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Talk:Speed of light/Archive 1 (Up to end of 2004) Talk:Speed of light/Archive 2 (2005-July 2006) Talk:Speed of light/Archive 3 (July-2006-end of 2006)

mnemonics

To be honest, I'm not convinced the mnemonics section is very encyclopaedic; do people feel that it enhances the quality of the article, or should we get rid? Thanks — Alan 21:15, 23 January 2007 (UTC)

I don't know, maybe leave just one, or maybe not. I'm for removal too, since people who wants to learn by heart the speed of light are rare (personal opinion). Certainly, this shouldn't be section 2 of the article. Dravick 01:42, 24 January 2007 (UTC)

It's kind of neat to have such a mnemonic. That being said, when I was a physics major in first-year physics we used rounded numbers. People who need the exact number probably have it programmed into their desktop Crays by now. But I think any mnemonics should be preserved for posterity in a footnote or other brief mention near the end of the message. P0M 02:28, 24 January 2007 (UTC)

Okay, thanks for the responses. I'll relegate them (or one of them) to a footnote. — Alan 20:48, 26 January 2007 (UTC)

On second thoughts -- even the existing footnote references are to serious material; a 'mnemonic' reference would be conspicuous among them by its comparative lack of seriousness. I'll just delete, as I can't see any way to include them that fits in with the content and style of the rest of the article. — Alan 20:58, 26 January 2007 (UTC)

The actual Speed of Light

It says in the main article that the speed of light is 299,792,458 metres per second... When in fact it's that many KILOmeters per second... Someone should really fix that.... Seriously.

Erm, not it isn't. The majority of physics undergrads knwo that the speed of light ~ 3E8 m/s. As the article says, 299,792,458. Which is the same as 299,792.458 km/s. Note comma in one, decimal point in the other. In localities where the comma is the decimal separator, then it would be written as 299.792.458 m/s or 299'792'458 m/s or 299.792,458 m/s or 299'792,458 m/s. To avoid decimal separator confusion, the speed of light is 299792458 m/s. Three hundred thousand kilometres per second, or three hundred million metres per second. Google Speed of light and look at Google Calculator if you don't believe me. Stannered 23:24, 28 March 2007 (UTC)

• Um, you're right, sorry about that. For some reason I thought the second comma was a decimal point. Feel free to delete this comment

Einstein's metaphysics implications HELP PLEASE !!!

In previous questions it is mentioned that if we assume that light has some kind of awareness, or if we place our selfs in the place of a photon (or simply if we would by some means travel at the speed of light) the speed would be not c but infinite. The Maths works out right i think. Light has infinite speed in it's own perspective (and maybe infinite mass as well...?). And time is 'stooped' at the speed of light. So, if for instance a photon leaves the Sun towards the Earth, get here and is reflected back, it would arrive at the Sun in time to see itself leave? Doesnt this mean that for light the Sun and the Earth are actually in the same place? Isnt the universe all in the same 'place' then? And can time stop and still exist at all? Doesnt all this mean that light is actually outside time and space? Meaning completely outside time and space as we know it, not in 3D or 4D but actually in zero dimensions. Should we (or can we) think of all this as some kind of Einstein's metaphysical implication of omnipresent and immortal light? Is one photon all the photons there are? Does all this make any sense? (Uanbiing 00:10, 7 February 2007 (UTC)).

Sorry, but your main idea of time is still tied to the age-old way of thinking about it. If you were rendered unconscious, and your wrist watch was stopped temporarily, and then you were put in the cargo compartment of a bus going from LA to NYC, (subjective)time would not have passed for you because your brain was effectively put in neutral just as your watch was turned off. When you were revived and your watch was restarted and put back on your wrist you would think that it was only a few minutes since the knock on your door in LA, but when you "went back outside" you would discover that a week or so had gone by, somehow. That's only an analogy. If travel were possible at the speed of light, not even in the dustiest cargo bin would dust settle on you. You would think that the trip had gone real fast, but only because you clock had been stopped. Your clock being stopped doesn't affect all the other clocks in the universe. They keep ticking, and when you wake up you have to deal with all that has changed in the interim.

Time is private and personal to the timekeeper. A perfectly good and highly accurate clock and its twin keep perfect time with each other while they are both on earth. Send one into orbit above the earth and, because it is whizzing around up there, it will slow down. The only way you can make sense of the words "slow down" is "slow down in comparison to its twin clock back on earth." So if it went so fast that it stopped then you could only say that "it slowed down to the point that it stopped -- from the standpoint of the twin clock on earth." The guy who was wearing this very expensive timepiece gets off the space ship after its return and says, "My, that trip hardly took any time at all, so why have my hamsters starved to death in my absence?" (Well, I hope we educate our astronauts well enough that they don't have such a naive reaction.)

Light is not inside and not outside of time. That is because time is not like a flowing river -- even though that is how we think of it most of the time. Being "in" time means being in motion, and being in motion can only happen if something is in space. If something were "not moving" at a radical level, if electrons don't even move around in that thing, then there would be no time for it because there would be no movement, no change, so it would be as though that thing were something in a motion picture and somebody stopped the projector. The clock showing on the screen is stuck at a certain hour, minute, and second until somebody starts the projector again. (Again, that's an analogy. Obviously we have no evidence that we are part of a movie. Some Buddhists have a kind of idea in which the universe exists as a sequence of little slices analogous to the frames of a movie but in 3-D, however.)

Going faster in space causes you to go slower in time. That's the key observation. We see that fact by observing the clocks on any GPS going slower, and we can formulate an equation that relates the time seen on that clock with the time seen on one's clock here on earth, and we now have made these measurements over and over and over again and always get the same answer, so we regard it as a very good bet that it will work the next time we try it. That kind of thing is the reality that we all see, and we have to take this basic and well-confirmed observation and theory and work out its consequences until we get used to it.

Probably if we all had to program GPS location finders and systems and had to deal with the time dilation on a regular basis, then it would cease seeming funny after a while. I suppose that the first human ever to regard the horizon as "the end of the world" and then try to get to the end of the world must have had some trouble explaining things to himself/herself. Even as kids we may have had the same impression. Eventually we all get over it. P0M 02:25, 7 February 2007 (UTC)

Huygens' and Newton's estimations

According to data in the article Huygens' value is "better" than Newton's. If we use equatorial Earth radius 6378135 m and assume data from article - Huygens (?): 2000 Earth radii per minute, we get about 212,600 km/s, Newton (Optiks, 1704): 33.2 Earth radii per second, we get about 211,750 km/s. Obviously Huygens' value is better, or am I missing something? Also Boscovich in 1758 gave estimated value 20,000 Earth radii per 1/8 hour (~ 450 s) and 283.470 km/s, what is even better. --xJaM 03:35, 8 February 2007 (UTC)

Newton and Huygens's values are virtually identical, according to the article. I don't think it is worth distinguishing one as "better" than the other, when they are within 1% of each other, yet more like 30% from the true value. You are right though, Huygens' value is slightly better. In regards to Boscovich's value, keep in mind that it was determined about 80 years after Newton's value. I have removed the incorrect wording from the article. Grokmoo 04:11, 8 February 2007 (UTC)

Yes, I uderstand that both values are almost identical - but just a sentence in the article said differently. I've added Boscovich's value not regarding directly to Huygens's or Newton's estimations, but just for an information (that might be included in the article). Yes, I know that it was estimated so late, but it is still a very good one, since a.u. at that time was not yet measured so good. Cassini measured a.u. in 1672 (109.8×10 m - parallax of Mars at oposition, Cassini, Flamsteed, 1672, 138.4×10 m - declination of Mars), and the first pretty satisfying measurement of a.u. was made in 1769 during the transit of Venus (153.9 and 148.2×10 m) - 11 years after Boscovich's estimation, which gives for a.u. around 127.6×10 m. --xJaM 01:56, 10 February 2007 (UTC)

Ok, everything looks good. Thanks for the heads up on the error. Grokmoo 17:13, 16 February 2007 (UTC)

Question on definition

Does anyone know why the definition went with 299792458 instead of a nice round 300000000? was it just to keep everyone from having to junk their exisiting meter sticks? —The preceding unsigned comment was added by 216.70.247.242 (talk) 22:42, 12 February 2007 (UTC).

You're exactly right. The metre was originally defined way back by the French. It was only in 1980s that the current (much more precise) definition was adopted as being a 1/299792458 of the distance light travels in a second. But they needed to choose the right number so that all existing distances were close to the new definition, so avoiding the need to remeasure everything and replace all rulers.Mralph72 11:45, 15 February 2007 (UTC)

Contradictory (?) sentence in Sec. 3.2 - Technical impossibility of hyper-light-speed travel

"If this is true, an object may travel at the speed of light, but it will not travel through space when this is done, as no time will pass while it is at the speed of light."

The above sentence, which appears in the article, is either contradictory in itself, or implies some kind of speed-of-light travel of objects which occurs, but not through space. If an object is traveling at the speed of light, but it is not traveling through space, where is it traveling? If "no time passes", it should be specified FOR WHAT or WHOM, or for the entire universe. Is this non-passage of time a way of saying that the event of faster-than-light travel doesn't actually occur? Or does it mean the TRAVELER or TRAVELING OBJECT experiences no passage of time? Also, if "no time passes", would not this object be permanently frozen in time, as no events could occur to it to change its circumstance? If no force can act upon it, since it is outside of time, would it therefor be an irresistable force? An unmoveable object? Is this an accepted theoretical phenomenon? Does the sentence mean that the object will in fact travel at the speed of light, but for a period of time exactly equal to zero? Does this make sense, or is it sophistry, or is it a kind of thought experiment? This sentence needs clarification, correction, or removal. --4.239.0.189 18:12, 23 February 2007 (UTC)

Ugh. I think this wording should be removed from the article entirely. If you look at the equations for the Lorentz transformations, you see that they indicate that no time will pass for an object traveling at the speed of light. This, in some sense, implies that for a light particle, all events actually occur simultaneously. Note that this is only for the light particle itself, not for an outside observer traveling at less than the speed of light (relative to what he is observing). The light particle is not really "outside of time" in the sense you are thinking, but yes, the wording in this section is not really clear. In fact, I think the whole section could really use a good rewrite. I'll mull this over for a couple days, and then hopefully have something better to put down. Any other contributions will be very welcome. Grokmoo 02:14, 24 February 2007 (UTC)

I agree. The idea of "traveling" when delta T = 0 and delta x,y,z = 0 is a total distortion of the idea of moving or traveling. If one observes cases of things that move at speeds closer and closer to c, one will observe their clocks going slower and slower, but their perception will be that less and less time passed while they were going a very great distance. P0M 02:39, 24 February 2007 (UTC)

Double Ugh! When I read the whole thing I realized that it is one of those pieces of writing that is "correct" when you know what the guy is trying to say, but is terribly confusing if you are coming at it from the outside. The basic idea is very simple, but gettint it out using everyday concepts like "velocity" to talk about time is bound to lead to problems because, for one thing, it hypostaticizes time. If you talk about clocks and how many ticks a clock makes between here an Alpha Centauri it is much clearer than saying that "the velocity of time slows" during this trip. I rewrote this section off the top of my head, depending on memory and visualization. I think I have not misrepresented the math, but this is one of the cases where more is less and less is more, i.e., if you take the reader through the math it's clearer than if you act as a "simplifier." Here is one case where good visual aids would be invaluable. Actually seeing the measured time bar shrink as the measured velocity bar increases would make things lots clearer. A little sweep second hand accompanying the measkured time bar that would slow down as the speedometer needle crawls up toward C would make things even clearer. And an accompanying equation could show the relevant calculations. Is anybody good at writing simulation software that will work on a website?

As it is, this section is still something that will make the unprepared reader conclude that relativity theory is incomprehensible. P0M 03:37, 24 February 2007 (UTC)

Is there an easy way to transfer an Excel spreadsheet to a Wiki page? I can do the simulation in numbers and static graphics, show the math, etc., but the easy way is to use a real spreadsheet. I don't feel like copying it cell by cell into a table. P0M 03:45, 24 February 2007 (UTC)

If what you are saying is true-- and "everything happens at once" from the point of view of a light particle, then is perhaps light a window on eternity?

Sean7phil 17:14, 17 June 2007 (UTC)

Another imperfection

The current text says that light moving through a medium other than a complete vacuum is actually a "light-like hybrid of electromagnetic waves and mechanical oscillations of charged or magnetic particles such as electrons or ions, whereas light in the strict sense is a pure electromagnetic wave." What is the reader supposed to make of this odd formulation? "Hybrid" is a biological term and is not helpful in guessing at what the writer was trying to say. The picture it paints in my mind is that of light in empty space hitting the window of a space craft (for instance) and suddenly turning into some sort of conjugal combination of electromagnetic waves traveling along with mechanical oscillations of various kinds of subatomic or atomic entities.

Somebody who has a thorough grounding on this subject should rewrite the quoted material so that it accurately communicates the correct understanding of the change in measured speed.P0M 05:21, 24 February 2007 (UTC)

Gravitational Lensing

"Changes of gravity, however, warp the space the light has to travel through, making it appear to curve around massive objects." Is it more accurate to say it "appears to curve" or simply that "it does"?69.119.13.218 20:59, 12 March 2007 (UTC)

The problem is that light always goes straight, and what "curves" is the space it travels through. Or, to put it another way, light always takes the shortest trip between two points. If we graph the positions of Earth and stars we think we know where to set our "gunsight" so that at a certain time of the day or night the gunsight will exactly line up on the star. Then we find that when the light has to skim by our sun on the way to earth the distant star is no longer where we expected it to be. Do we say that light in this case has selectively changed its nature so that it curves through empty space? Or do we say that light goes in a "shortest distance" path, but what is the "shortest distance" changes in the presence of a massive object like the sun? Light does not have mass, so we cannot say that the sun "pulls" the light out of its normal path.

The idea that gravity is an attraction explains why Mars stays in orbit around the sun, but it does not explain why something massless would be "attracted" by the mass of the sun. The idea that "gravity" is a misconception may be a very productive idea because space curvature explains what was previously called "mass attraction" but it also explains "curving" of light's path. P0M 03:46, 24 March 2007 (UTC)

"Massless" means that were the particle stationary, it would have no mass (see rest mass). You're not going to find a stationary photon, hence why it is massless (crudely). However, it has an energy, and that energy can be translated into a mass by E=mc^2 (or more accurately, E^2 = m^2 c^4 + p^2 c^2). So you can think of gravity applying to this equivalent mass, hence bending the light. Or, more accurately, you can indeed think of space-time curving. Mike Peel 05:14, 19 April 2007 (UTC)

Error in the main text

Because of my purr English I do not dare to mess with the original text of the article. However, somebody should take a look at the following erratic paragraph:

"At velocities at or approaching the speed of light, however, it becomes clear from experimental results that this rule does not apply. Two spaceships approaching each other, each travelling at 90% the speed of light relative to some third observer between them, do not perceive each other as approaching at 90% + 90% = 180% the speed of light; instead they each perceive the other as approaching at slightly less than 99.5% the speed of light.

This last result is given by the Einstein velocity addition formula:

${\displaystyle u={v+w \over 1+vw/c^{2}}\,\!}$

where ${\displaystyle v}$ and ${\displaystyle w}$ are the speeds of the spaceships as observed by the third observer, and ${\displaystyle u}$ is the speed of either space ship as observed by the other."

According to the explanation of the last sentence the formula should be:

${\displaystyle u={v-w \over 1-vw/c^{2}}\,\!}$,

since in the third observer's eye the two velocities are 0.9c and -0.9c. The formula results in the correct valule (0.995) only this way. If the formula does remain then the last sentece is incorrect:

"where ${\displaystyle v}$ and ${\displaystyle w}$ are the speeds of the spaceships as observed by the third observer, and ${\displaystyle u}$ is the speed of either space ship as observed by the other."

The right description is

"where ${\displaystyle v}$ is the speed of, say, the first paceship and ${\displaystyle w}$ is the speed of the other one if observed by the first spaceship."

Please consider my suggestion.

• • • • • • • • • • • • • • • Since no one had reacted I did the correction by myself.

I'm sorry that nobody responded to your comment before. I just checked the first textbook of relativistic physics I found on my bookshelf, Introduction to the special theory of Relativity, by Claude Kacser, Department of Physics and Astronomy, University of Maryland. Published by Prentice-Hall.

That textbook gives:

${\displaystyle v={v'+V \over 1+v'V/c^{2}}\,\!}$ (4.8)

The text explains: "Here v, v', and V are all parallel velocities, and Eq. (4.8) refers to their magnitudes. If V and v' are both positive, v is less than their algebraic sum. That is, parallel velocities do not add according to the laws of arithmetic. Of course, if both v and V are much smaller than c, the relativistic correction is negligible.

According to this writer's scheme, v' and V are velocities of two travelers as they experience their velocities. In other words, v' and V are velocities relative to each other or as measured by the two travelers, and v gives us the velocity relative to a "motionless" observer.

The author notes, "We have assumed that both v' and V are positive. All our equations hold independently of the signs of v' and V, and the modifications necessary for our discussion of negative v' or V are straightforward."

However, if you work through the formulas and use 0.9c and -0.9c you will indeed come up with a mathematical problem because you get a zero in the numerator.

A negative velocity is just a representation of motion in the opposite direction of some other motion given a positive velocity. Depending on where they are, the two are either approaching, passing, or departing from each other. Two positive or two negative velocities descriptive of some situation would indicate two ships going in the same direction. They could still have a velocity relative to each other, and they could still pass. On the other hand, what would happen if the two spaceships were both moving in the same direction at 0.9 c (or -0.9 c) according to an Earth observatory? They wouldn't have any velocity relative to each other. If we write the formula your way, then two identical velocities would calculate out as a 0 velocity between ships, which is what you want.

Now I'm puzzled as to why two different (?) sources have preferred a formula that looks like it needs a good spin doctor. The equations may "hold" regardless of the signs of v' and V, but the answers are different. P0M 07:07, 16 March 2007 (UTC)

I'll come back to this later. P0M 22:28, 16 March 2007 (UTC)

The equation works when you have a spaceship traveling at 0.9 c, and a very fast crew member runs at 0.9 c in the direction the ship is traveling and we get the 0.995c. If he runs in the other direction then it is at -0.9c, he appears stationary, and the zero numerator makes sense. These final values are seen by the fixed observer watching the ship come directly toward him. Now if the observer is on the ship, he appears fixed and the rest of the universe is coming directly toward him at 0.9c, in that universe is another ship moving at 0.9c wrt the universe. This is the same as observing a ship (the universe) and a crew memeber (the other ship.) Plugging the numbers in from that perspective makes the equation work as written, particularly since the original text said 0.995c was what the spaceships observed. I claim no knowledge of relativity, so correction would be welcome. John N. 00:29, 17 March 2007 (UTC)

The problem wasn't with the formula. I doubt that two reputable physics textbooks would get it wrong, and get it wrong the same way.

The problem was with applying formulae without thinking things through. The textbooks that give the formula that uses plus in the numerator and in the denominator talk about a situation wherein somebody is sitting in a train station, measures the speed of a departing train, knows somehow that some guy is running toward the locomotive at a speed relative to the floor of the train. He wonders how fast the guy is "really" going. In classical physics, if you knew the train was going 50 mph and the guy was running 15 mph you would conclude that his speed is 65 mph. The relativity formula follows that general scheme with a correction factor. (Either way, you have to do some addition.)

The situation involving two spaceships approaching each other is different. It's the guy sitting in the caboose who considers himself immobile, sees the train station receding in one direction and the runner receding in the other direction. Imagine that you are in the middle of the ocean on a boat. As far as you are concerned, you are immobile. You see one boat approaching you from the direction of your stern and another boat approaching from the direction of your prow. You would calculate the speed that they were moving toward each other by summing the absolute values of their apparent speeds (known by your handy radar speedometer). But if you were being a physicist about it, you would assign one of the velocities a positive number (because you've graphed its position along the x axis and it's going from lower to higher values of x, and you would give the other of its velocities a negative number (because it is moving from a higher position on the x axis to a lower position. If you add a positive number and a negative number you get something in between, but you want to get a larger result instead. (We all know what happens when cars going 70 mph have a head-on collision.) So you have to subtract the negative number. If you work a few examples you'll see. Three ships, Alph, Bell, and Cass are adrift. Bell sees Alph approaching at 7 mph. Bell also sees Cass approaching at -7 mph (i.e., it's going 7 mph in the opposite direction. They are approaching each other at 14 mph, not at 0 mph. If Alph is approaching at +7 mph and Cass is receding at +7 mph, they are not moving with respect to each other. (They way they see it, they are really "still" and Bell is moving at -7 mph between them.) 7 - 7 = 0.

P0M 03:58, 17 March 2007 (UTC)

Error in the main text

You completely misunderstand the velocity addition formula. The correct interpretation goes like this:

Take three observers. A, B, C. Let A observe the speed of B. Let this speed be VB. Also, Let A observe C, too and let the result be VC. Now, at the same time let B observe C and let the result be V. The velocity addition formula describes the relationship between these measurements:

VC=(VB+V)/(1+VB*V)(for the sake of simlicity c is 1 now.)

Your greatest mistake is that you identify the velocities of the right side of the equation with the measuremnts of A, the observer "in rest". No, only one of these variables belongs to A's observation, the other one belongs to B.

Let us now work out yor problem: Two space ships, B and C approach each other. Their speed measuremnts have to be the same, only the sign is and has to be different: Say V=-0.9, that is B sees C approaching him. The sign is negative because the direction of C is opposite. What does a A see? It depends on his relative motion. We may assume that VB=0,9. That is A sees B running towards C at 0.9. The velocity addition formula tells now that VC=0. Which is correct. This means that if A sees B running away at a speed 0,9 and B sees C approaching him at a speed of 0.9, then A must see C standing still.

But this is not what you wanted. What you wanted was: A sees B running in the same direction as C. A's measurement on B is VB=0.9 and B's measurement on C, V=0.9 then VC is not VB+V but (VB+V)/(1+VB*V)

That's not what the article wants.

Please reconsider your interpretation or I will change the whole paragraph. Then you can appeal to the saint moderators. —The preceding unsigned comment was added by Zgyorfi (talk • contribs) 10:06, 18 March 2007 (UTC).

Here is what the article says:

wo spaceships approaching each other, each travelling at 90% the speed of light relative to some third observer between them, do not perceive each other as approaching at 90% + 90% = 180% the speed of light; instead they each perceive the other as approaching at slightly less than 99.5% the speed of light.

So there are three spaceships. What the text wants is two spaceships, A and C that are approaching each other. B is between them. B measures the velocity of A as 0.9c and measures the velocity of C as -0.9 c. The article asks the question, what velocity will A and C measure relative to each other. (You assume that they measure their velocity, which is another matter.) So the formula says that to compute the speed of two things relative to each other based on measurements of both made by an observer, we use the formula given in the article. With your letters, that's asking to compute VC when you know VB and V.

VC = 0.9c + (-0.9 c)/(1PVB*V) = 0

That result is why the person who started this section wanted to change the formula in the article, claiming, rightly, that the universe is not so crazy that if I stand between two onrushing trains and measure a high velocity toward me for both of them it does not mean that the locomotives are not moving. I explained that we have to be careful about what we are doing with our numbers and not just apply formulae as they fall into our hands. (It turns out that the person who wanted to change the formula in the text was you.)

What do I do if I stand between locomotives and calculate their closing speed relative to each other? To save the remainder of my hair I am going to choose a coordinate system that helps keep the plusses and minuses easy to deal with. Somewhere is the station in Moscow. I am somewhere on the new train line heading due east. Behind me comes a locomotive. My little switching engine is stalled on the track. No fuel. My radar sees the train from Moscow approaching at 300 kph. Bad. Then I turn around and see a train approaching on the same track from the east, at 300 kph. Even worse. I get out of my little engine and stand on a hill some distance away. I wonder how fast their closing speed is. If they have radar speedometers, they will see each other as closing with a speed of 600 kph. If one of them is a physicist he may say that the other guy is going at +300 and he is going at -300, so adding their velocities their closing speed is 0. He should not be reassured. The reality is that with signed numbers in this case you have to subtract. It doesn't matter whether they hit at +600 kph or at -600 kph. It's all in their point of view.

What if on another day I am standing by the same track and observe one train heading east at 300 kph and a while later I see another train coming along at 600 kph. To know how fast they hit, I will have to subtract again.

So even when the speeds get up to large fractions of c the basic procedure has to be the same. The calculation that the article is looking for if it wants to get the answer it calculates is VC = VB-V/(1-VB*V)

Do the math. VC = 0.9 c - (-0.9 c)/ 0.9c - (-0.9c)/(1 - (-0.81)) = 1.8c/1.81 = .994 c

Please sign your postings. (~~~~) P0M 04:56, 19 March 2007 (UTC)

The citation form of the formula will calculate a 0, so it's not the right one to use in the word problem given in the article so I have added a little more explanation to the article. The formula the article quotes is the "citation form" of the formula, i.e., it is the way the physics textbooks give it. (It is not a mistake.) But when physics textbooks give this formula they explain what kind of a physical situation it is to be used to analyze. If we simply change the formula in the article then somebody with a physics textbook will change it back again. Probably what really needs to be done is to fix the article that is linked to that formula (assuming it doesn't make it clear that you can get in trouble by following the formula blindly). An easier way might be to advise people to use the absolute values of velocities. Physicists would probably get creeped out by that approach. The formula is already a simplification of a formula used when one cannot assume that the velocities are parallel. Maybe using absolute values would cause errors under some circumstances. I don't know off the top of my head. P0M 05:29, 19 March 2007 (UTC)

I don't know the history of the article, but it currently uses the word "speed" to describe all the variables, and speeds are always positive. That's more than enough for an article on the speed of light. Melchoir 02:03, 20 March 2007 (UTC)

The way you you fixed it looks fine to me. There are multiple occurrences of the word "velocity" in that section, but your way of phrasing things should make what is going on clear enough for the general reader. P0M 02:42, 20 March 2007 (UTC)

Dear fried, Patrick

I do not want to mess with your article but you'd better listen to me. You are wrong. Physics is not something that you quote, physics is something that you are supposed to understand.

As far as the cited article: That article is wrong too inasmuch it does not explain which velocity (vector) is observed by whom.... Now I go and organize a scandal over there too. But you understand quotations only the here is one from the article "Special relativity" from wikipedia:

Composition of velocities

If the observer in ${\displaystyle S\!}$ sees an object moving along the ${\displaystyle x\!}$ axis at velocity ${\displaystyle w\!}$, then the observer in the ${\displaystyle S'\!}$ system, a frame of reference moving at velocity ${\displaystyle v\!}$ in the ${\displaystyle x\!}$ direction with respect to ${\displaystyle S\!}$, will see the object moving with velocity ${\displaystyle w'\!}$ where

${\displaystyle w'={\frac {w-v}{1-wv/c^{2}}}.}$

This equation can be derived from the space and time transformations above.

Now, this is the formula that applies to your situation. You'd better try to understand rather than referrring to authorities.

Signature: I am not and never will be politically correct enough to follow your rules. My signature is my username, which is not hidden. Report to the inquisition if you like.

My, we've all expended quite a bit verbiage, and a wee bit of vitriol, to address something that didn't need correction. The main text clearly states (stated) that these are speeds. Signature: I am not politically correct, but that has nothing to do with my being civil, courteous, or considerate, or with keeping the agreement I made to abide by the guidelines when signing onto Misplaced Pages so I'll sign John N. 23:26, 26 March 2007 (UTC)

Speed of light no longer a constant

I'm rather new to contributing to wikipedia and i'm not quite bold enough to edit the article directly but there are recent studies that the speed of light has in fact been changing. While this is still a new thing, this has wild implications all over the physics world. While this is still too new of research to be redoing everthing, I think it at least merits a mention on the page and wanted to bring it up for discussion with more experienced editors. I'm sure anyone reading is skeptical so I'll refer to you an article from newscientist which can explain it far better than I could. http://www.newscientist.com/article.ns?id=dn6092 Dcpirahna 02:44, 27 March 2007 (UTC)

I understand why you (and, of course, lots of other people) are interested in the new debate that has opened up over the constancy of the speed of light over time. If the speed of light changes over long spans of time then that fact has implications for the history of the development of the universe in its present form.

I suspect that bringing up this contentious issue in the context of a beginning article on the speed of light for the average well-informed reader risks destabilizing the whole enterprise. It's hard enough keeping up with the consequences of speeds at substantial fractions of c not being additive, the consequences of time dilation, etc. For most people it is conceptually very challenging. For people like me, going over and over the calculations while making as sure as I can that I am not just blindly applying formulae is the only way to keep my head on straight.

If we bring in the idea that measurements of c taken at great time distances from each other could be different, then the reader may get confused and wonder if the speed of light could be changing or could be different for two observers living in close temporal proximity to each other. The greatest reason that c is important in modern physics is that anybody who measures it while moving at some constant velocity is going to get the same answer. People need to be given as much help as possible to see the consequences of this fact. Bringing in some finding that is possibly of great importance to cosmology but incidental to the things we are trying to inform readers about is likely to do a disservice to that reader.

It's an analogy, but possibly a helpful one, to say that the speed of a wave phenomenon is a function of the density of the medium through which the wave propagates, and if the density of the medium changes then the speed of wave propagation will also change everywhere in that medium. If we had to say something it might be best to say that there are indications, in line with this analogy, that the speed of light may have gradually changed everywhere as space has changed. P0M 04:10, 27 March 2007 (UTC)

Fair enough. It is a fairly confusing topic as is. Hopefully more research will be done on the changing speed over time and can always be included later once the ramifications of the research are better understood. Thank you for the response. Dcpirahna 05:59, 1 April 2007 (UTC)

Speed of Light and mathematical Notation

I am considering posing the following paragraph to the main page, so please reply with constructive criticism.

In mathematical notation we write concerning speed v=d/t where v, d and t are v velocity, distance and time respectively. v=d/t Implies that t=d/v. Since according to the Principle_of_relativity it is forbidden to travel faster than the speed of light, instead of writing t=d/v we could write t=d/(c-v_n). We define v_n =c-v. This would produce a singularity for time at the speed of light. This would imply that we could write (c-v_n) =d/t. This would symbolize that Velocity an object having Invariant mass is undefined at the speed of light. Where appropriate, c=299,792,458 m/s. 71.156.91.18 00:20, 3 April 2007 (UTC)

Hi, and thanks for posting this here first before trying to edit the article. Note that according to your derivation, with t = d / (c - v_n), and v_n = c - v, we end up with t = d / v, which is the same as the Galilean definition you use above. This does not produce a singularity for v = c. For the correct derivation of how this goes, you should check out Lorentz transformation. Also, note that there are many possible transformations that satisfy the singularity property you are talking about. However, the standard Lorentz transformation is picked, because it is in fact derived from the underlying Minkowski space metric. Grokmoo 18:14, 3 April 2007 (UTC)

Weber's Constant

Why does Weber's constant redirect here? It is part of psychology...used to determine the JND of various stimuli... while it can deal with light as a stimulus it has little to do with the speed of light and is not exclusive to light.

Speed of light in your top section is wrong

"In metric units, c is exactly 299,792,458 meters per second (1,079,252,848.8 km/h)."

hi, i was going over your page and i noticed that this part is wrong. your information on the speed of light is correct at the m/s stage, but your km/s calculation in wrong. it should be about 17,987,547.48 km/h not 1,079,252,848.8 km/h.

299,792,458 X 60 /1000 = 17,987,547.480

Ochosi 23:22, 4 April 2007 (UTC) Ochosi

You forgot to multiply by 60 again (times 60 for seconds to minutes, then again for minutes to hours). – mcy1008 (talk) 23:35, 4 April 2007 (UTC)

Removal of "Technical impossibility of travel faster than the speed of light" section

See this edit.

I removed the section because I found it to be highly unclear, and to the extent that I understood it a novel approach to the issue. As any novel synthesis is a violation of WP:NOR, I felt obliged to remove it. Note that my disagreement is with the apporoach, not the conclusion. If this approach is known and valid, then a reference should be provided.

If anyone stongly disagrees with this action, then please feel feee to reinstate it pending a decision on the issue by the editors of this article. In the meantime, I will be adding a note to the section of light as a limiting speed showing how the relativistic velocity addition formula gives this result. --EMS | Talk 04:21, 21 April 2007 (UTC)

Here is the section that EMS removed. In general I think it is better to discuss things first rather than making a drastic edit. P0M 00:45, 24 April 2007 (UTC)

To understand why an object cannot travel faster than light, it is useful to understand the concept of spacetime. Spacetime is an extension of the concept of three-dimensional space to a form of four-dimensional space-time. Having the classical concepts of height, width, and depth as the first three dimensions, the new, fourth dimension is that of time. Graphically it can be imagined as a series of static,three-dimensional 'bubbles', positioned along an arbitrarily chosen line, each bubble representing a separate position along one of the four dimensions. That graphical approach is analogous to using a sequence of two-dimensional cross-sections taken at some standard interval along the third dimension to represent a three-dimensional object on a two-dimensional surface. (Imagine a map of a multi-story building that is created by giving the floor plan for each story of the building on a new page.) The mapping of space and time can be rotated so that, e.g., the x dimension is replaced by the t dimension, and each "bubble" represents a cross-section taken along the x dimension. Supposing that travel is occurring along the y and or the z dimension, what one will observe is that change along the t dimension will decrease from "bubble" to "bubble" as change across the y-z plane increases from "bubble" to "bubble."

With this understood, there is a clear implication that an object has a total velocity through space-time at any instant, and for all particles of matter this velocity is equal to the speed of light. While this result may seem contradictory to the idea of speed-of-light travel being impossible, it in fact proves it, taking into account the fact that faster-than-light travel was a spatial, or three-dimensional concept, not a four-dimensional concept. In the case of four-dimensions, all of the total velocity of an object not accounted for in three-dimensional space is in the fourth dimension, or time. To go back to our bubble picture, if an object is remaining at the same x, y, z positions will make maximum progress in the t dimension. And that is just to say that any clock associated with whatever we are watching at x, y, z is ticking away at its maximum rate according to a static observer in the same frame of reference, e.g., somebody at x+3, y+4, z+5 or any other position that is not changing with respect to x, y, and z. But the greater the changes of x, y, and z according to the clock of the other observer, the smaller will be the changes in t. But using the Pythagorean theorem to calculate the distances between a point at x,y,z,t and some later point x', y', z', t', then those distances will always be the same.

While this may seem confusing, it shows that as displacement through space increases, measured time will decrease to maintain the overall space-time velocity. If this is the case, it makes speed-of-light travel impossible, since when as an object approaches the speed of light spatially, it will have to approach zero velocity temporally. Another implication is that an object might be said to travel through four-dimensional space-time at the speed of light, but only in cases wherein its velocity through space is zero. That statement is just a counter-intuitive way of expressing the idea that when one is motionless (according to another observer) one's clock is ticking away most rapidly, and that as one moves faster and faster (according to the other observer) one's clock is ticking at slower and slower rates that approach zero.

I noted at the end of February that I didn't know whether this part was worth saving. However, I think that if provided with some drawings it could help some readers, for whom other ways of explaining things are not easy to absorb, see the way things work. P0M 01:04, 24 April 2007 (UTC)

As someone who understands special relativity (SR), I find this to be a very complex bunch of handwaving which asserts a lot and shows the reader almost nothing. IMO, there are plenty of ways to show why faster-than-light travel cannot occur in SR which will make the point cleanly and concisely. One is what I did in the "constant velocity ..." section of the article, in which I showed that the velocity addition formula prevented it. The energy and momentum equations also show much the same thing, which infinite energy and momentum being needed for massive objects to reach the speed of light.

The point is certainly legitmate (which is why I preserved it), but the means are not. Do keep in mind that the topic of this article is the speed of light in general. Related topics such as the theories of relativity are better covered in their own articles. --EMS | Talk 04:02, 24 April 2007 (UTC)

Did you want others to not discuss this passage? I thought you had already made your point about not finding it helpful in an admirable way. Perhaps a third, fourth... opinion may be useful. P0M 04:15, 24 April 2007 (UTC)

LOL! (but at myself). You have a point there. I would like more opinions, in spite of the fact that I have my own, and am too ready to express then. --EMS | Talk 04:40, 24 April 2007 (UTC)

Earth/Moon photo makes no sense

The photo captioned "A line showing the speed of light on a scale model of Earth and the Moon" does not make any sense. Can someone come up with a better photo? Davidwr 00:29, 29 April 2007 (UTC)

I think it would be better if it showed a 3D globe of the earth with atmosphere, dark space and a 3D moon. Then it showed a finite beam of light leaving the earth's domain and hitting the moon. It also would be nice to have a clock running in micro-seconds. --Marvin Ray Burns 02:24, 29 April 2007 (UTC) P.S. I see that you almost have it that way.

I stole the existing photo and added a distance-scale so at least it makes sense now. Feel free to do a 3D one. Davidwr 04:50, 29 April 2007 (UTC)

It wasn't a photo, it was an animation. The new diagram isn't an animation (as far as I can tell). Stannered 09:47, 29 April 2007 (UTC)

I restored the old animation. EdwardLockhart 12:27, 29 April 2007 (UTC)

Thanks. I've reverted the other article that used this same image. Davidwr 21:52, 29 April 2007 (UTC)

299,792,458 m/s, but about 299 338 km per second?

the speed of light precisely 299,792,458 m/s ... the speed of light is about 186,000 miles (about 299 338 km) per second
This looks very odd. I suggest just saying about 186,282 miles per second . --Occultations 23:41, 18 May 2007 (UTC)

Good point. Done. Dravick 23:51, 18 May 2007 (UTC)

Dead Links

Footnotes 6 and 7 are dead links. Thus they should either be fixed or removed. --Trakon 10:46, 31 May 2007 (UTC)

Actually it looks like I meant 7 and 8. --Trakon 09:27, 7 June 2007 (UTC)

c is for celeritas

I deleted language to the effect that using c to denote the speed of light derives from constant. The one cited essay speculates that c may alternatively derive from constant, but acknowledges grudgingly that celeritas is the widely accepted derivation. Weber's constant isn't the speed of light. Further, with all the constants floating around, the idea that any one constant would appropriate the initial letter of constant as a symbol is far fetched. We should go with the consensus derivation. Finell (Talk) 04:49, 1 August 2007 (UTC)

I've reverted your change, although it might be reworded yet again to note that c for constant was only the original usage and now generally means celeritas. — Joe Kress 07:49, 1 August 2007 (UTC)

Standard nomenclature: speed of light in vacuum.

The standard physics nomenclature for this physical constant is "speed of light in vacuum", see for exampel NIST. I think wikipedia should adhere to to standardized physics nomenclature, to avoid endless discussions and redirects. I propose to rename this article. /Pieter Kuiper 18:11, 8 August 2007 (UTC)

Please do not rename the article. Misplaced Pages's naming convention on en:WP is to use the name that the typical English language Misplaced Pages user would be most likely to use. In this case, that name is Speed of light. Only a specialist might ever think to add "in vacuum". The current, very long-standing name of this WP:FA is also shorter, and therefore easier to type into a search box, than what you propose. A redirect from "speed of light in vacuum" is not necessary (although it wouldn't hurt); if someone were ever to enter "speed of light in vacuum" in the search box, this article tops the list of articles that is returned with 100% relevance. Also, the article in not exclusively or even primarily about the named physical constant; it is more broadly about the phenomenon, its history, underlying theory, etc. On the other hand, correct terminology should be used in the body of the article. Finell (Talk) 21:41, 8 August 2007 (UTC)

OK, I understand your points. I will create the redirect then, which will make a name change of these article more difficult. Thank you for taking the time to respond to my suggestion. /Pieter Kuiper 21:58, 8 August 2007 (UTC)

The Light Cone image does not show up

I'm using Linux & Firefox so it may be a upper-/lowercase problem. Does it look OK in Windows? Johnmuir 09:44, 11 August 2007 (UTC)

I've fixed it myself. It was a) wrongly spelt and b) marked as a thumbnail Johnmuir 09:49, 11 August 2007 (UTC).

I have no clue what your technical problem is. The image works fine as a thumb (I use Opera on a winXp - also works with Netscape) I'll try it on the computer lab machines w/ie later. Without thumbing the image the caption dissappears. Do you have any problems with other thumbed images? Does anyone else have the same problem? Thumbed images are preferred as it allows user pref to control the size when set at default. Vsmith 02:48, 15 August 2007 (UTC)

It's a common problem that .svg images don't display thumbnails. Unfortunately, the issue is hard to predict, reproduce, or fix. The move to a .svg in this article is fairly recent, which probably explains why more readers haven't complained. I recommend using raster files until the issue is solved (and I'll do so here). Melchoir 03:29, 15 August 2007 (UTC)

It looks fine now - the last change was successful. I'd be pleased to help on the general problem if I can. (Testing, programming or similar, but I'm not a graphics expert.) Johnmuir 19:25, 15 August 2007 (UTC)

That would be great! Unfortunately, I don't know if there's any centralized discussion yet. I'm sure the problem must lie in the MediaWiki software, not Misplaced Pages's implementation, and I don't do work outside of Misplaced Pages myself, but you could always ask Misplaced Pages:Village pump (technical) for guidance. Melchoir 21:59, 15 August 2007 (UTC)

graphic doesn't time 1.2 seconds

Using Opera, it's more like 3 seconds. IE timing seems more plausible. Since this is apparently browser dependant, I move that the comment should be corrected, or the graphic removed. — Preceding unsigned comment added by PoorLeno (talk • contribs)

I agree. Apart from the timing problems, the graph is annoying and distracting. Jaho 01:30, 23 August 2007 (UTC)

Possible to travel faster than speed of light

I found this article . Maybe we can add something like that into the article? Oidia (talk) 12:14, 21 August 2007 (UTC)

There must be a Disambiguation

Speed of Light is a song by Stratovarius. I'm not english and i've not time enought to make the desambiguation page. Please, could someone else do it? Thanks. —The preceding unsigned comment was added by 213.60.158.76 (talk) 07:36, August 23, 2007 (UTC)

Done. — Joe Kress 02:14, 25 August 2007 (UTC)

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