by Vaggelis Atlasis
Today I will be explaining one of the most well-known phenomenons of the special theory of relativity: time dilation. I learnt about this phenomenon while taking a course about the special theory of relativity on Coursera, by Stanford University. This information is based on what I learnt from lessons in Week 4 of the course.
What is a Light Clock?
To help explain time dilation we will use a light clock (see Figure 1). A light clock is a clock comprised of two parallel mirrors a length L away from each other. They reflect a beam of light between them. For a full tick of the clock, the light beam starts at the bottom mirror, reaches the top one and is reflected back at the bottom mirror.
A light clock; c indicates the speed of the light rays |
The Situation
Bob, stationary, has a light clock at his possession. Alice is in a spaceship, moving at a constant velocity v, and has an identical clock that is synchronised to Bob’s. Both Bob and Alice perceive themselves to be stationary and their respective counterpart moving in a certain direction; Bob sees Alice and her clock moving at v with respect to him and Alice sees Bob and his clock (as well as the Earth) moving at -v with respect to her.
Bob on Earth, with his clock, observing Alice travel to the right in her spaceship, with her clock, with velocity v. |
From Alice’s perspective, her clock is stationary and the time for a tick of her clock (ΔtickA) is equal to the distance travelled by light from the bottom mirror to the top one and back (2L) divided by its speed (c). However, from Bob’s perspective Alice’s clock is moving and the time for a tick of the moving clock (ΔtickB) will be different. As the light clock is constantly moving, by the time the light beam, travels the distance L, the top mirror will have moved some distance. Therefore, for the light to reach the top mirror it has to travel diagonally. Similarly to ΔtickA, ΔtickB will be equal to the distance travelled by light from the bottom to the top mirror and back divided by its speed. The distance travelled by light will not be 2L as light is travelling diagonally, not vertically. Instead, we will let this distance be 2D.
Alice's and Bob's clock from their respective frames of reference (perspective). For the moving clocks, snapshots have been taken at the points where light touches the mirrors (note how depending on the direction of travel the light rays travel in different directions). |
Calculating the Relationship Between ΔtickA & ΔtickB (& Deriving the Lorentz Factor)
Alice's clock as Bob sees it. L represents the distance between the mirrors, D represents the distance light travels from the bottom to the top mirror (and vice versa). Finally, X represents the horizontal distance the mirrors travel within one tick of the clock. |
We can calculate D using the Pythagorean theorem (a2 + b2 = c2). We know the vertical distance between the mirrors is L. The diagonal distance light travels from the bottom to the top mirror and vice versa is D, and the distance the bottom mirror moves in the time span of one tick of the clock (from Bob’s perspective), is X.
Hence, we end up with
D2 = L2 + X2.
By using the equation distance = velocity x time we find that
X = vΔtickB,
where v is the velocity of the spaceship and ΔtickB is the time for one tick of Alice’s clock from Bob’s perspective. We can also rearrange the previous equation
ΔtickA = 2L/c into L = c(ΔtickA)/2
and
ΔtickB = 2D/c into D = c(ΔtickB)/2.
Finally, by substituting these equations into the Pythagorean theorem equation, we end up with:
(c(ΔtickA)/2)2 + (vΔtickB/2)2 = (c(ΔtickB)/2)2
We can then simplify and rearrange this equation to find a relationship between ΔtickA and ΔtickB.
c2ΔtickA2/4 + v2ΔtickB2/4 = c2ΔtickB2/4
c2ΔtickA2 = c2ΔtickB2 – v2ΔtickB2
c2ΔtickA2 = ΔtickB2(c2 – v2)
ΔtickB = √(c2ΔtickA2/(c2 – v2))
ΔtickB = cΔtickA/√(c2-v2)
ΔtickB = ΔtickA x c/√c2(1 – v2/c2)
ΔtickB = ΔtickA x c/c√(1 – v2/c2)
ΔtickB = 1/√(1 – v2/c2) x ΔtickA
Finding the Equation for Time Dilation
From the section above, we’ve concluded that
Δtickmoving = γΔtickstationary .
By “moving” we mean an observer observing a moving clock and by “stationary” we mean an observer observing a stationary clock (Alice) from their perspective. The equation effectively tells us that an observer measuring the time on a moving clock will observe every tick of the clock take longer than an identical one at rest. We can find the relationship between the time elapsed of a moving clock and a stationary one. Let’s say γ = 4. In this case, the duration of a single tick of the moving clock corresponds to the duration of 4 ticks of the stationary clock. If the duration of one tick is 5 seconds, the elapsed time on the moving clock will be 1 x 5 = 5 seconds and the elapsed time on the stationary clock will be 4 x 5 = 20.
Hence, Δtmoving = xΔtstationary (t being time elapsed)
5 = 20x
x = ¼ = 1/γ
Therefore, we find that:
Δtmoving = (1/γ)Δtstationary.
Conclusion (Moving Clocks Run Slow)
Overall, we find that any “moving clock” (a moving object) in an observer’s perspective will run slower than an identical stationary clock, whether it is the observer’s or even if it is the same clock in another observer’s frame of reference where the clock is stationary. This is known as time dilation and is true for any moving object at a constant velocity, even if its effects are too small to notice.
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