4-Vectors and the Stress Tensor Flashcards
4-Velocity
Definition
V^α = dx^α/dλ
-where λ is the path parameter
4-Velocity
Transformation Rule
V’^α = dx’^α/dλ = Λ^α_β dx^β/dλ = Λ^α_β V^β
Basis Vectors
Transformation Rule
-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
Covariant Vectors and Dual Basis
Transformation Rule
-contravariant vectors:
Vα’ = Λ_α^β Vβ
-basis vectors for the dual basis transform like contravariant vectors
Geometrical Representations of Contravariant and Covariant Vectors
- contravariant vectors: gradients
- covariant vectors: contour lines
How are covariant vectors naturally generated?
- consider a scalar φ(X)
- on LT: φ’=φ
- derive it’s gradient
- the gradient of a scalar is a covariant vector
One-form
-another word for a covariant vector
Gradient
Notation
-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)
Inverse Gradient
∂^ν = η^νμ ∂μ = (-1/c ∂/∂t, ∂/∂x, ∂/∂y, ∂/∂z)
Massive Particle
4-Velocity
U^μ = dx^μ/dτ = dx^μ/dt * dt/dτ
= γ(u)(c,u_)
Massive Particles
Scalar Invariants
U^μ U_μ = η_αμ U^μ U^α = -c²
-scalar invariants can be calculated in ant frame - choose the simplest
Massive Particle
4-Momentum
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_)
Massive Particle
3-Momentum
-the three momentum is usually defined as:
p_ = γ(u) m u_
-where γ is added becase we are using the rest mass
Massive Particle
4-Momentum & 3-Momentum
-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_)
Massive Particle
What do we relate P^0 to?
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_)
Massive Particle
Mass Shell Constraint
-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
Massive Particle
4-Acceleration
A^μ = dU^μ/dτ = γ(u) dU^μ/dt
= (cγγ’, γ² a_ , γγ’u_)
-where a_ = u_’ = du_/dt
Massive Particle
Relationship Between 4-Velocity and 4-Acceleration
-4-velocity and 4-acceleration always orthogonal in all frames
Massless Particles
Overview
-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
Massless Particles
4-Momentum
P^μ = (E/c , E/c n_)
-where n_ is a unit vector in the direction of propagation
-trivially:
P^μ P_μ = 0
Number Density
Definition and Transformation
-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
Mass Density
Definition and Transformation
-mass density: ρ = nm -transforms like: ρ' = γ²nm -the extra γ arises from the mass -not scalar invariant
Energy Density
Definition and Transformation
-energy density is given by c²ρ = c²nm
-it transforms like:
c²γ²nm
Stress-Energy-Momentum Tensor
Definition
T^αβ
-the flux of the α component of the 4-momentum across a surface of constant x^β
Stress-Energy-Momentum Tensor
Interpretation
-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
Stress-Energy-Momentum Tensor
Perfect Fluid in Rest Frame
-for a perfect fluid in the rest frame is therefore:
T^αβ = diag{ρc², p , p, p}
Stress-Energy-Momentum Tensor
Perfect Fluid in General Frame - Description
-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
Stress-Energy-Momentum Tensor
Perfect Fluid in General Frame - A & B
A = ρ + p/c² B = p
Is T^αβ symmetric?
-yes (in general)
Stress-Energy-Momentum Tensor
Conservation Law
∂μ T^μν = 0
-since T is symmetric:
∂μ T^μν = 0 = T^μν,μ = T^μν,ν
