AQA A Level Physics

Revision Notes

12.3.5 Time Dilation

Time Dilation

  • When objects travel close to the speed of light, to an observer moving relative to that object, it looks as if the object has slowed down 
    • This is best demonstrated by clocks
  • Observer H is in a rocket moving close to the speed of light
    • They will see their clock ticking at a regular pace, say, it is reading 15:00
  • Observer G at rest on Earth, with remarkable eyesight, will measure the clock as ticking slower
    • They will observe that time has slowed down in the spaceship from their reference frame i.e. they may see the time as 14:45 instead of 15:00
  • However, the same occurs the other way around
  • For observer H on the rocket, it is observer G that is moving relative to them 
    • Therefore, observer H will measure observer G's clock as ticking slower i.e. they see time slow down on Earth from their reference frame

1-5-7-time-dilation-ib-2025-physics

A stationary observer in their reference frame views clocks as running slower in the moving reference frame. We're back to disagreeing

The Gamma Term

  • Recall from Einstein's Postulates that moving between frames of reference means the speed needs to be changed through simple addition or subtraction
    • When travelling at speeds close to the speed of light, this no longer applies, because the speed of light is independent of the frame of reference
  • This transformation from one reference frame to another is corrected by multiplying by a scaling factor called the gamma factorgamma

gamma space equals space fraction numerator 1 space over denominator square root of 1 space minus space v squared over c squared end root end fraction

  • As v  is always less than (the speed of light), this means gamma  is always greater than 1
  • This is especially important for time dilation and length contraction
    • We will use time dilation to explain where this gamma term comes from

Time Dilation Equation

  • Consider a light clock. This consists of two mirrors facing each other with a beam of light travelling up and down between them 
  • An observer is at rest relative to the clock on a train platform and watches the light reflect between the mirrors
  • The distance between the mirrors is L and the light travels at a speed c
  • Therefore, the time interval for the light to travel from the top mirror back down to the bottom is: 

straight capital delta t subscript 0 space equals fraction numerator 2 L space over denominator c end fraction

1-5-7-stationary-light-clock

  • Another light clock is on a moving train, relative to the initial observer on the platform, travelling at constant velocity v
    • This velocity v  is close to the speed of light 
  • The stationary observer on the platform sees the light clock on the train and watches the reflection of the rays between the mirrors
  • It appears that the light rays travel to the right at an angle to the direction of motion

1-5-7-moving-light-clock-ib-2025-physics

  • The observer on the train platform sees the mirror at:
    • Position 1, when the light leaves the bottom mirror
    • Position 2, when the light returns to it
  • The distance between the mirrors, L, is the same, but the light now travels a longer path, D
  • The distance travelled by the light ray is now 2D, and the time observed between the reflections is now

increment t space equals space fraction numerator 2 D over denominator c end fraction

  • The apparent distance horizontally travelled by the mirror is v straight capital delta t where v  is the speed of the train
  • Notice that this is part of a right-angled triangle, so using Pythagoras' theorem we can see that:

D squared space equals space L squared space plus space open parentheses fraction numerator v to the power of space increment t space over denominator 2 end fraction close parentheses squared 

  • Where: 

D space equals space fraction numerator c increment t space over denominator 2 end fraction

  • We want to find increment t, the time taken for the light ray to travel up and down in the reference frame of the stationary observer on the train platform, who is moving relative to the light clock on the train
    • Remember, although it is the train that is moving, in the reference frame of an observer on the train it is the observer on the platform that is moving

increment t space equals fraction numerator fraction numerator 2 L space over denominator c end fraction over denominator square root of 1 space minus space v squared over c squared end root end fraction space equals space fraction numerator increment t subscript 0 space over denominator square root of 1 space minus fraction numerator space v squared over denominator c squared end fraction end root end fraction

  • Remember the gamma factor gamma space equals space fraction numerator 1 over denominator square root of 1 space minus space v squared over c squared end root end fraction
  • Therefore:

increment t space equals space gamma increment t subscript 0

  • Where:
    • increment t = the time interval measured from an observer moving relative to the time interval being measured (s)
    • increment t subscript 0 = the proper time interval (s)
  • As gamma > 1, this means that the increment t space greater than space increment t subscript 0
    • In other words, a clock observed from a reference frame moving relative to it will be measured to tick slower than a clock that is at rest in its frame of reference
  • The observer on the platform will view the train clock as moving slower
  • The observer on the train will view the platform as moving slower
  • You may be wondering why it's the time that slows down for the light beam, and not the light beam just speeding up to hit each mirror at the same frequency
    • This is due to Einstein's second postulate
    • Both Observers G and H must measure the speed of light to be c, so it doesn't slow down or speed up according to either reference frame
  • It is important to note that the time has been measured at the same position
    • In other words, the time interval is the position at which the light leaves the first mirror and at which it returns to the second mirror in the reference frame of the mirror

Worked example

Alex's spacecraft is on a journey to a star travelling at 0.7 c. Emma is on a space station on Earth at rest. According to Emma, the distance from the base station to the star is 14.2 ly.

Show that Alex measures the time taken for her to travel from the base station to the star to be about 14.5 years.

Answer:

Step 1: List the known quantities:

  • Distance of space station according to Emma = 14.2 ly
  • Speed of Alex's spacecraft, v = 0.7c

Step 2: Analyse the situation

  • We are trying to find the time that Alex measures for her travel i.e. the time she would measure on her own clock in the spaceship which she is stationary relative to
  • This is the proper time, Δt0

Step 3: Calculate the time taken according to Emma, Δt

  • 1 ly (light year) is the distance, s, light (at speed c) travels in a year
  • Therefore it takes the light 14.2 years (time) to travel the distance at speed

s space equals space s p e e d space cross times space t i m e space equals space 14.2 c space straight m

  • Therefore, the time taken according to Emma is:

increment t space equals space s over v space equals space fraction numerator 14.2 c over denominator 0.7 c end fraction space equals fraction numerator 14.2 over denominator 0.7 end fraction space equals space 20.29 space years

Step 4: Substitute values into the time dilation equation

increment t space equals space gamma increment t subscript 0 space space space rightwards double arrow space space space increment t subscript 0 equals space fraction numerator increment t over denominator gamma end fraction

increment t subscript 0 equals space fraction numerator 20.29 over denominator fraction numerator 1 over denominator square root of 1 space minus space fraction numerator space open parentheses 0.7 c close parentheses squared over denominator c squared end fraction end root end fraction end fraction space equals space fraction numerator 20.29 over denominator fraction numerator 1 over denominator square root of 1 space minus space open parentheses 0.7 close parentheses squared end root end fraction end fraction space equals space 14.5 space years

Step 5: Check whether your answer makes sense

  • Since Emma (who is stationary) is viewing Alex's clock (which is moving) she would measure a longer time for Alex to reach the star than Alex will
  • As Emma records 20.29 years, but Alex only records 14.5 years, this time makes sense

Exam Tip

A nice way to remember this is 'moving clocks run slower'. The caveat is what is considered 'moving' depends on the reference frame.

You will not be expected to remember this derivation, but it's helpful to know where all the factors have come from. The time dilation equation is given on your data sheet.

The notion of 'proper time' is incredibly important here, as it depends on the reference frame the time interval is being measured from. 

The maths for the derivation is only using s p e e d space equals fraction numerator space d i s t a n c e over denominator t i m e end fraction and Pythagoras' theorem.

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Dan MG

Author: Dan MG

Dan graduated with a First-class Masters degree in Physics at Durham University, specialising in cell membrane biophysics. After being awarded an Institute of Physics Teacher Training Scholarship, Dan taught physics in secondary schools in the North of England before moving to SME. Here, he carries on his passion for writing enjoyable physics questions and helping young people to love physics.