Time dilatation between straight and curved lines in Minkowsi space

[thetafilippo]

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?

 

[PeroK]

In the inertial reference frame, who has the greater speed?

If you study the problem in that reference frame, why do you need a Lorentz Transformation?

 

[Ibix]

Inertial clocks always have the longest possible lapse of time between two events. This is the Minkowski geometry equivalent to the Euclidean "shortest distance between two points is a straight line".

However, I make the speed of your non-inertial astronaut $c\pi^2/8$, which is greater than $c$. So you will get nonsensical answers in this case. Adjust your scenario and try again.

 

[thetafilippo]

it is not specified, I think it is implicitly given by saying that it moves at constant speed along a semicircle of diameter d, therefore the centripetal acceleration $$ a_c = \frac {v ^ 2}{d/2} $$, could it be?
but in any case the centripetal acceleration is not given

 

[Ibix]

it is not specified, I think it is implicitly given by saying that it moves at constant speed along a semicircle of diameter d, therefore the centripetal acceleration $$ a_c = \frac {v ^ 2}{d/2} $$, could it be?
but in any case the centripetal acceleration is not given

Acceleration is not relevant. Only speed matters. However, note my previous answer.

 

[thetafilippo]

Inertial clocks always have the longest possible lapse of time between two events. This is the Minkowski geometry equivalent to the Euclidean "shortest distance between two points is a straight line".

However, I make the speed of your non-inertial astronaut $c\pi^2/8$, which is greater than $c$. So you will get nonsensical answers in this case. Adjust your scenario and try again.

The speed of non inertial astronaut is not given not $c\pi^2/8$, please could explain what you mean? i didn't get your reply!

 

[thetafilippo]

could someone explain to me what is the correct strategy to follow? I'm confused by this problem

 

[Ibix]

The speed of non inertial astronaut is $\frac{c \pi}{4}$ not $c\pi^2/8$, please could explain what you mean? i didn't get your reply!

Your original post says that the inertial astronaut travels at $c\pi/4$.

The first astronaut moved in a straight line at constant speed $$v=\frac{c \pi}{4}$$

I take it that was a mistake.

In that case, both astronauts are moving at constant speed, so all you need to do is calculate their $\gamma$ factors, calculate the elapsed coordinate time, and then use time dilation.

 

[Sagittarius A-Star]

The speed of non inertial astronaut is $\frac{c \pi}{4}$ not $c\pi^2/8$, please could explain what you mean? i didn't get your reply!

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.

Summary:: Special relativity - 2 astronauts syncronize their clocks and moves in different paths at different velocities, which clocks is left behind? and why?

Do both astronauts have the same speed?

 

[thetafilippo]

The speed of inertial astronaut is

Your original post says that the inertial astronaut travels at $c\pi/4$.I take it that was a mistake.

In that case, both astronauts are moving at constant speed, so all you need to do is calculate their $\gamma$ factors, calculate the elapsed coordinate time, and then use time dilation.

The speed of non inertial astronaut is not given and the speed of inertial astronaut is $c\pi/4$. How i could calculate the gamma factor of the non inertial one if the velocity is unknown?

 

[thetafilippo]

Do both astronauts have the same speed?

No, no the speed of non inertial astronaut is unknown

 

[Ibix]

The speed of inertial astronaut is

The speed of non inertial astronaut is not given and the speed of inertial astronaut is $c\pi/4$. How i could calculate the gamma factor of the non inertial one if the velocity is unknown?

How far does the non-inertial astronaut travel? How fast must he go if he is to meet the first astronaut at the other side of the semi-circle?

If the inertial astronaut is moving at $c\pi/4$ then the answer to my second question is $c\pi^2/8$, which is faster than light. Where did you get this question from?

 

[Sagittarius A-Star]

No, no the speed of non inertial astronaut is unknown

Do they arrive at the same time at A?

 

[Ibix]

Do they arrive at the same time at A?

Yes, according to the OP.

 

[Nugatory]

The speed of non inertial astronaut is not given and the speed of inertial astronaut is $c\pi/4$. How i could calculate the gamma factor of the non inertial one if the velocity is unknown?

First you calculate the speed of the non-inertial astronaut using the frame in which the starting point and the destination are at rest. You know the distance traveled, you know the time it took to cover that distance (because the astronauts arrive at the destination together), and that gives you the speed.

 

[Dale]

Summary:: Special relativity - 2 astronauts syncronize their clocks and moves in different paths at different velocities, which clocks is left behind? and why?

I've write down the Lorentz transformations

This problem is simple enough that no transform is needed. Just determine the time dilation for each path in the original frame.

No, no the speed of non inertial astronaut is unknown

You know the distance he travels (in the original frame) and the time it takes.

 

[Sagittarius A-Star]

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.

First you should double-check with the original problem description, if the given speed $v=\frac{c \pi}{4}$ really belongs to the first (inertial) or instead to the second (non-inertial) astronaut, with then the first astronaut's speed unknown (Reason). Then you should find out the ratio between the path lengths via the semi-circle and the straight line, to calculate the unknown speed from the known one.

 

[vanhees71]

Another hint: A clock by definition is a device which shows the proper time along its world line,
$$\tau=\frac{1}{c} \int_{\lambda_1}^{\lambda_2} \mathrm{d} \lambda \sqrt{\eta_{\mu \nu} \dot{x}^{\mu} \dot{c}^{\nu}}.$$
Here I use the west-coast convention with $\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)$. So just parametrize the world lines of the two astronauts and calculate their proper times when meeting again.

 

[thetafilippo]

First you should double-check with the original problem description, if the given speed $v=\frac{c \pi}{4}$ really belongs to the first (inertial) or instead to the second (non-inertial) astronaut, with then the first astronaut's speed unknown (Reason). Then you should find out the ratio between the path lengths via the semi-circle and the straight line, to calculate the unknown speed from the known one.

I checked and the velocity of the first astronaut is $v=\frac{c \pi}{4}$.
If i want to solve the problem i should evaluate the 2 gamma factors, then the 2 proper times and set the difference equal to 1 year, but how is possible that the velocity of the non inertial astronaut is greater than c? where is the mistake? this problem comes from an exam sheet of my professor, so the problem description is not wrong!

 

[Ibix]

this problem comes from an exam sheet of my professor, so the problem description is not wrong!

Professors make mistakes too (there were typos in one of my final exams - corrections were written on a blackboard in the exam room). The question cannot be solved as written.

To my mind, the most likely explanation is that the non-inertial astronaut is supposed to be doing $c\pi/4$, since then the inertial astronaut is doing $c/2$. Then the question can be solved as you proposed.