Galilean invariance proof

Question

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, according to my text book:

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.,

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

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!

Answer 1

Either I should turn in my medical marijuana card, or the author of your textbook should. The exercise doesn't make any sense.

Since we only have one particle, whose mass is fixed, we can set $m=1$. Also, the factor of 1/2 in the equation for KE is purely conventional, so let's drop that as well. We then have the following:

Given the definitions $p=v$ and $E=v^2$, prove that the relationship $E=p^2$ is Galilean invariant.

Solution: In the new frame, by definition $p'=v'$ and $E'=v'^2$. therefore $E'=p'^2$.

The author of the text seems to have convinced him/herself that this is a nontrivial proof, when in fact it's a trivial matter of definition. It would have held equally well if we had defined $p=1/v$ and $E=\sin\left(\exp({(v/v_o)^{142}})\right)$. The reason is that the three variables $v$, $p$, and $E$ are all related to one another by definitions, and the definition of the Galilean transformation only directly affects $v$. To get an example where it failed, you'd have to have some definition like $p=c_1v+c_2x$; then the Galilean transformation would affect the variables $v$, $p$, and $E$ in more than one way (through $x$ as well as $v$), and a nontrivial question of consistency would arise.

Answer 2

There's errors:

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

should be $$E'=\frac{1}{2}m\dot{x}'^2=\frac{1}{2}m(\dot{x}-V)^2=\frac{1}{2}m\dot{x}^2-m\dot{x}V+ \frac 1 2 mV^2$$

which is the Galilean transformation of kinetic energy used later.

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

should be $$E'=\frac{1}{2}m\dot{x}^2-m\dot{x}V+ \frac 1 2 mV^2$$

He's used the Galilean transformation of momentum with assuming the invariance of the kinetic/momentum relationship to get the correct expression for the Galilean transformation of kinetic energy. He therefore concludes that assuming the kinetic/momentum relationship to hold in $s'$ is therefore true.

The question is a contrived, and therefore confusing, use of the Galilean transformations to prove Galilean invariance in this case because $E' = 1/2m'v'^2$ and $p'=m'v'$ are not physical laws but mathematical definitions in S' and so $E'= p'^2/2m'$ independently of any transformation.