Thread: Physics! Hubble constant! What is km/s-1/Mpc-1?

1. Physics! Hubble constant! What is km/s-1/Mpc-1?

Not really a programming question, and I actually signed up at a physics forum and asked this question there this morning:

Calculating distance using redshift

But I'm dubious about the one reply and figured one of you geniuses must know.

What does it mean when units are specified to the power of -1? Eg:

The value of the Hubble constant that is determined using type 1a supernovae as standard candles is H0 = 65 km s-1 Mpc-1
-1 should be superscripted in both cases.

Am I correct in asserting that the value (65) should be multiplied by 0.1 to work in a formula which uses km/s (for c) and yields Mpc? Applying the formula from (see post) yields correct answers if I do, but is out by a factor of 10 if I don't.

If anyone can rearrange that algebraically to show how those units work out that'd be I have a math inferiority complex.

2. It just means that they are in the denominator. It really is preference, but sometimes it is more clear than trying to denote which units are in the denominator by using other methods. For instance in acceleration, ft/s2 or ft s-2.

http://i.imgur.com/lg20y.gif

3. Okay. Thanks for that explanation of the superscript.

So, I found an online calculator for this here:

Hubble law and the expanding universe

And H should not be divided by ten. My second example (the Large Magellanic Cloud) was way off in the other direction, but I think its redshift is anomalous. Presuming the other poster was right about the typo, that means the first two sets of data I tried were bad. O_O

But I got data for NGC2770, that came out properly. I dunno if I should trust this Hubble character or not.

4. Originally Posted by MK27
Okay. Thanks for that explanation of the superscript.

So, I found an online calculator for this here:

Hubble law and the expanding universe

And H should not be divided by ten. My second example (the Large Magellanic Cloud) was way off in the other direction, but I think its redshift is anomalous. Presuming the other poster was right about the typo, that means the first two sets of data I tried were bad. O_O

But I got data for NGC2770, that came out properly. I dunno if I should trust this Hubble character or not.
So i know nothing about this stuff, but isn't the Large Magellanic Cloud close enough to be affected by the gravity of the Milky Way? This would slow it down and make the hubble law inapplicable? Just a stab in the dark.

5. Originally Posted by Neo1
So i know nothing about this stuff, but isn't the Large Magellanic Cloud close enough to be affected by the gravity of the Milky Way? This would slow it down and make the hubble law inapplicable? Just a stab in the dark.
Hubble's law describes the expansion of space, not the movement of things. However, things which are extremely far from each other have so much space in between them that they appear to move relative to each other due to this expansion of space.

Imagine standing in a room and the room keeps getting wider. Your friend on the other side seems to be moving away from you, but this isn't because of what your friend is doing with his feet, it's because the room is changing. Your friend could walk toward you while the room expands, or he could walk away from you, too.

6. Originally Posted by brewbuck
Hubble's law describes the expansion of space, not the movement of things. However, things which are extremely far from each other have so much space in between them that they appear to move relative to each other due to this expansion of space.

Imagine standing in a room and the room keeps getting wider. Your friend on the other side seems to be moving away from you, but this isn't because of what your friend is doing with his feet, it's because the room is changing. Your friend could walk toward you while the room expands, or he could walk away from you, too.
Okay, but if he moves toward you at the same speed or faster than the speed of which the room is expanding, wouldn't he be either not moving or moving towards you? Since they are so close to each other (relatively speaking), the Milky Way and the LMC are being pulled towards each other (possibly faster than the space between them is expanding?), couldn't this explain why the numbers doesn't align with Hubble's law?

7. Originally Posted by Neo1
Okay, but if he moves toward you at the same speed or faster than the speed of which the room is expanding, wouldn't he be either not moving or moving towards you? Since they are so close to each other (relatively speaking), the Milky Way and the LMC are being pulled towards each other (possibly faster than the space between them is expanding?), couldn't this explain why the numbers doesn't align with Hubble's law?
Well, yes, but I don't see why that is a mystery. You observe things which don't follow Hubble's law all the time (in fact, it's pretty much the ONLY thing you see).

8. Originally Posted by Neo1
Okay, but if he moves toward you at the same speed or faster than the speed of which the room is expanding, wouldn't he be either not moving or moving towards you? Since they are so close to each other (relatively speaking), the Milky Way and the LMC are being pulled towards each other (possibly faster than the space between them is expanding?), couldn't this explain why the numbers doesn't align with Hubble's law?
That would makes sense except the redshift is much higher, not lower, than it should be according to Hubble's law.

I finally found an explicit explanation WRT the LMC and other galaxies in our local group:

www.worldnpa.org/pdf/abstracts/abstracts_570.pdf

Code:
```One of the most firmly established but least accepted results in
astronomy is that in groups dominated by a larger galaxy, the smaller
companion galaxies have systematically higher redshifts. Figure 1 in
this paper demonstrates that for the nearest, best known groups, the
Local Group and the M81 group, all the major companions are
redshifted with respect to the central galaxy. A total of 21 out of 21
permits a chance of only one in two million that the result could be
accidental. Every test of additional groups at larger distances confirms
the excess redshift of companions.```
This is attributed to the fact that the stars in the smaller companions are newer. I glanced through more of the paper and:

Code:
```The basic reasoning with respect to the magnitude of mass of an
elementary particle is that it must depend on the amount of material
with which it can exchange gravitons. That in turn depends on the
volume of the universe it sees, i.e. its light signal speed multiplied by
the time during which it has been in existence. It would seem absurd
to consider an electron to have a terrestrial value for its mass just after
it had appeared in a previously empty vacuum. Its mass would
dominate the whole universe which it saw.
[...]
The energy of the photon an atom emits or absorbs (and
hence inversely its redshift) is then proportional to the mass of the
electron making the orbital transition.
The theory leads to the conclusion that elementary particles have
masses which are a function of position and time,```

9. Originally Posted by carrotcake1029
It just means that they are in the denominator. It really is preference, but sometimes it is more clear than trying to denote which units are in the denominator by using other methods. For instance in acceleration, ft/s2 or ft s-2.

http://i.imgur.com/lg20y.gif
My typical notation is:

ft/s^2

I thought the ^ was a convention for exponentiation. Else I do ft/(s * s), which is more a programming convention.