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November 19

Time Dilation? Time speeding?

I think the following section in Special Relativity has a wrong conclusion, “This phenomenon is called time dilation.”, △t’=γ△t should mean time speeding, isn’t it? Please help. Thanks.

Time dilation and length contraction

Writing the Lorentz transformation and its inverse in terms of coordinate differences, where for instance one event has coordinates (x1,t1) and (x'1,t'1), another event has coordinates (x2,t2) and (x'2,t'2), and the differences are defined as Δx = x2 − x1, Δt = t2 − t1, Δx' = x'2 − x'1, Δt' = t'2 − t'1 , we get △t’=γ(△t-(v△x/c^2))， △x’=γ(△x-v△t)， and △t=γ(△t’+(v△x’/c^2)), △x=γ(△x’+v△t’)， Suppose we have a clock at rest in the unprimed system S. Two consecutive ticks of this clock are then characterized by Δx = 0. If we want to know the relation between the times between these ticks as measured in both systems, we can use the first equation and find: △t’=γ△t (for events satisfying Δx = 0) This shows that the time Δt' between the two ticks as seen in the 'moving' frame S' is larger than the time Δt between these ticks as measured in the rest frame of the clock. This phenomenon is called time dilation. Jh17710 (talk) 05:09, 19 November 2010 (UTC)

lengthy disscussion

I am having a bit of deja vu. Oh, yeah, that's because this question has been asked, by you, before, and answered already. To save everyone the trouble of answering it again, here are the responses from the first time it was answered. If there is something you need clarified from those responses, please feel free to ask that. --Jayron32 05:19, 19 November 2010 (UTC) Jayron, Thanks for you link. There were 3 responses and only the 2nd one responded to my explanation, however, there is no name to that 2nd response. This time, I did not put my explanation and just ask question. If △t’=γ△t and △t'=△t/γ are both for the same moving frame S' and rest frame S, even thay are for different events, △t’ is for the moving clock and △t is for the rest clock. My question is so simple, if we call △t'=△t/γ "time dilation" or "moving clock runs slow" then, we should not call △t’=γ△t the same names, right? I will respond to your comment and the comment from RedAct when I have time to study it. What I hope is that someone else could help me to understand my simple question.Jh17710 (talk) 06:22, 19 November 2010 (UTC) Actually, since neither clock has a privileged frame of reference, when each observer views the other person's clock, he sees it as running slower than his own. There is no time speeding because no observer will ever view another clock as running faster than his own; every observer takes their own frame of reference to be at rest. This is covered at Time_dilation#Relative_velocity_time_dilation. --Jayron32 06:57, 19 November 2010 (UTC) You're still making the same conceptual error I pointed out last time, by considering there to be "the moving clock" and "the rest clock". Each observer uses a system of synchronized clocks that are at rest according to that observer, not just one clock. In the example in the article, the clock being examined is a clock in the unprimed system. That clock is "moving", from the perspective of the primed observer. For the two events in the example, △t' is measured with two different clocks in the primed system, so there isn't a clock in the primed system that appears to be running fast according to the unprimed observer. △t' appears "too large" to the unprimed observer, not because there's a clock in the primed system that appears to be running fast, but because the unprimed observer considers the primed system's clocks to not be synchronized properly. Red Act (talk) 12:28, 19 November 2010 (UTC) Δt is the elapsed time, not the rate of the clock. These are reciprocals of each other. (The elapsed time might be measured in seconds, the clock rate in Hertz.) Δt' = γΔt means that Δt' is larger (since γ > 1), so the clock is running slower, in some sense, in the primed frame. Does that help? -- BenRG (talk) 00:24, 20 November 2010 (UTC) Jayron32, you clearly explained "moving clock runs slow". You also mentioned about one of the key point "If ... each is in communication with the other, ..." and according to Red Act, because the speed of light is finite there having to be a delay for light to travel the distance between the event and the clock, I think the delay is also true for your way of communication. However, my question is for the difference of two "equations". Red Act, let's have two systems of clocks, not just two clocks with two observers. When you said "...from the perspective of the primed observer." did you mean the primed system is considered the rest system? In that situation Δt' = γΔt is Δt = Δt'/γ , that is the same as Δt' = Δt/γ when S' is considered moving. Then, in your analysis, △t' is measured with two different clocks in the primed system, S'; since all clocks in the S' are synchronized, △t' is the actual time period of two ticks measured by the system of clocks in the S', so that "a" clock in the system of clocks should have same speed as the system of clocks. Finally, your main point is that the unprimed observer considers the primed system's clocks to not be synchronized; I think that could be one way to explain the difference between two systems of clocks if you can explain it in detail. That is why the observer in S considers clocks in S are synchronized and clocks in S' are not synchronized. Of course there isn't really any such thing as a "rest system" or a "moving system"; all inertial frames of reference can equally validly be said to be "at rest". However, for the purposes of illustrating the principle that "moving clocks run slow", in this particular example it's the unprimed system that needs to be considered to be "moving", and the primed system that needs to be considered to be "at rest". That's because the clock in question isn't moving (is at rest) according to the unprimed system, but it is moving according to the primed system. Unfortunately, the article referred to S' as the "moving" system, which is the opposite of the perspective needed to make the article's analysis mesh with the phrase "moving clocks run slow". I just this morning dropped the word "moving" from the article, because "moving" is basically meaningless when talking about a reference frame, and because it suggests adopting the opposite perspective of what's more appropriate for thinking about time dilation and length contraction. I also changed the article to point out that it's a clock in the unprimed frame that's running slow, as perceived from the primed frame (as opposed to a clock in the primed frame that's running fast). You said "Δt' = γΔt is Δt = Δt'/γ , that is the same as Δt' = Δt/γ when S' is considered moving". However, it's important to be mindful that Δt' = Δt/γ applies to a different set of circumstances than Δt = Δt'/γ does. It’s not just a matter of a choice as to which system is considered to be moving; the two equations apply to different pairs of events. (In the following, I'm presuming no motion in the y or z directions.) Δt = Δt'/γ only applies to pairs of events such that Δx=0, which is the case in the article's example. Δt' = Δt/γ would only apply to a pair of events such that Δx'=0, which is not the case in the article's example. The only way Δx=0 and Δx'=0 would both be true for a pair of events would be if S and S' were at rest with respect to each other, which gives the trivial case of γ=1, in which case there’s no contradiction between Δt = Δt'/γ and Δt' = Δt/γ. It isn't possible to explain the differences between the systems of clocks purely in terms of synchronization issues (which is formally known as relativity of simultaneity). Synchronization issues and time dilation both exist, and are distinct. Consider the equation from the article $\Delta t'=\gamma \left(\Delta t-{\frac {v\,\Delta x}{c^{2}}}\right)$ . Basically the phrase "time dilation" is referring to the γ factor not being 1, whereas the phrase "relativity of simultaneity" is referring to the $-{\frac {v\,\Delta x}{c^{2}}}$ term not being zero. You are correct in your observation that all clocks in the primed system run at the same speed. Both observers will agree that all clocks in the primed system all run at the same speed (although they will disagree about how fast it is that they are running). However, it still matters which specific clocks in the primed system you use, because according to the unprimed observer, the primed clocks do not all show the same time. "Showing the same time" means to look at all of the clocks simultaneously, and note that at that instant, the clocks have the same value. For example, if you look at all the clocks simultaneously, and they all say "noon", then the clocks all "show the same time". The problem is that the two observers do not agree on what "simultaneous" is. For more information, see relativity of simultaneity. Red Act (talk) 13:13, 20 November 2010 (UTC) $-{\frac {v\,\Delta x}{c^{2}}}$ is always 0 because the v is always 0 in Lorentz Transformation (LT). Why? If you apply the time equation to the spatial equation, you will get the spatial equation of inverse LT, x=γ(x'+vt'). That means, both of LT and inverse LT are coexist at the same time. For v and -v to coexist at the same time the speed v must be 0. Basically, there is a time equation missing in LT, that is t'=γt. Why? Let B(b,0,0) be a point in S. When b<0, let observers in S record the time that the origin point O' of S' moves from B to O by Δt = (0-t) and the distance between B and O by BO, we have BO = |v|Δt; let observers in S' measure the time O' moves from B to O by Δt' = (0-t') then we can calculate the distance between B and O as measured in S' by (BO)' = |-v|Δt'; based on hypothesis of ruler contraction, we have (BO)' = γBO so that |-v|Δt' = γ|v|Δt, then, △t’=γ△t. (0-t')=γ(0-t) so that t’=γt, that means when O' is at B, the time relationship between S' and S is t’=γt. By quite the same procedure, when O' is at B while b>0, t’=γt is also true. When b=0, t'=t=0, so that t’=γt. We have now derived that for any location of O', t’=γt is always true. I think the LT can be written as t’=γt, x'=γ(x-vt), y'=y, and z'=z.Jh17710 (talk) 23:20, 20 November 2010 (UTC) You are making the mistake of using (BO)' to mean two different things, and then assuming that the two values of (BO)' are equal. For the equation (BO)'=|-v|Δt' to be valid, (BO)' must be the distance as measured in the primed system between two events such that one event is at B, one event is at O, and the two events are simultaneous as measure in the primed system. But for the equation (BO)' =γBO to be valid, (BO)' must be the distance as measured in the primed system between two events such that one event is at B, one event is at O, and the two events are simultaneous as measure in the unprimed system. You wound up reaching the conclusion that v=0 because it is only when v=0 that those two different meanings of (BO)' have the same value. The problem again boils down to the fact that different observers do not agree on what "simultaneous" is. Please see relativity of simultaneity. Red Act (talk) 04:41, 21 November 2010 (UTC) Thanks for your response. I can understand that to measure a moving ruler, we need to snapshot both ends, so that simultaneity is required. However, for a given point B, when B is far away from O, we don't have long ruler in S' to snapshot the length of the moving BO. I think we can use some other way to get it like using the speed |-v| and the measured time period to calculate it. Could you tell me what is wrong to do so? Then, I think you understand that to me, the second (BO)' is also not a measured result. For (BO)' = γBO to be valid, I rely it on the "hypothesis of ruler contraction" but you said that the 2nd {(BO)' must be the distance as measured in the primed system between two events such that one event is at B, one event is at O, and the two events are simultaneous as measure in the unprimed system.}, do you mean, I cannot link the measured BO in S and the calculated (BO)' in S' for fixed points B and O by the "hypothesis of ruler contraction"? Why?Jh17710 (talk) 07:04, 21 November 2010 (UTC) Two (BO)' Yes, the distance between the moving points B and O as measured in S' is (BO)'=vΔt', where Δt' is the time elapsed on a clock stationed at O' as it passes from B to O. If you take BO to be the distance between B and O as measured in S, then the correct length contraction equation for measuring the distance between B and O as measured in S' is (BO)'=BO/γ. It does work to equate the two (BO)' if you define the two (BO)' as in this post. With the corrected length contraction equation, both versions of (BO)' are the distance as measured in S' between two events such that one event is at B, one event is at O, and the two events are simultaneous as measured in S'. Red Act (talk) 16:02, 21 November 2010 (UTC) Fine, just go by your (BO)'=(BO/γ), so that we have |-v|Δt' = |v|Δt/γ in LT, that means, the time equation in LT could be derived from Δt' = Δt/γ, and we have t'=t/γ follow my reasoning for all posible location of O'. That result still show that the inverse LT could have time equation of t=t'/γ; and the fact that the time equation in LT caused the inverse LT to coexist with LT can put both of t'=t/γ and t=t'/γ to coexist at the same time, and we still have the result that γ=1 and v=0 must be true in LT; mathematically.Jh17710 (talk) 17:29, 25 November 2010 (UTC) Δt' = Δt/γ is a correct equation if the Δt' and Δt are between two events which occur at the same location according to S' . Δt = Δt'/γ is a correct equation if the Δt' and Δt are between two events which occur at the same location according to S. If you use both equations, you are using Δt to mean two different things, and using Δt' to mean two different things. Red Act (talk) 08:00, 27 November 2010 (UTC) Red Act, this Δt' = Δt/γ means (0-t')=(0-t)/γ in our discussion, after you insisted (BO)'=(BO/γ). Follow (0-t')=(0-t)/γ, we have t'=t/γ, and the main point here is that, besides t'=γ(t-(vx/c^2)) LT also includes the time equation t'=t/γ. I did not put Δt' = Δt/γ and Δt = Δt'/γ together. I just added t'=t/γ into LT; and for the same reasoning, we have t=t'/γ in inverse LT. Then what happened is that within LT we find inverse LT, so that it is LT which puts t'=t/γ and t=t'/γ together. Now, you say that we could not use Δt to mean two different things, you should tell Lorentz that.Jh17710 (talk) 04:19, 28 November 2010 (UTC) The LT does not include the equation t'=t/γ. The equation t'=t/γ is incorrect, except for an event such that x=0, or in the case that v=0. Similarly, the equation t=t'/γ is incorrect, except for an event such that x'=0, or in the case that v=0. Red Act (talk) 13:41, 28 November 2010 (UTC) Please refer to your comment dated 16:02, 21 November 2010, you said (BO)'=vΔt' is fine and let BO to be the distance between B and O as measured in S, then (BO)'=BO/γ. Since we have BO=vΔt, follow above two (BO)'s, we should have vΔt'=vΔt/γ, then t'=t/γ, for all location of O' in LT. Isn't it? I don't understand the last paragraph in that comment: "With the corrected length contraction equation, both versions of (BO)' are the distance as measured in S' between two events such that one event is at B, one event is at O, and the two events are simultaneous as measured in S'." because the second (BO)' is based on the BO in S and the length contraction, it is not related to the measurement in S', isn't it?Jh17710 (talk) 16:20, 28 November 2010 (UTC) Δt' BenRG, thanks a lot. Your response is the one touched my question. Could you review the mathematics in "Δt' is larger (since γ > 1), so the clock is running slower,.." once again? Normally, Δt'=t'2-t'1, so that, when Δt' is larger, the clock should run faster so that we can count more units between time t'1 and t'2, isn't it? For the same event, Δt' = γΔt means the event is recorded as Δt' of S' units in S' and Δt of S units in S. If Δt' is larger, the clock in S' is quicker. That is why I said, the author of that section should use the clock in S' as the reference clock, same as what Einstein did in year 1905.Jh17710 (talk) 05:43, 20 November 2010 (UTC) Δt and Δt' are the time interval between the same two events, just measured with respect to different frames. They are talking about the same clock, not two different clocks. E.g., suppose you're stationary with respect to S, moving uniformly with respect to S'. You note down the times shown on nearby clocks of S and S' (event 1), then time 60 seconds with your stopwatch, then note down the times now shown on the clocks (event 2). Subtracting the unprimed and primed readings gives you Δt and Δt'. "Time dilation" refers to the fact that Δt' > 60 seconds. (Δt = 60 seconds.) The "dilated" clock is your stopwatch, not one of the clocks used to define the systems—although, since you're stationary with respect to S, you could use the nearest clock belonging to S as your stopwatch instead. The only way to measure the "speed of a clock" is using another clock, so could you say instead that the clocks of S' are running fast? Not exactly, because the two readings you subtracted to get Δt' were taken from two different clocks (you always take the reading from the clock that's nearest at the time of the reading). Thus the difference depends not only on the "speed of the clocks" but also on how they were synchronized. You can't exactly say that any clock is going fast, because which clock would that be? You never looked at any of them more than once. -- BenRG (talk) 08:11, 20 November 2010 (UTC) But according to Red Act, all of clocks in one system are running at the same speed, and I think we should assume that to make things easier. I like to hear more detail about your comment "..but also on how they were synchronized.", could you show me some ways they can be synchronized? Thanks.Jh17710 (talk) 23:20, 20 November 2010 (UTC) Again, how fast the clocks are running doesn't tell the whole story, as can be trivially illustrated with even a very low speed example. Assume for the purposes of illustration that all timing devices run at the same speed. However, assume that your kitchen wall clock is set to an hour later than your living room wall clock. Leave your living room when your wristwatch and the living room wall clock say 1:00. Walk to the kitchen, and notice that your wristwatch now says 1:01, but the kitchen wall clock says 2:01. If you incorrectly consider all wall clocks to be equivalent when measuring "wall clock elapsed time", then you would incorrectly conclude that wall clocks run 61 times faster than your wristwatch. The solution to avoid that incorrect conclusion is to instead say that wall clocks don't all show the same time (even though they all run at the same speed!), and be careful to specify which wall clock is being used when using a wall clock's time. You mean the kitchen wall clock says 2:02, because the moving wristwatch runs slow.Jh17710 (talk) 02:13, 21 November 2010 (UTC) Nope, I mean the kitchen wall clock says 2:01. I said "assume for the purposes of illustration that all timing devices run at the same speed", and that assumption also applies to the wristwatch. Red Act (talk) 02:59, 21 November 2010 (UTC) Then, "...wall clocks run 61 times faster than your wristwatch." should be 60 times faster than or as fast as 61 times of your wristwatch, right?Jh17710 (talk) 07:04, 21 November 2010 (UTC) A similar problem occurs in the relativistic situation. If the unprimed observer looks at all clocks in the primed reference frame at time t=0, the primed clocks will not be observed to all show the same time. Instead, at time t=0, the time shown on the primed clocks will show different times depending on the location of the clock, namely, t' = -γvx/c2. So you must be careful to consider which primed clock is being used when using a primed clock’s time. Since you mentioned about the term -γvx/c2 again here, could you comment to my reasoning of "v=0 is always true in LT"? After Lorentz assumed ruler contraction, the LT must be t'=γt, x'=γ(x-vt), y'=y, and z'=z; while the inverse LT must be t=γt', x=γ(x'+vt'), y=y', and z=z'. We can still play the same trick Lorentz did to combine spatial equation and inverse spatial equation so that we have t'=γ(t-(vx/c2)). However, that will mean we should combine t'=γt and t=γt' as well, and that will be γ=1 (since for c>|v|>0, γ>1); and that also means v=0. LT is almost a joke.Jh17710 (talk) 02:13, 21 November 2010 (UTC) See above for an explanation of why the derivation of v=0 was incorrect. Red Act (talk) 04:41, 21 November 2010 (UTC) Thanks a lot. I still have two questions asked above regarding your explanation. Could you help to provide more detail?Jh17710 (talk) 15:56, 21 November 2010 (UTC) See above. Red Act (talk) 23:05, 21 November 2010 (UTC) In the low speed example, you can avoid the problem by simply setting all wall clocks to have the same time. But that's not possible in the relativistic situation. Clocks at rest in the primed system will all have the same time according to the primed observer, but will not all have the same time according to the unprimed observer. For information on how clocks are theoretically synchronized in relativistic situations, see Einstein synchronization. Red Act (talk) 01:15, 21 November 2010 (UTC) Thanks, I will study it. Hope that is different from the one in his 1905-6-30 paper, because that one is not a complete one, it missed one condition that the distance between the emitting point and reflecting point is the same as the distance between the reflecting point and the ending point. Because of that mistake, Einstein did not prove LT in his paper. The equation in section 3, (t0+t2)/2 = t1 is actually an inequality (t0+t2)/2 < t1 so that follow the correct inequality, Einstein was unable to prove LT. His definition of synchronizatiion is fine if there is only one system so that we can assume that system is rest, but, there are two systems, so that we must add above condition to ensure two clocks are synchronized.Jh17710 (talk) 02:13, 21 November 2010 (UTC) You're absolutely right to say that in general, the (τ0+ τ2)/2= τ1 equation is incompatible with a system of synchronized clocks, with the only exception being that it will happen to work as long as the system is at rest. However, the "flaw" that you’re noticing isn't actually a bug, but a feature. Pick an arbitrary inertial system S. From the perspective of S, S is at rest, and all inertial systems are created equal, so when system S applies the Einstein synchronization procedure to its clocks, it will set the clocks to something that appears to be properly "synchronized" according to S. However, those clocks will not appear to be properly synchronized according to anybody else. Einstein did not "miss" the condition that the distance between the emitting point and reflecting point is the same as the distance between the reflecting point and the ending point. Indeed, that condition is not in general true. When system S is synchronizing its clocks, those two distances will be the same according to how S measures distances. But those distances will not be the same according to any other system. Red Act (talk) 23:05, 21 November 2010 (UTC) When we talk about a point, it is a fixed point in the universe. Let me tell you how to define the emitting point based on a book published recently in China. We just need some imagination, let a source of light shine once at point E, then the out going wave front will make a growing but thin out light ball. Do you agree that the center of that growing light ball is a fixed point as in the absolute rest universe? The reason is clear, because the speed of light is independent of the speed of the source of light so that the growing light ball is growing from a fixed point in the whole space, the point E. It is the same for the point R and F, for every event point. The only difficulty is that, unless the source of light stays at that same point while the light ball is growing, otherwise, we have no way to point out where is E, because our technology is unable to show us the actual growing light ball yet. Since now we have the points fixed, we could have the distance ER and RF fixed so that Einstein missed one condition for his two systems.Jh17710 (talk) 17:29, 25 November 2010 (UTC) It's incorrect to say "the center of that growing light ball is a fixed point as in the absolute rest universe". The phrase "the center of that growing light ball" has no meaning, without specifying which reference frame you are using to determine the "center" with. The "center" would be a point equidistant from a set of events on the growing light ball, where the events all occur at the same time. However, different inertial frames of reference disagree on what "occur at the same time" means (see relativity of simultaneity, again), which makes them disagree on what the "center" is. Red Act (talk) 06:33, 27 November 2010 (UTC) Red Act, if I hold a ball could I say the center point can be decided by the ball without mentioning about other reference frame? Yes, the ball itself is a reference frame. If we can see the expanding ball of light then that will solve our problem here. But, since it is there, even if we can not see it, it is still there, isn't it? Then we check the second issue about the "center". You had made the center point of a ball too complicated. If there is a ball, we can mark four points on the ball with the condition that they are not on one plane, then, draw perpendiculer planes at all middle points of six line segments and the intersection point is the center. The problem with this expanding ball of light is that we have no way to locate it. But it is real and since there must be one and only one center of it, we cannot deny its existence just because we are unable to locate it. Make sense? Jh17710 (talk) 04:56, 28 November 2010 (UTC) The ball of light topic is currently being discussed on the current "Absolute Rest Point" topic, so I won't duplicate the whole topic by also discussing it here. Red Act (talk) 13:41, 28 November 2010 (UTC) A fresh start with more precise notation This is really a continuation of the subsection called "Two (BO)'", but with more explicit notation. Up until now, I’ve been trying to stick to your notation as much as possible. However, there are two kinds of imprecision in what you're saying, that not only make it hard for me to follow you at times, but also lead to you making erroneous conclusions. The first problem is using a symbol that pertains to an event or pair of events, without being precisely clear about which event or pair of events the symbol pertains to. The second problem is using equations that are valid for only a subset of events or pairs of events, without being precisely clear as to what subset of events or pairs of event the equations are claimed to be valid for. The solution to the first problem is to from here forward always number all specific events involved, and use subscripts with those event numbers to indicate all coordinates or differences in coordinates. The solution to the second problem is to from here forward always explicitly state what set of events or pairs of events an equation is claimed to be valid for. I will label events Ep, where p is some integer. I.e., individual events in this problem will be called E0, E1, E2, etc. The coordinates for event Ep are tp and xp in the unprimed system, and t'p and x'p in the primed system. Given events Ep and Eq, I define Δxpq=xq-xp, Δtpq=tq-tp, Δx'pq=x'q-x'p and Δt'pq=t'q-t'p. The rigorous versions of all equations used above that I acknowledge to be valid are: Eq. 1: t'p=γ(tp-vxp/c2), which is valid for all events Ep. Eq. 2: x'p=γ(xp-vtp), which is valid for all events Ep. Eq. 3: tp=γ(t'p+vx'p/c2), which is valid for all events Ep. Eq. 4: xp=γ(x'p+vt'p), which is valid for all events Ep. Eq. 5: Δt'pq=γ(Δtpq-vΔxpq/c2), which is valid for all pairs of events Ep and Eq. Eq. 6: Δx'pq=γ(Δxpq-vΔtpq), which is valid for all pairs of events Ep and Eq. Eq. 7: Δtpq=γ(Δt'pq+vΔx'pq/c2), which is valid for all pairs of events Ep and Eq. Eq. 8: Δxpq=γ(Δx'pq+vΔt'pq), which is valid for all pairs of events Ep and Eq. Eq. 9: Δt'pq=γΔtpq, which is valid for all pairs of events Ep and Eq satisfying xp=xq. Eq. 10: Δx'pq=γΔxpq, which is valid for all pairs of events Ep and Eq satisfying tp=tq. Eq. 11: Δtpq=γΔt'pq, which is valid for all pairs of events Ep and Eq satisfying x'p=x'q. Eq. 12: Δxpq=γΔx'pq, which is valid for all pairs of events Ep and Eq satisfying t'p=t'q. Eq. 13: t'p=γtp, which is valid for all events Ep satisfying xp=0. Eq. 14: x'p=γxp, which is valid for all events Ep satisfying tp=0. Eq. 15: tp=γt'p, which is valid for all events Ep satisfying x'p=0. Eq. 16: xp=γx'p, which is valid for all events Ep satisfying t'p=0. Eq. 17: Δxpq=vΔtpq, which is valid for all pairs of events Ep and Eq satisfying x'p=x'q. Eq. 18: Δx'pq=-vΔt'pq, which is valid for all pairs of events Ep and Eq satisfying xp=xq. I’ll label the events of interest that have been explicitly or implicitly used above as: E0: O and O' at the instant they coincide, i.e., t0=0, t'0=0, x0=0, x'0=0. E1: The point B at time t=0, i.e., t1=0, x1=-d, where I'm introducing the symbol d to mean what had been called BO. We can use Eq. 1 to calculate t'1=γvd/c2. And because t1=0, we can use Eq. 14 to calculate x'1=-γd. This is also a good opportunity to illustrate a length contraction equation being used: Because t0=t1, we can use Eq. 10 to calculate Δx'10=γΔx10=γ(0 -(-d))=γd, which is consistent with calculating Δx'10 as Δx'10=x'0-x'1=0-(-γd)=γd. E2: The points B and O' at the time when they coincide. Being at O' gives x'2=0. And being at point B gives x2=-d. And because x'0=0 also, we have the x'2= x'0 criteria needed to be allowed to use Eq. 17 with the events E2 and E0. Doing so gives Δx20=vΔt20. Substituting in the definitions for Δxpq and Δtpq gives x0-x2=v(t0-t2). Substituting in the known values x0=0, t0=0 and x2=-d and solving for t2 gives t2=-d/v. If there are any other important events you want to talk about, assign them a number, define them, and derive coordinate values for them using the above equations, being explicit about why the validity conditions for the equations being used are met. Or if there are any other equations you want to talk about, state what the equation is, state the conditions under which you are claiming the equation is valid, and derive the equation from some of the 18 equations above, being careful to use an equation only if the validity conditions for the equation are met. Based on what you've said so far, it looks like your attempts to find an inconsistency in the Lorentz transformations are or will be along one of the following four similar approaches: Approach 1: Try to argue that it must be that γ=1 (i.e., v=0) is required in order for equations 13 and 15 to both hold. However, the validity conditions for equations 13 and 15 combined mean that the set of events for which a derivation using both equations is valid are only those events Ep such that both xp=0 and x'p=0. Assuming that v≠0, there's exactly one event that satisfies both of those conditions, namely E0. As proof, substitute xp=0 and x'p=0 into equation 2 to get tp=0. And substitute xp=0 and x'p=0 into equation 4 to get t'p=0. So when both equations 13 and 15 hold, the reason both equations hold isn't because γ=1, but because under the combined validity criteria, tp=0 and t'p=0. Approach 2: Try to argue that it must be that γ=1 (i.e., v=0) in order for equations 14 and 16 to both hold. However, the validity conditions for equations 14 and 16 combined mean that the set of events for which a derivation using both equations is valid are only those events Ep such that both tp=0 and t'p=0. Assuming that v≠0, there's exactly one event that satisfies both of those conditions, namely E0. As proof, substitute tp=0 and t'p=0 into equation 1 to get xp=0. And substitute tp=0 and t'p=0 into equation 3 to get x'p=0. So when both equations 14 and 16 hold, the reason both equations hold isn't because γ=1, but because under the combined validity criteria, xp=0 and x'p=0. Approach 3: Try to argue that it must be that γ=1 (i.e., v=0) in order for equations 9 and 11 to both hold. However, the validity conditions for equations 9 and 11 combined mean that the set of pairs of events for which a derivation using both equations is valid are only those pairs of events Ep and Eq such that both xp= xq and x'p= x'q. Assuming that v≠0, the only way for both of those conditions to be met is if Ep=Eq. As proof, substitute Δxpq=0 and Δx'pq=0 into equation 6 to get Δtpq=0. And substitute Δxpq=0 and Δx'pq=0 into equation 8 to get Δt'pq=0. So when both equations 9 and 11 hold, the reason both equations hold isn't because γ=1, but because under the combined validity criteria, Δtpq=0 and Δt'pq=0. Approach 4: Try to argue that it must be that γ=1 (i.e., v=0) in order for equations 10 and 12 to both hold. However, the validity conditions for equations 10 and 12 combined mean that the set of pairs of events for which a derivation using both equations is valid are only those pairs of events Ep and Eq such that both tp= tq and t'p= t'q. Assuming that v≠0, the only way for both of those conditions to be met is if Ep=Eq. As proof, substitute Δtpq=0 and Δt'pq=0 into equation 5 to get Δxpq=0. And substitute Δtpq=0 and Δt'pq=0 into equation 7 to get Δx'pq=0. So when both equations 10 and 12 hold, the reason both equations hold isn't because γ=1, but because under the combined validity criteria, Δxpq=0 and Δx'pq=0. Red Act (talk) 00:34, 30 November 2010 (UTC) Thanks for your analysis. It clearly listed most possible validity conditions and the results in LT and inverse LT. However, what I tried to do is to establish the time equation based on Galilean Transformation (GT) and "hypothesis of ruler contraction" so that when I started, there is no LT yet. Actually, the time equation I mentioned about is a required condition for the spatial equation of LT to be valid. I will provide more detail next time.Jh17710 (talk) 05:13, 1 December 2010 (UTC) The Galilean transformation? In the GT, the time equation is simply t=t', end of story. The GT and the LT are incompatible, except in the trivial case of v=0. You can't derive anything involving the LT in a consistent way, if your starting point involves assuming the GT is valid. To do anything in special relativity, you have to discard the GT as being inaccurate, except as a useful low-speed approximation. Red Act (talk) 07:28, 1 December 2010 (UTC) LT is a system modified from GT by assuming ruler contraction. However, he simply took the spatial equation x'=x-vt and assigned γ to the S portion (x-vt) in that equation so that x'=γ(x-vt) became the spatial equation of LT. Actually, that equation is not so simple. We will be able to see it easier if we limit all events in one dimensional space. Then, for t' and t < 0, we have (x'-OO')= x and when t' and t > 0, we have (x'+OO')= x. Since OO'=v|t'| as measured in the moving S', so that (x'+vt')= x for all t'. The actual ruler contraction should be (x'+vt')= γx, that means x' = γx-vt'. It is because of the time equation t'=γt actually exist when we modify GT by assuming ruler contraction, so that the x'=γx-vt' can be written as x'=γx-vγt=γ(x-vt) as well. I will derive the time equation t'=γt in as much detail like you wished above. That equation is for all events under the "ruler contraction version of GT". Not only for the limited condition of x=0 under LT.Jh17710 (talk) 05:05, 2 December 2010 (UTC) You say that "LT is a system modified from GT by ruler contraction". However, when you try to derive LT by applying length contraction to GT, you wind up getting t'=γt for all events, which is not LT, because it's inconsistent with LT's Eq. 1 above when x≠0. So when you play around with applying length contraction to GT, you get something different from LT. So according to your own equations, LT is not just a system modified from GT by ruler contraction. Red Act (talk) 06:08, 2 December 2010 (UTC) Thanks for your response. Let us go one sentence by one sentence. The first one is what I said. The second one, I am still working on deriving the t'=γt in more detail as you required, event after event; I showed the first step to derive LT as {However, he simply took the spatial equation x'=x-vt and assigned γ to the S portion (x-vt) in that equation so that x'=γ(x-vt) became the spatial equation of LT.} I did not use t'=γt. The second step to derive LT is just combine x'=γ(x-vt) in LT and x=γ(x'+vt') in inverse LT, then we get t'=γ(t-(xv/c^2)) and the derivation of LT by applying ruler contraction to GT is done. The third one, you are corerct. Based on the same starting condition as LT started, I can derive another result. The fourth one, according to my own equation, based on GT and ruler contraction, we can derive LT and other equation; logically speaking, LT is still just a system modified from GT by ruler contraction. If based on E and F we can derive G and H, G is still just based on E and F, isn't it? About the last sentence, could you tell me anything wrong in my two steps of deriving LT? If nothing wrong, then, LT is just a system modified from GT by ruler contraction. Based on LT, if people find something else, that is another issue, isn't it?Jh17710 (talk) 03:50, 3 December 2010 (UTC) OK, the statement that "LT is a system modified from GT by ruler contraction" can sort of be viewed as basically being correct, although that statement doesn't have a precise mathematical meaning, if you take x'=γ(x-vt) as expressing what you mean by ruler contraction. However, for clarity, that's quite different from saying that LT can be derived from GT and ruler contraction. A derivation is where you assume that some set of things are true, and logically deduce from those that something else must also be true. Importantly, the original set of things that are assumed to be true at the beginning of the derivation remain assumed to be true at the end of the derivation (I won't get into proof by contradiction here). That's different from what's going on above in arriving at the LT. In the above, what's happening is basically saying "discard all the GT equations as being invalid, and instead assume that x'=γ(x-vt), which happens to look similar to the x'=x-vt equation in the GT." In other words, the GT isn't really being used at all in deriving the LT. Instead, the initial assumptions being used to derive the LT are that x'=γ(x-vt), and that all inertial systems are equally valid (which lets you conclude that x=γ(x'+vt')). So there should be no expectation that the LT would be consistent with anything derived from the GT, or be consistent with anything that's a part of or derived from a different "modified version of the GT". In particular, note that the t'=γ(t-(xv/c2)) of the LT as derived above is inconsistent with the GT's equation t'=t (except in the trivial case of v=0), so the LT requires that the time portion of the GT must be discarded as well, in addition to discarding the spatial portion. Red Act (talk) 22:37, 3 December 2010 (UTC) You have a very good view point of the word "derivation". I should use modify to make things clear, LT is derived from modifying GT by ruler contraction. Does it look better now? THANKS FOR YOUR CORRECTION OF MY MISTAKE. I think LT uses everything of GT including when v=0, t'=t. Yes, you could start LT from x'=γ(x-vt) and x=γ(x'+vt') then combine them for t'=γ(t-(xv/c^2)), however, when you try to compare it with GT, ruler contraction will pop out and we are in the same situation. If you don't compare LT with GT, someone will, isn't it?Jh17710 (talk) 04:20, 4 December 2010 (UTC) I don't know what you mean by "LT uses everything of GT". LT disagrees with GT for almost all events and almost all speeds, with the only exceptions being the trivial case of v=0, or for the event for which x=0 and t=0. Red Act (talk) 16:14, 4 December 2010 (UTC) Besides the most different one, as you mentioned above, the time equation can fit in when v=0; LT used exactly the same arrangement of GT. And the main uasge is that LT modified the spatial equation in GT directly and inappropriately, but LT is lucky that x'=γ(x-vt) is still accurate after the careless modification as I said above on 12-2-10. I will derive t'=γt in detail later. The equation t'=γt backed up x'=γ(x-vt).Jh17710 (talk) 04:42, 5 December 2010 (UTC) Here's a very pertinent homework problem for you: There are three conceptual properties of the LT, that make the LT different from the GT. One is length contraction. One is time dilation. What is the third? Red Act (talk) 07:20, 2 December 2010 (UTC) I think there is no FIXED time dilation in LT. As in your equation 13 and 15 {derived from x'=γ(x-vt) and t'=γ(t-(xv/c^2)) in LT, don't need any equation in the inverse LT} the time in S' and S can have dilation or speeding depend on their observers are observing what kind of events. If they are observing events rest in S then the relation is 13, if they are observing events rest in S' then the relation is 15 under the same situation that S is the rest syatem and S' moves.Jh17710 (talk) 03:50, 3 December 2010 (UTC) This paragraph doesn't make sense. An event is one specific location at one specific time. Since events only have one time, it's meaningless to talk about an event being "at rest", or "moving", since both of those concepts require a discussion of more than one time, and hence, more than one event. An object is "moving" relative to some reference frame if it is at two different locations as measured in that reference frame at two different times. I.e., if the world line of an object includes two distinct events Ep and Eq such that xp≠xq, then the object is "moving" as measured in system S. Something is "at rest" relative to some reference frame if it is at the same location as measured in that reference frame at all times. I.e., if for all distinct pairs of events Ep and Eq on an object's world line, xp=xq, then the object is "at rest" as measured in system S. Red Act (talk) 22:37, 3 December 2010 (UTC) You are right, however, we can say that a baseball moves from A to B is an event, with a starting event time and ending event time without confusing. But, it is always good to be more precise, let me use Series Of Events (SOE), then there are rest SOE and moving SOE. Actually, in the real world, we are unable to observe an individual event yet. If a rest SOE happens in a very short time period, then we will treat it as an individual event, so that the key issue is if a SOE move or not. When you said "two events", those two events are actually two ends of a SOE, isn't it? Thanks.Jh17710 (talk) 04:20, 4 December 2010 (UTC) The standard term for a continuous time-like series of events, such as might be used to describe the path of the center of a baseball through spacetime, is world line. Note that an event is not unique to one world line. Consider a basketball which goes down through the precise center of a basketball basket's rim at time t=0 in the basket's frame of reference. The center of the basketball traverses a world line that moves, according to the basket's frame of reference. The center of the rim traverses a world line that is at rest, according to the basket's frame of reference. At time t=0, the two world lines intersect at the event whose time is at t=0, and whose location can be described both as being at the center of the basketball, and as being at the center of the rim. It's meaningless to say whether that event is moving or is at rest, since that event is just as much a part of one world line as it is of the other. Red Act (talk) 16:14, 4 December 2010 (UTC) I already changed the name to SOE and said moving SOE and rest SOE. Actually, it still make sense if we call an event in a moving SOE a moving event and an event in a rest SOE as rest event. When two events meet like in your case, there are still two events at same location. Before and after t=0, the SOE of the center of basket's rim is rest and the SOE of the center of basketball is moving so that I think it is two events happen at the same time at the same location, not just one event. However, this is a better practice to use SOE for rest or moving.Jh17710 (talk) 06:43, 5 December 2010 (UTC) The time dilation is claimed by SR, not LT. In LT, we do have time dilation when x'=0; but, when x=0, t' is not running slow any more. In SR, Einstein claimed that, no matter x'=0 or x=0, or any other situation, so long as S' is moving at constant speed v relative to S, then t'=t/γ is always true for any event. Am I wrong?Jh17710 (talk) 03:50, 3 December 2010 (UTC) Sentence 1 is at least inconsistent with Einstein's interpretation of the meaning of the LT equations. For sentence 2, the situation is symmetrical. A clock in S' which has x'=0 at all times will run slow as measured by S. A clock in S which has x=0 at all times will run slow as measured by S'. Sentence 3 is just wrong; Einstein never made any such claim. What he did claim was equation 1. Red Act (talk) 22:37, 3 December 2010 (UTC) Sentence 1, Lorentz did not claim time dilation, Einstein calimed it for LT to specify his SR in his 1905-6-30 paper. Sentence 3 is based on the section 4 of Einstein's paper, same paper published 1905-6-30, he said t'=t/γ is always true even if v changed from constant velocity to constant speed. That means, we don't have to care about the direction of v, to v or -v, t'=t/γ is alway valid.Jh17710 (talk) 04:20, 4 December 2010 (UTC) Section 4 of Einstein's 1905 paper says "Further, we imagine one of the clocks which are qualified to mark the time t when at rest relatively to the stationary system, and the time τ when at rest relatively to the moving system, to be located at the origin of the co-ordinates of k, and so adjusted that it marks the time τ" (emphasis mine). If the clock discussed in that section is at the origin (of the reference frame that we're calling S), then it is only used to measure the time of events such that x=0, which is exactly the validity criterion specified above for equation 13. Einstein does not claim that equation 13 is valid when x≠0. Red Act (talk) 16:14, 4 December 2010 (UTC) Thanks for providing the link. If you read further, Einstein stated "It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide." That is what I mentioned about.Jh17710 (talk) 06:52, 5 December 2010 (UTC) The clocks at A and B are only brought up after using (in our notation) t'=t/γ to describe the moving clock that passes through the common origin. The clock that starts at A is a different clock at a different location than the previously discussed clock that passes through the common origin at t=0. Red Act (talk) 06:09, 6 December 2010 (UTC) Since we cannot change Einstein's paper any more, we could work out what Einstein meant. On your response dated 12-4-10, the clock is at the origin of S' so that x'=0, not x=0. Equation 15 is valid in LT if x'=0. When Einstein brought up the later issue regarding "any polygonal line" that clock A is the clock with x'=0, except this time, Einstein let v change from constant velocity to constant speed. You can say the clock that starts at A is a different clock at a different location than the previously discussed clock, but, Einstein applied same equation t'=t/γ to that scenario.Jh17710 (talk) 17:15, 6 December 2010 (UTC) Yes, I made a mistake in my post of 12/4/10. Instead of saying S, x=0 and equation 13, I should have said S', x'=0 and equation 15. However, my point still stands that Einstein does not use t'=t/γ (our equation 15) when discussing the clock traveling from A to B. He does use the expression ½ tv2/c2, but in that expression, he defines t as "t being the time occupied in the journey from A to B". In other words, in that expression (and not before), he is using t to mean ΔtAB. What he is describing in that paragraph is equation 11, which can be expressed as Δt'AB=ΔtAB/γ. Equation 11 holds because x'A=x'B (even though x'A≠0 and x'B≠0). There is no implication for equation 15 in that paragraph. Red Act (talk) 18:45, 6 December 2010 (UTC) Right in the paragraph you selected above reasoning, Einstein already change the moving clock from constant velocity v to constant speed on polygonal line. That means, at that point, LT is out of the picture. No more equation 11. Another issue is about no t'=t/γ format in that paragraph. If you read the paragraph right before that paragraph, you will find out Einstein change t'=t/γ to t'=t-(1-(1/γ))t and use the difference amount,(1-(1/γ))t, to describe the difference of two clocks. That means, t'=t/γ is still in the picture.Jh17710 (talk) 05:49, 15 December 2010 (UTC) Lorentz transformations are by no means out of the picture in the paragraph where Einstein discusses A and B. A Lorentz transformation in general can be composed of an arbitrary rotation and a Lorentz boost in an arbitrary direction. It can also include a translation, if inhomogeneous Lorentz transformations are part of what's under consideration in a particular situation. It can even include space inversion and/or time reversal, if improper or non-orthochronous Lorentz transformations are under consideration. All of this is explained in the Lorentz transformation article. Rather than LTs being out of the picture in that paragraph, all that's happening is that Einstein is venturing a little bit beyond a standard configuration LT. The 18 numbered equations above were enumerated assuming that only a standard configuration LT was being dealt with, so most of them (all but 9, 11, 13 and 15) are invalid when considering a LT that involves a rotation. But equation 11 is valid for all orthochronous LTs, even inhomogeneous or improper ones, as long as x'p=x'q, y'p=y'q and z'p=z'q. In the paragraph where Einstein talks about moving a clock along a straight line from A to B, without putting any constraints on the coordinates of A and B, he basically is making the point that equation 11 holds (under the conditions x'p=x'q, y'p=y'q and z'p=z'q) even if the LT is not a standard configuration LT. The physical reason why equation 11 works for more general LTs is basically because two ideal clocks that are stationary with respect to each other will always run at the same speed, no matter where the clocks are positioned, or whether the clocks are oriented differently. Although equations 11 and 15 can both hold for LTs that involve a rotation, equation 15 does not hold as broadly as equation 11 does, because unlike equation 11, equation 15 is not valid for inhomogeneous LTs. LTs are still being used even in the case where the clock moves from A to B along a polygonal path. For every segment of the path, the clock uses a different coordinate system, so a different LT needs to be used for every segment. But equation 11 is valid along each segment of the path, so it winds up being valid for the whole path. Yes, Einstein is implicitly continuing to use the equations t'=t/γ and t'=t-(1-(1/γ))t after starting to talk about A and B. However, the meanings of the symbols t and t' (his τ) change at that point. With his original definition of terms, after he said that "…imagine one of the clocks…to be located at the origin of the co-ordinates of k…", t in those equations is taken to mean tp, and t' is taken to mean t'p, where the conversation is limited to a standard configuration LP, and limited to an event Ep such that x'p=0. After he changes his definitions with "…t being the time occupied in the journey from A to B…", he is still implicitly using the equations t'=t/γ and t'=t-(1-(1/γ))t, but t now means ΔtAB and t' now means Δt'AB, where events EA and EB are such that x'A=x'B, y'A=y'B and z'A=z'B. He is not at that point taking t'=t/γ to be valid with t meaning tp and t' meaning t'p, for any arbitrary event Ep, as you are attempting to do. Red Act (talk) 20:25, 18 December 2010 (UTC) Thanks a lot, Red Act. I feel very comfortable talking with you. A LT in general is just some theoretical reasoning. To make a useful result, let us focus on equation 11 when Δx'=0. The restriction of Δx'=0 means, the equation 11 is good only for events rest in S'. S' is the moving system as observed by observers in S. Now, in so many explanations of SR, how many of them put this restriction in their minds? As a matter of fact, in most physics textbooks, SR is true without any restriction; so long as Δt' is based on the time speed of the moving system and Δt is based on the time speed of the rest system, Δt'=Δt/γ is true for real, not just because of measurement effect. Am I wrong? That is not what I attempted to do, that is what already being done by most physicists. Now, let us think about the equation 11 once again, but this time from the angle of how do we pratically apply it. My question for you is how do observers in S' communicate with observers in S so that they can compare their measurements? What is your suggestion, practically? When physicists claimed that they verified SR or GR, how did they allow above communication to happen? Could you find just one sample, the best sample you know, and explain it briefly so that I can study it? Thanks again.Jh17710 (talk) 15:53, 19 December 2010 (UTC)

Wheatstone bridge

I'm supposed to find the change in voltage across the central voltmeter as Rx is moved away from equilibrium, assuming the first three resistors have a common resistance R.

However, I am obstructed by several conceptual problems.

Theoretically the voltmeter should have an infinite resistance, right? Yet it permits a finite current. Even for an ideal circuit? I have the relation ΔV / I_v = resistance of the voltmeter = ΔRx + R(I1/I_v - 2). John Riemann Soong (talk) 06:07, 19 November 2010 (UTC)

The current through an ideal voltmeter is zero. So an ideal voltmeter doesn't affect the circuit it's in at all. You basically analyze the voltages and currents in the circuit as if the voltmeter wasn't there. The voltage measured by the voltmeter is then the difference between the voltages of the two points in the circuit that the voltmeter is connected to. Red Act (talk) 12:51, 19 November 2010 (UTC)

The galvanometer used in a Wheatstone bridge is not an ideal voltmeter: it has a finite resistance. When the bridge is unbalanced, a current flows through the galvo and deflects it. You then adjust one of the resistors to balance the bridge so that the galvo does not deflect. Once the bridge is balanced, the resistance of the galvo doesn't matter since there is no voltage across it. --Heron (talk) 18:35, 19 November 2010 (UTC)

I'm presuming the current through the galvanometer is to be neglected in this problem, or else the galvanometer's resistance would have to have been specified as part of the problem. Red Act (talk) 19:33, 19 November 2010 (UTC)

Yes, I see what you mean. In which case we need to know where John gets his term I_v from, because that implies a non-ideal voltmeter. If I_v is non-zero then of course the equation in our article doesn't apply. --Heron (talk) 19:51, 19 November 2010 (UTC)

JRS gets a non-zero I_v because he doesn't understand the problem. The problem is completely soluble if I_v is taken to be zero, trivial even, the solution is given in the article. Physchim62 (talk) 20:00, 19 November 2010 (UTC)

From the article: "The direction of the current indicates whether R2 is too high or too low. Detecting zero current can be done to extremely high accuracy (see galvanometer)." 128.143.181.23 (talk) 20:41, 19 November 2010 (UTC)

The article is slightly confusing on this point. I've tried to clarify it. --Heron (talk) 11:52, 20 November 2010 (UTC)

：A galvanometer is not a voltmeter; it's an ammeter. An ideal ammeter has 0 resistance. --140.180.14.145 (talk) 08:01, 21 November 2010 (UTC)

Exactly. The problem with our article is that it confuses the two. It should make it clear that the original WB used a galvo to obtain zero current at balance, that you can also use a voltmeter to detect balance, and that if you have a high-impedance voltmeter then you can also do useful calculations with $V_{G}$ even when the bridge is unbalanced. --Heron (talk) 11:57, 21 November 2010 (UTC)

why isn't adenosine monophosphate used in alcoholic beverages?

I imagine that as an anti-bitterant it would mask the taste of alcohol really well -- or does it not? John Riemann Soong (talk) 10:14, 19 November 2010 (UTC)

Why would you want to mask the taste of alcohol? Some people actually enjoy the flavor of their drinks, indeed that is generally the idea. If you really wanted to, you could use a much less expensive alternatives, like fruit juices, simple syrup, colas, etc. A properly trained bartender can mix a drink to suit anyones taste without exotic chemicals. --Jayron32 16:16, 19 November 2010 (UTC)

Wikipedia:Reference desk/Archives/Science/2010 October 30#does citric acid actively mask the taste of ethanol? Nil Einne (talk) 18:52, 19 November 2010 (UTC)

I tweaked the wikilink to fix typo in anchor. Hope you don't mind the third-party mod, Nil! DMacks (talk) 19:13, 19 November 2010 (UTC)

Thanks! Nil Einne (talk) 06:15, 20 November 2010 (UTC)

I'm not sure I see the logic of using an anti-bitterant to mask the taste of something that most people don't describe as bitter and that is chemically unlike general bitter things or the more specific molecules noted as blocked by AMP. DMacks (talk) 19:08, 19 November 2010 (UTC)

Indeed. People are more likely to *add* bitter ingredients to drinks (e.g. tonic water of a gin and tonic, or even bitters themselves). -- 140.142.20.229 (talk) —Preceding undated comment added 19:50, 19 November 2010 (UTC).

The thing is, AMP is apparently a bitter reception antagonist. It's not a mere masker. I would describe the fire of vodka for example, as just really intense bitter. 128.143.181.23 (talk) 20:37, 19 November 2010 (UTC)

To me, ethanol just tastes like bitter like apple seeds, I find it repulsive. This is the reason for why I dislike strong liquors like spirits. I don't choose my liquors for the taste of the ethanol, ethanol is not the major contributor to the taste of the beverage. Ethanol is not meant to be anything other than a mind altering stimulant. Plasmic Physics (talk) 00:21, 20 November 2010 (UTC)

You don't like apple seeds? I quite enjoy them. They're a lot of work (you have to take off the endocarp I think it's called) but they have a very nice delicate flavor, reminiscent of almond delight. They're poisonous if you eat a lot of them (cyanogenic glucosides I think), but I only ever eat one or two at a time. --Trovatore (talk) 09:30, 20 November 2010 (UTC)

See also Apple seed oil. The key safety concern looks like amygdalin according to our apple article. DMacks (talk) 09:34, 20 November 2010 (UTC)

That is of course, a personal preference. I personally enjoy the taste of ethanol itself, good vodka is basically just water and ethanol. Ethanol is meant to be enjoyed, and for myself I find the flavor effects of ethanol to be more important than the mind altering effects. I actually find those somewhat of an anoyance, as getting drunk severly limits my ability to drink more ethanol. --Jayron32 01:42, 20 November 2010 (UTC)

OR here but I don't like the taste of most alcoholic drinks. However on those occasions when I've tried them, alcopops have seemed fine, no real different from a random soft drink (hence why I don't drink them a lot). I guess there are some people like JRS who still find alcopops bitter but I suspect they aren't that common and most people who just want to get intoxicated (which I presume is the case if you're trying to completely mask the taste of ethanol) and do find even alcopops bitter probably don't want to spend a lot of money. For those who don't mind alcopops, the fact that it's just masking the bitterness seems irrelevant unless perhaps your diabetic. Nil Einne (talk) 06:29, 20 November 2010 (UTC)

I would agree that ethanol doesn't seem bitter. But see PMID 12090789 - AMP simply isn't stable at room temperature, even in a pure solution. The organic phosphate bond is just too high in energy and too easily hydrolyzed especially in non-neutral pH. (I assume in the cell it spends most of its time bound to something or other that helps stabilize it, but I'm not recalling just what that might be at the moment) Wnt (talk) 13:17, 20 November 2010 (UTC)

Species identifcation for File:Sagittaria.jpg

The image in question

In order to expand on the image description, so the image can be moved to Commons, Is anyone on the Science Reference desk able to provide a more specific species identification?

It appears when the is uploaded, there was some discussion, it may have been mis-identifed.. Sfan00 IMG (talk) 12:46, 16 November 2010 (UTC)

Sagittaria sagittifolia

It looks very much like Sagittaria to me. There are lots of species though. Looie496 (talk) 17:58, 19 November 2010 (UTC)

shemales

Do shemales really exist? —Preceding unsigned comment added by 59.95.48.140 (talk) 13:28, 19 November 2010 (UTC)

Yes. See shemale, trans woman and transsexualism. Red Act (talk) 13:38, 19 November 2010 (UTC)

There are also people born with both genders, see hermaphrodite. --Jayron32 16:14, 19 November 2010 (UTC)

Also see intersex. Red Act (talk) 18:25, 19 November 2010 (UTC)

Definitive deletion of a phobia (classical conditioning)

By considering absence of the Renewal effect and the Spontaneous recovery a proof of a definitive deletion of a phobia, can one expect a definitive deletion of a phobia after sufficiently repeated extinctions?--Kooz (talk) 19:27, 19 November 2010 (UTC)

It's all a matter of probability, not certainty and proof. Any variety of phobias can be conquered, but in some people that means complete extinction and in others it means simply reduction of the response under an acceptable threshold. Time can make things better or worse, depending on the subject and the circumstances. Ideally a therapist will be able to help a patient come up with some exercises the patient can perform by themselves should the problem return. There's nothing exact in psychiatry. Ginger Conspiracy (talk) 04:03, 20 November 2010 (UTC)

Orgasmic sensation during pull-ups

When doing pull-ups or chin-ups, what is causing the tingling, sexual sensation in the groin area? This question is not looking for medical diagnoses or opinions. Thanks Reflectionsinglass (talk) 20:49, 19 November 2010 (UTC)

Orgasm#Spontaneous_orgasms covers some of this; it is entirely possible to experience orgasm, or orgasm-like sensations, without any direct stimulation of the genitals. --Jayron32 21:08, 19 November 2010 (UTC)

Chinups tend to be a very intense physical activity - it's possible that performing them gives you some type of endorphin rush, and combined with the increased peripheral blood flow that results from exercise, your body may be interpreting this as a sign of sexual release. There are suggestions that exercise in general helps to stimulate the libido, so this could be related. It would be interesting to attempt other strenuous exercises to see if it produces similar effects. --jjron (talk) 02:38, 20 November 2010 (UTC)

Are we allowed to talk about our own experiences in detail? If not, someone can delete this response, but consider this as simply my own experiment. I've been able to reach orgasm by lifting myself and holding myself up, since childhood, four or five years old (no emission, of course). Several years ago, much older and with other means of release, I'd suddenly remembered about this and I thought I would experiment. I used a chin-up bar, held myself up, and was able to experience full orgasm, including emission. It was extremely quick to achieve climax. I doubt I could do it again, I can barely hold myself up for 20 seconds these days. Thanks for the link to spontaneous orgasms. It was too brief a section, unfortunately. I may have to continue looking online. Reflectionsinglass (talk) 07:11, 20 November 2010 (UTC)

I tell you what, reflections and jjron, if there is anything in what you say, we've beaten the obesity epidemic right here. For my own case, I went to my doctor once and told him that in all my adult life, I felt strong orgasmic sensations when I sneezed. He said "Half your luck...". Myles325a (talk)