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Talk:Special Relativity/Simultaneity, time dilation and length contraction

Twin Paradox
I am thinking about the twin paradox, where Jim stays on Earth, Bill flies away to a distant place and comes back and find that their age are different, Bill is younger.

What I do not understand is that based on the relativity theory there is no absolute point, so we could also say that Bill stayed and Jim with the Earth traveled fast away from Bill and Jim came back. So I think when they reunite on Earth again they age should be the same. They age may be different because of the different gravity they were at, probably the gravity is higher at the Earth, so in this case Jim who stayed on Earth should be younger. It is very possible that I miss something. It would be good to understand this. I understand that time slows down relative to each other, when they going away from each other. But should not time also speed up when they going back to each other?

(Ervinn 18:51, 10 May 2006 (UTC))

 * Good point. The presentation will be improved to show explicitly what happens from the other traveller's viewpoint. RobinH 21:15, 10 May 2006 (UTC)

 * Gravity has nothing to do with it, in SR you pretend that gravity has no effect except to keep our example people on the ground. When he is moving, both people's clocks are both slowing down according to each other. However it is true that when Bill turns around, Jims clock jumps forward according to bill, and from jims perspective Bill's clock did nothing special except momentarily return to normal speed while he was stopped (turning around). EvanR 06:57, 17 July 2006 (UTC)

Thanks, for the clarification. I have one more question. What if there is no turning around? Let say, Bill stops and starts going back. Will there be any time gap then? Is the time gap there because Bill turns around in a high speed? Ervinn 23:59, 21 July 2006 (UTC)

 * The turn around is not required to demonstrate the twin "paradox". Suppose there were two travellers, Bill(1) who moves away from earth and Bill(2) who travels towards earth. If Bill(2) synchronises his clocks with the clocks on Bill(1) when they pass then the same difference in elapsed time between the clocks on Jim and Bill(2) will be observed as between Jim and Bill in the original example. RobinH 10:06, 22 July 2006 (UTC)

(I like the article, it is nice work.) I think, I understand now. The time gap is caused by the fact that when they synchronized their clocks they were moving away from each other and because of that they had different sets of things that are simultaneous. However if they had syncronized they clocks standing together on earth, and then one of them would move away and come back, there would be no time gap on their clocks after they reunited. Is that right? Ervinn 19:08, 25 September 2006 (UTC)

 * The twin paradox will still occur in your example of one of the twins accelerating away from the other. When the twin accelerates away from the other the plane of simultaneity for the accelerating twin becomes different from that of the nonaccelerating twin as their relative velocities change and the time gap appears again. If the acceleration were instantaneous there would be no difference between an accelerating twin and a twin who was just passing.

 * Thank you for raising this question, I will try to integrate it into the text. The text is possibly not clear enough about the way that phase, time dilation and length contraction are about different sections, or cuts, through a four dimensional universe. For instance, suppose you had a rod with a set of simultaneously flashing lights on it. If you give the rod to someone who moves away from you at high speed the lights will start flashing sequentially. The moving rod is a different section, or cut, of a four dimensional rod from your original stationary rod.  Each of us has a unique 3D universe that is a section of a larger 4D universe. If I hand you a 3D rod and you walk away from me your 3D rod is slightly different from the one that I had, yours is mounted slightly obliquely in space-time compared with mine. RobinH 09:57, 26 September 2006 (UTC)

(Thanks for your answer) I see, the acceleration cause the time to slow down for the one who moves away (they move away from each other on the time coordinate). Does not the time speeds up though, during deceleration and the two cancel each other out? (Don't they move back to each other on the time coordinate) Ervinn 14:53, 26 September 2006 (UTC)

 * RobinH - I like your presentation of "the nature of length contraction" and would like to ask:
 * Do you consider such a view of SR contraction (which I share) to be compatible with the notorious "Bell Spaceship Paradox" as regards Bell's claim that a string joining identical spaceships would behave differently (under identical acceleration) to the s'ships themselves ?
 * It seems "obvious" to me that the "differing simultaneity" and "world tube sections" description of "contraction" would falsify Bell's assertion that the string should break.
 * If you think otherwise, I would very much appreciate an explanation of how either differing simultaneity or sections of world tube could distinguish the "string distance" between string ends from the "space distance" between string ends ???? Rod Ball 15:55, 13 February 2007 (UTC)

 * This is a nice paradox although a bit advanced for the introductory part of the text. It definitely deserves a place in the advanced text.  If Bell had synchronised the clocks on the ships before accelerating them he would discover that neither ship would have clocks that agreed with those of the ground observer and this observer would see that both ships would have discrepant clocks ("phase" dependent upon their separation).  The problem framed by the paradox then becomes the mystery of how two ships could have a piece of string between them that connects two different times (time machine string)! Baez gives a fairly convincing answer at: http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html but the advanced section of the book should consider several opinions. RobinH 21:27, 13 February 2007 (UTC)

 * RobinH - I'm not quite clear about your 2nd sentence: "If Bell had synchronised .... and this observer would discover....discrepant clocks". Surely if this observer means the ground observer then he will not observe discrepant ship clocks until the s'ships interrupt their acceleration to synchronise clocks.

 * In accordance with your tilted axes presentation, during acceleration the ship clocks must remain synchronised from the ground but appear increasingly out-of-phase in the ships. Only if the acceleration ceases to allow ships to re-synchronise will the ship clocks then appear out-of-phase from the ground ( and in the opposite sense from before with the forward clock lagging the rear ).

 * This must be so not least because the same clock cannot read two different times to two differently moving coincident observers. Thus the front ship clock must read the same time on board as it does to a momentarily coincident ground observer. So those aboard the front ship "see" the rear ship clock lagging in phase. Those aboard the rear ship must agree with coincident ground observer on their own clock reading but will regard the front ship clock as "set ahead". Now two things we "know" fron SR are that the ground observers will always disagree about forward-rear phase with the ship observers and that after a period of acceleration the ships will need to re-synchronise by either setting rear clock forward or front clock backward.

 * What this means is that during acceleration the ships simply reckon that the other ships clocks are getting out of whack. Rear ship sees front clock going fast while front ship sees rear clock running slow. But ground observer sees both ship clocks staying in synchrony.

 * After constant velocity is achieved, the ships can re-synchronise their clocks ( front turned back, rear turned forward or some of both ) whence ground observers will then see forward clock lagging rear clock in phase - exactly in accordance with the length contraction simultaneity shift.

 * So the bottom line is that the string doesn't extend into future or past, merely that "perception of simultaneity" leads to some radical disagreements over clock synchronisation that can cause lengths to be measured shorter or longer. Consequently one is always left with the unavoidable identity of the spacetime geometry of two points that are (a) the ends of a string or (b) the attachments to which a string may be fixed.

 * I can't see any way Bell could reach his conclusion except by substituting pre-1905 Lorentz "contraction" for SR - which if you read his article in Chapter 9 of his "Speakable and Unspeakable in Quantum Mechanics" you will find is precisely what he appears to be doing !Rod Ball 13:18, 14 February 2007 (UTC)

 * How to include this in the book is interesting. The introductory text needs a fairly simple section on the acceleration of a rod with clocks at either end and the cessation of this acceleration resulting in constant velocity. Particular stress should be laid on a force acting at the position of the rear clock, then secondly at the front clock and then thirdly at both clocks simultaneously. The paradox can probably be presented in the introductory text with a bit of thought. The advanced text also needs to present the paradox and consider the interpretation of Bell's "constant acceleration".  Duplicate explanations using both hyperbolic geometry and standard space-time diagrams should be given. Finally the position of the ends and tension in the string needs to be calculated. RobinH 16:16, 14 February 2007 (UTC)

Why the need to introduce a whole load of unnecessary complications ? They only serve to obfuscate the basic principles and provide heaps of ways to go wrong and/or arrive at a desired answer. Your clear explanation of "einstein contraction" as different angles of simultaneity or differently angled slices of world tube shows simply and irrefutably that any measurement that reveals a shorter string must also reveal an equally shorter s'ship distance in the absence of string. Any measurement that yields a longer s'ship distance must yield an apparently longer string without creating tension. The measurement just determines which slice through the world tube is exhibited - the string does not "resist" changes in geometry ! The string, like the train passing the embankment, can be longer or shorter depending which observers "simultaneity slice" is used to fix the endpoints.

Please take a look at the source of all the confusion - Chapter 9 of "Speakable and Unspeakable in Q.M." - where Bell spends half a small page describing the setup before saying it's "obvious" the string must contract while the s'ships don't etc. No equations or formulae, no calculations, no spacetime diagrams - nothing - just that it's "obvious" !. If Bell had not been a famous quantum physicist no-one would have paid any attention. In fact Bell had never published anything in Relativity and was at the time flatly contradicted by all the theoreticians at CERN - who knew far more about SR than Bell did ! So why did Bell persist ? If you read the rest of Chapter 9 you will find that he actually prefers Lorentz's pre-1905 theory which he describes in detail, regarding it as equivalent to SR, and belittling Einstein's contribution. He actually didn't understand that SR is a radically different theory, and particularly that acceleration is the very thing that distinguishes them most sharply, making nonsense of Lorentz's approach.

Regarding your link to math.ucr.edu I would say that Michael Weiss's article has "plausibility" but is not convincing under scrutiny. "How did the ships get further apart ?" he asks - but he is confused about what he means by "distance" because he forgets that when oblique-axes moving co-ordinates are superimposed on a Minkowski diagram the units must be "elongated" by gamma and that the line of simultaneity between trajectories does not represent the distance in the moving frame. To estimate the distance the s'ships must first cease accelerating at the same pre-arranged time by on-board clocks ( to preserve the identicality ). Naturally this means the rear ship thinks the front ship cut engines first and front ship is consistent in thinking the rear cut engines later. The point is that the ground observer considers the speed of light constant in his own frame, so is led to expect the moving ships to measure a greater distance as a longer to-&-fro light time. This is the great weakness of Minkowski diagrams - they only capture the POV of a single inertial observer, so the term "worldline" is inappropriate and misleading because the fundamental postulate of SR insists that the moving inertial observer finds exactly the same to-&-fro time and thus the same distance for an interval that "looks longer" on the diagram.

So the s'ships will actually find the to-&-fro light time unchanged and will have thus remained at constant distance. By simply sticking to the two fundamental postulates of SR - the total equivalence of inertial frames and the same speed of light for all such - and not being misled by the "preferential frame" of Minkowski diagrams it is easy to see how "Bell's paradox" collapses and that there is nothing paradoxical since SR predicts no string-breaking, nor any string-tension or string-stresses at all ! What I like about the world-tube picture that you use is that it makes the kinematical nature of SR crystal clear - the geometry shows that the relationship between two points in different inertial frames has got nothing to do with whether a rod, a string, "silly putty", tenuous gas or empty space exists between them. Rod Ball 14:17, 15 February 2007 (UTC)

Question on twin paradox
I have one question regarding the time gap that Bill sees between the clocks in Jim's frame. I see that the actual calculation of the 8 second time gap was not actually shown. When I perform the calculation of the time difference between the two clocks, I do not get 8 seconds. The equation is gamma*(vx/c^2) to find the difference between the clocks as seen by bill. So with numbers this would be (.8*10/sqrt(1-.8^2)) correct? Where v=0.8, and x=10 which is the proper length between the two clocks. This computes to 13.33 seconds, not 8 seconds. Now, if you were to use the contracted length that bill sees between the clocks for x (i.e. gamma*(.8*6), then you get 8 seconds, but the equation clearly uses the proper length, not the contracted length (I have even gone through the derivation of the de-syncronization of the clocks to prove it to myself that it is indeed the proper length instead of the contracted length that is use in the equation).

I have been wrestling with this problem for about a year now, and cannot figure out the resolution. It seems that Bill will actually see Jim measure a longer time interval than Jim actually measures (4.5 + 13.33, instead of 4.5 + 8). The only answer I can think of is that bill will see a difference of 13.33 seconds, but there is some justification for bill to know that what he sees is not what Jim will measure, and you actually have to use the contracted length instead in order to find out what Jim actually measures. But I have yet to be able to justify that. Thanks for any help!

 * Consider the physical situation. Suppose Jim has a clock at the turnaround point that is observed by Bill.  It is synchronised with Jims other clocks and for Jim it reads 12:00 when bill passes him. When Bill turns round this clock reads 12:00:12.5.  Both Jim and Bill can agree on this - its the same clock for both of them.  But hang on, If Bill uses the time dilation formula alone to predict the elapsed time that should be measured by Jim at the turnaround point he works out that this is only sqrt(1-.8^2)*7.5 = 4.5 seconds!  But Jim seems to have measured 12.5 seconds and because they are both looking at the same clock Bill must agree that Jim's clock at the turnaround point records 12:00:12.5 which Jim regards as 12.5 seconds of elapsed time. Bill's calculation of Jim's elapsed time differs from Jim's observation by 8 secs - why? Its because he has not allowed for phase, the slippage of clocks with distance. Things get clearer if Jim and Bill are imagined as having lines of synchronised clocks on their x-axes that can be compared at each instant and place. If they use these clocks Bill discovers that Jim's clocks are actually set 8 secs in advance at the turnaround point even when he passes him at the start ie: $$(v/c^2)x$$ seconds.

 * The Lorentz transformations apply to a situation where each observer can actually observe the readings on the other's clocks. Bill can actually read the figures on Jim's clocks and use these to check relativistic calculations.  For instance he could use the LT and Jim's elapsed time (t) after synchronising clocks with Jim to predict his own timing of a distant light flash at x in Jim's frame:

 * $$t^' = \frac{t - (v/c^2)x}{\sqrt{(1 - v^2/c^2)}}$$

 * The term $$t - (v/c^2)x$$ arises in the LT because at the location of the event Jim's clocks are seen by Bill to already read $$(v/c^2)x$$ seconds at the moment of synchronisation, when they "should" have been set by Jim to zero.

 * You can see that all the LT involves is a time dilation equation that uses the phase adjusted elapsed time at a distant point.

 * I think the main reason for your problem is that Bill actually sees Jim's clock readings to be 8 seconds out of phase, $$(v/c^2)x$$ at the turnaround point, not $$\gamma (v/c^2)x$$ seconds.

 * Perhaps some of the problem here is due to the difference between elapsed times and absolute times at a particular point in space. The reason why the Andromeda paradox is important in understanding the twin paradox is that an observer who is moving away from you in a given direction has times that are in your future in his present.  This is the import of the relativity of simultaneity.  This observer would calculate that distant events start much earlier than you calculate. If he had an array of clocks he would observe that this was the case. If he is your twin in the twin paradox then, in our example he observes your clocks at the turnaround point to be 8 seconds in advance when he passes you. This 8 seconds is not time dilated because it is not a comparison of time differences between frames of reference any more than an observation by him of two different times on your own clock would be a comparison between frames. RobinH 14:25, 30 November 2007 (UTC)

I agree that the main problem is that we are using different formulas to calculate the difference between two clocks moving relative to my frame. The reason for this is because when I derived the formula, it came out with the gamma in front, and my modern physics book (Modern Physics by Kenneth Krane, 2nd edition) also derived the same equation as I did. Let me explain how I derived my formula and perhaps you could explain why mine doesn't match yours.

I started by setting up two clocks which are stationary to each other (frame C) and a light pulse is generated half way between the two clocks. That system is moving at a velocity as viewed from another frame (frame B). In frame C, the light pulse reaches each clock simultaneously. As viewed by frame B, the trailing clock receives the pulse prior to the leading clock. I then calculated the time it takes for the light to reach each clock as viewed from frame B and subtracted the times. This should be the time difference between the two clocks as seen from frame B.

 * Ideally the "moving" observer should consist of an array, or line, of synchronised clocks that can be coincident with the "stationary" observer's clocks at all times because we are comparing events at two fixed points in the stationary observer's frame of reference.

So, with respect to frame B, the two clocks are separated by a distance L (this would be the contracted length, not the proper length), and are moving with velocity v.

 * The two moving clocks move out of alignment with the "stationary" clocks. As they move out of alignment the planes of simultaneity of the moving and stationary observers diverge so you are going to calculate the extent of this divergence over a distance. But what we are calculating is the desynchronisation that would occur in the rest frames clocks over a distance $$L_0$$ in the rest frame as observed by a moving observer . ...

The distance the light has to travel to reach the trailing clock would be L/2 - v*t1, where t1 is the time for the light to reach the trailing clock. That distance must also be equal to the distance light would travel in that amount of time which would be c*t1, so we have

 * The moving observer is examining the clocks in his reference frame that have the same separation as those he observes in the rest frame. He has to set his clocks $$L=L_0/\gamma$$ apart. So in the moving observers frame:

L/2 - v*t1 = c*t1

or, solving for t1:

t1 = L/(2(c+v))

Similarly, for the other clock we would have the distance light has to travel would be

L/2 + v*t2 = c*t2

Or

t2 = L/(2(c-v))

Subtracting we have:

t2-t1 = (L/2)(1/(c-v) - 1/(c+v))

getting a common denominator gives:

(L/2)((c+v)/(c^2-v^2) - (c-v)/(c^2-v^2))

simplifying:

(L/2)(2v/(c^2-v^2) = L*v/(c^2(1-v^2/c^2))

 * or time difference = $$\frac{Lv/c^2}{1-v^2/c^2}$$

 * $$= \frac{Lv\gamma^2}{c^2}$$

Now, the L is the contracted length between the two clocks, so if we want the equation in terms of the proper length we have:

L=L0*(1-v^2/c^2)^.5

plugging this in, and realizing that 1-v^2/c^2 = ((1-v^2/c^2)^.5)^2 gives:

(L0*v/c^2)*1/sqrt(1-v^2/c^2), or gamma(L0*v/c^2)

which should equal the time difference between the two clocks in frame B. This is the same thing my modern physics book calculated directly from the Lorentz transformations.

 * The moving observer calculates his own time difference as $$\gamma \frac{L_0v}{c^2}$$ but this was not the question. The question was what would the moving observer see as a time difference on the rest frame's clocks, when his own clocks read the same time. It is actually very difficult to use your experimental setup to calculate the desynchronisation of stationary clocks as seen by a moving observer because you cannot make the two sets of clocks coincide for more than an instant.  Without rehearsing the whole calculation it is clear that your t1 and t2, if applied to the inverse situation (observing the stationary frame from the moving frame when the events are simultaneous in the moving frame), will be time dilated so getting rid of the gamma term in the observed desynchronisation but you would need some very circuitous arguments to avoid comparing clocks at different places.

So, I guess my main problem is that I don't know what I (or my book for that matter) has done wrong the derivation. It is possible that I am misinterpreting what this formula means, but it seems pretty clear to me that this would be the amount of time difference as seen from frame B, between two events that are seen to be simultaneous in frame C (e.g., the two clocks striking 12 o'clock).

I also agree that this would not be a consequence of time dilation. Both clocks will be seen to be slow compared to mine. Also, the trailing clock will be seen to be ahead of the leading clock by (what I thought would be) what I just derived.

Any ideas? Thanks again!

 * It is maybe easier to use the Lorentz transformation to calculate the discrepancy in clock times with distance. Suppose clocks can be set so that when a left hand moving clock meets a left hand stationary clock the moving and stationary clocks read the same time, t=t'=0, as they coincide, what does the moving observer observe as the time on the right hand, stationary clock at that instant? What value of t, the time of an event at the right hand clock (at x) as seen by a "stationary" observer, would be regarded by the "moving" observer as simultaneous with his own left hand clock at t'=0?  The answer is clear from the LT for time:

 * $$0 = \frac{t - (v/c^2)x}{\sqrt{(1 - v^2/c^2)}}$$

 * so $$t = (v/c^2)x$$

 * In other words an event that is $$(v/c^2)x$$ secs in the future for the stationary observer, at a distance x away from him in the direction of relative motion, is seen as simultaneous with the present for the moving observer.  I hope this helps. RobinH 13:30, 4 December 2007 (UTC)

 * I wonder why the textbook used $$\gamma \frac{L_0v}{c^2}$$ as an example of desynchronisation, it is correct, it shows the relativity of simultaneity, but it doesn't instantly tie in with wider problems like the twin paradox or the Andromeda paradox. Most importantly it doesn't compare clocks at the same location so is a real brain teaser for applying to other scenarios!  Perhaps it should be included in the current textbook as a worked example... RobinH 15:57, 4 December 2007 (UTC)

 * Your equation is correct for the experimental setup you describe but this is not quite the same as the diagram shown in the link. In your setup there are 4 events (the light strikes each moving clock at a different moment and position from when it strikes the stationary clocks) whilst there are only two in the diagram.  Yours is the equation for the moving observer's estimation of the time difference between two events in his own reference frame when an analogous set of events (not the same events) are simultaneous in the rest frame.  The twin paradox is explained in the article here in terms of a different time difference.  The 8 secs "time gap" is the time difference in the rest frame as observed by the moving observer when the same events are simultaneous in the moving frame. This is shown in the diagram for the twin paradox that accompanies this article. The time gap expresses how the plane of simultaneity of the moving observer slopes upwards on the time axis of the stationary observer. RobinH 00:22, 5 December 2007 (UTC)

I think I have finally put it together! Imagine we are watching the two clocks go by, and each clock will start ticking when the light pulse reaches it. At the moment the trailing clock receives the light pulse, it will start ticking off time. Then imagine that 2 seconds later, by our watch, we see the other pulse reach the leading clock, and it starts ticking. Now, although 2 seconds has gone by on our watch since the first clock started ticking, less than 2 seconds will have gone by on the first clock because we see it as running slow due to time dilation! Therefore, the true question is, what does the first clock read when our clock reads 2 seconds. All we have to do is apply time dilation to our two seconds (2/gamma) and there we go, we have the reading on the first clock when the second clock starts! It would be $$\frac{\gamma \frac{L_0v}{c^2}}{\gamma}$$ and the gammas cancel out to leave just $$\frac{L_0v}{c^2}$$ !

Hurray!!! :) Thanks for all the help!

Image:Constudrelmag.gif @ Evidence for length contraction
To me, it seems that there is something slightly wrong with the diagram Constudrelmag.gif. Multiplying the binomial approximation vor the excess positive charge density with a 2 in the denominator with the formula from the Biot-Savart law, with also a 2 in the denominator, I would expect a 4 in the denominator of the final expression. This would match with the formula for Coulomb's law (http://en.wikipedia.org/wiki/Electric_field), but not with what is given on the diagram. If I am wrong, which is probably the case, please explain my mistake and don't flame me, as I am new to life, wiki, and special relativity.

 * Well spotted! The illustration was taken whole from elsewhere, with permission. It was not actually wrong but the text in it was not anywhere near sufficient so I have updated this section. The original author gave permission for the carve up of his image. The treatment is based on that due to Purcell but he seems to gloss over the thinning out of the moving charges, as does everyone else. As always, there is a set of publications discussing whether this treatment is valid or incomplete etc., etc. and what it really means! But this is the fun of physics. RobinH 17:49, 27 February 2007 (UTC)

Geometrical assumptions
I think you are using geometrical assumptions about relativity which have been superseded by new work in the history of mathematics. Specifically, your geometry is based on an acceptance of Einstein's use of a "natural" coincidence of points, which he makes explicit in RELATIVITY in the train experiment.

It is a concept which is not justified logically in the argument. However, Einstein included it because he subscribed to natural mathematics, which he read about in Poincare's SCIENCE AND HYPOTHESIS. The gist of natural mathematics is mathematical arguments cannot be internally consistent or they will lead to paradoxes.

So you have to insert in your argument somewhere the idea that mathematics is an inherent human characteristic. Einstein does this by including the idea, in the train experiment, that point M "naturally coincides" with point M'.

As has been recently pointed out in the Stachel and Howard book on Einstein's early years, Einstein's incorporation of natural mathematics also caused problems in the 1905 paper on Brownian motion.

Poincare had a very pernicious effect. As Grattan-Guinness points out in THE SEARCH FOR MATHEMATICAL ROOTS, Poincare had a VERY poor understanding of the set theory arguments out of which Poincare formulated his version of natural mathematics. Einstein knew nothing of Poincare's lack of understanding, and never questioned it.

So, we are in the midst of a reassessment of the fundamental mathematics out of which special relativity emerged. A good point to start is A. Garciadiego's BERTRAND RUSSELL AND THE ORIGINS OF THE SET-THEORETIC 'PARADOXES.' Here is my own summary of current research:

Ryskamp, John Henry, "Paradox, Natural Mathematics, Relativity and Twentieth-Century Ideas" (May 19, 2007). Available at SSRN: http://ssrn.com/abstract=897085

 * I have also had some difficulty with this problem. Special relativity became highly geometrical in its early years (1908-1918) and many authors used an 'imaginary time' model of the universe.  SR is, of course, a theory that was eclipsed when GR was produced. I believe that Einstein himself felt that SR was a mistake.  This leaves us with a problem, do we teach SR as a separate theory of flat spacetime or do we forget SR and go for GR straight.  If SR had been THE physical theory for 200 years (like Newtonian/Hamiltonian physics) there would be no problem:  obviously we would teach SR as a geometrical theory and then move on to how it was modified by GR.

 * The principle advantage of the geometrical approach is that it gives students an intuitive grasp of simultaneity and it is the physics of simultaneity that defeats the average student.

 * So, what to do.... It would certainly be premature to raise the "hole argument" at the beginning of an SR course. We have come across a similar problem in this book with the problematical concept of "relativistic mass" - how do you explain that there are problems with the concept of mass in relativity without introducing the problematical concepts?  The solution to the interpretation of SR is probably the one that has been adopted, the book should be in two parts, the introductory part should be designed to give physics students an intuitive grasp of flat spacetime and the second part should explore the whole subject again from a more mathematical and philosophical viewpoint for advanced physics students and mathematicians. What do you think? Would you be prepared to add a section to the advanced text on the problems that you identify in your paper, perhaps mentioning alternative approaches to spacetime? RobinH 09:35, 6 July 2007 (UTC)

JOHN RYSKAMP RESPONSE:

I think you are making a mistake when you assume that general relativity can be taught as if it does not contain a "natural" coincidence. It does. If you cannot get beyond the relativity of simultaneity because it contains a "natural" coincidence of points, you certainly cannot get to general relativity. A spacetime point is Einstein's "natural" coincidence.

I think you have to proceed much more carefully. I certainly think you ought to read Garciadiego and then reassess whether you yourself believe in natural mathematics. As I say in the paper, natural mathematics long predates the set theory controversy and relativity. It appears as early as Aristotle, who felt that ancient "paradox" had to be avoided or solved.

I think you have to come to terms with your own approach to "paradox." Do you think there are any? Do you think Zeno's paradox has any logical content? I do not. I have not found any of the paradoxes to have any logical content. But it would be very good for you to explore them, because it would tell you a lot about your own conception of mathematics. I think you must have problems with your concept of mathematics if you think QED doesn't contain "natural" coincidence. It most certainly does.

I think you best approach general relativity understanding that Einstein had a natural mathematics orientation in formulating it. Above all, this approach stands for the proposition that arguments CANNOT be internally consistent--otherwise they lead necessarily to paradoxes. This is an odd way to view relativity, because its defenders have always asserted that it has no discernable internal inconsistency.

Now that point of view cannot any longer be maintained--the identification of the "natural" coincidence error has completely destroyed it: and look how blind the scientific community was: it has been sitting on the page in Einstein's book RELATIVITY for EIGHTY years. Now that's what I call a scientific community of morons.

If you feel that "natural" coincidence disposes of special relativity, which it does, then you must look for some OTHER internal inconsistency in general relativity, because, as a natural mathematician, Einstein PUT one in general relativity. But where is it? I think it is impossible to proceed even with the assumptions of general relativity without granting that "natural" coincidence is part of general relativity.

There is no getting over "natural" coincidence as an objection. It is pervasive. I don't go where you are going in this argument. I don't proceed to general relativity, because I don't think one can identify "natural" coincidence in the relativity of simultaneity and then get, logically, to general relativity. I think that after some reflection, everyone will agree about that.

No, to me the argument goes next to the Pythagorean theorem. Where is "natural" coincidence in that theorem? That is the state of the art question. Frankly, I don't know yet, but as I say in the paper, I think it is there, and invalidates the theorem. If it is, then we do not need recourse to general relativity to disprove it, as you will note.

As for spacetime, it is not a concept which has any internal consistency because it is "natural" coincidence. ALL the conceptual framework leading up to, and of, relativity becomes irrelevant as the role of "natural" coincidence is identified. That invalid term posed the issues. Because it is gone, all its issues are gone.

We simply have to think anew.

 * I have transferred your text to Special Relativity/The interpretation of special relativity and inserted a sentence to point readers to the advanced text in the section on the Andromeda Paradox. The advanced text now has a separate section on the interpretation of SR with your approach as a major subsection in the "Contents". I hope that you will find this acceptable.

I HAVE CHANGED ONE PHRASE. YOU HAD

Note that the geometrical interpretation

AND I HAVE CHANGED IT TO

Note that the metrical interpretation.

THE REASON IS THE 1905 PAPER, IN WHICH THE GEOMTRICAL APPROACH IS IMPLICIT, NOT EXPLICIT. THAT IS, EVEN IN METRICAL DISCUSSIONS, "NATURAL" COINCIDENCE IS IMPLIED. IN OTHER WORDS, THE CLOCK AND ROD EXPERIMENT IS IDENTICAL TO THE TRAIN EXPERIMENT.

 * We have a problem here because the book must transmit an understanding of relativity that is suitable for undergraduates, especially with respect to answering questions on simultaneity, but there is a good deal of philosophical and physical debate about the whole structure of relativity (and physical science). It is important that we do not pull up the rope ladder of knowledge behind us when we teach this subject, instead I am in favour of describing historical approaches and then their amendments.  I would like to see you flesh out the section on the problems with the interpretation of relativity in the advanced text so that readers can get some of the excitement of considering leading edge and alternative thinking.  Most importantly I would like to see a clear description of how the a priori assumption of a Minkowski space-time disguises a natural mathematics approach so that people will be able to reflect upon whether this assumption is valid. I would also like to see the advanced text becoming the main text with sufficient quality and diversity to be of interest to graduates. RobinH 11:44, 23 July 2007 (UTC)

MINKOWSKI SPACE-TIME IS AN EXTENSION OF THE APPROACH OF LORENTZ. FROM THE POINT OF VIEW OF SPECIAL RELATIVITY, THE LORENTZ TRANSFORMATION IS ARBITRARY AND CAPRICIOUS, IT IS AD HOMINEM AND UNJUSTIFIED. IT IS ON SUCH A LOW LEVEL OF SYNTHESIS THAT I DON'T THINK WE CAN EVEN SAY THAT 'NATURAL' COINCIDENCE OCCURS IN IT--IT DID NOT OCCUR TO LORENTZ TO CAPITALIZE ON THIS PARTICULAR PREJUDICE.

IF YOU LOOK AT RYCKMAN'S THE REIGN OF RELATIVITY, YOU WILL SEE THAT EINSTEIN--ALONG WITH EVERYONE ELSE--SIMPLY DANCED AROUND THE GEOMETRICAL PROBLEM, NOT BEING QUITE AWARE OF IT, UTTERLY DEVOTED TO NATURAL MATHEMATICS WITHOUT BEING ABLE TO RESOLVE THE LOGICAL ANOMALIES (I SUSPECT RYCKMAN IS IN THIS BOAT AS WELL--IN RELATIVITY, IT IS TRULY A CASE OF THE BLIND LEADING THE BLIND). MINKOWSKI IS PART OF THIS CROWD. I DON'T EVEN THINK MINKOWSKI IS JUSTIFIED IN HIS PROCEDURE WITH RESPECT TO SPECIAL RELATIVITY--I DON'T THINK HE UNDERSTOOD IT VERY WELL. IF EINSTEIN DIDN'T--AND HE DIDN'T--HOW ARE WE SUPPOSED TO SAY MINKOWSKI UNDERSTOOD ANYTHING ABOUT IT? THESE PEOPLE ARE SO LOST IN VARIOUS MATHEMATICAL PREJUDICES THAT THEY SEEM LIKE A BUNCH OF GOBBLING GHOSTS. SAYING THAT 'NATURAL' COINCIDENCE OCCURS IN MINKOWSKI SPACE-TIME IS GIVING MINKOWSKI MORE CREDIT THAN HE DESERVES. HE FUMBLED THE CONCEPT--THERE WAS NO WAY HE COULD HAVE DONE ANYTHING ELSE, APPARENTLY BECAUSE HE WAS IMPRISONED IN HIS INTELLECTUAL ERA. AS WAS EINSTEIN. AS WE ARE, NOT DOUBT, BUT WE DON'T KNOW YET THE PRISON IN WHICH THE REALIZATION OF 'NATURAL' COINCIDENCE, OCCURS. THE NEXT ISSUE IS, WHERE DOES 'NATURAL' COINCIDENCE OCCUR IN THE PYTHAGOREAN THEOREM? THAT IS THE QUESTION OF THE MOMENT. ONCE WE FIND IT, WE WILL HAVE GIVEN POSTERITY...A NEW ILLUSION OF FREEDOM!! AND SO IT GOES.

AS FOR PULLING UP THE LADDER, YOU ARE ASSUMING THAT WE 'KNOW' SOMETHING. THE POINT OF SPECIAL RELATIVITY IS THAT IT CLAIMS RELATIONSHIPS. BUT IT DOES NOT PROVE THOSE RELATIONSHIPS. THUS IT IS NOT SO MUCH A MATTER OF RETAINING KNOWLEDGE AS IT IS A MATTER OF DECIDING WHETHER THERE IS AN ARGUMENT TO BEGIN WITH. THERE MAY BE RELATIONSHIPS, AND PIGS MAY FLY. BUT SO FAR, THE QUESTION WE HAVE FOR EINSTEIN IS THE SAME QUESTION WE HAVE FOR GODEL: WHAT IS YOUR ARGUMENT? ONLY THE IDENTIFICATION OF 'NATURAL' COINCIDENCE MADE THIS QUESTION LOGICALLY POSSIBLE. TO ASSUME KNOWLEDGE IS TO ASSUME THERE IS AN ARGUMENT TO BEGIN WITH. BUT THERE IS NO ARGUMENT. SO WHAT EXACTLY IS THE KNOWLEDGE YOU ARE SO ANXIOUS WE NOT JETTISON?

 * I am not one of those who believe that a textbook should be purely directed at providing students with the tools for answering exam questions. If we structure the book correctly we should be able to both help students achieve their grades and, having done this, take them into interesting areas of speculation, research and re-consideration of the subject.  At the end of a course directed at undergraduates there should be a connection with the speculative world of postgraduate life, whether this involves pure physics, the philosophy of science etc... RobinH 09:09, 31 July 2007 (UTC)

THEN THERE SHOULD ALWAYS BE A SIDEBAR CALLED

REVISITING SETTLED QUESTIONS

THIS IS BECAUSE, NO MATTER HOW IMPORTANT THE POINT AT ISSUE MAY BE, THERE ARE ALWAYS UNDERLYING ASSUMPTIONS. LOOK HOW RECHERCHE SET THEORY HAS BECOME. IT'S EASY TO FORGET THE UNDERLYING QUESTION: IS THIS EMPEROR WEARING ANY CLOTHES? THAT IS, BENEATH ALL THE MATH WALLPAPER ARE SIMPLE, PLAIN-LANGUAGE ASSUMPTIONS WHICH STUDENTS NEED TO ADDRESS. IF YOU READ GARCIADIEGO, YOU WILL SEE THE WELL-NIGH UNBELIEVABLY VAGUE NONSENSE WHICH CANTOR USES TO UNDERGIRD HIS SET THEORIES.

THE PROBLEM IS, WHERE WE ARE NOW, I DOUBT WHETHER THOSE EXAM QUESTIONS ARE WORTH ANSWERING, OR IF THEY ARE WORTH ANSWERING, WHETHER THOSE ANSWERS WOULD NOT NECESSARILY INVOLVE REVISTING SETTLED QUESTIONS. LOOK AT HOW COMPLEX SOME STRING THEORY EXAM QUESTION WOULD BE. AND YET IT NECESSARILY INVOLVES A 'NATURAL' COINCIDENCE WHICH MEANS IT ISN'T A QUESTION AT ALL.

IS THAT WHAT THE STUDENT SHOULD WRITE ON THE EXAM PAPER, 'THIS QUESTION INVOLVES A 'NATURAL' COINCIDENCE OF POINTS AND SO IS NOT INTERNALLY CONSISTENT. I THINK YOU HAVE TO REALIZE THAT WHERE WE ARE NOW, INVOLVES PUTTING PEOPLE IN OPPOSITION TO THE PHYSICS ESTABLISHMENT. HOW DO WE PREPARE THEM FOR THAT? SEND THEM TO THE THEORY OF EVOLUTION, WHICH ALSO CONTAINS A 'NATURAL' COINCIDENCE?

Quantum Mechanics explained by the twin paradox??
I have a theoretical question: I know that the speed of light can not be reached, but let suppose that we are traveling on a space ship going at the speed of light. At the speed of light time stops. What happens when time stops? Does it mean that we are frozen, unmoved objects for eternity until we slow down, and time start again?

Lets say that we collide to an object in that high speed and our space ship split two halves, each continue going at the speed of light. Because time stopped for us, does it meant the split did not happened for us, we still be in one piece. For the observer however, we are two pieces. But when the observer ask a property of one of our halves, one halves also represent the whole, since the split did not happened for us.

Is this similar to the twin paradox?

The reason I am asking, it is not what we see when in Quantum Mechanics, in an accelerator, a particle is split apart and each travel a distance, and when we measure the property of one, the property of the other halves get known instantly, even if they are far away to each other. And It is unknown how could that information be there faster than the speed of light. Can it be explained by the twin paradox? The particle was split apart for us, but it was still one piece for the particle point of view, when the measurement of the property was made Ervinn (talk) 06:12, 17 September 2008 (UTC)

 * The "speed" of light that we are discussing is actually a constant that relates units of time to units of space in such a way that the Minkowski metric applies (ie: the square of distance minus the square of c times time in the equation is zero). It is not really a speed at all. Aligning the direction of motion with the x axis:

 * $$0 = x^2 - (ct)^2$$

 * This equation describes a particular section of a 4D spacetime rather than a motion. The "zero" is an invariant interval, seen by all observers as zero. This suggest a conservation law exists (for every invariant there is a conservation law).  The conservation law in this case is the "conservation of centre of energy" (see: http://arxiv.org/PS_cache/physics/pdf/0501/0501134v1.pdf Illustrations of the Relativistic Conservation Law for the Center of Energy, Timothy H. Boyer).  This bothers me because it must be one of the most important conservation laws in physics, being the direct result of the metric, but it has contributed almost nothing to modern theoretical physics. Boyer mentions how the interaction with the solenoid involves the centre of energy but what about particles? Why does an electron act as a point when its energy is distributed in space and time?  Given that an electron can interfere with its historical self (see attosecond laser double slit experiments http://physicsworld.com/cws/article/news/21623) and so is a time-extended entity, why is so much of its energy available at a point during collisions? Is there a common centre of energy for an electron-positron pair? etc. etc. I dont know the answers to these questions but they are fascinating! RobinH (talk) 11:05, 19 September 2008 (UTC)

 * "Yet another interpretation of this phenomenon is that quantum entanglement does not necessarily enable the transmission of classical information faster than the speed of light because a classical information channel is required to complete the process." - http://en.wikipedia.org/wiki/Quantum_entanglement

Paradoxes
Sorry, the page above is sort of a mess without clear structure. -- First it says a rocket fired from moving car will have head start of several days. I am too lazy to think that over completely, but I doubt it. The head start is an artefact of the silly inertial frames (as is the huge E_kin Andromeda galaxy acquires when I speed up the car) and will evaporate as the distance between rocket and fleet shrinks, leaving only the base speed bonus from the car (presumably small versus the rocket's speed itself). The mention of "observers" should be BANNED anyway. -- You can have twin paradox without any acceleration. It is possible to establish something like Galactic Normal Time over distances by using harmless speeds or syncing clocks. It can be tested by laser interferometry or Doppler or something whether stations are at rest against each other. Now someone flying at relativistic speed along these stations will experience much less time on his clock than the stations. (He can ask by radio, or can hit a planet and the Galactic Normal Time people can analyze the debris and find out how much C-14 and U-238 is left.) The effect is not symmetric, although he can make up all silly kinds of warped "simultaneity". To actually watch those in action, he would need at least a second spaceship flying in distance at the same speed (i.e. at rest in his frame). AFAIK that has never been tested. J / k. --88.74.180.101 (talk) 18:37, 1 June 2009 (UTC)
 * I think it might be rewarding to think "over completely" the part of the chapter on reference frames. The head start is indeed due to the slice of the universe that the car driver considers to be simultaneous, the car driver considers that the future of distant parts is simultaneous with his own present. It is this difference in planes of simultaneity that is central to the theory.  The text agrees that you can have a twin "paradox" without acceleration. RobinH (talk) 15:02, 26 November 2009 (UTC)

The whole notion that there is some logical content to the idea of "paradox," has been severely undercut by analysis of known paradoxes. It is irresponsible not to discuss this and mention Alejandro Garciadiego's landmark contribution, BERTRAND RUSSELL AND THE ORIGINS OF THE SET-THEORETIC 'PARADOXES.'

Above all, reexamination of the "paradoxes" leads us to consider the point of view which proceeds from them: constructivism. Constructivism, with its idea that some arbitrary intervention in an argument must be made in order to "save" it from "paradox," is so entirely without any foundation that it seems rather absurd to posit paradoxes, refuse to examine them closely because they seem alluring, and then go on to "decide" what they mean or what to do about them.

You are FAR behind the research. Wake up.

5.2 De Broglie waves
This section appears to be a bit on its own with very little lead in and not a great deal of descriptive body.

I find it difficult to see how it fits with the previous discussion and how it adds to further development of topic.

Would suggest either a fuller introduction and expansion or remove to more advanced text section for fuller description.

Otokia (discuss • contribs) 22:11, 25 April 2011 (UTC)

Relativistic electromagnetism
I think there is something missing in the proof (e.g. approximation?)
 * $$\lambda = \frac{q}{l} ( \sqrt{1 - v^2/c^2} - \frac{1}{\sqrt{1 - v^2/c^2}})$$

Using the binomial expansion:
 * $$\lambda = \frac{q}{l} (1 - \frac{v^2}{2c^2} - 1 - \frac{v^2}{2c^2} + \dots) \approx \frac{qv^2}{l c^2}$$

Can somebody revise it and correct if possible?--Almuhammedi (discuss • contribs) 05:05, 29 October 2011 (UTC)

Orthogonality
The discussion with coordinates for John and Bill provides a detailed basis for relativity of simultaneity. Each of them has a time coordinate and a space coordinate. The relationship of these coordinates can be made complete by introducing Hyperbolic orthogonality: this concept generalizes perpendicularity to the context of a plane including both space and time. Review Calculus/Hyperbolic angle for treatment of both "rapidity" and hyperbolic orthogonality, which is the invariant relationship of space to time in special relativity. Rgdboer (discuss • contribs) 21:55, 12 July 2019 (UTC)


text:
Programmable Logic/Verilog Operators

This page is going to talk about some of the verilog operators.

Arithmetic Operators
The arithmetic operators are as follows:

 * + (addition)
 * - (subtraction)
 * * (multiplication)
 * / (division)
 * % (modulus, or remainder).

In practice, the division and modulus operators are typically only usable in simulation, not synthesis. Division is a particularly complicated operation, and most programmable chips do not have dedicated divider modules.

Logical Operators
There are a number of logical operators. Logical operators act on an entire value (multiple bits), and treat the values of "zero" as being "false", and "non-zero" as being "true".

 * ! (Logical NOT)
 * && (Logical AND)
 * || (Logical OR)

The reduction operators are

 * ! NOT
 * && AND
 * || OR

What happens is that Verilog converts the whole number into either a 1 (if it contains a nonzero bit) or 0 (if it only contains zero), then performs the equivalent bitwise operation. Thus, the answer is also one bit.

Examples:

!0000 = ~0 = 1

!1101 = ~1 = 0

!1000 = ~1 = 0

!(any nonzero binary number) = ~1 = 0

0000 && 1101 = 0 & 1 = 0

0010 && 1101 = 1 & 1 = 1

0000 || 0110 = 0 | 1 = 1

0000 || 0000 = 0 | 0 = 0

Bitwise Operators
There are a number of bitwise operators to perform boolean operations on each individual bit in a bus.

 * ~ (Bitwise NOT)
 * & (Bitwise AND)
 * | (Bitwise OR)
 * ^ (Bitwise XOR)
 * ~^ (Bitwise XNOR)

Assignment Operators
there are three assignment operators, each of which performs different tasks, and are used with different data types:

 * assign (continuous assignment)
 * <= (non-blocking assignment)
 * = (blocking assignment)

The continuous assignment is typically used with wires and other structures that do not have memory. A continuous assignment is happening continuously and in parallel to all other computational tasks. The order of continuous assignments, or their location in the code do not matter.

Non-blocking assignments are assignments that occur once, but which can all happen at the same time. These assignments are typically used with registers and integer data types, and other data types with memory. The following two code snippets with non-blocking assignments are equivalent:

a <= a + b; b <= b + a;

and

b <= b + a; a <= a + b;

All non-blocking assignments in a single code block occur simultaneously. They happen only once, and the input values for all such assignments are read before the operation takes place (which requires the use of additional latch structures in synthesis).

Blocking assignments are also used with registers and integers (and other memory data types). Blocking assignments occur sequentially, and the code after the assignment will not execute until the assignment has occured.

Shift and Rotate
if you want to shift right assign shr = a >> 1 // shift a right by 1 bit.

example a=8'b10011011 then x=8'b01001101

assign shr = a >> 2 // shift a right by 2 bits.

example a=8'b10010101 then x=8'b00100101

if you want to shift left assign shl = a << 1 // shift a left by 1 bit.

example a=8'b10011011 then x=8'b00110110

assign shl = a << 2 // shift a left by 2 bits.

example a=8'b10011011 then x=8'b01101100

Concatenation and Replication
Combines 2 or more than 2 operands

Reduction Operators
These are the same as the bitwise operations above.

Conditional Operator
? : ;

If the Condition is true, the value of will be taken, otherwise the value of will be taken. Example:

assign EDB = ram_rd_en ? ram[EAB[9:0]] : 32'bz;

ram[EAB[9:0]] will be assigned to EDB in case ram_rd_en is true.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: Programmable Logic/Verilog Operators


text:
Programmable Logic/Parallel Execution

This page is going to talk about parallel program execution, one of the biggest issues when programming an HDL.

Inline Execution
Inline execution is what normally occurs in a computer program. Instructions are performed sequentially, with one instruction being performed to completion before the next operation is performed to completion. This is not the way HDL programming works, although the HDLs that we will be studying here allow some functionality for sequential operations. It is typically highly inefficient to attempt to force a hardware design to be completely sequential, and in such instances it will typically be better design to simply use a micro controller, and perform your task in software instead of hardware.

Continuous Execution
Verilog, VHDL, and even SystemC are methods for describing electric circuits, as an alternative to large and complicated schematics. Electric hardware typically has many components that operate in parallel with one another, simultaneously.

As a result of this fact, the statements in an HDL are typically seen as being executed in parallel, unless specified otherwise. This is a very different paradigm from traditional computer programming, and it can be difficult for accomplished computer programmers to translate their talents to HDL.

Problems and Pitfalls
Many traditional computer algorithms rely on the fact that code is executed sequentially. For this reason, many such algorithms cannot be applied to HDL with any ease or certainty.

Attempting to make operations in hardware to be sequential is not only a difficult task (takes longer to program), but also requires more hardware resources in the form of latches, registers, and timing circuits. Abandoning sequential operations entirely can also cause problems with timing and program complexity.

Relation to Software
Programmers with experience in multi-threaded software may quickly relate to the idea of synchronous parallel execution, although there are some fine points that are worth discussing. In software, all instructions are considered to be executed sequentially, one after another in a predefined order. The illusion of concurrent processes or "threads" occurs because the different threads switch back and forth so quickly as to appear to be running in parallel. Even in multi-core processors, it is not really possible for all threads to be running parallel, as there are typically more threads of active execution scheduled then there are cores available to process them all simultaneously.

In hardware, separate processes do happen simultaneously with no illusions. Because multiple processes happen at the same time, and their actions are synchronized to a global clock signal, it is easier to get timing correct between processes, and to avoid race conditions or deadlock. In software, the task scheduler may schedule individual threads in any order, or spread them out among separate processing cores in an arbitrary way (it is typically not known by the application programmer what algorithm the scheduler uses).

Another issue is the fact that hardware modules do not load or unload dynamically, as software modules do. This means that all hardware to be used must be created at once and is effectively static. Additional modules cannot be loaded at runtime. Once a chip is produced, the only way to add functionality to it is to design and produce a new chip, and then interface the chips together. Another facet to this concept is that the hardware modules still exist, and still operate even if the programmer doesnt take that operation into explicit account.

Consider a chip design with 3 modules: an input module, a processing module, and an output module. Signals are read from an external bus by the input module, processed, and then written back to the bus by the output module. Let's say that the signal is currently in the processing module: What then are the other two modules doing at that same time? The operations of all modules needs to be taken into account at all times, and if some modules should act "dormant" at some times, that functionality needs to be added.

title: Programmable Logic/Parallel Execution


text:
Programmable Logic/Verilog Always and Initial

This page is going to look at the Verilog keywords always and initial

Synchronous and Asynchronous
always blocks can be either synchronous or asynchronous structures. The instructions inside these structures are said to occur synchronously with the given condition, or asynchronously (if no condition is specified).

Initial Blocks
Initial blocks can be used in either synthesizable or non-synthesizable blocks. They are commonly used in test benches. Initial blocks cause certain instructions to be performed at the beginning of the simulation, before any other instructions operate. Initial blocks only operate once.

A synthesizable initial block can be used to set the power-on value of registers, RAM, and ROM within FPGAs. However, initial blocks cannot be synthesized in CPLDs or ASICs.

Always Blocks
Always blocks describe things that should repeat indefinitely, or things that should repeat on a given synchronization condition. An always block with no condition is asynchronous:

always begin ... end

This block of code will continuously execute regardless of the clock signal of the circuit.

Restrictions
Hardware modules may not be instantiated inside an always block. All of the always blocks in a module are considered to be parallel blocks of hardware. Likewise, modules are considered to be separate hardware blocks that operate in parallel. Functions, however, can be instantiated in an always block.

A wire or register may not be written to by more than one always block. A wand or wor can, however, be written by multiple blocks.it assigns the values to

Always @
The always @ structure is a synchronous condition structure. It takes the form:

always @

Where  can be either posedge, negedge, or omitted. The  parameter is the signal to be synchronized against (usually a clock signal). For instance, if we want a particular block of code to execute at the positive edge of every clock pulse, we would write:

always @(posedge clock)

If we do not put in either posedge or negedge qualifiers, the always block will be triggered at every change (every positive and negative edge of the signal). For example:

always @(clock)

will perform at the positive and negative edges of the clock signal.

Implementations
Always blocks can be considered to be circuit constructs where the input to that circuit is controlled by a latch. The event that triggers the always block is considered to be the clock input to the latch. The circuit only receives new input when the correct condition is received on the latch's trigger input.

Edge-Triggers
The keywords posedge and negedge are edge trigger conditions, and are only used with always blocks.


text:
Blender 3D: Noob to Pro/Tutorial Links from all over the web

Here you can find useful Blender Tutorial Links in English language. Feel free to add some cool tutorial, but here only in English. For tutorials in other languages please read About. Tutorial about editing this pages is described here.

New in 3D world? Read To Those Learning 3D.

Official Blender Documentation
Almost all Manuals you will find in latest official Blender Manual Books. For additional tutorials look below this topic.

 * Blender MediaWiki * exposed
 * Documentation at Blender3D.org

Interface

 * User Interface
 * Blender Interface Theme Repository
 * Basic Editing
 * Blender Hotkeys
 * Blender Hotkeys II
 * Basic Hotkeys (Java tutorial)
 * Navigating in 3D Space
 * Product Presentation
 * A short overview of functionality
 * Basic Blender
 * Rotation, Scaling and Layers Tutorial
 * Appending Textures
 * Basic Blender
 * Basic Blender Interface (Java tutorial)
 * Basic Interface
 * Basic Loading and Saving (Java tutorial)
 * Basic Parenting (broken pictures links: 7 March 2005)
 * Basic Scene (Java tutorial)
 * Blender Multimedia
 * Changing Cameras
 * Changing Views
 * First Impression
 * Getting Started
 * Removing Orphan Vertices
 * Seeing Textures with Alt-Z
 * The Camera
 * Stereoscopic camera for editing

Mesh Modeling

 * DupliVerts Tutorial
 * Heightmaps
 * Mesh Editing Techniques
 * Object Extrusion and Procedural Objects
 * Loops Corkscrews Problem
 * Loopings Problem
 * Knots in Curves Problem
 * ZeTool
 * DupliFrames Modeling I
 * DupliFrames Modeling II
 * Dupliframes
 * Spin and Spin Dup
 * Automatized BevOb
 * Extrusion along a Path using BevOb
 * Path Extrusion
 * DupliVerts
 * Limit DupliVerts
 * Types of Handles for Bezier Curves
 * Blender Booleans (broken pictures links: 7 March 2005)
 * Beveled Cube (pdf tutorial)
 * Joining/Separating Parts of a Mesh
 * Extrusion Controlled by IPO
 * Basic objects
 * Modelling Techniques and Strategies
 * To Remove Easily Orfan Edges and Vertices
 * Non-destructive bevel effect

Nurbs and Subsurface Modeling

 * IPOs, Lattices, Nurbs & Stuff
 * Cross-Sections
 * Curve Resolution
 * Weight Parameter
 * OrderU, UniformU and EndpointU
 * Curves -n- Bevels
 * Subsurf Modeling
 * Making a Hole in Subsurf
 * Subsurf Edge Rolling – Round
 * Subsurf Edge Rolling – Square
 * Modeling a fork (broken pictures links: 7 March 2005)
 * Modeling with lattices – a fork
 * Subsurf modeling 2
 * Metaballs
 * Abstract SubSurf modeling I

Specific Object Modeling

 * Double-Helix
 * Sword ''(Missing)
 * Turtle
 * Leaf Shader Tutorial
 * Balcony and Fire Escape ''(Broken picture links, 2/1 2006)
 * Oscilloscope ''(MISSING?)
 * Sears-Roebuck Dairy Barn
 * Golf Ball
 * Celtic Knot
 * Simple Tree
 * Making a pedestal
 * Building a Castle
 * Ice Cube (pdf tutorial)
 * Water I (pdf tutorial)
 * Water II (pdf tutorial)
 * Space Pod
 * Grass
 * Bongo Creature
 * Spiral Stairs
 * Waste Basket
 * Beveled Cube
 * Simple Box
 * Fountain with Moving Water
 * Track Creation
 * Dolphin
 * Dice
 * Logo
 * Cutting through steel
 * Roller Coaster
 * Roller Coster Cobra
 * Modelling a Glass
 * Ocean View
 * Solar Systems for beginners
 * Chair
 * Spiral Stair (pdf tutorial)
 * Lighthouse
 * Landscape
 * Landscapes Easy
 * Cloth Tutorial (broken link)
 * Cup
 * Cola cane
 * Candle
 * Volcano
 * Orbital Logo
 * Realistic Planet
 * Landscape Cartoon (pdf tutorial)
 * Text

Human

 * Head Modeling
 * Head Modeling & Texturing Tutorial
 * Head (Part 1 of 12-part Flash Tutorial)
 * Get a Head
 * Hand Palm
 * LowPoly Leg Modeling
 * Make Human: Modeling a New Target
 * Eyes
 * Hair Tutorial
 * Real Hair with Blender 3D
 * Female Character
 * Clothes for humans (broken link)
 * Texturing Skin using Vertex Painting and Repeating Image Textures

Cars

 * Porsche 550 * exposed
 * Car IV small (broken link 4 May 2007)
 * Car V F1 (broken link 4 May 2007)
 * Creating a Toon Car

Texture Mapping

 * Texturing Tutorial (pdf tutorial) * exposed
 * Making a Lightsaber with Halos
 * The Unofficial Texturing Tutorial
 * Material Indice Tutorial
 * Raytraced Transparency and Refraction
 * Chrome and Shiny Metal Surfaces
 * Texturing a castle
 * Texturing Tutorial
 * Textures in OrCo mode
 * Textures for Allosaurus
 * Chrome Effect
 * Texturing Part I
 * Texturing Part II
 * Texturing a Ship
 * Material Indices
 * Textures Channels
 * Textures Mapping
 * Specular Color
 * The Color
 * Materials in Blender
 * Alpha, Bump, and Specular image textures
 * Image Textures
 * Decals
 * Chrome
 * Alpha Masks
 * Skies I
 * Skies II
 * Shockwave (broken link 1 March 2006)
 * Texture Types in Blender (broken link 2.23.2006)
 * Basic Texturing
 * Basic Texturing 2 (Java tutorial)
 * Textures with Alpha
 * Textures with Bumpmapping
 * Using More Than One Color
 * Adding Color To Your Shapes
 * Using Textures and Bumpmapping Them
 * Texture Mapping Tutorial (pictures ok now 18/03/2006)
 * Environment Mapping (pictures ok now 18/03/2006)
 * Mapping Caracter
 * Creation and Mapping
 * Displacement Mapping
 * Game Sprites
 * Stencilling Textures
 * Using Texture Stencils

UV Mapping

 * Intermediate and Advanced UV Mapping * exposed
 * UV Mapping & Texturing
 * Sub-Surf and UV-Texturing problem
 * UV Texture Skinning
 * Character UV Textures
 * UV Mapping a Head
 * UV Texturing in Blender
 * Tuhopuu UV editor mapping
 * UV Mapping
 * Complex material for a sword

Animated Textures

 * Blender TexMesh Tutorial * exposed
 * animating masks for simulating ice freeze
 * Animated Procedural Textures (pdf tutorial)

2D Texture Painting Techniques
This part isn't about using Blender, but about necessary 2D knowledge for advanced Blender users.

 * Steven Stahlberg's Tutorials * exposed
 * Dirty Metal Texture
 * Gritty Pipelines
 * Digital Painting
 * Advance Painting and Weathering Techniques for Train Textures
 * NWN Texturing Tutorials
 * Fixing Lighting Irregulaties in Self-Tiling Maps
 * Texture Map for the Iris of the Eye
 * Airplane Texture Tutorial
 * Game Skinning Tips
 * Game Lighting Basics I
 * Game Lighting Basics II
 * Texturing with Gimp

Lighting, Shadows and Rendering

 * Light – a detailed tutorial
 * Yafray as an Integrated External Renderer
 * Basic 3 Point Lighting
 * Radiosity I
 * Radiosity II
 * Ramp Shaders
 * Simulating Radiosity
 * Caustic Sampler Tutorial
 * Shadows Control
 * Light Types (Broken Link 5 May 2009)
 * The 'World' buttons within Blender
 * Mastering Shadows
 * Ambient Occlusion
 * Area Lights
 * Batmanish Spot Logo (broken pictures links; 7 March 2005)
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Maybe someday ...

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CG artists must see or read:

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About
This links list is language filtered and extended version of personal collection originally provided by IamInnocent. So if you looking for tutorials in other languages check this link at www.elYsiun.com.

 * German Blenders should have a look at http://de.wikibooks.org/wiki/Blender_3D:_Tutorial_Linkliste

If you want to add Blender tutorial link in some other language you can add it temporary on talk page. We will later made such WikiBooks in other languages too.

Thank you all who contributed to this nice and useful links collection! Feel free to add your name or link if you think you need to be mentioned here.


text:
Portuguese/Contents/BPL1/Terceira lição

Verb "ir"/ Verbo ir
The verb to go is an important verb in Portuguese. It's an irregular verb:

See some examples:
 * Nós vamos para a escola. ("We go to the school.")
 * Eles vão para a casa deles. ("They go to their house.")

Compound future/Futuro composto
The verb "ir" can work as an auxiliary verb for the (near) future tense (just like "to be going to" in English). This structure is made using the "ir" plus the infinitive of an other verb.

See some examples:
 * Nós vamos ser amigos. ("We are going to be friends.")
 * Eles vão estar aqui? ("Are they going to be here?")
 * Ele não vai estudar português, mas eu vou. ("He is not going to study Portuguese, but I am going to.")

Present indicative of regular verbs/Presente do indicativo de verbos regulares
The conjugation form for each stem vowel in the present indicative tense is much alike. There is a table below with all three of them (stem vowels are highlighted):

Conjugation rules:
 * the first person singular (eu) always take -o.
 * the second person singular (você) takes the stem vowel (ir-verbs take -e instead).
 * the third person singular (ele/ela) takes the stem vowel (ir-verbs take -e instead).
 * the first person plural (nós) takes the stem vowel plus -mos.
 * the second person plural (vocês) is equals to the third person singular (você) plus -m.
 * the third person plural (eles/elas) is equals to the third person singular (ele/ela) plus -m.

There is a table with three examples below: "cantar" ("to sing"), "comer" ("to eat") and "partir" ("to leave").

There are below six examples of common verbs (with regular present indicative tense) for each conjugation.

title: Portuguese/Contents/BPL1/Terceira lição


text:
User:Rycanada

Future suggestions for Mandarin Chinese Simplified wikiproject: Each lesson should have a stroke order worksheet, pinyin should be beneath each line of zhongyin text from chapters 1-8, and there should be tips on pronunciation in each lesson. Lesson 1 might be too much info, breaking it into lessons 1.0 and 1.5 might be better particularly given how much info is needed to understand even the most basic Mandarin sentence (at least, from a unilingual anglophone's perspective).

title: User:Rycanada


text:
User:TheListUpdater in Wikibooks

 * 1) REDIRECT User:TheListUpdater

title: User:TheListUpdater in Wikibooks


text:
Introduction to Physical Science/2.3

=2.3 The Density of A Solid= Density calculations for regular solids are very simple. As with all density calculations, you need to know both the volume of the object, and its mass. Finding the volume of regular solids (solids of a recognizable form: cube, rectangular prisim, etc..) is very easy because you can measure the sides and calculate the volume using a formula (see apendix: Formulae). Finding the Volume of an irregular solid is more difficult. You must use water displacement to find the volume of an irregular solid. (for instructions see chapter 1.4: Experimentation). For this experiment, take the objects your teacher gives you, and find their density by using the formula D=M/V.

Calculating and converting
When converting densities (a mass divided by a cubic length (such as kg/m3)), remember to include the cubic nature of the length.

Q: How many cubic meters in a cubic kilomter?

A: While a kilometer is 1000 (1x3) meters, a cubic kilometer is equivalent to one million cubic meters (1x109 m3), because there are three sides of 1000 meters that make a cubic kilometer, and volume is calculated from length x length x length, or in this case 1000 x 1000 x 1000 = 1x109.

Equivalencies
When comparing certain metric cubic volumes, they will be equivalent when both units for length and mass are changed together. So in this case 1 kg/m3 will equal 1 g/cm3.

title: Introduction to Physical Science/2.3


text:
Introduction to Physical Science/A.1

= Appendix A1: Practice Problems Set 1

Example problems

Answer: NO Di=M/V=157/10=15.7g/cm^3 because 15.7g/cm^3<19.3g/cm^3,so we can judge this article of handicraft is not pure gold.
 * There is a article of handicraft,it has a V=10cm^3 and M=157g. Is it pure gold?(Note: D Of pure gold is 19.3g/cm^3)

Answer [m]-length [kg]-Mass [s]-time [a]-electrical current [k]-temp [mol]-amount of substance [cd]- liminous intensity
 * What are the SI unit base symbols and what do they mean?

Practice Problems and Solutions
Students - please insert your questions and solutions in the appropriate categories below and make sure you sign your problems. Use a # sign to number the problem and put a space before each line in your answer to put it in a box. --J.M.Pearce (talk) 12:55, 11 February 2008 (UTC)

Displacement

 * You are on a trip to visit your grandparents in Myrtle Beach (719 mile trip) for spring break. You drive half way there and realize that you forgot your toothbrush. So you turn around and drive back to get it. Then you complete the drive and spend the week with your grandparents and then go home.  What is your displacement?

Solution. Your displacement is 0 because displacement is final position minus initial position.

 * Dawn is driving to her new job today. She left at 8am this morning. She drove 10 miles from her house in the small town of Cranston to the town of Meadenrock. From Meadenrock, she drove 20 miles to Richport. From Richport, She drove 40 miles back through Meadenrock and Cranston to Porterville. When she reached Porterville she realized she was very lost. What is Dawn's total displacement?

Displacement = total change in position Start - Cranston 10mi - Meadenrock 20mi - Richport (this is the farthest point before she makes a wrong turn) 40mi backwards - Porterville 30 - 40 = -10 so...her total displacement is only 10 miles Ten miles will equal the quality of light years from the units of space and volume which = 206.

Force

 * A linebacker runs towards a Quarterback who is standing in the pocket looking for a receiver. The linebacker tackles the stationary quarterback in the pocket. What is the linebackers acceleration if we know the Linebackers weight is 312kg and the force of the impact on the QB is 342 Newtons.

Solution: F=ma First you must manipulate the equation for this particular problem A=f/m Once the equation has been manipulated, then you simply plug your information in. A=f/m A=342kg/312N A=1.096m/s^2

 * A 10.0 kg log was pulled along the ground (assuming with no friction) by a force of 200 N. What is the acceleration?

We know: F=ma F= 200 N m= 10 kg

200 N = 10.0 kg (a) 200 N/10.0 kg = a 20 m/s^2= a

 * What is the force of a 6kg baby sitting in your lap?

Force (F) = Mass (m) x Accleration (a) Here the a is due to gravity = 9.8 m/s^2 or about 10 m/s^2 So F = ma = 6kg x 10 m/s^2 = 60 N

You know that F=ma 1 lbs = 0.4356 kg 180 * .4356 = 78.41 Kg a= force of gravity 9.8 m/s^2 so... F= 78.41Kg (9.8m/s^2) F=768.42 N
 * If a person who weighs 180lbs falls out of a tree, with how much force does he hit the ground with.

 * A hockey puck is sent flying over the ice. If the air resistance and friction of the hockey puck are ignored, what is the  force required to keep that puck moving at a constant velocity?

The solution would be zero newtons.

 * A 32.1 kg log is raised off the ground with a rope. The acceleration of the gravity is 9.8m/s^2.  If the upward acceleration of the log is 3.9m/s^2, find the force excreted by the rope on the log. Please use units of newtons.

F=MA F=(32.1kg)(9.8m/s^2+ 3.9m/s^2) F=(32.1kg)(13.7m/s/s) F= 439.77 N

 * A 15.0 kg mass pulled along a frictionless surface by a horizontal force of 100 N will have what acceleration?

Use F=ma (100N)=(15.0kg)a a=6.67m/s^2

 * An object sits on a frictionless surface. There is a 16 N force applied to an object parallel to the surface and its acceleration is at 2m/s/s. What is its mass?

Use F=ma 16N=m(2m/s^2) m=8kg

 * The pressure inside of a bottle is 7 atm, and the pressure outside of is 3 atm. The inner radius of the bottle is 2cm. so what is the frictional force on the cork?

P=P(inside)-P(outside)=4 atm F=(3.14r^2)P=3.14x(0.02)^2x1.013x10^5 x4 atm =508.9N

 * there is an object of mass 2kg on the table. and there is a force pushes it to move forward and this force is 5N, but the object is still there. it doesn't move at all. find out what forces apply on this object.

answer: G(gravity)=mg=2kgx9.8n/kg=19.6N F=G=mg=19.6N F(push)=5N Friction force=F(push)=5N and G and F are opposite directions, F(push) and friction are opposite directions.

 * What is the force due to gravity of a hamburger sitting on a table? Assume that each bun weighs 1/4 kg and the beef weighs 1/2 kg.

F=ma a=g=9.8s/m^2 m= 1/4 + 1/4 + 1/2 m= 1 kg F= (1kg)(9.8s/m^2) F= 9.8 N mmmm tasty :)

 * A constant force is applied to an object, causing the object to accelerate at 8.0m/s^2. What will the acceleration be if the objects mass is doubled?

answer: 4m/s^2 a=f/m

 * A baseball player is sliding into 2nd base, identify the forces on the player.

Weight, normal force by ground, kinetic friction force by ground

 * Suzan pulled a wagon that weighed 5 kg along the street at a force of 100 N, what is the acceleration? Assume she was pulling the wagon directly in the direction of motion (no angle).

F=ma a=f/m a=100N/5kg a=20N/kg F=ma F=(67,500 kg)(5m/s^2) F=337,500 N
 * What is the force needed to pull a car that weighs 67,500 kg at an acceleration of 5m/s^2?

 * If you have a box that is pulled at an acceleration of 10 m/s^2 with a force of 98 N. What is the boxes mass?

F=ma m=F/a m= 98 N / 10 m/s^2 m= 9.80 kg

 * Jose pushes a boulder that weighs 60 kg and exerts a force of 60 N. What is the acceleration of the boulder?

F=ma 60 N= 60kg x a a= 1 m/s^2

 * Which Law states that if no force acts on an object, then its speed and direction do not change?

Newton's 1st Law

Velocity

 * If you are given the maximum velocity of 20m/s for a car. What would be its acceleration if it requires 500 seconds to reach max speed?

To solve you would first see that acceleration is $$change in V/change in T$$ = $$Vf-Vi/Tf-Ti$$ Therefore, the acceleration would be $$20m/s-0m/s/500s-0s$$ = 0.04m/s2

 * A car is traveling down the Highway and it takes the car 20 minutes to travel from Exit 1 to Exit 2. The car was traveling 5 mph over the posted speed limit which is 55 mph. What is the distance between these two exits in miles.

V=d/t t= 20min/60min/hr = 1/3 hour -- To solve for time you must first convert minutes into hours v= 60 mph d= vt = 60 mph x 1/3 hour - Once you have found time and distance all you have to do is plug those answers into the equation d=vt, and then solve. The final answer is d=20 miles

 * If the change in the velocity of vector A = 5m/s to the right and the change in the velocity of vector B = 8m/s to the right, then what is the total change in velocity to the right?

The correct answer is 13 m/s. Remember to add vectors from head to tail. Therefore, Vector A + Vector B = Total change in velocity. 5 m/s + 8 m/s = 13 m/s

The correct answer is 0.81. Remember to plug the numbers into the SSF formula. t/2h=v^2/rg. The correct answer is V=10.91m/s. Remember to use part one's answer in place of t/2h. Also remember that G= 9.8m/s.
 * A.Given that a van has a height of 1.1m and 1.8 is the track width, what is the SSF?(static stability folder)
 * Part B. Given R=15m, what is the fastest you can take a corner without tipping over the van?

 Answer: 1.34 m/s v = d/t v = 885 meters/11 minutes 11min.x60sec.=660seconds v = 885 meters/660 seconds v = 1.34 meters/second
 * 1) Joe drove 885 meters in 11 minutes at a constant velocity. At what velocity was Joe traveling?

Answer: 159 meters d = df-di * 318 meters / 2 d = 318 meters - 159 meters d = 159 meters
 * Sarah has drive 318 meters to the grocery store to get ingredients to bake a cake for her mom's birthday. When she left the store she realized she forgot to buy milk. So she stopped at the gas station convenient store on the way home. The convenient store is located exactly halfway between the grocery store and her house. What was her displacement when she stopped at the convenience store?

 * A professional baseball pitcher throws a curveball to the number four hitter of the opposing team. The curveball travels 60ft to homeplate. The batter hits the curveball for a homerun which went 409ft to dead center field. What is the total displacement of the baseball from home plate?

The correct answer would be 409ft Remember that displacement is the Δx therefore the equation is Δx= xf - xi Δx= 409-60  Δx= 369ft

the correct answer is 14.29 m/s v= d/t v= 25 meters/ 1.75 seconds v= 14.29 m/s
 * A football quarterback throws a touch pass of 25 meters and it takes the ball 1.75 seconds to reach the receiver.  How fast did the QB throw the ball.

 * A toddler pushes "Block A" to the right toward "block B"(which is at rest). "Block A"(which is 1kg), has a velocity of 6 m/s moving to the right toward "Block B". "Block B" is 2kg. Now "Block A" is at rest and "Block B" moves to the right at an unknown velocity. Find this unknown velocity.

∑ MV0 = ∑ MVf (1kg)(6m/s)+(2kg)(0m/s)=1kg 6kgm/s = 2kgV 2kg 2kg V=3m/s

Speed Limits
The runner's average speed is given by: average speed= distance/time= 800/86= 9.3 m/s
 * An Olympic runner ran the a 800 m in 86 s. What was his average speed?

We know that acceleration is a change in velocity over a given changed in time A=V/T So we plug 95 km/h in for V, but V is measured in m/s so we must convert. 94.5 km * 1000m/km = 95,000km and 1 hr * (60m/1hr) * (60s/1m)=3,600 s so after converting V=95,000m/3,600s V= 26.39 m/s then we plug in T which is 8s so A=(26.39 m/s)/ 7.8s A=3.3 m/s2
 * A car is going 94.5 km/h. A deer runs out infront of the car, causing it to come to a complete stop in 7.8 s. What is the acceleration of the car?

Using the equation V=d/t we can manipulate it to solve for time for each person. So using the equation t=d/v we plug in the numbers for Julie's trip t= 100mi/50mph. We get t=2hrs for Julie's time. Now we use the same equation to solve for Matt's time. t=d/v. t=100mi/60mph. Matt's time equals 1.67hrs. So if you take Julie's time and subtract Matt's time from it 2hrs-1.67hrs=0.33. '''Therefore, Matt has to wait ~20 mins for Julie to arrive in Pittsburgh.'''
 * Julie leaves Clarion at 8:00 am to drive to Pittsburgh, 100 miles aways. She travels at a constant speed of 50mph. Matt leaves Clarion at 8:00am and travels a constant speed of 60mph. How long does the first person to arrive have to wait for the second person?

ΔV = ΔX/Δt (change in velocity equals the change in displacement divided by the change in time) Δt = Δx/Δv 15m/ 50m/s = 0.3 s
 * A speedboat with a velocity of 50 m/s only has 15 m until it crosses the finish line. How long will it take the boat to cross the finish line?

a) changeV= changeX /changet = Xf - Xi / tf - ti b) V=2(pi)r / T correct answer : a)
 * What is the correct equation for velocity?

v=d/t v=800 meters /20 seconds v=40 meters per second
 * What is the speed of a vehicle that travels 800 meters in 20 seconds?

t=d/v t=425,000 meters /70,000 meters t=6.07 hours
 * How long will your trip take (in hours) 425 km at a speed of 70 km per hour?

V=change in distance/change in time V of the first car = 23.8 (given) V of second car = 89m/2.7s V of second car = 32.96m/s Therefore, the second car has a higher velocity than the first.
 * One car is traveling 23.8 m/s. Another car traveled 89 meters in 2.7 seconds.  Which car has a higher velocity?

Circular Motion

 * A baseball with a 3 in diameter is rolled on the ground with no slipping and it revolves 50 times how far does it go?

The correct answer is ~ 471 inches Distance Traveled = 50*C Diameter/(d)= 3in C=d(pie) pie= 3.14 50*C 50d(pie) 50(3in)(3.14)= ~ 471 inches

The correct answer is 131.88in or 10.99 feet C=2TTR d=2r C=dTT distance=7c 7dTT TT is equal to pi or 3.14. 7(6in)TT=42inTT=42*3.14=131.88in or 10.99ft.
 * If a baseball is rolled on the ground and revolves seven times, how far does it go in feet? Given the diameter is six inches.

Formula: C=2(TT)(r) D=2(r) C=(D)(TT) =(10)(D)(3.14) =(10)(4)(3.14) Solution: 125.6 inches or 10.5 feet
 * If a child rolls a ball on the ground and resolves ten times. How far does the ball travel? The diameter of the ball is 4 inches.

Circumference of the bowling ball: 21.5*3.14=67.51cm Length of the bowling path in cm: 19.15*100=1915cm Rounds of the bowling ball rolled:1915/67.51=28.37 rounds Frequency: 28.37/8= 3.55 r/sec
 * A regular bowling path is 19.15m long; the diameter of a bowling ball is 21.5cm. If the bowling ball goes the whole way straightly in 8seconds, what is its frequency in this situation?

1 revolution takes 3 minutes. 5 dollars times 1 revolution = 5 revolutions. 5          revolutions times 3 minutes a revolution = 15 minutes.
 * If a ferris wheel takes 3 minutes to go one revolution and you want to pay the man running it to get an extra long ride with your significant other. He tells you that he will make your ride longer for each dollar you give him by 1 revolution.  You only have 5 dollars in your wallet.  How long will your ride be?

D=7C C=2TTrsquared= TTd D=7(TTx9.07in)=199.46 in The distance traveled by the ball was approximately 199.46 in.
 * A basketball is 9.07 inches in diameter. It rolls across the court 7 times before a player stops it.  How far did it roll?

Static Stability Factor
The formula for (SSF)= t/2h t= 2.2 m     h= .95 m      2.2 m / 2 (.95 m)      2.2 m / 1.9 m      SSF = 1.16
 * If you are driving a Jeep Laredo with a track width (t) of 2.2 m, with a height (h) of .95 m, what is the static stability the factor (SSF)?

 * If a vehicle is 1.2m high, to the center of gravity, and the distance between the tires is 2.3, at what velocity can you take a curve with the radius being 12m.

t=2.3 m h=1.2 m  r=12 m  g=9.8 m/s² v=sq. root of rgt/2h v=12(9.8m/s²)(2.3)/2.4 v=sq.root of 112.7m/s² v= 10.61m/s²

Formula: t/2h=v^2/rg 2.5/2(2.8)=2.5/5.6 Solution: .45m Part B. Given R=25m. Coming into turn three, what is the fastest that the racecar can take the turn without tipping over? Formula: t/2h=v^2/rg (Substitute t/2h with the previous answer of .45m) .45m=v^2/rg (Substitute rg with 25 (being r) and 9.8 (being g) .45m=v^2/(25)(9.8) Multiply both sides by (25)(9.8) .45(25)(9.8)=v^2 Take the square root of .45(25)(9.8) Solution: 10.5m
 * Given that a racecar has a height of 2.5m and the width of the racetrack is 2.8m. What is the static stability folder (SSF)?

Remember tipping point is t/2h=v2/rg Rearrange so that v2=trg/2h SSF=t/2h v2=(SSF)rg v2=(1.25)(12 m)(9.8 m/s2) v2=147 m/s2 v=12.1 m/s A Hummer can go around the curve r=12 m at a speed of 12.1 m/s without flipping.
 * A Hummer with a SSF of 1.25 is rounding a curve with r=12 m. How fast can the Hummer go around the curve without flipping?

57.7inches= 1.47m 25inches= 0.64m 215 km/h= 59.72m/second t/2h=v^2 /rg 1.47/ (2*0.64) = 59.72^2/ r(9.8) 1.15=3655.48/r(9.8) 11.27= 3655.48/r 3566.48/11.27=r r=316.45m
 * A mini cooper has a track width (t) of 57.7 inches, with the central mass height (h) of 25 inches. Suppose it is running with 215 km/h, what would be the max radius of the curve can this mini copper turn?

t=1.5m SSF=1.25 SSF=t/2h solve the SSF=t/2h equation for "h"-- (2h)SSF=t Divide both sides by (2)(SSF) h=t/2(SSF) Plug the known values of "t" and the SSF into this equation to get the height. h=1.5m/2(1.25) h=0.6m The height of the vehicle's center of mass is 0.6 meters.
 * If a vehicle has a track width of 1.5m and a SSF of 1.25, what is the height of its center of mass?

the equation for SSF is SSF=t/2h SSF=1.5(2*.75) SSF= 2.25
 * If we know that the track width of a van is 1.5m, the height is .75m, what is the SSF?

Torque
The magnitude of the force, and where the force is applied or the point of application.
 * Torque depends on what two things?

Torque
 * What is the measure of effectiveness of a force to turn or twist an object?

Torque is T=rF(perpendicular), which is measured in Nm. Torque= (.10m)(6)= 0.6Nm
 * A pinwheel is turned with the force of 6N at a distance of 10cm from the rotation. What is the torque?

Remember acceleration of gravity= 9.8 m/s2 F=ma=(50 kg)(9.8N/kg) T=Fr=(50 kg)(9.8N/kg)(4m) T= 1960 Nm
 * A child of mass of 50 kg is sitting on a seesaw 4m from the pivot- What is the magnitude of the torque created by his weight?

 * A mother and child are playing on a see-saw. The child is sitting 3 m from the pivot.  The magnitude of torque created by the child is 999.6 Nm.  How much does the child weigh?

The equation for torque is T=rF. Remember F=ma. So, T=rma. Solve equation for m, m=T/ra. m=T/ra m=(999.6 Nm)/(3 m * 9.8 N/kg) m=(999.6)/(29.4 kg) m=34 kg A child sitting 3 m from the pivot, creating a torque of 999.6 Nm, weighs 34 kg.

 * If somebody is using a wrench with the F of 30N at a distance of 18cm from the rotation axis, what is the magnitude of the torque?

The equation for torque is T=Fr. Remember to change cm to m. T=Fr T=(30N)(.18m) T=5.4Nm

C=2pir 20=2pir 10/pi=r 3.18 ft=r feet must be changed to meters so 3.18(30.5)= 96.99 cm= .9699 m T=rF(parallel) 740 Nm= .9699m(F) F= 762.9N T= + rF(perpendicular) - (46kg)x(9.8N/kg)<--Gravity (46kg)x(9.8N/kg)x(2.4)=1081Nm
 * A circular silo is a used to store grain for feeding cattle. A man is trying to turn a silo unloader by hand. Each arm of the unloader is parallel to the axis of the unloader. How much Force should he use if the silo has a circumference of 20 feet and is using a torque of 740 Nm?
 * A little girl of 46Kg is being pushed on a merry go round counter clockwise. She is 2.4m from the center. What is the magnitude of the Torque?

T=Fr T=30N x 62 cm / 100cm/m T=18.6 Nm
 * Phil took his car to get his tires aligned. The mechanic elevated the car and spun the tire with a force of 30 N.  The force he exerts is 62 cm from the rotational axis.  What is the magnitue of the torque?

T=Fr T=15N x 85cm / 100cm/m  T=12.75Nm
 * A door is pushed open with a force of 15N. The force is applied perpendicular to the door at a distance of 85 cm from the rotation axis.  What is the magnitude of the torque?

States of Matter
It is a change in pressure.
 * When talking about Pascal's Principal, a change in WHAT? at any point in a confined fluid is transmitted everywhere throughout the fluid?

Solids, they have a rigid/fixed shape, and the strongest bond of the states of matter.
 * Which of the three states of matter (solid, liquid, and gas) has the strongest bond?

All states of matter are described by their DENSITY. This is mass per unit volume. The equation for this is D=m/V. The category is FLUIDS and the two states of matter that are in FLUIDs are gases and liquids.
 * All of the states of matter,(solids, liquids, and gases), are described by what? And what is the equation for the description?
 * There are three states of matter (solid, liquid, and gas) two of them fall into a category that has no fixed shape and is easily deformed. What is that category and what two fall into that category.

V= md V= (78g)(1 cm^3/19.3g) V= 4.04 cm^3
 * Gold has a density of 19.3g/cm^3. What is the volume of a gold bar that has a mass of 78g?

D=m/v D=37kg/12m^3 D= 3.083kg/m^3
 * The volume of a mystery fluid is 12 cubic meters with a mass of 37kg. What is it's density?

Gas, it is compressible, and when heated, its molecules moved farther apart.
 * Which out of the three states of matter [Solid, Liquid, Gas]is the only one able to have a change in volume?

F=P*A P=3*1.013*10^5=3.039*10^5Pa F=3.039*10^5pa*12*(1/1000)m^2=364.68N
 * If the pressure on the box is 3atm, the area of box on the surface is 12cm^2.What is the force on the box?

Elasticity and Oscillations
Means that a object returns to it's original shape and size after it has been deformed.
 * What is the definition of elasticity?

2.5in=6.35cm, 6in=15.24cm Strain= change in length/length (15.24cm-6.36cm)/6.35cm = Strain Strain = 1.4cm
 * Steve picks up a rubber band, which at rest measures 2.5 inches long. He then streches the rubber band across two fingers, the rubber band is now 3.5 inches longer than it was at rest. What is the strain Steve is placing on the rubber band?

stress/strain = stiffness strain = ΔL/L 3.5 / .30-.20/.20 =  7 [Pa]
 * The original length of a spring is .20 m long. When the spring is stretched to .30 m long with the stress=3.5, what is the stiffness of the spring?

1. Force of gravity- graviton 2. Strong Nuclear Force- holds protons and neutrons together in an atom 3. Weak Nuclear Force- radioactive decay 4. Electromagnetic Force- contact forces
 * What are the Four Fundamental Forces?

strain=ΔL/L strain=1.2-.5/.5 strain=.7/.5 strain=1.4
 * Debbie is exercising with an elastic cord which when at rest measures .5m. When she pulls it it stretches to a length of 1.2m.  What is the strain on the cord?

 * 1) A Frisbee is 25.4 cm in diameter.  If it is thrown, not caught, and hits a wall the new diameter is 24.6 cm.  What is the strain on the Frisbee?

Strain=deltaL/L =-0.08/ 25.4 cm= -0.03

title: Introduction to Physical Science/A.1


text:
Portuguese/Contents/BPL1/Sexta lição

 * Não falo português = i don't speak portuguese (Pron. naum falu portuges)

 * Você fala inglês? = do you speak english?

 * Onde é o hotel? = Where is the hotel? (Pron. on-de é o otel?)

 * Onde? = Where?

 * Quem? = Who?

 * Quanto? = How much?

 * Por que? = Why?

 * Quando? = When?

 * Como? = How?

 * Quantos? = How many?

title: Portuguese/Contents/BPL1/Sexta lição