Galaxy one is located in a dense area of the universe and galaxy two is located in a less dense part of the universe. Would galaxy one appear red-shifted to galaxy two? Is the mass density at our position in the universe less dense than most other positions?
6
-
12$\begingroup$ Jen, I get the feeling at time that you think science is a collection of unconnected facts and are just flailing around asking a lot of questions without much structure. To really make progress you have to stop jumping from subject to subject and focus on getting some of the basics down firmly. Once you've done that these things will be much easier for you, but you need to start from the beginning. A side effect of that is that you need to take you focus off the "big problems" and learn the tools that put those big problems into focus. $\endgroup$– dmckee --- ex-moderator kittenCommented Dec 10, 2016 at 2:18
-
1$\begingroup$ Does this mean that the question is now off the table thanks to the advice of @dmckee? $\endgroup$– freecharlyCommented Dec 10, 2016 at 2:59
-
7$\begingroup$ If I meant the question couldn't go ahead I would have closed it. That was by way of advice to the avid student. $\endgroup$– dmckee --- ex-moderator kittenCommented Dec 10, 2016 at 3:09
-
1$\begingroup$ @Muze considering density I wonder how much time dilation would play apart. If density caused Time to move slower from your point of view, how would you view distant galaxies? And then considering the time that light took to cross the void where time moves quicker. It seems to me with the right combination you could even get red shift if the universe were collapsing. $\endgroup$– Bill AlseptCommented Dec 10, 2016 at 3:36
-
4$\begingroup$ I'm a bit troubled by the downvotes to this question. Yes it's a beginners question, but then how many working physicists have heard of the Sachs Wolfe effect? $\endgroup$– John RennieCommented Dec 11, 2016 at 6:05
|
Show 1 more comment
3 Answers
$\begingroup$
$\endgroup$
3
The answer to your question is yes, but the gravitational redshift could not be confused with cosmological redshift because it is small.
The basics of gravitational redshift can be grasped by the simple approximation that the redshift is given by $$z \simeq \Delta \Phi /c^2,$$ where $\Delta \Phi$ is the difference in gravitational potential between where the photon was emitted and where it is absorbed/detected. This approximation is valid, where the ratio of gravitational potential energy to rest mass energy is small - so practically all circumstances except near black holes and neutron stars.
What this means is that if a photon is emitted from deep inside a potential well and then is detected higher in a potential well, a positive $z$ or redshift would be observed. If you reverse the experiment, then a blue shift would result.
If a photon travels into a galaxy, through it and then out the other side, there is no net redshift (unless the gravitational potential somehow changes whilst the photon is travelling, which is actually the Sachs-Wolfe effect referred to byJohn Rennie).
Now, coming to your scenario. You are specifying that the photons are emitted in a deep potential well, then detected in a shallower potential well. This results in a positive $\Delta \Phi$ and a net redshift. However, as the photon travels across the universe it is affected by the expansion and gets further redshifted. Some orders of magnitude would be helpful:
Suppose a galaxy is in a dense, spherically symmetric cluster, with $10^{14}$ solar masses interior to it, and at a distance from the centre of $\sim 1$ Mpc. If $\Phi \sim GM/r$, then a photon emitted by the galaxy will have an observed redshift at infinity of $GM/r c^2 = 5\times 10^{-6}$, equivalent to an apparent recession velocity of only 1.5 km/s. Meanwhile, the apparent recession velocity due to the expansion of the universe is $H_0 d$, where $H_0 = 70$ km s$^{-1}$/Mpc and $d$ is the distance in Mpc. Thus, even if the cluster of galaxies was only a fraction of a Mpc away (like the Andromeda galaxy), the gravitational redshift would be completely swamped by the redshift due to the expansion of the universe. In addition, we have not subtracted the blue shift caused by the photons falling into the gravitational potential of our own Galaxy (maybe 0.1 km/s) or the fact that galaxies have peculiar motions within a cluster that are of order hundreds of km/s!
The Milky Way is not in a particularly dense part of the universe. The local group is quite sparse, and we are about 20 Mpc from the nearest dense cluster (Virgo).
-
1$\begingroup$ This answer suggests that the descending photon somehow gains energy. It does not. If you send a 511keV photon into a black hole, the black mass increase is 511keV/c². $\endgroup$ Commented Dec 11, 2016 at 14:14
-
1$\begingroup$ There is no contradiction in that. I will change my wording slightly to avoid giving the impression that the redshift is something "experienced" by the photon. $\endgroup$– ProfRobCommented Dec 11, 2016 at 15:35
-
1$\begingroup$ Rob : the important point is that the redshift is observed. When I accelerate you in free space away from a photon source you observe a redshift because you changed, not because the photons changed. In similar vein when I lift you up in a gravitational field away from a photon source you observe a redshift because you changed, not because the photons changed. $\endgroup$ Commented Dec 11, 2016 at 16:55
$\begingroup$
$\endgroup$
1
Yes, you're quite correct that galaxies in a region of the universe with greater than average density would appear redshifted to observers outside that region.
In fact this is the origin of the Sachs-Wolfe effect, which is an important way that we study various properties of the universe.
Actually doing the calculation is hard because of course the universe is expanding and the rate of expansion changes with time. We need to take this into account and make sure that we exclude the cosmlogical red shift from our calculation. However almost everywhere in the universe the gravitational fields are weak enough that we can use an approximation called the weak field limit.
One of the ideas that physics students get taught early on is gravitational potential energy. You have probably been taught that for small distances it's:
$$ \text{PE} = mgh $$
though in GR we're normally interested in the potential energy per unit mass i.e. we set $m=1$ to get:
$$ \Delta \Phi = gh $$
And for large distances from a spherical body it's given by Newton's equation:
$$ \Delta \Phi = -\frac{GM}{r} $$
This also applies to galaxies though the equations get more complicated, and it also applies to groups of galaxies and galaxy clusters. In fact measuring the potential energy of galaxy clusters was how Fritz Zwicky first discovered the existence of dark matter.
The point of all this is that in the weak field limit the difference in gravitational potential between two places is directly related to the difference in the time dilation between those places by:
$$ \frac{d\tau_a}{d\tau_b} = \sqrt{1 - \frac{2\Delta \Phi}{c^2}} \tag{1} $$
So if you calculate the difference in the potential energy between the overdense region and the underdense region and feed it into equation (1) then it will give you the relative time dilation. Since in general $\Delta U$ will not be zero that means in general this is a difference in the time dilation and one galaxy will appear red shifted relative to the other. The reverse is of course also true - one galaxy will appear blue shifted relative to the other.
Lastly you ask:
Is the mass density at our position in the universe less dense than most other positions?
And the answer is that we seem to be about average. We're in the outer edges of the Virgo galaxy cluster, so we're neither in the overdense region at the centre of a cluster nor in the underdense region in a void.
answered Dec 10, 2016 at 11:30
-
$\begingroup$ @John Rennie. Actually the Sachs Wolfe effect winds up being a third of the naive analysis. See for instance the Ned article at ned.ipac.caltech.edu/level5/Glossary/Essay_sachswolfe.html. There's other more formal papers, and the original paper. The basic reason is that there's actually two effects, the gravitational one and the change in scale of the universe, a, which winds up subtracting 2/3 of the former for a matter dominated flat universe. Otherwise this is a nice answer and good enough for people who want to understand why those are observed. $\endgroup$– Bob BeeCommented Dec 13, 2016 at 5:13
$\begingroup$
$\endgroup$
11
Can a photon that is emitted from a denser part of the universe to a less dense part be redshifted?
I'm not quite clear what you're asking, but I'll say no. One example of a photon moving from a denser part of the universe to a less dense part is gravitational redshift, which is also called Einstein shift. But people who don't understand relativity often get this wrong. See what Einstein said:
"An atom absorbs or emits light at a frequency which is dependent on the potential of the gravitational field in which it is situated".
The photon is emitted at a lower frequency when it's lower down. It is emitted already redshifted. It doesn't get redshifted as it ascends. In similar vein an intergalactic photon doesn't reduce in frequency when it moves to a less-dense part of the universe.
Galaxy one is located in a dense area of the universe and galaxy two is located in a less dense part of the universe. Would galaxy one appear red-shifted to galaxy two?
Yes it would. Because photons in that denser region are emitted at a lower frequency than in less-dense regions.
Is the mass density at our position in the universe less dense than most other positions?
Not as far as I know. If it was, we might expect to see the same redshift for galaxies at different distances. We don't, we see redshift increasing with distance, in line with an expanding universe. By the by, my understanding of general relativity tells me that space just has to expand. It can't contract, or stay in a steady state. The mystery of course is why Einstein didn't predict an expanding universe.
answered Dec 10, 2016 at 15:55
-
8$\begingroup$ What do you mean by this: "The photon is emitted at a lower frequency when it's lower down". The emitted frequency is lower than what? What is correct and phrased in terms of observations is: an observer who sees the photon emitted will record a different (higher) frequency then an observer who sees the photon after it has climbed out of the gravitational well. $\endgroup$– AndrewCommented Dec 10, 2016 at 16:11
-
1$\begingroup$ @Andrew : lower than the frequency of a photon emitted by an identical emitter at a higher location. What you say about observation is correct, but the 2nd observer sees the photon redshifted because his clocks are running faster at the higher location, not because the photon frequency reduced during its ascent. There is no mechanism by which the E=hf photon loses energy as it climbs. However we do work on the observer when we lift him up to the higher location. We add energy to him. So to him, the photon appears to have lost energy. But it hasn't. Energy is conserved. $\endgroup$ Commented Dec 10, 2016 at 16:24
-
2$\begingroup$ I could use the peaks of the wavefronts of a laser as the basis of a clock (for example if I trap the laser in a box). You say clocks higher up in a gravitational well run slower than clocks in the well. My clock is counting how many laser periods have passed. So higher up in a well the period of the light in my clock must Increase--or the frequency must decrease. You have to be careful about energy in GR, for the Schwarzcschild metric the conserved energy of a photon is $E=\frac 1 2 \hbar \omega (1-(r_s/r)^2)$. For $E$ to be conserved as $r$ increases, $\omega$ must decrease. $\endgroup$– AndrewCommented Dec 10, 2016 at 21:18
-
1$\begingroup$ Apologies I was working from memory, the energy (for $r\gg r_s$) is $E=\frac 1 2 \hbar \omega (1-(r_s/r))$ (no 'squared'). The point about the gravitational redshift is the same. $\endgroup$– AndrewCommented Dec 10, 2016 at 21:39
-
1$\begingroup$ @Andrew : clocks run slower when they're lower, and we use the peaks of the wavefronts of Caesium hyperfine spinflip microwaves to define our second. When 9,192,631,770 microwaves have passed you by, you declare that a second has elapsed. The frequency of the microwaves is then 9,192,631,770 Hz by definition. Repeat this at a higher elevation, and the frequency of the microwaves is still 9,192,631,770 Hz by definition. At the higher location the second is shorter because the microwaves go faster. You could do this in the guise of a maser in a box and the same principle applies. $\endgroup$ Commented Dec 11, 2016 at 14:00