Conservation of Momentum

Here’s something that’s in textbooks, but they tend to leave out lots of little bits and pieces, the way I used to when I made sandwiches for Arby’s one summer. Not that you’ll get the full story here, either. But you’ll get a more satisfying hunk of disgusting, gray, dampish meat clumps, and a little piece of metaphorical lettuce, too.

When two particles interact, Newton’s third law postulates

F_{12} = - F_{21},

where F_{12} means, “the force particle ‘1’ exerts on particle ‘2’.” This is useless knowledge unless you have some sort of interpretation of force. Force is defined by the second law

F = ma.

Someone once tried to tell me that Newton’s second law is not just a definition of force, but has some deeper meaning. I think they were lying because they wanted to seduce me. (No luck there, Grandpa!)

So Newton’s second law defines force, and is meaningless without some rules about what force should do. For example, if you say that a particle with absolutely nothing around to interact with must have no force on it, you’ve said something about force and now Newton’s second law can step in. In this case it says

0 = ma

so that a free particle does not accelerate. (That’s Newton’s first law. However, there are philosophical problems with such a conclusion. If there is nothing around for the particle to interact with, then how could you tell whether or not it’s accelerating?)

The third law, a rule about force, is lame without a definition of force. The second law, a definition of force, is lame without any rules. They were made for each other, like rabbits and lawn mowers (but with less of those annoying screaming sounds). By combining Newton’s second and third laws for two interacting particles, we get

m_1a_1 = F_{21} = -F_{12} = -m_2a_2

By the transitive law

m_1a_1 = -m_2a_2

or

m_1a_1 + m_2a_2 = 0

and assuming that mass is constant

\int_{t_a}^{t_b}dt \left(m_1a_1 + m_2a_2\right) = \int_{t_a}^{t_b} dt*0 = 0

for arbitrary times t_a and t_b. Using the fundamental theorem of calculus and the definition

a = \frac{dv}{dt}

yields

\left(m_1v_1(t_b) + m_2v_2(t_b)\right) - \left(m_1v_1(t_a) + m_2v_2(t_a)\right) = 0

again for arbitrary times t_a, t_b. What we’ve discovered is that if you take measure the quantity

m_1v_1 + m_2v_2

at any two times, you will always get the same answer. That quantity is called “momentum”, and the fact that it doesn’t change is called “conservation of momentum.”

We haven’t proved it to be true. Science doesn’t prove anything to be true. What we’ve proved is that it follows from certain assumptions. If we make some measurements and find that the “law” of momentum conservation doesn’t hold, there are a few possibilities that I can think of:

  1. We made a mistake with the measurements. Our apparatus is broken, or we did something dumb like converting units wrong, etc.
  2. Newton’s laws are wrong. They do not accurately represent the interaction of particles.
  3. We were not doing an experiment with exactly two particles. (That is the only situation for which we did the proof. Maybe the theorem failed because there was some third particle around that we didn’t see, or maybe the objects in our experiment were not particles, but instead more complicated composite things that are not bound by Newton’s laws.)
  4. The mass of the particles is not constant. (Remember that this was an assumption used in the proof).

Maybe you can think of other explanations. I can’t at the moment. But it turns out that these explanations can account for a lot of situations. Item (1) comes up frequently enough – it’s just a fact that people make misteaks.

Explanation (2) is sometimes correct as well; Newton’s laws aren’t true. Special relativity modifies them. General relativity pretty much scraps them (er, don’t quote me on that). In quantum mechanics, momentum is important, but no longer has an interpretation as mass*velocity. In fact it (mathematically) no longer has any “interpretation” – instead it is its own primary quantity, equally as fundamental to the theory as the concept of “position”. It even steals “position”‘s claim to the letter ‘p’. Momentum is nobody’s bitch.

Complication (3), that we aren’t using two isolated particles, arises in practice as well. There are obvious examples, such as everything. When I drop my spoon, it starts gaining momentum until it hits the floor, when it loses momentum. Then I pick it up and lick it clean, and its momentum bounces all around as I lick more and more violently. All this occurs because a spoon is not a system of two particles.

There are more interesting (but less tasty) examples where the “not-two-particles” explanation manifests. Take two charged nonrelativistic, non-quantum particles and let them interact. They won’t conserve momentum. The reason is charged bodies generate electromagnetic fields, and our assumption that the only things around are the charged bodies fails. The electromagnetic field can carry its own momentum, although technically in order to break the proof all it would have to do is exist. In another example, Wolfgang Pauli was thinking about another case in which momentum is not conserved – beta decay. He decided options (1), (2), and (4) were not for him, and instead guessed that beta decay must involve some previously-unseen stuff. That stuff is the neutrino.

Finally, explanation (4), that the mass is variable, is not something that occurs in practice to my knowledge, but it could. Of course, if a meteor shooting through space hurls off some of its rock-junk when it get near the sun and heats up, then the meteor’s mass decreases. But that doesn’t count because it’s not two particles, and also momentum actually is conserved in that situation if you consider the momentum of the space junk, the meteor, and the sun altogether. What I mean is that I’m not aware of any evidence that fundamental particles can have variable mass.

What if there are three particles? Can we prove that momentum is still conserved if we define momentum to be

p = m_1v_1 + m_2v_2 + m_3v_3?

No. We can’t because we could only prove anything by getting some knowledge about force from Newton’s third law. But Newton’s third law is only telling us the story for two lone particles. When there’s a third, all bets are off. However, there is another assumption that we usually take along with Newton’s laws, often implicitly. This is that forces add linearly.

Imagine conducting an experiment with particles 1 and 2, and no particle 3 around. Measure the force on particle 1. Now conduct a new experiment where particle 1 does the same thing it did before, but particle 2 is absent, and particle 3 is around doing whatever it wants. Again, measure the force on particle 1.

We assume that if we conduct a third experiment with particles 1, 2, and 3 all together, the force on particle 1 will be the sum of the forces in the first two experiments.

With this law that forces add linearly, we can prove momentum conservation for three particles. And if we assume forces continue to add in the simple manner for any number of particles, then momentum conservation also holds for any number of particles.

tomorrow: energy

Advertisements

Tags: , , , , , , ,

2 Responses to “Conservation of Momentum”

  1. Conservation of Energy « Arcsecond Says:

    […] Arcsecond As Simple As Improbable, But No Less Unconfusing « Conservation of Momentum […]

  2. Irodov707 Says:

    Nice read. Was trying to settle a debate today with a cosmologist and a HEP theorist. I’m a cond. matt physicist. I was of the opinion that Newton’s 2nd law in itself is nothing more than definition of force. Your article reinforces my opinion. Thanks again!

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s


%d bloggers like this: