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February 13

Travelling to the future

I believe that the following is correct:

If you approach the speed of light, then relative to a slower moving body, time will seem to move more slowly
Therefore, if you circled the Earth at close to the speed of light, time on Earth would appear to pass more quickly than time on your "spaceship"
So, if you left now, got up to the speed of light quickly and then can back in five years, far more than five years could have elapsed on Earth (so you are in a sense travelling into the future)

Also:

With the lack of friction in space, you could potentially use a form of propulsion to gradually accelerate yourself to the speed of light
It would be within the reach of current technology to create a spaceship that could withstand being accelerated to close to the speed of light (because the accelaration forces would be very low) - although if it hit anything it could be destroyed
With the like of ion drives and a nuclear power source it is feasible to build a propulsion system that could accelate the spacecraft to close to the speed of light

My questions are:

Are all of the above assumptions correct?
Is there anything else that makes this completely unfeasible?
If it is unfeasible, what kind of technological advances would be needed to make it feasible? (or will it likely always be impossible)?
Could the spacecraft be accelerated close enough to the speed of light, quickly enough to mean that somebody launching tomorrow, would be able to travel a significant way into the future, and how far could they travel?
- For example, if a 20 year old man, with a life expectancy of 80, left today, how far could he travel into the future, taking into account that he would have to accelerate and decelerate again, with the 60 years of his remaining lifespan.

Any thoughts on this would be really appreciated. Thanks Blooper Watcher (talk) 01:23, 13 February 2010 (UTC)[reply]

I did some quick calculations myself, based upon a thrust figure of 88,000 mN (which seems to be the highest achieved by an ion engine). Assuming a mass of 10,000 kg for the spacecraft, I reckon that over 1,000 years you would accelarate to approximately 283,000 km/s (which is just shy of the speed of light). The calculation being:

(88 N * (1,000 * 365 * 24 * 60 * 60)) / 10,000 = 283,824,000 m/s

Obviously that is too long for a human lifespan (although I am not sure how close you need to get to the speed of light to see significant variations in relative time), and the engine would need to be 100 times more powerful to get to the speed of light in 10 years (and 10 more years to decelerate). But a 100 fold increase in power does not seem beyond the realms of possibility. Blooper Watcher (talk) 01:42, 13 February 2010 (UTC)[reply]

There is no doubt that if you could somehow accelerate yourself to somewhere close to the speed of light - then return to Earth again - then you would have "travelled to the future" - the faster you went, and the longer you stayed at that speed, the further into the future you'd go. In as much as you would seem to others to have lived more than your expected lifespan - and you would be "in the future" - complete with flying cars, personal rocket packs and a small fortune sitting in your bank account thanks to the one penny you left in there and the actions of compound interest!

That's a scientific fact that almost all scientists would agree on.

However, we don't have a way to get something as large as a human up to anything remotely close to that speed...just look at the hardware in the Large Hadron Collider, needed just to get some teeny-tiny sub-atomic particle up to close to light speed! But it's been calculated that the Apollo astronauts (who travelled faster than anyone else on earth ever has) did travel a tiny amount into the future. Apollo 8 astronaut Frank Borman is said to have actually demanded 400 microseconds of overtime pay from NASA because of the time dilation incurred in his trip.

So yeah - we have the technology to push someone 400 microsecond into The World Of The Future! SteveBaker (talk) 01:54, 13 February 2010 (UTC)[reply]

Thanks Steve. Do my figures for the acceleration using the most powerful current ion drive look correct? It would seem to me that a 100 improvement in the technology would be feasible which, again, if my figures are correct would be acceleration up to the speed of light in 10 years, and then decelarating back down in another 10 years. The thing is, how close do you need to get to the speed of light to get a noticable effect?

Let's say that we used the current drive, and over 10 years got up to 1% of the speed of light, stayed there for 40 years and then decelrated back down to stationary. How far, beyond the 60 year journey, would we have travelled? If we could accelarate up to 99% using a more powerful drive, and did the same 10 year accelarate, 40 years travel, 10 years decelerate, how much time travelling would we have done?

Also, aside from getting up to speed. Is there anything else that would always make this unfeasible? Blooper Watcher (talk) 02:04, 13 February 2010 (UTC)[reply]

One more thing, shouldn't NASA have been chasing Borman to pay some money back? Seems to me that he worked 400 microseconds *less* than the guys on the ground. Blooper Watcher (talk) 02:08, 13 February 2010 (UTC)[reply]

You should remember that mass is also relative. As you accelerate, the mass increases and you'll need more force to achieve the same acceleration (that's why you cannot reach the speed of light - see Speed of light#Upper limit on speeds about that). Your calculations do not seem to take that into account. Also, about "Therefore, if you circled the Earth at close to the speed of light, time on Earth would appear to pass more quickly than time on your "spaceship"" - it is going to be very hard to circle the Earth at such speed as it is going to be much higher than the escape velocity... --Martynas Patasius (talk) 02:42, 13 February 2010 (UTC)[reply]

For sure there is no way you could do this in orbit. The effects of relativity as you approach the speed of light make it increasingly difficult to push your speed that little bit faster. Even ion drives require reaction mass - and if you made them 100 times better, they'd need 100 times the amount of reaction mass - which would drastically increase the mass of your spacecraft and thereby eat most of the benefit you got from better engines. This has been thought about many times before by many smarter people than us - and it's quite utterly impractical. SteveBaker (talk) 05:03, 13 February 2010 (UTC)[reply]

As you say, there would be a big problem if the spacecraft hit anything when going at relativistic velocities. Unfortunately, it is guaranteed to hit something - the interstellar medium. In the vicinity of the Sun, the interstellar medium contains about 50,000 atoms (we'll assume they are all atoms of hydrogen, which is close enough to the truth) per metre cubed. If we assume a speed of 0.99c, that's a gamma factor of about 7. That means that, from the perspective of the spacecraft 50,000*7=350,000 atoms per metre cubed (due to length contraction) and each has a mass of 7 times that of hydrogen (due to relativistic mass) and they are all travelling at 0.99c. That means that, in every second, for every metre squared of cross sectional area, the ship will hit 350,000*300,000,000*7 u of gas moving as 0.99c, that applies a force of 0.00037 N. That compares to a car travelling at 30 m/s (70mph) having air resistance of 1080 N. Clearly, that isn't a big problem (although it is still a problem, since ion drives typically produce far far less force than a car's engine), but if we increase the velocity to 0.999999997c, it becomes roughly the same resistance as the car. So, as you can see, the resistance from the interstellar medium does provide a limit to how fast you can travel, although that limit is pretty high. (All of these calculations are assuming perfectly elastic collisions - if the hydrogen doesn't just bounce off the ship then you need to look at the energy of the atoms, which is extremely high since it is proportional to velocity squared, so we would probably find the ship being significantly eroded by the constant collisions.) --Tango (talk) 13:26, 13 February 2010 (UTC)[reply]

Tango, all that hydrogen can be seen as an aset instead of a liability. In principle it could be collected and used to power a fusion reactor that suplies energy to the ion drive. Dauto (talk) 15:40, 13 February 2010 (UTC)[reply]

Bussard ramjet perhaps? 220.101.28.25 (talk) 16:36, 13 February 2010 (UTC)[reply]

That article discusses the feasibility. Most studies seem to show that it wouldn't work (you get greater drag from the collector than you can generate thrust). Our article gives an example where it would work, but assumes an extremely high exhaust velocity with no justification for it being possible. --Tango (talk) 17:09, 13 February 2010 (UTC)[reply]

The now-cancelled Project Orion was to get a spacecraft to a theoretical upper limit of 10% of the speed of light. ~A H 1^(TCU) 23:25, 13 February 2010 (UTC)[reply]

As a torchship scuts through space at near the speed of light, a scoop on the front funnels any molecules encountered into the reaction chamber where they are accelerated to provide additional thrust. The scoop and the channel the molecules pass through are made of Unobtanium, a metal which is not readily available at present (but neither is Plutonium), but which has 157 hits at Google Book Search. Edison (talk) 04:35, 14 February 2010 (UTC)[reply]

See Unobtainium. Amazing the things we have articles on! --220.101.28.25 (talk) 06:00, 14 February 2010 (UTC)[reply]

isnt this similar to the question I had asked earlier about time travel, when I experienced " time travel" in flying between Dubai and Doha in 55 minutes and coming to a time zone 1 hour behind, thereby experiencing 5 minutes of the evning again... but as one person pointed out, this is not time travel, this is basically readjusting the watch. similarly your question - "if you left now, got up to the speed of light quickly and then can back in five years, far more than five years could have elapsed on Earth (so you are in a sense travelling into the future" is flawed becos while you feel you travelled for 5 years its merely readjusting the cclocks with regard to what the time is on earth. so this actually doesnt qualify as time travel... and all this if someone mages to fly almost the speed of light. —Preceding unsigned comment added by 213.130.123.12 (talk) 11:23, 16 February 2010 (UTC)[reply]

but the problem you have to realy consider, is that even with artificial gravity, im space you loose weight. thats why astronauts workout and still come back lighter then they where when they left. the thing just is, ower bodys arnt made to be in space. --Talk Shugoːː 20:51, 16 February 2010 (UTC)[reply]

Carbon dioxide and plants at night

A question has come up on a sleep forum, in regard to plants in the bedroom. I've tried to read the photosynthesis articles (difficult here, too simple at simple), and haven't found an answer. Do (some? all?) plants release carbon dioxide in the dark? Thank you. - Hordaland (talk) 10:36, 13 February 2010 (UTC)[reply]

Plants release carbon dioxide continuously, since they respire in the same way animals do. However, during the day (when they are in light) they also photosynthesise, and this converts carbon dioxide to oxygen. Therefore, it is generally true that the amount of carbon dioxide they take in during the day exceeds the amount they give out. This is not true at night time. However, the amount of CO2 they release should not be enough to cause any problems to people in the same room. --Phil Holmes (talk) 11:20, 13 February 2010 (UTC)[reply]

Yes, all plants do this - but the amount they produce is completely negligable compared to the amount a person sleeping in that room would produce. There are families in poorer countries who sleep 10 people to a room - and they don't suffocate! This is SO far from being a possible problem that might concern a "sleep forum" that it's almost laughable - so tell the people there "Don't worry - it's such a tiny effect that it's completely unimportant." SteveBaker (talk) 15:09, 13 February 2010 (UTC)[reply]

Thank you both! The forum (listserve, actually) is for sufferers of severe Delayed sleep phase syndrome, and we take each others' questions seriously. There is no effective treatment, so we question any little thing that might help -- or not. I will politely explain that the questioner needn't consider the CO2 from the "humungus bamboo tree" next to the bed. Thanks again! Hordaland (talk) 16:24, 13 February 2010 (UTC)[reply]

Steve, "suffocate" is your word. We haven't contemplated any such thing. Hordaland (talk) 16:31, 13 February 2010 (UTC)[reply]

The room would have to be airtight if something like hypercapnia were to occur. ~A H 1^(TCU) 23:20, 13 February 2010 (UTC)[reply]

That's the point. Virtually no rooms are airtight, also due to building regulations. --Ayacop (talk) 18:15, 14 February 2010 (UTC)[reply]

I have seen architectural plans of hospitals from the 1920's which had a "plant room" on each floor. Plants from patient rooms were stored in the plant room at night for fear they might deprive the patient of some oxygen if left in the patient room. I do not know what year this practice arose or when it was phased out. Schaum's Outline of Biology(2009) Q. 20.3 says that plant removal from hospital rooms at night is still done in some locations. Edison (talk) 20:59, 14 February 2010 (UTC)[reply]

magnetic vector potential actual meaning?

whats exactly ment by magnetic vector potential?is it only a conceptual idea? what tha physical meaning of magnetic vector potential. —Preceding unsigned comment added by Dakshu (talk • contribs) 10:51, 13 February 2010 (UTC)[reply]

The vector potential is definatly real. read for instance Aharonov–Bohm effect. Dauto (talk) 13:54, 13 February 2010 (UTC)[reply]

Well... the Aharonov–Bohm effect depends only on the integral of A around a closed loop. E and B tell you the integral of A around all infinitesimal loops. Assuming space is simply connected, you can integrate the infinitesimal loops inside the finite loop to get the integral around the finite loop, unless there's a singularity inside. So the only case in which the Aharonov–Bohm effect depends on something beyond the E and B fields is when there's a singularity in the field. Even then you can get it from the E and B fields alone if you treat them as generalized functions.

On the other hand, to calculate the Aharonov–Bohm effect from the E and B fields you have to integrate them over a region where the electron doesn't go. This isn't even a quantum-mechanical takes-all-paths situation, because this region is excluded from the path integral too—the electron really isn't there. But if you use A, the effect depends only on the region where the electron does go. In that sense, A is closer to the reality. But A has a large gauge freedom, meaning that only part of it is real, while the E and B fields do a perfect job of capturing the part of A that's real. What Aharonov–Bohm shows is that they capture it in a way that's nonlocal with respect to (some of) the physics.

There's a GR counterpart to Aharonov–Bohm in lensing around cosmic strings. The exterior gravitational field of a cosmic string is zero. Spacetime is flat around the string; objects do not fall towards the string. But if two objects moving in parallel pass on opposite sides of a cosmic string, they'll end up heading towards each other on the other side, because the integral of the metric on a loop around the string is less than a full circle. Instead of using the metric, you can get the same result by integrating the scalar curvature over a surface that intersects the cosmic string, as long as you don't mind that you're integrating over a region where the particles don't go. -- BenRG (talk) 23:07, 13 February 2010 (UTC)[reply]

Think of it as the magnetic field's version of a voltage (or, electric potential). As the voltage is just a mathematically different way to express an electric field, the vector potential is simply another way to represent the magnetic field, in terms of relative potential energy. Then there are just different mathematical conveniences and uses than the B and H fields. Magnetic potential goes into more detail with math and stuff if you haven't read it yet. —Akrabbim^talk 14:15, 13 February 2010 (UTC)[reply]

The reason a magnetic field's energy must be described with a vector potential, as opposed to a scalar potential, is because the force imparted by a magnetic field does not only depend on position (it also depends on relative velocities and orientations). In order to quantify this field in a way that satisfies conservation of energy, the potential field must be described by a vector. This is a direct consequence of the Lorentz force law, which is empircally observed. It helps to thoroughly understand what a potential field means in general before trying to apply that mathematical concept to the somewhat pathological case of magnetism (whose force is defined by a vector cross product and a time derivative - not difficult concepts, but enough to make the math much harder than the electrostatic potential case!) Nimur (talk) 16:44, 13 February 2010 (UTC)[reply]

Nimur, does an isolated positive electric charge move if you place it in a scalar electric potential field? Cuddlyable3 (talk) 17:49, 13 February 2010 (UTC)[reply]

A meaning of pathological is Relating to or caused by a physical or mental disorder.[1]. Nimur, is there another word you could use to characterise the case of magnetism? Cuddlyable3 (talk) 17:19, 13 February 2010 (UTC)[reply]

Sorry. I meant it in the sense that the mathematical convention seriously breaks from the normal form of a scalar potential field. I use "pathological" in the sense of any instance that diverges from the norm in a way that breaks simplifications. I think this usage is very common in math, physics, and computer programming - it has evolved and is distinct from the meaning in the medical field. I don't think I'm coining a neologism - let me find some prior usage. Nimur (talk) 17:27, 13 February 2010 (UTC)[reply]

See pathological (mathematics). -- BenRG (talk) 23:07, 13 February 2010 (UTC)[reply]

There's nothing wrong with using the word 'pathological' per si. But I don't agree with what Nimur is saying since the electric field expression also depends on the vector potential. Dauto (talk) 17:25, 13 February 2010 (UTC)[reply]

Sure - but we can construct a totally scalar electric potential field - that's the simplification I'm referring to above. It is possible (although it limits the situations you can consider) to construct the electric potential in scalar form. No possible simplification exists to represent the magnetic potential in scalar form. Nimur (talk) 17:29, 13 February 2010 (UTC)[reply]

\mathbf {E} =-\nabla \phi -{\frac {\partial \mathbf {A} }{\partial t}}.

Dauto (talk) 17:29, 13 February 2010 (UTC)[reply]

Electrostatics - therefore, there is no time variation. I think I mentioned this before. Nimur (talk) 17:30, 13 February 2010 (UTC)[reply]

Read Magnetic potential#Magnetic scalar potential.Dauto (talk) 17:43, 13 February 2010 (UTC)[reply]

Nimur, does an isolated positive charge move if you place it in a scalar electric potential field? Cuddlyable3 (talk) 17:49, 13 February 2010 (UTC)[reply]

Yes. That does not mean the field is time-varying - that does not mean a full electrodynamics treatment is necessary. See test particle if you don't understand why. This is called electrostatics and it is a well-developed, mathematically rigorous approach. Obviously, it is inapplicable in situations where the time-variance is non-negligible - that is, by definition, electrodynamics - and it is more general, requires harder math, and requires the vector potential Dauto has spelled out above. Nimur (talk) 00:07, 14 February 2010 (UTC)[reply]

Nimur, I read Test particle#Test particles in plasma physics or electrodynamics and don't understand how there can be a force to move the positive charge without a vector to define the force direction. Is not every electric potential field a vector field? Cuddlyable3 (talk) 18:42, 14 February 2010 (UTC)[reply]

Force is the gradient of the scalar potential. The gradient operator can create a vector field out of a scalar field. This is a very common formulation of energy-force relationships, even outside the realm of electricity and magnetism. The extension of the potential to a full vector field throws some non-trivial kinks into the matter, which is what the OP was asking about in the first place. As I've mentioned above, because magnetic force is defined by the lorentz force (which involves a vector cross product and a time derivative), the corresponding potential field is most succinctly described as a vector potential. Dauto has listed above a magnetic scalar potential - I have never seen this used in practice. But, what do I know, I'm only a professional numerical physicist... Nimur (talk) 18:46, 14 February 2010 (UTC)[reply]

Thus an electric potential field (except in the degenerate case of a uniform field where all points have the same or zero potential) is inseparable from a vector field. Yes the vectors are the gradients of the potentials at each point. This is we agree a common formulation. So Nimur I don't know how you claim "we can construct a totally scalar electric potential field". Cuddlyable3 (talk) 23:18, 14 February 2010 (UTC)[reply]

I think you should review the gradient article. Nimur (talk) 03:41, 15 February 2010 (UTC)[reply]

Yes. And? Cuddlyable3 (talk) 16:08, 15 February 2010 (UTC) who is also familiar with the Jacobian.[reply]

The vector potential does have a physical meaning, but it's somewhat difficult to visualize. I'll try to explain it by a toy example. Imagine you have a cylinder of wire mesh, kind of like this. But it's twisted: the circles are still circles, and they're still parallel to each other, but they've been rotated relative to each other so the wires running between them aren't straight any more. No matter how twisted the cylinder is, you can always untwist it if you're strong enough. Now imagine you have a torus of wire mesh, kind of like this. Again, it's twisted. You can always locally untwist any part of it, but in contrast to the cylinder, you can't necessarily untwist the whole thing. If the sum of all the relative rotations of adjacent circles all the way around the torus is zero, then you can untwist the whole thing; otherwise, you can't.

This "freedom to twist" is called "gauge freedom". The only properties of the wire mesh that we care about are those that you can't untwist. This means that in the case of the cylinder, none of the twisting matters; it's all "just gauge". In the case of the torus, the only thing that matters is a single number giving the total amount of twist around the whole torus.

Now (here's where it gets hard) increase the number of dimensions. You can think of it this way: if you take a line and replace each point with a circle, you get a cylinder. If you take a loop and replace each point with a circle, you get a torus. What you want to do now is take space and replace each point with a circle. As you go from point to point in space, there's a corresponding twisting of the circles, and the question is to what extent you can untwist them and to what extent you can't. The answer is that what you can't untwist is completely captured by the total twist around all closed loops in space (the total twist around the loop being defined in the same way as it was for the torus). You can get the twist around large loops by adding the twist around smaller loops inside the large loop, as illustrated at Stokes' theorem#Underlying principle. This means that if you know the twisting around infinitesimal loops then you can find the twisting around all loops, and hence you know everything there is to know about the twisting. This is what the B field tells you. The magnitude of the B field at a point is the maximal amount of twisting of any infinitesimal loop at that point, and the direction of the B field tells you which loop is maximally twisted (namely, the one in the plane perpendicular to the field).

What the A field tells you is the amount of twisting as you move from one point to another. The direction of the A field at a point is the direction in which the twisting is maximal, and the magnitude of the A field is the amount of twisting in that direction. The A field has a gauge freedom corresponding to your freedom to twist. The B field doesn't; it tells you just the part of A that matters.

This analogy is much closer to reality than it might seem. If you imagine there really is a circle at every point in spacetime, twisted in the way I described, and write down the general relativistic equations for 5D gravity in that 5D space, you actually get Maxwell's equations and 4D gravity, the A and B fields literally mean what I just said they mean, and electric charge corresponds to momentum around the extra dimension. This idea dates back to the 1920s and is called Kaluza–Klein theory. -- BenRG (talk) 23:07, 13 February 2010 (UTC)[reply]

Paranormal

Did Jung have any faith in supernatural ? As far as I know Freud did not beleive that any entity does survive bodily death. But Jung's collective unconscious is some what near and dear to spooks...no ?

Jon Ascton (talk) 11:39, 13 February 2010 (UTC)[reply]

Many interpretations of Jung's writing do exist - it wouldn't be too far to consider his archetypal unconscious as a "supernatural" concept - even though he surely considered it "science". Because the collective unconscious was not based on empircal evidence, though, it's not really a scientific conclusion. Later generations of psychologists categorically discredit that line of reasoning in favor of observation-based theories of psychology. Nimur (talk) 16:39, 13 February 2010 (UTC)[reply]

See our article on synchronicity -- most people think of that as supernatural. Looie496 (talk) 18:15, 13 February 2010 (UTC)[reply]

Also look at collective unconscious. ~A H 1^(TCU) 23:17, 13 February 2010 (UTC)[reply]

Why did the Challenger shuttle explode in a white cloud?

When watching the explosion again recently, I was struck by the fact that the "smoke" was all white, rather than black/grey that I would expect to see from an explosion. Is this because the white stuff is actually refrigerated fuel that did not combust and instead turned to a ball of vapour that obscured any of the smoke? Blooper Watcher (talk) 14:42, 13 February 2010 (UTC)[reply]

The fuel in the external tank is about 800 tons of liquid hydrogen and liquid oxygen. When that exploded, the reaction produced 800 tons of water vapor - which is white. What you're seeing is a small cloud - in the sense of a "rain cloud". (Um - actually, less than 800 tons - it would have used a good chunk of that during the launch sequence - but still a heck of a lot.) SteveBaker (talk) 15:00, 13 February 2010 (UTC)[reply]

(ec)Small quibble. The Shuttle itself did not explode, it was the external fuel tank that ~~exploded~~ 'deflagrated'(?). Aerodynamic forces broke up the orbiter. See Shuttle challenger disaster. The long continuous trails are exhaust from the solid rocket boosters. As per SteveBaker & cloud See Shuttle_challenger_disaster#Post-breakup_flight_controller_dialog 220.101.28.25 (talk) 15:09, 13 February 2010 (UTC)[reply]

This NASA link can probably answer any questions about the Challenger STS-511 --220.101.28.25 (talk) 15:19, 13 February 2010 (UTC)[reply]

As an interesting aside, since it produces so much water, the launches of shuttles like these cause sudden downpours a few hours later. Vimescarrot (talk) 16:27, 13 February 2010 (UTC)[reply]

^{(citation needed?)} Eight hundred tons of water spread over a path 100 km long and 100 meters wide (let's say) works out to a lousy 80 grams per square meter -- less than a tenth of a millimeter of liquid water, so I don't imagine it's just the Shuttle exhaust falling out of the sky. What I could believe is that the long trail of water vapour could nucleate cloud formation and trigger rainfall that way.... TenOfAllTrades(talk) 17:02, 13 February 2010 (UTC)[reply]

A (probably overly-dramatic) show presented by James May. Probably James May's 20th Century. Vimescarrot (talk) 17:07, 13 February 2010 (UTC)[reply]

Even smaller quibble. Water vapor itself is basically colorless in the visible spectrum. The white stuff, like any cloud, is condensed water (i.e. tiny droplets of liquid). Buddy431 (talk) 16:33, 13 February 2010 (UTC)[reply]

drag vs Reynolds #

There is a relationship between coefficient of drag and Reynolds number. When the object is sphere, the relationship can be pictured like this. I've seen versions of this figure in a number of chemical engineering textbooks. What I haven't been able to find out is whether this relationship is an empirical correlation, or an analytical solution, i.e. is this a relationship that is just observed to be true in experiments, or is it a relationship that must be just so according to the theory in this area? ike9898 (talk) 19:23, 13 February 2010 (UTC)[reply]

The limiting behavior (i.e. the behavior at very high and very low Reynolds number) can be derived analytically. The intermediate regimes can probably be modeled computationally for idealized fluids, but are more likely to be measured empirically for any actual fluid. Dragons flight (talk) 08:11, 14 February 2010 (UTC)[reply]