What is the time dilation relative to a outside observer if you were going at 99.9999999999 percent the speed of light?

Answer 1

in what follows, please consider sqrt() the square root function, and sqr() the square function (sqr(x) = x * x).

Let's suppose we have a bus, that is travelling at 99.9999999999% of the speed of light (let's call it c, which is more or less 3e8 meters/second)

Let's suppose further that the bus is 3 meters high.

Let's suppose we have a beam of light travelling from the floor of the bus to the ceiling.

If I'm going inside the bus, I'll see the beam reaching the ceiling in some time t.

This time would be t= 3/c.

For an outside observer, s/he would see the same beam performing a diagonal trajectory, rather than a vertical, like me.

From the outside observer's point of view, we have the beam travelling with a speed that has both a vertical and an horizontal component. The horizontal one would equal bus's speed which is (0.999999999999 * c).

But Einnie proved that the beam speed is always constant and always equal to c, both for me and for the outside observer. So, the vertical component of the speed of the beam would be sqrt(sqr(c) - sqr(0.999999999999 * c)), according to Pitagoras Theorem Let's call it cy

As the bus' heigth is 3 meters we have that for the outside observer the time that the beam took to reach the ceiling would be 3 / cy, or

3 / sqrt(sqr(c) - sqr(0.999999999999*c)).

The ratio of the outside observer time to my time will be c / (sqrt(sqr(c) - (sqr(0.999999999999 * c))

if we consider c = 3e8, we'd have this ratio equal (if I computed correctly) to 707106,78118672430109614106527721.

For me, one day in the bus would mean 707106,78 days for the outside observer, or

1935,9528 years. Almost 2 milleniums...

Hope This Helps,

Best,

Oliveiros Cristina

System SW Engineer

Lisbon, Portugal