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Galilean invariance verification (kinetic energy and momentum)

  • Thread starter Tiddo
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  • #1
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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 [itex]x' = x - Vt[/itex] than [itex]F_x = F'_x[/itex], 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, [itex]E = p^2/2m[/itex], 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:

[itex]E = \frac{1}{2}m\dot{x}^2;[/itex] [itex]p=m\dot{x}.[/itex]

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

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

In S':

[itex]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[/itex]

[itex]p'=m\dot{x}'[/itex]

Assume the relationship holds: i.e.,

[itex]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,[/itex]

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!
 

Answers and Replies

  • #2
cepheid
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There is a problem with the computation of E' in the solutions. It should read:

[tex] 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) [/tex]

[tex] = \frac{1}{2}m\dot{x}^2 - m\dot{x}V + \frac{1}{2}mV^2 [/tex]

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.
 
  • #3
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thank you sp much! I completely missed that. now it'd all clear to me:)
 

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