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A heavy star should look redshifted, due to gravitational time dilation. How is that incorporated into the calculations of distances to the stars, or is it negligible?

How about an entire region of space that is denser or more energetic than our region? Would that not appear further away, than another region that is less dense than our region?

If we were in a region of low density and energy, and the density of space increases as we look further away, is that something that could theoretically explain Einsteins "blunder"? Can one be sure that this is not the case?

If the universe were collapsing, and gravity does indeed propagate. Wouldn't all distant regions seem to be affected more by gravity than the region of the observer, and thus ever more redshifted? (this due to less gravity in "front" of an approaching region of space than "behind" it, relative to us)

I'm curious of how these things are found to be important or unimportant on a cosmic scale.

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That's a lot of not quite trivial questions! I'll try to answer part of it. First, the red-shift can be composed of relativistic Doppler-effect and gravitational red-shift. When neglecting the gravitationl part, we get a higher radial velocity. The radial velocity can be used to calculate a distance estimate via the Hubble "constant". So for low velocities the gravity part is not negligible in respect to relative errors.

Stars are moving more or less randomly. Therefore it's necessary to look at a sufficiently large population of stars or galaxies to get a distance estimate that way. For low red-shifts this doesn't work reliably. For larger red-shifts, the gravity part plays a minor relative role, as long as one looks at usual stars.

A region of high gravity can be detected by gravitational lensing effects. So this source of error can be avoided by working properly.

Einstein's "blunder" was the assumption, that the universe needs to be static on large scales. Therefore he induced a non-zero cosmological constant to avoid the universe to expand or collapse. He could have predicted the big bang by assuming the constant to be zero.

Looking further away means looking into the past, when the universe has been denser.

A hollow-world scenario with a dense shell, heavy enough to cause the observed redshift would probably collapse rapidly to the shell. If the shell as a whole doesn't collapse, kind of anti-gravity provided by a cosmological constant or function would have to be provided. But this would probably also annihilate the red-shift, thus not consistent with observation.

In a collapsing universe objects would look blue-shifted instead of red-shifted. The degree of blue-shift would be dependent of the way the collapse would take place.

These things are relevant to be considered to exclude an option of an alternate spacetime which could explain observations. I could recommend reading more about the Planck results, that many options are actually considered, e.g. starting with this blog, then continuing with the original Planck papers.

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In a hollow world, wouldn't distant objects be closer to the denser shell, and would look redshifted - due to gravitational time dilation? Also, distant objects would seem to move faster relatively to us (higher kinetic energy). Finally, since looking into the denser past of the universe, we should actually see an "illusion" of a hollow world with a dense shell? 13 billion years ago, the universe must have been extremely dense. I'm not necessarily suggesting that the universe itself is collapsing, but at least some of the redshift must be attributed to those those assertions? –  frodeborli Jan 9 at 19:50
They would look redshifted in a hollow world, if the cosmologic constant is set to zero. But to keep it stable, kind of anti-gravity is needed for balance; this would annihilate the red-shift. –  Gerald Jan 9 at 19:54
I also believe that a region of space can be more energetic - more photons, gamma rays - which we won't see but which will encapsulate stars in a large region of "slower than our clocks". The difference might be small though, but can we be sure? –  frodeborli Jan 9 at 19:57
But why can't the universe be collapsing and the cosmological constant zero? Especially if the cosmological constant was introduced to explain the expansion. "A scientific theory should be as simple as possible, but no simpler" –  frodeborli Jan 9 at 20:00
We see regions of space with higher density by gravitational lensing, meaning light does take a different path than without the additional mass. Stars and galaxies behind the massive object look different. –  Gerald Jan 9 at 20:05
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You have many questions. I only answer the first. It's doesn't only matter how heavy a star is, but also how big. For ordinary stars, the effect is neglible (work it out by yourself -- it's a useful exercise). Even for compact stars, such as white dwarves or neutron stars, the effect is small.

However, what astronomers commonly refer to as (stellar-mass) black holes, may actually be strange stars consisting of some quark-gluon plasma (a white dwarf is like a big crystal, a neutron star like a big atomic nucleus, a strange star like a big neutron). These stars would have a high gravitational redshift (1000 or more) at their surface, such that the surface is effectively invisible. This makes them very hard/impossible to distinguish form "real" black holes.

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Also hard to distinguish from very distant, ordinary stars I suppose, but I guess one is in those cases looking at other stars in the same galaxy to determine the distance? –  frodeborli Jan 9 at 19:34
@frodeborli At cosmological red-shift 1000 there are no stars. –  Walter Jan 10 at 15:44
Is Doppler redshift distinguishable from Einstein shift? Could not cosmological redshift be Einstein shift, and not Doppler shift? Looking 13 billion years into the past, one would expect a lot of Einstein shift due to a very dense universe? –  frodeborli Jan 10 at 16:04
@Walter: For WDs the effect is far from being small when it comes to spectroscopical velocities, check my numbers in this thread. –  Alexey Bobrick Jan 11 at 16:48
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I will make a small calculation here, but please proceed to the results if you may like to.


Stars are spherical and static, so metric near their surface (photosphere) and outside on is Schwarzschild. Hence time-time metric component on the surface is:


where $R_*$ is the radius of the star and $R_{grav,*}$ is its gravitational radius.

Then, if the velocity of the star is much smaller than the speed of light, gravitational redshift in the lowest order does not depend on this velocity. Therefore, the emitting star can be assumed at rest.

The light from the star propagates along isotropic geodesic in Schwarzschild metric. The geodesic is described by lagrangian: $$ \mathcal{L} = \dfrac{1}{2}g_{\mu\nu}k^{\mu}k^{\nu}$$, where $k^\mu=(\vec{k},\omega/c)$ is the 4-vector of the light wave and $\omega$ is the frequency of light. Since the metric is static $\dfrac{d\mathcal{L}}{dk^4}=g_{\mu 4} k^4=g_{44}k^4=\textrm{const}$. Therefore:

$$(1-\dfrac{R_{grav,*}}{R_*})\omega = \textrm{const}$$

for the light as it travels towards us. So:

$$\omega_{obs}=\omega_{emitted}(1-\dfrac{R_{grav,*}}{R_*})\Longleftrightarrow \lambda_{obs}=\dfrac{\lambda_{emitted}}{(1-\dfrac{R_{grav,*}}{R_*})}$$, where $\lambda$ is the wavelength.

Redshift is simply $z=\dfrac{\lambda_{obs}-\lambda_{emitted}}{\lambda_{emitted}}$. Assuming $z\ll 1$ one has a simple formula: $$ z_0=\dfrac{R_{grav,*}}{R_*} $$

If $z_0$ turns out to be comparable to unity, one should compute $$ z=\dfrac{1}{1-z_0}-1, $$ which then gives the correct value of redshift. Note that redshift does not depend on $\lambda$.

Nice numerical forms for this would come from $R_{grav,*}=2.95\textrm{km}\dfrac{M_*}{M_\odot}$:

$$z_0=0.295\dfrac{10\textrm{km}}{R_*}\dfrac{M_*}{M_\odot}\Longleftrightarrow z_0=4.24\cdot 10^{-6}\dfrac{R_\odot}{R_*}\dfrac{M_*}{M_\odot}$$

It is also nice to express the redshift in $\textrm{km}/\textrm{s}$:

$$ z_0=8.84\cdot 10^4 \dfrac{10\textrm{km}}{R_*}\dfrac{M_*}{M_\odot} \textrm{km/s}\Longleftrightarrow z_0=1.27 \dfrac{R_\odot}{R_*}\dfrac{M_*}{M_\odot}\textrm{km/s} $$

Summary and discussion

In summary, when redshift $z$ is small, it is approximated by $z_0$, the numerical expressions for which are given just above. If $z_0$ turns out to be not small, one can calculate $z=\dfrac{1}{1-z_0}-1$, which then gives correct redshift.


One can see that:

  • For normal stars like the Sun ($R_*\sim R_\odot, M_*\sim M_\odot$) the redshift is of order $1\textrm{km/s}$. It is almost important since stars in solar neighbourhood move normaly at the velocity of a few tens of $\textrm{km/s}$
  • For white dwarfs ($R_*\sim 10^4 \textrm{km}, M_*\sim M_\odot$) redshift is a few times $100 \textrm{km/s}$ and gets very important when doing proper spectroscopy. Therefore one typically accounts for it.
  • For neutron stars ($R_*\sim 10 \textrm{km}, M_*\sim M_\odot$) redshift is very important $z\sim 0.4$, but neutron stars are general relativistic objects anyway, so it would be expected beforehand.

So, in summary, when measuring light from individual stars one does have to account for gravitational redshifts to get accurate results, and particularly so when studying white dwarfs.

Groups of objects

Now, the same formulae are correct to an order of magnitude when applied for larger volumes of space, with $R_*$ and $M_*$ now meaning the size of the volume and the mass inside it. However, as typical interstellar distances are of order of parsec and $\textrm{pc}=3\cdot10^{13}\textrm{km}$, resulting $z$ will be very small even for such dense groups as globular clusters ($z$ is of order $10^{-8}$ in this case). So, groups of objects do not affect the redshift.

Cosmological overdensities

Nevertheless, cosmological scale underdensities of order few $100\textrm{Mpc}$ in size may affect the apparent redshift of distant objects, as we would be inside the underdensity. However, such an underdensity would have to be significantly symmetric around us in order to explain the lack of corresponding anisotropy in cosmic microwave background. Therefore, it is considered unlikely.

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Thank you for your calculations. I assume we distinguish the types of stars by looking at their color spectrum? It clarifys things. What is your comment on the idea that distant objects, being from a earlier and ofcourse denser universe, will due to this be Einstein shifted as a function of their distance to us? –  frodeborli Jan 9 at 21:38
@frodeborli: Yes, spectroscopy is the most informative type of measurements for stars, so it has to be rather robust. Cosmological redshift (that you are mentioning here), doppler effect and gravitational redshift are all important. And then when a component is small, it can be just added up. When two components are large, they have to be combined in trickier ways. –  Alexey Bobrick Jan 9 at 22:07
How about the assertion that a star that is 13.6 billion lightyears away will appear to be very close to every other object in the universe (since the universe was much smaller), and it should therefore be showing a lot of gravitational redshift? –  frodeborli Jan 9 at 22:18
@frodeborli: Well, the following statements are true. 1) Everything emitting from early universe get significantly redsifted (you can get a feel for it here en.wikipedia.org/wiki/Distance_measures_%28cosmology%29). 2) There were stars early on, one possible candidate being so called Population III stars (see the corresponding section in en.wikipedia.org/wiki/Metallicity) 3) Additional redshift from the stars being massive comes from the surface of individual stars, not from them being in groups. These stars were around few $100 M_\odot$ in mass and several $10 R_\odot$ in radius. –  Alexey Bobrick Jan 10 at 14:50
@frodeborli: so additional redshift for one star was comparable to the one from typical stars nowadays. The stars were closer indeed, but collective effect is small. Consider some extreme case of two such Population III stars at $10 R_*$ separation: $M_*/R_*$ will become $2M_*/(10R_*)=M_*/(5R_*)$, smaller than for individual stars. So at high redshifts corrections due to stellar gravity will be much smaller than cosmological redshifts. –  Alexey Bobrick Jan 10 at 14:55
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