Evidence for Universe Expansion Found

Yahoo (AP) ^ | 3/16/2006 | MATT CRENSON

[PatrickHenry]

Nice prime!

[VadeRetro]

Well, either support your claim line by line with a math analysis, or else retract your claim as being mere talk.

OK. I was referring to your post 365, in which you respond to Doctor Stochastic giving you a value for the Hubble Constant and telling you that you can do the math. He was obviously mistaken.

Hubble's Constant gives you an observed average acceleration per distance, about 71 km/s per megaparsec. In the context of the conversation on the thread at that time, you were expected to compute how many megaparsecs one AU is and figure what the observed acceleration of Earth away from the Sun would be in the absence of constraining forces.

The distance in parsecs is (1 [AU] / 63,240 [AU per parsec]). The distance in megaparsecs is that result divided by a million, or 1.58 x 10^-11. That's the number you multiply by 71 meters/second megaparsec to get your meters/second acceleration.

This means that if there were no gravity, a distant observer would see the Earth at its current distance start drifting away at the breakneck speed of .0000000011 km/s, or .0000011 m/s, or .0011 mm/s, or 1.1 mm/s. But then you have to figure that the Earth actually has a very non-trivial orbital velocity, and the Sun has a non-trivial gravitational pull on the Earth. The real question is thus whether the tiny accelerative pseudo-force perturbs the equilibrium of Earth's stable orbit enough to be noticeable or it is swamped by the real forces. Various papers linked on this thread have attempted that calculation with a sophistication I cannot emulate.

Now, I'm not sure I plugged all the little numbers into my calculator right, but I'm pretty sure I'm at least doing the right calculation. That is, I have addressed what Hubble's Constant means in the context of whether we should see the Earth drifting away from the Sun. Your calculation is so far from the right ballpark than an analysis of what the heck you're even trying to show there is best left to you.

Wrong units, yes. Or, maybe it's right units, wrong formula. Or, maybe it's wrong units, wrong formula, wrong day.

Thus ThinkPlease's question to you, "What the Sam Hill are you trying to do?" You have literally gone so lost it is frankly impossible to tell what planet you're on, although my money would be on Number Seven.

[VadeRetro]

Should probably have pinged you two to 802 per normal FR courtesies which I typically forget/ignore. (Forgore?)

[VadeRetro]

This means that if there were no gravity, a distant observer would see the Earth at its current distance start drifting away at the breakneck speed of .0000000011 km/s, or .0000011 m/s, or .0011 mm/s, or 1.1 mm/s. But then you have to figure that the Earth actually has a very non-trivial orbital velocity, and the Sun has a non-trivial gravitational pull on the Earth.

Indeed, me screw up, spuriously starting with kilometers instead of meters. Thus, the final result acceleration is 1.1 nanometers/second. I can feel the neck-snapping whoosh even as I type.

[VadeRetro]

It has already been pointed out in freepmail that I conflate recessional velocity and acceleration. More bad.

Credit longshadow.

[Southack]

Your claim was that I used the wrong units, not that my calculations were in error. In the end, you managed to prove neither (wrong units nor wrong calculations) as well as screw up your own math (which longshadow was apparently so frightened by that he freepmailed you privately rather than post a simple note on this thread where I'd see it).

Here's the meat of my post #381 math:
The Hubble Constant gives us the increase in velocity of a distant object.

For the Earth to Sun distance (the original question) of 1 AU we first have to translate 1 AU into Mpc's.

1 Mpc = 3.08568025E22 meters
1 AU = 149,598,000,000 meters

So 1 Mpc = 206,265,000,000 AU
Thus, 1 AU = 4.84813681 × 10-12 Mpc

Now, if I've got the wrong value for AU to Mpc, speak up here.

So the Hubble Constant of 71,000 m/s/Mpc * 4.84813681 × 10-12 Mpc will give us 3.44217714 × 10-6 m/s

So, if I've got the expected Hubble acceleration wrong for 1 AU, speak up here.

3.44217714 × 10-6 m/s * 17 Billion years * 365 days/year * 24 hours/day * 60 minutes/hour * 60 seconds/minute gives us 1.84539247E13 meters (123.356761 AU) of space expansion in between the Earth and Sun for a 17 Billion year timeframe.

Note: this is a deliberate over-simplification. All that I'm doing is saying that if you had a *constant* acceleration (rather than one that increases as expected), that multiplying Time in seconds (the seconds cancel out, revealing that I've used the correct units -- contrary to some ill-informed claims) by that acceleration will give us a minimum expected Distance in meters.

More usefully, it would show that the Earth/Sun distance should increase 1.08552498 kilometers every year.

This is actually a large-enough distance for our existing level of science to detect and confirm...except, the confirmation is that no such increase is occurring.

And that scientific confirmation leads us to conclude that the Hubble Expansion does not occur in any way, shape, or form inside areas bound by Gravity.

[VadeRetro]

I have to retract one of my error confessions. It was itself an error, as the Hubble Constant is indeed in km/s and not m/s as I brainfarted out loud.

But now let me add a new one. I used the conversion factor AU/lightyear, nor AU/parsec.

Now, if I've got the wrong value for AU to Mpc, speak up here.

The good news is you didn't. But an additional factor of 1/3.262 just makes that recessional velocity smaller.

So, if I've got the expected Hubble acceleration wrong for 1 AU, speak up here.

You seem to have just conflated recessional velocity with acceleration. That was the subject of that freepmail I just got. Velocity is distance/time. Acceleration is velocity/time. You were apparently too busy looking for cheap shots to score to notice what was being said.

3.44217714 × 10-6 m/s * 17 Billion years * 365 days/year * 24 hours/day * 60 minutes/hour * 60 seconds/minute gives us 1.84539247E13 meters (123.356761 AU) of space expansion in between the Earth and Sun for a 17 Billion year timeframe.

Here is where I perhaps recall some people telling you that you need to take a calculus course. This is incomprehensible garbage. One has to actually ask what you think happened 17 billion years ago. There is no 17 billion years, an inconvenience you wave away rather oddly. What does the Hubble Constant tell you? It tells you that the recessional velocity changes with distance, and in particular changes in such a way that the closer the thing is, the less such velocity it will have.

Thus, if you are trying to extrapolate back to a time when the Earth-Sun distance would have been even smaller, the recessional velocity would have been smaller as well. If you really don't do calculus, you have to do a first-order numerical approximation with an initial distance, an appropriately small time-slice, a calculation of a recessional velocity, a calculation of the resulting new distance at the end of the time-slice, a new velocity, a new distance traveled, a new velocity, and so proceed.

Your attempt is so wrongheaded as to be unrecognizeable. Did you start with zero distance? What's the recessional velocity at zero distance?

[Doctor Stochastic]

You're correct. I was wrong. Barbie was right.

[Doctor Stochastic]

I had a car like that once. It did take more than a day to waltz across Texas in it.

[VadeRetro]

More usefully, it would show that the Earth/Sun distance should increase 1.08552498 kilometers every year.

Beyond horsemanure. The Earth is not moving freely. You have to compute the expected perturbation of the Earth's orbit. Show your work, or critique the papers linked already on this topic.

[VadeRetro]

I made it across the Texas panhandle pretty fast on I-40 once. The real joy was finding that vegetables replaced white gravy biscuit in restaurants once you get to New Mexico.

[Southack]

"The Earth is not moving freely. You have to compute the expected perturbation of the Earth's orbit. Show your work, or critique the papers linked already on this topic."

Hubble's Constant makes no such orbital exceptions. Hc is also only useful for non-Gravity-bound systems, and we *know* this because the Hubble math doesn't give us the observed results for the Earth-Sun distance.

For the Earth to Sun distance (the original question) of 1 AU we first have to translate 1 AU into Mpc's.

1 Mpc = 3.08568025E22 meters

1 AU = 149,598,000,000 meters

So 1 Mpc = 206,265,000,000 AU

Thus, 1 AU = 4.84813681 × 10-12 Mpc

So the Hubble Constant of 71,000 m/s/Mpc * 4.84813681 × 10-12 Mpc will give us 3.44217714 × 10-6 m/s

Well, 3.44217714 × 10-6 m/s * 1 year * 365 days/year * 24 hours/day * 60 minutes/hour * 60 seconds/minute gives us an expected distance increase of 1.086 kilometers every year.

This is actually a large-enough distance for our existing level of science to detect and confirm...except, the confirmation is that no such increase is occurring.

And that scientific confirmation leads us to conclude that the Hubble Expansion does not occur in any way, shape, or form inside areas bound by Gravity.

[VadeRetro]

Hubble's Constant makes no such orbital exceptions.

What stupid lawyer's game is this? There is no need to put "orbital exceptions" in Hubble's constant. The space is assumed to expand uniformly. We still get the universe we see.

Your proferred calculations are in no way, shape or form an expected perturbation of the Earth's orbit. They are at this point only bizarrely robotic repetition. I will answer in kind. You *have to* *have to* *have to* compute an expected perturbation of the Earth's orbit.

The Earth has a huge velocity vector. The Earth has a huge gravity vector. The Hubble expansion is a tiny, tiny, tiny, velocity vector. I do not see these considerations in your numbers. That is a problem.

[Southack]

"There is no need to put "orbital exceptions" in Hubble's constant. The space is assumed to expand uniformly. We still get the universe we see."

Yes and no. Yes, there is no need to put "orbital exceptions" in Hubble's constant. Not coincidentally, my math above doesn't.

No, the space does *not* expand uniformly by observation (though the math doesn't distinguish and in fact assumes uniformity). For instance, we observe that the Earth-Sun distance does not increase by 1.086 kilometers each year as Hubble's Constant would dictate.

Thus, space is not expanding in regions bound by Gravity. Hubble's Constant does hold true for vast distances not bound by Gravity...but does not hold true for distances inside solar systems and galaxies at all.

[longshadow]

"He's dead, Jim."

[AntiGuv]

unique physics placemarker

[azhenfud]

"When space itself moves, deforms, contracts, expands, the physics of our day-to-day world does not apply because there is no reference, although relativity attempts to address some of the observations."

That's exactly what Job realized.

Job 26:7 "He stretcheth out the north over the empty place, and hangeth the earth upon nothing."

[VadeRetro]

Yes and no. Yes, there is no need to put "orbital exceptions" in Hubble's constant. Not coincidentally, my math above doesn't.

Not coincidentally, your math is wrong in precisely that way. The expansion of space wants to carry the Earth away. The Sun's gravity is trying to pull the Earth in. If there were no orbital velocity, you would have a war between gravity and the Hubble velocity. Gravity would win and the Earth would fall into the Sun. I hope you can see that.

So what happens is that the Hubble expansion is canceling some tiny, tiny percentage of the Sun's gravity. Does that throw the Earth's orbit out of equilibrium? Nah. Stable orbits are possible at a variety of intensities of graviational fields. We see this. Kepler's Laws do not depend on a specific value of G. Precise outcomes do, but the existence of stable orbits does not.

The question is whether we should see anything at all. Again, that linked paper attempted to model the problem and several related ones accurately. You have refused to address it, persevering in strawmanning.

Look at this hash of deliberate fallacy from you.

For instance, we observe that the Earth-Sun distance does not increase by 1.086 kilometers each year as Hubble's Constant would dictate.

Thus, space is not expanding in regions bound by Gravity...

Failure to see tightly bound objects move in precisely the manner predicted for utterly unbound objects is not an observation of the failure of space to expand. This cannot be interpreted as an innocent mistake at this point on the thread.

[VadeRetro]

How many times on this thread have I disparaged wasting time on the attention-trolls? The spirit is willing but the flesh is weak.

[Southack]

"Failure to see tightly bound objects move in precisely the manner predicted for utterly unbound objects is not an observation of the failure of space to expand."

What would be your acceptable observation of the failure of space to expand?