Why physicists refuse to measure the one-way speed of light directly?
in [Physics]
|
From: Da Do Ron Ron on 14 Jan 2010 13:32 E Gisse asked: >What part do you have difficulty with? Well, that was given up front, but I don't mind repeating it, as follows: Dr. Tom claimed (correctly) that Einstein's clock synch procedure or process GUARANTEES that the one-way speed of light will be measured to be c; all I asked him was HOW does this happen? (What is the physical process?) Indeed, I don't mind _further explaining_ the query, as follows: As we all know, c-invariance means that all observers in all inertial frames get c for the speed of light. Thus, as Andie did NOT know, the only way to show c-invariance is by using two (or more) frames. And, as I have said, we must not let each frame have its own light source as this is merely repeating Frame A over and over; we must let the frames SHARE the same source so that they are not repeated. Only in this way can we have Frames A, B, C, etc. Of course, we need only two frames, so let's use Frame A and Frame B. As Tom (again correctly) said, each frame's observers MUST USE TWO CLOCKS in the case of the one-way speed of light. We now have the extremely simple setup of two frames with two clocks each, and a single light source, as follows: Frame A [clockA1]---------x----------[clockA2] Source S [clockB1]---------x----------[clockB2] Frame B x is of course the distance measured in each frame by a standard ruler laid at rest relative to the frame in which the measurement is being made. No mystery here! We can let Frame B move to the right relative to Frame A, and we can attach the source to Frame A. We can keep "simplicity mode" turned on by letting the two origin clocks (A1 & B1) read ZERO at the start, as follows: Frame A [0]---------x----------[clockA2] Source S~~>light [0]---------x----------[clockB2] -->v Frame B As shown above, whenever the light is emitted from S towards the two distant clocks (A2 & B2), the two origin clocks will start on time zero. At this point, the two distant clocks are not yet started. Indeed, we have yet to give the time readings for these two clocks. Here is what we have so far: Frame A [0]---------x----------[?] Source S~~>light [0]---------x----------[?] -->v Frame B My original question can now be rephrased as follows: What exactly does Einstein's definition call for in the case of the two distant clocks? That is, what times must be placed on their faces prior to the release of the light from S? This is the part that I am having difficulty with. Hope this answers your question. ~~RA~~ (You may still need Tom's or PDiddy's help!?) (But I would strongly suggest _not_ asking for help from either Andie-boy or Prof. Seto!)
From: PD on 14 Jan 2010 13:58 On Jan 14, 12:32 pm, Da Do Ron Ron <ron_ai...(a)hotmail.com> wrote: > E Gisse asked: > > >What part do you have difficulty with? > > Well, that was given up front, but I don't mind repeating it, > as follows: > > Dr. Tom claimed (correctly) that Einstein's clock synch > procedure or process GUARANTEES that the one-way speed > of light will be measured to be c; all I asked him was > HOW does this happen? (What is the physical process?) It's simpler than a physical process. Einstein's synchronization procedure requires some signal that is guaranteed to go at the same speed in either direction. The procedure is this: 1. Note the time t1 at clock A. 2. Send a signal from clock A to clock B at a speed v. 3. When the signal arrives, note the time t2 at clock B. 4. Send a signal from clock B to clock A at a speed v. 5. When the signal arrives, note the time t3 at clock A. 6. If t3-t2 = t2-t1, then the clocks are synchronized. Note that this procedure only requires a signal that is guaranteed to be the same speed in either direction. If you could be sure you could WALK at constant speed, that would work too. An obvious choice is using light, because light has never been observed to travel locally in a vacuum at any speed other than c. I'll leave it to you now to figure out that if two clocks have been so synchronized, the measured OWLS will always be c. PD > > Indeed, I don't mind _further explaining_ the query, as > follows: > > As we all know, c-invariance means that all observers in > all inertial frames get c for the speed of light. Thus, > as Andie did NOT know, the only way to show c-invariance > is by using two (or more) frames. And, as I have said, > we must not let each frame have its own light source as > this is merely repeating Frame A over and over; we must > let the frames SHARE the same source so that they are not > repeated. Only in this way can we have Frames A, B, C, etc. > Of course, we need only two frames, so let's use Frame A > and Frame B. As Tom (again correctly) said, each frame's > observers MUST USE TWO CLOCKS in the case of the one-way > speed of light. We now have the extremely simple setup of > two frames with two clocks each, and a single light source, > as follows: > > Frame A > [clockA1]---------x----------[clockA2] > Source S > [clockB1]---------x----------[clockB2] > Frame B > > x is of course the distance measured in each frame by a > standard ruler laid at rest relative to the frame in which > the measurement is being made. No mystery here! > > We can let Frame B move to the right relative to Frame A, > and we can attach the source to Frame A. > > We can keep "simplicity mode" turned on by letting the two > origin clocks (A1 & B1) read ZERO at the start, as follows: > > Frame A > [0]---------x----------[clockA2] > Source S~~>light > [0]---------x----------[clockB2] -->v > Frame B > > As shown above, whenever the light is emitted from S towards the > two distant clocks (A2 & B2), the two origin clocks will start > on time zero. At this point, the two distant clocks are not yet > started. Indeed, we have yet to give the time readings for these > two clocks. > > Here is what we have so far: > > Frame A > [0]---------x----------[?] > Source S~~>light > [0]---------x----------[?] -->v > Frame B > > My original question can now be rephrased as follows: > > What exactly does Einstein's definition call for in the case > of the two distant clocks? That is, what times must be placed > on their faces prior to the release of the light from S? > > This is the part that I am having difficulty with. > > Hope this answers your question. > > ~~RA~~ > (You may still need Tom's or PDiddy's help!?) > (But I would strongly suggest _not_ asking for > help from either Andie-boy or Prof. Seto!)
From: eric gisse on 14 Jan 2010 14:57 Da Do Ron Ron wrote: [...] > What exactly does Einstein's definition call for in the case > of the two distant clocks? That is, what times must be placed > on their faces prior to the release of the light from S? The times are dictated by what the respective observers see, which is further dictated by light speed. Read PD's response. > > This is the part that I am having difficulty with. > > Hope this answers your question. > > ~~RA~~ > (You may still need Tom's or PDiddy's help!?) > (But I would strongly suggest _not_ asking for > help from either Andie-boy or Prof. Seto!)
From: Androcles on 14 Jan 2010 18:17 "Henry Wilson DSc" <..@..> wrote in message news:5p6vk55ehpu6na7q78jdufmhjeafc6i6rv(a)4ax.com... > On Mon, 11 Jan 2010 05:59:13 -0800 (PST), kenseto <kenseto(a)erinet.com> > wrote: > >>Why physicists refuse to measure the one-way speed of light directly? >>The answer: >>The one-way speed of light is physical distance dependent. >>BTW that's why they invented a new definition for a meter length: 1 >>meter=1/299,792,458 light second >>Using this definition the one-way speed of light is c by definition. >> >>Ken Seto > > Why all the confusion? > > 'c' is a universal constant with dimensions L/T Bullshit. When Michelson measured it the speed depended on the voltage across the carbon arc. UV is faster than IR. You'll never be an engineer.
From: eric gisse on 14 Jan 2010 18:59
...@..(Henry Wilson DSc) wrote: [...] > For instance, if a moving source (MS) OWLS experiment was to be carried > out over 3000 metres the travel time would be around 10^-5 seconds. If the > source were subsequently made to move at 30 m/s, the travel time would > change by only 1 part in 10^7...so the travel time has to be measured to > considerably better that 1 part in 10^12 to be meaningful. > Since this type of experiment involves TWO synched clocks, this is asking > quite a lot. Its' been done, a fact that you are clearly unaware. Such precision is easily attainable, and you gave a really slow source velocity. That you aren't interested in the literature simply means you don't care about science. [snip rest] |