Wize University Physics Textbook (Master) > Modern Physics

Basics of Special Relativity

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Lorenz Transformations

Wize Concept
Basis of Special Relativity:
The laws of physics are the same in all inertial reference frames.
The speed of light is constant (in vacuum) in all inertial reference frames, and is approximately equal to c=3×108m/sc=3\times 10^8 m/sc=3×108m/s

Inertial reference frames are systems of coordinates that move with a constant speed relative to each other, but do not accelerate with respect to each other.

In special relativity, every event is represented in the 444-dimensional spacetime as (x,y,z,t)(x,y,z,t)(x,y,z,t), where xxx, yyy and zzz are the spatial coordinates and show the point at which this event happens, and ttt is the time at which it happens. All 444 coordinates are measured from an arbitrary "origin" at which they're all zero.

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Consider an event shown by (x,y,z,t)(x,y,z,t)(x,y,z,t) in frame KKK. 

The coordinates (x′,y′,z′,t′)(x',y',z',t')(x′,y′,z′,t′) describe the same event in another frame K′K'K′, which is moving along the x−x-x−axis with a constant speed vvv relative to KKK. 

The Lorentz Transformations between these two reference frames are: 

 t′=γ(t−vxc2) \boxed{ \ t'=\gamma\bigg(t-\dfrac{vx}{c^2}\bigg) \ } t′=γ(t−c2vx​) ​

 x′=γ(x−vt)\boxed{ \ x'=\gamma(x-vt)} x′=γ(x−vt)​

 y′=y \boxed{\ y'=y \ } y′=y ​

 z′=z \boxed{\ z'=z \ } z′=z ​

where Gamma is defined to be:      
 γ=11−v2c2 \boxed{\ \gamma=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}} \ } γ=1−c2v2​​1​ ​

Wize Concept
Gamma is also called the dilation factor, and since v<cv<cv<c  always, we have γ>1\gamma>1γ>1.
When the speed vvv is very small, γ\gammaγ is close to 111.
As the speed vvv approaches the speed of light, γ\gammaγ approaches infinity.

Exam Tip
If the other reference frame is moving along the y−y-y− or z−z-z−axis instead, then that particular coordinate will be transformed, and the other two will stay the same.

Wize Concept
The Inverse Lorentz Transformations from K′K'K′ to KKK can be found by switching  x↔x′\bf{x\leftrightarrow x'}x↔x′ ,  y↔y′\bf{y\leftrightarrow y'}y↔y′ ,  z↔z′\bf{z\leftrightarrow z'}z↔z′ ,  t↔t′\bf{t\leftrightarrow t'}t↔t′  and  v↔−v\bf{ v\leftrightarrow -v }v↔−v  in the equations above.

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Time Dilation and Length Contraction

Time Dilation
The time between two events in not the same in inertial reference frames that are moving with respect to each other.

Clocks run slower in a moving frame compared to a stationary frame. That is why the time they measure between two events is longer than the time between the same two events where the clocks are stationary:

 Δt=Δt01−v2c2=γ Δt0 \boxed{ \ \Delta t=\frac{\Delta t_0}{\sqrt{1-\dfrac{v^2}{c^2}}}=\gamma \ \Delta t_0 \ } Δt=1−c2v2​​Δt0​​=γ Δt0​ ​

where Δt0\Delta t_0Δt0​ is the time interval between the two events in the rest frame, and is called the proper time.

Since γ>1\gamma>1γ>1, the time interval in the moving frame is greater than the proper time (time dilation).

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Length Contraction

Length is also relative. The length of a moving object seen from an inertial reference frame at rest is:

 l=l0 1−v2c2=l0γ \boxed{ \ l=l_0\ \sqrt{1-\frac{v^2}{c^2}}=\frac{l_0}{\gamma} \ } l=l0​ 1−c2v2​​=γl0​​ ​

where l0l_0l0​is the length of the object at rest, and is called the proper length. 

Since γ>1\gamma>1γ>1, the length in the moving frame is smaller than the proper length (length contraction).

Exam Tip
Length contraction can only happen along the direction of motion.

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Example: Moving Train

A train travels at  80%80\%80% of speed of light relative to a stationary observer. The observer closed his eyes for 111 minute according to his wristwatch. How long were his eyes closed from the point of view of a passenger on the train? 

We have v=0.8 cv=0.8\ cv=0.8 c

Using the time dilation formula we get:

t=γ t0=t01−v2c2 t=\gamma\ t_0=\dfrac{t_0}{\sqrt{1-\dfrac{v^2}{c^2}}}t=γ t0​=1−c2v2​​t0​​

                 =11−0.82c2c2= \dfrac{1}{\sqrt{1-\dfrac{0.8^2\cancel{c^2}}{\cancel{c^2}}}}=1−c20.82c2​​1​

                 =1.67=1.67=1.67  (min)

                 =100=100=100  (s)

Practice: Car in a Garage

A car has length of 444 m at rest. At what speed should the car move to fit in a garage of length 3.93.93.9 m measured by a person that's stationary with respect to the garage? Answer in terms of ccc.

Enter the number only (2 decimal places, don't enter the c):

I don't know

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Relativistic Velocity Transformations

Imagine an inertial reference frame K′K'K′ that is moving with a constant speed vvv relative to another inertial reference frame KKK. We have an object moving with respect to K′K'K′with velocity of u′u'u′. 

The velocity of this object with respect to the KKK reference frame is called uuu and is given by: 

 u=u′+v1+u′vc2 \boxed{ \ u=\frac{u'+v}{1+\dfrac{u'v}{c^2}} \ } u=1+c2u′v​u′+v​ ​

The reverse of this transformation can be found by switching  u↔u′\bf{u\leftrightarrow u'}u↔u′  and  v↔−v\bf{ v\leftrightarrow -v }v↔−v  in the equation above, and is given by:

 u′=u−v1−uvc2 \boxed{ \ u'=\dfrac{u-v}{1-\dfrac{uv}{c^2}} \ } u′=1−c2uv​u−v​ ​

Relativistic Momentum and Energy

Relativistic Momentum is given by: 
 p=γ mv \boxed{ \ p=\gamma\  mv \ } p=γ mv ​

with  γ=11−v2c2\gamma=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}γ=1−c2v2​​1​   as before. 

Relativistic Energy is given by:

 E2=p2c2+m2c4 \boxed{\ E^2=p^2c^2+m^2c^4 \ } E2=p2c2+m2c4 ​

If the object is at rest, then p=0p=0p=0 and we have the rest mass energy:

 E0=mc2 \boxed{\ E_0=mc^2 \ } E0​=mc2 ​

The relativistic kinetic energy could be found using:

 KE=mc2 (γ−1) \boxed{\ KE=mc^2\ (\gamma-1) \ } KE=mc2 (γ−1) ​

Watch Out!
These only apply to massive particles.  For massless particles such as photons we still have  p=hλp=\dfrac{h}{\lambda}p=λh​  and  E=hfE=hfE=hf.

Practice: Relativistic Velocity, Momentum and Energy

A spaceship is moving away from Earth at a speed of 0.8 c0.8\ c0.8 c relative to Earth when the spaceship captain shoots a bullet of mass 555 g with a speed of 0.2 c0.2\ c0.2 c relative to the spaceship, in the direction along the motion of the spaceship. Calculate: 

a) The speed of the bullet relative to someone stationary on Earth. 
b) The momentum of the bullet relative to someone stationary on Earth. 
c) The rest mass energy of the bullet.
d) The kinetic energy of the bullet according to the captain.

Speed (in m/s, 2 decimal places, don't include units):

Momentum (in kg m/s, 2 decimal places, don't include units):

Rest mass energy (in J, 2 decimal places, don't include units):

Kinetic energy (in J, 2 decimal places, don't include units):

I don't know