3

I'm studying Friedman and Susskind's Special Relativity and Classical Field Theory and follow them in using $c=1$.

They derive the above relation by first using Lagrangian of a free particle $\mathcal L=-m\sqrt{1-v^2}$ to show that conjugate momenta are given by $P^i = mU^i$ (where $U^\mu$ is the 4-velocity). Then they write out the Hamiltonian for the free particle using this Lagrangian and show that $H=mU^0$. Then writing $E$ for $H$ and since $U^\mu$ is a 4-vector and $m$ a scalar, they conclude that the quadruple $(E, P^1, P^2, P^3)$ forms a 4-vector. Then they use the invariance of the norm of the this 4-vector to show that $E^2 - P^2 = m^2$ (by observing the free particle in its rest frame, which is possible only since the particle was free to start with, otherwise there'd be no inertial frame in which the particle is seen at rest always).

Now the entire argument above was for a free particle. My question is whether the oft-used relation $E^2-P^2=m^2$ (or a modification thereof) also valid for a non-free particle (by which I mean a particle whole Lagrangian differs by that of a free particle).

Edit:

I realized (thanks to probably_someone in a comment below) that the original argument that the authors give is slightly different than mine. They multiply the equation $(U^0)^2-(U^i)^2 = 1$ with $m^2$ to yield $E^2-P^2=m^2$. Nevertheless, I still like my argument better. :)

Atom
  • 1,931
  • No it is true for stable particles, which in the litterature, we call on the mass-shell. There is a great deal of detail in this answer to this post: https://physics.stackexchange.com/q/4349/ – thecakeisLie Jul 08 '20 at 20:28
  • I just read through the relevant chapter, and I really don't see how you're concluding that "they use the invariance of the norm of the this 4-vector to show that $E^2-P^2=m^2$ (by observing the free particle in its rest frame)". Maybe quoting the specific passages you are getting this from, rather than paraphrasing them, would be more helpful? – probably_someone Jul 09 '20 at 03:08
  • @probably_someone Turns our my argument was true but different from what they originally write (which can be found on page 110 under the section 3.5.2). – Atom Jul 09 '20 at 03:21
  • Ok, so I still don't see the logic in what you now call "your" argument; specifically, you need to elaborate on the step that I quoted before. – probably_someone Jul 09 '20 at 03:32
  • @probably_someone This is what I mean. After having shown that $(E, P^i)$ is a 4-vector, we know that $E^2-(P^i)^2 = E^2-P^2$ is invariant in all inertial frames. Then in the rest frame of the free particle, since $v=0$, we get $E = m/\sqrt{1-v^2} = m$. And hence it follows. – Atom Jul 09 '20 at 03:59
  • 1
    At any instant, a "well-behaved" non-inertial frame has a corresponding comoving inertial frame (just as a smooth curved path always has a tangent line at any point). In any of these inertial frames, Lorentz scalars have the same value. So, at any instant in the non-inertial frame, the Lorentz scalars must also have the same value, correct? – probably_someone Jul 09 '20 at 04:10
  • @probably_someone Oh! I see what you mean! That was brilliant! Thanks! – Atom Jul 09 '20 at 04:14
  • But still we used in our derivation the Lagrangian and Hamiltonian of a free particle. How to generalize that? – Atom Jul 09 '20 at 04:15

2 Answers2

3

There used to be an answer here that incorrectly claimed that in the electromagnetic field, this equation is modified as $(p_{\mu} - q A_{\mu})(p^{\mu} - q A^{\mu}) = m^2$. The author of that answer realized it was incorrect after seeing my comments and deleted the answer.

In short, $p_{\mu} p^{\mu} = m^2$ is the fundamental kinematical relation that is valid for any massive object in special relativity. There may be external forces exerted on that object, which doesn't change this relation.

In the electromagnetic field with potential $A_{\mu}$, the definition of the canonical momentum shifts by $$ p_{\text{canonical}}^{\mu} = p^{\mu} + q A^{\mu}. $$

Hence, the relation $p_{\mu} p^{\mu} = m^2$ can be rewritten as $$ (p_{\text{canonical}\;\mu} - q A_{\mu})(p_{\text{canonical}}^{\mu} - q A^{\mu}) = m^2. $$

However, the canonical momentum, though a good variable for describing the system with Hamiltonian mechanics, is not a meaningful variable physically. For instance, it isn't gauge invariant.

The physical momentum $p_{\mu} = m u_{\mu}$ still obeys the same relation $p_{\mu} p^{\mu} = m^2$. In fact, it follows trivially from its definition together with $u_{\mu} u^{\mu} = 1$, which in turn follows from the definition of 4-velocity $u^{\mu}$:

$$ u^{\mu} = \frac{dx^{\mu}}{ds}, $$ $$ u^{\mu} u_{\mu} = \frac{dx^{\mu} dx_{\mu}}{ds^2} = 1. $$

Prof. Legolasov
  • 15,888
  • Thanks Prof. That satisfied me! – Atom Jul 12 '20 at 11:20
  • But one more thing. Is $p^0$ always equal to $E$? Or is it true only because we assumed the particle to be free? – Atom Jul 12 '20 at 11:27
  • @Atom for two interacting objects the total energy consists of free energies of the objects, and the energy of the interaction (note that is is defined only for a collection of interacting objects, not for individual objects). Free energies obey the said kinematic relationship. – Prof. Legolasov Jul 12 '20 at 13:06
  • So $E^2-(p^i)^2=m^2$ is indeed true only for free particles? Cuz for interacting particles, $E$ might not be equal to $p^0$. Right? – Atom Jul 12 '20 at 13:23
  • @Atom what exactly do you mean by “interaction”? Which physical theory are you considering? Classical electrodynamics with point charges? In that case my answer explains it. Quantum Field Theory? – Prof. Legolasov Jul 12 '20 at 15:30
  • Yes, I'm only studying classical electrodynamics for now. Is what I said correct for it? – Atom Jul 12 '20 at 15:39
  • @Atom a charged particle’s energy and momentum do satisfy the kinematic reaction, as indicated in my answer, even if that particle is in the electromagnetic field. The interaction energy is not part of the interacting particle energy, but stored in the field itself. – Prof. Legolasov Jul 12 '20 at 15:49
0

It does depend on what is meant by P, and E. If $E^2=P^2-m^2$, even with forces, the particle would have a constant velocity. I think that the canonical forms are more commonly used, so $(E-V)^2=({\bf P}-q{\bf A})^2+(m+S)^2$, with $\bf P$ the canonical momentum, and $(V,{\bf A})$and S vector and scalar potentials.

Jerrold Franklin
  • 1,748