My understanding is that supernovae are used as standard candles, whose spectral lines indicate the recession velocity of the host galaxy.

But the material from the supernova is ejected at a significant fraction of the speed of light, which would apply an additional Doppler shift which is (presumably) impossible to predict. If this "explosion Doppler shift" exists, how does it not wash out the data used to determine the cosmological Doppler shift?


2 Answers 2


TL; DR you measure the redshift of the host galaxy, not the supernova.

Indeed, the intrinsic blue-shifts or complicated P-Cygni profiles of expanding supernova matter do not make for precise or accurate redshift determination.

If you look at some of the early Type Ia supernovae discovery papers of (relatively) low redshift supernovae, you will see that the redshifts used in the redshift-distance relationship are the redshifts of the host galaxies, often resolved from the supernova itself, and have uncertainties $<0.001$ (e.g. Perlmutter et al. 1997). i.e. When a supernova goes off in a (relatively) nearby galaxy, you can easily spatially distinguish the SN from the rest of the galaxy and get a separate spectrum of the galaxy.

The process is similar in principle for very high redshift ($z>1$, and therefore very faint) supernovae that are spatially unresolved from their host galaxies, but there you might be looking for distinct absorption features/edges or Lyman alpha emission (type Ia SNe don't have hydrogen) due to gas in the host galaxy, superimposed on the supernova spectrum, to provide the redshift.

The example below (from Curtin et al. 2019) illustrates this. The top panel shows a full spectrum (in black) with observed wavelelength on the x-axis. (The red line is the 1-sigma error in each spectral point.) The supernova features are mainly on the red side of the spectrum. Ihe blue part of the spectrum contains various absorption features caused by the host galaxay that are in the rest-frame ultra-violet part of the spectrum. The lower panel shows a zoom-in of this, now labelled in rest-frame wavelength, using a redshift of $z=2.399$, and with various absorption features from the host galaxy labelled. The blue line is a template spectrum of an absorbing galaxy component that can be (roughly speaking) cross-correlated with the observed spectrum to yield the redshift of the host galaxy. The full modelling process is a little more complex and results in a redshift of $z=2.399 \pm 0.004$ in the case shown.

Supervova spectrum

There will be some additional source of error because you don't know how fast the supernova progenitor was moving as part of its host galaxy. There is also some uncertainty because the host galaxy gas might be moving within the galaxy, but these are going to be fairy negligible contributors (hundreds of km/s at most) at high redshift.

  • $\begingroup$ Thank you for the very detailed answer! Can you explain what you mean by "redshifts of the host galaxies, often resolved from the supernova itself"? $\endgroup$
    – RC_23
    May 8, 2022 at 14:45
  • $\begingroup$ @RC_23 I mean you can spatially resolve the SN in the galaxy and place your spectrograph slit across the galaxy without getting light from the SN. $\endgroup$
    – ProfRob
    May 8, 2022 at 15:48
  • $\begingroup$ Very cool! I'm always amazed at what can be gleaned simply from analyzing the light in different ways $\endgroup$
    – RC_23
    May 8, 2022 at 17:31

Since a supernova tells the observer how bright the event is seen to be at its distance when it is always (approximately) the same standard actual brightness, this is the method used to calculate the distance to the galaxy showing this observed nova.

  • $\begingroup$ But this is the other way around, not using it for a standard candle(?). $\endgroup$ May 9, 2022 at 14:26
  • $\begingroup$ Hi @Peter Mortensen: I do not get what you are seeing as "the other way around". The standard candle means that the particular kind of nova has a very uniform brightness. Therefore when seen at a distance, the observed brightness gives the clue about the distance. $\endgroup$
    – Buzz
    May 16, 2022 at 23:31

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