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November 3

Time dilation? Time speeding?

The beginning of the section of {Time dilation and length contraction} under Special Relativity is confusing to me. Hope someone can help. It states:

See also: Twin paradox

Writing the Lorentz transformation and its inverse in terms of coordinate differences, where for instance one event has coordinates (x1,t1) and (x'1,t'1), another event has coordinates (x2,t2) and (x'2,t'2), and the differences are defined as Δx = x2 − x1, Δt = t2 − t1, Δx' = x'2 − x'1, Δt' = t'2 − t'1 , we get

△t’=γ(△t-(v△x/c^2)) △x’=γ(△x-v△t) and △t=γ(△t’+(v△x’/c^2)) △x=γ(△x’+v△t’)

Suppose we have a clock at rest in the unprimed system S. Two consecutive ticks of this clock are then characterized by Δx = 0. If we want to know the relation between the times between these ticks as measured in both systems, we can use the first equation and find: △t’=γ△t (for events satisfying Δx = 0). This shows that the time Δt' between the two ticks as seen in the 'moving' frame S' is larger than the time Δt between these ticks as measured in the rest frame of the clock. This phenomenon is called time dilation.

I am confused by the last two sentences. To explain △t’=γ△t in more detail, we can assume the unit time in S is Ut and in S' is Ut', then the time of the {two consecutive ticks event} measured in S is △t*Ut=(t2-t1)Ut, the t2-t1 can be 1; while measured in the S' is △t'*Ut'=(t'2-t'1)Ut', the t'2-t'1 may not be 1. Since they are for the same event, we have △t*Ut=△t'*Ut', then, put △t’=γ△t in it, we have Ut=γUt', that means the clock in S' is running quicker than the clock in S. This phenomenon should be called time speeding, isn't it?

When Einstein used x'=0 to derive t’=t/γ, so that △t’=△t/γ, we have time dilation; how can we call △t’=γ△t as time dilation too? Please help me to understand that paragraph. Thanks.

Regards, John HuangJh17710 (talk) 04:53, 3 November 2010 (UTC)[reply]

Relativity of simultaneity actually has some nice, non-mathematical, explanations of the problem. If there are any two events, seperated by some arbitrary distance, which are perceived in one reference frame to occur at the same time, there will exist other reference frames which perceive them not to be simultaneous, some in which one event occurs first, and some in which the other event occurs first. This perception is dependent not only on the location of the reference frames but on their relative motion to each other. In other words, for your problem, lets assume that you and I are moving relative to one another. If we both start our stop watches at what we perceive at the same time, there will exist three different classes of reference frames relative to us:

Some will perceive our clocks to run at the same speed.
Some will perceive your clock to run faster than mine
Some will perceive my clock to run faster than yours

That's the key problem with Special Relativity, there is no reference frame which is "right". All are equally valid. By convention, we hold the reference frame of the designated observer to be "stationary" and measure time changes relative to that one; thus since everyone else is moving relative to THAT observer, everyone elses clock is running a little bit slower. But this convention holds regardless of who you choose as the observer. If two spaceships are moving away from each other at some fraction of c, but each is in communication with the other, each will perceive the other's clock to be slowed down. If they then, at some later time, change speeds to be stationary relative to each other, the perception of which clock slowed down will depend on which ship had to speed up or slow down to "match speeds". If the two ships performed identical, mirror image paths and speed changes, then the clocks will match when they reach the same relative motion. If one ship decelerated while the other maintained a constant speed, you will get different results. Special_relativity#Lack_of_an_absolute_reference_frame and Consequences of special relativity also discusses some of this. --Jayron32 05:34, 3 November 2010 (UTC)[reply]

The OP is makingn a crucial mistake. He states "Since they are for the same event, we have △t*Ut=△t'*Ut'". That's wrong. In fact both observers will be using similarly constructed clocks which will be using the same time unit, say a second: Ut=Ut'= 1 second.67.78.137.62 (talk) 15:22, 3 November 2010 (UTC)[reply]

Yeah, the phrase "time dilation" is potentially confusing, because that phrase by itself doesn't do a great job of helping you remember which observer will measure a greater amount of time between two events under which circumstances. Fortunately, there's another phrase that expresses the time dilation phenomenon that's more helpful in that regard, namely "moving clocks run slow". That phrase is particularly useful if you think of a "clock" as being a light pulse bouncing between two mirrors, as described at time dilation#Simple inference of time dilation due to relative velocity. In this case, the clock in question is stationary in the unprimed system, so it's moving from the perspective of the primed system. Therefore, the "seconds" as ticked out by the clock in question will appear to be longer than a "real" second, according to the primed observer. That's time dilation.

The reason you're thinking of this problem as being a "time speeding" is essentially because you're thinking that the clock in the primed system that's measuring the time between the two events in question is running faster than it ought to, as considered from the perspective of the unprimed observer. The crucial flaw in that line of reasoning is that there is no such clock in the primed system.

You're making the mistake of thinking that there's a clock in the unprimed system that measures the time of events in the unprimed system, and a clock in the primed system that measures the time of events in the primed system. But the time of events can't be accurately measured that way. Because the speed of light is finite, trying to measure the time of all events with one clock would run into the problem of there having to be a delay between when an event occurs, and when news about the event reaches the clock, informing the clock to measure the current time. You might think you could just take the time delay into account, but that would involve needing to know the distance that the light traveled in between the event and the clock, which opens up a whole other can of worms. So what's done instead is that times of events are measured by a system of synchronized clocks, stationed at every location of interest. It is possible to synchronize a system of clocks, as long as all of the clocks in the system are stationary with respect to each other. However, to an observer that's moving with respect to those clocks, the clocks won't appear to be properly synchronized.

In the problem in question, the two events involved can indeed be measured by a single clock that's stationary in the unprimed system, because the two events occur at the same location as measured by the unprimed observer. However, the two events are not at the same location as measured by the primed observer, so the primed observer, who measures events with clocks that are stationary with respect to the primed observer, must measure the two events with two different clocks. The reason the unprimed observer thinks that the primed observer measured too much time in between the two events is not at all because the unprimed observer thinks that the primed observer's clocks are running too fast. Indeed, the unprimed observer will think that the primed observer's clocks are running too slow. Instead, the unprimed observer will consider the primed observer to have measured too much time between the two events because the primed observer's clocks aren't properly synchronized. Red Act (talk) 18:37, 3 November 2010 (UTC)[reply]

Heart muscles

What property of the heart muscles are exploited in the heart transplant?
In blood groups,why is it imposible for one individual to carry all 3 alleles?
why is the oxygen dissociation curves of animals living at high altitudes located to the left of most of those of other animals?

— Preceding unsigned comment added by 196.0.7.4 (talk • contribs)

Hmm, these look suspiciously like homework questions. Try reading "Heart", particularly the "Functioning" section. Axl ¤ [Talk] 10:41, 3 November 2010 (UTC)[reply]

"ABO blood group system", section "Inheritance". Axl ¤ [Talk] 10:44, 3 November 2010 (UTC)[reply]

"Oxygen-haemoglobin dissociation curve", section "Factors that affect the standard dissociation curve", will point you in the right direction, although it does not contain the answer that you seek. Axl ¤ [Talk] 10:50, 3 November 2010 (UTC)[reply]

Well an individual could carry all 3 groups if they were triploid... 128.143.170.201 (talk) 19:26, 3 November 2010 (UTC)[reply]

...or is a Chimera (genetics). DMacks (talk) 01:23, 5 November 2010 (UTC)[reply]

The proportion of swine flu to all flues

Of those people having flu in late 2009 in the UK, what proportion of them would have had swine flu, diagnosed or not? 2009 flu pandemic. Thanks 92.24.178.95 (talk) 18:41, 3 November 2010 (UTC)[reply]

It's a bit vague ('late 2009') yet also specific (UK). Not sure of that exactly. If it's any help the WHO Centre on Influenza in Melbourne says 90% of flu viruses in Australia and NZ in 2010 were H1N1 2009. The WHO influenza page may hold the information somewhere, but I could only see data going back for this year - the map there for January 2010 for example indicates about 80-90% of flu cases in Western Europe were H1N1 2009, but it doesn't seem to refine down to countries. The European influenza section of the WHO site may contain some information, as may the WHO Global Atlas of Infectious diseases, but that seems to want a login. --jjron (talk) 02:33, 4 November 2010 (UTC)[reply]

Try Google Flutrends, which should show the influenza cases and past predictions for individual countries. ~A H 1^(TCU) 18:27, 7 November 2010 (UTC)[reply]

limestone make-up and other soft stone

(Copied from the Earth Portal's talk page)

I don't know if this will reach anyone but am going to try anyway. I am an authenticator of North American Indian artifacts. I have been sent several items recently by a man that are no brainers to identify but have one really odd piece that I have not come accross before. It is a statuette of an Aztec or Central Amer. pre-columbian art. It appears to me to be Peperino Tuff. Whis is not from this hemisphere I don't believe. More likely Italy. My question is, could I be mistaking this for some sort of Limestone. Under high powered magnification it shows fine crystal like clusters covering the surface with isolated black pepper like crystal inclusions. It also appears to have a pinkish hue deep with-in the stone. It is porous with a conglomerate look to it. It also has a slight sodium taste to it when touching your tounge to it. Can anyone help itentify this stone type??? How long does it normally take to get a response for questions? Many thanks, Bob 11/03/10 —Preceding unsigned comment added by 74.197.184.126 (talk) 18:51, 3 November 2010 (UTC)[reply]

Your description sounds very similar to that in our Peperino article (which needs to be merged with our peperite article I see) and much less like a limestone, but note that that rock does contain fragments of limestone. Peperites are found all over the world however, Nigeria, Spain, Wales, France, Italy, California, Australia, Argentina, Canada, Chile, Japan, Antarctica, Kenya, South Africa, Scotland (just a quick selection from [1]), so it may not tell you much about provinance. Mikenorton (talk) 21:21, 3 November 2010 (UTC)[reply]

If you want more confirmation, maybe try contacting one of these geologists at your local university (I worked that out from checking where your IP locates). If you register an account here and take a photograph of the piece and upload it, then someone may be able to help further as well. SmartSE (talk) 11:52, 4 November 2010 (UTC)[reply]

if the moon and moon landings are real

why didn't they run around the moon to prove it is a sphere and not a disc like it looks? 85.181.145.78 (talk) 08:58, 3 November 2010 (UTC)[reply]

We can observe 59% of the moon's surface, thus it is not a disk Tidal_locking#Earth.27s_Moon. Also there have been plenty of satellites (inclusing the moon landers themselves) that orbited the moon, observing it completely. See e.g. SELENE. Furthermore, the moon is simply too large to run around, especially with a limited oxygen supply. 157.193.175.207 (talk) 09:28, 3 November 2010 (UTC)[reply]

To put it in context, the equatorial circumference of the moon is 10,921 kilometres (6,786 mi). Ghmyrtle (talk) 09:35, 3 November 2010 (UTC)[reply]

The Apollo rover had a top-speed of 18 miles per hour (29 km/h), and a range of 57 miles (92 km). IIRC, NASA didn't want the rovers to go out of walking-distance of the lander, incase they broke down. CS Miller (talk) 09:52, 3 November 2010 (UTC)[reply]

The following is imported from a duplicate question

It's rather a long way! They did fly round it - will that not suffice to convince you? (later) sorry edit conflict. The much better answer above wasn't there when I started to edit. Dbfirs 09:31, 3 November 2010 (UTC)[reply]

See Apollo 10. BTW that also confirmed that the Earth is (roughly) a sphere in case you doubt that too. Cuddlyable3 (talk) 13:38, 3 November 2010 (UTC)[reply]

Actually Apollo 8 of 1968 was the first mission in which humans circled the Moon and observed its far side; and, a decade previously, the Soviet Luna 3 mission of 1959 was the first probe to circle the Moon and send us photos of the far side. Comet Tuttle (talk) 17:17, 3 November 2010 (UTC)[reply]

Besides, the nature of lunar phases is sufficient to demonstrate that the moon is spherical. This has been understood for thousands of years. — Lomn 13:42, 3 November 2010 (UTC)[reply]

In other words, the moon doesn't look like a disc. Simple observation of the moon with the naked eye makes it abundantly clear that it is nearly spherical. Nimur (talk) 15:42, 3 November 2010 (UTC)[reply]

Trolling questions are best ignored. Looie496 (talk) 17:41, 3 November 2010 (UTC)[reply]

Observe this ~~film~~ animation of the Libration of the Moon and ask yourself whether a flat disk could do that. Cuddlyable3 (talk) 21:30, 3 November 2010 (UTC) I restored the question, see Talk page.Cuddlyable3 (talk) 21:30, 3 November 2010 (UTC)[reply]

Technically, that's not a film, it is a simulation generated from composite texture photographs by Clementine (spacecraft) (you can plainly see the orbit-track artifacts!) But the point stands. Careful long-term observation of the moon with the naked eye reveals exactly the same behavior. Moon enthusiasts regularly photograph this behavior. Nimur (talk) 21:43, 3 November 2010 (UTC)[reply]

Thank you Nimur for the correction. The animation is for a theoretical observer at the center of the Earth.Cuddlyable3 (talk) 08:03, 4 November 2010 (UTC)[reply]

Why does the moon get larger and and smaller in that animation? Does the apparent size from earth change that way over a month? Ariel. (talk) 22:28, 3 November 2010 (UTC)[reply]

Yes. As you can see in the infobox at Moon, the apparent angular size of Moon as observed from Earth fluctuates between about 29.3 to 34.1 arc-minutes - a visible difference of around 20% ! This is because its orbit, while "nearly" circular, actually has a perigee of 360,000 km and an apogee of 400,000 km - a pretty huge difference! Nimur (talk) 22:31, 3 November 2010 (UTC)[reply]

Why on earth would they run around the moon? That is very nearly the most useless thing they could do with their brief and incredibly expensive time on the moon. And it wouldn't have occurred to them that anyone would doubt that they were there. -FisherQueen (talk · contribs) 22:35, 3 November 2010 (UTC)[reply]

If they had "run around the moon", why would any of the moon-landing-deniers believe them?! I can't see why that would be "proof" of anything to anyone for whom their having orbited around the moon isn't "proof" of anything. Do you really think that would have "proven" to you that the moon is a sphere[oid], OP...? WikiDao ☯ (talk) 01:12, 4 November 2010 (UTC)[reply]

The OP's question caused some initial confusion because its premise "if the moon and moon landings are real" has nothing to do with the real question and instead provokes some to think this is about Moon landing conspiracy theories. The question itself is reasonable because (i) the full Moon looks disc-like because of its even illuimination, and (ii) until the late 1950s the Far side of the Moon was an unseen mystery. On October 7, 1959 the Soviet probe Luna 3 did indeed "run around the Moon" and took the first photographs of the lunar far side. Comet Tuttle already mentioned this. 08:20, 4 November 2010 (UTC)Cuddlyable3 (talk)

A flat-Earther would point out that a single circumnavigation will not prove that the Moon is a sphere. You've got to circumnavigate it longitudinally and latitudinally. That would be a lot of running. APL (talk) 14:30, 4 November 2010 (UTC)[reply]

And people have circled the earth and the flat-earth people don't believe them, so I don't see circling the moon as convincing to people dedicated to believing the moon landings were a conspiracy. Googlemeister (talk) 14:34, 4 November 2010 (UTC)[reply]

They have an explanation for longitudinal circumnavigation of a flat Earth, but if someone flew from South America to Australia via the south pole, then their theory would be in trouble. —Arctic Gnome (talk • contribs) 16:14, 4 November 2010 (UTC)[reply]

The mere act of doing that would not disprove the Frisbee model of the Earth. It would be entirely possible to start at South America, fly out to the outer edge of the world (Known to round-earthers as "the south pole"), fly about halfway around the edge, and then fly back inwards to arrive in Australia.

Admittedly, the observations made along the way would pretty much prove it (Man, We've been flying for hours and were still over the south pole?), but who trusts observations made by scientists?!? APL (talk) 20:23, 5 November 2010 (UTC)[reply]

The manned probes that used a Lunar orbit rendezvous course did indeed complete a loop around the Moon. ~A H 1^(TCU) 18:25, 7 November 2010 (UTC)[reply]

From childhood, it never seemed reasonable to me to perceive the Moon as a disk when looking at it with the naked eye. After all, it looks like a ball, and being a ball, the pattern of light and dark makes sense, whereas otherwise, why would these patterns be there? So cartoon depictions of somebody setting on a crescent moon, etc. seemed only like some bizarre comic convention to me when I was growing up. I'm curious how the OP ended up seeing it the other way, and if it's actually common for people to do so. Wnt (talk) 12:04, 8 November 2010 (UTC)[reply]