- #1
arydberg
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Why is it that in SR we always seem to jump to the Lorenz equations when there is a simpler way. This is the concept of an interval. The interval is defined as the square root of ( T squared minus X squared) . In Special relativity the time and distance are different for different observers but the interval is always conserved.
For instance here is a relativity problem solved by both methods. The units used are feet, and nanoseconds. With these units the velocity of light is equal to 1 ( with a 2% error) G = gamma or 1/square root (1-V^2/C^2)
Problem:
X and T are the platform coordinates and X' and T' are moving train coordinates.
A speeding train passes a 500 foot station platform. It's velocity is equal to 0.6 . At the entry end of the platform X = 0 and T = 0 . and T' = 0 . How old is the train engineer as the train passes the exit end of the platform.
The time to transit platform = T = D/V = 500/ .6 = 833.3333 nanoseconds
Lorenz equations:
X' = G * ( X - V*T) ( not used)
T' = G * ( T - X*V )
G = 1.25 ( for V = .6 )
now T' = 1.25 * ( 833.333 - 500 * .6) = 666.6666 nanoseconds for the age of the engineer
Interval method :
By The interval method the interval on the platform between the train entering one end and exiting the other end is square root ( T ^ 2 - X^2 ) or I = square root (833.333 ^2 - 500^2 ) = 666.6666 nanoseconds. For the moving train the interval is equal to the time as X' is always equal to zero as the train engineer is always in the cab in front of the train.
For instance here is a relativity problem solved by both methods. The units used are feet, and nanoseconds. With these units the velocity of light is equal to 1 ( with a 2% error) G = gamma or 1/square root (1-V^2/C^2)
Problem:
X and T are the platform coordinates and X' and T' are moving train coordinates.
A speeding train passes a 500 foot station platform. It's velocity is equal to 0.6 . At the entry end of the platform X = 0 and T = 0 . and T' = 0 . How old is the train engineer as the train passes the exit end of the platform.
The time to transit platform = T = D/V = 500/ .6 = 833.3333 nanoseconds
Lorenz equations:
X' = G * ( X - V*T) ( not used)
T' = G * ( T - X*V )
G = 1.25 ( for V = .6 )
now T' = 1.25 * ( 833.333 - 500 * .6) = 666.6666 nanoseconds for the age of the engineer
Interval method :
By The interval method the interval on the platform between the train entering one end and exiting the other end is square root ( T ^ 2 - X^2 ) or I = square root (833.333 ^2 - 500^2 ) = 666.6666 nanoseconds. For the moving train the interval is equal to the time as X' is always equal to zero as the train engineer is always in the cab in front of the train.