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Doppler effect distinguish whether the red/blue shift

  1. Oct 8, 2004 #1
    when observing heavenly objects, there is an important role of doppler effect. but is there a way to distinguish whether the red/blue shift is because of translational, rotational motion or perhaps thermal motion of the atoms?
  2. jcsd
  3. Oct 8, 2004 #2
    In general, thermal and rotational (on the source's own axis) motion will cause a broadening of the line in the spectrum. Translation motion will cause a shift in the line altogether.
  4. Oct 9, 2004 #3
    yes but think about rotational motion:

    if you check the edge of the object, it will seem to have only a tangentional motion. there will be a red/blue shift depending on which side you look.

    generally what i want to know is whether it is clear for us that the wavelength of EM we observe is because of the motion of the star/galaxy or because of some internal factors depending on the composition(or anything else) of it
  5. Oct 9, 2004 #4


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    Hi serali, you forgot to mention the cosmological redshift. Most of the redshift we see is not a doppler effect: it is not due to "translational motion" of the source object. or any of the things you mentioned.

    For example, objects with redshift 6 are observed.
    this means that the wavelengths are longer by a factor of 7.
    Light which was emitted with a wavelength of 100 nanometers has,
    when it gets to us, a wavelength of 700 nanometers.

    Try to explain that by a doppler effect :smile:

    the basic method is essentially pattern recognition-----see the same pattern of spectral lines but shifted over to longer wavelengths.

    I agree with what Moose352 said:
    that is typically how it is, typically you just see a dot and you cannot resolve one edge from the other, so you can't separate out the light from the edge coming towards from the edge going away-----all you see is a fraction of a percent broadening of the line

    if you can image a galaxy sharply enough to resolve one edge from the other then of course you have a good handle on its rotation, but the doppler shift due to rotation is still going to only be like a small fraction of a percent because the rotation speed at the edge is typically a small fraction of a percent of the speed of light!

    a tiny fraction of a percent splitting or broadening of a line is WAY different from seeing the line shift by a factor of 2, or 3 or 7---like you get from cosm. redshift.

    it is easy for people to tell the difference!

    You ask:
    I guess the simple answer is yes, typically it is clear what to attribute the redshift to---at least the bulk of it. there is bound to be some small percent uncertainty and disagreement about the cause in some ambiguous cases but interpreting the main bulk of redshift is not at issue.
    (indeed some little bit of shift can be thermal, gravitational, due to rotation or radial motion etc, but except for nearby objects the effect of the expansion of space typically swamps these effects)
  6. Oct 9, 2004 #5
    The object which emitted the light is moving away from us at 0.96c.

    (Sorry, I'm feeling in an argumentative mood) :rolleyes:
  7. Oct 9, 2004 #6


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    Better recalibrate that red shift calculator. z = 6 equates to a recession speed of about 3.3 'c' at the time that light was emitted.
  8. Oct 10, 2004 #7
    That depends on how you define your coordinate system. (General relativity allows you a lot of freedom in doing this). I realise that cosmologists usually use a coordinate system which reflects the homogeneity of the universe. However if you decide to use a coordinate system in which the speed of light is constant then you would find that the velocity is 0.96c. My feeling is that this coordinate system makes more sense when you are comparing velocities at different places.

    (I told you I was feeling argumentative)

    In case you're interested, I calculated the velocity from the formula on
  9. Oct 10, 2004 #8


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    bravo to both of you gentlemen, both right in a certain sense

    Have to say that overall I remain in agreement with Chronos about this. it is simpler to stick with standard cosmology. Chronos has used a standard cosmology calculator and gotten the usual numbers corresponding to that redshift. easier to imagine. when I try to think along chronon universe lines my mind gets boggled.

    However it is nice to have the relativistic doppler formula recalled here!
    and we can try to see if coordinates you propose would be useful for cosmology (personally I would find it awkward or unusable, will say why later)

    the doppler formula of special rel is just what your link says:

    [tex]1 + z = \text{ratio of wavelenths} = \sqrt{\frac{1 + \beta}{1 - \beta}}[/tex]

    and for convenience call the ratio = r and solve

    [tex]\beta = \frac{r^2 - 1}{r^2 + 1}[/tex]

    so we were considering the ratio = 7 , that is light emitted at 100 nm was arriving with wavelength 700 nm and that means
    beta should be 48/50

    this is just what chronon said, 96 percent of speed of light!

    But chronon, if I understand, according to you we should imagine we live in a non-expanding space, the flat Minkowski space of special relativity.
    the entire redshift is caused by the speed the galaxy had when it emitted the light and space did not expand while the light was traveling to us (otherwise it would have redshifted it some more still besides what we assumed)
    Also the length of time the light took is exactly represented by the distance that the galaxy was at the time it emitted the light.

    BTW what distance would you associate with that z = 6 redshift?
    I think we need an alternative cosmology calculator. something that you type in a redshift and it tells you the object's distance and the speed away from us at the time the light set out. and which assumes no expansion in the meantime because it is working in the 1905 special relativity space.

    I guess in this case the Hubble Law, v = Hd, would not be true.
    It would only be valid as an approximation at small distances.
    there would be no FRW metric--such as cosmologists are fond of using.
    And there would be a center, away from which the galaxies were scattering. Perhaps WE would be at the center. where we are would be assumed as the point where the "big bang" occurred and it would, in this picture, be a kind of material explosion.

    Actual big bang theory and inflation scenarios would not work because they depend on a different theory of space-time, 1915 General.
    And one would not know how to have accelerating expansion.
    there would be no place for the cosmological constant or dark energy.

    So a lot of usual topics go out the window, but one could depict an expanding universe in a crude way, as galaxies scattering apart in ordinary non-expanding flat Minkowski space. One could make a portrait of the present moment, even though it might not handle stuff like big bang or dark energy or acceleration very well.

    Well, ....chronon you say you wrote this in an argumentative mood. I am not feeling at all argumentative. I just cannot cope with this because the only good theory of gravity I know is Gen Rel and it does not work in Minkowski space. So altho you can make a trompe l'oiel
    lifelike picture of the present it doesnt work for me.

    I guess you have to use newtonian gravity in it and the galaxies are constantly slowing down by their own gravity. A cloud of galaxies with us at the center perphaps that somehow got started flying apart.

    how did they ever get started moving 96 percent of speed of light?
    hope I am not misunderstanding you, chronon, and that my response not overly naive
  10. Oct 10, 2004 #9
    What I am saying is that saying that space is expanding and saying that the galaxies are flying apart are just different ways of saying the same thing. More precisely the spacetime metric (and so the physics) is the same in both cases, it is just the coordinate system that differs.

    The point I am trying to make is that people often see physical effects in this choice of the coordinate system which is wrong. For instance, we can think of the space stretching leading to the redshift of light, and so reducing the energy of photons from a galaxy. That may seem the best way of thinking about it. However, suppose a space probe was sent from the galaxy towards us at 0.99c (from the galaxy's point of view). When it reaches us it is travelling much slower than this with respect to us. Do we say that it has always been travelling at the slower speed with respect to us or has its energy somehow been 'sucked out' of it by the expanding space - implying a physical effect which isn't really there.
  11. Oct 16, 2004 #10
    ok now a related question(i think its unnecessary to start a new thread):
    what is the difference between emission and absorption spectra. i read smt in some astrophysics books but they were very confusing for me.
    Last edited: Oct 16, 2004
  12. Oct 16, 2004 #11


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    If you heat a material until it glows, and take a spectrum from that, that's an emission spectrum. Typically it consists of a series of bright, separated lines in the different colors. Suppose now that you vaporize that same material, turning it into a gas, and arrange to have that gas between a source of white light (say an arc) and the spectrograph. Then you will see a continuous spectrum crossed by a series of dark lines. This is the absorption spectrum. And the dark lines will be in the same place as the bright ones were in the emission spectrum.
  13. Nov 9, 2004 #12
    Where [tex]\beta[/tex] = v as % of c
    and [tex]r[/tex] = z+1 (redshift + 1)

    Marcus, I have to admit it is easier for me to understand Red Shift using SR as chronon pointed out, and your ratio formula.
    But then I'm only starting to extend my understanding from SR to GR.

    Maybe better understanding the " Red Shift Calculator" will help see where I'm need to go. For a red shift of z = 6 my Newtonian math tells me that the "apparent" speed to stretch the wavelength from 100 nm to 700 nm needs to be 6.0'c'.

    Yet Chronos points out that a better "red shift calculator" (GR based I assume) equates a z = 6 to a recession speed of about 3.3 'c'.
    Does that leave me with a 2.7'c' motion speed to account for?
    I must be missing something.

    Can you help me understand a Cosmology (ie. GR?) based "red shift calculator" ?
    Assuming a linear red shift of z=6,
    And assuming a motion speed of 0.8c.
    How do I calculate the recession speed?
    Or do I need to consider a range of speeds for these since a Red Shift by it self would not be enough information to detail exact distances and speeds.

    On a side note what is a realistic color (wavelength) to use in Cosmology?

    Do we know what the starting generated wavelength from the origin of the background microwave radiation was? ... The wavelength of glowing hydrogen maybe?
  14. Nov 9, 2004 #13


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    For discussion of super luminal recession see

    Superluminal Recession Velocities

    If super luminal recession is unpalatable or just too confusing, try these alternatives:

    A New Redshift Interpretation

    Remarks on the "New Redshift Interpretation"

    The New Redshift Interpretation Affirmed

    A Relativistic Description of Gentry's New Redshift Interpretation
  15. Nov 9, 2004 #14


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    RandallB, I think I agree with Chronos here. I have to go to the redshift calculator to make sure.
    Yes, the redshift calculator does employ GR-------as I imagine you realize, SR is not so useful over long distances, it's more of a local geometric tool.

    I also agree with Chronos when he says to look at this:

    Superluminal Recession Velocities
    Tamara M. Davis, Charles H. Lineweaver
    4 pages, 2 figures, Cosmology and Particle Physics 2000 Conference Proceedings

    Abstract: "Hubble's Law, v=HD (recession velocity is proportional to distance), is a theoretical result derived from the Friedmann-Robertson-Walker metric. v=HD applies at least as far as the particle horizon and in principle for all distances. Thus, galaxies with distances greater than D=c/H are receding from us with velocities greater than the speed of light and superluminal recession is a fundamental part of the general relativistic description of the expanding universe. This apparent contradiction of special relativity (SR) is often mistakenly remedied by converting redshift to velocity using SR. Here we show that galaxies with recession velocities faster than the speed of light are observable and that in all viable cosmological models, galaxies above a redshift of three are receding superluminally."

    what Chronos and also Lineweaver/Davis are telling you is consensus cosmology. Lineweaver is on the short list of worldclass mainstream cosmologists.

    There's no law that says you have to believe today's consensus of working cosmologists, but it is a good idea to understand it so that if you try some variation on it you do so consciously.

    So I suggest you learn how to use the redshift calculator
    the URL is in the A and C reference thread----one of the stickythreads at the head of the forum
  16. Nov 9, 2004 #15


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    quick path to Siobhan Morgan's calculator!

    Try this!
    Just google the two words
    Cosmos calculator

    I tried it and I got the best redshift calculator I know which is
    of a professor of astronomy at an Iowa university.
    She is named Siobhan Morgan

    You have to type in today's estimates of dark-energy-density Lambda (0.73)
    and the matter-density Omega (0.27)
    and the Hubble parameter (71)

    But typing in three numbers is no big deal. you get used to it.

    then whenever you type in a redshift, symbol z, you get information about the galaxy that emitted the light.

    Please see if googling "cosmos calculator" works for you. I just tried and it hit.
  17. Nov 9, 2004 #16


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    well I tried Morgan calculator on redshift z = 6
    and it gave that the recession of the thing (when it emitted the light)
    was 2.75 c

    so that is roughly same as Chronos-----he said something around 3 c.

    It matters what parameters you put into the calculator
    with Morgan you should put in

    and for z you put in 6 if you want redshift 6.
  18. Nov 9, 2004 #17


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    If you forget to change the default setting, you get 3.3c... :blushing:
  19. Nov 10, 2004 #18


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    The formulas to calculate the recession speed for a given redshift are not difficult to obtain, but you should be a little bit familiar with the Friedmann equation:

    [tex]H^2 = \frac{8 \pi G} {3 c^2} (\rho_m + \rho_r) - \frac{k c^2} {a^2} + \frac{\Lambda c^2} {3}[/tex]

    with [tex]\inline{H = \frac{\dot{a}}{a}}[/tex] the Hubble parameter, [tex] \inline{\rho}[/tex] the energy density of matter and radiation (the energy density of radiation can be neglected), [tex] \inline{k}[/tex] the curvature parameter and [tex] \inline{\Lambda}[/tex] the cosmological constant.

    The steps to obtain the expressions are, first, define the parameters which determine the cosmological model:

    [tex]\Omega_m = \frac{8 \pi G} {3 c^2 H^2} \rho_m[/tex]

    [tex]\Omega_r = \frac{k c^2} {a^2 H^2}[/tex]

    [tex]\Omega_\Lambda = \frac{\Lambda c^2} {3 H^2} [/tex]

    Substitute in the formula and evaluate H for the current cosmological time as H0 (which does also determine the cosmological model).

    Further on, consider the evolution of matter as:

    [tex]\frac{\rho_{m0}}{\rho_{m}} = (\frac{a}{a_0})^3[/tex]

    and consider the definition of redshift [tex] \inline{1+z = \frac{a}{a_0}}[/tex], to express da as a function of dz and, further on, dt as a function of dz considering [tex]\inline{dt = da/\dot{a}}[/tex]...

    One arrives to an equation:

    [tex]dt = \frac{-dz}{(1+z) H_0 \sqrt{f(z)}}[/tex]

    You can integrate this between z = 0 and the given redshift of the galaxy to get the look-back time. The distance can be obtained measuring along the path of a radial photon considering that in the RW metric:

    [tex]0 = - c^2 dt^2 + a^2dr^2[/tex]

    [tex]dr = \int \frac{c dt}{a}[/tex]

    substituting dt with the equation above for dt, and substituting a with the definition of redshift you should get:

    [tex]D = \frac{c}{H_0} \int \frac{dz}{\sqrt{f(z)}} [/tex]

    For the general case, this must be integrated numerically. For special cases (such as a flat space) one can simplify things and calculate the integral.

    If you have the distance, the recession speed can be calculated with the Hubble law v = H D, with H the current value of the Hubble parameter (H_0).

    If you want to make a simple script to compute this an other values for the general case, another (similar) derivation can be used, which simplifies things. If you are interested, you may ask me. I have a javascript which is shorter, far more simple and more intuitive than Ned Wrights one – which is the one used by Siobahn Morgan (I also have no problem to post the code or the link here if others are also interested).
  20. Nov 10, 2004 #19
    Yes, well I'm one of the people who invoke special relativistic calculations. The point is that GR allows choice of the coordinate system for a given metric. Cosmologists seem over-attached to one coordinate system, and so believe that there is one 'correct' velocity of a receeding galaxy. I think that when you are considering velocities, an SR like coordinate system might be more appropriate. Note that a similar problem arises when mapping the globe, one projection preserves area, another one preserves angles.

    This overattachment to one coordinate system leads people to see a physical effect where there isn't one. I have written about this on the following webpage: www.chronon.org/articles/stretchyspace.html

    Moving on to fig 2 in the above paper, we are all agreed (I hope) that its a myth that the subluminal/superluminal boundary is some sort of horizon. The events happening within the light gray area will eventually be seen by us.
    I would say that there is a problem with the way the graph is presented, in that it seems to suggest that events in the white area are behind the particle horizon, i.e. we will never see them, and galaxies suddenly appear across this horizon. This is not the case. We will also eventually see the events in the white area as well.
    Last edited: Nov 10, 2004
  21. Nov 10, 2004 #20


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    it's neat you made your own. If I had the computer gumption I would try it.
    As it is I only use other people's javascripts that are accessible online.

    I saw somewhere that Morgan's calculator was one that was published in Sky and Telescope magazine. do you know if that is right?
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