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

AI Thread Summary
Time dilation indicates that moving clocks run slower compared to stationary ones, which can be confusing when considering simultaneous events. In the example given, a stationary clock shows 5 PM after an hour, while the clock on a fast-moving spacecraft will show less time elapsed due to the effects of time dilation. Observers on the spacecraft perceive distances differently due to length contraction, which affects their time measurement. The key point is that the faster the movement, the more pronounced the time difference becomes, as described by Einstein's equation. Understanding these concepts clarifies how time is experienced differently by observers in motion versus those at rest.
logearav
Messages
329
Reaction score
0
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
 
Physics news on Phys.org


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
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top