A constant speed of light in all reference frames? Surely you can't be serious.
in [Relativity]
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From: mpc755 on 13 Mar 2010 09:57 On Mar 13, 3:30 am, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > If they are not directly across from each other you are making > assumptions about the speed of light carrying the signal. If you use > two clocks you made assumptions about the speed of light when you > synchronized the two clocks. > > _____________________________ > When solving problems in SR or LET, you don't have to assume the speed of > light is constant. You know it is. Light waves propagate at 'c' with respect to the aether.
From: ben6993 on 13 Mar 2010 09:59 On Mar 12, 11:56 pm, "Inertial" <relativ...(a)rest.com> wrote: > "ben6993" <ben6...(a)hotmail.com> wrote in message > > news:716c1760-a5db-4ce8-b116-7a739eaae397(a)o30g2000yqb.googlegroups.com... > > > > > > > On Mar 12, 6:16 pm, PD <thedraperfam...(a)gmail.com> wrote: > >> On Mar 12, 12:00 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > >> > On 11 Mar, 20:57, PD <thedraperfam...(a)gmail.com> wrote: > > >> > > On Mar 11, 2:15 pm, Ste <ste_ro...(a)hotmail.com> wrote: > > >> > > > On 11 Mar, 15:12, PD <thedraperfam...(a)gmail.com> wrote: > > >> > > > > On Mar 11, 6:43 am, Ste <ste_ro...(a)hotmail.com> wrote: > > >> > > > > > On 11 Mar, 01:51, "Peter Webb" > >> > > > > > <webbfam...(a)DIESPAMDIEoptusnet.com.au> > >> > > > > > wrote: > > >> > > > > > > No, perhaps you didn't understand. As I say, this is *not* > >> > > > > > > the twins > >> > > > > > > paradox, because in the twins paradox only *one* twin leaves > >> > > > > > > Earth. > > >> > > > > > > ________________________ > >> > > > > > > Its functionally the same. It is exactly the twins paradox, > >> > > > > > > but with two > >> > > > > > > twins apparently doing exactly the same thing. > > >> > > > > > > Even if you cannot see that, the explanation on the Wikipedia > >> > > > > > > page of the > >> > > > > > > Twins Paradox is trivially adapted for two twins. > > >> > > > > > > I assume that you do not understand the Wikipedia twins > >> > > > > > > paradox page, or > >> > > > > > > else you would know the answers to your questions already. > >> > > > > > > Which parts don't > >> > > > > > > you understand? > > >> > > > > > Let's just go through it step by step Peter, as we have been > >> > > > > > doing. > >> > > > > > It's pointless spending 10 more postings arguing about how the > >> > > > > > Wikipedia page does or does not answer the question, or how it > >> > > > > > is or > >> > > > > > is not relevant. As I've just said in a post to Inertial, the > >> > > > > > only > >> > > > > > analogy between my scenario and the twins paradox is that, in > >> > > > > > my > >> > > > > > scenario, both twins leave Earth, and both return the same age > >> > > > > > as each > >> > > > > > other - hence no paradox, and hence bearing no resemblance at > >> > > > > > all to > >> > > > > > the twins paradox. > > >> > > > > First of all, let's establish what you think is paradoxical at > >> > > > > all > >> > > > > about the description of the twins in the twin puzzle. Then let's > >> > > > > see > >> > > > > whether this paradox is present in the case you mention. > > >> > > > As I understand it, the supposed "paradox" in the twins paradox was > >> > > > that one returned younger than the other. It was, of course, not a > >> > > > paradox at all, but that's besides the point. > > >> > > No, then you do not understand the paradox, because there is nothing > >> > > contradictory in that statement at all. It may be surprising, but > >> > > it's > >> > > not contradictory, not paradoxical. Disagreement of clocks is not a > >> > > paradox. > > >> > > The paradox, which is what is perceived (normally) by freshmen when > >> > > first introduced to this statement, is embodied in their immediate > >> > > classroom question: "But in the frame of the traveling twin, it is > >> > > the > >> > > earth twin that is moving away and returning. Since this is symmetric > >> > > to the case of the traveling twin moving away and returning, then > >> > > shouldn't the traveling twin expect the earth twin to be younger when > >> > > they meet again?" Now perhaps the paradox is more apparent to you. > > >> > > However, the puzzle is specifically designed to emphasize the danger > >> > > of oversimplifying. In fact, the two twins are NOT symmetric, because > >> > > one unambiguously experiences acceleration and the other > >> > > unambiguously > >> > > experiences no acceleration. This then leads to a discussion of what > >> > > produces the asymmetry in the time. > > >> > I know Paul. I know. > > >> You can imagine my surprise, since what you said explicitly above was > >> that the paradox was that the twins aged differently. > > >> > > Perhaps if you had started out by asking, "Since I don't see any > >> > > obvious paradox here at all, perhaps someone could illuminate me as > >> > > to > >> > > why this is called the twin paradox?" Then at least you would have > >> > > been on square one. > > >> > Really I just wanted to avoid going off on a long tangent about the > >> > twins paradox. As I said, the scenario that were were addressing is > >> > different from the twins paradox, in that we have three clocks, and > >> > the two clocks with which we are now concerned (B and C) both return > >> > to the origin point *synchronised* (albeit both lagging behind A), > >> > whereas the twins' ages are not synchronised on the return of the > >> > astronaut twin. > > >> Well, yes, it is a different result from the application of the same > >> principle. > > >> If I calculate the angle that I can tip a TV tray before the coffee > >> cup on it starts to slide, I find that I'm using the same principle > >> (equilibrium of forces) that I would use to determine the tension in > >> picture-hanger wire when mounting a photo on the wall. > > >> Different situation. Very same principle. > > >> Fine example of losing the forest for the trees. As you've done here. > > >> > So let me say again. The twins paradox would be applicable if we were > >> > talking about A and B, or A and C. In the event, we are talking about > >> > what B and C observe of each other from their own reference frames. > >> > There is, therefore, no correspondence with the twins paradox, because > >> > unlike the twins, B and C return synchronised with each other.- Hide > >> > quoted text - > > >> - Show quoted text -- Hide quoted text - > > >> - Show quoted text - > > > I have written below a twin plank paradox of length. I know the > > explanation must really be very straightforward, but I don't see it > > yet. This is not a time paradox, but it looks to be a similar issue. > > > A plank has four units of length (----) in its own frame and a garage > > has two units of length (--) in its own frame. > > There are two similar planks and garages in relative motion such that > > plank P and garage G are moving fast towards the other plank p and > > garage > > > g. The relative speed is such that lengths are approximately halved > > in relativistic contraction. > > > <--- direction of motion > > p ---- G' - > > g -- P' -- > > > Where ' sign indicates motion in the other twins' frame. > > > Here, p does not fit within G', but P' fits approx. within g. > > > Looked at from the other frame, below, a similar result occurs: p' > > fits approx. within G, but P does not fit within g'. > > ----> > > p' -- G -- > > g' - P ---- > > > As the same two events are looked at twice, i.e. from two frameworks, > > there are four outcomes in total. The shed doors are destroyed in two > > outcomes and the planks fit into the sheds in the other two outcomes. > > So it looks like one of the two sheds is damaged and the other is > > safe. But that cannot be true as the motion is relative not absolute, > > and the planks and garages are twins, and so the two outcomes should > > be identical. > > They are identical if you do identical things . If you do the 'close both > garage doors simultaneously in the garage's rest frame of reference when the > pole is fully within' for both garages, both are safe. > > If you close the door at different times in their frame, then they may not > be safe (depending on when you do)- Hide quoted text - > > - Show quoted text - Thanks very much for your help, Inertial. I have introduced time into my calculations now and have relative speed is approx. 0.866c; contraction is to half the length; and, time slows to double for the moving plank and garage. The time dilation for the moving garage enables the two planks to fit through the garage doors in the appropriate, equivalent time intervals. I still need to work more on fitting the stationary plank within the moving garage, but am resolving that by taking a break from physics for a one week contract job! Physics paradoxes are very tiring!
From: Bruce Richmond on 13 Mar 2010 13:28 On Mar 13, 3:30 am, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > If they are not directly across from each other you are making > assumptions about the speed of light carrying the signal. If you use > two clocks you made assumptions about the speed of light when you > synchronized the two clocks. > > _____________________________ > When solving problems in SR or LET, you don't have to assume the speed of > light is constant. You know it is. In response to both you and Inertial, mpc755 wrote, "The clock which ticks the fastest is most at rest with respect to the aether." So we were not discussing SR. And since he thinks he can detect the ether frame we are not discussing LET either.
From: mpc755 on 13 Mar 2010 13:31 On Mar 13, 1:28 pm, Bruce Richmond <bsr3...(a)my-deja.com> wrote: > On Mar 13, 3:30 am, "Peter Webb" > > <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > > If they are not directly across from each other you are making > > assumptions about the speed of light carrying the signal. If you use > > two clocks you made assumptions about the speed of light when you > > synchronized the two clocks. > > > _____________________________ > > When solving problems in SR or LET, you don't have to assume the speed of > > light is constant. You know it is. > > In response to both you and Inertial, mpc755 wrote, "The clock which > ticks the fastest is most at rest with respect to the aether." So we > were not discussing SR. And since he thinks he can detect the ether > frame we are not discussing LET either. I did not say the aether frame can be detected, what I wrote and what you quote above stands on its own. The aether is most at rest with respect to the clock which ticks the fastest.
From: Bruce Richmond on 13 Mar 2010 13:34
On Mar 13, 1:31 pm, mpc755 <mpc...(a)gmail.com> wrote: > On Mar 13, 1:28 pm, Bruce Richmond <bsr3...(a)my-deja.com> wrote: > > > > > > > On Mar 13, 3:30 am, "Peter Webb" > > > <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > > > If they are not directly across from each other you are making > > > assumptions about the speed of light carrying the signal. If you use > > > two clocks you made assumptions about the speed of light when you > > > synchronized the two clocks. > > > > _____________________________ > > > When solving problems in SR or LET, you don't have to assume the speed of > > > light is constant. You know it is. > > > In response to both you and Inertial, mpc755 wrote, "The clock which > > ticks the fastest is most at rest with respect to the aether." So we > > were not discussing SR. And since he thinks he can detect the ether > > frame we are not discussing LET either. > > I did not say the aether frame can be detected, what I wrote and what > you quote above stands on its own. > > The aether is most at rest with respect to the clock which ticks the > fastest.- Hide quoted text - > > - Show quoted text - Then all you would need to do is determine which clock is ticking fastest and you have found the ether frame. Most people would consider that detecting the ether. |