A cube of proper edge length ( and mass m is moving past you at speed 0.8c. Dete

Question

A cube of proper edge length ( and mass m is moving past you at speed 0.8c. Determine the volume for the object that you observe if the velocity is directed (a) along an edge, (b) along a face diagonal, and (c) along the body diagonal of the cube. Note: The body diagonal is difficult. What density do you observe the cube to have in each situation How do these densities compare to nuclear densities? At what speed wll the cube's density become an appreciable fraction (say, 40%) of nuclear densities?

Explanation / Answer

1. given cube of proper length l

mass m

v = 0.8 c

a. if v is along an edge

that edge contracts to l' = lo*sqrt(12 - v^2/c^2) = 0.6lo

the other two lengths are unaffected

V = 0.6lo^3

b. if v is along a face diagonal

the face diagonal length becomes

d' = 0.6d

but

d = sqrt(l^2 + l^2) = l*sqrt(2)

hence

d' = 0.6*l*sqrt(2) = 0.84852*l

length perpendicul;ar to this remains same, d = l*sqrt(2)

the figure would become a rhombus

Area A = dd'/2 = l^2*2*0.6/2 = 0.6l^2

V = 0.6l^3

c. along body diagonals

two diagolnals remain of same length, the third one contracts

originally

body diagonal length d

d = l*sqrt(3)

Vo = l^3 = d^3/3^1.5 = (d/sqrt(3))^3

hence

for new volume

V' = d*d*d*0.6/sqrt(3)^3 = 0.6l^3

in each situation the density increases by the sdame amount as the volume has decvreased by the same amount

also

rho' = rho/0.6 ( where rho is original density)

nuclear density rhon = 2.3*10^17 kg/m^3

density of average matter = 5*10^3 kg/m^3

hence

0.4*2.3*10^17 = 5*10^3/x

x = 5.4347*10^-14 = sqrt(1 - v^2/c^2)

v = 0.99999999999999999999999999852316c

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