Momentum in Galilean transformation

[Pengwuino]

I need to show that the definition of linear momentum p=mv, has the same form p'=mv' under a Galilean transformation. What does it mean to "show" such a thing? I have no idea where to start :(

 

[Triss]

It means that you have to apply the Galileo transformation to your momentum p and end up with p' in your transformated system.

 

[Pengwuino]

How would I go about doing that...

 

[Triss]

Say you have a primed system moving with a velocity v_0 relative to your unprimed system. What is the Galileo transformation then? What does that tell you about the velocity transformation? The mass m is a constant, so what does that tell you about the momentum?

 

[Pengwuino]

I'm not sure what the velocity transformation is. I know x'=x-vt but i don't konw where to go from there if that's even the right place to start.

 

[Triss]

It is. Recall that x=x(t), now what do you get when you apply d/dt to it?

 

[Pengwuino]

Ok my professor gave me a little and I got here…

\begin{array}{l}<br /> x&#039; = x - ut \\ <br /> p = mv \\ <br /> p = m\frac{d}{{dt}}(x&#039; + ut) \\ <br /> p = m(v&#039; + u) \\ <br /> p = mv&#039; + mu \\ <br /> p = p&#039; + mu \\ <br /> \end{array}

So what exactly does this tell me? What does p' mean in actuality? If K is the stationary frame, is it the momentum as seen by the person in the reference frame K'? And is p simply the momentum from the K frame?

 

[cepheid]

To answer your questions: yes and yes.

What you are trying to show is that if a person in the K frame sees an object with momentum p = mv, where v is the velocity of the object in the K frame, then will a person in the K' frame see the object as having a moment p' = mv', where v' is the object's velocity as measured in the K' frame? If he doesn't, then the laws of physics (specifically the equation for momentum) are not *invariant* under a Galilean transformation. But it turns out they are.:

If the object has velocity v = dx/dt in K, then in K' its velocity is dx'/dt = d(x - ut)/dt = dx/dt - u = v - u. So we have established that v' = v - u in a Galilean transformation.

And since in your result you got p' = m(v-u) = mv', you know you're ok.

 

[Pengwuino]

So p is the momentum in the rest frame? And is the p' the momentum as seen from the K' frame?

 

[cepheid]

So p is the momentum in the rest frame? And is the p' the momentum as seen from the K' frame?

To answer your questions: yes and yes.

Umm...the notation does seem pretty unambiguous to me. Primed quantities are measured relative to the primed coordinate system. Unprimed quantities are measured relative to the unprimed coordinate system.