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 $x' = x - Vt$ 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':



Assume the relationship holds: i.e.,


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?


2 Answers 2


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.

  • $\begingroup$ For the mass, we don't have one particle, I have to verify that this holds for any mass in any frame. I think it'll become a little more complicated than. And by the way, why can we drop the factor of 1/2? I don't really understand why you can do that? $\endgroup$
    – Tiddo
    Aug 10, 2011 at 8:40
  • $\begingroup$ For the mass, you can simply take units such that in those units, the mass of the particle equals 1 unit. For the 1/2, as I said it's purely a matter of convention. If you double all your definitions of forms of energy, you still get a valid conservation law. $\endgroup$
    – user4552
    Aug 10, 2011 at 16:52

There's errors:


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.


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.

  • $\begingroup$ Ok thank you very much! I didn't see that error and that's why I didn't understand it. But now it's clear to me! $\endgroup$
    – Tiddo
    Aug 10, 2011 at 8:30
  • $\begingroup$ @Tiddo are the errors from the book, or yours? $\endgroup$ Aug 10, 2011 at 13:58
  • $\begingroup$ The fist one is from the book, the second one is a type from me, but I already corrected that in my original post. $\endgroup$
    – Tiddo
    Aug 10, 2011 at 15:37

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