ff at a lookback time of 11.3billion years and Z=1.7 which, because it was fainter than was expected for its Z, led to the discovery that space expansion is accelerating. Also, note that the relation between redshift and distance is model-dependent -- look at distance vs. Z charts for the pre-1998 version of the model (omega-sub-M = 1, omega-sub-lambda = 0) and charts for the current standard model (sub-M = 0.28, sub-lambda = 0.72). Apparent magnitude of a type Ia SN, on the other hand, is not model-dependent.

4. Nov 15, 2012

### CSSlemaker

P.S. - Time dilation is a simple function of Z. A time period of N days, as measured by an observer on Earth for a galaxy at redshift Z, would actually have been N/(1+Z) days as measured in that remote galaxy.

5. Nov 15, 2012

### Lino

I think that I agree with everything that your saying re redshift & distance - I'm not sure of the difference.

In relation to time dilation, I also agree. But what I was hoping for was a pointer to a paper were it was measured / discussed. For example, one would expect some comment on the observed timeline of the light curve associated with sn1997ff (and others) versus the observed timeline for closer sn's, but I can't find anything! I assume that I'm not looking in the right way / place!

Regards,

Noel.

6. Nov 16, 2012

### George Jones

Staff Emeritus
Sorry, I don't think I understand your question.

I could try and run through the maths using the observed apparent magnitude and redshift of SN 1997ff, if you (or anyone else) think that this might be useful or interesting.

Steps:

1) use the apparent and absolute magnitudes to calculate luminosity distance;

2) use the luminosity distance to calculate an estimate for the dark energy component of the universe;

3) using a model that has no dark energy, has the observed z, and has the luminosity distance calculated in 1), calculate apparent magnitude and compare to observed apparent magnitude.

I don't know if these calculations will help to illuminate a "physical explanation".
In another thread, poster George Jones was going to post a reference for this, but, before he could post, silly moderator George Jones locked the thread. Now, George Jones has forgotten the reference.

7. Nov 16, 2012

### CSSlemaker

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? Then:

1. There are several "flavors" of distance: Luminosity, Radial Comoving, Light Travel, Angular Diameter. Is the distance determined from the SN distance modulus (m-M) the Luminosity Distance, or the Light Travel Distance?

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! 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".

So, there are my questions, but #1 is by far the most important. I am scheduled to give a talk on the subject in ten days and I don't want to misstate the matter. Thanks for any help you can offer.

8. Nov 16, 2012

### Lino

George, Any chance of a (oblique) reference now?

Regards,

Noel.

9. Nov 17, 2012

### George Jones

Staff Emeritus
Yes.
Right, in cosmology, there are several operational definitions of distance, with no one definition being, in general, best.
Distance modulus is defined in terms of luminosity distance.
Yes, this is what I get when I do the calculation for our standard model of the universe.
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.

Last edited: Nov 17, 2012
10. Nov 18, 2012

### CSSlemaker

Thank you very much, George. But if you don't mind, could you please indulge me in one last clarification?

You state: "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."

In accordance with your "yes" answer to the following, my 1st question:

"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?"

I assume that the apparent magnitude 23.8 is the value after redshift correction and the -19.2 absolute magnitude is the value after time-dilation correction of the light curve.

Here's my remaining confusion: These two corrections, it seems to me, already compensate for the fact that the universe is expanding. And the distance of 4000 Mpc (13 Billion light years) does not seem to be the extreme value that I see in published charts of luminosity distance vs. redshift for redshifts above 1. A luminosity distance of 4000 Mpc appears to correspond to a redshift of roughly 0.6 or 0.7. Would that be a reasonable redshift for the example SN you describe?

Best regards,
Carroll Slemaker

11. Nov 19, 2012

### clamtrox

Hmm, can you explain a little what you mean by this? Time doesn't really come into the calculations at all; what you observe is redshifts and luminosities, whereas time is an unobservable model parameter. The calculations are totally independent of the definition of the model time coordinate.

By redshift dimming, do you mean the fact that photons lose energy when space expands? If so, then yes it is indeed taken into account. In fact, it's how we know the universe is expanding in the first place.

12. Nov 19, 2012

### George Jones

Staff Emeritus
Echoing what clamtrox wrote, I, too, would like some clarification. Apparent magnitude $m$ is a measure of the flux $F$, the power per square meter, that we (i.e., at the Earth's location) receive from an object, is a model-independent observation, and does not, in general, have anything to do with the expansion of the universe. Apparent magnitude is given by

$$m = \frac{5}{2} \log_{10} \left( \frac{F_{Vega}}{F} \right).$$

13. Nov 19, 2012

### CSSlemaker

Hi Clamtrox. Expansion of space during the time it takes the radiation from the SN to reach Earth has stretched each radiation wave by a factor of (z+1). But it also has stretched the entire wave train, from explosion to peak to disappearance, by this same factor. In other words, the observed time from peak to some specified fraction of peak luminosity will be (z+1) times the actual time it took as measured in the vicinity of the SN. It is my understanding that the absolute magnitude of a type Ia SN is determined from the rate at which its luminosity declines, and because absolute magnitude is an intrinsic property (not an observed property) of the SN, this decline rate must be determined for the epoch and location of the SN.

Also, after all of that rambling, I have read in several places that time dilation is a factor for which correction is required and that the correction is simply 1/(1+z). Note that the only dependence upon a model is the assumption that space is expanding and that redshift is a direct reflection of the magnitude of that expansion.

Best regards,
Carroll

14. Nov 19, 2012

### CSSlemaker

Hi George. Sort of along the lines of my reply to clamtrox, I have read that a model-independent adjustment must be made to the observed apparent magnitude (only in the case of a light emitter which is at a cosmologically significant distance/redshift). In a static universe, the relation of apparent magnitude to absolute magnitude is determined solely by the surface area of the expanding sphere of radiation - thus, the "inverse-square law". But in an expanding universe, an additional factor contributes to the observed apparent magnitude, the fact that expansion has shifted the radiation to a lower-energy region of the spectrum. In other words, the observed apparent magnitude of a strongly redshifted source is not a simple, direct indicator of distance to that source.

So, if the uncorrected apparent magnitude were used to calculate the distance modulus, I can understand why that would lead to an exceptionally large "distance". But my question would then become, Why not perform such a relatively simple and model-independent correction before calculating the distance modulus???

Best regards,
Carroll

15. Nov 20, 2012

### clamtrox

I see, so you are talking about the Reciprocity Theorem? It is true that it holds, but I find it very confusing to talk about "time dilation" as the result is entirely geometrical, and therefore independent of your coordinate choices. So the Reciprocity Theorem relates two different distance measures: the luminosity distance and the area distance. The result is just
$$D_L = (1+z)^2 D_A.$$
This result holds always when geodesic equations behave well. Neither of these two distance measures are better than the other one. They just represent different ways you can define distances on curved manifolds. Both of them can be easily defined and calculated theoretically. Observationally, it's very simple to calculate relative luminosity distances to supernovae just from their relative fluxes.

16. Nov 20, 2012

### CSSlemaker

Thanks clamtrox. Unfortunately, the dialogue is straying away from my original concern: the algorithm for converting a type Ia SN light curve as observed into a "distance" (not how to convert one type of "distance" into another), and which type of "distance" is the immediate result based presumably upon a distance modulus, m-M.

As for the phrase "time dilation", that is the term I have encountered at numerous Web sites on this subject and those sites state that it is an effect which must be corrected in determining the peak absolute magnitude of the SN.

Finally, my understanding - and please correct me if I am wrong - is that the part of the universe which is visible to us appears flat (but is probably only a tiny fraction of the total universe, which probably is curved) and, therefore, distance measures are calculated on that basis.

Thanks again.

17. Nov 20, 2012

### clamtrox

The other end of the luminosity distance coin is the relation
$$D_L = \sqrt{\frac{L}{4 \pi F}}$$ where L is the absolute luminosity and F is the observed flux. Now, L is not observable at all, but F is, and it's completely independent of any cosmological considerations. Of course the actual calculation is a bit more complicated, as the flux is not constant and so forth, but that should give you the general picture.

Since you don't know what L is, you can't measure absolute distances but only relative distances (supernova A is twice as far as supernova B or whatever). That allows you to determine luminosity distances to supernovae modulo some multiplicative constant, which you can then find afterwards using some other methods.

That is something that you can do, however the determination that the universe is roughly flat is not independent of supernova observations. What you do is that you take all possible data (SN,CMB,BAO,Hubble) and see what kind of model fits better. Currently it appears that all observations are fitted well by a model which has no spatial curvature.

18. Nov 20, 2012

### Lino

CSSlemaker, Are you sure that it is not peak / absolute power of the SN that the websites discussed? The absolute power being the area under the light curve, after the "width" of the curve has been adjusted (divided) by 1+z.

Regards,

Noel.

19. Nov 20, 2012

### CSSlemaker

To both clamtrox and Lino: Thank you very much. I had composed a detailed reply to both of you, but the damned site timed out (with no warning) and I lost it all! I just don't have the energy to recompose it.

Thanks again and have an enjoyable Thanksgiving (if you're in or from the U.S.)

Best regards,
Carroll Slemaker

20. Nov 20, 2012

### George Jones

Staff Emeritus
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