<p>There's only one kind of redshift in general relativity. The cosmological redshift, gravitational redshift, and special-relativistic redshift formulas are special cases of it, which apply to spacetimes with certain symmetries.</p>

<p>If you put approximate Minkowski coordinates on a patch of spacetime that's small enough to be approximately flat, you'll find that objects in that patch that are moving with the Hubble flow are moving away from each other in the special-relativistic sense with respect to to those coordinates. If you use the special-relativistic redshift formula to calculate the redshift between objects A and B on that patch, then do the same in an approximately flat patch containing B and C, and keep doing that until you get to a very distant object Z, and multiply all those redshift factors together, you'll get the correct cosmological redshift between A and Z, up to an error arising from the deviation of each patch from flatness. In the limit of very small patches, this becomes exact.</p>

<p>So the answer to your question is that cosmological redshift and redshift due to relative motion don't add together because they're the same thing. Adding them would count the same redshift twice.</p>

<p>It's an extremely common misconception, found in many textbooks and even in <a href="https://arxiv.org/abs/astro-ph/0310808" rel="nofollow noreferrer">Davis and Lineweaver</a>, that there is some sort of fundamental difference between ordinary relative motion and "expansion of space" in general relativity. In reality, there is no generally covariant way to distinguish them. Spacetime is just a manifold, and worldlines are just worldlines. As a (quite close) analogy, consider lines of constant longitude on a globe. On a local map (small enough that there's negligible distortion when flattening it), they converge to a point at the poles and there's a nonzero angle (angle ~ <a href="https://en.wikipedia.org/wiki/Rapidity" rel="nofollow noreferrer">rapidity</a>) between them everywhere except near the equator. But you could also say that they are "at rest" at fixed longitudes but the metric distance between them increases as you move away from the poles. These descriptions are equivalent. You can calculate the coordinates of a <a href="https://en.wikipedia.org/wiki/Rhumb_line" rel="nofollow noreferrer">rhumb line</a> in the latter picture by integrating the reciprocal of the latitudinal scale factor, just as in cosmology. If you look at the distance between rhumb lines of the same slope at different latitudes, you'll find that it grows in proportion to the latitudinal scale factor, just like cosmological redshift. This does not mean the distance metric is expanding in any objective sense. It's just a simple consequence of the global symmetries of the manifold.</p>

<p>I've written some previous answers to similar questions that go into more mathematical detail. <a href="//physics.stackexchange.com/questions/228492/redshift-of-distant-galaxies-why-not-a-doppler-effect/504804#504804">Here's one</a>; <a href="//physics.stackexchange.com/questions/211797/interpretation-of-cosmological-redshift/504962#504962">here's another</a>.</p>

CC BY-SA 4.0
2019-11-11T04:24:53.720Z
2019-11-11T04:24:53.720Z
benrg
<p>I think the answers so far are predicated on a flat, that is Euclidean, universe with no curvature.</p>
<p>If the universe has curvature it is quite possible that the direction of expansion is perpendicular to the three spatial dimensions. In this case there will be no cosmological Doppler effect.</p>
<p>In addition it seems to be assumed that redshift due to gravitational time dilation is a peculiar or local effect only, which might be true for a flat universe. But a curved universe can have gravitational time dilation on a cosmological scale.</p>
<p>In a curved universe it seems possible that all of cosmological redshift is due to gravitational time dilation and none is due to the Doppler effect. This is the case for the FLRW curvature-only metric.</p>
<p>I admit that the full reasoning behind this has not been peer reviewed. I would very much welcome such a review.</p>

CC BY-SA 4.0
2024-11-30T19:22:01.523Z
2024-11-30T19:22:01.523Z
John Hobson
<p>There are several sorts of redshifts and they are not always easy to separate from one another. First you have cosmological redshift. These redshifts tend to be so high they could not be caused by anything else. Then you have gravitational redshifts, which when they originate with distant quasars having a mass equivalent to trillions of suns might become entangled with their cosmological redshifts. We also have redshift (sometimes blueshift) which arises from the proper motion of the galaxy in question. These red or blue shifts tend to mask any cosmological redshift for nearby galaxies, which redshift is in any case trivial. The Andromeda galaxy, for example, is slightly blue shifted, as it is coming towards us and is 'only' 2.5 million light years away. The distant quasars I referred to are supermassive black holes many billions of light years away. </p>

CC BY-SA 4.0
2019-09-16T00:20:43.937Z
This doesn’t address the question; also, quasars do not have masses of “trillions of suns”.
Peter Erwin
CC BY-SA 4.0
2019-09-15T20:18:36.510Z
2019-09-15T20:18:36.510Z
Michael Walsby
<p>After considering @benrg's comments, I realize that my first answer contained too strong statements about the relation between the two redshifts. I try here to moderate my answer, but you might want to accept their answer instead.</p>

<p>It is common to think of the two redshifts as having nothing to do with each other. Doppler shifts arise when the observer and/or the emitter moves through space, whereas the cosmological redshift can be derived considering stationary emitters and stationary observers in an expanding space.</p>
<p>Because the cosmological redshift doesn't involve movement through space, it is often considered completely different from the Doppler. However, it is also possible to derive the cosmological redshift by considering it as infinitely many infinitesimally small Doppler shift (e.g. <a href="https://ui.adsabs.harvard.edu/abs/2016arXiv160508634L" rel="nofollow noreferrer">Lewis 2016</a>). I admit that I'm not well enough versed in general relativity to be certain about my statements, but just because an infinitesimally small patch of spacetime is flat doesn't necessarily mean that infinitely many such patches add up to be flat. However, as @benrg says, in GR there is only one redshift.</p>
<h3>Different or the same?</h3>
<p>The reason I think it makes sense to view the Doppler shift and the cosmological redshift as two separate mechanisms is the following:</p>
<p>In principle you could have a universe (non-capitalized, since it's not <em>our</em> universe, the Universe) that were static when a distant galaxy emitted a photon, then at some point expanded quickly by a factor of 2, and then again is static. In this hypothetical case, the observer would still measure the photon to have been redshifted by a factor of 2 (i.e. <span class="math-container">$z=1$</span>).</p>
<p>That this is true can bee seen from considering the mathematical derivation of the cosmological redshift (see e.g. <a href="https://en.wikipedia.org/wiki/Redshift#Mathematical_derivation" rel="nofollow noreferrer">here</a>) which involves an integral, the result of which only depends on the initial and the final state, not on the expansion <em>history</em>.</p>
<p>In contrast, if you and your friend stand still with respect to each other while your friend shines her flashlight at you, then run away from each other with a relative velocity of <span class="math-container">$0.6c$</span>, then stand still before you receive the light (i.e. analogously to the hypothetical universe above), then you would measure <em>no</em> redshift; you wouldn't measure the special relativistic Doppler shift of <span class="math-container">$z+1 = \sqrt{\frac{1+v/c}{1-v/c}} = 2$</span> that you would if you were receding from each other while either emitting or observing.</p>
<p>In the real Universe, galaxies move through space (i.e. they change their comoving coordinate <span class="math-container">$\chi$</span>), and space expands (i.e. the scale factor <span class="math-container">$a$</span> evolves). If the physical distance to a galaxy is
<span class="math-container">$$
d = a \chi,
$$</span>
then the change in this distance gives their total velocity wrt. us, and is obtained through differentiation:</p>

<p><span class="math-container">$$
\begin{array}{rcl}
v_\mathrm{tot} &amp; =      &amp; \dot{a}\chi + a\dot{\chi} \\
               &amp; \equiv &amp; v_\mathrm{rec} + v_\mathrm{pec},
\end{array}
$$</span>
where dots denote differentiation with respect to time, and the two terms have been identified as the cosmological recession velocity, and the peculiar, &quot;normal&quot; velocity. Each of these terms give rise to a redshift, but through two very different mechanisms. Only the latter term is called a Doppler shift.</p>

CC BY-SA 4.0
2019-09-16T14:57:38.260Z
The last part, however, makes it seem like the cosmological redshift would be an instantaneous thing again and you could just plug v_rec in the standard formula for the doppler shift. Therefore giving rise to a redshift that only depends on the da/dt at the moment of emission.
SpaceCore
CC BY-SA 4.0
2019-09-16T23:06:06.360Z
Thanks for this great, detailed answer pela! So is it correct to say that a recession velocity is not an ordinary velocity we would encounter on Earth? In the sense that recession velocity doesn't really describe the motion between two bodies but rather the expansion of space between them? And therefore recession velocity does not cause a Doppler redshift? Peculiar velocities on the other hand are like the velocities we encounter on Earth. And the total redshift would be the sum of a cosmological redshift due to rec. velocity and Dopple redshift due to the pec. velocity of the observed object?
user120112
CC BY-SA 4.0
2019-09-17T10:29:38.373Z
Up vote because the reference to semantics. For many cosmological red shift is a doppler shift. I think focusing on comoving and proper coordinates remove the confusion or whatever the need to call things a way or another.
Alchimista
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2019-09-17T15:08:18.710Z
@user120112 Yes, exactly! :)
pela
CC BY-SA 4.0
2019-09-17T15:29:08.470Z
@SpaceBread Yes, maybe it's a bit confusing to write it like this, but it's just Hubble's law. Without (or with negligible) peculiar velocities, $v_\mathrm{rec} = \dot{a}\chi = \dot{a}d/a \equiv H_0 d$.
pela
CC BY-SA 4.0
2019-09-19T11:03:16.357Z
@SpaceBread I think I misunderstood your comment. The "$\dot{a}$" doesn't, in general, refer to the change in $a$ at the time of emission, but to the general value at the time you wish to know the recession velocity. So in the hypothetical universe that is static except for a brief, intermediate expansion, you have $\dot{a}(t=0) = 0$ and $\dot{a}(t=t_\mathrm{em}) = 0$, meaning that $v_\mathrm{rec}=0$ at $t=0$ and at $t_\mathrm{em}$. In _our_ Universe, however, $\dot{a}(t=0) = H_0 \ne 0$.
pela
CC BY-SA 4.0
2019-11-11T04:59:31.777Z
There's only one kind of redshift in GR: the ratio of the distances between differentially separated null geodesics at emission and detection. There's no generally covariant way to separate it into "cosmological" and "relative motion" redshift; it's all part and parcel of the same thing. There's also no way to define a covariant concept of "expanding space" in GR (e.g. there's no space-expansion tensor that's nonzero where space is expanding and zero where it isn't). Expanding space is just a particular coordinate description of a manifold that can be described in other ways. See my answer.
benrg
CC BY-SA 4.0
2019-11-11T05:10:06.140Z
If you plug your spontaneously doubling cosmology into the GR field equation, you'll find that the expansion is due to the manifestation of exotic matter that fills the universe, changes its energy density from positive to negative for no reason, then vanishes for no reason. GR technically allows this, just like it technically allows escaping from a black hole in an Alcubierre drive. The redshift in your example can be understood as due to local gravitational interaction with the exotic matter. In more mundane cosmologies, redshift can be understood in more mundane ways.
benrg
CC BY-SA 4.0
2019-11-12T23:53:53.917Z
@benrg I agree that cosmological redshift can be thought of as infinitely many infinitesimally small Doppler shift, but I'll have to think more about the rest of your statements at a time when it's not so late. Thanks for your comments :)
pela
CC BY-SA 4.0
2019-09-16T12:23:31.817Z
2020-09-05T07:36:57.357Z
2020-09-05T07:36:57.357Z
pela
<p>I'm reading Harrison's "Cosmology: Science of the universe" because Harrison focuses on the distinction between cosmological redshift (he calls it expansion redshift) and the Doppler redshift. </p>

<p>He states that "they [Doppler redshifts] are produced by peculiar and not by recession velocities"
and
"[expansion redshifts] are produced by recession and not
peculiar velocities".</p>

<p>I understand the concepts of both kinds of redshifts but have a hard time understanding this strict separation. Please correct me, where/if I am wrong:</p>

<p>Suppose a galaxy has no peculiar movement. This means that its position will stay (approximately) the same in comoving coordinates. In fact though ( proper distance), it IS moving away from Earth with recession velocity V, caused by the expansion of the universe. So it should have a Doppler effect based on this recession velocity AND a cosmological (expansion) redshift should also take effect because the light gets stretched with the expansion of the universe. Even though the recession velocity is not due to a peculiar movement, it means that the source of light is moving away from the observer and hence the light should be redshifted and on top the light gets redshifted on its way through expanding space.</p>

<p>Please correct me or tell me if I am right or wrong, I have spent a lot of time reading but still don't fully understand this. Thanks.</p>

CC BY-SA 4.0
2019-09-16T02:02:14.747Z
An idea that I have, is that my assumption, that recession velocities cause a Doppler redshift is wrong and that they cannot be compared to other "ordinary" or  peculiar velocities. Maybe recession velocities don't cause Doppler redshifts because the galaxy moving away due to recession velocity is not moving compared to its environment. Objects in a Doppler scenario (moving light source on Earth or galaxy with peculiar motion), are moved against their environment (or their environment is moved, if in the objects rest frame). Maybe this solves my problem. I'd love to discuss/get feedback.
user120112
CC BY-SA 4.0
2020-01-13T10:40:58.430Z
This article might be of interest https://arxiv.org/abs/astro-ph/0310808v2
usernumber
CC BY-SA 4.0
2020-09-01T00:21:30.607Z
The current accepted answer by pela is incorrect. Davis and Lineweaver (linked in the previous comment) are also incorrect on this point, unfortunately, though I like their paper in general. Since the upshot of the paper is that even famous physicists can be and have been wrong about basic properties of FLRW cosmology, I hope you can accept that there may be lingering misconceptions that they didn't correct, and may even still suffer from themselves. This is one of them. Please see my comments on pela's answer, my answer, and (for more details) the other answers linked from it.
benrg
<cosmology><expansion><redshift><doppler-effect>
Cosmological redshift vs doppler redshift
user120112
# Cosmological redshift vs doppler redshift

(Asked by: username_0 on 2019-09-15)

I'm reading Harrison's "Cosmology: Science of the universe" because Harrison focuses on the distinction between cosmological redshift (he calls it expansion redshift) and the Doppler redshift.

He states that "they [Doppler redshifts] are produced by peculiar and not by recession velocities"
and
"[expansion redshifts] are produced by recession and not
peculiar velocities".

I understand the concepts of both kinds of redshifts but have a hard time understanding this strict separation. Please correct me, where/if I am wrong:

Suppose a galaxy has no peculiar movement. This means that its position will stay (approximately) the same in comoving coordinates. In fact though ( proper distance), it IS moving away from Earth with recession velocity V, caused by the expansion of the universe. So it should have a Doppler effect based on this recession velocity AND a cosmological (expansion) redshift should also take effect because the light gets stretched with the expansion of the universe. Even though the recession velocity is not due to a peculiar movement, it means that the source of light is moving away from the observer and hence the light should be redshifted and on top the light gets redshifted on its way through expanding space.

Please correct me or tell me if I am right or wrong, I have spent a lot of time reading but still don't fully understand this. Thanks.

**Tags:** cosmology, expansion, redshift, doppler-effect

> **Comments:**
> **username_0**: An idea that I have, is that my assumption, that recession velocities cause a Doppler redshift is wrong and that they cannot be compared to other "ordinary" or  peculiar velocities. Maybe recession velocities don't cause Doppler redshifts because the galaxy moving away due to recession velocity is not moving compared to its environment. Objects in a Doppler scenario (moving light source on Earth or galaxy with peculiar motion), are moved against their environment (or their environment is moved, if in the objects rest frame). Maybe this solves my problem. I'd love to discuss/get feedback.  
> **username_1**: This article might be of interest https://arxiv.org/abs/astro-ph/0310808v2  
> **username_2**: The current accepted answer by username_6 is incorrect. Davis and Lineweaver (linked in the previous comment) are also incorrect on this point, unfortunately, though I like their paper in general. Since the upshot of the paper is that even famous physicists can be and have been wrong about basic properties of FLRW cosmology, I hope you can accept that there may be lingering misconceptions that they didn't correct, and may even still suffer from themselves. This is one of them. Please see my comments on username_6's answer, my answer, and (for more details) the other answers linked from it.  

---

## 4 answers

### Answer by username_6 on 2019-09-16. Score: 12 | Accepted answer

After considering @username_2's comments, I realize that my first answer contained too strong statements about the relation between the two redshifts. I try here to moderate my answer, but you might want to accept their answer instead.

It is common to think of the two redshifts as having nothing to do with each other. Doppler shifts arise when the observer and/or the emitter moves through space, whereas the cosmological redshift can be derived considering stationary emitters and stationary observers in an expanding space.

Because the cosmological redshift doesn't involve movement through space, it is often considered completely different from the Doppler. However, it is also possible to derive the cosmological redshift by considering it as infinitely many infinitesimally small Doppler shift (e.g. [Lewis 2016](https://ui.adsabs.harvard.edu/abs/2016arXiv160508634L)). I admit that I'm not well enough versed in general relativity to be certain about my statements, but just because an infinitesimally small patch of spacetime is flat doesn't necessarily mean that infinitely many such patches add up to be flat. However, as @username_2 says, in GR there is only one redshift.

### Different or the same?

The reason I think it makes sense to view the Doppler shift and the cosmological redshift as two separate mechanisms is the following:

In principle you could have a universe (non-capitalized, since it's not *our* universe, the Universe) that were static when a distant galaxy emitted a photon, then at some point expanded quickly by a factor of 2, and then again is static. In this hypothetical case, the observer would still measure the photon to have been redshifted by a factor of 2 (i.e. $z=1$).

That this is true can bee seen from considering the mathematical derivation of the cosmological redshift (see e.g. [here](https://en.wikipedia.org/wiki/Redshift#Mathematical_derivation)) which involves an integral, the result of which only depends on the initial and the final state, not on the expansion *history*.

In contrast, if you and your friend stand still with respect to each other while your friend shines her flashlight at you, then run away from each other with a relative velocity of $0.6c$, then stand still before you receive the light (i.e. analogously to the hypothetical universe above), then you would measure *no* redshift; you wouldn't measure the special relativistic Doppler shift of $z+1 = \sqrt{\frac{1+v/c}{1-v/c}} = 2$ that you would if you were receding from each other while either emitting or observing.

In the real Universe, galaxies move through space (i.e. they change their comoving coordinate $\chi$), and space expands (i.e. the scale factor $a$ evolves). If the physical distance to a galaxy is
$$
d = a \chi,
$$
then the change in this distance gives their total velocity wrt. us, and is obtained through differentiation:

$$
\begin{array}{rcl}
v\_\mathrm{tot} & = & \dot{a}\chi + a\dot{\chi} \\
& \equiv & v\_\mathrm{rec} + v\_\mathrm{pec},
\end{array}
$$
where dots denote differentiation with respect to time, and the two terms have been identified as the cosmological recession velocity, and the peculiar, "normal" velocity. Each of these terms give rise to a redshift, but through two very different mechanisms. Only the latter term is called a Doppler shift.

> **Comments:**
> **username_7**: The last part, however, makes it seem like the cosmological redshift would be an instantaneous thing again and you could just plug v_rec in the standard formula for the doppler shift. Therefore giving rise to a redshift that only depends on the da/dt at the moment of emission.  
> **username_0**: Thanks for this great, detailed answer username_6! So is it correct to say that a recession velocity is not an ordinary velocity we would encounter on Earth? In the sense that recession velocity doesn't really describe the motion between two bodies but rather the expansion of space between them? And therefore recession velocity does not cause a Doppler redshift? Peculiar velocities on the other hand are like the velocities we encounter on Earth. And the total redshift would be the sum of a cosmological redshift due to rec. velocity and Dopple redshift due to the pec. velocity of the observed object?  
> **username_8**: Up vote because the reference to semantics. For many cosmological red shift is a doppler shift. I think focusing on comoving and proper coordinates remove the confusion or whatever the need to call things a way or another.  
> **username_6**: @username_0 Yes, exactly! :)  
> **username_6**: @SpaceBread Yes, maybe it's a bit confusing to write it like this, but it's just Hubble's law. Without (or with negligible) peculiar velocities, $v_\mathrm{rec} = \dot{a}\chi = \dot{a}d/a \equiv H_0 d$.  
> **username_6**: @SpaceBread I think I misunderstood your comment. The "$\dot{a}$" doesn't, in general, refer to the change in $a$ at the time of emission, but to the general value at the time you wish to know the recession velocity. So in the hypothetical universe that is static except for a brief, intermediate expansion, you have $\dot{a}(t=0) = 0$ and $\dot{a}(t=t_\mathrm{em}) = 0$, meaning that $v_\mathrm{rec}=0$ at $t=0$ and at $t_\mathrm{em}$. In _our_ Universe, however, $\dot{a}(t=0) = H_0 \ne 0$.  
> **username_2**: There's only one kind of redshift in GR: the ratio of the distances between differentially separated null geodesics at emission and detection. There's no generally covariant way to separate it into "cosmological" and "relative motion" redshift; it's all part and parcel of the same thing. There's also no way to define a covariant concept of "expanding space" in GR (e.g. there's no space-expansion tensor that's nonzero where space is expanding and zero where it isn't). Expanding space is just a particular coordinate description of a manifold that can be described in other ways. See my answer.  
> **username_2**: If you plug your spontaneously doubling cosmology into the GR field equation, you'll find that the expansion is due to the manifestation of exotic matter that fills the universe, changes its energy density from positive to negative for no reason, then vanishes for no reason. GR technically allows this, just like it technically allows escaping from a black hole in an Alcubierre drive. The redshift in your example can be understood as due to local gravitational interaction with the exotic matter. In more mundane cosmologies, redshift can be understood in more mundane ways.  
> **username_6**: @username_2 I agree that cosmological redshift can be thought of as infinitely many infinitesimally small Doppler shift, but I'll have to think more about the rest of your statements at a time when it's not so late. Thanks for your comments :)  

---

### Answer by username_2 on 2019-11-11. Score: 3

There's only one kind of redshift in general relativity. The cosmological redshift, gravitational redshift, and special-relativistic redshift formulas are special cases of it, which apply to spacetimes with certain symmetries.

If you put approximate Minkowski coordinates on a patch of spacetime that's small enough to be approximately flat, you'll find that objects in that patch that are moving with the Hubble flow are moving away from each other in the special-relativistic sense with respect to to those coordinates. If you use the special-relativistic redshift formula to calculate the redshift between objects A and B on that patch, then do the same in an approximately flat patch containing B and C, and keep doing that until you get to a very distant object Z, and multiply all those redshift factors together, you'll get the correct cosmological redshift between A and Z, up to an error arising from the deviation of each patch from flatness. In the limit of very small patches, this becomes exact.

So the answer to your question is that cosmological redshift and redshift due to relative motion don't add together because they're the same thing. Adding them would count the same redshift twice.

It's an extremely common misconception, found in many textbooks and even in [Davis and Lineweaver](https://arxiv.org/abs/astro-ph/0310808), that there is some sort of fundamental difference between ordinary relative motion and "expansion of space" in general relativity. In reality, there is no generally covariant way to distinguish them. Spacetime is just a manifold, and worldlines are just worldlines. As a (quite close) analogy, consider lines of constant longitude on a globe. On a local map (small enough that there's negligible distortion when flattening it), they converge to a point at the poles and there's a nonzero angle (angle ~ [rapidity](https://en.wikipedia.org/wiki/Rapidity)) between them everywhere except near the equator. But you could also say that they are "at rest" at fixed longitudes but the metric distance between them increases as you move away from the poles. These descriptions are equivalent. You can calculate the coordinates of a [rhumb line](https://en.wikipedia.org/wiki/Rhumb_line) in the latter picture by integrating the reciprocal of the latitudinal scale factor, just as in cosmology. If you look at the distance between rhumb lines of the same slope at different latitudes, you'll find that it grows in proportion to the latitudinal scale factor, just like cosmological redshift. This does not mean the distance metric is expanding in any objective sense. It's just a simple consequence of the global symmetries of the manifold.

I've written some previous answers to similar questions that go into more mathematical detail. [Here's one](//physics.stackexchange.com/questions/228492/redshift-of-distant-galaxies-why-not-a-doppler-effect/504804#504804); [here's another](//physics.stackexchange.com/questions/211797/interpretation-of-cosmological-redshift/504962#504962).


---

### Answer by username_3 on 2024-11-30. Score: -3

I think the answers so far are predicated on a flat, that is Euclidean, universe with no curvature.

If the universe has curvature it is quite possible that the direction of expansion is perpendicular to the three spatial dimensions. In this case there will be no cosmological Doppler effect.

In addition it seems to be assumed that redshift due to gravitational time dilation is a peculiar or local effect only, which might be true for a flat universe. But a curved universe can have gravitational time dilation on a cosmological scale.

In a curved universe it seems possible that all of cosmological redshift is due to gravitational time dilation and none is due to the Doppler effect. This is the case for the FLRW curvature-only metric.

I admit that the full reasoning behind this has not been peer reviewed. I would very much welcome such a review.


---

### Answer by username_4 on 2019-09-15. Score: -3

There are several sorts of redshifts and they are not always easy to separate from one another. First you have cosmological redshift. These redshifts tend to be so high they could not be caused by anything else. Then you have gravitational redshifts, which when they originate with distant quasars having a mass equivalent to trillions of suns might become entangled with their cosmological redshifts. We also have redshift (sometimes blueshift) which arises from the proper motion of the galaxy in question. These red or blue shifts tend to mask any cosmological redshift for nearby galaxies, which redshift is in any case trivial. The Andromeda galaxy, for example, is slightly blue shifted, as it is coming towards us and is 'only' 2.5 million light years away. The distant quasars I referred to are supermassive black holes many billions of light years away.

> **Comments:**
> **username_5**: This doesn’t address the question; also, quasars do not have masses of “trillions of suns”.  

---

