If a train is not aging faster than the station
in [Relativity]
|
Prev: Vereinheitlichungsmodell von allgemeiner Relativität und Quantenmechanik
Next: The Ether-nal Ether-eal SpaceTime
From: Mathal on 12 Aug 2010 11:45 On Aug 12, 8:11 am, "Inertial" <relativ...(a)rest.com> wrote: > "Mathal" wrote in message > > news:4caa41b7-8b88-4efb-9c24-6100a7ed11d9(a)x18g2000pro.googlegroups.com... > > On Aug 11, 4:13 pm, "Inertial" <relativ...(a)rest.com> wrote: > > > > > > > "Mathal" wrote in message > > >news:dbd3420d-cbc3-45d1-a76d-97638747506d(a)u31g2000pru.googlegroups.com.... > > > On Aug 11, 7:08 am, "Inertial" <relativ...(a)rest.com> wrote: > > > > "Mathal" wrote in message > > > >news:f60dfe9a-c079-4399-9c3f-ae055a0e0469(a)s24g2000pri.googlegroups.com.... > > > > >On Aug 10, 11:57 pm, "Inertial" <relativ...(a)rest.com> wrote: > > > >> "Mathal" wrote in message > > > > >>news:9424dbce-5320-4adc-a119-279eed017661(a)g21g2000prn.googlegroups.com... > > > >> On Aug 9, 5:28 pm, "Inertial" <relativ...(a)rest.com> wrote: > > > > >> >> You really need to think some more here .. you don't yet get SR.. > > > > >> > When it is said that there is no preferred frame in SR what is > > > >> > being > > > >> >referred >to is the FACT that, with two isolated frames in motion > > > >> >WRT > > > >> >each > > > >> >other, neither frame can deduce who is in motion WRT the other. > > > > >> Yes .. we know what relative motion is and what the principle of > > > >> relativity > > > >> is. > > > > >> >Both frames will perceive the other frame as operating at a slower > > > >> > pace than their own BUT only by taking their own frame as > > > >> > motionless, WRT the other frame, will the information gathered > > > >> > from the other frame be in accord with what the SR calculates > > > >> > and what physics stipulates is viable. > > > > >> No .. it doesn't matter WHAT frame you use .. what you get is in > > > >> accord > > > >> with > > > >> what SR calculates > > > > >In this scenario there are 4 calculations that can be made. > > > > At least > > > > > 1. Frame 1 can treat itself at rest and Frame 2 moving. > > > > 2. Frame 1 can treat itself as moving and Frame 2 as > > > > motionless > > > > 3. Frame 2 can treat itself at rest and Frame 1 moving. > > > > That's the same as 2, and the opposite of 1 > > > > < 4. Frame 2 can treat itself as moving and Frame 1 as > > > motionless > > > > That's the same as 1 and the opposite of 2 > > > > > Only 2 of these calculations will conform with what the frame > > > > observes in the other frame. > > > > No .. they all will 'conform' because they are all equivalent. > > > > > If the objects are moving away from each > > > > other the incorrect use of SR will tell you the other frame is sending > > > > blue-shifted light. > > > > You really are confused. Perhaps you've had troubles in correctly using > > > SR > > > in the past. That is your problem .. not that of SR. > > > > [snip] > > > > > My understanding of SR is much clearer than yours is, clearly. > > > > Nope.. But keep trying > > >If the observing frame makes the calculations for time using himself > > >as the rest frame what he measures of the other frame will conform > > >with the SR calculations of what should be observed. > > > Of course > > > >If the observing frame uses the other frame as the rest frame his > > >calculations for time will have him expecting blue shifted light. > > > They will give exactly what the moving frame would see.. > > > As you've not specified where the light source is, we don't know if it is > > red or blue shifted (or not shifted at all). > > > > The > > >observer always has to treat himself as motionless > > > ... in his own rest frame .. that's pretty much obvious. but he doesn't > > need to treat his frame as at rest. He DOES need to know what the hell > > he's > > doing, however (ie which frame he IS treating as at rest) in order to make > > sense of what he calculates. > > > > for his > > >observations of the other frame to conform with what SR stipulates > > >that he will SEE. > > > As he knows (trivially) that he IS motionless in his own frame .. if he > > wants to calculate what he will see he would want to calculate what a > > motionless observer in his frame will see. It doesn't matter which frame > > is > > treated as at rest or which it moving, or if both are treated as moving > > etc. > > Its all relative and the lLorentz Transforms ensure you get the correct > > result regardless. > > > > I'm not sure whether it's English or SR or both that you are > > >having difficulty with. > > > Neither. English and SR are just fine for me.. Its just your ignorance > > that > > I'm having troubles with. You really shouldn't attempt to speak with > > authority about things you don't fully understand. SR is one of those > > things for you. > > Given- two frames traelling directly away from each other (to > > eliminate transverse velocity from the tableau) > > Fine .. so assuming a 1D case .. that's quite common. > > > with knowledge of their actual velocity WRT > > each other. They start out in the same place each > > with one of two synchronized clocks. One frame A > > moves away at velocity X. X is greater than 0. > > And the other with opposite velocity > > > They each have a laser pointing device and the > > means to measure the frequency of the laser light. > > Fine > > > They know the frequency is the same for both lasers. > > In the rest frame of said laser > > > They point their laser at the other frame. > > Observers in both frames measure the same > > red-shift in frequency. > > Yes they do > > > The observer in the "rest" frame > > What rest frame? You've not given that label to any frame yet > > > knows the other frame is moving WRT his/her > > frame so the red-shift is no surprise. > > Of course > > > The observer in the moving frame > > What moving frame? You've not given that label to any frame yet > > > knows he is moving WRT the other frame and > > understands SR. This means he knows his frame is > > operating at a slower rate than the frame that is > > at rest. > > Not exactly .. both frame operate at the correct rate. Neither is 'slow'. > But observers in one frame will measure the rate of a clock in the other > frame as being slower. > > But yes .. he would know that observers in the other frame will measure his > clock as ticking slower. Just as observers in his frame measure the other > frame as slower > > > His clock is moving slower than the other frame's > > clock. > > It depends on who measures it > > > He knows this but when he uses SR to interperet what he is > > seeing he has to ignore this information and treat his/her > > frame as stationary and let the velocity be in the other frame. > > No .. he doesn't *have* to do that .. he can calculate things from whatever > frame he wants. As long as he does the appropriate transformations to what > he would see (being at rest in his frame) > > > This allows what he is perceiving to conform to > > how he knows the universe operates. > > At the same time he is aware that the other > > frame is NOT operating in a slower time pace > > Neither is operating at a slower time pace. But each MEASURES the others > clocks as ticking slower. > > > because he knows he is moving WRT this frame. > > But he MEASURS the clocks in that other frame as ticking slower. > > > The mechanics of what is seen. I hope I can dummy this down enough > > for you. > > I already understand it, thanks. But it means I can see your confusion. You > think that 'slower time pace' is something that you can talk about > absolutely (i.e. that one is absolutely slower than the other). Its not. > That's your point of confusion > > >From the "real" rest frame's point of view > > What 'real' rest frame? There are not real rest frames in SR. > > > the moving frame is > >operating in a slower frame > > Every frame is measured as 'operating slower' by every other frame (so every > other frame. > > Every frame measure every other frame as operating slower than their own. > > > and so the light from the frame will be red-shifted WRT his frame. > > The light from a source at rest in each frame is measured as red shifted by > observers in the other frame .. as we have the light sources and observers > moving away from each other > > > No problem.From the moving frame's point of view SR stipulates he must > > calculate from a rest frame so he can't take into account the slower rate > > of his clock. > > He doesn't have a slower rate clock . All the clocks tick at their correct > rate in their own frame. > > He will measure other frames' clocks as ticking slower than his, and other > frames' clocks will measure his clock as ticking slower than theirs > > > He has to treat the other frame as moving and therefore operating at a > > slower pace than his time. > > You are very confused here. > > > He is aware that this is an illusion because he understands SR. > > When each sends and receives signals with a time check the > > reality of who is in the slower frame > > There is no 'in reality .. is in the slower frame. Neither is absolutely > slower than the other > > > is confirmed. After taking out the transit time for the signal > > to arrive the real moving frame will see he is moving in a > > slower time WRT the other frame, but he already knew that > > because he understands SR. > > This clearly applies to the train/station scenario. > > You are very confused here. You really need to get a 'handle' on what > "mutual" and "relative" mean. You just don't seem to get it yet. One frame moves away from the other- This moving frame will have a clock which measures it's time rate. It's time rate will be necessarily slower than the frame that is motionless WRT it. Clocks don't lie, in any frame they measure the rate that time passes in that frame but you don't want to get that because your house of cards would all fall down. It is interesting though, how easily you deceive yourself. I tried. You will remain clueless until you die in all probability. Mathal |