# 5.3 Time dilation

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By the end of this section, you will be able to:
• Explain how time intervals can be measured differently in different reference frames.
• Describe how to distinguish a proper time interval from a dilated time interval.
• Describe the significance of the muon experiment.
• Calculate time dilation given the speed of an object in a given frame.

The analysis of simultaneity shows that Einstein’s postulates imply an important effect: Time intervals have different values when measured in different inertial frames. Suppose, for example, an astronaut measures the time it takes for a pulse of light to travel a distance perpendicular to the direction of his ship’s motion (relative to an earthbound observer), bounce off a mirror, and return ( [link] ). How does the elapsed time that the astronaut measures in the spacecraft compare with the elapsed time that an earthbound observer measures by observing what is happening in the spacecraft?

Examining this question leads to a profound result. The elapsed time for a process depends on which observer is measuring it. In this case, the time measured by the astronaut (within the spaceship where the astronaut is at rest) is smaller than the time measured by the earthbound observer (to whom the astronaut is moving). The time elapsed for the same process is different for the observers, because the distance the light pulse travels in the astronaut’s frame is smaller than in the earthbound frame, as seen in [link] . Light travels at the same speed in each frame, so it takes more time to travel the greater distance in the earthbound frame. (a) An astronaut measures the time Δ τ for light to travel distance 2 D in the astronaut’s frame. (b) A NASA scientist on Earth sees the light follow the longer path 2 s and take a longer time Δ t . (c) These triangles are used to find the relationship between the two distances D and s .

## Time dilation

Time dilation is the lengthening of the time interval between two events for an observer in an inertial frame that is moving with respect to the rest frame of the events (in which the events occur at the same location).

To quantitatively compare the time measurements in the two inertial frames, we can relate the distances in [link] to each other, then express each distance in terms of the time of travel (respectively either $\text{Δ}t$ or $\text{Δ}\tau$ ) of the pulse in the corresponding reference frame. The resulting equation can then be solved for $\text{Δ}t$ in terms of $\text{Δ}\tau .$

The lengths D and L in [link] are the sides of a right triangle with hypotenuse s . From the Pythagorean theorem,

${s}^{2}={D}^{2}+{L}^{2}.$

The lengths 2 s and 2 L are, respectively, the distances that the pulse of light and the spacecraft travel in time $\text{Δ}t$ in the earthbound observer’s frame. The length D is the distance that the light pulse travels in time $\text{Δ}\tau$ in the astronaut’s frame. This gives us three equations:

$2s=c\text{Δ}t;2L=v\text{Δ}t;2D=c\text{Δ}\tau .$

Note that we used Einstein’s second postulate by taking the speed of light to be c in both inertial frames. We substitute these results into the previous expression from the Pythagorean theorem:

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