If a train is not aging faster than the station
in [Physics]
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From: BURT on 11 Aug 2010 17:49 On Aug 10, 11:50 pm, "Inertial" <relativ...(a)rest.com> wrote: > "BURT" wrote in message > > news:a8d25f35-1c15-454e-9ca3-a705c65d391f(a)v6g2000prd.googlegroups.com... > > On Aug 10, 7:33 pm, "Inertial" <relativ...(a)rest.com> wrote: > > > > > > > "BURT" wrote in message > > >news:10c5468c-e7e1-4c01-9f80-57af901e7899(a)b4g2000pra.googlegroups.com... > > > >On Aug 10, 6:43 pm, "Inertial" <relativ...(a)rest.com> wrote: > > >> "BURT" wrote in message > > > >>news:1cab682b-e1dd-4c18-9909-e885fa5677ec(a)w15g2000pro.googlegroups.com... > > > >> >The idea of the paradox > > > >> There is no paradox > > > >> >is that they have mutual slowdowns but one > > >> > ages more than the other. > > > >> That doesn't happen with a train and station scenario. > > > >Any motion at all plugs into Gamma for time. > > > Both train and station are moving relative to each other. And relative to > > an observer halfway between them, they both move at the same speed but in > > opposite directions. Its all relative. > > > >As a thought experiment if the train could move at near light speed > > > ... relative to the station > > > Then the station is also moving at near light speed relative to the train. > > > > the train would age less. > > > Not exactly .. the train would be measured by observers at rest wrt the > > station as aging less > > And the station would be measured by observers at rest wrt the train as > > aging less > > Its mutual. Hence the term MUTUAL time dilation > > > > So how can such an imaginary train see when > > > passing the station the station aging less? > > > That's what i asked you to think about .. over a dozen times .. how can an > > observer on the train measure the ticking rate of the station clock. You > > refused to even attempt to answer.- Hide quoted text - > > > - Show quoted text - > >If the im aginary train moves near the speed of light its clock slows > >down compared to the station. > > You mean it is measured as slower by the station-frame observers > And the station clock is measured as slower by train-frame observer. > It all depends on who is doing the measuring > As I just told you > > > So how can it see the station age less > >as it passes(if it does?) > > It does, and I already explained it. If it is aging less how does it see the other aging less? Mitch Raemsch
From: Mathal on 11 Aug 2010 18:54 On Aug 10, 9:20 pm, Sam Wormley <sworml...(a)gmail.com> wrote: > Mathal > > Multiple frame are not requited... an observer frame can be > defined. All observed objects or events in relative motion > to the observer will exhibit phenomena such as Doppler shift, > time dilation, etc. Multiple -how many 2 or more? frame -typo should have been frames are not- you got that part right requited- anthropomorphic and nebulous. This portion didn't add anything. >an observer frame can be > defined. All observed objects or events in relative motion > to the observer will exhibit phenomena such as Doppler shift, > time dilation, etc. Precisely. As I've stated elsewhere SR in the context it is being used in this and most threads always describes how the 'moving' frame can be seen to conform to the physics of this universe from the 'rest' frame perspective. Do remember the caveat though that gravity can throw monkey wrenches into what is observed. SR still applies but it applies to the absolute velocity difference between the frames. Geosynchronous satellites are an example of that. Anyone that can see the satellite always sees it at the same distance and position and so, out of context no time dilation might be mistakenly inferred. Mathal
From: Inertial on 11 Aug 2010 19:13 "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.
From: Mathal on 12 Aug 2010 10:20 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 transvere velocity from the tableau) 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. They each have a laser pointing device and the means to measure the frequency of the laser light. They know the frequency is the same for both lasers.They point their laser at the other frame. Observerers in both frames measure the same red-shift in frequency. The observer in the "rest" frame knows the other frame is moving WRT his/her frame so the red-shift is no surprise. The observer in the moving frame 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. His clock is moving slower than the other frame's clock. 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. 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 because he knows he is moving WRT this frame. The mechanics of what is seen. I hope I can dummy this down enough for you. From the "real" rest frame's point of view the moving frame is operating in a slower frame and so the light from the frame will be red-shifted WRT his frame. 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 has to treat the other frame as moving and therefore operating at a slower pace than his time. 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 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. Mathal
From: Inertial on 12 Aug 2010 11:11
"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. |