- #1

- 940

- 69

- Homework Statement
- I was trying to do a simple problem related to special theory of relativity and need to clarify few points. It's not a homework.

- Relevant Equations
- I'm including all the equations in the post. Thanks.

Hi,

It's not a homework but still thought to post it here as advised in the past.

A rocket is going to leave Earth's surface and it is decided that a data pulse encoding emission time of pulse will be sent every second from Earth station to the rocket, and the rocket would do the same.

The rocket's speed is 0.5c where 'c' is speed of light, i.e., 300000 km/s. Assuming the rocket accelerates to 0.5c instantaneously. (Not sure if it's legitimate assumption. I think while acceleration the gravitational time dilation will be quite dominant and should also be factored in for more accuracy but ignoring it in favor of simplicity.)

t₀ is proper time in Earth's frame of reference. From Earth's reference, the first pulse will get to the rocket when time on rocket is 2 seconds after departure, as the calculation below shows, and by that time the rocket would have traveled distance of c(sec) according to Earth station. I hope I have it correct this far.

$$

\begin{aligned}

&\mathrm{c}\left(\mathrm{t}_{0}-1\right)=0.5 \mathrm{ct}_{0} \\

&\mathrm{ct}_{0}-\mathrm{c}=0.5 \mathrm{ct}_{0} \\

&\mathrm{t}_{0}-1=0.5 \mathrm{t}_{0} \\

&\mathrm{t}_{0}-0.5 \mathrm{t}_{0}=1 \\

&\mathrm{t}_{0}=\frac{1}{1-0.5} \\

&\mathrm{t}_{0}=2

\end{aligned}

$$

The rocket's crew would think that their clock is running fine but the clock on rocket is time dilated as shown below.

$$

t^{\prime}=\frac{t_{0}}{\sqrt[2]{1-\frac{v^{2}}{c^{2}}}}

$$

taking $t_{0}=1$ and $\mathrm{c}=300000 \mathrm{~km} / \mathrm{s}$, and $\mathrm{v}=0.5 \mathrm{c}$

$$

\begin{aligned}

t^{\prime} &=\frac{1}{\sqrt[2]{1-\frac{(0.5 c)^{2}}{c^{2}}}} \\

t^{\prime} &=1.1547

\end{aligned}

$$

The result above means that for every 1.1547 seconds on earth, the rocket's clock will read 1 second. Please correct me if I'm wrong.

But by the time t₀=2 the rocket has traveled distance of 346410 km as result of length contraction as is shown below.

\begin{aligned}

L^{\prime} &=\frac{L_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \\

L^{\prime} &=\frac{300000}{\sqrt{2 / 1-\frac{(0.5 c)^{2}}{c^{2}}}} \\

L^{\prime} &=346410 \mathrm{~km}

\end{aligned}

Do I have it correct this far? I think it's better to confirm this before I proceed with the main queries. Thanks for the help, in advance!

It's not a homework but still thought to post it here as advised in the past.

A rocket is going to leave Earth's surface and it is decided that a data pulse encoding emission time of pulse will be sent every second from Earth station to the rocket, and the rocket would do the same.

The rocket's speed is 0.5c where 'c' is speed of light, i.e., 300000 km/s. Assuming the rocket accelerates to 0.5c instantaneously. (Not sure if it's legitimate assumption. I think while acceleration the gravitational time dilation will be quite dominant and should also be factored in for more accuracy but ignoring it in favor of simplicity.)

t₀ is proper time in Earth's frame of reference. From Earth's reference, the first pulse will get to the rocket when time on rocket is 2 seconds after departure, as the calculation below shows, and by that time the rocket would have traveled distance of c(sec) according to Earth station. I hope I have it correct this far.

$$

\begin{aligned}

&\mathrm{c}\left(\mathrm{t}_{0}-1\right)=0.5 \mathrm{ct}_{0} \\

&\mathrm{ct}_{0}-\mathrm{c}=0.5 \mathrm{ct}_{0} \\

&\mathrm{t}_{0}-1=0.5 \mathrm{t}_{0} \\

&\mathrm{t}_{0}-0.5 \mathrm{t}_{0}=1 \\

&\mathrm{t}_{0}=\frac{1}{1-0.5} \\

&\mathrm{t}_{0}=2

\end{aligned}

$$

The rocket's crew would think that their clock is running fine but the clock on rocket is time dilated as shown below.

$$

t^{\prime}=\frac{t_{0}}{\sqrt[2]{1-\frac{v^{2}}{c^{2}}}}

$$

taking $t_{0}=1$ and $\mathrm{c}=300000 \mathrm{~km} / \mathrm{s}$, and $\mathrm{v}=0.5 \mathrm{c}$

$$

\begin{aligned}

t^{\prime} &=\frac{1}{\sqrt[2]{1-\frac{(0.5 c)^{2}}{c^{2}}}} \\

t^{\prime} &=1.1547

\end{aligned}

$$

The result above means that for every 1.1547 seconds on earth, the rocket's clock will read 1 second. Please correct me if I'm wrong.

But by the time t₀=2 the rocket has traveled distance of 346410 km as result of length contraction as is shown below.

\begin{aligned}

L^{\prime} &=\frac{L_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \\

L^{\prime} &=\frac{300000}{\sqrt{2 / 1-\frac{(0.5 c)^{2}}{c^{2}}}} \\

L^{\prime} &=346410 \mathrm{~km}

\end{aligned}

Do I have it correct this far? I think it's better to confirm this before I proceed with the main queries. Thanks for the help, in advance!