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Iggy

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Everything posted by Iggy

  1. The choice of which case to pick belongs to the player. When the player has three cases remaining they choose one without knowing which of the three amounts is in the case. There is an equal (1/3) probability that the case contains each of the three remaining denominations. The case is opened and the amount revealed and everyone now knows which of the three amounts it contains: the lowest amount, the middle amount, or the highest amount. Apparently the banker then offers the player the option to switch the remaining two cases if and only if the previous case contained the lowest amount. Everyone, including the player, knows if the previously opened case held the lowest amount so the banker reveals nothing. In Deal or no Deal the act of switching does not increase the odds of opening a better valued case. In Monty Hall, switching gives better odds. It is not enough for the banker to know what's in a case. It might be most helpful if I just describe the Monty Hall problem and it should be clear how Deal or no Deal is different. The Monty Hall problem has a contestant choose one of three doors. Behind one of the doors is a car. Behind each of the other two doors is a goat. The player is trying to choose the door which hides a car and they make a choice 1, 2, or 3. The chosen door is not immediately opened. The host instead opens a different door which he chooses himself (out of the remaining two) revealing one of the two goats. The host knows where the car is and that knowledge allows him to choose a door with a goat. There are now two closed doors one of which the player has already chosen and an open door which revealed a goat. The host then asks the player if they want to switch their initial choice and instead pick the other unopened door. In this situation it is advantageous for the player to switch. Going with the original choice will have a one in three chance of getting the prize. Switching results in a two in three chance. Most people intuitively think that switching will make no difference. But, think of it like this: When the player first picks a door each of these situations is as likely as the other two: The chosen door holds the car The chosen door holds goat A The chosen door holds goat B If (1) happens then the host will reveal either goat A or goat B. Switching means the player loses. If (2) happens then the host must reveal goat B since he never reveals the car. Switching then results in a win for the player. If (3) happens then the host must reveal goat A. A switch then results in a win. Without switching only one of situations 1, 2, and 3 get the car. With switching two of the situations get the car. Switching therefore increases the chance of success from 1/3 to 2/3. Knowing where the prize is means the probability of winning is 1. The choice is always successful no matter how many unopened boxes there are. Knowing that a box is not the prize is only that advantageous if there are only two options. In Monty, the box in (1) is not revealed. In Deal, (2) is not always the case. In Monty, (3) is an advantageous move. The wikipedia page: http://en.wikipedia.org/wiki/Monty_Hall_problem might explain better than I have.
  2. I didn't know that, but I don't watch the show. It seems odd--only if the third to last case is the lowest amount Hum. No. Neither the banker or host are choosing cases to open or making any move based on secret knowledge. Even if the banker only offers a switch if the 3rd to last case was the small amount--the player has that info as well, so it doesn't reveal anything. It doesn't help constrain the probability of the last two cases. Monty Hall would only work if the host decided which case to open knowing that it could not be the larger amount--effectively narrowing down the possible location of the largest amount to 2 out of 3 cases rather than 3 out of 3. Edited To Add-- Here is a comparison I found on wikibooks: When only three cases remain, Deal or No Deal might seem like a version of the Monty Hall problem. Consider a Deal or No Deal game with three cases (similar to the three doors in the Monty Hall problem). The contestant has one case. Then, one of the two other cases is opened. Finally, the contestant is given the option to trade his or her case for the one unopened case remaining. The Monty Hall problem gives the contestant a 2/3 chance of winning with a switch and a 1/3 chance of winning by keeping his or her case. However, there is a critical difference between Let's Make a Deal and Deal or No Deal. In the Monty Hall problem, the host has used his secret knowledge of what lies behind each of the three doors to cause a bad choice to always be revealed. This non-random selection of a bad choice is what causes the difference in odds of winning between switching and not switching on Let's Make a Deal. In Deal or No Deal the odds behave exactly as you would instinctively expect them to: 2 boxes with a fifty-fifty chance of the top prize being in either of them. Introduction to Game Theory/Deal Or No Deal
  3. That sounds somewhat like asking how far 1 cubic meter is. The answer is that a cubic meter is not a measure of distance but is rather a measure of volume. c3 would have dimensionality of volume / cubic time which is not a velocity. When quantities (be they like-dimensioned or unlike-dimensioned) are multiplied or divided by each other, their dimensional symbols are likewise multiplied or divided; this corresponds to vector addition or subtraction (on the exponents). When dimensioned quantities are raised to a rational power, the same is done to the dimensional symbols attached to those quantities; this corresponds to scalar multiplication on the exponents. http://en.wikipedia.org/wiki/Dimensional_analysis#Mathematical_properties A physical quantity is expressed as the product of a numerical value and a physical unit, not merely a number. It does not depend on the unit distance (1 km is the same as 1000 m), although the number depends on the unit. http://en.wikipedia.org/wiki/Scalar_%28physics%29
  4. The odds go up in the Monty Hall problem because the host knows which door has the better prize and eliminates a different option purposefully. In Deal or no Deal the host doesn't know which case has the larger amount. Even if he did, the host doesn't eliminate a choice on that game show. There is no advantage to switching when 2 cases are left. Both cases have a .5 probability of being the larger amount before and after switching.
  5. My vote is for 9,255,858,616,106,894,221,940,637,969,821,537,730,502,421,652 Planck times.
  6. Here is an experimental demonstration of the dimming of a star by the atmosphere: http://spiff.rit.edu/classes/phys445/lectures/atmos/atmos.html One airmass means looking at the star when it is directly above and an airmass of 2 is about 30 degrees off the horizon. A larger apparent magnitude means the thing is less bright. As expected, the larger the path length through the atmosphere, the dimmer the star. It looks roughly linear with path length which I believe Beer's law would attest to.
  7. I might just disagree with the way you are saying it. I wouldn’t say, for example, that c = 1 in the international system of units. In that system of units c = 299,792,458 m/s. In other unit systems it is normalized to unity. But, I think you mean that c is always equal to 1 in natural units even if it has a different value in other unit systems. Absolutely. Ok, I think I see where you’re coming from. An example would be that beta of the Lorentz factor is unitless in any system of units: [math]\gamma = \frac{1}{\sqrt{1 - \beta^2}}[/math] And if you had a velocity in m/s (1.5E8 for example) and wanted beta you would: [math]\beta = 1.5 \times 10^{8} \; m/s \left( \frac{1}{3 \times 10^{8} \; m/s} \right ) = 0.5[/math] I agree. I would just avoid saying that c=1 is the same in any unit system which would mean to me, for example, that the equality’s lhs should be unity in SI or any other units: [math]\displaystyle { c } = \frac{1}{\sqrt{\mu_0 \epsilon_0 }} [/math]
  8. By my understanding, unity does not imply non-dimensionality. In physics, a dimensionless physical constant (sometimes fundamental physical constant) is a universal physical constant. Because it is a dimensionless quantity, its numerical value is the same under all possible systems of units. Fundamental physical constant may also refer (as in NIST) to universal but dimensional physical constants such as the speed of light c, vacuum permittivity ε0, Planck's constant ħ, or the gravitational constant G. http://en.wikipedia.org/wiki/Dimensionless_physical_constant
  9. Maybe I don't follow your meaning, but using SI, for example, won't let you set c=1. One unit distance per one unit time in SI is 1 m/s.
  10. Equal to one, yes. Dimensionless, no. A dimensionless quantity is the same in any system of units.
  11. Your answer of 20 N is correct. F = Pt/d F is force in Newtons, P is power in watts (100), t is time (1 sec), and d is distance (5 m). F = 100*1/5 = 20 Newtons.
  12. I’m sorry you find my post/posts distasteful, Akhenaten. I was asking Swansont to clarify something very specific that he had said, which he did. I was not answering, or trying to answer, the opening post--and not directing my comments toward your position or your argument. If you think there is any way for me to help satisfy your contentions on this issue of the moon's rotation then please let me know.
  13. I agree completely and that is exactly what I was referring to in saying that the original question in this thread had been answered very well.
  14. I think the original question in this thread has been answered very well and I post in fear of looking like a persistent pest on this issue, but I really cannot understand this it-rotates-or-it-doesn't thing. What makes this effect so interesting and important is that while other effects that I have described in this book, including the geodesic precession, have to do with such concepts as gravitational fields, curved space-time, and nonlinear gravity, this effect tells us something about the inertial properties of space-time. If you ask yourself, “Am I rotating?” and you wish an answer with more accuracy than you get simply by seeing if you are getting dizzy, you usually turn to a gyroscope, for the axis of a gyroscope is assumed to be nonrotating relative to inertial space. If you were to build a laboratory whose walls were constructed to be lined up with the axes of three gyroscopes arranged to be perpendicular to each other, you would conclude that your laboratory was truly inertial (and if it were in free fall, that would be even better). However, if your laboratory happened to be situated outside a rotating body, the gyroscopes would rotate relative to the distant stars because of the dragging effect I have just described. Therefore, your laboratory can be nonrotating relative to the gyroscopes, yet rotate relative to the stars. In this way, general relativity rejects the idea of absolute rotation or absolute nonrotation, just as special relativity rejected the idea of an absolute state of rest. --"Was Einstein right?", Clifford Martin Will, ch. 12 If you set up a coordinate system by some gyroscope axes (call them A) in the manner of the quote above and you are not rotating in that coordinate system then you are not rotating relative to A. At the same time someone else may set up a different coordinate system by gyroscope axes (call them B) while they are, let's say, a little closer to the sun. Now the person is rotating relative to A and not rotating relative to B. If you are rotating relative to one gyroscope and not rotating relative to another and you define rotation by the action of a gyroscope then I fail to understand "either it rotates or it doesn't". I think the gut reaction to my argument would be to point out that one set of gyroscopes is local and the other is not and say that absolute rotation is determined by the local set of gyroscopes. But, consider time. Nobody would say that time is absolute because local clocks always run the same rate. Time dilates the closer a clock gets to a massive body the same way rotation precesses the closer a gyroscope gets to a spinning body. If clocks measure time and gyroscopes measure rotation and we say that time is relative because different clocks run different rates then rotation should be relative because different gyroscopes measure different rates of rotation. In the same respect, nobody would claim that seeing the stars move means you have velocity. If you see a star move then you have velocity relative to that star. you may not have velocity relative to a different star.
  15. Yes. It would be like the ergosphere of a rotating black hole. Speeds are, and must be, faster than c because spacetime is being dragged around with the mass. I looked for a paper this morning supporting the Machian view of the relativity of rotation with GR and this is the first I found: I think Swansont is correct that Mach's principle is a conjecture. The paper I was just reading sounds sure of itself, but I'm sure other authors have found different results and would disagree. Maybe "rotation is relative" and "rotation is not relative" are both conjectures, and mostly philosophical.
  16. I don’t think I follow what you mean that freefall in a uniform field can’t be distinguished from any other linear acceleration. Someone in a rocket in deep space who is accelerating relative to the background stars can surely determine that they are accelerating relative to the background stars. I am definitely not arguing that. The first quote in my last post means to me that the acceleration is relative--not absolute. The person in the rocket cannot say that they are absolutely undergoing a change in velocity while the rest of the universe is at rest because it is just as valid, although nowhere near as convenient, for the person to say “I am at rest in a uniform gravitational field that has the background stars accelerating--in freefall.” The person can tell that they are accelerating relative to the bulk mass of the universe, but which is accelerating--the ship or the mass of the universe--is not absolute but relative. I do believe rotation can be made to look like that. The second quote of my last post appears to me to bear that out. I certainly agree that the Foucault pendulum can reveal rotation relative to the background stars. I’m not arguing that. But, if the earth were at rest while all the mass of the background stars circled around it then the rotating mass should create a gravitational field which moves the Foucault pendulum in the expected way. There is a Flash animation showing this. It shows the pendulum while letting you choose what rotates ‘earth’ or ‘stars’ as a demonstration of Mach's principle: http://www.upscale.utoronto.ca/PVB/Harrison/Flash/ClassMechanics/Foucault/Foucault.html It would appear that GR confirms this at least where the universe is approximated by a shell of rotating mass (which I think is what Rusty was describing): If my understanding is correct, rotation is truly relative. It would not strictly be correct to say “earth rotates” as much as it would be to say “earth rotates relative to the background stars”. Which of the two is rotating and which is at rest is a coordinate choice--both equally valid--if not incredibly inconvenient. I want to stress that I'm not trying to make an argument--only to explain my understanding which is very tenuous. I'm not a physicist and I defer to the expertise of those here who are.
  17. That is correct. In either frame while light moves dL an object with velocity vo will move vo(dL/vL). Distance is velocity times time and (dL/vL) is the time that light travels. Hence, do = vo(dL/vL) is equivalent to do = vo*to if tL = to which is exactly what you mean when you say "while light moves...". You mean that both the light and the object have moved for the same duration. For example, in your thought experiment: BM |----O-----------------------------------T v<-O' if the velocity of O' is .6c relative to O then O' may find that the light moves 2 light seconds while O moves 1.2 light seconds between the two events (the emission and co-location with the burn mark). Between the same two events O would find light moves 2.5 light seconds while O' moves 1.5 light seconds. As you say: do = vo(dL/vL) in the frame of O': 1.2 = 0.6(2/1) in the frame of O: 1.5 = 0.6(2.5/1) Here is a space time diagram of this example where v = .6c: O and O' are co-located at the black dot when the light is emitted and O' is co-located with the burn mark at the green dot when both frames calculate the distance the light has traveled. O calculates 2.5 light seconds and O' calculates 2 light seconds. I understand you would rather O and O' agree on the spatial distance the light travels between events, but special relativity does not work that way. I suggest that your disagreement with relativity does not present a contradiction in the theory, but instead a contradiction between your expectations and the results of the theory. Since the results of the theory are shown to be experimentally accurate, you may want to examine your expectations--I would humbly suggest.
  18. Velocity involves distance and time. If you want to know how far something travels with a given velocity between two events then you need to consider the time between events. It is a contradiction to say "We are not using clocks" and to assign velocity to things and talk about "when" things happen.
  19. My understanding, which is admittedly tenuous, has always been that rotation is relative. I understand that an observer can tell if they are in an inertial frame by the forces they feel, but whether the cause of the force is acceleration or gravitational is not absolute but relative. If the observer considers themselves at rest then they are gravitational and if they consider the fixed stars at rest then they are inertial. The quote is page 150 of the linked document--Einstein's "The Foundation of the General Theory of Relativity": I think it can be said that earth rotates relative to the background stars. But, to say which, the earth or the background stars, rotates and which is at rest in an absolute sense would not be consistent with the freedom of coordinate choice in Einstein's gtr. Choosing coordinates where the earth is at rest would result in a gravitational field caused by the rotating background mass indistinguishable form the centrifugal force expected if we choose the background stars to be at rest. There is another quote of Einstein from 1918 that he writes as a dialog between a critic and a relativist--"Dialog about Objections against the Theory of Relativity"--that I think would be relevant to the thread: Perhaps I misunderstand your meaning, Swansont. Or, I misunderstand, or don't fully understand, the issue.
  20. It can. Look at a space time diagram: Does a ray of light need to be measured the same spatial distance in both frames? I get the feeling you are either not very familiar with relativity or not too fond of it. Either way, this tutorial might be helpful: http://casa.colorado.edu/~ajsh/sr/paradox.html because it starts with a 'paradox' like your example and works through the solution.
  21. You're talking about "when" things happen. In SR if, BM |----O-----------------------------------T v<-O' BM and T are simultaneous for O and O' is moving toward BM then O' must expect BM to happen before T. That is consistent with relativity. It is actually demanded by it. The order of events is relative in relativity, so I don't think there is any contradiction.
  22. Two events that are simultaneous in O may not be simultaneous in O'. http://en.wikipedia.org/wiki/Relativity_of_simultaneity
  23. The space shuttle as a point of reference has around 3.3E12 J of kinetic energy. Assuming there are 100,000 shuttles reentering and all the energy is converted to heat I find, Q = m x Cp x dT dT = 3.30E17 J / (5.15E+18 kg * 1000 J/kg-K) = 6.4E-5 K 100 thousand would raise the temperature by 0.00006 Kelvin. To raise the atmosphere an average of one degree would take about a billion reentering shuttles
  24. I agree. Sorry it took me so long to respond, although I'm glad I waited--excellent explanations Martin Think of velocity as being measured with a measuring tape and clock. For example, a baseball might go around 30 meters along a measuring tape while a clock advances 1 second. The thing about relativity is that clocks and measuring tapes change the rate they run and the distance they measure as they change velocity. So while the person throwing the baseball might measure it at 30 m/s someone passing the field in a rocket might measure the baseball at 10 m/s relative to the pitcher with their tape and clock. The distance and time are relative to the velocity of the person measuring them. Your thought experiment has two things moving away from a central observer, A <----- B -----> C There are three frames of reference. Intuition would say that A measures the velocity of C the same as B measures the velocity of A plus the velocity of C. In other words, A measures VC the same as B measures |VA| + |VC|. If B measures VA at -1c (negative meaning it’s moving to the left) and VC at 1c like your thought experiment then that would be: VC' = |-1c| + |1c| = 2c But, that would ignore time dilation and length contraction which are key to special relativity. Just because B says that C is moving at some speed that does not mean C is moving that speed relative to B in A’s reference frame where clocks and measuring tapes work differently. The correct relativistic formula which accounts for relative time and distance is: [math]{V_{C}}'=\frac{V_{C}-V_{A}}{1-\dfrac{V_{C} \times V_{A}}{c^2}}[/math] where [math]{V_{C}}'[/math] is the velocity of C relative to and measured from A. Both [math]V_{C}[/math] and [math]V_{A}[/math] are relative to and measured from B. If A and C are photons like you say where c=1 (in light units) you’ll get: [math]{V_{C}}'=\frac{1-(-1)}{1-\dfrac{1 \times -1}{1}}[/math] [math]{V_{C}}'=\frac{2}{2}=1[/math] So it is not paradoxical if you add velocities using the relativistic velocity addition formula. The formula works because space and time are relative to velocity. In other words, taking into account that two observers in different reference frames will not measure distance and time equally. I hope this helps, and also this really is a great page: http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html and also: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel2.html#c2 I wouldn't know how to answer if infinity actually exists in the universe. I am philosophically of the opinion that if something is real and exists then it can be measured and I don't think infinity, as a quantity, can be measured. But, I really don't feel qualified to say. As far as the universe being infinitely big I would definitely say that is not certain. It is implied by general relativity and standard cosmology if the universe is homogeneous and flat or negatively curved. But, we don't know and may never know (as far as I know ) the geometry of the universe outside of our visible neighborhood. If I understand then I agree. The ideas of actually measuring something infinitely large or infinitely small seem problematic. That's not to say that infinity isn't a very useful concept, just that actually measuring such a quantity seems kind of like a contradiction in term.
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