Register to reply

Velocity and redshift

by TrickyDicky
Tags: redshift, velocity
Share this thread:
TrickyDicky
#1
Oct7-10, 04:45 PM
P: 3,035
What is the formula used to convert the measured redshift into a velocity?, not the approximated formula for low speeds v=cz , but the more general and accurate one.

Thanks.
Phys.Org News Partner Science news on Phys.org
Mysterious source of ozone-depleting chemical baffles NASA
Water leads to chemical that gunks up biofuels production
How lizards regenerate their tails: Researchers discover genetic 'recipe'
atyy
#2
Oct8-10, 02:50 AM
Sci Advisor
P: 8,533
http://nedwww.ipac.caltech.edu/help/cosmology_calc.html
Passionflower
#3
Oct8-10, 02:56 AM
P: 1,555
Quote Quote by TrickyDicky View Post
What is the formula used to convert the measured redshift into a velocity?, not the approximated formula for low speeds v=cz , but the more general and accurate one.

Thanks.
Do you want the answer for special relativity or cosmology or both?

TrickyDicky
#4
Oct8-10, 03:01 AM
P: 3,035
Velocity and redshift

Quote Quote by Passionflower View Post
Do you want the answer for special relativity or cosmology or both?
For cosmology, the one used to get a velocity from the redshift and plug it in the Hubble Law formula.
TrickyDicky
#5
Oct8-10, 11:01 AM
P: 3,035
Quote Quote by TrickyDicky View Post
For cosmology, the one used to get a velocity from the redshift and plug it in the Hubble Law formula.
I think , this is the one

v=[((1+z)^2-1)/((1+z)^2+1)]c=Ho*D

c=light speed constant
Ho=Hubble constant
D=distance
v=velocity
George Jones
#6
Oct8-10, 12:49 PM
Mentor
George Jones's Avatar
P: 6,242
Quote Quote by TrickyDicky View Post
I think , this is the one

v=[((1+z)^2-1)/((1+z)^2+1)]c=Ho*D

c=light speed constant
Ho=Hubble constant
D=distance
v=velocity
No, this isn't correct. See section 3 from

http://arxiv.org/abs/astro-ph/0310808.

It is fairly easy to derive equation (1) from this paper.
Calimero
#7
Oct8-10, 01:15 PM
P: 256
Quote Quote by TrickyDicky View Post
I think , this is the one

v=[((1+z)^2-1)/((1+z)^2+1)]c=Ho*D

c=light speed constant
Ho=Hubble constant
D=distance
v=velocity
I don't think it is correct. For zero density universe it is:

[tex]v=H_{0}D[/tex]

[tex]D=(c/H_{0})ln(1+z)[/tex]
TrickyDicky
#8
Oct8-10, 01:43 PM
P: 3,035
Quote Quote by George Jones View Post
No, this isn't correct. See section 3 from

http://arxiv.org/abs/astro-ph/0310808.

It is fairly easy to derive equation (1) from this paper.
The one I wrote is exactly equation (2) from that paper.

Quote Quote by Calimero View Post
I don't think it is correct. For zero density universe it is:

[tex]v=H_{0}D[/tex]

[tex]D=(c/H_{0})ln(1+z)[/tex]
This is not exactly what I wanted. I asked for the way to translate from z to velocity for high z or at least >1, this must be a very common formula for cosmologists, I'd say.
The formula I used maybe is not correct for the Hubble law but I'm interested in the first part, express v as a function of z, is that so difficult?
TrickyDicky
#9
Oct8-10, 02:04 PM
P: 3,035
Ok, I see what you mean, after looking at the paper and the formula again, I see what you mean, but according to some cosmologists the formula that doesn't give superluminal velocities is alright too, and anyway this is a cosmology debate that I find artificial and tiresome and I don't really wanna get into it , I think it's been discussed enough in these forums, just remember that people as prestigious as David Hogg supports the view of cosmological redshift as Doppler.
George Jones
#10
Oct8-10, 02:05 PM
Mentor
George Jones's Avatar
P: 6,242
Quote Quote by TrickyDicky View Post
The one I wrote is exactly equation (2) from that paper.
Yes, but this is not the correct equation to use for cosmology.
Quote Quote by Calimero View Post
I don't think it is correct. For zero density universe it is:

[tex]v=H_{0}D[/tex]

[tex]D=(c/H_{0})ln(1+z)[/tex]
This expression and the expression that TrickyDicky gave in post #5 are both true in special relativity, i.e., in an empty universe. The conventions used for distance, however, are different in posts #5 and #7, and this leads to differing expressions for speed.
Calimero
#11
Oct8-10, 02:56 PM
P: 256
Quote Quote by George Jones View Post
This expression and the expression that TrickyDicky gave in post #5 are both true in special relativity, i.e., in an empty universe. The conventions used for distance, however, are different in posts #5 and #7, and this leads to differing expressions for speed.
Yes, for empty universe [tex]D=(c/H_{0})ln(1+z)[/tex] gives distance that goes into Hubble's law. Equation (1) you pointed at is general one, and [tex]\dot{R}[/tex] would depend on particular values of [itex]\Omega_{\lambda}[/itex] and [itex]\Omega_{m}[/itex] you choose.
Calimero
#12
Oct8-10, 05:20 PM
P: 256
Quote Quote by TrickyDicky View Post
...and anyway this is a cosmology debate that I find artificial and tiresome and I don't really wanna get into it , I think it's been discussed enough in these forums, just remember that people as prestigious as David Hogg supports the view of cosmological redshift as Doppler.

What debate?

Quote Quote by TrickyDicky View Post
What is the formula used to convert the measured redshift into a velocity?, not the approximated formula for low speeds v=cz , but the more general and accurate one.
Quote Quote by TrickyDicky View Post
For cosmology, the one used to get a velocity from the redshift and plug it in the Hubble Law formula.


Register to reply

Related Discussions
Cosmological redshift and doppler redshift Cosmology 16
Expansion redshift VS gravitational redshift? Cosmology 57
Help with velocity/redshift/distance law Cosmology 2
Gravitational redshift and phase velocity Special & General Relativity 0
Redshift > 1.46 means Recessional Velocity over C? Astronomy & Astrophysics 8