Year 1 Sem 1/LC Classical Mechanics and Relativity 1/Lecture 07
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Time Dilation
We concluded the previous lecture with:
We rearrange to get:
Hence:
This tells us that the time recorded for the photon to travel by the two observers is different. The moving clock runs slower, so the stationary observer measures a longer duration than the moving clock records. For non-relativistic speeds, $\gamma \approx 1$ so the difference is negligible; however, at larger speeds the disparity becomes much larger and grows without bounds. This makes physical sense, as for faster speeds, the trolley will have travelled a larger horizontal distance, therefore (A) will measure a longer path, and hence require a larger time.
Generally, we have:
Where $t_0$ is the “proper time” and is defined as the time interval taken between two events that take place in the same frame, by an observer in that frame.
Length Contraction
We have the same identical setup, except the setup is now horizontally on the trolley. Observer (B) is again attached to the cylinder, with photons again bouncing along the length of the cylinder, just with the left/right instead of top/bottom surfaces. The cylinder is still moving on a trolley, and observer (A) is still stationary.
We define $\Delta t$ as the interval between emission and detection, again with a subscript to denote who is making the measurement.
In (B)’s frame of reference:
In (A)’s frame of reference:
The cylinder has moved to the right as the photon travels, this adds some extra length that the photon must travel.
This extra length (between the two dotted lines) is $u \Delta t_1$ where $\Delta t_1$ is the time for the photon to hit the right wall. The photon now hits the wall and bounces back, while the cylinder is still moving to the right. The cylinder will move $u \Delta t_2$, where $\Delta t_2$ is the time taken for the photon to travel back and hit the left wall.
The time taken between emission and detection $\Delta t$ is given by:
For the first part of the trip, the distance is $L_A + u \Delta t_1$, so:
For the second part of the trip, the distance is less as the cylinder “catches up” with the photon as it moves, giving us a distance of $L_A - u \Delta t_2$. This gives us:
Hence the round-trip time is:
So:
And we know that:
The fact A and B don’t agree is fine, we can apply the time dilation formula:
This tells us that measurements of lengths for a stationary vs moving observer also do not agree. Just like time runs slower for a moving object, a moving object will be measured to be smaller by a stationary observer. Effectively, moving lengths shrink. This is called length contraction.
Generally, we have:
Where $\Delta L_0$ is the “proper length” of an object, i.e. the length of an object measured by an observer at rest relative to it.
Example
We have someone on a spaceship, coming back to earth. The spaceship is travelling rightwards, directly towards the earth in 1D.
The spaceship has an astronaut, and we consider one person on the earth in Mission Control. MC spots the spaceship at distance $l = 3,000 \text{km}$ and is at speed $u$ corresponding to $\gamma = 10$ (so $u \approx c$). The earth is stationary.
When the spaceship is at this distance $l$, the spaceship sends earth a distress signal saying $\Delta t_r = 10^{-3}$s, where $\Delta t_r$ is how long the spaceship’s oxygen supply lasts. We assume the astronaut cannot hold their breath and dies immediately if the oxygen supply runs out before the ship makes it back to Earth. Will the astronaut survive?
MC’s child, who has no knowledge of relativity, calculates the travel time:
The child compares this to the oxygen supply ($10^{-3}s$) and concludes the astronaut dies. This isn’t quite accurate however, as we need to treat the oxygen time relativistically. To MC, the clock on the spaceship runs slow (Time Dilation):
Comparing the relativistic oxygen duration ($10^{-2}s$) to the travel time ($10^{-2}s$), we see that the astronaut (just barely!!) survives.
We solved this in the Earth frame, where the distance $l$ is a proper length. We could alternatively solve this in the Astronaut’s frame, where the distance to Earth is length contracted to $L = l/\gamma = 300\text{km}$. In that frame, the travel time is $10^{-3}s$, which matches the proper time of the oxygen supply, so gives the same result.