I was wondering about Time-dilation in Special Relativity. I am still a middle school student who wonders so please excuse me if I missed any important aspects.

Let us assume we have a system of coordinates where an point $A$ and $B$ which are located upon the walls of an object where $A$ is located in the axis of $(X_1, Y_1)$ and $B$ upon axis of $(X_2, Y_1)$ are stationary and let us assume the time is the time an beam of light takes to reach from $A$ till $B$ and back.

Using the following coordinates given above we can calculate the objects length (no need for width) as being $L = X_2 - X_1$

Using the assumption we can say the ray must travel the distance of $2AB$ in order for an unit of our time to pass or since $AB$ are the line till the start and edge of object the light ray must only travel $2L$ in an stationary system of coordinates (frame of reference).

Now let us incorporate this concept of our "time unit" and apply this to a new system of coordinates where the object is travelling at an velocity $v_0$ in the $X$ axis of our system of coordinates, using the derivations of special relativity of length contraction (Lorentz-Fitzgerald Contraction) we can say the object will begin to contract in the direction of velocity ($X$ axis) with the following factor:

$$L' = \gamma L$$ $$L' = L * (\sqrt{1 - \frac{v^2}{c^2}})$$ $$L' = (X_2 - X_1) * (\sqrt{1 - \frac{v_0{^2}}{c^2}})$$

Using the equation of the time unit I derived for time, we can say now a unit of our time for an observer will be the distance (light will cover) for a unit of time to pass for us (observer):

$$2L'$$ $$2 * ((X_2 - X_1) * (\sqrt{1 - \frac{v_0{^2}}{c^2}})) $$

This clearly shows as the object attain velocity $L'$ gets smaller super-exponentially and from Michelson-Morley experiment we know that $c$ is always constant in all frames of reference and time-frames, now using basic calculations from classical mechanics we can say that light takes less time to cover $L'$ as:

$D = \frac{L}{c}$ will always be bigger than $D' = \frac{L'}{c}$, therefore as solving this further we will get:

$$D = \frac{(2 * (X_2 - X_1))}{c}$$

while $D'$ gives us: $$D' = \frac{(2* ((X_2 - X_1) * (\sqrt{1 - \frac{v_0{^2}}{c^2}})))}{c}$$

so we can say $c$ has to cover less distance in $D'$ as $D' > D$which leads to us as a observer to calculate a quicker time on an moving body on contrary to Einsteins actual time-dilation which shows complete opposite, that shows time slows down as you increase velocity.

Not only has Einstein been proven correct but using logical reasoning I am still confused as I still believe I should be correct yet I am not, why is my reasoning wrong? Am I missing an important factor of Special Relativity that should have been added in this thought experiment? If so what is it? and why is my line of thought incorrect?

  • $\begingroup$ Actually, time goes slower when you accelerate (or "decelerate", which is just acceleration in a different direction). In your example, both frames are moving at a constant velocity. Remember, if someone whizzes by you at .90c, they actually think they are fixed and you are moving at .90c in the other direction. In other words, you both see each other the same way. It's only if one of you accelerates to the other's frame will there be non-symmetry. This is one of the fundamental mistakes made in the book "Time For the Stars" $\endgroup$ – barrycarter Oct 29 '14 at 1:15
  • $\begingroup$ @barrycarter - Although it's true the views of constant-velocity observers are symmetrical, it's also true that each measures the other one's clock to be running slower than their own, so in a frame-dependent sense it's perfectly correct to say that moving clocks run slower (relative to whatever frame you're using). $\endgroup$ – Hypnosifl Oct 29 '14 at 1:48

"Using the equation of the time unit I derived for time, we can say now a unit of our time for an observer will be the distance (light will cover) for a unit of time to pass for us (observer)"

Is this observer meant to be the one who sees the object with the walls A and B moving at velocity v? If so, then although it's true this observer will see the length contracted to $ L' = L \sqrt{1 - v^2/c^2} $, it's not true that she'll see the time for the light to go from A to B and back as $2L'/c$. You're forgetting that when light is emitted from A, B is moving away from the point of emission in the observer's frame, which adds to the time on this leg of the trip, and when light is reflected from B back towards A, the A is moving towards the point of reflection, which shortens the time on this leg.

For example, say A and B at either ends of a ruler which is 50 light-seconds long in the rest frame of A and B, and they are moving at v=0.6c in my frame. Then in my frame, the ruler is only 40 light-seconds long, so that's the distance between A and B at any given instant. But this does not mean I measure the time for light to go from A to B and back to be 80 seconds. Suppose at t=0 in my frame, A is at position x=0 and B is at position x=40, and at that moment a light flash is emitted at the position of A. Then the light will not actually catch up with B until t=100 in my frame, because after 100 seconds B has moved 100*0.6c = 60 light-seconds, and since it was at x=40 at t=0, at t=100 it will be at x=40+60=100; meanwhile of course the light moves 100 light-seconds in 100 seconds, so at t=100 it will also be at x=100. And at t=100, if B is at x=100 then A must be at x=60 at the same moment, since the distance between them is always 40 light-seconds in my frame.

Then 25 seconds later at t=125 in my frame, A must have moved 25*0.6c = 15 light-seconds, so it will be at position x=60+15=75 light-seconds; and in those 25 seconds the light reflected at position x=100 will have moved 25 light-seconds in the -x direction, so it'll be at x=75 light-seconds as well. Thus, we conclude that in my frame it takes a time of 125 seconds for the light to travel from A to B and back to A, in spite of the fact that the distance between A and B at any given instant is only 40 light-seconds in my frame. And since the clock at A is only running at 0.8 the rate of my own clock (as measured in my frame), it will have only elapsed 125*0.8 = 100 seconds in between the light departing A and the light returning to A, which is exactly what you'd expect given that the distance from A to B is 50 light-seconds in the A/B rest frame.

If you're interested, I analyzed this example in more detail, showing how both frames manage to agree about the one-way speed of light as well as the two-way speed of light, in this answer to another question.


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