4-Vectors and the Stress Tensor Flashcards Preview

PHYS55390M General Relativity > 4-Vectors and the Stress Tensor > Flashcards

Flashcards in 4-Vectors and the Stress Tensor Deck (30)
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1
Q

4-Velocity

Definition

A

V^α = dx^α/dλ

-where λ is the path parameter

2
Q

4-Velocity

Transformation Rule

A

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

3
Q

Basis Vectors

Transformation Rule

A

-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

4
Q

Covariant Vectors and Dual Basis

Transformation Rule

A

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

5
Q

Geometrical Representations of Contravariant and Covariant Vectors

A
  • contravariant vectors: gradients

- covariant vectors: contour lines

6
Q

How are covariant vectors naturally generated?

A
  • consider a scalar φ(X)
  • on LT: φ’=φ
  • derive it’s gradient
  • the gradient of a scalar is a covariant vector
7
Q

One-form

A

-another word for a covariant vector

8
Q

Gradient

Notation

A

-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)

9
Q

Inverse Gradient

A

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

10
Q

Massive Particle

4-Velocity

A

U^μ = dx^μ/dτ = dx^μ/dt * dt/dτ

= γ(u)(c,u_)

11
Q

Massive Particles

Scalar Invariants

A

U^μ U_μ = η_αμ U^μ U^α = -c²

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

12
Q

Massive Particle

4-Momentum

A

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_)

13
Q

Massive Particle

3-Momentum

A

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

14
Q

Massive Particle

4-Momentum & 3-Momentum

A
-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_)
15
Q

Massive Particle

What do we relate P^0 to?

A
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_)
16
Q

Massive Particle

Mass Shell Constraint

A

-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

17
Q

Massive Particle

4-Acceleration

A

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

18
Q

Massive Particle

Relationship Between 4-Velocity and 4-Acceleration

A

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

19
Q

Massless Particles

Overview

A

-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

20
Q

Massless Particles

4-Momentum

A

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

21
Q

Number Density

Definition and Transformation

A

-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

22
Q

Mass Density

Definition and Transformation

A
-mass density:
ρ = nm
-transforms like:
ρ' = γ²nm
-the extra γ arises from the mass
-not scalar invariant
23
Q

Energy Density

Definition and Transformation

A

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

24
Q

Stress-Energy-Momentum Tensor

Definition

A

T^αβ

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

25
Q

Stress-Energy-Momentum Tensor

Interpretation

A

-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

26
Q

Stress-Energy-Momentum Tensor

Perfect Fluid in Rest Frame

A

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

T^αβ = diag{ρc², p , p, p}

27
Q

Stress-Energy-Momentum Tensor

Perfect Fluid in General Frame - Description

A

-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

28
Q

Stress-Energy-Momentum Tensor

Perfect Fluid in General Frame - A & B

A
A = ρ + p/c²
B = p
29
Q

Is T^αβ symmetric?

A

-yes (in general)

30
Q

Stress-Energy-Momentum Tensor

Conservation Law

A

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