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thetafilippo
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Moved from technical forums so no template
Summary:: Special relativity - 2 astronauts syncronize their clocks and moves in different paths at different velocities, which clocks is left behind? and why?
Hi everyone, i have the following problem and I'm not understanding if my strategy to solve it is correct:
Two astronauts synchronize their clocks starting at time zero from the origin O of an inertial reference frame S. They simultaneously reach point $$A:{ x_A=d, y_A=0}. $$
The first astronaut moved in a straight line at constant speed $$v=\frac{c \pi}{4}$$ while the second traveled at constant velocity (in absolute value) over a semicircle in the xy plane of diameter d. Upon their meeting, the astronauts notice a 1 year difference between their clocks. Which clock is left behind, and why?
Which value has d?
Now, I've write down the Lorentz transformations
$$x'=\gamma(ct-\beta x)$$
$$ct'=\gamma(x-\beta ct)$$
And then i substitute $$x'=d$$, $$\beta=\frac{\frac{\pi c}{4}}{c}$$, $$x=dcos\theta$$ and because of the first reference frame reach the point d and the second reach the point $$x=dcos\theta$$
$$d=\gamma(ct-\frac{\frac{\pi c}{4}}{c} d cos \theta)$$
$$0=\gamma(d cos \theta-\frac{\frac{\pi c}{4}}{c} ct)$$
And i get from the second $$d=\frac{\pi}{4}ct$$. It's correct? how i could properly say which clock is left befind? I think that the one that travels in a straight line needs more time to reach the same point,cause the straight line in the Minkowski space maximize the time. Anyone can help me?
Hi everyone, i have the following problem and I'm not understanding if my strategy to solve it is correct:
Two astronauts synchronize their clocks starting at time zero from the origin O of an inertial reference frame S. They simultaneously reach point $$A:{ x_A=d, y_A=0}. $$
The first astronaut moved in a straight line at constant speed $$v=\frac{c \pi}{4}$$ while the second traveled at constant velocity (in absolute value) over a semicircle in the xy plane of diameter d. Upon their meeting, the astronauts notice a 1 year difference between their clocks. Which clock is left behind, and why?
Which value has d?
Now, I've write down the Lorentz transformations
$$x'=\gamma(ct-\beta x)$$
$$ct'=\gamma(x-\beta ct)$$
And then i substitute $$x'=d$$, $$\beta=\frac{\frac{\pi c}{4}}{c}$$, $$x=dcos\theta$$ and because of the first reference frame reach the point d and the second reach the point $$x=dcos\theta$$
$$d=\gamma(ct-\frac{\frac{\pi c}{4}}{c} d cos \theta)$$
$$0=\gamma(d cos \theta-\frac{\frac{\pi c}{4}}{c} ct)$$
And i get from the second $$d=\frac{\pi}{4}ct$$. It's correct? how i could properly say which clock is left befind? I think that the one that travels in a straight line needs more time to reach the same point,cause the straight line in the Minkowski space maximize the time. Anyone can help me?