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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.
  • Explain why the twin paradox is not a contradiction.
  • 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.

Figure a shows an illustration of an astronaut in the space shuttle observing an analog clock with an elapsed time Delta tau. The details of the clock experiment are also shown as follows: There is a light source, a receiver a short distance to its right, and a mirror centered above them. The vertical distance from the receiver and light source to the mirror is labeled as D. The path of the light from the source, up to the mirror, and back down to the receiver is shown. Figure b shows an observer on earth with an analog clock showing a time interval Delta t. Above the observer are three diagrams showing the clock experiment on the space shuttle at three different times and the path of the light. The light source in the diagram on the left is labeled “beginning event.” The receiver in the diagram on the right is labeled “ending event.” The path of the light forms a straight line going diagonally up and to the right, from the source in the diagram on the left to the mirror in the center diagram, and then another straight line going diagonally down and to the right, from the mirror in the center diagram to the receiver in the diagram on the right. The vertical distance from the receiver to the mirror is labeled D. The horizontal distance from the beginning event to the clock location in the center diagram is labeled L= v Delta t over 2. The horizontal distance from the clock location in the center diagram to the ending event is labeled L. Figure c shows an isosceles triangle with a horizontal base. The triangle is divided by a vertical line from its apex to its base into two identical right triangles with the vertical line forming a side that is shared by the two right triangles. This side is labeled D. The base of the triangle on the left is labeled L= v Delta t over 2. The base of the triangle on the right is labeled L. The hypotenuse of each of the right triangles is labeled s. Above the diagram is the equation s equals the square root of the quantity D squared plus L squared.
(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 Δ t or Δ τ ) of the pulse in the corresponding reference frame. The resulting equation can then be solved for Δ t in terms of Δ τ .

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 Δ t in the earthbound observer’s frame. The length D is the distance that the light pulse travels in time Δ τ in the astronaut’s frame. This gives us three equations:

2 s = c Δ t ; 2 L = v Δ t ; 2 D = c Δ τ .

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:

Questions & Answers

A round diaphragm S with diameter of d = 0.05 is used as light source in Michelson interferometer shown on the picture. The diaphragm is illuminated by parallel beam of monochromatic light with wavelength of λ = 0.6 μm. The distances are A B = 30, A C = 10 . The interference picture is in the form of concentric circles and is observed on the screen placed in the focal plane of the lens. Estimate the number of interference rings m observed near the main diffractive maximum.
Jyoti Reply
A Pb wire wound in a tight solenoid of diameter of 4.0 mm is cooled to a temperature of 5.0 K. The wire is connected in series with a 50-Ωresistor and a variable source of emf. As the emf is increased, what value does it have when the superconductivity of the wire is destroyed?
Rupal Reply
how does colour appear in thin films
Nwjwr Reply
in the wave equation y=Asin(kx-wt+¢) what does k and w stand for.
Kimani Reply
derivation of lateral shieft
James Reply
total binding energy of ionic crystal at equilibrium is
All Reply
How does, ray of light coming form focus, behaves in concave mirror after refraction?
Bishesh Reply
Refraction does not occur in concave mirror. If refraction occurs then I don't know about this.
What is motion
Izevbogie Reply
Anything which changes itself with respect to time or surrounding
and what's time? is time everywhere same
how can u say that
do u know about black hole
Not so more
Radioactive substance
These substance create harmful radiation like alpha particle radiation, beta particle radiation, gamma particle radiation
But ask anything changes itself with respect to time or surrounding A Not any harmful radiation
explain cavendish experiment to determine the value of gravitational concept.
Celine Reply
 Cavendish Experiment to Measure Gravitational Constant. ... This experiment used a torsion balance device to attract lead balls together, measuring the torque on a wire and equating it to the gravitational force between the balls. Then by a complex derivation, the value of G was determined.
For the question about the scuba instructor's head above the pool, how did you arrive at this answer? What is the process?
Evan Reply
as a free falling object increases speed what is happening to the acceleration
Success Reply
of course g is constant
acceleration also inc
which paper will be subjective and which one objective
normal distributiin of errors report
normal distribution of errors
acceleration also increases
there are two correct answers depending on whether air resistance is considered. none of those answers have acceleration increasing.
Acceleration is the change in velocity over time, hence it's the derivative of the velocity with respect to time. So this case would depend on the velocity. More specifically the change in velocity in the system.
photo electrons doesn't emmit when electrons are free to move on surface of metal why?
Rafi Reply
What would be the minimum work function of a metal have to be for visible light(400-700)nm to ejected photoelectrons?
Mohammed Reply
give any fix value to wave length
40 cm into change mm
Arhaan Reply
40cm=40.0×10^-2m =400.0×10^-3m =400mm. that cap(^) I have used above is to the power.
i.e. 10to the power -2 in the first line and 10 to the power -3 in the the second line.
there is mistake in my first msg correction is 40cm=40.0×10^-2m =400.0×10^-3m =400mm. sorry for the mistake friends.
40cm=40.0×10^-2m =400.0×10^-3m =400mm.
this msg is out of mistake. sorry friends​.
what is physics?
sisay Reply
Practice Key Terms 2

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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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