# Relativistic momentum definition

1. Apr 29, 2006

### bernhard.rothenstein

many textbooks start teaching relativistic dynamics by defining the relativistic momentum as
p=dx/dtau

2. Apr 29, 2006

### Meir Achuz

It should be p=m dx/dtau.
U^i=dx^i/dtau are the three spatial components of a 4-vector.
The basic differential space-time 4-vector is dx^mu=(dt;dx,dy,dz)
Its invariant length squared (a 4-scalar) is
dtau^2=dt^2-dx^2-dy^2-dz^2.
To get a 4-vector that behaves like a velocity, the 4-vector is divided by the scalar: U^mu=dx^mu/dtau.
Then a 4-vector momentum is just p^mu=mU^mu. It's x,y,z components are seen to behave for small v like the NR momentum. Its time-like component is the energy.

3. Apr 29, 2006

### bernhard.rothenstein

the authors i mention say "Mermin It's about time]and many others ar far from using four vectors as far as I am. Do ou now an other explanation or it is a simple guess work?

4. Apr 29, 2006

### nrqed

The justification for defining a momentum vector is the same as in classical physics, as far as I know. It is a quantity that is conserved in a collision. This is *the* reason for defining it this way. After all one could define a vector $m {\vec a }$ in Newtonian physics or $m {d^2 x\over d \tau^2 }$ in SR? Of course these could be defined but they would be useless quantities.

EDIT I meant ''One could define a vector $m^2 {\vec a }$ in Newtonian physics or $m^2 {d^2 x\over d \tau^2 }$ in SR but this would be useless...''

Last edited: Apr 29, 2006
5. Apr 29, 2006

### bernhard.rothenstein

you think that we should start with the classical defnition of the momentum p=mdx/dt and to search what should we do in order to bring it in accordance with special relativity? but the author I quote start with
p=mdx?dtau without to offer an explanation for it.

6. Apr 29, 2006

Staff Emeritus
Taylor and Wheeler, in their Spacetime Physics, do specifically derive fourmomentum from collisions under relativistic conditions.

7. Apr 29, 2006

### nrqed

I think that using collisions and looking for a quantity that is conserved is the *logical* way to introduce the expression for relativistic momentum. If the author just quotes the final result it's because he/she did not want to take the time to justify where it comes from.

exactly the same thing happens in Newtonian physics, you know. Why does one introduce this strange vector $m {\vec v}$?? it is because one can show that the total momentum is conserved in a collision. If that was not the case, one would never introduce this quantity in the first place.

Regards

Patrick

8. Apr 29, 2006

### nrqed

it depends what you mean by ''bringing in accordance with SR''. If you just use the argument ''just replaces dt by dtau'' to get to SR, I would say this is unsatisfactory (my personal opinion)

But if the motivation is ''now let's look at a collision in different frames taking into account the Lorentz transformations..then we see that $m {\vec v }$ is not conserved in a collision. Can we figure out a quantity which *is* conserved? (and that quantity will reduce to the usual expression in the nonrelativistic limit). Yes, and here it is.

That's all my personal preference. If I was teaching it, this is the way I would introduce the concept.

9. Apr 29, 2006

### Staff: Mentor

And this is exactly what Mermin takes great pains (and several pages) to do in his nice little book "It's About Time". He certainly doesn't just state the relativistic definition and forget it.

To bernhard.rothenstein: What book were you quoting? The only one you mentioned was Mermin's. (True, Mermin doesn't mention 4-vectors, but it's meant as a popular exposition.)

10. Apr 29, 2006

### bernhard.rothenstein

I quote from Jon Ogborn Introducing special relativity Physics Education 40 (3) 213 2005
In Newtonian mechanics momentum p=mdx/dt. The relativistic idea is that the right way to clock the motion of a particle is to use the proper (or "wristwatch time tau. So Einstein replaced dt by dtau and redefined momentum as p=mdx/dtau. The velocity v is still dx/dt so in the new definition p=m(dx/dt)(dt/dtau). We know that t=gamatau so dt/dtau. Thus the newly defined momentum is given by p=gamamv.
Is that guesswork or intuition?

11. Apr 29, 2006

### Staff: Mentor

I would say brilliant intuition. But I think most of us here agree that just stating that is not sufficient, certainly not if teaching is your goal. I was just responding to a perceived slight against Mermin's book. (Mermin is an pedagogical master!)

12. Apr 29, 2006

### bernhard.rothenstein

I think in the same way. where could I find your response?

13. Apr 29, 2006

### robphy

In introductory physics, momentum is usually introduced by the impulse-momentum theorem... where it appears in an interesting difference of a quantity characterizing the system before and after the impulse. (Kinetic energy is introduced in a similar way in the work-kineticEnergy theorem.) The discussion of conservation laws comes later. Based on this, it would be great if the relativistic motivation retold this story. Otherwise, I'd suggest that the galilean-newtonian story be rewritten to parallel the relativistic one.

From a top-down view, I'd look at the more abstract interpretations of momentum (say, from a Lagrangian or Hamiltonian view), then try to formulate a pedagogical story that could be paralleled in the galilean-newtonian and relativistic cases.

14. May 3, 2006

### bernhard.rothenstein

15. May 4, 2006

### robphy

The "impulse-momentum theorem" essentially says $$\int_{t_i}^{t_f} dt \vec F_{net}=\vec p_f - \vec p_i$$, where the left-hand side is called the impulse. When introducing momentum for the first time [given Newton's Law in the constant-mass form $$\vec F_{net}=m\frac{d\vec v}{dt}$$ for a point particle], I carry out the integration of the left and define momentum as an interesting quantity associated with the system before and after the impulse.

Looking over the relatively new introductory books on my shelf, I see that most books actually state a definition of $$\vec p=m\vec v$$ first, then possibly discuss its conservation, before doing the impulse-momentum theorem. Maybe the approach I used comes from [an old edition of?] Halliday-Resnick? I can't find my [old] copy now.

In any case, my motivation is that it parallels the work-[kinetic-]energy theorem, which [I think] is usually introduced before defining kinetic energy and before discussing any notion of energy conservation.

Until I can find my reference: http://www.google.com/search?q=+impulse+momentum+theorem