Time & Cosmology: Observing Past Quantities & Effects on Equations

[Arman777]

I know that If we look further we seethe past...So we are measuring some quantities like a mass of a far galaxy or dark matter mass of a galaxy...All in general does this mean we are observing past quantities ?
These time dependent quantities can change with time so How this affect our Cosmological equations ?

And If there's an galaxy like a 2 million light years away ,the speed will be V=HD but it was 2 million years ago speed,so speed in right now will be much more faster.I tried to calculate it but I don't know its correct or not.
(I found for this example is $V=H.1.23 Mpc$).Twice of the pre result.

Some ideas would be nice,

Thanks

 

[PeterDonis]

All in general does this mean we are observing past quantities ?

Yes.

If there's an galaxy like a 2 million light years away ,the speed will be V=HD but it was 2 million years ago speed,so speed in right now will be much more faster

Only if the expansion of the universe is accelerating. It is now, but before a few billion years ago it wasn't.

 

[Arman777]

Yes.
Only if the expansion of the universe is accelerating. It is now, but before a few billion years ago it wasn't.

Is my equation true ?, actually Hubble constant will also change..We think universe is accelerating.Right now, its acceleration is faster then the observable one then ?

And past quantities can lead wrong results ?

Is that mean couple billion years ago V(d)=constant ??

 

[Orodruin]

One thing to note is that the Hubble constant is not necessarily increasing in an accelerating universe. In a cosmological constant dominated universe, it goes to a constant. However, the distance to objects increases exponentially.

 

[Jorrie]

Is my equation true ?

Yes, we calculate the present recession rate of a galaxy by taking the present Hubble constant and multiply it by the comoving distance D.

If we want to know what the recession rate was when the light has left the galaxy, we have to calculate what the Hubble constant was then and multiply it by the comoving distance divided by (z+1), i.e. $V_{rec}(t) = H(t) D/(z+1)$.

H(t) is obtained from the first Friedmann equation.

 

[Arman777]

One thing to note is that the Hubble constant is not necessarily increasing in an accelerating universe. In a cosmological constant dominated universe, it goes to a constant. However, the distance to objects increases exponentially.

There were some H-t graphs that I can remember,and in there H were constant.I understand now

 

[Arman777]

Yes, we calculate the present recession rate of a galaxy by taking the present Hubble constant and multiply it by the comoving distance D.

If we want to know what the recession rate was when the light has left the galaxy, we have have to calculate what the Hubble constant was then and multiply it by the comoving distance divided by (z+1), i.e. $V_{rec}(t) = H(t) D/(z+1)$.

H(t) is obtained from the first Friedmann equation.

I was meant $V(D)=H.1.23Mpc$ but 2 million years ago $V(D)=H0.615Mpc$ these two equatons are true ? First equation is the speed of a galaxy right now that we observe that its 2 million ly away.The other one is If we think 2 million ly is now.

In wiki says

D is proper distace (wiki is not a good source so I don't know https://en.wikipedia.org/wiki/Hubble's_law in the interpretation part )

Here let's suppose there's a galaxy $10^6$ parsec away.But we know that It was there $3.3.10^6$ years ago.($10^6 pc=3.2 10^6 ly$)So by now the distance of it must be a much more higher and from there the $V(D)$ must be much more higher.I tried to calculate it but I am not sure for my result.

If $3.3.10^6$ year ago the speed was $500\frac {km} {pc}$.So after $3.2 10^6$ year later from now we will see the real $V(D)$ which that corresponds to $2.10^6$ pc so simply ;
If we observe an object at its distance is $10^6$ pc from us its speed is not $500\frac {km} {pc}$ but its $1000\frac {km} {pc}$

 

[Jorrie]

I was meant $V(D)=H.1.23Mpc$ but 2 million years ago $V(D)=H0.615Mpc$ these two equatons are true ? First equation is the speed of a galaxy right now that we observe that its 2 million ly away.The other one is If we think 2 million ly is now

You should write the equations more clearly, e.g. did you mean $V(D) = H_0 D$ where D=1.23 Mpc? Do not use decimal points to indicate multiplication.

The chart that you have shown seems to be the original Edwin Hubble result, which was about a factor 7 off the scale of today's Ho. But the point to take out of it is that it was a linear relationship and it is linear for any present proper (i.e. comoving) distance. And V(D) is the present recession rate, not the recession rate when the light has left the galaxy.

I do not understand what you are asking in your last paragraph, e.g. what is $500 \frac{km}{pc}$?

 

[Arman777]

Lets forget everything.

Lets suppose there s a galaxy, $10^6 ly$ away from us ? Whats the $V(D)=?$
This is the question.

So we can say the distance from us is $1ly=0.30 Mpc$ and so $V=HD$ so $V=0.30H$
But galaxy had $0.30H$ speed, $10^6$ years ago not now,so the current speed will be,
$V=0.60H$

 

[PeroK]

Lets forget everything.

Lets suppose there s a galaxy.$10^6 ly$ away from us ? Whats the V(D)= ?
This is the question.

So we can say the distance from us is $1ly=0.30 Mpc$ and so $V=HD$ so $V=0.30H$
But galaxy had $0.30H$ speed, $10^6$ yeears ago not now,so the current speed will be,
$V=0.60H$

First, $1$ light-year is about $0.3$ parsecs (not Mega Parsecs).

Second, the Andromeda Galaxy is about $2 \times 10^6$ light-years away, but that is gravitationally bound to us. I.e. it's moving towards us. To take away the effects of gravity, you would have to consider galaxies at much greater distances.

You probably should read this Insight on the expanding universe:

https://www.physicsforums.com/insights/inflationary-misconceptions-basics-cosmological-horizons/

 

[Arman777]

I ll ask my prof thanks all

 

[Arman777]

First, 111 light-year is about 0.30.30.3 parsecs (not Mega Parsecs).

I made a typo in there it would be 10^6

Second, the Andromeda Galaxy is about 2×1062×1062 \times 10^6 light-years away, but that is gravitationally bound to us. I.e. it's moving towards us. To take away the effects of gravity, you would have to consider galaxies at much greater distances.

It was just an example to illustrate my example.

My explanation is not enough clear I guess

 

[Jorrie]

Lets forget everything.

Lets suppose there s a galaxy, $10^6 ly$ away from us ? Whats the $V(D)=?$
This is the question.

OK, let's start over and change your scale to: "Lets suppose there is a galaxy, $D=10^3$ Mpc (i.e. 1 Gpc) away from us ? Whats the $V(D)$?

Easy: $V(D) = H_0 D = 68 \times 10^3$ km/s, or 0.226c

This is the recession rate now, not when the light has left the galaxy. See my post (#5) above.

 

[Arman777]

OK, let's start over and change your scale to: "Lets suppose there is a galaxy, $D=10^3$ Mpc (i.e. 1 Gpc) away from us ? Whats the $V(D)$?

Easy: $V(D) = H_0 D = 68 \times 10^3$ km/s, or 0.226c

This is the recession rate now, not when the light has left the galaxy. See my post (#5) above.

I see know...

 

[Chronos]

It is helpful to remember we can only observe the effects of redshift on light when it was emitted and during its long journey to our detectors. Reconstructing these effects can be a complicated process.

 

[Arman777]

Just I thought something...but I don't know cosmology..They are complicated yeah..