How Fast and Slow Does Time Flow on a Speeding Spaceship?

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SUMMARY

The discussion centers on calculating the speed parameter β of a spaceship and the time dilation factor in special relativity. The spaceship's length is given as 1/3 of its proper length, leading to a derived Lorentz factor γ of 3. Consequently, the time dilation factor indicates that the spaceship's clocks run 3 times slower than those in the observer's frame. The correct calculation for β results in a speed of approximately 0.9428c, confirming the relationship between length contraction and time dilation in special relativity.

PREREQUISITES
  • Understanding of special relativity concepts, particularly Lorentz transformations.
  • Familiarity with the equations for length contraction and time dilation.
  • Knowledge of the Lorentz factor (γ) and its significance in relativistic physics.
  • Basic algebra skills for manipulating equations involving square roots and fractions.
NEXT STEPS
  • Study the derivation and applications of the Lorentz factor (γ) in special relativity.
  • Learn about the implications of time dilation in various relativistic scenarios.
  • Explore the concept of simultaneity in special relativity and its effects on observers in different frames.
  • Investigate the relationship between velocity, time dilation, and length contraction through practical examples.
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Students of physics, educators teaching special relativity, and anyone interested in understanding the effects of high-speed travel on time and space.

FaraDazed
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I was not too sure if this was the correct forum, so feel free to move if needed.

1. Homework Statement

A spaceship is measured to be exactly 1/3 of its proper length.

(a) What is the speed parameter β of the spaceship relative to the observer's frame?

(b) By what integer factor do the spaceship's clocks run slow, compared to clocks in
the observer's frame?

Homework Equations


<br /> L=\frac{L_0}{\gamma} \\<br /> t=\frac{t_0}{\gamma} \\<br /> \gamma = \frac{1}{\sqrt{1-\beta^2}} \\<br /> \beta = \frac{v}{c}<br />

The Attempt at a Solution


For A i did:
<br /> L=L_0 \sqrt{1-\beta^2} \\<br /> \frac{L_0}{3}=L_0 \sqrt{1-\beta^2} \\<br /> \frac{1}{3}= \sqrt{1-\beta^2} \\<br /> \frac{1}{9}=1-\beta^2 \\<br /> -\frac{8}{9}=- \beta^2 \\<br /> \frac{8}{9}=\beta^2 \\<br /> \beta = \sqrt{\frac{8}{9}}<br />

I am not to sure that is correct. But for part B I was stuck but during typing this up managed to get an integer answer so hopefully it is correct.
<br /> t=\frac{t_0}{\sqrt{1-\beta^2}} \\<br /> \frac{t}{t_0}=\frac{1}{\sqrt{1-\beta^2}} \\<br /> \frac{t}{t_0}=\frac{1}{\sqrt{1-\frac{8}{9}}} \\<br /> \frac{t}{t_0}=\frac{1}{\frac{1}{3}} =3 \\<br />

Would appreciate any help/advice/feedback, thanks :)
 
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Your answers are fine, but it sounds like you don't understand the equations.

##L_0=\gamma L\\
T=\gamma T_0##
Where the 0's index the proper time, for this case.

You were told: ##L=\frac{1}{3}L_0##. This means that ##\gamma = 3##.
... so you have automatically answered the second question without any further working out: ##T=3T_0##

(Pretty much the first thing you want to know in any SR problem is ##\gamma##.)

... for the first question, you want ##\beta##: $$\gamma = \frac{1}{\sqrt{1-\beta^2}} \implies \beta = \sqrt{1-\frac{1}{\gamma^2}}$$ ... it is best practice to do the algebra with the symbols before putting numbers in.
$$\beta = \sqrt{1-\frac{1}{9}} = \sqrt{\frac{8}{9}} \implies v= 0.9428c $$ ... can you see how much easier that was that what you did?
No worries though everyone does it the hard way at first ;)
 
Many
Simon Bridge said:
Your answers are fine, but it sounds like you don't understand the equations.

##L_0=\gamma L\\
T=\gamma T_0##
Where the 0's index the proper time, for this case.

You were told: ##L=\frac{1}{3}L_0##. This means that ##\gamma = 3##.
... so you have automatically answered the second question without any further working out: ##T=3T_0##

(Pretty much the first thing you want to know in any SR problem is ##\gamma##.)

... for the first question, you want ##\beta##: $$\gamma = \frac{1}{\sqrt{1-\beta^2}} \implies \beta = \sqrt{1-\frac{1}{\gamma^2}}$$ ... it is best practice to do the algebra with the symbols before putting numbers in.
$$\beta = \sqrt{1-\frac{1}{9}} = \sqrt{\frac{8}{9}} \implies v= 0.9428c $$ ... can you see how much easier that was that what you did?
No worries though everyone does it the hard way at first ;)
Many thanks for your help and feedback, much appreciated :)
 

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