Jump to content

redshift


michel123456

Recommended Posts

Here below a picture from the Wikipedia article about redshift:

 

Redshift.png

 

So I made a screenshot of this picture and imported into a Cad program.

Then I joinded graphically the end of the absorption lines, as accurately as possible, and ended with the following graph:

 

redshift.jpg

 

So I thought the Wiki picture is not good.

 

I found this other one from Brian P. Schmidt Nobel lecture page 5 ( http://www.nobelprize.org/nobel_prizes/physics/laureates/2011/schmidt-lecture_slides.pdf )

.

dopplershift1.jpg

 

And I did the same thing, importing in Cad, joigning the lines, and obtaining this below:

 

dopplershift.jpg

 

 

 

Then I did something else: i scaled the image down and put it above the original one:

 

scaledshift.jpg

 

With a spectacular result: the B line above the A is an exact match.

 

Just as if the redshift was not a shift, but a scale.

Edited by michel123456
Link to comment
Share on other sites

Found this http://tap.iop.org/astronomy/astrophysics/702/file_47550.doc page 2.

--------------------------------------------------------------

 

ScrShot056.jpg

 

-------------------------------------------------------------

Where it seems that I am correct.

 

if x axis represents wavelength then a change of scale is the expected result.

 

Why do you say that? For me a "shift" is a translation, not a scaling.

Edited by michel123456
Link to comment
Share on other sites

Why do you say that? For me a "shift" is a translation, not a scaling.

The Wikipedia link about Redshift posted in the OP gives the following formulae for cosmological redshift in an expanding universe:

 

[math] 1 + Redshift = \frac {Observed_{wavelength}} {Emitted_{wavelength}} [/math]

 

If we pick a few numbers like: 10, 20 and 40 for emitted wavelengths at the absorption lines and choose a redshift of 2 then we can see that:

 

Emitted

---------10--------20------------------40--

 

Observed

-----------------------------30----------------------------60----------------------------------------------------------120-------

 

When space is expanding and bringing objects more apart then the first and the last photon in a beam will get equally separated too.

(In a twice as long beam the photons will get separated twice as much as in the shorter beam.)

 

When the expansion is scalar then the difference between the absorbtion lines will also scale equally.

(The *shift* in the separation between the lines has equal rate determined by the redshift factor.)

Edited by Spyman
Link to comment
Share on other sites

  • 1 month later...

For me a "shift" is a translation, not a scaling.

 

This ambiguity is one of the reasons we rely on equations rather than descriptions. When one consults the equation, one sees the exact nature of the shift. "Shift" likely being used because we already had experience with changes in frequency/wavelength from the Doppler effect, which is called the Doppler shift.

Link to comment
Share on other sites

If I have to suppose that a single sound has multiple frequencies, then for a moving source the answer is yes.

 

But if one has to compare to what we are observing in the sky, one has to consider 2 objects at 2 different distances from the observer, going away at the same velocity compared to the observer and emitting a same sound of the same frequency.

 

In this case If I understand correctly the 2 sounds will have the same frequency shift (as heard by the observer), because the doppler effect is a function of velocity and not a function of distance.

 

---------

edit

 

So in the sky the doppler effect is caused by a difference in velocity. Please correct me if I am wrong.

Edited by michel123456
Link to comment
Share on other sites

If I have to suppose that a single sound has multiple frequencies, then for a moving source the answer is yes.

 

But if one has to compare to what we are observing in the sky, one has to consider 2 objects at 2 different distances from the observer, going away at the same velocity compared to the observer and emitting a same sound of the same frequency.

 

In this case If I understand correctly the 2 sounds will have the same frequency shift (as heard by the observer), because the doppler effect is a function of velocity and not a function of distance.

 

---------

edit

 

So in the sky the doppler effect is caused by a difference in velocity. Please correct me if I am wrong.

If you ONLY consider the doppler effect then yes.

 

But I think it has to be noted that when discussing objects in the "sky" and in the context of your latest threads on the subject, that there are other effects that also can cause a frequency shift. When talking about very distant objects, then cosmological redshift have a much larger impact on the observed shift than relativistic doppler, because galaxies normally move with speeds that are a fraction of light speed while the expansion brings them away with multiples of light speed.

Link to comment
Share on other sites

Realizing someone of you (at least on a short while) that starlight is spread through intergalactic lightconductive environment that is not superconductive and therefore has its "operational" losses? Realizing someone of you that no real physical environment is not lossless? In this case, is unnecessary Doppler effect and the redshift is explained a natural way, as "operational" mass/frequency losses of light passing through a lightconductive environment.

No doubt you were close to reality.

Link to comment
Share on other sites

But no cigar.

You seem to be talking about this

http://en.wikipedia.org/wiki/Tired_light

which is already known to be wrong.

 

Vain I wonder what mental process you've got to cigars? Congratulations. Fritz Zwicky, really hadn't their day in the formulation causes redshift. This is not because of the scattering of light in intergalactic gas, but the loss in inhomogeneous propagation medium. "Fatigue of light" of scattering doesn't cause crashes, but its passage through gravitational field. Any gravitational field, in principle, also by A. Einstein, distorts the Euclidean geometry, and the structure of intergalactic space homogeneity and thus the conductivity of the light! Absolutely lossless is just a Euclidean environment with complete homogeneity. Such, however, does not exist in the real universe. Fritz Zwicky at that time knew nothing of relict radiation.

So: As a result of the space gravitational environment it logically generated energyloss for EM waves. Detained energy with this environment logically "heats" and becomes to a secondary source. lowheat radiation - Relic Radiation. If Zwicky had knowledge of existence RR certainly would this connection realized. They are obviously two sides of the same coin: the primary "grinding" = redshift, and secondary radiation RR. If you ask of what elements constitutes that "grinding" environment, they are still hypothetical Smolin's and Socrates discrete elements of the physical vacuum. No formless ether.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.