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Monday, May 29, 2017

Special Relativity

I was so bored on a flight, so I borrowed a pen from my neighbor passenger and wrote a quick derivation of a few results in special relativity:

Suppose an observer is doing a light speed measurement on a train by measuring the time it takes to go upwards and then reflect back and measure the time elapsed, he measures the time elapsed to be $ 2t $ and the distance traveled to be $ 2h $.

But the train is moving, so to an observer outside of the train, he measures the light moved in diagonal manner, let the time spent to be $ t_0 $, and the overall distance the light traveled is $ 2\sqrt{(vt_0)^2 + h^2} $. We assumed the two observers measure the same $ h $ and the same $ v $.

Both measurements are speed of light, so they equals:

$ \frac{2h}{2t} = \frac{2\sqrt{(vt_0)^2 + h^2}}{2t_0} = c $

First, we cancel out all the $ 2 $.

$ \frac{h}{t} = \frac{\sqrt{(vt_0)^2 + h^2}}{t_0} = c $

Square the second equality to get

$ (vt_0)^2 + h^2 = c^2t_0^2 $

Rearranging, get 

$ h^2 = (c^2 - v^2)t_0^2 $

Dividing by $ c^2 $ get

$ \frac{h^2}{c^2} = (1 - \frac{v^2}{c^2})t_0^2 $

But $ \frac{h^2}{c^2} $ is simply $ t^2 $ by the first equality, therefore

$ t^2 = (1 - \frac{v^2}{c^2})t_0^2 $

or 

$ t = \sqrt{1 - \frac{v^2}{c^2}} t_0 $.

The equation shows a few things:

First, $ v \leq c $, for it does not make sense to have imaginary time. That indicates nothing can run faster than the speed of light.

Second, $ 0 \leq \sqrt{1 - \frac{v^2}{c^2}} \leq 1 $, therefore, the time interval between the same events is shorter for the observers on the train. Suppose $ t_0 = 1 $ second, then $ t $ is less than a second, so if there is a clock there, it doesn't tick yet. So for the observer outside the train, it appears that a moving clock go slower. This is called time dilation.

Third, since the both observers agrees on the relative velocity, the distance traveled, measured by the two observers are $ vt $ and $ vt_0 $ respectively. We know $ t \leq t_0 $, therefore, to the observer on the train, distance appears shorter as well. This is called length contraction.


Sunday, May 7, 2017

Chemistry (2)

Problem:


Solution:

The electron configuration of potassium is $ 1s^22s^22p^63s^23p^64s^1 $, therefore the unpaired electron is occupying the 4s orbital.

Chemistry (1)

Problem:


Solution:

Oxygen has 8 protons always
The isotope with mass number 16 has 8 neutron
The isotope with mass number 17 has 9 neutron
The isotope with mass number 18 has 10 neutron