Four vector representation
Kirit Makwana
September 17, 2020
Kirit Makwana Four vector representation September 17, 2020 1 / 21 Frame transformation
The rotation transform is
x0 = x cos θ + y sin θ (1) y 0 = −x sin θ + y cos θ (2)
Kirit Makwana Four vector representation September 17, 2020 2 / 21 Consider the Lorentz transform of 2 events (x1, t1)& x2, t2
0 0 2 x1 = γ(x1 − vt1) t1 = γ(t1 − (v/c )x1) (3) 0 0 2 x2 = γ(x2 − vt2) t2 = γ(t2 − (v/c )x2) (4)
Then the transformation of the intervals between these events are ∆t = t2 − t1, ∆x = x2 − x1
∆x0 = γ(∆x − v∆t) (5) ∆t0 = γ(∆t − (v/c2)∆x) (6)
So these intervals also transform as the Lorentz transform. This is a consequence of the linearity of the transform This also implies that they obey Lorentz invariance
∆x2 + ∆y 2 + ∆z2 − c2∆t2 = ∆x02 + ∆y 02 + ∆z02 − c2∆t02 (7)
Kirit Makwana Four vector representation September 17, 2020 3 / 21 Space-time diagram
We can plot events on a 2D map of x − t (ignoring y and z) As the origin of this map can be shifted to the position of the event (x1, t1), the x-axis is ∆x and y-axis is ∆t For convenience we take units of length and time, such that speed of light is 1. In these units a light ray starting from the origin will travel along 45o straight lines. These form the “light cone”
Kirit Makwana Four vector representation September 17, 2020 4 / 21 Space-time diagram
With respect to the first event, ∆t > 0 represents the future, while ∆t < 0 is the past A particle moving with velocity v will follow a trajectory of ∆t = ∆x/v Consider the Lorentz invariant quantity ∆τ 2 ≡ ∆x2 − c2∆t2
Kirit Makwana Four vector representation September 17, 2020 5 / 21 Minkowski diagram
How do the axis of the S0 frame look? For ∆x0 axis, set ∆t0 = 0. This means ∆t = v∆x. This makes an angle ψ = tan−1(v) w.r.t. the ∆x axis Similarly the ∆t0 axis makes an angle ψ with the ∆t axis We have transformed to a non-orthogonal axis setup
Kirit Makwana Four vector representation September 17, 2020 6 / 21 Minkowski diagram
It can be shown that α = sin ψ(∆t − ∆x tan ψ) =⇒ β = α/ cos(2ψ) (8) 1 =⇒ ∆x0 = (∆x/ cos ψ) − β = (∆x − v∆t) (9) cos ψ(1 − tan2 ψ) √ 1 + v 2 = f (v) (∆x − v∆t) (10) 1 − v 2
Kirit Makwana Four vector representation September 17, 2020 7 / 21 Minkowski diagram
To find f (v) consider a rod of length L0 stationary in frame S, with on end at origin. It’s trajectory is vertical. It cuts the ∆x0 axis at distance L0 s f (V )L L 1 − v 2 L0 = 0 = 0 =⇒ f (V ) = (11) cos(ψ) γ 1 + v 2 This recovers the Lorentz transform. Similarly the time transformation can also be derived. Kirit Makwana Four vector representation September 17, 2020 8 / 21 Intervals Intervals with ∆τ 2 < 0 are called time-like. Such an interval lies inside the light cone of the event For such intervals, the physical separation is less than the distance light can travel during the time interval If ∆t > 0 for a time-like interval, then you can never find a frame where ∆t0 = 0, because that would violate Lorentz invariance. This also implies that ∆t > 0 in all frames This means that such events always occur in the same order in all reference frames and there can be a causal relation between them For such intervals, it is always possible to find a frame where the 2 events occurred at the same space-point
Kirit Makwana Four vector representation September 17, 2020 9 / 21 Intervals
Intervals with ∆τ 2 > 0 are called space-like. Such an interval lies outside the light cone This implies that no light ray (or any information) can travel across this interval No frame can be found where these events occur at the same point. However, one can always find a frame where the 2 events occurred at same time There can be no causal relation between such intervals, as the time ordering of such events can change between frames Intervals with ∆τ 2 = 0 are called “light-like” or “null”. Such intervals can only be connected with light rays
Kirit Makwana Four vector representation September 17, 2020 10 / 21 4 vectors
Define a 4-component vector in 3 space + 1 time dimension
ct µ x X = , µ = 0, 1, 2, 3 (12) y z
i.e. X 0 = ct, X 1 = x, X 2 = y, X 3 = z Lorentz transform can be written as a matrix multiplication
0 −γv ct γ c 0 0 x0 X −γv γ 0 0 X 0µ = = ΛµX ν;Λµ = c (13) y 0 ν ν 0 0 1 0 ν z0 0 0 0 1
Here µ represents row number while ν represents the column number
Kirit Makwana Four vector representation September 17, 2020 11 / 21 Tensors Tensors can be thought of as objects that follow multi-linear maps going from one space to another space (warning: not a mathematically rigorous statement, tensors have a rigorous mathematical definition that I am not going into) Practically they are scalars, vectors, matrices, and even higher dimensional objects that obey certain transformation laws under change of co-ordinate systems A co-ordinate system consists of basis vectorse ˆi and any vector can be expressed as a linear combination of these basis vectors X A = Ai eˆi (14) i 0 In another coordinate system, the unit vectors might be differente ˆi , and the same physical vector might have different co-ordinates
X 0 0 A = Ai eˆi (15) i
Kirit Makwana Four vector representation September 17, 2020 12 / 21 Tensor transformation
0 The vector A will be a tensor if the co-ordinates Ai are related to Ai by a multi-linear map
ˆ ˆ 0 0 ˆ0 0 ˆ0 The position vector r = x1i + y1j will transform to r = x1i + y1j , 0 0 where x1 = x1 cos θ + y1 sin θ & y1 = −x1 sin θ + y1 cos θ. This is a linear map 2ˆ 2ˆ We could also define a vector w = x1 i + y1 j. Its x-component will 02 2 2 2 2 transform as x1 = x1 cos θ + y1 sin θ + 2x1y1 sin θ cos θ. This is not a linear map
Kirit Makwana Four vector representation September 17, 2020 13 / 21 Einstein summation convention
The rotation and Lorentz tranforms both can be written in matrix form as
x0 cos θ sin θ x X = =⇒ x0i = T i xj (16) y 0 − sin θ cos θ y j j 0µ X µ ν X = Λν X (17) ν Einstein convention - an index which comes twice (once upper and one lower) is summed over (without explicitly writing the summation symbol) 0i i j 0µ µ ν x = Tj x ; X = Λν X (18)
Kirit Makwana Four vector representation September 17, 2020 14 / 21 Contravariant tensors
The 4-vector X = (ct, x, y, z) is transformed to X 0 = (ct0, x0, y 0, z0) By the nature of linearity of the Lorentz transform we can see that
∂X 0µ ∂X 0µ X 0µ = X ν;Λµ = (19) ∂X ν ν ∂X ν
This is how a contravariant vector transforms and hence the space-time 4-vector is a contravariant vector (written with upper indices) In general a Contravariant tensor tranforms as
∂x0i ∂x0j ∂x0n T 0ij...n = ... T ab...f (20) ∂xa ∂xb ∂xf
−1 µ The inverse tranform is given by the inverse matrix (Λ )ν , which is just given by replacing v with −v
Kirit Makwana Four vector representation September 17, 2020 15 / 21 Covariant tensors
The transformation rule of a covariant tensor is given by
∂xν Y 0 = Y (21) µ ∂x0µ ν
It is written with lower indices. A covariant tensor transforms in general as
∂xa ∂xb ∂xf T 0 = ... T (22) ij...n ∂x0i ∂x0j ∂x0n ab...f
For the Lorentz transformation, the covariant transformation matrix is γv γ c 0 0 ν γv ∂x γ 0 0 −1 ν = c = (Λ ) (23) ∂x0µ 0 0 1 0 µ 0 0 0 1
Kirit Makwana Four vector representation September 17, 2020 16 / 21 Metric tensor
Define a special 2nd rank “metric” tensor in 4 dimension, g, such that g 00 = −1, g 11 = g 22 = g 33 = 1, and rest all components are zero Check that this obeys (remains invariant) under the laws of transformation of a contravariant tensor
0ab a b cd a b a b a b a b g = Λc Λd g = −Λ0Λ0 + Λ1Λ1 + Λ2Λ2 + Λ3Λ3 (24)
If a 6= b check that g 0ab = 0. If a = b, we can check that g 000 = −1, g 011 = g 022 = g 033 = 1 So g is invariant under the contravariant transformation. Now consider the metric tensor as a covariant tensor with the same elements, i.e. g00 = −1, g11 = g22 = g33 = 1, and rest all components zero In the same manner it can be verified that this tensor remains invariant under the covariant transformation rules
Kirit Makwana Four vector representation September 17, 2020 17 / 21 Metric tensor
Also see that ab a g gbc = Ic (25) a a where Ic is the identity tensor, with Ic = 0 if a 6= c, and it is equal to unity if a = c
Kirit Makwana Four vector representation September 17, 2020 18 / 21 Metric tensor
Now consider a contravariant vector with components T a. Now b consider the object Ya = gabT . This object looks like a covariant vector. Lets check if it transforms like that
0 0 0b 0b Ya = gabT = gabT (26) b c b c k = gabΛc T = gabΛc Ik T (27) b ce k b ce = gabΛc g gek T = gabΛc g Ye (28)
b ce The product gabΛc g includes a sum over indices b and c. This sum will contain only one non-zero element when b = a and c = e. If e a = e, then it is just Λa. If a 6= e and one of them is 0, then it is e −1 e −Λa. This is nothing but (Λ )a. And Eq. 20 is the transformation b law of a covariant vector. So the operation gabT creates a covariant vector out of a contravariant vector. This is called index lowering
Kirit Makwana Four vector representation September 17, 2020 19 / 21 4-gradient
Consider the derivative of a scalar field ∂φ ∂ φ = (29) µ ∂xµ Consider its transformation ∂φ ∂ 0 φ = (30) µ ∂x0µ ∂φ ∂xν = (31) ∂xν ∂x0µ ∂xν = ∂ φ (32) ∂x0µ ν
This transforms like a covariant vector. So the 4-gradient ∂µφ is a covariant vector.
Kirit Makwana Four vector representation September 17, 2020 20 / 21 Covariant position vector
β Define a covariant position vector as Xα = gαβX . It can be shown that this is a covariant vector (try it like done 2 slides back...) Consider the derivative w.r.t. this
α ∂ ∂ ∂ ≡ = β (33) ∂Xα ∂(gαβX ) H.W. - Prove that ∂α transforms as a contravariant vector
Kirit Makwana Four vector representation September 17, 2020 21 / 21