Galilean invariance verification (kinetic energy and momentum)

[Tiddo]

Hi all,

First of all, sorry for not using the template, but I think in this situation it's better to explain my problem right away:
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 proof that if you consider 2 frames S and S' in standard configuration that the second law of Newton is Galilean invariant by proofing that if x' = x - Vt than F_x = F'_x, 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 = p^2/2m, is Galilean invariant. I couldn't really figure it out by myself so I looked at the answers. The answer is as followed:

In S:

E = \frac{1}{2}m\dot{x}^2; p=m\dot{x}.

Substitute \dot{x} = p/m in the equation for the energy:

E = \frac{1}{2}m(\frac{p}{m})^2=p^2/2m

In S':

E'=\frac{1}{2}m\dot{x}'^2-\frac{1}{2}m(\dot{x}-V)^2=\frac{1}{2}m\dot{x}^2-m\dot{x}V^2

p'=m\dot{x}'

Assume the relationship holds: i.e.,

E'=\frac{p'^2}{2m}=\frac{1}{2m}(m\dot{x}-mV)^2=\frac{1}{2}(\dot{x}^2-2\dot{x}V+V^2)=\frac{1}{2}m\dot{x}-m\dot{x}V+\frac{1}{2}mV^2,

in agreement with the Galilean transformation of the kinetic energy

Source: McComb, W. D., 1999. Dynamics and Relativity. New York: Oxford University Press Inc.

I understand all of the equations, however I just don't understand why this verifies that this relationship is Galilean invariant.

Can someone explain this to me?

Thanks!

 

[cepheid]

There is a problem with the computation of E' in the solutions. It should read:

E^\prime = \frac{1}{2}m(\dot{x}^{\prime})^2 =\frac{1}{2}m(\dot{x}-V)^2 = \frac{1}{2}m(\dot{x}^2 - 2\dot{x}V + V^2)

= \frac{1}{2}m\dot{x}^2 - m\dot{x}V + \frac{1}{2}mV^2

You can see that this expression for E' is the same as what is obtained if you assume that E' = (p')²/2m, showing that this relation between kinetic energy and momentum does indeed hold in S' as well.

 

[Tiddo]

thank you sp much! I completely missed that. now it'd all clear to me:)