Wikipedia:Reference desk/Archives/Science/2010 September 2

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September 2

Thrust vectoring

Does anyone know of a flat, rather than round, thrust vectoring system that can vector in the yaw besides the X-36? And does anyone have a picture of the X-36's thrust vectoring system? --The High Fin Sperm Whale 03:00, 2 September 2010 (UTC)

Some scant pics here. --Sean 18:24, 2 September 2010 (UTC)

Shrinking helium

In last Sunday's Ask Marilyn article, Marilyn vos Savant discussed helium escaping from balloons. I understand the microscopic porous nature of the balloon, but I'm skeptical of the following statement: "Helium contracts as the temperature drops, which allows gas to slip through the pores of a balloon more easily."

I know that the gas as a whole would contract with decreasing temperature (assuming a constant pressure), but is this true at the atomic level? -- Tom N (tcncv) talk/contrib 04:39, 2 September 2010 (UTC)

It's a complex process. When a gas contracts because of temperature dropping, the molecular density increases; that is there are more molecules per unit volume of the gas, which may mean more collisions with pores. Of course, as the temperature drops, so does Root mean square speed of the molecules, which may mean less collisions with the pores. Of course, under a first approximation (ideal gas), these effects should exactly cancel. Since real gases behave slightly differently than ideal gasses, its quite more complex, and one effect may predominate over the other. The relevent discussions could be found in the articles Effusion and Graham's law, though the effusion article seems to lack some. --Jayron32 04:52, 2 September 2010 (UTC)

• Minor nitpick: you're talking about helium atoms; helium doesn't form molecules. --Anon, 04:08 UTC, September 3, 2010.

Speaking in general terms, its tough to choose a term which readers will recognize, and which is a stand in for "atoms and/or molecules and/or other particles". Chemists use the term "moety" sometimes to mean this, but its not a term in widespread use. When speaking in general terms over the behavior of something like gases in general, given that there are about 7 gases which exist as lone atoms, and about 1 billion that exist as molecules, just using the term molecule is, statistically atleast, close enough. --Jayron32 04:41, 3 September 2010 (UTC)

I know; in fact I just used "molecule" myself that way in answering another question, below. But on this question we're talking specifically about helium. No big deal. --Anon, 06:42 UTC, September 3, 2010.

The rubber around the balloon also contracts, and presumably, so do the pores. The walls should actually be slightly thicker as the balloon contracts. Gas transmission rate through the balloon-wall is much more complicated than a simple application of the ideal-gas law. I'd stake my bet on an empirical measurement of gas escape rate as a function of temperature; there are so many relevant factors (gas thermal velocity; pore size; collision rate; rubber material properties, etc) that trying to model them from fundamental physics is unlikely to be accurate. Nimur (talk) 05:35, 2 September 2010 (UTC)

Also, Helium is about as close to an ideal gas as you can really get, so ideal gas laws are pretty relevant. Googlemeister (talk) 13:05, 2 September 2010 (UTC)

Energy from pressure difference

How do you calculate energy from a pressure difference?

For example a hurricane has an eye 100km by 1km high, so a volume of 1.6×10^11 cubic meters, and a pressure of, say, 950 millibars, compared to 1013 millibars for the air around it. How much energy is contained in that pressure difference? Ariel. (talk) 05:40, 2 September 2010 (UTC)

You need a volume and a pressure to determine energy stored by a gas; and it simplifies if you assume an isothermal process for the energy release. Nimur (talk) 20:19, 2 September 2010 (UTC)

There is a discussion at Potential_energy#Relation_between_potential_energy.2C_potential_and_force which, while it focuses more on gravity, does deal with the mathematical relationship between forces and potential energy, but not in a simple "Here's a nice equation to use to calculate this." --Jayron32 05:49, 2 September 2010 (UTC)

Another related idea is Elastic_potential_energy#Elastic_Internal_Energy_in_Compressible_Gases_and_Liquids. Remember that the Gas constant can be expressed in Volume*Pressure OR as Energy, Joules and pascal*cubic meters are (I think) equivalent units. I think the relevent issue is that the potential energy is proportional to either PdV or VdP; that is under a constant pressure changes in volume represent a change in energy; while under a constant volume, pressure differential would be the relevent relationship to energy. So, I think (and I could be wrong here) that this is as simple as volume of the eye * pressure differental across the eyewall; corrected for units. Since m³*mbar is technically an energy unit, you could leave it as that, or you can do the dimmensional analysis, and change it to joules or kilojoules or whatever. --Jayron32 06:00, 2 September 2010 (UTC)

Doing that gives me an energy of 1×10^15 joules (about a quarter of a megaton of TNT), which seems low (this says hurricanes release 6×10^14 J/second, and hurricanes can last at least 2 or 3 days on land). But I'm surprised that it doesn't depend on the density of the air. Wouldn't a heavier fluid contain more energy? Moving a given volume of water takes more energy than moving a given volume of air. Maybe not - I guess with a heavier fluid you need to move fewer atoms to produce a given pressure? Ariel. (talk) 06:07, 2 September 2010 (UTC)

The actual masses of the particles are irrelevent for this particular calculation; pressure and temperature already account for mass in their inherant formulation. The pressure is merely the force acting on a surface; that force doesn't care whether its a smaller particle moving fast or a heavier particle moving slow; its the same force so its effect on the energy should be the same. Also, that potential energy you have is instantaneous potential, that is the energy of the storm right now Besides dissipating energy, a storm also accumulates energy over time; that 1E15 Joules of energy you just calculated is not the total energy availible to the storm, since it will continue to gather energy even over land. On the balance, the storm loses more energy than it gains over land, but that's only because the rate of dissipation is greater than the rate of accumulation; the rate of accumulation is not zero. --Jayron32 06:22, 2 September 2010 (UTC)

I thought the rate of accumulation over land was zero since the energy source is latent heat of condensation of water. Ariel. (talk) 06:27, 2 September 2010 (UTC)

Tropical_cyclone#Mechanics and Tropical cyclogenesis discuss this in detail. There's even numbers in there. Yes, the primary source of energy is condensing water vapor. However, the process is far too complex to declare that this primary source of energy accounts for 100% of the energy in the storm... --Jayron32 06:34, 2 September 2010 (UTC)

That page says "When a tropical cyclone passes over land, it is cut off from its heat source and its strength diminishes rapidly." Seems to me that hurricanes must store energy somewhere else besides the pressure gradient, maybe in all the moisture they carry with them. Ariel. (talk) 06:39, 2 September 2010 (UTC)

Again, that doesn't mean that the energy accumulation over land is zero, only that it is much smaller than the energy accumulation over water. The energy is stored in both the pressure gradient, and in the internal energy of the gasseous water molecules, as compared to the internal energy of liquid water. If you want to know ALL of that energy, you'd also need to know the temperature of the air in the defined volume; that temperature should equal the dew point (the air should be saturated) so you can then compare that to the vapor pressure of water at that temperature, which can then be used to find the partial pressure of the water vapor, which can be used to calculate the total mass (or moles, whatever) of water vapor in the air. You can then use that to calculate the energy differential between the gasseous and liquid water; i.e. the latent heat of vaporization. However, your question only dealt with the energy stored as the pressure gradient. --Jayron32 06:48, 2 September 2010 (UTC)

Yes it did because I thought that was all of it. I'm going to try calculating the water energy tomorrow (unless someone wants to do it for me :) BTW thank you for the replies. Ariel. (talk) 06:56, 2 September 2010 (UTC)

Another idea (just popped into my head); a third, probably non-negligible source of energy is gravitational potential energy. After all, you just raised a HUGE mass of water rather high into the air, this height differential should itself be a possible source of potential energy. --Jayron32 07:00, 2 September 2010 (UTC)

Indeed it is. This gravitational potential is extremely significant. It is responsible for things like flash-condensation - which you might know as a lenticular cloud or a mushroom cloud (things with real wide, flat-bottomed layer structures). As air masses convect and are forced to different altitudes by fluid-dynamics (e.g. by momentum and viscous forces), they are working against an energy gradient, and they must cool rapidly to account for this energy-change. Rapid temperature decreases will lead to a change of state, and vaporous water will condense. That can result in a flat-bottomed cloud - this represents a gravitational equipotential surface (perturbed by local changes in humidity and temperature); but you can often see very sharp, crisp "bottom layers" when this occurs. Nimur (talk) 20:15, 2 September 2010 (UTC)

Pressure-volume work is defined as the integral of $\int {PdV}+\int {VdP}$. So, to calculate the work capability of a system with a pressure gradient, you can assume the hurricane is a giant carnot engine; the pressure gradient defines the limits of pressure in a thermodynamic cycle; and the volume can be estimated based on geographic scale of the hurricane. If there is net energy influx from solar radiation (or indirectly via ground or water re-radiation of heat), you can account for this by moving the adiabat. The work done (rather, the energy released by the hurricane), is the difference between two curves (just like any other heat engine). This work done is released as energy; it takes the form of kinetic energy of wind and water; as state-changes as water evaporates and re-precipitates; as changes to the temperature of massive volumes of air, water, and land, and so on. Nimur (talk) 20:06, 2 September 2010 (UTC)

It's only the first of those two integrals: if V is constant, no work is done because no interface is moving, regardless of the (change in) pressure. --Tardis (talk) 20:30, 2 September 2010 (UTC)

Well, it's not in a confined area like a piston cylinder; but the volume doesn't have to be fixed. The air-mass could be expanding. But you're right, you can make simplifying assumptions. Trying to model a hurricane from first-principles will require "some approximation." Nimur (talk) 21:26, 2 September 2010 (UTC)

The $\int VdP$ term is simply wrong and should be dropped; my example of constant V was just to prove that. --Tardis (talk) 22:35, 2 September 2010 (UTC)

Are you saying that the total derivative $dW=d(PV)$ does not equal $P\partial {V}+V\partial {P}$ ? Or that the term $V\partial {P}=0$ for this case? In either case, I don't understand your reasoning. (The former is a mathematical fact; the latter is because pressure can change in weather systems, and volume is non-zero, so V dP is not negligible). There is additional discussion about path-independent work in our article. Increasing the pressure for a fixed volume requires work; I don't know why you would say "no work is done." Nimur (talk) 23:04, 2 September 2010 (UTC)

Falling stick

A straight vertical rod of length l, when pushed, slides for a little while and then topples from the vertical position. What will be the velocity of the upper end when this end hits the ground?

My approach was to use the work due to torque (∫τdθ = (1/2)Iω²) about the end touching the ground, but I don't think this is right. If axis I'm using is accelerating (and if there's friction, it should be accelerating), then won't I have to take into account ficticious torques? But that would mean knowing something about the friction, no? 74.15.136.172 (talk) 10:45, 2 September 2010 (UTC)

To me this looks like a much simpler question. The horizontal component seems to be neglected, what with the "push" and the "sliding" and all. But for the vertical component, you know that the stick starts with the potential energy g*(l/2)*m (m=its mass, irrelevant) and no kinetic energy (neglecting horizontal). When it hits the ground it has zero potential energy and (neglecting any energy dissipated in friction) kinetic energy 1/2 m v^2. So v (on average) = sqrt ( 2/m * m/2 * l * g) = sqrt (lg). If one end is stationary, then the other end should be twice that. Unless I forgot something. Wnt (talk) 14:10, 2 September 2010 (UTC)

Well, I think you forgot about rotational kinetic energy...but anyways, if I were to ignore friction, then what I wrote above would be right, except that I don't think that friction can be ignored; otherwise, why mention that it slides for a little while? 74.15.136.172 (talk) 20:37, 2 September 2010 (UTC)

I don't believe I have to take rotational versus translational into account, if I simply average up the energy in all the particles of the rod. However, when I said "twice that", I think I needed to consider more carefully just what was twice what, since kinetic energy goes according to the square of the velocity... Wnt (talk) 00:55, 3 September 2010 (UTC)

OK. To take this your way, the rotational kinetic energy = 1/2 I ω². For a rigid rod spinning around its center, I = 1/12 m L². But here let's look at it spinning around one end, so all the kinetic energy is rotational. Using the integral from that page, I work out that its I = 1/3 m L². So Lgm/2 = 1/6 m L² ω². So ω²=3g/L, and ω=sqrt(3g/L). The top of the rod moves at speed sqrt(3gL) --- note that this differs from the 2 sqrt(gL) I said before because of the error in averaging kinetic energy I mentioned above. Hope I'm right this time... Wnt (talk) 01:24, 3 September 2010 (UTC)

Both our ways give the same answer, so I guess they're equivalent...but the question seems to suggest that we take friction into account. Any thoughts? 74.15.136.172 (talk) 01:51, 3 September 2010 (UTC)

I think the implication of "slides for a little while" is that the stick must have been pushed at its base. It stayed vertical until it came to stop. Then it toppled from stationary, meaning that there is no trace momentum of the original push. I think we have to assume the stick is very thin and ignore air friction, although that maeans the base is a point which makes sliding difficult. So we also assume the surface is frictionless. But then the stick would never stop sliding. I am not here and did not think any of this. Cuddlyable3 (talk) 13:16, 3 September 2010 (UTC)

Mathematical sequence

Hello PPL, whats the best approach for solving mathematical sequences like 1 4 17 54 145 368 945 ___? , the ones that are in IQ tests?Is there any kind of logical one? Or just trying blindly? TY much and sorry for my english.194.138.12.146 (talk) 12:41, 2 September 2010 (UTC)

For integer sequences, taking forward differences is a good place to start. If you don't see a pattern in the first forward differences, take forward differences again, rinse and repeat. Or, if you get impatient, you can look up the sequence at OEIS. Gandalf61 (talk) 12:55, 2 September 2010 (UTC)

I didn't see the pattern in the forward differences, even after looking at the entry - maybe I needed to compare something other than adjacent terms? But the ratios tell you something is up that is more complicated than an x_n+1=ax_n+b sort of thing:

4 4.25 3.17 2.68 2.53 2.56 ...

I'm actually surprised that this "Titan test" is so public. And to think the UK newspapers are griefing us over posting the ending to The Mousetrap! Wnt (talk) 14:18, 2 September 2010 (UTC)

Sounds interesting. Got a link? Vimescarrot (talk) 16:27, 2 September 2010 (UTC)

Assuming you mean the Mousetrap thing, see User_talk:Jimbo_Wales/Archive_64#Cyclopia.27s_unexpected_promotion_to_.22spokesman.22_for_Wikipedia_by_journalists_of_the_Independent, with lots more links. - Jarry1250 ^{[Humorous? Discuss.]} 18:34, 2 September 2010 (UTC)

Cheers! Vimescarrot (talk) 19:42, 2 September 2010 (UTC)

Actually I was asking the question from work, was not signed in, after 5 hours and 5 papers I solved it. After a while I began to try combinations with 2^, as it didnt work I tryed 3^, and there I saw a pattern, the algorithm is n^3+3^n but I was not following any rule for solving it.And its not "that" public, I was looking specifically for"Titan Test"DSTiamat (talk) 16:08, 2 September 2010 (UTC)

These sorts of sequences can be fit by an infinite number of functions, so there is not one "correct" pattern. Providing several integers and asking to fit "any" function leaves a null space of infinite size - there is literally no limit to the number of different, unique solutions that can yield those numbers. Typically, there's an implicit "integer-coefficients" assumption; there is usually the assumption that the sequence will be a linear sum of simple terms (like a polynomial in n), but unless these requirements are formally laid out for the sequence, there is no correct answer. Proper IQ tests do not use "guess the next number in this sequence" questions, because there is no objective way to define "correctness." If you have smarter and smarter participants (especially those who have formal training in mathematics, but not necessarily), they will solve the problem more effectively than the test-provider intended; and the test becomes invalid; so well-structured IQ tests will never ask this kind of question. Nimur (talk) 19:59, 2 September 2010 (UTC)

While there are indeed an infinite number of functions that'll work, it's generally clear that what is required is the simplest function that'll do the job - and it's usually very clear what that is once you find it.

Of course some sequences are tougher:

• 0,1,2,5,8,11...is the first N numbers that look the same if viewed upside down on a calculator display.
• 3,3,5,4,4,3,5,5,4,3,6...is the number of letters in the English word for 'N'.

If someone is using a criterion like that then no amount of messing around with numerical stuff will get you the "correct" answer you really do have to just have the answer pop into your head. But some are completely ambiguous:

• 4,8,12,16,20,24,28,32... has got you thinking 4N - when in fact I was thinking of the years in our calendar that were leap years and you wouldn't have had a hope in hell of realizing that until I said...92,96,104,108,112 and you were forced to figure out why I skipped '100'. Of course if I'd said 2004,2008,2012,2016... then you might have jumped at "Leap year" immediately rather than 2000+4N - but if I'd said 1234,1238,1242,1246,...then again, you might not have thought "leap years" and would have gone back to 1230+4N.
• Even if the answer is "simple" and "numerical", you might never find it other than by luck - but I bet you get: 3,1,4,1,5,9,2,6,5,3,5... which is certainly "simple" and numerical/arithmetic - but trying to fit a function to it would take you forever!

(10^n*pi)%10-(10^n*pi)%1 That wasn't so hard. —Preceding unsigned comment added by DanielLC (talk • contribs) 18:36, 4 September 2010 (UTC)

This clearly isn't about mathematics so much as the very human idea of "simplicity".

SteveBaker (talk) 06:13, 4 September 2010 (UTC)

What's funny is that all these sequences are in OECD. It actually lists four other answers for the 0,1,2,5,8,11 sequence! Wnt (talk) 17:49, 4 September 2010 (UTC)

Ant?

What is this? It looks like a queen ant from the species Myrmica rubra. Is it the ant flight time of year for this species in the UK? 82.44.55.25 (talk) 15:16, 2 September 2010 (UTC)

It's a bit hard to tell for sure but yes, it looks like a Myrmica rubra queen. AntWeb is your friend in all things ant and Wikipedia's too since they have released all of their images under CC-BY-SA. I guess it's still nuptial flight time in the UK for these guys, yes. Sean.hoyland - talk 16:22, 2 September 2010 (UTC)

". . .for these guys"?!! Richard Avery (talk) 19:40, 2 September 2010 (UTC)

Does "guys" not refer to a group consisting of either/both genders? 82.44.55.25 (talk) 19:55, 2 September 2010 (UTC)

It can, loosely; most people will understand it correctly. --jpgordon^{::==( o )} 19:59, 2 September 2010 (UTC)

guys, plural noun used to address people: used to address a group of people of either sex Hey, guys, where are you off to?

Microsoft® Encarta® 2009. © 1993-2008 Microsoft Corporation. All rights reserved. --Chemicalinterest (talk) 20:37, 2 September 2010 (UTC)

Yes, "guys" as in "Hey guys, let's watch Sex in the City and talk about why the liquids in sanitary napkin adverts are never red". Also, I believe Myrmica rubra males have long campaigned for a move to grammatically genderless semiochemicals. Sean.hoyland - talk 03:20, 3 September 2010 (UTC)

Microsoft Encarta? Why not Collins or Chambers or Oxford. They support the case, and I stand a little wiser. Up the workers!! Richard Avery (talk) 06:59, 3 September 2010 (UTC)

End organ

In your discussions of cutaneous senses and mechanoreceptors, reference is made to the "end organs" (e.g. Pacinian Corpuscles end organ[s]). What does that mean: "end organ" in terms of these sensory cells? I know cell body, and terminal buttons (at the synapse gap) and dendrites (on the receiving cell). I do not understand what "end organ[s]" is/are. Please explain this to me. (I am studying Neuroscience and have used the text and many web searches. Thank you.97.126.243.41 (talk) 17:07, 2 September 2010 (UTC)

As you stated, Pacinian corpuscles (together with Merkel discs, Ruffini endings, Meissner's corpuscles, end bulbs of Krasue and others) are examples of modified dendrite tips known collectively as special cutaneous receptors. There are also free nerve endings. DRosenbach ^{(Talk | Contribs)} 19:29, 2 September 2010 (UTC)

(after EC) They call the cutaneous sensory receptors "organs" because they are generally complex arrangements of different types of cells that together serve the purpose of transducing a given sensation to the nerve ending. Take a close look at the pictures in the Pacinian corpuscle, Meissner's corpuscle, organ of corti, olfactory bulb, and taste bud articles -- these structures are complicated collections of nerve fibers, connective tissue cells, blood vessels, etc. and they clearly meet the definition in organ (anatomy): "a collection of tissues joined in structural unit to serve a common function". The structures you are listing (cell body, terminal boutons, and dendrites) are all integral parts of the neuron itself, enabling communication of the sensory reception into the nervous system, but the overall process of sensory reception requires the more complex "end organ". Does that help? --- Medical geneticist (talk) 19:38, 2 September 2010 (UTC)

Where find actuarial statistics?

Where could I find data to answer questions like what percentage of woman lived past age 50 circa 1850? RJFJR (talk) 17:07, 2 September 2010 (UTC)

Which country do you want data on? --TammyMoet (talk) 17:57, 2 September 2010 (UTC)

Try life table. 92.29.120.223 (talk) 19:37, 2 September 2010 (UTC)

lifting an average passenger train car

I am not sure how heavy the average North American passenger train car is, but I am wondering if one were to want to lift said vehicle off the ground (say even 10 feet) using a helium-style balloon, how big would this balloon need to be? How would this compare to the Hindenberg in size?

Related to this, if one were to fill all available space (except where passengers sit, of course)with helium, how significant would this lighten the train car? 142.46.225.77 (talk) 17:54, 2 September 2010 (UTC)

You're going to need about a cubic meter of helium per kilogram of train car (And per kilogram of balloon!)

So if you imagine a 75ton train car, you're going to need about 70,000 cubic meters of helium.

So, Maybe a spherical balloon 110 meters across? Very roughly. APL (talk) 18:05, 2 September 2010 (UTC)

Our article suggests the Hindenberg had a volume of about 200,000 cubic meters, but then again it did weigh a significant amount in itself. I have no idea how much that is. - Jarry1250 ^{[Humorous? Discuss.]} 18:17, 2 September 2010 (UTC)

Our article also gives a "useful lift" of 10,000 kg, well below that of a rail car. On the other hand, I'm sure that you could build an airship comparable in size to the Hindenburg both much lighter, and that actually had light gas envelopes filling more of the volume. Buddy431 (talk) 18:59, 2 September 2010 (UTC)

Also, the Hindenburg (famously) used hydrogen rather than helium as its lifting gas - so it would have had a somewhat higher payload than a helium balloon. SteveBaker (talk) 05:40, 4 September 2010 (UTC)

I also see that a company known as CargoLifter wanted to construct an airship capable of carrying 160 tons, which would have had a volume of 550,000 cubic meters. Perhaps unsurprisingly, the company no longer exists. You can visit their former hanger, though, as the Tropical Islands resort in Germany. Buddy431 (talk) 19:06, 2 September 2010 (UTC)

First, if an Amfleet car is taken as the typical passenger car today, it weighs 116,000 pounds (58 short tons or 52.6 metric tons) according to Wikipedia. (Traditional passenger cars were often heavier, up to 70-80 tons, while subway cars are often lighter, under 40 tons.)

In the book The Golden Age of the Great Passenger Airships: Graf Zeppelin & Hindenburg by Dick and Robinson, details are given of the loading of the Hindenburg for its first transatlantic flight, a year before the disaster. I may as well reproduce the entire table for interest. All weights are in kilograms; divied by 1,000 for metric tons.

         2,960   37 passengers
         1,840   23 engine personnel
         1,680   21 deck crew
           800   10 stewards
           600   passenger baggage
         1,080   crew baggage
         3,000   provisions
         1,269   freight
            84   mail
        55,230   fuel oil
         4,000   lubricating oil
         1,200   reserve parts
         1,400   reserve radiator water
         1,150   drinking water and liquors
        11,300   trim ballast
         3,000   emergency ballast
         1,120   miscellaneous
         2,500   bedding, utensils, etc.
         1,000   moisture
         1,000   lightness
        96,213   SUBTOTAL
       118,000   dead weight (approx.)
       214,213   TOTAL

Clearly a bunch of these items apply only because we're talking about a passenger airship. And likewise for some of the "dead weight" -- if it wasn't carrying passengers you wouldn't need passenger cabins and all the things that go with feeding the passengers. And you could also go with a lighter structure (a blimp rather than a dirigible) if you were willing to have a less capable airship. So in short I think it's fair to say that while a Hindenburg-size airship wouldn't be quite big enough, it's too small by not all that large a margin.

Of course, the Hindenburg was inflated with hydrogen; with helium you get about 8% less lift, so you need a slightly bigger airship for that reason. (The calculation is (29-4)/(29-2) = 0.926, where 29, 4, and 2 are the respective average molecular weights of air, helium, and hydrogren.)

On the other question, the enclosed volume of a passenger car is something like 70 x 10 x 8 feet, which is 5,600 ft³ or say 160 m³. That is less than 1/1,000 of the gas capacity of the Hindenburg, so you can see that even if you were willing to suffocate the passengers by filling it with helium, you would not get any significant lift.

--Anonymous, 04:45 UTC, September 3, 2010.

Earth's Atmosphere

Coming back to wake you up again! If Earth had no atmosphere and we were still alive somehow; how can we see things around us look like, including the sky, horizon, sunset, sunrise, ..etc?--Email4mobile (talk) 21:55, 2 September 2010 (UTC)

The sky looks blue due to Rayleigh scattering - without an atmosphere it'd just be black. It doesn't answer any of your other questions, though. (I don't know whether you'd be able to see the stars during the day, but I doubt it.) Vimescarrot (talk) 21:58, 2 September 2010 (UTC)

You'd surely be able to see the stars so long as your eyes were dark-adapted at all (which would just involve being in a hole behind a wall). --Tardis (talk) 23:08, 2 September 2010 (UTC)

There would be no pre-dawn brightening of the sky. It would be completely dark until the sun first appeared over the horizon. Similarly at dusk, the sky would go dark as soon as the sun had set. Rojomoke (talk) 23:36, 2 September 2010 (UTC)

If it's not obvious, it would be exactly what it is like on the moon. One more thing I would add to the above is that shadows would be much sharper and darker because there would be no light scattering from the atmosphere or clouds. Vespine (talk) 23:49, 2 September 2010 (UTC)

However, contrary to popular belief, shadows would not be pitch-black. There's still diffuse interreflection to light things up. --Carnildo (talk) 01:20, 3 September 2010 (UTC)

While you might be able to see the stars, they wouldn't twinkle as that is caused by turbulence in the atmosphere. Confusing Manifestation(Say hi!) 01:56, 3 September 2010 (UTC)

No mirages. No Aurora. No Meteor showers to see until they make craters around you. No Rain, Snow, Wind or Weather at all. During a Lunar eclipse the occluded area of the Moon would be black not dark reddish. Your Barometer pointer is hard against the stop. You might notice that the tires of your bicycle and car are bulging because they are over-inflated, though I don't know where you got the gas to pump into them. If you had anything held by a Suction cup, well it isn't held. You would see neither the Space Shuttle nor parachutists land more than once (painfully). If you break an ordinary lamp bulb it may still work. Anyway everybody without a space suit would be dead which makes it a horrible idea. Cuddlyable3 (talk) 12:37, 3 September 2010 (UTC)

We didn't cover sunrises and sunsets. In both cases, the sun would appear as a sharp white circle - there would be no change in the color of sky or sun as it rose or set. Also no rainbows (because no rain). Kinda boring really. SteveBaker (talk) 05:38, 4 September 2010 (UTC)

The Sun would also be a sharp white circle at all the other times of the day. Without the sky scattering the blue, it wouldn't be yellow anymore. Also, the stars and planets would look a bit bluer. The oceans would freeze dry, and it would look like it's boiling and freezing at the same time. Not really an optical effect, but it's still cool. — DanielLC 17:11, 4 September 2010 (UTC)

Theory of Relativity and Time Travel

How does Einstein's theory explain time travel, and more importantly Grandfather's Paradox ?  Jon Ascton  (talk) 23:46, 2 September 2010 (UTC)

It doesn't, really. There are various interpretations of Special and General Relativity that seem to allow backward time travel from some reference frame (see Time_travel#In_physics), but they are pretty speculative and my understanding is that most physicists think they wouldn't have much of a chance of working. Relativity certainly has nothing to say in particular to say about the grandfather paradox. --Mr.98 (talk) 00:08, 3 September 2010 (UTC)

See however tachyonic antitelephone. If it were possible to send information faster than light by a frame-independent and direction-independent process, it would also be possible to send information backwards in time. That gets you the grandfather paradox. --Trovatore (talk) 00:15, 3 September 2010 (UTC)

See closed timelike curve, especially Tipler cylinder. Wnt (talk) 00:51, 3 September 2010 (UTC)

Right, but my point is, there are clever (and probably not workable ways) is come up with some kind of possible backwards time travel using SR/GR, but SR/GR itself doesn't say anything about how the paradoxes would be resolved, if they would be. --Mr.98 (talk) 12:09, 3 September 2010 (UTC)