Astronomy Trivia Challenge: Can You Answer These Questions About the Night Sky?

[marcus]

another riddle

Well I seem to have answered my own question in the course of explaining how to answer it. So here is another

The temp of the CMB has been measured with fine accuracy to be 2.726 kelvin

But Bill Unruh determined that simple acceleration gives space a temperature. A moving box, if it is accelerating, would have inside it some thermal radiation with a temperature proportional to the amount of accelaration.

So I ask you----how fast would you need to accelerate (meters per second per second, feet per second per second, gees, whatever)
in order to produce an Unruh temp of 2.726 kelvin?

Aaargh ! Nobody likes to calculate! Labguy has made very plain that he loathes to arithmeticize. But shouldn't we know how much acceleration it would take to duplicate the CMB temp?
For sentimental reasons if for no other. The CMB temp is the most prevalent temp in the universe---it IS the temp of the universe.

 

[Labguy]

Aaargh ! Nobody likes to calculate! Labguy has made very plain that he loathes to arithmeticize. But shouldn't we know how much acceleration it would take to duplicate the CMB temp?

The word is "Matherize", not arithmeticize, you bumbling human calculator...

I get 37.14159265 cm/sec./sec. How?; heck, I don't know, I just made it up. But, someone should easily recognize the eight decimal places as something they use a lot.

Sorry for the "No-Answer" post.

Labby

 

[marcus]

Originally posted by Labguy
The word is "Matherize", not arithmeticize, you bumbling human calculator...

I get 37.14159265 cm/sec./sec. How?; heck, I don't know, I just made it up. But, someone should easily recognize the eight decimal places as something they use a lot.

Sorry for the "No-Answer" post.

Labby

Some how----a lucky shot in the dark?----you got the right answer! Your turn Labguy.

 

[marcus]

The forumula for the temperatue is T = g/2pi

where g is the acceleration.

It is sort of like the relation of radius and circumference of a circle
the temperature is the radius of the acceleration. whaaaaa?

So I ask you----how fast would you need to accelerate (meters per second per second, feet per second per second, gees, whatever)
in order to produce an Unruh temp of 2.726 kelvin?

Aaargh ! Nobody likes to calculate! Labguy has made very plain that he loathes to arithmeticize. But shouldn't we know how much acceleration it would take to duplicate the CMB temp?
For sentimental reasons if for no other. The CMB temp is the most prevalent temp in the universe---it IS the temp of the universe.

The CMB has a perfect thermal spectrum that a black body would radiate at a temperature of 1.93 x 10^-32 natural.

Only need to multiply that by 2pi to get the desired acceleration.
which is 12 x 10^-32 Planck.

Evidently this is the acceleration (the answer to the question, cause I said any units you want) which Labguy must have calculated.

Labguy your only mistake was in trying to convert it to metric, where because of the awkwardness of the system you went way wrong in the fifth decimal place---maybe other places too.

However this will be overlooked in your favor.

Your turn!

 

[marcus]

I was curious to see how it turned out in metric so I
converted the acceleration

12 x 10^-32

into metric and it came out

67 x 10¹⁹ meters per second per second.

tho your answer was a few order of magnitudes off I am
delighted to overlook this in the interest of the game

 

[Shadow]

Oh, sorry marcus I have been busy and was unable to reply to your post.

 

[Labguy]

Originally posted by marcus
The forumula for the temperatue is T = g/2pi

where g is the acceleration.

It is sort of like the relation of radius and circumference of a circle
the temperature is the radius of the acceleration. whaaaaa?



The CMB has a perfect thermal spectrum that a black body would radiate at a temperature of 1.93 x 10^-32 natural.

Only need to multiply that by 2pi to get the desired acceleration.
which is 12 x 10^-32 Planck.

Evidently this is the acceleration (the answer to the question, cause I said any units you want) which Labguy must have calculated.

Labguy your only mistake was in trying to convert it to metric, where because of the awkwardness of the system you went way wrong in the fifth decimal place---maybe other places too.

However this will be overlooked in your favor.

Your turn!

So which is it; wrong at the 5th decimal place, or off by "a few orders of magnitude?? To four decimal places seems Ok, but "a few orders of magnitude" sounds way off. Are you giving me "actual" credit as right or wrong?

 

[Labguy]

Originally posted by marcus
I was curious to see how it turned out in metric so I
converted the acceleration

12 x 10^-32

into metric and it came out

67 x 10¹⁹ meters per second per second.

tho your answer was a few order of magnitudes off I am
delighted to overlook this in the interest of the game

Is that a typo, or did you mean 67 x 10^-19 m/s/s instead of the ⁺¹⁹ ??

 

[marcus]

Originally posted by Labguy
Is that a typo, or did you mean 67 x 10^-19 m/s/s instead of the ⁺¹⁹ ??

Hello Labguy and Shadow! Glad to hear you. I want Labguy to have the next turn and ask the next question because he at least came up with some answer.

It was not numerically correct and I was just kidding about being only a "little bit" off. But even if waaaayyy off, some answer is better than none at all! Even one meant jokingly (as I think Labguy's) is better.

So your turn to ask.

(I will tell you about calculating the temperature of acceleration---as in Bill Unruh's 1976 paper----in a later follow-up post. The quantum gravity people are often referring to 3 things from the Seventies: Bekenstein's 1973 BH entropy, Hawking 1975 BH temperature, Unruh 1976 acceleration temperature. It is a famous and surprising result that helped stimulate current directions of research. So you might like to know. But right now the game is more important.)

 

[marcus]

Footnote about T = g/2pi

Unruh discovered that merely accelerating gives space around you a temperature

best to think of an observer in a box. the box is given acceleration so there is thermal radiation inside and some equilibrium temperature

this effect is very very very tiny. Good old mother nature does not confront us with really gross freakishness here. The effect is so tiny that you will never detect it. But it is based on standard physics and is a solid, tho surprising, result.
(essentially just like Hawking's BH temp but in a different context)

No reasonable acceleration is enough to produce a macroscopic temperature, even a small one like 2.726 kelvin!

Anyway the formula for the temperature seen in space by an accelerating observer is T = g/2pi

So to produce a temp of 10^-32
requires an acceleration of 2pi x 10^-32

You always multiply by 2pi to get the acceleration.

Now Planck temperature is sort of Big Bang temperature and in metric terms it is 1.417 x 10³² kelvin.
So when I say a temp of "10^-32 " I mean 1.417 kelvin.

And in metric terms, Planck unit acceleration is 5.56 x 10⁵¹ meters per sec per sec. So "10^-32 " of acceleration is 5.56 x 10¹⁹ meters per sec per sec

You just have to subtract 32 from 51 to get 19.

And 2pi times that is 35 x 10¹⁹ meters per sec per sec

That is for making the temperature be 1.4 kelvin or so, earlier I did it for the CMB temperature which is almost twice that.

A more attainable temperature to get by acceleration is a femtokelvin---a quadrillionth of a kelvin

To get 1.4 femtokelvin you need the acceleration to be 35 x 10⁴ meters per second

 

[Labguy]

Well, since I didn't actually get the correct answer on Marcus' last question, I don't feel I should be asking one here. But, I will and it should be a slam-dunk for anyone with eyes.

Easy One:

My last answer was 37.14159265 cm/sec/sec. (That is a statement)
QUESTION:
Based on the number above (forget cm/sec/sec);
(a) Pick one number only to remove: remove the #___ ?
(b) Place the decimal after the number____ ?
(c) Name the resulting number_____ ?

It is a very common number, constant, ratio, measure, etc., used in astronomy every day.

 

[Labguy]

Easy One:

My last answer was 37.14159265 cm/sec/sec. (That is a statement)
QUESTION:
Based on the number above (forget cm/sec/sec);
(a) Pick one number only to remove: remove the #___ ?
(b) Place the decimal after the number____ ?
(c) Name the resulting number_____ ?

It is a very common number, constant, ratio, measure, etc., used in astronomy every day.

Come on, guys. This one is too easy to sit for more than a few hours!

 

[marcus]

Originally posted by Labguy
Come on, guys. This one is too easy to sit for more than a few hours!

OK but you better reciprocate and answer the easy one I ask back!

37.14159265 remove the 7, leave the point after the 3, and
get pi.

My question for you is-------suppose a distant galaxy is observed with cosmological redshift z = 1

then what's the prevailing estimate of the light travel time come out to be?

It depends on values assumed for some parameters and the prevailing assumptions are spatial flatness, Hubble value of around 71 (in the usual units), and a cosmological constant representing 73 percent of the energy density. These will do fine and are the default settings in the calculator provided to calculate just that, among other things, at one of the cosmology websites.

I meant this question especially for Labguy but anyone else is welcome to answer.

 

[Labguy]

I told you (anyone) that the answer to my question was as easy as pi. You did it easily, I see.

For Z=1, the light travel time would be about 7.731 billion years if you keep Omega_M at 0.27, and allow your other assumptions to stand.

 

[marcus]

Originally posted by Labguy
I told you (anyone) that the answer to my question was as easy as pi. You did it easily, I see.

For Z=1, the light travel time would be about 7.731 billion years if you keep Omega_M at 0.27, and allow your other assumptions to stand.

Bravo Labguy! Your turn to ask.

I guess your answer means that in the past 7.731 billion years the universe has expanded by a factor of two.

 

[Labguy]

Another easy one:

If:
("Blank") = 4.8-[2.5*log(Lstar/Lsol)]
where "L" stands for luminosity;

What is "Blank"? ( Definition, not a number)

 

[chroot]

Absolute magnitude -- the apparent brightness of a star from a standard distance of 10 parsecs.

- Warren

 

[Labguy]

Originally posted by chroot
Absolute magnitude -- the apparent brightness of a star from a standard distance of 10 parsecs.

- Warren

That was too quick, I'll have to start throwing in some tougher ones if I get another go.

Your turn, speed-breath...

 

[chroot]

Nearly all models of inflation considered today are "slow-roll" inflationary models. Lots of evidence has been piling up (from the Wilkinson Microwave Anisotropy Probe, for example) that we're at least on the right track.

What does the modifier "slow-roll" mean in the context of the space of inflationary models?

- Warren

 

[Labguy]

Originally posted by chroot
Nearly all models of inflation considered today are "slow-roll" inflationary models. Lots of evidence has been piling up (from the Wilkinson Microwave Anisotropy Probe, for example) that we're at least on the right track.

What does the modifier "slow-roll" mean in the context of the space of inflationary models?

- Warren

It can get quite complicated, as there are currently 23 different inflationary models being perused.

But, in simple terms, the "slow roll" term is one model (inflation) that predicts that the universe is flat, and that the spectroscopic variances seen in the CMBR indicate that there was an inflation (singular) where the potential energy slowly "rolled down" with the kinetic energy being damped by the Hubble expansion. Other models predict that this damping (slow rolling) happened more than once, and each had its own "starting and stopping" phase based on the total energy available at the time. In other words, several periods of inflation instead of just the one originally proposed by Guth.

This is a huge simplification.

 

[chroot]

Sorry Labguy, let me see if I can make my question more specific.

The inflationary model first proposed by Guth involved a particular kind of "exit mode," while slow-roll inflationary models involve a very different one. What is this essential, fundamental difference between the "exit mode" of Guth's model and slow-roll models?

Hint: the fundamental difference can be stated in only one sentence.

- Warren

 

[Labguy]

Originally posted by chroot
Sorry Labguy, let me see if I can make my question more specific.

The inflationary model first proposed by Guth involved a particular kind of "exit mode," while slow-roll inflationary models involve a very different one. What is this essential, fundamental difference between the "exit mode" of Guth's model and slow-roll models?

Hint: the fundamental difference can be stated in only one sentence.

- Warren

I would have to say that Guth's original proposal for inflation had no explanation available for the originating and exiting times of the inflationary period. He defined the period, but not why it began or ceased. This is important, because it is why Hawking, and others, doubted inflation to the point that the numerous other models have been developed, even two more by Guth himself.

As I said, there are ~23 models currently being proposed and considered. The only one I know of proposes that a single (scalar) field can be used with no consideration needed for the coordinate system(s). This model assumes that only the Hubble field (size) and the inflationary field (combined wavelength of energies) need to be considered. The "Exit mechanism" (slow roll) occurs when the wavelength of the scalar field becomes larger than the "Hubble Radius", at that time, and that inflationary period stops. Then, the kinetic and potential energy can recombine, only to separate again and (sometimes) reach the point where kinetic energy "damps" and causes another, short inflationary period, which again will "exit" when the wavelength once more reaches the new Hubble scale.

That wasn't one sentence, and if not what you have in mind, it is all I'll be able to add to this topic.

 

[chroot]

Hrm, well... :-/

The answer I was looking for is rather simple. I'll go ahead and give it.

Inflationary models that are NOT slow-roll depend upon a quantum-mechanical tunneling mechanism to move from a classically stable minimum of the inflation potential to a classically stable state with zero inflation potential.

Slow-roll models do not depend on quantum-mechanical tunneling. Instead, the inflation potential is classically unstable, and, just like a ball rolling down a hill, the inflation potential slowly rolls to zero.

So the difference between slow-roll and non-slow-roll models is essentially this: slow-roll models do not require a quantum-mechanical tunneling mechanism to end inflation.

- Warren

 

[Labguy]

Ask a new one!

 

[chroot]

Okay, I will. Was my last question malformed? Was the answer I gave incorrect or unusual in some respect? I apologize if my question sucked, I thought it was an good one! :)

- Warren

 

[Labguy]

Originally posted by chroot
Okay, I will. Was my last question malformed? Was the answer I gave incorrect or unusual in some respect? I apologize if my question sucked, I thought it was an good one! :)

- Warren

No, I thought that it was a good one. But, everybody and his brother have a "theory" of inflation, so it is tough to decide which one is "best" as of this week...

 

[chroot]

Okay, here's one that hits close to home. Someone in my introductory adult astronomy classes asks me this question about once every six months or so. The question is "can we see the Lunar rover with your telescope?"

So my question to you is a simple one: how big would my telescope have to be to be able to see the Lunar rover from the surface of the earth? (Neglecting the ever-obnoxious atmosphere, which actually makes it impossible.)

- Warren

 

[stuffy]

The rover is about 3m long and the moon is 3.82x10^8 m away. Using trig you get an angle of 7.85x10^-9 radians.

Using Resolution = wavelength/diameter (I'll use 500nm)

You get d = (500x10^-9/7.85x10^-9)

Which is about 64 meters across. The Keck is only 10 meters!

 

[chroot]

Excellent work, stuffy. :)

I usually just tell my students it would take a scope about the size of a football field, and that I just don't have room for one in my two-seat roadster. :)

Your turn.

- Warren

 

[Labguy]

Originally posted by chroot
Okay, here's one that hits close to home. Someone in my introductory adult astronomy classes asks me this question about once every six months or so. The question is "can we see the Lunar rover with your telescope?"

So my question to you is a simple one: how big would my telescope have to be to be able to see the Lunar rover from the surface of the earth? (Neglecting the ever-obnoxious atmosphere, which actually makes it impossible.)

- Warren

Using the so-called Dawes' Limit at 4.56/diameter of telescope, I get an angular resolution for the 10-foot Rover of 0.00164 arc seconds. 4.56/0.00164 = ~2780 inches; about 70 meters.

But, to see it, as your question asks, we would also need to know its albedo (reflectivity), and I don't think anyone can figure that.

EDIT: Too Late!