Different methods of deriving the energy-momentum equation

[albertrichardf]

Hello,
In deriving the energy-momentum equation:
$$E^2 = (pc)^2 + (mc^2)^2$$
the following equations are used:
$$p = ymv$$
$$E = ymc^2$$

But both equations are equations that depend on mass, while the final result does not and applies to massless particles. Besides the energy-momentum equation is often cited as the equation relating mass and relativity rather than the one from which it was derived. This leads me to think that this equation is more fundamental than either of the ones used to derive it, so there should be a way to derive it without these two. However, I cannot find any such method. Does anyone knows of how I could derive it without using those two equations?

Thank you.

 

[SiennaTheGr8]

This came up in a thread here just a few days ago.

Peter Donis and some other users indicated that it is indeed possible to show that $\sqrt{E^2 - (pc)^2}$ is Lorentz-invariant without recourse to the mass concept. Maybe someone would be kind enough to chime in here and show us how this would be done. My guess is that you'd start with Minkowski spacetime and the translational symmetries of space and time, then use Noether's Theorem to define energy and momentum as the associated conserved quantities, and then do some analysis to show that those conserved quantities constitute the components of a four-vector. One would then define "mass" as the vector's magnitude.

But I'm not sure really.

 

[albertrichardf]

Sorry for the reposting.
Would you happen to have a link to the thread? Or the title?

 

[PeterDonis]

My guess is that you'd start with Minkowski spacetime and the translational symmetries of space and time, then use Noether's Theorem to define energy and momentum as the associated conserved quantities, and then do some analysis to show that those conserved quantities constitute the components of a four-vector.

You could do that, but it would only apply to inertial motion (since Noether's theorem only gives you constants of the motion for inertial motion). But you can always define a 4-momentum vector for any object, even one moving non-inertially; it's the tangent vector to the object's worldline at a given event, normalized such that its components in any inertial frame are the energy and momentum of the object at that event as measured by an observer at rest in that frame. This vector is a 4-vector by definition.

One would then define "mass" as the vector's magnitude.

Yes, after defining the vector as above.

 

[SiennaTheGr8]

Thanks, PD.

 

[SiennaTheGr8]

Sorry for the reposting.
Would you happen to have a link to the thread? Or the title?

https://www.physicsforums.com/threads/massless-photon.900960/

 

[albertrichardf]

Thank you for the input and the link

 

[Ibix]

The thing I don't get (you can detect when I realized I didn't completely understand this from the date of my last post in the thread Sienna linked) is how one gets to the four-momentum in the first place.

I can see how to get to the four velocity - it's the tangent vector to the particle's worldline. But the four-momentum (for a massive particle) is that times the mass. And as far as I'm aware the Lagrangian is $g_{\mu\nu}P^\mu U^\nu$ (give or take a square root), which implies knowledge of the four momentum.

Do you just observe that the sum of four-velocities is conserved in certain types of collision (spoiler: between particles of equal mass), and observe that there is some scalar property of particles (which I will call m for no reason at all) that can multiply the four velocity such that the sum of $m U^\mu$ is conserved for all collisions? And then take the Newtonian limit and show that m is the Newtonian mass? Then study the interaction between particles on null worldlines and massive particles to derive $P^\mu$ for null paths?

 

[PeterDonis]

I can see how to get to the four velocity - it's the tangent vector to the particle's worldline

So is the 4-momentum. Both are tangent to the particle's worldline (as is any scalar multiple of either). To choose a particular tangent vector you have to pick a normalization. The 4-velocity is normalized to be a unit vector--physically that means its components in a particular inertial frame are $\gamma$ and $\gamma v$, where $v$ is the object's ordinary velocity in that frame and $\gamma$ is the gamma factor associated with $v$. The 4-momentum is normalized so that its components in a given inertial frame are the particle's energy and momentum in that frame. Both normalizations are perfectly valid physically, and neither one is logically prior to the other.

 

[SiennaTheGr8]

I'm still confused too, Ibix.

The step I'm missing is what comes after defining energy and momentum in a general, non-mass-dependent way (using Noether's theorem?). How would you then demonstrate that these conserved quantities do indeed constitute the components of a 4-vector? Equivalently, how would you determine that $E^2 - (pc)^2$ is Lorentz-invariant?

 

[PeterDonis]

How would you then demonstrate that these conserved quantities do indeed constitute the components of a 4-vector?

To apply Noether's theorem, we use the time and space translation symmetries of Minkowski spacetime to define four conserved quantities for inertial motion. So we have four Killing vectors, which I'll call $t^\mu$, $x^\mu$, $y^\mu$, and $z^\mu$, and they define four conserved quantities by contraction with $p_\mu$, which is a vector tangent to the particle's worldline. (Note that this works for any tangent vector, so we still have to choose a normalization as I described in my previous post. I'll assume that we've chosen the normalization so that the values of the conserved quantities are in fact the energy and momentum that would be measured in an inertial frame.)

So we have four conserved quantities of the form $X = \xi^\mu p_\mu$, where $\xi^\mu$ is one of the above Killing vectors. But each one of these conserved quantities is a Lorentz scalar--it doesn't change when we change frames! What is going on here?

What is going on is that we forgot that the translation symmetries of Minkowski spacetime are not just four Killing vector fields; they are a four-parameter group of Killing vector fields. Picking four mutually orthogonal Killing vectors $t^\mu$, $x^\mu$, $y^\mu$, $z^\mu$ at an event is equivalent to picking a particular inertial frame (and using these Killing vectors as its coordinate basis vectors). So transforming between frames means transforming from one set of four Killing vectors to another. Given a fixed tangent vector $p_\mu$, this transformation will change the four conserved quantities derived by contracting $p_\mu$ with the four Killing vectors; but it should be obvious that the four conserved quantities must transform under this change as the components of a 4-vector, because the Killing vectors are just the coordinate basis vectors, and that's how the components of a 4-vector are defined--by contracting a chosen vector with the basis vectors.

 

[SiennaTheGr8]

Thank you. I'll have to put in some legwork to understand all that, but that's what I was looking for.

 

[Mister T]

But both equations are equations that depend on mass, while the final result does not and applies to massless particles.

The relations $E=\gamma mc^2$ and $p=\gamma mv$ are valid only for massive particles, so it's no surprise that you can't use them to derive a relation that applies to massless particles.

Note that if you start with $mc^2 = \sqrt{E^2-(pc)^2}$ you can show that it's valid for both massive and massless particles.

You can do the same with the relation $p=(\frac{E}{c^2})v$.

 

[Mister T]

But you can always define a 4-momentum vector for any object, even one moving non-inertially; it's the tangent vector to the object's worldline at a given event, normalized such that its components in any inertial frame are the energy and momentum of the object at that event as measured by an observer at rest in that frame.

Okay, so let's see if I understand this correctly. When you normalize you are, in essence, defining $m$ as the scalar magnitude?

 

[PeterDonis]

When you normalize you are, in essence, defining $m$ as the scalar magnitude?

No. Normalizing is picking a scalar magnitude (out of the infinite number of possible ones) that meets whatever requirement you are trying to meet.

 

[Mister T]

No. Normalizing is picking a scalar magnitude (out of the infinite number of possible ones) that meets whatever requirement you are trying to meet.

But does that choice not serve as the definition of $m$?

 

[PeterDonis]

does that choice not serve as the definition of $m$?

I'm not sure why, unless you want to describe "choosing a normalization that makes the components of the 4-vector match the energy and momentum measured in a given inertial frame" as a definition.

 

[Orodruin]

Both normalizations are perfectly valid physically, and neither one is logically prior to the other.

I would give the 4-momentum a slight advantage since it also works for light-like world lines whereas the 4-velocity does not.

 

[Ibix]

So, what you are doing is, for some particle:

1. Defining a quantity called energy, which is related to the heat released bringing the particle to a stop relative to me
2. Defining a quantity called momentum, which is related to how much a standard target, initially at rest relative to me, recoils when struck by the particle.
3. Noting that we can make these definitions even if "the particle" is a light pulse (although my wording was sloppy for the energy in this case).
4. Noting that there exists an affine parameter $\lambda$ along the worldline of the particle such that $dx^0/d\lambda$ behaves like energy and the three $dx^i/d\lambda$ behave like the components of the momentum. The modulus of that vector is naturally a Lorentz scalar, and can easily be shown to be the Newtonian mass in the appropriate limit.

Right? Sorry for extreme pedantry.

Except for the last step we could, in fact, make the same case in Newtonian physics. Then Newton would have defined mass as the constant of proportionality between the recoil velocity of our standard target and the initial velocity of the particle. But it's not so natural because we don't need to consider massless things in Newtonian physics - their status isn't clear due to electromagnetism being a relativistic theory.

 

[PeterDonis]

Defining a quantity called energy, which is related to the heat released bringing the particle to a stop relative to me

This won't do because it requires you to already know the rest mass (since otherwise all you will know from the above experiment is the kinetic energy relative to you). That's why SiennaTheGr8's method of defining energy and momentum using Noether's theorem is a better choice.

 

[Ibix]

That's why SiennaTheGr8's method of defining energy and momentum using Noether's theorem is a better choice.

Fair point about total vs kinetic energy. But doesn't the Noether's theorem approach start with the action, which includes the four momentum so assumes the normalisation? Or does it show you which normalisation to pick?

 

[PeterDonis]

doesn't the Noether's theorem approach start with the action, which includes the four momentum so assumes the normalisation?

Noether's theorem requires a Lagrangian, but there's nothing that requires a particular normalization of that Lagrangian. You can write the Lagrangian in terms of whatever generalized coordinates you want.

 

[Ibix]

Noether's theorem requires a Lagrangian, but there's nothing that requires a particular normalization of that Lagrangian. You can write the Lagrangian in terms of whatever generalized coordinates you want.

So we could start with the action being $g_{\mu\nu}U^\mu U^\nu$, and the conservation of stress-energy drops out somehow?

 

[PeterDonis]

we could start with the action being $g_{\mu\nu}U^\mu U^\nu$,

Not if the particle is massless--unless by $U^\mu$ you just mean a general tangent vector without requiring it to be a unit vector. But to be fully explicit I would write the tangent vector as $dx^\mu / d\lambda$ or even $dq / d \lambda$ to make it clear that we are making no assumptions about the form or units of the generalized coordinates.

the conservation of stress-energy drops out somehow?

Noether's theorem does not give you the conservation of stress-energy. It gives you a constant of the motion for a particle whose worldline is a geodesic.

 

[Mister T]

I'm not sure why, unless you want to describe "choosing a normalization that makes the components of the 4-vector match the energy and momentum measured in a given inertial frame" as a definition.

The reason why is because I'm getting at the way we define mass. If, as mentioned previously, we start with $E=\gamma m$ and $p=\gamma mv$ we can derive $m=\sqrt{E^2-p^2}$ and even show that this last relation is also valid for massless particles. But as mentioned that derivation has the drawbacks of being based on relations that already make use of $m$ and moreover are valid only for massive particles.

 

[PeterDonis]

I'm getting at the way we define mass

Any tangent vector is going to have an invariant norm. So we can always find an invariant that describes the particle. The only choice we have to make is normalization (basically, what units we use for the norm). So all we're really doing when we "define mass" is saying that we want to normalize to mass units (or energy units, or momentum units, they're all the same in relativity). We don't need to "define mass" to know that the tangent vector has an invariant norm; that's a matter of geometry.

 

[Ibix]

I still feel I'm missing something here.

Noether's theorem tells me there's a conserved quantity along a geodesic ($dq/d\lambda$ as you noted Peter - thanks). But I still don't see how you normalise to mass units. I can't just normalise to 1kg (or whatever unit) because that doesn't work for particles of different masses, let alone worrying about null paths.

So do I do something like measure the 3-momentum through the recoil of a unit mass under impact from the particle, observe that it behaves like the spatial component of a four vector, rather like the one that fell out of Noether's theorem in fact, construct the four vector and off I go?

 

[PeterDonis]

I can't just normalise to 1kg (or whatever unit) because that doesn't work for particles of different masses

Sure it does. An object with 1 kg invariant mass will have a tangent vector whose norm is 1. An object with 2 kg invariant mass will have a tangent vector whose norm is 2.

You're right that this by itself doesn't tell you how to normalize null vectors, but see below.

do I do something like measure the 3-momentum through the recoil of a unit mass under impact from the particle, observe that it behaves like the spatial component of a four vector, rather like the one that fell out of Noether's theorem in fact, construct the four vector and off I go?

That would work, yes. And you can also measure kinetic energy, and find one of two cases:

(1) Kinetic energy is the same as (the 3-norm of) 3-momentum (in units where $c = 1$). In that case you have a massless particle.

(2) Kinetic energy is smaller than (the 3-norm of) 3-momentum (in units where $c = 1$). In that case you have a massive particle, and you can find its invariant mass by figuring out the dispersion relation between kinetic energy and momentum (how much larger K.E. is than momentum as a function of speed).

 

[Ibix]

Sure it does. An object with 1 kg invariant mass will have a tangent vector whose norm is 1. An object with 2 kg invariant mass will have a tangent vector whose norm is 2.

You misunderstood me. I was just meaning that I can't normalise all (or even all non-null) four-momentum vectors to 1kg in the same way I can normalise all four-velocities (where defined) to 1. Obviously I can pick any time-like four-momentum and define its length as 1, but others would have different lengths - and it was how to get those lengths where I was stumbling.

But I think I do understand now. I can relate the easily-measured 3-momentum or momentum-and-KE to a four-vector, which gets me an invariant quantity, which is also conserved (thanks, Noether). And I can easily show that it has the properties I expect of mass.

So I see how to do it now. Thanks (once again), Peter!