-------------------------------------------------------------
show that the length of rod is invariant under Galilean transformation.
Be fast fast .
____________________________________
Answers
SOLUTION :-
suppose the ordinates of two points A and B in two inertial frames S and S' are (x1, y1, z1), (x2, y2, z2), (x1', y1', z1'),(x2' , y2' , z2') respectively.
If S' moves with velocity v relative to S along x' axis, then according to Galilean transformation.
X1'= X1 - Vt , Y1'= y1 z1'= z1
X1'= X1 - Vt , Y1'= y1 z1'= z1X2'= x2 - vt, y2 '= y2, z2'= z2
The distance between the points A and B in the frames S'
= [( X2' - X1')² + (Y2' - Y1')² + ( Z2' - Z1')²]½
[{(X2 - vt)-(X1 - vt)}² +(y2 - y1)² + (z2 - z1)²]½
= [(x2 - x1)² + (y2 - y1)² + (z2 - z1)²]½
= Distance between the points in the frame S. Consequently the length of rod in invariant under Galilean transformation.
The consequences research work of Galileo on the motion of the projectile led him to formulate transformations which later on ,were called after his name 'Galilean transformations'. These are used to describe the motions which are observed by two observers in two different inertial frames.
✴ the motion of a particle projected at any angle maybe derived from the motion of the particle thrown vertically upward.
✴ if a particle is thrown straight up from a cart which is moving with uniform speed, the observer on the cart may see the particle moving up and down but the motion observed by an observer on the ground maybe described by superimposing the motion of the cart into that of projectile.
I'm studying for a physics test, but I think I don't really understand Galilean invariance. In my textbook there is an example in which they prove that if you consider 2 frames S and S' in standard configuration that the second law of Newton is Galilean invariant by proving that if x′=x−Vtx′=x−Vt than Fx=F′xFx=Fx′, so this law holds in both frames. So far I understand this.
However, in the book there is one assignment in which they ask me to verify that the relationship between kinetic energy and momentum, E=p2/2mE=p2/2m, is Galilean invariant. I couldn't really figure it out by myself so I looked at the answers. The answer is as followed, according to my text book:
In S:
E=12mx˙2;E=12mx˙2; p=mx˙.p=mx˙.
Substitute x˙=p/mx˙=p/m in the equation for the energy:
E=12m(pm)2=p2/2mE=12m(pm)2=p2/2m
In S':
E′=12mx˙′2−12m(x˙−V)2=12mx˙2−mx˙V2E′=12mx˙′2−12m(x˙−V)2=12mx˙2−mx˙V2
p′=mx˙′p′=mx˙′
Assume the relationship holds: i.e.,
E′=p′22m=12m(mx˙−mV)2=12m(x˙2−2x˙V+V2)=12mx˙2−mx˙V+12mV2,E′=p′22m=12m(mx˙−mV)2=12m(x˙2−2x˙V+V2)=12mx˙2−mx˙V+12mV2,
in agreement with the Galilean transformation of the kinetic energy