4-Vectors and the Stress Tensor

Question 1

4-Velocity

Definition

Answer 1

V^α = dx^α/dλ

-where λ is the path parameter

Question 2

4-Velocity

Transformation Rule

Answer 2

V’^α = dx’^α/dλ = Λ^α_β dx^β/dλ = Λ^α_β V^β

Question 3

Basis Vectors

Transformation Rule

Answer 3

-since V is frame independent, it can be expressed in terms of both a transformed and a non-transformed basis
-‘cancel’ V^β
=>
Λ_β^α eα’ = eβ
-i.e. the basis vector transforms using the inverse of the Lorentz transformation

Question 4

Covariant Vectors and Dual Basis

Transformation Rule

Answer 4

-contravariant vectors:
Vα’ = Λ_α^β Vβ
-basis vectors for the dual basis transform like contravariant vectors

Question 5

Geometrical Representations of Contravariant and Covariant Vectors

Answer 5

• contravariant vectors: gradients

- covariant vectors: contour lines

Question 6

How are covariant vectors naturally generated?

Answer 6

• consider a scalar φ(X)
• on LT: φ’=φ
• derive it’s gradient
• the gradient of a scalar is a covariant vector

Question 7

One-form

Answer 7

-another word for a covariant vector

Question 8

Gradient

Notation

Answer 8

-the gradient is often written more simply as:
∂φ/∂x^ν = φ,ν or ∂ν φ
-the actual components are:
∂ν = (1/c ∂/∂t, ∂/∂x, ∂/∂y, ∂/∂z)

Question 9

Inverse Gradient

Answer 9

∂^ν = η^νμ ∂μ = (-1/c ∂/∂t, ∂/∂x, ∂/∂y, ∂/∂z)

Question 10

Massive Particle

4-Velocity

Answer 10

U^μ = dx^μ/dτ = dx^μ/dt * dt/dτ

= γ(u)(c,u_)

Question 11

Massive Particles

Scalar Invariants

Answer 11

U^μ U_μ = η_αμ U^μ U^α = -c²

-scalar invariants can be calculated in ant frame - choose the simplest

Question 12

Massive Particle

4-Momentum

Answer 12

P^μ = mU^μ
-where m is the rest mass

-then the 4-velocity becomes:
U^μ = γ(u) (c, u_)
=>
P^μ = γ(u) (mc, mu_) = (γmc, p_)

Question 13

Massive Particle

3-Momentum

Answer 13

-the three momentum is usually defined as:
p_ = γ(u) m u_
-where γ is added becase we are using the rest mass

Question 14

Massive Particle

4-Momentum & 3-Momentum

Answer 14

-from 3-momentum:
p_ = γ(u) m u_
-then the 4-velocity can be written:
U^μ = γ(u) (c, u_)
-giving the 4-momentum as:
P^μ = γ(u) (mc, mu_) = (γmc, p_)

Question 15

Massive Particle

What do we relate P^0 to?

Answer 15

P^0 = γmc = mc [1 - u²/c²]^(-1/2) 
-binomial expansion:
~ mc [1 + 1/2 u²/c² + ... ]
=>
c² P^0 ~ mc² +  mu²/2
-i.e. rest mass energy plus kinetic energy
-therefore equate cP^0=E where E is total energy:
P^μ = (E/c , p_)

Question 16

Massive Particle

Mass Shell Constraint

Answer 16

-in the rest frame:
P^μ P_μ = -m²c²
-and in general:
P^μ P_μ = -(P^0)² + p_ . p_ = -E²/c² + |p_|²
-equating the two gives the mass shell constraint:
E² = m²c^4 + |p_|²c²
-in the rest frame, as expected, the total energy equates to the rest mass energy

Question 17

Massive Particle

4-Acceleration

Answer 17

A^μ = dU^μ/dτ = γ(u) dU^μ/dt
= (cγγ’, γ² a_ , γγ’u_)
-where a_ = u_’ = du_/dt

Question 18

Massive Particle

Relationship Between 4-Velocity and 4-Acceleration

Answer 18

-4-velocity and 4-acceleration always orthogonal in all frames

Question 19

Massless Particles

Overview

Answer 19

-photons have no rest mass
-since ds²=0, there is no “observer’s” or proper time
=> no DEFINED 4-velcocity
-but they do have a 4-momentum

Question 20

Massless Particles

4-Momentum

Answer 20

P^μ = (E/c , E/c n_)
-where n_ is a unit vector in the direction of propagation
-trivially:
P^μ P_μ = 0

Question 21

Number Density

Definition and Transformation

Answer 21

-if there are N objects in volume V and the number is conserved, the number density is:
n = N/V
-due to length contraction along the direction of boost, this transforms like:
n’ = γn

Question 22

Mass Density

Definition and Transformation

Answer 22

-mass density:
ρ = nm
-transforms like:
ρ' = γ²nm
-the extra γ arises from the mass
-not scalar invariant

Question 23

Energy Density

Definition and Transformation

Answer 23

-energy density is given by c²ρ = c²nm
-it transforms like:
c²γ²nm

Question 24

Stress-Energy-Momentum Tensor

Definition

Answer 24

T^αβ

-the flux of the α component of the 4-momentum across a surface of constant x^β

Question 25

Stress-Energy-Momentum Tensor

Interpretation

Answer 25

-T^00 is the flux of P^0 across x^0, flux of E/c across a surface of constant ct, =ρc² in the rest frame
-in the rest frame P^μ = (ρc, 0) and U^μ = (c , 0)
=>
T^0i = T^i0 = 0 for all i=1,2,3
Tij = 0 when i≠j
-for i=j, you have momentum flux normal to a spatial surface, pressure p in the rest frame

Question 26

Stress-Energy-Momentum Tensor

Perfect Fluid in Rest Frame

Answer 26

-for a perfect fluid in the rest frame is therefore:

T^αβ = diag{ρc², p , p, p}

Question 27

Stress-Energy-Momentum Tensor

Perfect Fluid in General Frame - Description

Answer 27

-the only additional factors that can be important are U^α (bulk motions) and η^αβ (background space)
-must have something like:
T^αβ = A U^α U^β + B η^αβ
-where A and B are functions of ρ and p

Question 28

Stress-Energy-Momentum Tensor

Perfect Fluid in General Frame - A & B

Answer 28

A = ρ + p/c²
B = p

Question 29

Is T^αβ symmetric?

Answer 29

-yes (in general)

Question 30

Stress-Energy-Momentum Tensor

Conservation Law

Answer 30

∂μ T^μν = 0
-since T is symmetric:
∂μ T^μν = 0 = T^μν,μ = T^μν,ν