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Luminosity distance

by Myslius
Tags: distance, luminosity
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Myslius
#1
Aug25-13, 03:56 PM
P: 98
Explain the difference between distances.



Red lines indicating light travel path. Green lines indicating distance.
Why there is length difference between green lines?
In this diagram object moves at 0.8c away from the observer in t FOR.

How far away observer in t' FOR sees an object (t FOR) when t'=3?
Which green line indicates luminosity distance? I know that luminosity distance should be measured by the bottom green line. But... If the object moves away from us and we emit light for example when t=1 (in this diagram), the number of photons hitting the moving observer (t') should be proportional to the green line above by R^-2 law, because it takes time for light to catch moving observer (t'). Also, luminosity distance should not depend on 3+1 rotation (velocity) at the moment when light hits it (ignoring redshift), moving observer can change his rotation just before light hits him (at t=2.9999), for example making relative velocity v=0.

Here's a similar diagram where t' makes a rotation just before light hits him:

speed is 0.6c in this case, and rotation just before t'=2.
At t'=2, t' observer sees t 1.5 away and t=1 :) FTL, none the less.

If we take that luminosity distance is not a green line below, but a green line above we get following formulas:


Which is really close to what we see with the universe expansion.
No configurable parameters, except time which is unavoidable. Static universe, relativistic motion. Also it explains few phenomenas how some objects (like quasars) become so big so fast. t_observed is always equal to t/(z+1) in relativistic motion.

Strangely enough, the ratio between two green lines is the change of scale factor or = (z+1)=√((1+β)/(1-β))
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Mentz114
#2
Aug25-13, 05:32 PM
PF Gold
P: 4,087
Are you talking about this

Luminosity distance - Wikipedia, the free encyclopedia or radar distance ?
Myslius
#3
Aug25-13, 05:41 PM
P: 98
Yes, not radar distance. Radar distance measurements include light coming back. There is no problem with that. All according to standard relativistic understanding. For example, in first diagram, if we send a beam of light at t=1, we will receive a signal back at somewhere around t=9 and we will see a distance of around 4 and t1's time of 3, All good.

Mentz114
#4
Aug25-13, 06:52 PM
PF Gold
P: 4,087
Luminosity distance

Quote Quote by Wiki
Luminosity distance DL is defined in terms of the relationship between the absolute magnitude M and apparent magnitude m of an astronomical object.
I think what you have derived is the doppler shift between moving bodies. Obviously this is related to z, but I don't see any connection with the definition above.
Myslius
#5
Aug25-13, 07:31 PM
P: 98
I'm arguing about, that you need doppler^2 for luminosity distances which is not standard understanding. Because there is an paradox there, simply by looking from different frame of references.
yuiop
#6
Aug25-13, 08:35 PM
P: 3,967
Have you taken relativistic beaming into account? This is also known as the headlamp effect. If a bright object is moving in a given direction, the light is focused towards the front due to relativistic aberration. In our case the bright objects are moving away from us and the relativistic aberration makes the photons spread out and more dilute/dimmer than a simple 1/r^2 and relativistic Doppler effect would suggest.
Myslius
#7
Aug26-13, 12:31 PM
P: 98
This one explains quite good:

Relativistic aberration results in brightness/luminosity reduction by a factor (1+z)^-2 due to deflection of light rays; reddening of photons together with time dilation (elongation of time interval between the photons arriving to the detector) give another factor (1+z)^-2, totaling in luminosity reduction proportional to 1/D2(1+z)^4 and resulting in apparent luminosity distance increase DL = D (1+z)^2. It means that with respect to the distance D΄ in the restframe and, correspondingly, to the distance to the disk at the moment of detection the flux reduction factor is proportional to 1/ D΄^2(1+z)^2 and the luminosity distance DL = D΄(1+z). Thus, the effect of relativistic aberration “shifts” an object from its distance from the apparent size D to the distance D(1+z) that corresponds to the distance to it at the moment of detection equal to the distance in the restframe D΄, and the redshift together with time dilation shift it even farther by another factor (1+z).
As for the non-resolvable objects, their geometric images are much smaller then the focal spots produced by the diffraction and optical aberrations of optical systems. From the point of view of optical system, these objects have no size and their visible angular size has no sense together with the effect related to the increase of their visible size. With no angular size there is no apparent angular size enlargement. For a pointlike object located on a line of sight, all original rays are directed out of the line of sight and, therefore, all the rays are further deflected from it by the relativistic aberration effect resulting in actual diminishing of apparent luminosity measured by the illuminance of focal spot with its size determined by diffraction and optical aberrations of optical system. The effect of relativistic aberration results in reduction of apparent luminosity of the receding source, in decreased flux of photons incident on the telescope’s aperture, in decreased illuminance of the diffraction spot of an optical system, and in our judgment from this illuminance reduction about the apparent increase of the luminosity distance to the source by a factor (1+z). Together with redshift and time dilation the luminosity distance DL = D (1+z)^2.
I have seen very unrealistic graphs like:


This one looks closer, but still, i have no idea how they calculated 24.2 for z=1:

My calculations show that it should be 24.775 which is way closer. Giving a 5-6 Glyr luminosity distance difference from the results provided here.
Mentz114
#8
Aug26-13, 03:32 PM
PF Gold
P: 4,087
This post belongs in the Astronomy sub-forum. Why don't you ask a mentor to move it ?
yuiop
#9
Aug26-13, 08:29 PM
P: 3,967
Quote Quote by Myslius View Post
Explain the difference between distances.



Red lines indicating light travel path. Green lines indicating distance.
Why there is length difference between green lines?
In this diagram object moves at 0.8c away from the observer in t FOR.

How far away observer in t' FOR sees an object (t FOR) when t'=3?
Sticking to the strictly Special Relativistic aspects of the question, I think you are reading this diagram wrong.

In the t FOR the light left the blue source at approx t=1.6 (and t'=1) and arrived at t=3 having travelled a distance of approx Δx=1.3 (The length of the lower green line.) t can estimate if the source continued at a constant speed that the source is at a distance of 2.4 units away 'now' even thought it cannot directly see it now.

In the t' FOR, the light left the black source at approx t'=1.6(and t=1) and arrived at t'=3 having travelled a distance of Δx'=1.3 (This is not the length f the upper green line.) t' can estimate if the source continued at a constant speed that the source is at a distance of 2.4 units away 'now' even thought it cannot directly see it now.

Δx' is obtained by drawing a line parallel to the tilted d' axis and through t=1,x=0, or by transforming the whole diagram to the t' FOR. The upper green line is the distance to the blue object when t=5 in the t FOR and not the distance as measured by the t' FOR at t'=3. If you look at the measurements I have quoted above for the t and t' FORs, you will see they are identical except for swapping the primed and unprimed variables and swapping black for blue. There is no sense of one observer being the one that is actually moving away or being the one that is actually stationary.
Myslius
#10
Aug27-13, 04:05 AM
P: 98
That's exactly what i read, when i read distances. But I'm interested in luminosity distances.
Suppose t emits light when t=1, in t FOR light catches up moving object at t=5 and d=4. In t' FOR it's t'=3, d'=1.33.
Let's say the number of photons hitting an object at d=1 is 1 photon per second. In t FOR the number of photons hitting the moving object is (1/16)/s due to the distance, and (1/9) due to Doppler effect, resulting in a total of (1/144). Luminosity distance is 12. t' should observe that luminosity distance. Light travel path in t' FOR is equal to the bottom green line in t FOR. In this case, light travels a distance of 1.33, Doppler effect 3, resulting in luminosity distance of 4. Also if we take additional Doppler effect because of this aberration effect, we get a symmetry here, and t' sees luminosity distance of 12 too.
LumD=D*(z+1)^2, both observers at t=3 sees each other D=1.33 away, and both measure luminosity distance LumD=12
yuiop
#11
Aug27-13, 05:13 AM
P: 3,967
Quote Quote by Myslius View Post
That's exactly what i read, when i read distances. But I'm interested in luminosity distances.
As Mentz mentioned, the question is perhaps better asked in the astrology (or cosmology) sub forums where they are more familiar with the terminology you are using. To help us here, could you define what you mean by T(now), T(observed), D and LumD in terms of your t/t'/d/d' diagram at the start of this thread. For example is T(observed) the time the light was emitted as measured in the t' FOR? Is T(now) the coordinate time that the light is received? Is T an interval of time (eg the time between initial observation of a supernova and the time of its peak luminosity) or simply a coordinate time? How do you measure time emitted directly? Usually what is measured is the frequency received and the redshift. Is D measured in t or t' FOR?

Quote Quote by Myslius View Post
Suppose t emits light when t=1, in t FOR light catches up moving object at t=5 and d=4.
Agree.
Quote Quote by Myslius View Post
In t' FOR it's t'=3, d=1.33.
If you mean d'=1.33 then yes.

Quote Quote by Myslius View Post
Let's say the number of photons hitting an object at d=1 is 1 photon per second. In t FOR the number of photons hitting the moving object is (1/16)/s due to the distance, and (1/9) due to Doppler effect, resulting in a total of (1/144). Luminosity distance is 12. t' should observe that luminosity distance. Light travel path in t' FOR is equal to the bottom green line in t FOR. In this case, light travels a distance of 1.33, Doppler effect 3, resulting in luminosity distance of 4. Also if we take additional Doppler effect because of this aberration effect, we get a symmetry here, and t' sees luminosity distance of 12 too.
LumD=D*(z+1)^2, both observers at t=3 sees each other D=1.33 away, and both measure luminosity distance LumD=12
This does not agree with your equations for D in your first post, but I assume that is because you have since added in a factor for aberration effect. I think it might help if you posted a transformed diagram boosted to the t' FOR.
Myslius
#12
Aug27-13, 05:33 AM
P: 98
yuiop, I've asked to move this thread to Astronomy sub-forum.
Yes, i mean d'=1.33, I've made a fix just before you posted your answer.
And you are right about D, i have added one Doppler to it. D, in this case, means light travel distance from emitter's FOR to the moving observer. Less confusing and more convenient formula would be by dividing that D by the Doppler effect.
yuiop
#13
Aug27-13, 05:53 AM
P: 3,967
Quote Quote by Myslius View Post
yuiop, I've asked to move this thread to Astronomy sub-forum.
Yes, i mean d'=1.33, I've made a fix just before you posted your answer...
Just be warned that they will probably be more comfortable in discussing luminosity in terms of the FRW metric where space itself is expanding and all objects (without peculiar motion) are stationary with respect to that metirc and not too comfortable with discussing relativistic motion in a flat universe. Redshift in the FRW metric occurs due to stretching of space and stretching of wavelengths during travel time, rather than due to classic Doppler shift and time dilation. So if have you have any questions directly relating to the relativistic aspects you should ask them here.
Mentz114
#14
Aug27-13, 09:06 AM
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P: 4,087
Quote Quote by yuiop View Post
As Mentz mentioned, the question is perhaps better asked in the astrology (or cosmology) sub forums
I did not write that ! I write 'astronomy'. Your planet must be in opposition to mine

But you're right to say that luminosity distance is model dependent.
yuiop
#15
Aug27-13, 11:58 AM
P: 3,967
Quote Quote by Mentz114 View Post
I did not write that ! I write 'astronomy'. Your planet must be in opposition to mine

But you're right to say that luminosity distance is model dependent.
LOL! Oops! Sorry for the typo ... but good response I have a friend who is heavily into astronomy and has a large telescope in a observatory he built himself. It annoys him intensely when we ask how he is getting on with his 'astrology' in his 'conservatory', so of course we always do ... but I guess that little joke has become a habit that has backfired on me now.


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