Exploring the Effects of Relativity on Clocks

[Achillez]

Now according to special relativity when in the case of two people traveling at different speeds have their own clocks running differently depending on their velocity

I have understood the theoritical part of this that as c=d/t and as d= space and t= time and c= velocity of light . thus to keep c constant the value of space-time gets adjusted among itself

so my question is that in a practical situation how will the different velocities will affect the physical clock (a wrist watch) that the two people are carrying with them??how will they have their clocks show different times ??

 

[russ_watters]

I'm not sure I understand the question. Clocks are devices that show time. So if time is relative, clocks may show different times. That's it.

Are you looking for some "force" that causes a clock to move slower or faster? That isn't how it works. Time itself is varying.

 

[Achillez]

I'm not sure I understand the question. Clocks are devices that show time. So if time is relative, clocks may show different times. That's it.

Are you looking for some "force" that causes a clock to move slower or faster? That isn't how it works. Time itself is varying.

Ok let me reframe my question
When we say that there is time dilation or length contraction and also in the twin paradox wherein the twin that travels near the speed of light will age slower
so my question is that do all these things just APPEAR to an external observer or do these things really happen ??

So are these things just an APPEARANCE or a REALITY??
if reality how will physically a twin traveling near the speed of light age slower ??

 

[jtbell]

So are these things just an APPEARANCE or a REALITY??

How would you distinguish, in practice, between "appearance" and "reality" in this situation?

 

[Achillez]

How would you distinguish, in practice, between "appearance" and "reality" in this situation?

meaning in case of twin paradox as the twin on Earth sees that the other is aging slower
so my ? is that is the otr twin actually aging slower or is he just appearing to age slower to the twin on the Earth ??

similarly when we say that for an observer outside a fast train , the length of the train decreases so is it that the length of the train actually (physicaly) decreases or its just an illusion to the external observer ??

Thanks for the replies

 

[jtbell]

Sorry, my question apparently wasn't clear. I'll try again with different words:

How does one determine whether the traveling twin is "actually" aging slower or only "apparently" aging slower, from the point of view of the twin on earth? Similarly, how does one determine whether the twin on Earth is "actually" aging slower or only "apparently" aging slower, from the point of view of the traveling twin?

If the traveling twin turns around and returns so that the two twins can compare their wristwatches while standing next to each other, then they will find that the traveling twin has actually aged less. But while the traveling twin is still traveling and the twins are still separated, there is no way to compare their wristwatches directly.

 

[Mentz114]

meaning in case of twin paradox as the twin on Earth sees that the other is aging slower
so my ? is that is the otr twin actually aging slower or is he just appearing to age slower to the twin on the Earth ??

Biological clocks are subject to the same relativistic effects as mechanical ones. If the twins counted their heartbeats, then the 'away' twin would count fewer - assuming their environments were much the same. It is a real lasting effect.

Another poster suggested using a (sedated ?) rat as a clock !

similarly when we say that for an observer outside a fast train , the length of the train decreases so is it that the length of the train actually (physicaly) decreases or its just an illusion to the external observer ??

If this observer measured the length of the train using some device, they would measure it to be shorter than the passenger. They measure different lengths, you have to decide which is 'real', and to whom.

 

[belliott4488]

Something that always causes trouble with the twin paradox is that it necessarily involves acceleration, when the "moving" twin first goes into his moving frame, and then again when he comes to rest with his brother (and also midway, if he turns around to return). Accelerations are not really a part of Special Relativity, though, so this is always difficult to explain in a very satisfactory way.

The answer to your question is in two parts, as a result: for true inertial frame, i.e. no accelerations, observers will measure different times between events and different distances between them, yet they will all be correct. There is no illusion, no one of them is more "correct" than any other. Time and space simply work that way. An important note, however, is that each observer sees the other's clock running more slowly than his own - it's a symmetric situation.

The second part of the answer to your question has to do with the twin who returns home to find himself younger than his brother. That is not a symmetric situation, i.e. they both agree on who aged less, but in this case it's due to the fact that one of them underwent several episodes of acceleration and the other did not.

 

[russ_watters]

meaning in case of twin paradox as the twin on Earth sees that the other is aging slower
so my ? is that is the otr twin actually aging slower or is he just appearing to age slower to the twin on the Earth ??

The twins paradox is a thought experiment. The "paradox" is in the very real possibility of having a pair of identical twins who are of different ages -- and they are standing right next to each other. We don't have the propulsion technology to make that happen for real.

But there are physical measurements that show exactly the same thing. Clocks placed in orbit on satellites tick at a different rate than those on Earth and if you bring them back to earth, they show that a different amount of time has elapsed for one than the other. It is a very real effect.

Length contraction is a little more complicated, but put it this way - the closest star is about 4.5 light years away. If you were to accelerate to a high fraction of the speed of light, you would measure the distance to be shorter and you could arrive (according to your clock) in less than 4.5 years, despite never exceeding the speed of light. But since length is not like time, there is no accumulated effect like with time dilation. You can't have two formerly identical objects sitting next to each other and being different sizes.

 

[arbol]

Now according to special relativity when in the case of two people traveling at different speeds have their own clocks running differently depending on their velocity

I have understood the theoritical part of this that as c=d/t and as d= space and t= time and c= velocity of light . thus to keep c constant the value of space-time gets adjusted among itself

so my question is that in a practical situation how will the different velocities will affect the physical clock (a wrist watch) that the two people are carrying with them??how will they have their clocks show different times ??

Let S’ be a stationary one-dimensional coordinate system, and let the x’-axis of S’ lie along the x-axis of another stationary one-dimensional coordinate system S. Let S’ move along the x-axis of S in the direction of increasing x with the constant velocity v. Each system is provided with a meter stick and a number of clocks, and the meter stick and the clocks of one system are in all respects like the meter stick and the clocks of the other system.
If a ray of light moves from the origin of the moving system S’ to the point (x’,t’), then
x = x’ + v*t’ (1), and
if the ray of light moves from the origin of the stationary system S to the point (x,t), then
x’ = x – v*t (2).
x’ = c*t’, and x = c*t.
If we substitute c*t’ for x’ in equation (1) and c*t for x in equation (2), then
x = c*t’ + v*t’ (3), and
x' = c*t – v*t (4).
If we add equations (3) and (4), then
x + x' = c*(t’ + t) + v*(t’ – t).
If we keep c constant in equation (5), as long as
c = 299,792,458m/s = k*299,792,458m/k*(1s) = m*299,792,458m/m*(1s),
it does not matter what the value of v is.
For example, if
x’ = 299,792,458m,
t’ = 1s,
x = 2*299,792,458m, and
t= 2s, these values will not be affected regardless of what the value of v is.
But if x’/t’ or x/t is less than c, then their values depend on the value of v.

 

[Hernik]

The twins experience time the same way.

Answer to Archillez. I think what you might be asking is how does it feel for the two twins when time is fast for one and slower for the other.

The basic idea in special relativity is that the laws of physics are the same for any frame of reference. The very fast spaceship with twin number one on board is one frame of reference. The Earth where the other twin waits is a second frame of reference. Both twins will experience time exactly the same. It will pass just as it does for everybody. None of them will feel time differently. It is what is outside the frame of reference which changes compared to inside the frame of reference. The higher the velocity of the spaceship, the shorter distances in the slower moving surroundings outside the spaceship becomes in the direction of movement of the spaceship. And the faster time flows outside the spaceship.