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## Main Question or Discussion Point

Introduction

Special relativity in the most simple of definitions holds that each inertial frame of reference can be considered stationary. As a consequence, calculations concerning an object in one frame of reference must be congruent with another, for which ever inertial frame produces a calculation that is fitting with experimental results can be considered zero. As all inertial frames should be able to be considered at rest, they must produce results that are equivalent to each other.

Example of congruency

An object with a clock on-board is observed passing by two stationary points, 1 light second apart with a relative velocity to the observer of 0.5C . This observer then calculates the elapsed time, taken to cover the distance between the two points, according to the object.

(1 light second/0.5C)x(1-(0.5)^2)^0.5 = 2(0.75)^0.5 = 1.732 seconds

A second observer observes this object with a velocity of 0.8C pass by two moving points, both with a velocity of 0.5C and a distance apart of (0.75)^0.5 light seconds. This observer then calculates the elapsed time, taken to cover the distance between the two points, according to the object.

((0.75)^0.5 light seconds/(0.8C – 0.5))x(1-(0.8)^2)^0.5 = 2(0.75)^0.5 = 1.732 seconds

As can be seen each observer calculated the same result, for the elapsed time according to the object.

Relativity of Simultaneity and Different Observers

A long rod that is (0.75)^0.5 long has a clock on each end that are synchronised according to an observer, that observes the rod to have a velocity of 0.5C. This observer then calculates the difference in clocks is for the rod.

(((0.75)^0.5light seconds/(1-(0.5)^2)^0.5))x(0.5C))/C^2 = leading clock 0.5 seconds ahead

A second observer observes the rod to have a length of 0.6 light seconds and to have a velocity of 0.8C, with the trailing clock ahead by 0.5 seconds in comparison to the leading. This observer then calculates the difference in clocks for the rod.

(((0.6 light seconds/(1-(0.8)^2)^0.5)x(0.8C))/C^2) – 0.5 seconds = leading clock 0.3 seconds ahead

As the observers have reached different answers there is something wrong.

Conclusion

After reading this an individual will arrive at 4 possible conclusions.

(1) congruency is not required, the relativity of simultaneity calculations are wrong, Special relativity is valid.

(2) congruency is not required, the relativity of simultaneity calculations are correct, Special relativity is valid

(3) congruency is required, the relativity of simultaneity calculations are wrong, Special relativity is valid

(4) congruency is required, the relativity of simultaneity calculations are correct, Special relativity is invalid.

If one can explain why they have arrived at a particular conclusion, then it may help me to find one also. Thank you

Special relativity in the most simple of definitions holds that each inertial frame of reference can be considered stationary. As a consequence, calculations concerning an object in one frame of reference must be congruent with another, for which ever inertial frame produces a calculation that is fitting with experimental results can be considered zero. As all inertial frames should be able to be considered at rest, they must produce results that are equivalent to each other.

Example of congruency

An object with a clock on-board is observed passing by two stationary points, 1 light second apart with a relative velocity to the observer of 0.5C . This observer then calculates the elapsed time, taken to cover the distance between the two points, according to the object.

(1 light second/0.5C)x(1-(0.5)^2)^0.5 = 2(0.75)^0.5 = 1.732 seconds

A second observer observes this object with a velocity of 0.8C pass by two moving points, both with a velocity of 0.5C and a distance apart of (0.75)^0.5 light seconds. This observer then calculates the elapsed time, taken to cover the distance between the two points, according to the object.

((0.75)^0.5 light seconds/(0.8C – 0.5))x(1-(0.8)^2)^0.5 = 2(0.75)^0.5 = 1.732 seconds

As can be seen each observer calculated the same result, for the elapsed time according to the object.

Relativity of Simultaneity and Different Observers

A long rod that is (0.75)^0.5 long has a clock on each end that are synchronised according to an observer, that observes the rod to have a velocity of 0.5C. This observer then calculates the difference in clocks is for the rod.

(((0.75)^0.5light seconds/(1-(0.5)^2)^0.5))x(0.5C))/C^2 = leading clock 0.5 seconds ahead

A second observer observes the rod to have a length of 0.6 light seconds and to have a velocity of 0.8C, with the trailing clock ahead by 0.5 seconds in comparison to the leading. This observer then calculates the difference in clocks for the rod.

(((0.6 light seconds/(1-(0.8)^2)^0.5)x(0.8C))/C^2) – 0.5 seconds = leading clock 0.3 seconds ahead

As the observers have reached different answers there is something wrong.

Conclusion

After reading this an individual will arrive at 4 possible conclusions.

(1) congruency is not required, the relativity of simultaneity calculations are wrong, Special relativity is valid.

(2) congruency is not required, the relativity of simultaneity calculations are correct, Special relativity is valid

(3) congruency is required, the relativity of simultaneity calculations are wrong, Special relativity is valid

(4) congruency is required, the relativity of simultaneity calculations are correct, Special relativity is invalid.

If one can explain why they have arrived at a particular conclusion, then it may help me to find one also. Thank you