What Is the Effect of Length Contraction on Time Dilation in Relativity?

[hprog]

Suppose we have two frames of references A and B moving in a linear motion v where v is a number between 1 and 0.
Let each frame have two clocks, x and y, where y is the rear clock in the other frame, with a distance L between clocks.
Each frame K will claim that the space between the other frames clocks contracts, so the situation is symmetric.
Let the x clocks of both frames meet together at time t=0, each frame H will claim that the others y clock is set ahead with Lv/c².
Now what will happen if for each of the y clocks when their time shows t=0 they each send out another object that will travel with such a speed that when the clock will read t=Lv/c² it should arrive by the x clock of the opposite frame K^'.
(Since according to other frame K^' he is at rest while the current frame K is moving we don't have to worry about any additional motion here. )
Such a velocity is to be (if we denote the space contraction factor - and we will use here a number between o and 1 for that - with f) u=fL/(Lv/c²) = fL/( L(v/c²) ) = f/v/c² = f(c²/v) = f1/v = f/v.
Since each of them has to arrive when the y clock of the frame reads t=Lv/c² which is what the opposite frame K^' claims to be the time that y will show in the event of the meeting of the x clocks, so which one will actually arrive there?
Any one that will arrive will disprove the current frame's claim that the x and y clocks of this frame are synchronised.
On the other hand one that will not arrive will disprove the opposite frame's claim that the clock is set ahead.
So what will happen?

 

[JesseM]

Suppose we have two frames of references A and B moving in a linear motion v where v is a number between 1 and 0.
Let each frame have two clocks, x and y, where y is the rear clock in the other frame, with a distance L between clocks.
Each frame K will claim that the space between the other frames clocks contracts, so the situation is symmetric.
Let the x clocks of both frames meet together at time t=0, each frame H will claim that the others y clock is set ahead with Lv/c².
Now what will happen if for each of the y clocks when their time shows t=0 they each send out another object that will travel with such a speed that when the clock will read t=Lv/c² it should arrive by the x clock of the opposite frame K^'.

You haven't specified what frame you want the event of the y clock reading t=LV/c² to be simultaneous with the event of the object arriving at the x clock of the opposite frame. Do you want these events to be simultaneous in the frame of the y clock that sent the object, or simultaneous in the frame of the x clock that receives the object?

Since each of them has to arrive when the y clock of the frame reads t=Lv/c² which is what the opposite frame K^' claims to be the time that y will show in the event of the meeting of the x clocks, so which one will actually arrive there?

This claim is poorly-defined for the reason above. You can't talk about one event occurring "when" another distant clock reads a certain time without specifying what definition of simultaneity you are using.

Such a velocity is to be (if we denote the space contraction factor - and we will use here a number between o and 1 for that - with f) u=fL/(Lv/c²) = fL/( L(v/c²) ) = f/v/c² = f(c²/v) = f1/v = f/v.

If you want the event of the object arriving at an x-clock and the event of the y-clock in the opposite frame reading Lv/c² to be simultaneous in the frame of the y-clock, then this calculation is wrong because it ignores the fact that in the y-clock's frame the x-clock is moving so the distance between the y-clock and the x-clock at that moment will not simply be fL where f is the contraction factor (normally the contracted length would be denoted by L/gamma). In the frame of the y-clock, the distance between the y-clock and the opposite x-clock was fL at time t=0, so by time t=Lv/c² it is different. On the other hand if you want the events to be simultaneous in the frame of the x-clock, then you have ignored the fact that the y-clock was moving in this frame so the distance to the y-clock at the moment it actually sent out the object (when it read t=0) was different from fL. So in either frame, the total distance the ball travels is different from fL, meaning your calculation is incorrect either way.

edit: I messed up a little in what I said about the distance between the y-clock and the x-clock if you wanted the events to be simultaneous in the y-clock's frame. I said the distance at t=0 would be fL, but actually in this frame the distance between the y-clock and the opposite x-clock was L at time t=0 (because the opposite x-clock coincided with the y-clock's partner x-clock at t=0). From this, we can also figure out that at the time t=Lv/c² in this frame, the opposite x-clock has traveled an additional distance of v*(Lv/c²) so the distance at that moment is only L - Lv²/c² = L(1 - v²/c²) = f²L (where f is the length contraction factor $$\sqrt{1 - v^2/c^2}$$). Thus the ball sent from the y-clock to the opposite x-clock would have traveled a distance of L(1 - v²/c²) in time Lv/c² in this frame, so its speed in this frame must have been (1 - v²/c²)/(v/c²) = (c²/v) - v = (c² - v²)/v.

 

[darkhorror]

Suppose we have two frames of references A and B moving in a linear motion v where v is a number between 1 and 0.
Let each frame have two clocks, x and y, where y is the rear clock in the other frame, with a distance L between clocks.
Each frame K will claim that the space between the other frames clocks contracts, so the situation is symmetric.

distance L in which frame? It won't be the same depending on the frame of reference you chose.

 

[JesseM]

distance L in which frame? It won't be the same depending on the frame of reference you chose.

hprog must have meant each pair of x-y clocks has a distance L in their own mutual rest frame, since this will mean that in the frame of the other pair the two clocks will be out of sync by Lv/c².

 

[Dale]

Any one that will arrive will disprove the current frame's claim that the x and y clocks of this frame are synchronised.
On the other hand one that will not arrive will disprove the opposite frame's claim that the clock is set ahead.

Draw an accurate spacetime diagram of the situation. You will see that nothing is "disproved" here.

 

[hprog]

Let me clarify the situation again, this time with a more simplified situation.

Suppose we have to frames A and B.
A has one clock x, and B has two clocks y and z separated by distance L in B's frame.
The x and y clocks meet at time t=0 and see their clocks synchronised, while the z clock is yet to meet x.
So according to x then z is set ahead, and on the z clock t=0 has already passed according to x.

So according to frame A and the x clock, if another object H starts at the z clock when t=0 on the z clock, and H is moving toward x and y, there would according to B and x be a possibility that he will arrive when x and y are meeting (or at least close after).
While according to frame B (and the y and z clock) this may never happen, since they claim y and z to be synchronised, any object starting at z at t=0 already started at the time x and y had meet.

What I wanted to show with the calculations in the original posts is just to show that such motion (of H) should not have to be greater then the speed of light, and if one want to deduce the motion of the clocks then it will be even a smaller distance to cover and a smaller velocity to use, which just proves again that such situation is indeed possible.

 

[Dale]

Draw an accurate spacetime diagram as I suggested above. The resolution will be clear and you will learn a lot. Specifically, figure out the slope of the worldline for H.

 

[JesseM]

So according to frame A and the x clock, if another object H starts at the z clock when t=0 on the z clock, and H is moving toward x and y, there would according to B and x be a possibility that he will arrive when x and y are meeting (or at least close after).

Try calculating the actual velocity that would be needed for H to get from the event of z clock reading t=0 and the event of x and y meeting. Your scenario is only "possible" if it could work without H's velocity needing to be faster than light! And as it turns out, if two events are simultaneous in some inertial frame (like z reading t=0 and x&y meeting, as seen in the B frame), then that means they have a spacelike interval in all frames, so all inertial frames will agree that the distance in light-years between the events is greater than the time in years between them.

 

[John232]

Length contraction doesn't affect time dialation. A beam of light sent from an object in constant motion will have the same trajectory regardless of how much that object contracted. The beam sent perpendicular from the direction of motion will continue in a straight line from the contracted object. So then the added distance traveled of the photon relative to the observer at rest remains the same since it will move forward at the same rate compared to the objects velocity that emmitted it.