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Expansion redshift VS gravitational redshift?

  1. Jan 12, 2010 #1
    While objects closer to us tend to shift both in direction red or blue, depending on their movement in relation to us, distant objects such as galaxies tend to only shift to the red.

    As I understand this is the base of the idea that the universe is expanding. But how are we sure that is the case, and redshifts are not due to the gravitational pull of all those objects that lie between us and the observed objects?
     
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  3. Jan 12, 2010 #2

    sylas

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    Because gravitation effects of stuff in between the source and the observer cancel out.

    It is the difference in gravitational potential between source and observer that matters. Dropping a little bit into and then out of the gravitational well along the way may alter the direction of light, but that's all. This change in direction is measured and used in study of gravitational lensing.

    Cheers -- sylas
     
  4. Jan 12, 2010 #3
    Yep, that makes sense, thanks a lot

    The only gravity that would not cancel out is that of the observed object, as it is the starting point it can only pull light back

    I am not sure but I think I've read that the CMBR has shifted uniformly, but if space indeed expanded in such a high rate - shouldn't different regions of the CMBR be redshifted by a different amount, depending on their position relative to our point of observation?
     
  5. Jan 12, 2010 #4

    Ich

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    No, it results in a blueshift.
    If it'd cancel out, there would be no deceleration of expansion either.
     
  6. Jan 13, 2010 #5

    Chalnoth

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    Huh? Sylas is correct. On average, the effects completely cancel, because compared to the average density, there are just as many voids as collapsed objects (by some appropriate measure).
     
  7. Jan 13, 2010 #6

    Wallace

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    I think we probably all agree and are just using the words differently, but in fact gravitational effects play a crucial role in cosmological redshift. The distance vs redshfit relationship is a key cosmological probe and the very reason it tells us about the composition of the Universe is because the gravitational effects at play can be modelled and that tells us how much stuff (and how much of different kinds of stuff) are around. This is true even if we ignore structure (i.e. the simplest homogenous FRW model).

    To see how gravity is important, try modelling a matter only Universe using Newtonian physics only. You will see that even for quite large distances, you get pretty close to the correct answer by modelling the redshift as a combination of a Doppler redshift plus a blueshift due to the gravitation matter enclosed in a sphere centred on the observer with a radius equal to the distance to the emmitter (Gauss's law lets us ignore everything outside in a homogenous universe). In the Newtonian case you can work out the gravitational blueshift by thinking about the potential energy difference between the emmitter and observer.

    Now, in the full relativistic case, there is an inherent ambiguity in dividing up the redshift into the 'doppler' and 'graviational' parts, and it depends on the co-ordinates you choose as to which label gets what. There have been various bun fights in the literature about this, but the bottom line is that both motion and gravity are at play in determining what redshift is observed.
     
  8. Jan 13, 2010 #7

    Ich

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    Thanks, Wallace.

    If we have, as a toy model, a whole universe filled with static dust, the gravitational effects between any two points do not "cancel out". Instead, they make the whole thing collapse. This concernes photons also, they get blueshifted.


    Exactly.
    But "potential" is not a basic feature of GR; for example, it is not defined in
    homogeneous coordinates. That does not mean that there are no gravitational effects.
    To define a potential, you have to use static coordinates. The potential is then [itex]\sqrt{g_{tt}} \simeq g_{tt}/2[/itex].
    Static coordinates are centered around one (arbitrary, of course) point r=0. The potential is then (in Newtonian limit)
    [tex]\frac{2 \pi G}{3 c^2} \rho r^2[/tex]
    As long as there are no significant density changes during the light travel time, you can decompose photon redshift unambiguously into gravitational blueshift and doppler redshift.
    In case of a static spacetime, like de Sitter, an unambiguous decomposition is always possible.
     
  9. Jan 13, 2010 #8

    sylas

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    Not so. The simplest such model, the Milne model, is a flat universe filled with dust at critical density, and it keeps expanding indefinitely, though slowing down indefinitely as well. Photons in this universe are always redshifted.

    The only way you get blue shift is with supercritical densities, which can reverse expansion into contraction and a Big Crunch. You can get blue shifts once contraction gets underway, which is perfectly obviously not going on in our universe.

    As Wallace points out, you can, depending on how you work with co-ordinates, regard the redshift (or blueshift, in a contracting universe) as a gravitational effect, associated with the difference in density between emission and observation of a photon in different regions of the dust filled universe. Whether you do this with Newtonian approximations or the more correct relativistic methods only makes a difference on sufficiently large scales.

    If the dust is inhomogenous on small scales, then you end up with something a bit more like our own universe, with local motions and clusters of galaxies and so on. If a photon passes by clumps of matter between emission and observation, this makes no difference, which is the point I was trying to make. What counts is the state at emission, and at observation. Going into and out of a localized gravitational well along the way has no effect, except perhaps on directions, which is the basis of gravitational lensing.

    Cheers -- sylas
     
  10. Jan 13, 2010 #9

    Ich

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    No, the Milne Model is massless, filled only with "expanding" test particles. Consequently, it has zero density and is negatively curved. Redshift is purely doppler, there are no gravitational effects.
    I'm talking ablout homogeneous static dust, like a closed universe at maximum expansion. Just to illustrate that gravitational effects definitely do not cancel out.
    I'm not talking about a net blueshift. I said that gravitation results in a blueshift, which is outweighed by doppler redshift in an expanding universe.
    That's not how I read Wallace's post, and I wouldn't agree either. As i understand it, Wallace and I are claiming that the "potential" approach is valid in an homogeeous universe. (quote:"This is true even if we ignore structure (i.e. the simplest homogenous FRW model).")
    Yep, but the underlying physics is more accessible in the Newtonian formulation. You can see easily that it's exactly the matter between two points which accelerates them, not some other dubious effect.
    Again: perfectly homogeneous dust (or DE, for that matter) does not mean that there is no potential difference. It means that, by choosing an origin for static coordinates, you can define where the global minimum is. Every photon being observed at r=0 comes from a higher potential in that picture.
     
  11. Jan 13, 2010 #10

    Wallace

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    Edit: Crossed posts with Ich. We seem to be in good agreement though...

    Right, but when you add a homogenous matter distribution what do you find? (Edit: missed seeing that you suggest Milne model is at critical density, as Ich points out it is empty, there is no dust in it. In the Milne model nothing ever slows down, all proper velocites remain fixed). You find that the more matter you add, the less redshift you see for objects a fixed distance from you (however you define that). This is because of the effects of gravity "adding a blueshift" in some loosely defined way as the photon travels.

    I think we are talking at cross purposes. Ich explained how you get a component of the redshift which is gravitational (and in fact this componet reduces the redshift), not how any gravity gives you a net blueshift.
     
    Last edited: Jan 13, 2010
  12. Jan 13, 2010 #11

    Chalnoth

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    After the collapse has begun, sure. But our universe is expanding.
     
  13. Jan 13, 2010 #12

    sylas

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    Sorry! You are quite right; I mixed up the names of my models. I meant what is sometimes called the "Einstein-de Sitter" model, which is confusing because neither Einstein nor de Sitter proposed it. I meant "dust", with mass, at critical density; not the massless test particles of the Milne model. My mistake.

    Ah! I had take the "static" to mean no peculiar motions, sometimes the taken as the defining quality of "dust". My apologies again. Yes, this model will contract from that static starting point, and you will have blue shifts. The gravitational effects can be considered as gravitational in the sense Wallace described, and I think we are all on the same page with that.

    My original remark about "cancellation" was strictly intended to refer to the effect of a photon passing near a massive object, in a inhomogeneous universe. Fritz Zwicky considered whether something like this could work (it is a form of "tired light" model). But it doesn't work. A localized patch of higher density matter along the photon's path has no net effect. You can think of the photon being blueshifted as it moves into the denser local region, and then redshifted as it moves back out again, with net cancellation as if that intervening clump of matter had not been there at all. Apart from a change in direction, possibly, as in gravitational lensing.

    Cheers -- sylas
     
  14. Jan 13, 2010 #13

    Chalnoth

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    Well, actually this is only the case for matter domination. In the case of some form of dark energy, there is some net effect because gravitational potentials decay with time: the photon has less of a well to climb out of on the way out than in.

    This is why I added the point that on average, due to the various underdensities and overdensities of the universe, these effects tend to cancel. In a more detailed analysis, they don't cancel exactly, but instead have some extra directional variation as a result (there's still no average effect when taken over the entire sky).
     
    Last edited: Jan 13, 2010
  15. Jan 13, 2010 #14

    sylas

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    That's an interesting idea... I had not thought of that. The effect would have to be phenomenally tiny in our universe, but I see how it could work. I wouldn't like to try and calculate it, however!

    Cheers -- sylas
     
  16. Jan 13, 2010 #15

    Chalnoth

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    Good stuff! It's known as the Integrated Sachs-Wolfe Effect, and basically it slightly increases the fluctuations in the CMB at large scales (at small scales the effect cancels more).
     
  17. Jan 13, 2010 #16

    Ich

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    That's why doppler redshift dominates. In fact, there is additional redshift due to a negative dark energy potential.

    Now I understand why you meant that the effects cancel on average. What cancels are the inhomogeneities (except our own cluster, of course). I'm talking about the total matter distribution, which adds a blueshift component to incoming light.
     
  18. Jan 13, 2010 #17

    Chalnoth

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    That doesn't make any sense to me. If you take, for instance, a closed universe, and take two times equidistant from the turnover point, there will be no net redshift or blueshift between them, whereas by your claim, one would expect a net blueshift.
     
  19. Jan 13, 2010 #18

    Ich

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    Sorry, I'm not sure I understand that phrase.
    If you mean a photon emitted dt before maximum expansion and received the same dt after maximum expansion:
    The distance r is then 2dt*c.
    In this case, you have a gravitational blueshift of
    [tex]
    \frac{2 \pi G}{3 c^2} \rho r^2
    [/tex]
    The coordinate acceleration of the emitter is
    [tex]
    \frac{4 \pi G}{3} \rho r
    [/tex]
    Since emitter and observer were at relative rest at turnaround, and the signal was sent dt = r/2c before, the relative velocity at the time of emission was
    [tex]
    dv=\frac{4 \pi G}{3} \rho r*dt = \frac{2 \pi G}{3c} \rho r^2
    [/tex]
    giving a redshift of
    [tex]
    \frac{2 \pi G}{3c^2} \rho r^2
    [/tex]
    which exactly cancels the blueshift above.

    Really, I'm not claiming new physics. This is simply a local Newtonian approximation to an FRW metric - weak field, small velocity, no pressure.
     
  20. Jan 13, 2010 #19

    Chalnoth

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    How do they have a net relative velocity, though? At emission, the emitter would have been moving away from the observer. But at the same time, since the system is symmetric, the observer would be moving towards the emitter by the same amount when the photon was observed, canceling that redshift.
     
  21. Jan 14, 2010 #20

    Chronos

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    Gravity works both ways, matter on the far side counters gravitational effects from the near side. A net zero effect. Expansion is the only logical explanation.
     
  22. Jan 14, 2010 #21

    Wallace

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    You are double counting somehow.

    Lets look at this in two ways. The simplest way is to place to origin of some co-ordinates at the reciever such that they remain fixed. Imagine a spherical region around them with the emmitter at the edge of that region. When they fire the photon towards the centre they are moving away from the reciever. Since the reciever is always fixed, this means there is a redshift from the original motion so it doesn't matter that later on the emmitter starts moving towards the observer when the Universe begins contracting. The gravitational blueshift, in this case, exactly cancels this original redshift. It looks like this:

    Motion at emmission causing a Doppler redshift
    Obs . . . . . . Em ->

    Photon is falling towards the bottom of the potential well, causing a blueshift
    Obs . . . . . . << Photon

    We can instead define the co-ordinates to be centred on the emmitter. In this case it remains fixed. If you think about this it means that compared to the rest frame of the emmitter, the observer will be moving towards the emmitter when the photon is observed. Thus you will have a blueshift due to motion. This might be confusing, until you realise that in these co-ordinates, the photon is moving away from the origin, climbing out of the potential well we have define, and therefore in this system the effect of gravity is to cause a redshift, in this case exactly cancelling the Doppler blueshift. It looks like this:

    Motion at reception, causing Doppler blueshift
    Em . . . . . . <- Obs

    Photon is climbing out of potential well, causing gravitional redshift
    Em . . . . . >> Photon

    We could also place the origin between the emmitter and observer. In this case the relative motion cancels out, so there is no Doppler contribution. But also, we now define the bottom of the potential well to be between the two, so the photon picks up a blueshift falling in, which exactly cancels the redshift of it climbing out. It looks like this

    Motion at emmission

    <-Obs . . . . . . O . . . . . . Em ->

    Motion cancelled at reception, no net Doppler effect

    Obs -> . . . . . . O . . . . . . <- Em

    Photon falls into potential well, gaining energy

    . . . . . . O . . . . . . << Photon

    But then loses the same amoung climbing out again

    << Photon . . . . . . O . . . . . .

    This might sound like a bit of mathemagic, but it is all just co-ordinate tricks with classical physics. As with any problem to do with energy, you have to be very careful about where you are defining the arbitary zero point, and make sure you are referencing everything consistantly with respect to that.
     
  23. Jan 14, 2010 #22

    Chalnoth

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    In any case, these things are vastly easier to understand if you just take them in co-moving coordinates, where both the emitter and observer are stationary (up to local peculiar velocities). In co-moving coordinates, the only source of redshift is the overall expansion, and so the redshift is simply:

    [tex]z + 1 = \frac{a_{obs}}{a_{emit}}[/tex]
     
  24. Jan 14, 2010 #23

    Wallace

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    But hang on, we know that we can always just use these co-ordinates. The question is what the hell do they mean? The OP asked how motion and or gravity is responsible for causing redshift, which is a very reasonable question. Simply stating the above equation tells you how to calculate it, but it doesn't tell you what that means and doesn't answer the question. Reducing everything to the effect of 'the overall expansion' leaves you at sqaure one; what precisely is that motion, and how does it cause redshift? In fact the 'motion' implied by looking at da/dt is nothing like the intuitive motion we see in day to day life, since it encodes gravitational effects as well. This is very very convenient for cosmologists, since it reduces everything to the single function a(t), but it is horrible for people new to the area trying to work out what that function means in terms that are familiar.

    Ich and I explained how you can understand the interplay between motion and gravity by looking at how the more familiar Newtonian physics gives you the same answer, but more obviously demonstrates how both motion and gravity are both at work, even in a homogenous universe.

    Writing down a simple relation, and really understanding what that means are two vastly different things.
     
    Last edited: Jan 14, 2010
  25. Jan 14, 2010 #24

    Chalnoth

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    I guess I just don't see those sorts of questions as very productive. There are so vastly many ways of looking at the situation that one can't say that they mean any one particular thing in these terms. So I'd rather just go by the simplest explanation, which is that the photons are expanded along with space.
     
  26. Jan 14, 2010 #25

    Wallace

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    Well then I have to disagree. When you say 'photons are expanded along with space' you are talking about something that is only true for one specific set of co-ordinates and you also imply a false causality; that there is a physical effect called 'expansion of space' which causes photons to stretch.

    Simply saying 'there are many ways of looking at this, so none of them mean anything' is not very useful. In fact, as has been explained, the physics is universal, and can be seen readily by looking at the Newtonian picutre, to which all co-ordinate descriptions will converge to for small distances. The co-ordinates are what are malleable, yet you want to fix on just one co-ordinate system and force the physics to conform to that (since you remove gravity and motion and invent a new placeholder fictious effect which acts for both). I'm afraid that is bass-ackwards.

    As can be readily evidenced in this forum, blanket use of this phrase without context leads to much wailing and nashing of teeth, such as 'why don't galaxies get expanded by space?' 'does the expansion of space drive electrons further from the nucleus of atoms?'. These are reasonable questions to ask when you've been told to just think of everything in terms of some ill-defined 'expansion of space' but the are easily done away with when you break it down into the simple underlying physics.

    Again, I go back to the OP. It was asked whether motion and/or gravity is responsible for the observed redshift of galaxies. How does writing down 1 + z = a/a_0 and saying 'the photons get stretched by expanding space' answer this question? Redshift can be understood in simple well understood terms like motion and gravity, I see no reason to force people to abandon these intuitive notions in favour of a co-ordinate dependant mathematical function which has no universal physical meaning.

    It depends on what we are trying to help people with. If you want to learn how to calculate cosmological quantities, then you need to learn the maths behind co-moving co-ordinates, and learn the easiest way to make calculations. If someone wants to a good non-mathematical intuitive understanding in terms of familiar concepts, then this is clearly not the best way to go.
     
    Last edited: Jan 14, 2010
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