# Is there a way to have momentum without energy

I've been exploring some philosophies regarding movement and energy. One of the fun questions I've been able to play with is how momentum fits into those philosophies. I have to ask "can you have momentum without energy?" I know the opposite is true: you can have potential energy without momentum. It's the question of momentum without energy that I'm interested in.

From what I learned from school, it is impossible to have momentum without energy. Momentum is $mv$ and energy is $\frac{1}{2}mv^2$ when using their common definitions, so as long as you have momentum, it would appear you must also have energy. In the normal easy-mode physics world, it would be clear that I can answer my question with an emphatic "no, you cannot."

I'm less versed in the more exotic realms of physics like general relativity, quantum mechanics, and others of the like. Do we ever see a concept that we call "momentum" that can be non-zero when energy is zero, or one where momentum exists but no corresponding concept of energy exists?

• You are correct if you only refer to kinetic energy, because $E_k=p^2/2m$, but potential energy can become negative, and thus you can have a moving body with zero total energy.
– user126422
Commented Feb 17, 2017 at 0:49
• Answer here: physics.stackexchange.com/questions/90231/… Commented Feb 17, 2017 at 3:39

How about virtual particles in the Breit-frame (the frame in which no energy is transferred)? For example, consider a head on collision between 2 electrons with 4-momenta:

$p^1_{\mu} = (E, {\bf \vec{p}})$

and

$p^2_{\mu} = (E, {\bf -\vec{p}})$

(with, of course, $E^2 = p^2 + m^2$).

So they backscatter 180 degrees in the CoM frame:

$p'^1_{\mu} = (E, {\bf -\vec{p}})$

$p'^2_{\mu} = (E, {\bf \vec{p}})$

In the $t$-channel, the virtual photon has 4 momentum:

$q_{\mu} = p'^1_{\mu}-p^1_{\mu} = (0, -2{\bf \vec{p}})$.

So if you consider virtual particles, then the answer is "yes".

In special relativity, energy and momentum of a particle form the energy-momentum 4-vector, to which certain constraints apply:

Massive particles come with a rest frame where their momentum is zero and their energy corresponds to their mass.

Massless particles have nonzero energy and momentum in all frames, equal in magnitude if 'natural' units where $c=1$ are used.

There are also hypothetical particles called tachyons which, instead of a rest frame, come with a critical frame where they have non-zero momentum, but zero energy. Due to the way special relativity works, tachyons would have some rather peculiar properties and wouldn't really look all that particle-like. There is no experimental evidence for their existence.

• their energy corresponds to their mass always. A faster-moving particle will also appear to be more massive. Commented Feb 17, 2017 at 1:12
• @DepressedDaniel: in my experience, that notion of relativistic mass has largely been abandoned; John Roche via Wikipedia in 2005: "Today, nuclear and particle physics make no reference to relativistic mass. It has also been estimated that about 60% of authors now writing on special relativity do not introduce it", where the reference for the figure is a 1990 article in Physics Today Commented Feb 17, 2017 at 12:04

Well, that would be warp drive kind of thing where space is making you move without you having to spend energy and space makes you stop without you having to spend energy. By without, I mean not as much as needed to gain that much momentum.

Then the question would be - will that kind of movement will even be called momentum? For example, universe is expanding with space. Can we say that the universe has momentum? I doubt. Even if the expansion qualified as momentum, then, universe is expanding in every direction. Momentum is a vector thing. What would be direction of that momentum?

So, any movement through space is momentum and not without energy. But movement with space, if defined as momentum, can be without energy. Time to invent it!