Experiment to test mutual time dilation
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From: BURT on 13 Aug 2010 00:35 On Aug 12, 8:33 pm, "Inertial" <relativ...(a)rest.com> wrote: > "Tony M" wrote in message > > news:82a8d94c-94e9-4496-8d3f-9fa338036613(a)x21g2000yqa.googlegroups.com... > > >The same goes for the other combinations. If we apply your theory to 3 > >observers none will agree with (all) each-other's observations. > > Ken's 'theory' is self-contradictory nonsense. But he's an ignorant troll > and won't admit it even when it is so clearly pointed out (some of it can > 'work' for 2 bodies, but put in a third and it falls apart). Please explain Relativity along with everyone else. Mitch Raemsch
From: kenseto on 13 Aug 2010 09:06 On Aug 12, 11:01 pm, Tony M <marc...(a)gmail.com> wrote: > On Aug 4, 1:48 pm, kenseto <kens...(a)erinet.com> wrote: > > > > > > > On Aug 4, 11:33 am, Tony M <marc...(a)gmail.com> wrote: > > > > On Aug 4, 9:57 am, kenseto <kens...(a)erinet.com> wrote: > > > > > So every observer > > > > does not know if the observed clock is running slow or fast compared > > > > to his clock. This means that he must include both possibilities when > > > > predicting the rate of an observed clock as follows: > > > > Observed clcok runs slow: > > > > Delta(t')=gamma*Delta(t) > > > > Observed clock runs fast: > > > > Delta(t')=Delta(t)/gamma > > > > Ken, why not Delta(t)/gamma <= Delta(t') <= gamma*Delta(t)? Think > > > about it! > > > No....Delta(t')=gamma*Delta(t) means that the passage of > > Delta(t') on the t' clock is equal to the passage of > > gamma*Delta(t) on the t clock....that means that the t' clock is > > running slower than the t clock. > > Similarly...Delta(t')=Delta(t)/gamma means that the passage of > > Delta(t') on the t' clock is equal to the passage of > > Delta(t)/gamma on the t clock....that means that the t' clock is > > running faster than the t clock. > > Ken, I believe that's backwards, but let's try the following exercise. No it is not backward. > > Let's use 3 observers, A, B and C, with their time intervals tA, tB > and tC, all moving with the same relative velocity v, so we have the > same gamma between each pair of observers. (Yes, that's possible.) No....this is not possible. Why? Because the passage of an A second does not correspond to the passage of a B second and the passage of a C second. Each observer must make his won determination using his clock second to determine velocity and the gamma factor. In other words you can't compare gamma factor between frames. A would predict B's rate as follows: T_b=T_a/gamma_ab OR T_b=(gamma_ab)T_a A would predict C's rate as follows: T_c=T_a/gamma_ac OR T_c=Gamma_ac(T_ac) What this mean is that each observer must use his own clock second to determine the rate of an observed clock. > > From your theory the below observations should all be true: > Observer A would measure: > (1) tB=tA*gamma or (2) tB=tA/gamma > (3) tC=tA*gamma or (4) tC=tA/gamma > Observer B would measure: > (5) tC=tB*gamma or (6) tC=tB/gamma > > Now, if (1) and (3) are true that means tB=tC, which contradicts both > (5) and (6); observers A and B would disagree on their observations. > If we take (1) and (4) as true then tB=tC*gamma^2, which contradicts > both (5) and (6); observers A and B would disagree on their > observations. > The same goes for the other combinations. If we apply your theory to 3 > observers none will agree with (all) each-other's observations. At > least in SR they agree to disagree.- Hide quoted text - > > - Show quoted text -
From: Tony M on 13 Aug 2010 12:02 On Aug 13, 9:06 am, kenseto <kens...(a)erinet.com> wrote: > On Aug 12, 11:01 pm, Tony M <marc...(a)gmail.com> wrote: > > > > > > > On Aug 4, 1:48 pm, kenseto <kens...(a)erinet.com> wrote: > > > > On Aug 4, 11:33 am, Tony M <marc...(a)gmail.com> wrote: > > > > > On Aug 4, 9:57 am, kenseto <kens...(a)erinet.com> wrote: > > > > > > So every observer > > > > > does not know if the observed clock is running slow or fast compared > > > > > to his clock. This means that he must include both possibilities when > > > > > predicting the rate of an observed clock as follows: > > > > > Observed clcok runs slow: > > > > > Delta(t')=gamma*Delta(t) > > > > > Observed clock runs fast: > > > > > Delta(t')=Delta(t)/gamma > > > > > Ken, why not Delta(t)/gamma <= Delta(t') <= gamma*Delta(t)? Think > > > > about it! > > > > No....Delta(t')=gamma*Delta(t) means that the passage of > > > Delta(t') on the t' clock is equal to the passage of > > > gamma*Delta(t) on the t clock....that means that the t' clock is > > > running slower than the t clock. > > > Similarly...Delta(t')=Delta(t)/gamma means that the passage of > > > Delta(t') on the t' clock is equal to the passage of > > > Delta(t)/gamma on the t clock....that means that the t' clock is > > > running faster than the t clock. > > > Ken, I believe that's backwards, but let's try the following exercise. > > No it is not backward. > > > > > Let's use 3 observers, A, B and C, with their time intervals tA, tB > > and tC, all moving with the same relative velocity v, so we have the > > same gamma between each pair of observers. (Yes, that's possible.) > > No....this is not possible. Why? Because the passage of an A second > does not correspond to the passage of a B second and the passage of a > C second. > Each observer must make his won determination using his clock second > to determine velocity and the gamma factor. In other words you can't > compare gamma factor between frames. > A would predict B's rate as follows: > T_b=T_a/gamma_ab > OR > T_b=(gamma_ab)T_a > > A would predict C's rate as follows: > T_c=T_a/gamma_ac > OR > T_c=Gamma_ac(T_ac) > What this mean is that each observer must use his own clock second to > determine the rate of an observed clock. > > > > > > > From your theory the below observations should all be true: > > Observer A would measure: > > (1) tB=tA*gamma or (2) tB=tA/gamma > > (3) tC=tA*gamma or (4) tC=tA/gamma > > Observer B would measure: > > (5) tC=tB*gamma or (6) tC=tB/gamma > > > Now, if (1) and (3) are true that means tB=tC, which contradicts both > > (5) and (6); observers A and B would disagree on their observations. > > If we take (1) and (4) as true then tB=tC*gamma^2, which contradicts > > both (5) and (6); observers A and B would disagree on their > > observations. > > The same goes for the other combinations. If we apply your theory to 3 > > observers none will agree with (all) each-other's observations. At > > least in SR they agree to disagree.- Hide quoted text - > > > - Show quoted text -- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text - "Yes, thats possible." I included that comment in anticipation of your objection. Ken, relative velocity is mutual, v_ab=v_ba, regardless of the length of ones second. Why? Because velocity is distance/time, and both distance and time vary by the same factor. If one measures longer time he also measures longer distance; the ratio stays the same, right? And yes, you can have 3 observers moving at the same relative velocity, v_ab=v_ac=v_bc and therefore the same gamma; gamma_ab= gamma_ac= gamma_bc. Gamma is a function of only the relative velocity. In fact, in 3 spatial dimensions, you can have 4 observers moving at the same relative velocity. Try harder!
From: kenseto on 13 Aug 2010 14:18 On Aug 13, 12:02 pm, Tony M <marc...(a)gmail.com> wrote: > On Aug 13, 9:06 am, kenseto <kens...(a)erinet.com> wrote: > > > > > > > On Aug 12, 11:01 pm, Tony M <marc...(a)gmail.com> wrote: > > > > On Aug 4, 1:48 pm, kenseto <kens...(a)erinet.com> wrote: > > > > > On Aug 4, 11:33 am, Tony M <marc...(a)gmail.com> wrote: > > > > > > On Aug 4, 9:57 am, kenseto <kens...(a)erinet.com> wrote: > > > > > > > So every observer > > > > > > does not know if the observed clock is running slow or fast compared > > > > > > to his clock. This means that he must include both possibilities when > > > > > > predicting the rate of an observed clock as follows: > > > > > > Observed clcok runs slow: > > > > > > Delta(t')=gamma*Delta(t) > > > > > > Observed clock runs fast: > > > > > > Delta(t')=Delta(t)/gamma > > > > > > Ken, why not Delta(t)/gamma <= Delta(t') <= gamma*Delta(t)? Think > > > > > about it! > > > > > No....Delta(t')=gamma*Delta(t) means that the passage of > > > > Delta(t') on the t' clock is equal to the passage of > > > > gamma*Delta(t) on the t clock....that means that the t' clock is > > > > running slower than the t clock. > > > > Similarly...Delta(t')=Delta(t)/gamma means that the passage of > > > > Delta(t') on the t' clock is equal to the passage of > > > > Delta(t)/gamma on the t clock....that means that the t' clock is > > > > running faster than the t clock. > > > > Ken, I believe that's backwards, but let's try the following exercise.. > > > No it is not backward. > > > > Let's use 3 observers, A, B and C, with their time intervals tA, tB > > > and tC, all moving with the same relative velocity v, so we have the > > > same gamma between each pair of observers. (Yes, that's possible.) > > > No....this is not possible. Why? Because the passage of an A second > > does not correspond to the passage of a B second and the passage of a > > C second. > > Each observer must make his won determination using his clock second > > to determine velocity and the gamma factor. In other words you can't > > compare gamma factor between frames. > > A would predict B's rate as follows: > > T_b=T_a/gamma_ab > > OR > > T_b=(gamma_ab)T_a > > > A would predict C's rate as follows: > > T_c=T_a/gamma_ac > > OR > > T_c=Gamma_ac(T_ac) > > What this mean is that each observer must use his own clock second to > > determine the rate of an observed clock. > > > > From your theory the below observations should all be true: > > > Observer A would measure: > > > (1) tB=tA*gamma or (2) tB=tA/gamma > > > (3) tC=tA*gamma or (4) tC=tA/gamma > > > Observer B would measure: > > > (5) tC=tB*gamma or (6) tC=tB/gamma > > > > Now, if (1) and (3) are true that means tB=tC, which contradicts both > > > (5) and (6); observers A and B would disagree on their observations. > > > If we take (1) and (4) as true then tB=tC*gamma^2, which contradicts > > > both (5) and (6); observers A and B would disagree on their > > > observations. > > > The same goes for the other combinations. If we apply your theory to 3 > > > observers none will agree with (all) each-other's observations. At > > > least in SR they agree to disagree.- Hide quoted text - > > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text - > > "Yes, thats possible." I included that comment in anticipation of > your objection. > Ken, relative velocity is mutual, v_ab=v_ba, regardless of the length > of ones second. Why? Because velocity is distance/time, and both > distance and time vary by the same factor. If one measures longer time > he also measures longer distance; the ratio stays the same, right? No....In that case you can't compare Tc as predicted by A with Tc as predicted by B. Even if you assume that A, B and C measure the same velocity.... A predicts the following for Tc and Tb: Tc<Ta OR Tc>Ta Tb<Ta OR Tb>Ta This means that you can't assume that Tc=Tb. This means that your assumed contradiction based on what B predicts for Tc is not valid. In any case how does your gedanken support SR? SR predicts from A's point of view: Tc=Ta/gamma Tb=Ta/gamma Therefore Tc=Tb From B's point of view: Tc=Tb/gamma This means that Tc=/=Tb This means that B's point of view is directly contradicting A's point of view. Ken Seto > And yes, you can have 3 observers moving at the same relative > velocity, v_ab=v_ac=v_bc and therefore the same gamma; gamma_ab= > gamma_ac= gamma_bc. Gamma is a function of only the relative velocity. > In fact, in 3 spatial dimensions, you can have 4 observers moving at > the same relative velocity. Try harder!- Hide quoted text - > > - Show quoted text -
From: Tony M on 13 Aug 2010 15:36
On Aug 13, 2:18 pm, kenseto <kens...(a)erinet.com> wrote: > On Aug 13, 12:02 pm, Tony M <marc...(a)gmail.com> wrote: > > > > > > > On Aug 13, 9:06 am, kenseto <kens...(a)erinet.com> wrote: > > > > On Aug 12, 11:01 pm, Tony M <marc...(a)gmail.com> wrote: > > > > > On Aug 4, 1:48 pm, kenseto <kens...(a)erinet.com> wrote: > > > > > > On Aug 4, 11:33 am, Tony M <marc...(a)gmail.com> wrote: > > > > > > > On Aug 4, 9:57 am, kenseto <kens...(a)erinet.com> wrote: > > > > > > > > So every observer > > > > > > > does not know if the observed clock is running slow or fast compared > > > > > > > to his clock. This means that he must include both possibilities when > > > > > > > predicting the rate of an observed clock as follows: > > > > > > > Observed clcok runs slow: > > > > > > > Delta(t')=gamma*Delta(t) > > > > > > > Observed clock runs fast: > > > > > > > Delta(t')=Delta(t)/gamma > > > > > > > Ken, why not Delta(t)/gamma <= Delta(t') <= gamma*Delta(t)? Think > > > > > > about it! > > > > > > No....Delta(t')=gamma*Delta(t) means that the passage of > > > > > Delta(t') on the t' clock is equal to the passage of > > > > > gamma*Delta(t) on the t clock....that means that the t' clock is > > > > > running slower than the t clock. > > > > > Similarly...Delta(t')=Delta(t)/gamma means that the passage of > > > > > Delta(t') on the t' clock is equal to the passage of > > > > > Delta(t)/gamma on the t clock....that means that the t' clock is > > > > > running faster than the t clock. > > > > > Ken, I believe that's backwards, but let's try the following exercise. > > > > No it is not backward. > > > > > Let's use 3 observers, A, B and C, with their time intervals tA, tB > > > > and tC, all moving with the same relative velocity v, so we have the > > > > same gamma between each pair of observers. (Yes, that's possible.) > > > > No....this is not possible. Why? Because the passage of an A second > > > does not correspond to the passage of a B second and the passage of a > > > C second. > > > Each observer must make his won determination using his clock second > > > to determine velocity and the gamma factor. In other words you can't > > > compare gamma factor between frames. > > > A would predict B's rate as follows: > > > T_b=T_a/gamma_ab > > > OR > > > T_b=(gamma_ab)T_a > > > > A would predict C's rate as follows: > > > T_c=T_a/gamma_ac > > > OR > > > T_c=Gamma_ac(T_ac) > > > What this mean is that each observer must use his own clock second to > > > determine the rate of an observed clock. > > > > > From your theory the below observations should all be true: > > > > Observer A would measure: > > > > (1) tB=tA*gamma or (2) tB=tA/gamma > > > > (3) tC=tA*gamma or (4) tC=tA/gamma > > > > Observer B would measure: > > > > (5) tC=tB*gamma or (6) tC=tB/gamma > > > > > Now, if (1) and (3) are true that means tB=tC, which contradicts both > > > > (5) and (6); observers A and B would disagree on their observations.. > > > > If we take (1) and (4) as true then tB=tC*gamma^2, which contradicts > > > > both (5) and (6); observers A and B would disagree on their > > > > observations. > > > > The same goes for the other combinations. If we apply your theory to 3 > > > > observers none will agree with (all) each-other's observations. At > > > > least in SR they agree to disagree.- Hide quoted text - > > > > > - Show quoted text -- Hide quoted text - > > > > - Show quoted text -- Hide quoted text - > > > > - Show quoted text - > > > "Yes, thats possible." I included that comment in anticipation of > > your objection. > > Ken, relative velocity is mutual, v_ab=v_ba, regardless of the length > > of ones second. Why? Because velocity is distance/time, and both > > distance and time vary by the same factor. If one measures longer time > > he also measures longer distance; the ratio stays the same, right? > > No....In that case you can't compare Tc as predicted by A with Tc as > predicted by B. Even if you assume that A, B and C measure the same > velocity.... A predicts the following for Tc and Tb: > Tc<Ta OR Tc>Ta > Tb<Ta OR Tb>Ta Don't change the equations Ken, Tc<Ta OR Tc>Ta is not the same as Tc=Ta/gamma OR Tc=Ta*gamma. In your original equations you CAN compare the times and see the contradictions. > This means that you can't assume that Tc=Tb. This means that your > assumed contradiction based on what B predicts for Tc is not valid. > In any case how does your gedanken > support SR? It's not trying to, just pointing out the flaws in your theory. >SR predicts from A's point of view: > Tc=Ta/gamma > Tb=Ta/gamma > Therefore Tc=Tb > > From B's point of view: > Tc=Tb/gamma > > This means that Tc=/=Tb > This means that B's point of view is directly contradicting A's point > of view. > Use that thought process on your theory. You just proved it wrong yourself. > Ken Seto > > > > > And yes, you can have 3 observers moving at the same relative > > velocity, v_ab=v_ac=v_bc and therefore the same gamma; gamma_ab= > > gamma_ac= gamma_bc. Gamma is a function of only the relative velocity.. > > In fact, in 3 spatial dimensions, you can have 4 observers moving at > > the same relative velocity. Try harder!- Hide quoted text - > > > - Show quoted text -- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text - |