Resolução Marion [Mec.Clássica] - Capítulo 14

Resolução Marion [Mec.Clássica] - Capítulo 14

(Parte 1 de 6)

The Special Theory of Relativity

14-1. Substitute Eq. (14.12) into Eqs. (14.9) and (14.10):

From (1)

From (2)

vx c γ


c γ γ or

14-2. We introduce cosh,sinhyyvcαα≅≅ and substitute these expressions into Eqs. (14.14); then cosh sinh

cosh sinh ; x ct x t a x x

Now, if we use cosh α = cos (iα) and i sinh α = sin (iα), we can rewrite (1) as ( ) ( )

cos sin sin cos x i ict i ict x i ict iαα α α

Comparing these equations with the relation between the rotated system and the original system in ordinary three-dimensional space,

cos sin sin cos x x x x x

We can see that (2) corresponds to a rotation of the 1xict− plane through the angle iα.

14-3. If the equation

xi ct xi ct is Lorentz invariant, then in the transformed system we must have

xi ct xi ct where

We can rewrite (2) as


Now, we first determine how the operator 2

We know the following relations:


2xx x xµν µλ µν µλ νλ ν λ xµ νλ ν


x x µν λ µ x∂ ∂µ νλ

Since µ and λ are dummy indices, we see that the operator 2x2µ∂∂∑ is invariant under a Lorentz transformation. So we have

This equation means that the function ψ taken at the transformed point (x′,ict′) satisfies the same equation as the original function ψ (x,ict) and therefore the equation is invariant. In a Galilean transformation, the coordinates become x v t y v t

z zv t

Using these relations, we have x y xt x x t x v t

y v t zy v t

Therefore, xy z

xy z x y z c t xyz c t v v v t vx t v y t v z t

This means that the function ψ (x′,ict′) does not satisfy the same form of equation as does (,)xictψ, and the equation is not invariant under a Galilean transformation.

14-4. In the K system the rod is at rest with its ends at and . The K′ system moves with a velocity v (along the x axis) relative to K. 1x 2x x x

If the observer measures the time for the ends of the rod to pass over a fixed point in the K′ system, we have v t

v t


14-5. The “apparent shape” of the cube is that shape which would be recorded at a certain instant by the eye or by a camera (with an infinitesimally short shutter speed!). That is, we must find the positions that the various points of the cube occupy such that light emitted from these points arrives simultaneously at the eye of the observer. Those parts of the cube that are farther from the observer must then emit light earlier than those parts that are closer to the observer. An observer, looking directly at a cube at rest, would see just the front face, i.e., a square.

When in motion, the edges of the cube are distorted, as indicated in the figures below, where the observer is assumed to be on the line passing through the center of the cube. We also note that the face of the cube in (a) is actually bowed toward the observer (i.e., the face appears convex), and conversely in (b).

(a)Cube moving toward the observer. (a)Cube moving away from the observer.


x x

We transform the time t at the points and in the K system into the K′ system. Then, 1x2x vx t

vx t

From these equations, we have

1xx t t v v x


K K′ v

Suppose the origin of the K′ system is at a distance x from the origin of the K system after a time t measured in the K system. When the observer sees the clock in the K′ system at that time, he actually sees the clock as it was located at an earlier time because it takes a certain time for a light signal to travel to 0. Suppose we see the clock when it is a distance A from the origin of the

K system and the time is t in K and 11t′ in K′. Then we have v t ct t

(Parte 1 de 6)