Imagine a light at point A in Fig. 1. It sends a pulse of light to a mirror, M_{A}, 0.5 light seconds away, from whence the light is reflected back to A, triggering another pulse of light. Giving us a ‘light clock’ that has one second ‘ticks’.

Another identical light clock, C, travelling at 0.6c (c is the speed of light), passes A; and the two clocks are synchronised at the moment they pass.

What will an observer who is stationary at A see as he observes C pass him?

First, that the path of the light to C’s mirror MC will no longer be vertical but will be rotated by the relative velocity; as shown by the line A-M_{C} in fig. 1.

The clock C will have travelled a distance horizontally (0.6ct) by the time its pulse of light has reached its mirror. Where t = 0.5 seconds. = 0.3 light secs

And here lies the problem that Special Relativity resolved.

We know that the light in each identical clock will travel to its mirror, a distance of 0.5 light seconds, in 0.5 seconds – – – [1]

But what of the moving light observed by A? The distance A-M_{C} is greater than 0.5 light seconds so it has to travel faster than c (impossible) or take longer than 0.5 seconds. Which we know it doesn’t.

Pythagoras tells us that h = √(ct’)^{2} – (vt’)^{2} = ct’√(1 – v^{2}/c^{2}) then

t’ = h/√(1 – v^{2}/c^{2})

Where 1/√(1 – v^{2}/c^{2}) is the Lorentz Factor. – – – – – – [2]

Now c = 1, v = 0.6, h = 0.5

therefore t’ = 0.5/√0.64 = 0.5/0.8 = 0.625 seconds

Yet we established in [1] above the h = 0.5 light seconds and 0.5 seconds in distance and time.

So we have an apparent anomaly for the time cannot be both 0.5 seconds AND 6.25 seconds can it?

Well yes it can! Einstein’s Special Relativity asks us if: the times and distances/lengths measured in the stationary system are the same units as those measured in a moving system and gives us the answer as a resounding NO.

That they should be, was but an assumption that had always been made. In this example one can clearly see that they are not the same.

That 0.5 seconds on the stationary clock has the same duration as 0.625 seconds observed on the moving clock.

We can see how that works in Fig. 2. In which diagram we can see how the relative velocity between the two clocks rotates the axis of the light path from A-M_{A} to A-M_{C} so that the 0.1 second coordinate of the moving clock lies on the 0.08 second coordinate of the stationary clock, the unit of time (and length as they are seconds and light seconds) is contracted, from 0.1 to 0.8; while the totals of time and length are dilated – the number of units being increased from 0.5 to 0.625.

Imagine a light at point A in Fig. 1. It sends a pulse of light to a mirror, M_{A}, 0.5 light seconds away, from whence the light is reflected back to A, triggering another pulse of light. Giving us a ‘light clock’ that has one second ‘ticks’.

Fig. 1

Another identical light clock, C, travelling at 0.6c (c is the speed of light), passes A; and the two clocks are synchronised at the moment they pass.

What will an observer who is stationary at A see as he observes C pass him?

First, that the path of the light to C’s mirror MC will no longer be vertical but will be rotated by the relative velocity; as shown by the line A-M_{C} in fig. 1.

The clock C will have travelled a distance horizontally (0.6ct) by the time its pulse of light has reached its mirror. Where t = 0.5 seconds. = 0.3 light secs

And here lies the problem that Special Relativity resolved.

We know that the light in each identical clock will travel to its mirror, a distance of 0.5 light seconds, in 0.5 seconds – – – [1]

But what of the moving light observed by A? The distance A-M_{C} is greater than 0.5 light seconds so it has to travel faster than c (impossible) or take longer than 0.5 seconds. Which we know it doesn’t.

Pythagoras tells us that h = √(ct’)^{2} – (vt’)^{2} = ct’√(1 – v^{2}/c^{2}) then

t’ = h/√(1 – v^{2}/c^{2})

Where 1/√(1 – v^{2}/c^{2}) is the Lorentz Factor. – – – – – – [2]

Now c = 1, v = 0.6, h = 0.5

therefore t’ = 0.5/√0.64 = 0.5/0.8 = 0.625 seconds

Yet we established in [1] above the h = 0.5 light seconds and 0.5 seconds in distance and time.

So we have an apparent anomaly for the time cannot be both 0.5 seconds AND 6.25 seconds can it?

Well yes it can! Einstein’s Special Relativity asks us if: the times and distances/lengths measured in the stationary system are the same units as those measured in a moving system and gives us the answer as a resounding NO.

That they should be, was but an assumption that had always been made. In this example one can clearly see that they are not the same.

That 0.5 seconds on the stationary clock has the same duration as 0.625 seconds observed on the moving clock.

Fig. 2.

We can see how that works in Fig. 2. In which diagram we can see how the relative velocity between the two clocks rotates the axis of the light path from A-M_{A} to A-M_{C} so that the 0.1 second coordinate of the moving clock lies on the 0.08 second coordinate of the stationary clock, the unit of time (and length as they are seconds and light seconds) is contracted, from 0.1 to 0.8; while the totals of time and length are dilated – the number of units being increased from 0.5 to 0.625.

Imagine a light at point A in Fig. 1. It sends a pulse of light to a mirror, M_{A}, 0.5 light seconds away, from whence the light is reflected back to A, triggering another pulse of light. Giving us a ‘light clock’ that has one second ‘ticks’.

Fig. 1

Another identical light clock, C, travelling at 0.6c (c is the speed of light), passes A; and the two clocks are synchronised at the moment they pass.

What will an observer who is stationary at A see as he observes C pass him?

First, that the path of the light to C’s mirror MC will no longer be vertical but will be rotated by the relative velocity; as shown by the line A-M_{C} in fig. 1.

The clock C will have travelled a distance horizontally (0.6ct) by the time its pulse of light has reached its mirror. Where t = 0.5 seconds. = 0.3 light secs

And here lies the problem that Special Relativity resolved.

We know that the light in each identical clock will travel to its mirror, a distance of 0.5 light seconds, in 0.5 seconds – – – [1]

But what of the moving light observed by A? The distance A-M_{C} is greater than 0.5 light seconds so it has to travel faster than c (impossible) or take longer than 0.5 seconds. Which we know it doesn’t.

Pythagoras tells us that h = √(ct’)^{2} – (vt’)^{2} = ct’√(1 – v^{2}/c^{2}) then

t’ = h/√(1 – v^{2}/c^{2})

Where 1/√(1 – v^{2}/c^{2}) is the Lorentz Factor. – – – – – – [2]

Now c = 1, v = 0.6, h = 0.5

therefore t’ = 0.5/√0.64 = 0.5/0.8 = 0.625 seconds

Yet we established in [1] above the h = 0.5 light seconds and 0.5 seconds in distance and time.

So we have an apparent anomaly for the time cannot be both 0.5 seconds AND 6.25 seconds can it?

Well yes it can! Einstein’s Special Relativity asks us if: the times and distances/lengths measured in the stationary system are the same units as those measured in a moving system and gives us the answer as a resounding NO.

That they should be, was but an assumption that had always been made. In this example one can clearly see that they are not the same.

That 0.5 seconds on the stationary clock has the same duration as 0.625 seconds observed on the moving clock.

Fig. 2.

We can see how that works in Fig. 2. In which diagram we can see how the relative velocity between the two clocks rotates the axis of the light path from A-M_{A} to A-M_{C} so that the 0.1 second coordinate of the moving clock lies on the 0.08 second coordinate of the stationary clock, the unit of time (and length as they are seconds and light seconds) is contracted, from 0.1 to 0.8; while the totals of time and length are dilated – the number of units being increased from 0.5 to 0.625.

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