A constant speed of light in all reference frames? Surely you can't be serious.
in [Relativity]
|
From: PD on 9 Mar 2010 09:31 On Mar 8, 11:53 pm, Bruce Richmond <bsr3...(a)my-deja.com> wrote: > On Mar 8, 1:58 am, "Inertial" <relativ...(a)rest.com> wrote: > > > > > "Bruce Richmond" <bsr3...(a)my-deja.com> wrote in message > > >news:29fd9f7d-c9b4-41e6-b33a-585c3e0e7acf(a)q23g2000yqd.googlegroups.com.... > > > > On Mar 7, 9:48 pm, "Inertial" <relativ...(a)rest.com> wrote: > > >> "Bruce Richmond" <bsr3...(a)my-deja.com> wrote in message > > > >>news:1c6a2640-39f5-4ea4-9c85-127e71f4e6a2(a)33g2000yqj.googlegroups.com.... > > > >> > On Mar 7, 6:58 pm, "Inertial" <relativ...(a)rest.com> wrote: > > >> >> "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote in message > > > >> >>news:4b943853$0$11336$afc38c87(a)news.optusnet.com.au... > > > >> >> > "Inertial" <relativ...(a)rest.com> wrote in message > > >> >> >news:4b942bcf$0$27789$c3e8da3(a)news.astraweb.com... > > >> >> >> "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote in > > >> >> >> message > > >> >> >>news:4b93bf73$0$28464$afc38c87(a)news.optusnet.com.au... > > > >> >> >>>>> 1) your statement: "For clarity, both effects are purely > > >> >> >>>>> observational - SR presumes (ideal) clock mechanisms are > > >> >> >>>>> completely > > >> >> >>>>> unaffected by a clock's motion." -- I agree the physical > > >> >> >>>>> mechanism > > >> >> >>>>> of > > >> >> >>>>> the clock is unaffected, but this is a really misleading > > >> >> >>>>> statement, > > >> >> >>>>> since the amount of proper time that the clock consumes is > > >> >> >>>>> affected > > >> >> >>>>> by > > >> >> >>>>> its motion. Are you trying to say > > > >> >> >>>> I was "trying to say" exactly what I did say. If you didn't find > > >> >> >>>> it > > >> >> >>>> clear enough, try this: relative slow-downs/speed-ups observed in > > >> >> >>>> the > > >> >> >>>> readings of SR's ideal clocks aren't due to changes in the tick > > >> >> >>>> mechanisms of those clocks. > > > >> >> >>> I still don't find it clear, as it begs the question - it says > > >> >> >>> what > > >> >> >>> doesn't cause the change, not what does cause the change. > > > >> >> >>> The standard SR answer is much more direct - the clocks slow down > > >> >> >>> due > > >> >> >>> to > > >> >> >>> relativistic time dilatation from them being in different > > >> >> >>> reference > > >> >> >>> frames. > > > >> >> >>> Is that standard position of SR also your position? Or is your > > >> >> >>> somehow > > >> >> >>> different? > > > >> >> >> SR says that the difference in clock sync (clock settings) cause > > >> >> >> the > > >> >> >> measurement of length to be contracted and measurement of clock > > >> >> >> ticking > > >> >> >> rates to be dilated. > > > >> >> > More or less. > > > >> >> That's what it is :) > > > >> >> > But I asked you about *your* position, not SR's position. > > > >> >> My position is SR's position > > > >> >> > Do you agree that that the clocks slow down due to relativistic time > > >> >> > dilation, as predicted by SR, or not? > > > >> >> They are measured as slower, just as a rod is measured as shorter.. > > >> >> This > > >> >> is > > >> >> due to the difference in simultaneity. They don't slow down because a > > >> >> moving observer is looking at them any more than a rod shrinks because > > >> >> a > > >> >> relatively moving observer is looking at it. > > > >> >> Here's a little example you might follow .. with time differences > > >> >> exagerated > > >> >> for clarity > > > >> >> Here are six clocks, in tow rows (S and S'), all ticking at the > > >> >> correct > > >> >> rate, but set with different times... > > > >> >> S' 10:30 11:00=A 11:30 <--v > > >> >> S 11:30=C 11:00=B 10:30 -->v > > > >> >> Clocks B sees the A is synchronized with it. > > > >> >> Now .. the clocks are moving in opposite directions so after an hour > > >> >> we > > >> >> have > > > >> >> S' 11:30 12:00=A 12:30 > > >> >> S 12:30=C 12:00=B 11:30 > > > >> >> Clock A has moved away from clock B .. but another clock (C) in S can > > >> >> see > > >> >> the time on it. Clock C sees that clock A is half an hour slow (A > > >> >> shows > > >> >> 12:00 when C shows 12:30). So according to the clocks in S, clock A > > >> >> is > > >> >> ticking slower. We also note that clock B now sees a *different* S' > > >> >> clock > > >> >> next to it as being fast (it shows 12:30 when B shows 12:00) > > > >> >> If you look at the same scenario but from the point of view of the > > >> >> other > > >> >> row > > >> >> of clocks, you get symmetric results. > > > >> >> This is how clock synch affects measured ticking rates for moving > > >> >> clocks > > >> >> in > > >> >> SR. Even though the clocks themselves do NOT change their intrinsic > > >> >> ticking > > >> >> rates.- Hide quoted text - > > > >> > Looks good, but let's take it one step further. The observer with > > >> > clock A jumps to frame S" which is traveling in the same direction as > > >> > S relative to S' but at twice the velocity. > > > >> > S" 1:00 12:00=A 11:00 -->2v > > >> > S' 11:30 12:00=A 12:30 <--v > > >> > S 12:30=C 12:00=B 11:30 -->v > > > >> > Clocks A and B continue to tick at there same intrinsic ticking rate > > >> > and an hour later A has overtaken B. > > > >> > S" 2:00 1:00=A 12:00 -->2v > > >> > S' 12:30 1:00=A 1:30 <--v > > >> > S 1:30=C 1:00=B 12:30 -->v > > > >> > The above provides the same situation as the twins paradox. Clock A > > >> > left clock B and returned. So why doesn't clock A show less time > > >> > elapsed than B? (Note the clocks in S" are further out of sync than > > >> > those in S due to the higher velocity.) > > > >> The three clock situation cannot be so easily drawn .. bit like trying to > > >> drawing a three dimensional figure in 2d :) This sort of diagram only > > >> really works for a single pair of clocks looking from a third frame in > > >> which > > >> they move with the same speed. Things are trickier when there is frame > > >> jumping going on :):)- Hide quoted text - > > > >> - Show quoted text - > > > > The question still remains, if there is no change in the tick rate of > > > the clock, how can clock A have fewer ticks recorded when it is > > > brought back to clock B? > > > Look at the Lorentz transforms to see. Its all due to clock synch.- Hide quoted text - > > > - Show quoted text - > > You are going to give yourself a head cold with all that arm waving. > You are saying the tick rate doesn't change, yet SR says that the > returning twin's clock will show less elapsed time. You don't see any > conflict there? No, there is no conflict. When you say that the tick rate does not change, this is a LOCAL statement. What it means is that a process measured locally with this clock will still have the same duration. For example, if the half-life of a muon that is slow in frame A and is measured with a clock at rest in A is 2.2 us, then if you make the same measurement of the half-life of a muon that is slow in frame B and is measured with a clock at rest in B, clock B will still show the half-life to be 2.2 us. In this sense, we say that the clock tick rate has not changed, because measurements of local phenomena are unchanged. However, this does NOT mean that the tick rate of clock B will agree with the clock rate of all other clocks, nor that it will read the durations of nonlocal processes to be the same. Do you see the distinction? PD
From: Sue... on 9 Mar 2010 09:41 On Mar 9, 9:26 am, PD <thedraperfam...(a)gmail.com> wrote: > On Mar 9, 7:49 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > On 8 Mar, 22:55, PD <thedraperfam...(a)gmail.com> wrote: > > > > On Mar 6, 5:32 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > What confuses me is that, if the clocks run slow by 2% for all the > > > > time that they are moving, how does one reconcile this with the fact > > > > that, if one uses the frame of one of the moving clocks, say clock B, > > > > then it seems to be to be your argument that there is no slowdown at > > > > all for B, and it is the other clocks, A and C, that slow down (i.e.. > > > > *disregarding* both acceleration and propagation delays). > > > > Be careful. The acceleration profiles are common between B and C, but > > > they are not common to A. So while there is no difference between B > > > and C due to the acceleration, you CANNOT say that the acceleration > > > has no effect whatsoever. In fact, it is the indisputable fact that B > > > and C accelerate and A does NOT accelerate that makes the situation > > > nonsymmetric for A. This is what makes the worldline for A straight, > > > and the worldline for B and C kinked. > > > Yes, but we're supposed to have isolated the effect of acceleration, > > and disregarded it. > > No, we did not. We said that it cannot account for the DIFFERENCE > between B and C, but this does not discount or remove acceleration > from further consideration, particularly with regard to how clock A's > rate is seen by B. > > > > > And in any event, the more important question is > > the discrepancy between B and C. > > > > Two places where I will try to intercept misconceptions. > > > 1. The first temptation is to say, well, if the kink is what's > > > responsible for the time dilation, then all the dilating must happen > > > during the acceleration. That is not the case. Note the time dilation > > > is different for B and C, even though they have the same kink (the > > > same acceleration profile). The fact that there IS a kink is what > > > makes the elapsed time less on B and C than it is on A (where there is > > > no kink), but how much less depends on the steepness and length of the > > > straight parts of the worldline on either side of the kink. > > > As I say, I've stipulated that we are measuring on the outbound > > journey, before any of the clocks have turned back. So we've had one > > episode of acceleration and now B and C are travelling at the same > > speed away from the origin point, but in opposite directions. What > > amount of time dilation does C suffer relative to B? Nil? 2%? 4%? > > It's not that simple, and to get a number, you need to use the math. Indeed! The maths! << Einstein's relativity principle states that: All inertial frames are totally equivalent for the performance of all physical experiments. In other words, it is impossible to perform a physical experiment which differentiates in any fundamental sense between different inertial frames. By definition, Newton's laws of motion take the same form in all inertial frames. Einstein generalized[1] this result in his special theory of relativity by asserting that all laws of physics take the same form in all inertial frames. >> http://farside.ph.utexas.edu/teaching/em/lectures/node108.html [1]<< the four-dimensional space-time continuum of the theory of relativity, in its most essential formal properties, shows a pronounced relationship to the three-dimensional continuum of Euclidean geometrical space. In order to give due prominence to this relationship, however, we must replace the usual time co-ordinate t by an imaginary magnitude sqrt(-1) ct proportional to it. Under these conditions, the natural laws satisfying the demands of the (special) theory of relativity assume mathematical forms, in which the time co-ordinate plays exactly the same rôle as the three space co-ordinates. >> http://www.bartleby.com/173/17.html << where epsilon_0 and mu_0 are physical constants which can be evaluated by performing two simple experiments which involve measuring the force of attraction between two fixed charges and two fixed parallel current carrying wires. According to the relativity principle, these experiments must yield the same values for epsilon_0 and mu_0 in all inertial frames. Thus, the speed of light must be the same in all inertial frames. >> http://farside.ph.utexas.edu/teaching/em/lectures/node108.html Sue... [...] > Sorry, but that's the facts. The 2% is the Lorentz dilation factor, > and that is given by a function 1/sqrt(1-v^2/c^2), and you can see > that doubling the speed will not double the Lorentz dilation factor. > Furthermore, the speed of C relative to B is not twice the velocity of > B with respect to A, because the relation for combining velocities is > (v1+v2)/(1+v1*v2/c^2). > > To find out what the Lorentz dilation factor is for C relative to B, > then you simply need to put in the numbers and crank. > > > > > > 2. The second temptation is to say, well, in B's frame it does not > > > accelerate, and A does accelerate. This is not correct. > > > I know. > > > > Now, it IS fair to say that while the clock B is on its outward > > > journey at constant speed, it can look back at clock A and discern > > > that clock A is running slow relative to B. And the same goes on the > > > inbound journey. And so it's a fair question to say, how does it > > > happen then that by the time B lands back at A, it is BEHIND clock A? > > > And here is where things do get interesting in terms of what B sees in > > > terms of the clock reading on A in the turnaround. > > > No, the question is between B and C, not B and A. I'm using B and C > > specifically in order to allow us to disregard any effects of > > acceleration. > > > And we've already stipulated that, if B and C have the same > > acceleration profiles and travel the same distance before returning to > > A, then they will be in agreement with each other, but both will have > > slowed relative to A. The question is, while they are still on the > > outbound journey, what do B and C report about each other? Do they > > both agree with each other still? Or not? > >
From: bert on 9 Mar 2010 09:48 On Mar 9, 9:41 am, "Sue..." <suzysewns...(a)yahoo.com.au> wrote: > On Mar 9, 9:26 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > On Mar 9, 7:49 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > On 8 Mar, 22:55, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Mar 6, 5:32 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > > What confuses me is that, if the clocks run slow by 2% for all the > > > > > time that they are moving, how does one reconcile this with the fact > > > > > that, if one uses the frame of one of the moving clocks, say clock B, > > > > > then it seems to be to be your argument that there is no slowdown at > > > > > all for B, and it is the other clocks, A and C, that slow down (i..e. > > > > > *disregarding* both acceleration and propagation delays). > > > > > Be careful. The acceleration profiles are common between B and C, but > > > > they are not common to A. So while there is no difference between B > > > > and C due to the acceleration, you CANNOT say that the acceleration > > > > has no effect whatsoever. In fact, it is the indisputable fact that B > > > > and C accelerate and A does NOT accelerate that makes the situation > > > > nonsymmetric for A. This is what makes the worldline for A straight, > > > > and the worldline for B and C kinked. > > > > Yes, but we're supposed to have isolated the effect of acceleration, > > > and disregarded it. > > > No, we did not. We said that it cannot account for the DIFFERENCE > > between B and C, but this does not discount or remove acceleration > > from further consideration, particularly with regard to how clock A's > > rate is seen by B. > > > > And in any event, the more important question is > > > the discrepancy between B and C. > > > > > Two places where I will try to intercept misconceptions. > > > > 1. The first temptation is to say, well, if the kink is what's > > > > responsible for the time dilation, then all the dilating must happen > > > > during the acceleration. That is not the case. Note the time dilation > > > > is different for B and C, even though they have the same kink (the > > > > same acceleration profile). The fact that there IS a kink is what > > > > makes the elapsed time less on B and C than it is on A (where there is > > > > no kink), but how much less depends on the steepness and length of the > > > > straight parts of the worldline on either side of the kink. > > > > As I say, I've stipulated that we are measuring on the outbound > > > journey, before any of the clocks have turned back. So we've had one > > > episode of acceleration and now B and C are travelling at the same > > > speed away from the origin point, but in opposite directions. What > > > amount of time dilation does C suffer relative to B? Nil? 2%? 4%? > > > It's not that simple, and to get a number, you need to use the math. > > Indeed! The maths! > > << Einstein's relativity principle states that: > > All inertial frames are totally equivalent > for the performance of all physical experiments. > > In other words, it is impossible to perform a physical > experiment which differentiates in any fundamental sense > between different inertial frames. By definition, Newton's > laws of motion take the same form in all inertial frames. > Einstein generalized[1] this result in his special theory of > relativity by asserting that all laws of physics take the > same form in all inertial frames. >>http://farside.ph.utexas.edu/teaching/em/lectures/node108.html > > [1]<< the four-dimensional space-time continuum of the > theory of relativity, in its most essential formal > properties, shows a pronounced relationship to the > three-dimensional continuum of Euclidean geometrical space. > In order to give due prominence to this relationship, > however, we must replace the usual time co-ordinate t by > an imaginary magnitude > > sqrt(-1) > > ct proportional to it. Under these conditions, the > natural laws satisfying the demands of the (special) > theory of relativity assume mathematical forms, in which > the time co-ordinate plays exactly the same rôle as > the three space co-ordinates. >>http://www.bartleby.com/173/17.html > > << where epsilon_0 and mu_0 are physical constants which > can be evaluated by performing two simple experiments > which involve measuring the force of attraction between > two fixed charges and two fixed parallel current carrying > wires. According to the relativity principle, these experiments > must yield the same values for epsilon_0 and mu_0 in all > inertial frames. Thus, the speed of light must be the > same in all inertial frames. >>http://farside.ph.utexas.edu/teaching/em/lectures/node108.html > > Sue... > > [...] > > > > > Sorry, but that's the facts. The 2% is the Lorentz dilation factor, > > and that is given by a function 1/sqrt(1-v^2/c^2), and you can see > > that doubling the speed will not double the Lorentz dilation factor. > > Furthermore, the speed of C relative to B is not twice the velocity of > > B with respect to A, because the relation for combining velocities is > > (v1+v2)/(1+v1*v2/c^2). > > > To find out what the Lorentz dilation factor is for C relative to B, > > then you simply need to put in the numbers and crank. > > > > > 2. The second temptation is to say, well, in B's frame it does not > > > > accelerate, and A does accelerate. This is not correct. > > > > I know. > > > > > Now, it IS fair to say that while the clock B is on its outward > > > > journey at constant speed, it can look back at clock A and discern > > > > that clock A is running slow relative to B. And the same goes on the > > > > inbound journey. And so it's a fair question to say, how does it > > > > happen then that by the time B lands back at A, it is BEHIND clock A? > > > > And here is where things do get interesting in terms of what B sees in > > > > terms of the clock reading on A in the turnaround. > > > > No, the question is between B and C, not B and A. I'm using B and C > > > specifically in order to allow us to disregard any effects of > > > acceleration. > > > > And we've already stipulated that, if B and C have the same > > > acceleration profiles and travel the same distance before returning to > > > A, then they will be in agreement with each other, but both will have > > > slowed relative to A. The question is, while they are still on the > > > outbound journey, what do B and C report about each other? Do they > > > both agree with each other still? Or not?- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text - It would not be a photon if it did not go at c It would not be a photon if it bounced. TreBert
From: Dono. on 9 Mar 2010 10:02 On Mar 9, 1:33 am, "Inertial" <relativ...(a)rest.com> wrote: > "Dono." <sa...(a)comcast.net> wrote in message > > news:9a5eb50c-6ed1-4435-9493-0a0fef9039df(a)v20g2000prb.googlegroups.com... > > > On Mar 8, 6:06 pm, "Inertial" <relativ...(a)rest.com> wrote: > > >> In LET the speed of light is only isotropically c in the aether frame. > > > ...then all the modern tests that put severe constraints on light > > speed anisotropy falsify LET. > > No .. because we measure the speed as isotropic due to the distorted rulers > and malfunctioning clocks that movement thru the aether causes (according to > LET). > You are contradicting yourself , in the earlier post you claimed (correctly) that LET predicts light speed to be isotropic ONLY in the preferrential frame of the "aether". In ALL other frame, light speed is ANISOTROPIC. So, you are now contradicting yourself and your new post is wrong.
From: Dono. on 9 Mar 2010 10:05
On Mar 9, 2:49 am, "Inertial" <relativ...(a)rest.com> wrote: > "Jerry" <Cephalobus_alie...(a)comcast.net> wrote in message > > > No reason they cannot. No reason why gravity waves and EM waves cannot both > travel at c. Any test showing that the do both travel at c will simply > refute your assertion that they must be different .. it won't refute aether > theory itself. And if necessary, aether will be modified yet again to > account for it. Bad answer: EM waves are TRANSVERSE whereas gravitational wave are LONGITUDINAL waves. The same medium cannot propagate both, so, you need AT LEAST two different "aethers" |