3. Spacetime and Special Relativity

Question 1

Space, Time and Motion in Newtonian Physics

Outline

Answer 1

• absolute Euclidian space
• absolute time agreed on by all observers
• absolute free motion of a particle in the absence of forces
• these three points lead to the concept of an inertial frame which describes the laws of physics

Question 2

Geodesics in Euclidian Space

Answer 2

-straight lines

Question 3

Relativity

Answer 3

-each type of relativity describes a way of relating different reference frames / observers

Question 4

Galilean Relativity

Definition

Answer 4

-in all inertial reference frames, the laws of physics should agree

Question 5

Galilean Relativity

Position

Answer 5

-have reference frame O and reference frame O’ moving along the x axis with relative velocity v such that at t’=t=0, O=O’
-Galilean transformations:
t = t’
y = y’
z = z’
x = x’ + vt

Question 6

Galilean Relativity

Velocity

Answer 6

-differentiate the position Galilean transforms:
dy/dt = dy’/dt’
dz/dt = dz’/dt’
dx/dt = dx’/dt’ + v

Question 7

Galilean Relativity

Speed of Light

Answer 7

-in the frame O:
w = (dx/dt, dy/dt, dz/dt)
-and in frame O' :
w' = (dx'/dt', dy'/dt', dz'/dt')
-then:
w = w' + v
-what if w'=c, speed of light
-then:
w = w' + v = c + v > c
-for v>0

Question 8

Space, Time and Motion in Special Relativity

Outline

Answer 8

• Einstein’s principle of relativity: “all laws of physics are the same in all inertial frames”
• speed of light principle: “the speed of light is the same in all inertial frames

Question 9

Special Relativity

Position Derivation

Answer 9

-start with the most general possible linear transformation:
x’ = Ax + Bt
t’ = Cx + Dt
-where A, B, C and D are at most functions of v
-using conditions:
x’=0, x=vt
x=0, x’=-vt’

Question 10

Special Relativity

Position in terms of Av and Ev

Answer 10

y' = y
z' = z
x' = Av (x - vt)
t' = Av (Ev x + t)

Question 11

Special Relativity

Lorentz Transformation Derivation

Answer 11

• suppose there is a third frame O’’
• O’ is moving with speed v with respect to O and O’’ moves with speed w with respect to O’
• find an expression for {x’‘,t’’} in terms of {x,t}
• all of these relations should have the same structure since they are the same transformation

Question 12

Special Relativity

Lorenzt Transformations

Answer 12

y' = y
z' = z
x' = [x - vt] \ √[1 - v²/c²]
t' = [-xv²/c² + t]\√[1 - v²/c²]

Question 13

Minkowski Metric

Answer 13

-spacetime requires a non-Eulician metric e.g. Minkowski metric
gm = -c²dtxdt + dxxdx + dyxdy + dzxdz

Question 14

Minkowski Metric and the Lorentz Transformation

Answer 14

-the Minkowski metric is invariant under the Lorentz transformation

Question 15

Lorentz Factor Definition

Answer 15

γ = 1 / √[1 - v²/c²]

γ≥1

Question 16

Lorentz Transforms in Terms of the Lorentz Factor

Answer 16

t = γ (t' + v/c² x')
x = γ(x' + vt')
y = y'
z = z'

Question 17

Special Relativity

Speed of Light

Answer 17

-consider a particle moving with speed w’=dx’/dt’ along x’ axis in frame O’ and w=dx/dt in frame O:
-using the Lorentz transforms to find expressions for dx’ and dt’
=>
w = [w’ + v] / [1 + w’ v/c²]
-suppose w’=c, then w=c
=>
-the Lorentz transforms are consistant with the speed of light principle

Question 18

Time in Special Relativity

Answer 18

t=t’

-each inertial frame has its own time

Question 19

Special Relativity

Relativity of Simultaneity and Temporal Order

Answer 19

-consider two events, suppose in frame O’ they are separated by time interval Δt’ and their x’ coordinates differ by Δx’
-then Δt and Δx in frame O can be determined using the Lorentz transforms
=>
Δt = γv/c² Δx’
-assuming v>0, this is >0 f0r Δx’>0, =0 for Δx’=0 and <0 for Δx’<0
-hence, the order of events is different in different inertial frames

Question 20

Special Relativity

Time Dilation Effect

Answer 20

-consider a standard clock at rest in O’, denote τ as the time of this clock in O’ (the proper time since the clock is at rest in this reference frame)
-then for this clock, Δx’=0 and Δt’=Δτ
-using the Lorentz transforms,
Δt = γΔτ
-since γ≥1, Δt ≥ Δτ
-similarly, any standard clock at rest in O will appear to run slower when observed in frame O’
-the rate of a moving clock slows down compared to time in the inertial frame where the clock is observed

Question 21

Special Relativity

Relativity of Spatial Order

Answer 21

-consider two events, suppose in frame O’ they are separated by time interval Δt’ and their x’ coordinates differ by Δx’
-then Δx in frame O is given by the Lorentz transform
=>
Δx = γ(Δx’ + vΔt’)
-suppose Δx’=0, then Δx=γvΔt’
-assuming v>0, this is >0 for Δt’>0, =0 for Δt’=0 and <0 for Δt’<0
=>
-spatial order of events is different in different inertial frames

Question 22

Special Relativity

Length Contraction Effect

Answer 22

-consider a bar at rest in frame O, denote l as its length measured in this frame (proper length) and suppose it is aligned with the x axis:
l = xb - xa = Δx
-it doesn’t matter when the ends of the rod, xa and xb, are measured in frame O since the bar is at rest in that frame
-in frame O’ the bar is moving so the coordinates need to be measured simultaneously e.g. at Δt’=0:
l’ = xb’ - xa’ = Δx’
-using the Lorentz transforms:
Δx=γΔx’ or l’=l/γ
-since γ≥1, l’ ≤ l
=>
-the length contraction effect, in frame O’, the coordinate grid of frame O appears contracted along the x-axis
-each inertial frame has its own space / coordinate grid

Question 23

Spacetime in Special Relativity

Answer 23

• special relativity insists that absolute space and absolute tie don’t exits, to describe any physical process an inertial frame must first be specified
• spacetime is absolute, whichever frame is used to measure space and time, the spacetime description will be the same

Question 24

Spacetime Interval General Formula

Answer 24

ds² = -c²dt² + dx² + dy² + dz²
-or separating into space and time components:
ds² = -c²dt² + dl²

Question 25

Normalising Minkowski Spacetime Coordinates

Answer 25

ds² = -c²dt² + dx² + dy² + dz²
-time coordinate is not normalised, let:
xo = ct
x1 = x
x2 = y
x3 = z
=>
ds² = -dxo² + dx1² + dx2² + dx3²

Question 26

What are the three types of spacetime interval?

Answer 26

1) space-like
2) time-like
3) light-like

Question 27

Describe the future and past light cones diagram

Answer 27

• the observer is at the centre
• a flat surface with the observer at the centre describes a hypersurface of simultaneity, a surface on which time is constant
• the vertical axis represents time
• the past light cone is a cone down the negative time axis
• the future light cone is a cone up the positive time axis
• the cones make a 45’ angle with the hypersurface
• light can only travel on the surface of the cone
• events are represented by points
• any event that has a chance of effecting you in the present lies in your past light cone, any event that you can affect lies in your future time cone

Question 28

Space-Like Spacetime Intervals

Answer 28

ds² = -c²dt² + dl²
-if ds² > 0, there exists a frame where dt=0
=>
ds² = dl²
-this represents lines on the hypersurface of simultaneity

Question 29

Time-Like Spacetime Intervals

Answer 29

ds² = -c²dt² + dl²
-if ds² < 0, there exists a frame where dl=0
=>
ds² = -c²dt²

Question 30

Light-Like Spacetime Intervals

Answer 30

ds² = -c²dt² + dl²
-if ds²=0 but dt=0, then:
|dl/dt| = c
-represents lines on the cone surface
-two events that can be connected via a light signal e.g. two events in the life of a photon