geistkiesel
Valued Senior Member
Absolute Velocity of Inertial Frames Using Time Signatures
ArrangementA and B moving toward each other with a relative velocity V(a) – V(b)= Vab. The clocks on A and B were synchronized wrt, C the common point of origin of A and B which were launched in opposite directions from the earth surface with known equal absolute velocities wrt C. The programmed trajectories were originally equal circles and where now A and B are headed toward each other in the configuration indicated below.
The Experiment
A emits signal at T0 = 10 with included time signature, “emitted at T0 = 10”.
A |-->----------------------------------------------------| B.
d1
A |-------------------------------------------->| B Rec’d A message at T1 = 11 B replies at T1 = 11 with included message: “B reply at 11;d1 = c(T1 - T0) = c(11-10)”
d2
|<---------------------------------------| A Rec’d the B reply at T2 = 11.5 and calculates, d2 = c(T2 – T1) = c(11.5 – 11) = .5c.
| VT |<-- A has moved a distance VT since first emitting the time/signed signal at T0 = 10.
Now, d1 – d2 = .5c = VT where T = the time of travel for A from the original emission time from A at T0 = 10 to current time T2 = 11.5. Hence, V(T2 – T0) = V(1.5) = .5(c) and ergo,
V = .5c/1.5 = 1/3c,
the absolute velocity of A wrt zero. QED.
This is the absolute velocity of the A inertial frame measured from an actual initial velocity equal to zero.
Explanation (if needed)
When B receives the A signal that included the message, “sent at 10”, B knows the distance between B’s postion at T1 wrt the A location when the A message was sent at T0; B thinks, “The light left A at 10 and received by me at 11, hence the time of flight for the pulse is T1 – T0 = 1 and the distance the light traveled was c(T1 – T0)”.
A does the same calculation when receiving the B reply. A could have calculated d1 from the information in the reply from A knowing what time his original message was emitted and the time the message was received by B.
ArrangementA and B moving toward each other with a relative velocity V(a) – V(b)= Vab. The clocks on A and B were synchronized wrt, C the common point of origin of A and B which were launched in opposite directions from the earth surface with known equal absolute velocities wrt C. The programmed trajectories were originally equal circles and where now A and B are headed toward each other in the configuration indicated below.
The Experiment
A emits signal at T0 = 10 with included time signature, “emitted at T0 = 10”.
A |-->----------------------------------------------------| B.
d1
A |-------------------------------------------->| B Rec’d A message at T1 = 11 B replies at T1 = 11 with included message: “B reply at 11;d1 = c(T1 - T0) = c(11-10)”
d2
|<---------------------------------------| A Rec’d the B reply at T2 = 11.5 and calculates, d2 = c(T2 – T1) = c(11.5 – 11) = .5c.
| VT |<-- A has moved a distance VT since first emitting the time/signed signal at T0 = 10.
Now, d1 – d2 = .5c = VT where T = the time of travel for A from the original emission time from A at T0 = 10 to current time T2 = 11.5. Hence, V(T2 – T0) = V(1.5) = .5(c) and ergo,
V = .5c/1.5 = 1/3c,
the absolute velocity of A wrt zero. QED.
This is the absolute velocity of the A inertial frame measured from an actual initial velocity equal to zero.
Explanation (if needed)
When B receives the A signal that included the message, “sent at 10”, B knows the distance between B’s postion at T1 wrt the A location when the A message was sent at T0; B thinks, “The light left A at 10 and received by me at 11, hence the time of flight for the pulse is T1 – T0 = 1 and the distance the light traveled was c(T1 – T0)”.
A does the same calculation when receiving the B reply. A could have calculated d1 from the information in the reply from A knowing what time his original message was emitted and the time the message was received by B.
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