A problem regarding time dilation

THE HARLEQUIN
Messages
52
Reaction score
4
two spacecraft s travel parallel to each other along straight lines AB and CD as shown in the picture here ...
while s1 moves with a simple harmonic motion , s2 moves with a constant speed and only at points A and B it gets enough thrust within a very negligible time to reverse its direction of velocity ...
s1 and s2 travel the same AB=CD distance after a same amount of time ..
now both of the spacecraft s fire rockets R1 and R2 at the same time from points D and B respectively and they come back after time t .
now if S1,R1, S2,R2 carry 4 persons of the same age then after time t when they meet
1. what will be the difference between their ages ?
2. on what condition it is possible for all of them to have the same age when they meet ?
[ note: you can take v of S2 and f of S1 arbitrarily ]
answers with proper visualizing power will be appreciated ... mathematical analysis is also appreciated but not needed ...

THE HARLEQUIN
20150221_132729.jpg
 
Last edited:
Physics news on Phys.org
Draw a space-time diagram. Spaceship A's world line will be a sinusoidal curve, except oriented vertically instead of horizontally because the t axis is vertical. Spacecraft B's world line will be a zigzag line. The worldlines of the two rockets will be straight line segments, starting at the event where they're launched. The change of age of each person is given by the spacetime distance along their worldline between the starting point and the end point.

That's your visualization tool. To actually calculate the changes in age, you'll need to do a (straightforward) line integral to calculate the change of age along the sinusoidal world line. For the straight segments, we don't need the calculus - ##\Delta{\tau}=\sqrt{\Delta{t}^2-\Delta{x}^2}## will do the trick.
 
Nugatory said:
Draw a space-time diagram. Spaceship A's world line will be a sinusoidal curve, except oriented vertically instead of horizontally because the t axis is vertical. Spacecraft B's world line will be a zigzag line. The worldlines of the two rockets will be straight line segments, starting at the event where they're launched. The change of age of each person is given by the spacetime distance along their worldline between the starting point and the end point.

That's your visualization tool. To actually calculate the changes in age, you'll need to do a (straightforward) line integral to calculate the change of age along the sinusoidal world line. For the straight segments, we don't need the calculus - ##\Delta{\tau}=\sqrt{\Delta{t}^2-\Delta{x}^2}## will do the trick.
thanks .. that was helpful ... i also want to know if the spacecraft 2 has the same velocity as the rocket will the age of the man in in the spacecraft and the man in the rocket differ ? if it does then by what factor ?
 
Draw your spacetime diagram, label the end points of the relevant line segments, calculate the amount of aging on each path... And let us know what you find.
 
OK, so this has bugged me for a while about the equivalence principle and the black hole information paradox. If black holes "evaporate" via Hawking radiation, then they cannot exist forever. So, from my external perspective, watching the person fall in, they slow down, freeze, and redshift to "nothing," but never cross the event horizon. Does the equivalence principle say my perspective is valid? If it does, is it possible that that person really never crossed the event horizon? The...
ASSUMPTIONS 1. Two identical clocks A and B in the same inertial frame are stationary relative to each other a fixed distance L apart. Time passes at the same rate for both. 2. Both clocks are able to send/receive light signals and to write/read the send/receive times into signals. 3. The speed of light is anisotropic. METHOD 1. At time t[A1] and time t[B1], clock A sends a light signal to clock B. The clock B time is unknown to A. 2. Clock B receives the signal from A at time t[B2] and...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...

Similar threads

Replies
14
Views
2K
Replies
88
Views
7K
Replies
27
Views
3K
Replies
55
Views
3K
Replies
11
Views
2K
Back
Top