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mcastillo356
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- TL;DR
- A proper reference frame in the theory of relativity is a particular form of accelerated reference frame, in which an accelerated observer can be considered as being in rest. Well, the context isn't Special Relativity, this is, inertial frames of reference?
MOVING CLOCKS
In this section, we show that clocks moving at high speeds run slowly.
We construct a clock, called a light clock, using a stick of proper lenght ##L_0##, and two mirrors. The two mirrors face each other, and a pulse of light bounces back and forth betweem them. Each time the light pulse strikes one of the mirrors, say the lower mirror, the clock is said to tick. Between successive ticks the light pulse travels a distance ##2L_0## in the proper reference of frame of the clock. Thus, the time between ticks ##T_0## is related to ##L_0 ##by
$$2L_0=cT_0\quad{R_1}$$
Next, we consider the time between ticks of the same light clock, but this time we observe it from a reference frame in which the clock is moving perpendicular to the stick with speed ##v##. In this reference frame, the clock moves a distance ##vT## between ticks and the light pulse moves a distance ##cT## between ticks. The distance the pulse moves in travelling from the bottom mirror to the top mirror is
$$\displaystyle\sqrt{L_0^2+\bigg (\frac{1}{2}vT\bigg )^2}$$
The light pulse travels the same distance in traveling from the top mirror to the bottom mirror. Thus,
$$2\displaystyle\sqrt{L_0^2+\bigg (\frac{1}{2}vT\bigg )^2}=cT\quad{R-2}$$
Because the speed of light is the same in all inertial reference frames, we have the same symbol ##c## for the speed of light in Equations ##R-1## and ##R-2##. Solving Equation ##R-1## for ##L_0## and substituting into Equation ##R-2## gives
$$\displaystyle\sqrt{\bigg (\frac{1}{2}cT_0\bigg )^2+\bigg (\frac{1}{2}vT\bigg )^2}=\frac{1}{2}cT$$
Solving for T gives
$$T=\displaystyle\frac{T_0}{\sqrt{1-(v^2/c^2}}\quad{R-3}$$
According to Equation ##R-3##, the time between ticks in the reference frame in which the clock moves at speed ##v## is greater than the time between ticks in the proper reference frame of the clock.
Attempt at a solution
The proper reference frame both observers can be considered as being in rest.
Post without preview.
In this section, we show that clocks moving at high speeds run slowly.
We construct a clock, called a light clock, using a stick of proper lenght ##L_0##, and two mirrors. The two mirrors face each other, and a pulse of light bounces back and forth betweem them. Each time the light pulse strikes one of the mirrors, say the lower mirror, the clock is said to tick. Between successive ticks the light pulse travels a distance ##2L_0## in the proper reference of frame of the clock. Thus, the time between ticks ##T_0## is related to ##L_0 ##by
$$2L_0=cT_0\quad{R_1}$$
Next, we consider the time between ticks of the same light clock, but this time we observe it from a reference frame in which the clock is moving perpendicular to the stick with speed ##v##. In this reference frame, the clock moves a distance ##vT## between ticks and the light pulse moves a distance ##cT## between ticks. The distance the pulse moves in travelling from the bottom mirror to the top mirror is
$$\displaystyle\sqrt{L_0^2+\bigg (\frac{1}{2}vT\bigg )^2}$$
The light pulse travels the same distance in traveling from the top mirror to the bottom mirror. Thus,
$$2\displaystyle\sqrt{L_0^2+\bigg (\frac{1}{2}vT\bigg )^2}=cT\quad{R-2}$$
Because the speed of light is the same in all inertial reference frames, we have the same symbol ##c## for the speed of light in Equations ##R-1## and ##R-2##. Solving Equation ##R-1## for ##L_0## and substituting into Equation ##R-2## gives
$$\displaystyle\sqrt{\bigg (\frac{1}{2}cT_0\bigg )^2+\bigg (\frac{1}{2}vT\bigg )^2}=\frac{1}{2}cT$$
Solving for T gives
$$T=\displaystyle\frac{T_0}{\sqrt{1-(v^2/c^2}}\quad{R-3}$$
According to Equation ##R-3##, the time between ticks in the reference frame in which the clock moves at speed ##v## is greater than the time between ticks in the proper reference frame of the clock.
Attempt at a solution
The proper reference frame both observers can be considered as being in rest.
Post without preview.