Hubble sphere is the limit of seeing objects from Earth. But if it wasn't like that would it be possible to see a supernova just because of luminosity over that distance. In simple words would a distance so large make the supernova light too weak to be seen from Earth?

  • $\begingroup$ I assume you're asking about seeing a supernova by telescope, it's not likely that we could see a supernova with the naked eye in another galaxy unless it's very close, eg in the Magellanic Clouds, which are satellite galaxies of the Milky Way. BTW, the Hubble radius is about 14.4 billion lightyears, but the radius of the observable universe is almost 46.6 billion lightyears. See en.wikipedia.org/wiki/Observable_universe and en.wikipedia.org/wiki/Hubble_volume $\endgroup$
    – PM 2Ring
    Feb 17, 2020 at 11:25
  • $\begingroup$ Ofcourse by telescope.... As I understand the past picture of the edge of the observable universe is hubble sphere... which is the luminosity we can detect. $\endgroup$ Feb 17, 2020 at 12:14
  • $\begingroup$ Actully, if you completely disregard the cosmological expansion and just consider static universe this may be a fun question if you think about it. If you know at which wavelength supernova have a peak luminosity you may get the best angular resolution you can have e.g. within a solar system. At some distance the mean luminosity of all other objects within this angle will become comparable to the luminosity of a typical supernova. This gives you a maximum distance you ever can detect it using anything within a solar system $\endgroup$
    – OON
    Feb 17, 2020 at 12:43
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    $\begingroup$ No, we can see a lot further than the Hubble sphere. Using radiotelescopes, we can "see" the cosmic microwave background, and the surface of last scattering which emitted those photons is now roughly 45.6 billion lightyears away (but it was a lot closer to us when those photons started on their journey over 13 billion years ago). See this great answer (by an astrophysicist) on this topic: physics.stackexchange.com/a/293212/123208 $\endgroup$
    – PM 2Ring
    Feb 17, 2020 at 12:47
  • $\begingroup$ To be fair, the notion of distance on the cosmological scale can get confusing, due to the expansion of space. There's an introduction to the topic of cosmological distance scales on Wikipedia that might be helpful: en.wikipedia.org/wiki/Distance_measures_(cosmology) And I guess it can't hurt to plug (yet again) the excellent article about the expansion of space by Davis & Lineweaver: people.smp.uq.edu.au/TamaraDavis/papers/SciAm_BigBang.pdf $\endgroup$
    – PM 2Ring
    Feb 17, 2020 at 13:11

1 Answer 1


The limit is set not by our ability to detect supernova, but by when the first stars appeared.

The most distant supernova observed, SN 1000+0216 (Cooke et al. 2012), had a redshift of $z \simeq 3.9$, corresponding to a distance modulus of $\mu = 47.8$ (note that it also corresponds to a distance of 23.7 Glyr, well outside the Hubble radius of 14.4 Glyr).

It was a superluminous supernova reaching a peak absolute magnitude of $M = -21.5$, so the apparent magnitude was $$ m = \mu + M = 26.3. $$

The first stars in the Universe emerged around redshift $z\sim20$, roughly 180 million years after the Big Bang (Bowman et al. 2018). If a similar supernova goes off at this redshift, its apparent magnitude would be $m = 30.3$, which would be observable with Hubble (albeit on the border). Future JWST will reach even higher limiting magnitudes. Moreover, gravitational lensing can magnify events in some cases by a factor of 1000 or more, so in principle we should be able to see the first stars explode.

  • $\begingroup$ Are there supernovas that due to redshift emit highest intensity below visible light? $\endgroup$ Feb 17, 2020 at 16:06
  • $\begingroup$ @KrešimirBradvica Yes, definitely. If a supernova peaking in the optical (λ ~ 5-6000 Å) is observed at z ~ 4, its light will have been redshifted to ~3 μm, which is in the optical. $\endgroup$
    – pela
    Feb 17, 2020 at 16:23

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