Relativistic Energy Dispersion Relation: Explained

[Parcival]

I'm in the process of learning special relativity (SR), and I'm a bit confused as to why the relativistic energy dispersion relation $E^{2}=m^{2}c^{4}+p^{2}c^{2}$ gives the energy for a free particle? I get that it is the sum of (relativistic) kinetic energy plus the rest mass term (a constant), but where in the derivation of this expression does one assume that the particle is free?

From what I've read, one way that one can deduce that $E^{2}=m^{2}c^{4}+p^{2}c^{2}$ is from the relativistic 3-momentum of a particle, $\mathbf{p}=\gamma m\mathbf{v}$ (where $\gamma$ is the Lorentz factor, $m$ the rest mass of the particle, and $\mathbf{v}=\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}$, with $t$ the coordinate time), and the line element, $-c^{2}\mathrm{d}\tau^{2}=-c^{2}\mathrm{d}t^{2}+\mathrm{d}\mathbf{x}^{2}$ (where $\tau$ is the proper time). Indeed, from the latter, we can imply that $(mc)^{2}=(\gamma mc)^{2}-p^{2}$. Taylor expanding $(\gamma mc)^{2}$, we have $$(\gamma mc)^{2}=\frac{1}{c^{2}}\left(mc^{2}+\frac{1}{2}mv^{2}+\cdots\right)^{2}$$ where the dots denote higher order terms in $v/c$. Recognising $\frac{1}{2}mv^{2}$ as the non-relativistic kinetic energy of a particle, we identify the expression in brackets with the energy of the particle, such that $$\frac{E^{2}}{c^{2}}=m^{2}c^{2}+p^{2}$$ I realize that one can also arrive at this expression, by noting that $\gamma mc$ is the zeroth element of the 4-momentum vector. The thing is, in neither of these ways does one assume that the particle is free, so is the fact that it describes the energy of a free particle something that is determined a posteriori?

 

[PeroK]

What do you mean by a free particle in this context?

$E$ in your equations is defined to be $\gamma mc^2$, which is generally seen as the particle's rest energy, $mc^2$, plus its kinetic energy, $(\gamma - 1)mc^2$. This equation is purely kinematic: it says nothing, in particular, about how the particle came to be moving at speed $v$ in a particular reference frame.

 

[Parcival]

What do you mean by a free particle in this context?

I mean, not subject to any external forces.

EE in your equations is defined to be γmc2\gamma mc^2, which is generally seen as the particle's rest energy, mc2mc^2, plus its kinetic energy, (γ−1)mc2(\gamma - 1)mc^2. This equation is purely kinematic: it says nothing, in particular, about how the particle came to be moving at speed vv in a particular reference frame.

This was my understanding, however, I have read several notes (of which I can't find of hand, but I'm currently looking for an example) in which they refer to this expression for $E$ as giving the energy of a free particle, leaving me confused.

 

[PeroK]

I mean, not subject to any external forces.

This was my understanding, however, I have read several notes (of which I can't find of hand, but I'm currently looking for an example) in which they refer to this expression for $E$ as giving the energy of a free particle, leaving me confused.

There's no implication that the particle is not subject to external forces. Particles in a particle accelerator, an accelerating spaceship etc. are all covered.

It, obviously, doesn't include any potential energy. How could it?

In that sense, it is the total energy for a free particle. But that seems an odd way to look at it, unless you have been explicitly considering motion under a potential.

 

[Parcival]

There's no implication that the particle is not subject to external forces. Particles in a particle accelerator, an accelerating spaceship etc. are all covered.

That's what I thought. It comes from purely kinematical considerations, without any mention of a potential.

In that sense, it is the total energy for a free particle. But that seems an odd way to look at it, unless you have been explicitly considering motion under a potential.

I guess maybe the author is just observing the fact that the energy only contains kinetic energy plus a constant term, and thus can be interpreted as the energy of a free particle (however, this would implicitly assume that the kinetic energy is constant)? If one explicitly considers the dynamics, in which the particle is subject to a potential, then the expression for energy would include an additional potential energy term, right?!

 

[PeroK]

That's what I thought. It comes from purely kinematical considerations, without any mention of a potential.
I guess maybe the author is just observing the fact that the energy only contains kinetic energy plus a constant term, and thus can be interpreted as the energy of a free particle (however, this would implicitly assume that the kinetic energy is constant)? If one explicitly considers the dynamics, in which the particle is subject to a potential, then the expression for energy would include an additional potential energy term, right?!

Take a look here, for example:

http://www.reed.edu/physics/courses/Physics411/html/page2/files/Lecture.11.pdf

I think you have to differentiate between cases where you are interested in the external forces and those where you are not. You will get questions talking about the "maximum" energy achieved in a particle accelerator. In which case you don't care about any additional potential energy, as it doesn't contribute to the energy available in a particle collision.

Likewise, if you look at the relativistic rocket equation, your rocket is not in a potential but it is subject to an external force.

In terms of what is and is not a "free particle", I think you have to pay attention to the context.

 

[Parcival]

Take a look here, for example:

http://www.reed.edu/physics/courses/Physics411/html/page2/files/Lecture.11.pdf

I think you have to differentiate between cases where you are interested in the external forces and those where you are not. You will get questions talking about the "maximum" energy achieved in a particle accelerator. In which case you don't care about any additional potential energy, as it doesn't contribute to the energy available in a particle collision.

Likewise, if you look at the relativistic rocket equation, your rocket is not in a potential but it is subject to an external force.

In terms of what is and is not a "free particle", I think you have to pay attention to the context.

Thanks for the details.

Should one view $E^{2}=m^{2}c^{4}+p^{2}c^{2}$ then, as describing the energy of a particle without regard of any potential that may be present?

In which case you don't care about any additional potential energy, as it doesn't contribute to the energy available in a particle collision.

Why is this? Is it because a particle collision occurs at a point, and since only potential energy differences are physical (at least in SR), there is no potential energy available in the collision.

 

[PeroK]

Thanks for the details.

Should one view $E^{2}=m^{2}c^{4}+p^{2}c^{2}$ then, as describing the energy of a particle without regard of any potential that may be present?

That's probably a good way to put it.

Thanks for the details.

Why is this? Is it because a particle collision occurs at a point, and since only potential energy differences are physical (at least in SR), there is no potential energy available in the collision.

Basically, that equation is one of the basics for particle collisions and particle decay, which are instantaneous.

Are you learning SR on your own?

 

[Parcival]

Basically, that equation is one of the basics for particle collisions and particle decay, which are instantaneous.

Ah ok.

Are you learning SR on your own?

Yes, I am. Trying to be as rigorous as I can.

 

[PeroK]

Yes, I am. Trying to be as rigorous as I can.

Good luck!

 

[Parcival]

Good luck!

Thanks. And thanks for your help here.