Understanding Time Dilation: Explained with an Everyday Example | Help Needed"

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SUMMARY

The discussion centers on the concept of time dilation, specifically the phenomenon where moving clocks run slow relative to stationary observers. A scenario is presented involving a spacecraft traveling at high speeds, illustrating that while a stationary clock shows 5 PM after one hour, the clock on the spacecraft will show an earlier time due to the effects of time dilation. The equation used to calculate time distortion is Einstein's t' = (1-velocity/speed of light)^-(1/2). The discussion emphasizes that observers on the spacecraft perceive distances differently due to length contraction, leading to discrepancies in time readings between moving and stationary clocks.

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Time Dilation --- Help needed

Homework Statement



Time dilation says, " Moving clocks run slow".

Homework Equations





The Attempt at a Solution


For example, a stationary clock shows 4 PM and at the same 4 PM a spacecraft moves with a clock inside it showing the same 4 PM. Now after an hour, the spacecraft returns to earth. The stationary clock will show 5 PM. I presume the clock in the spacecraft will also show 5 PM. Then how can we say moving clocks run slow. Forgive if i sound silly. I need clarity on this concept. Thanks in advance
 
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The idea of time dilation is that light moves at the same constant rate seen from any perspective and that other moving objects don't. As you move faster you're perception of reality has a decreased rate. Everything else appears slower to you. To find the time distortion you would use Einstein's equation:

t' = (1-velocity/speed of light)^-(1/2)

Whichever clock is moving faster is going to be the clock that reads the earlier time.
 


You are assuming that the observers on the spaceship will say that they spent 1 hour traveling. But this does not happen; the spaceship observers will see the distance between the Earth and themselves to be contracted, which is another affect of traveling at high speeds (length contraction). So when the spaceship observers see themselves reaching the Earth by traveling a different distance than the Earth observers say, they will show a different time than the Earth observers (depending on the speed they travel at).
 


Thanks everyone for your replies
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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