colp, why did AE use the word "relativity"?
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From: Daryl McCullough on 22 Jun 2010 17:27 colp says... > >On Jun 22, 11:48=A0am, stevendaryl3...(a)yahoo.com (Daryl McCullough) >wrote: >> colp says... > >> >At this point I'll ask a question with respect to the second premise: >> >What elapsed time will be shown by an clock traveling at constant >> >speed of 0.866c when a stationary local clock shows an elapsed time of >> >1 second? >> >> To compute the elapsed time on a clock, you have to specify the >> end points: elapsed time from what event to what other event? >> You need to specify the end points in order to compute elapsed time. > >OK, how about this one? > >What elapsed time will be shown by an clock traveling at constant >speed of 0.866c when a stationary local clock shows an elapsed time of >1 second, and the two clocks are at the same place at the beginning of >that second? You've only specified one endpoint for the traveling clock, namely the point at which both clocks are together at the same location. To compute the elapsed time for the traveling clock, you have to give a starting point and and endpoint. Then you can compute the elapsed time between those two points. I assume that what you mean is something like this (in the language of events): Let E1 be the event at which both clocks are at the same location. (For simplicity, let's assume that both clocks show time 0 seconds at this event). Let E2 be the event at which the "stationary" clock shows 1 second. Let E3 be some event taking place at the "traveling" clock that is simultaneous with E2. The problem with this is that you have to say which coordinate is used to decide simultaneity. If you give a coordinate system, then we can use that to find the event E3 that is simultaneous with E2, and then we can compute the elapsed time on the "traveling" clock between events E1 and E3. -- Daryl McCullough Ithaca, NY
From: colp on 22 Jun 2010 18:51 On Jun 23, 9:15 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > colp says... > > >Right. The thing is that observations of time that are made in one > >inertial frame can be used as the basis of of an inference which > >involves a different inertial frame. For example, a twin can know > >whether on not time is dilated for the other twin (with respect to his > >own time) if he can measure radio pulses from that twin that are sent > >each time the other twin's clock ticks and he knows his velocity > >relative to the other twin. That knowledge can can then then be used > >as the basis of an inference which involves knowledge of different > >inertial frame. > > That certainly is true, but you have to remember that time dilation > is a relationship between a clock and a coordinate system. It is *not* > a relationship between two clocks. So why is a coordinate system relevant in terms of the premises of SR? After all, co-ordinate systems aren't explicitly referred to by your second premise. <quote> 2. An ideal clock traveling at speed v for time period t will show an elapsed time of T = t square-root(1-(v/c)^2). </quote> I mean, coordinate systems are only descriptions of reality. They don't change anything in terms of the real elements, namely that two clocks travelling in inertial frames are at one point in time located at the same point, and are later separated by a known distance when the time on one of the clocks shows that one second has elapsed from the time when the clocks were co-located.
From: colp on 22 Jun 2010 19:22 On Jun 23, 9:27 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > colp says... > >What elapsed time will be shown by an clock traveling at constant > >speed of 0.866c when a stationary local clock shows an elapsed time of > >1 second, and the two clocks are at the same place at the beginning of > >that second? > > You've only specified one endpoint for the traveling clock, namely > the point at which both clocks are together at the same location. > To compute the elapsed time for the traveling clock, you have to > give a starting point and and endpoint. Then you can compute the > elapsed time between those two points. > > I assume that what you mean is something like this (in the > language of events): > > Let E1 be the event at which both clocks are at the same location. > (For simplicity, let's assume that both clocks show time 0 seconds > at this event). O.K. Let's call this location x = 0. > > Let E2 be the event at which the "stationary" clock shows 1 second. > > Let E3 be some event taking place at the "traveling" clock that > is simultaneous with E2. O.K. Lets define E3 to be when the traveling clock reaches location x = d d = vt v = 0.866c t = 1 second > > The problem with this is that you have to say which coordinate is > used to decide simultaneity. How does that work? I mean, simultaneity occurred at time t = 0, so why can't we say that it also occurs when the travelling clock reaches x = d at event E3?
From: Androcles on 22 Jun 2010 19:37 <paparios(a)gmail.com> wrote in message news:1c2e51fb-9e4e-4c37-bc87-0dea187a8ddd(a)i28g2000yqa.googlegroups.com... On 22 jun, 14:44, colp <c...(a)solder.ath.cx> wrote: > On Jun 22, 11:48 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) > > > To compute the elapsed time on a clock, you have to specify the > > end points: elapsed time from what event to what other event? > > You need to specify the end points in order to compute elapsed time. > > OK, how about this one? > > What elapsed time will be shown by an clock traveling at constant > speed of 0.866c when a stationary local clock shows an elapsed time of > 1 second, and the two clocks are at the same place at the beginning of > that second? The Lorentz equations are: x=(x�+vt�)/(sqrt(1-v^2/c^2)) ; t=(t�+vx�/c^2)/(sqrt(1-v^2/c^2)) x'=(x-vt)/(sqrt(1-v^2/c^2)) ; t�=(t-vx/c^2)/(sqrt(1-v^2/c^2)) So if (x,t) are the coordinates of the stationary clock and (x',t') are the coordinates of the traveling clock, then we have the following: When x=0 and t=0 both clocks are colocated and then x'=0 and t'=0 (from the second group) When t=1 sec, x=0, so from the second equation group we have x'=(-259980)/0.5=-519960 km t'=1/0.5=2 sec Miguel Rios ============================================ beta = 1/sqrt(1-v^2/v^2) = 1/sqrt(1-0.866^2) = 1/sqrt(1-0.75) = 1/sqrt(0.25) = 1/0.5 = 2 Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img54.gif tau = t * sqrt(1-v^2/c^2) tau = t * 1/beta tau = 1 * 1/2 tau = 0.5 seconds Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img61.gif Moving clocks run slow in relativity. papafuckingmoron can't manage simple schoolboy algebra!!!!! ROFLMAO!!!!
From: Daryl McCullough on 22 Jun 2010 19:43
Daryl McCullough says... >Let E1 be the event at which both clocks are at the same location. >(For simplicity, let's assume that both clocks show time 0 seconds >at this event). > >Let E2 be the event at which the "stationary" clock shows 1 second. > >Let E3 be some event taking place at the "traveling" clock that >is simultaneous with E2. > >The problem with this is that you have to say which coordinate is >used to decide simultaneity. If you give a coordinate system, >then we can use that to find the event E3 that is simultaneous >with E2, and then we can compute the elapsed time on the "traveling" >clock between events E1 and E3. Let me illustrate how simultaneity completely changes the judgment of which twin is older. Once again, let (x,t) be the coordinate system used by the "stationary" twin, and (x',t') be the coordinate system used by the "traveling" twin. Let the relative velocity between the twins be 0.866c, with the traveing twin moving in the positive x-direction, as measured by the "stationary" twin. That means the relationship between the two coordinate systems is: x' = gamma (x-vt) = 2 (x - 0.866 t) = 2x - 1.732 t t' = gamma (t - vx/c^2) = 2 (t - 0.866 x/c^2) = 2t - 1.732 x (measuring distance in light-seconds, so that light has speed 1) Now, as above, we let E1 be the event at which both twins are at the same location. The coordinates for this event are: x_1 =0, t_1 =0 x_1'=0, t_1'=0 E2 is the event at which the "stationary" twin's clock shows t=1. So the coordinates for this event are: x_2=0, t_2=1 x_2'= -1.732, t_2' = 2 Let E3 be the event that takes place at the traveling twin, but is simultaneous with E2, as measured in the stationary twin's frame. We have to do a little calculation to compute the coordinates for E3. Let x_3, t_3 be its coordinates in the stationary twin's frame, and let x_3', t_3' be its coordinates in the traveling twin's frame. What we know is that it takes place at the location of the traveling twin. That means that x_3' = 0. We also know that in the frame of the stationary frame, it is simultaneous with E2. That means that t_3 = t_2 = 1 second. So we have: x_3' = 2 x_3 - 1.732 t_3 So, since x_3' = 0, t_3 = 1, we have: 0 = 2 x_3 - 1.732 So x_3 = 0.866 Now, we can compute t_3': t_3' = 2 t_3 - 0.866 x_3 = 2 - 0.866 * 0.866 = 2 - 0.75 = 1.25 So E3 has coordinates x_3=0.866, t_3=1 x_3'=0, t_3'=1.25 Now we can compute the elapsed time for the traveling twin in going from E1 to E3: it is given by square-root(1-(v/c)^2) t_3 = 1/2 So the traveling twin ages 1/2 second between events E1 and E3. The stationary twin ages 1 second between events E1 and E2. So the stationary twin could reason: "At the same time that I have aged 1 second, the other twin has aged only 1/2 second". WE ARE NOT DONE YET! The choice of event E3 was made by the criterion that E3 and E2 must be simultaneous in the frame of the stationary twin. We could, instead, choose an event, E4 that is simultaneous with E2 in the frame of the *traveling* twin. That means we want an event with coordinates (x_4, t_4) and (x_4', t_4') such that: x_4' = 0 t_4' = t_2' = 2 seconds We can compute x_4 and t_4 via: x_4' = 2 x_4 - 1.732 t_4 So we have: 0 = 2 x_4 - 1.732 t_4 ==> t_4 = 1.155 x_4 t_4' = 2 t_4 - 1.732 x_4 = 2 * 1.155 x_4 - 1.732 x_4 = 0.578 x_4 So we have: 2 = 0.578 x_4, so x_4 = 3.46 t_4 = 1.155 * 3.46 = 4 seconds Now we can compute the elapsed time for the traveling twin in going from E1 to E3: it is given by square-root(1-(v/c)^2) t_4 = 1/2 * 4 = 2 seconds. Since in the frame of the traveling twin, the traveling twin has velocity 0, we conclude that: Between events E_1 and E_4, the traveling twin ages 2 seconds. Between events E_1 and E_2, the stationary twin ages 1 second. So the traveling twin could reason: "At the same time that I have aged 2 seconds, the other twin has aged only 1 second". So, there we have it: the traveling twin thinks that E_2 occurs at the same time as E_4, and the stationary twin thinks that E_2 occurs at the same time as E_3. The two twins agree that: 1. The elapsed time for the stationary twin between E_1 and E_2 is 1 second. 2. The elapsed time for the traveling twin between E_1 and E_3 is 1/2 second. 3. The elapsed time for the traveling twin between E_1 and E_4 is 2 seconds. They *both* agree with all those facts. Where they disagree is whether E_2 and E_4 are simultaneous, or E_2 and E_3 are simultaneous. -- Daryl McCullough Ithaca, NY |