# Momentum of a particle? [closed]

I really need help to understand what is momentum of a particle (of a photon, proton, an electron...) I see so many definitions! My main questions are:

•What exactly is momentum

•What are the differences between momentum and spin of a quantum particle.

•What is the relation between momentum of a photon and it's wave.

I just wanted simple explanations. Because after months hearing about momentum, I still don't know what is it.

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## closed as not a real question by Alfred Centauri, akhmeteli, Waffle's Crazy Peanut, Qmechanic♦Jun 18 '13 at 11:21

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

The problem, I think, might be this: "I just [want] simple explanations." The fact is, there are many simple explanations. What you're asking for is, evidently, an explanation that requires little or no effort on your part to understand. Because, honestly, it doesn't take that much effort to, at least, develop some relatively sophisticated notion of momentum. But then, there is the question of whether or not momentum is something fundamental in which case, the question of "what exactly is momentum" cannot actually be answered in terms of anything "more" fundamental. – Alfred Centauri Jun 18 '13 at 0:25
Every time I try to read about it, I don't understand anything =x – user2489690 Jun 18 '13 at 0:30
Then you need to remedy that problem first. Math is the language of physics. Anyone that claims to understand physics "in words" is, well... telling a tall one. – Alfred Centauri Jun 18 '13 at 0:34
The simplest answer is, it's mass multiplied by velocity. I know that may sound almost like a non-answer, but that's really the simplest way it can be defined. A more comprehensive way to define it would involve describing its properties. – Ataraxia Jun 18 '13 at 1:18
Seriously, why did someone close this topic? What is wrong with my question? I made it pretty clear what I was asking, and I also got a good answer! I don't understand... – user2489690 Jun 18 '13 at 16:12

When you hear momentum, one means the linear momentum of a particle. It's a measure for how much a particle moves. Mathematically (and according to classical mechanics): $p=mv$, in words: mass times velocity.

It's intuitively a very useful concept: comparing two objects with the same velocity, but different mass, one would easily say that the one with the greater mass is "more moving", one needs more work to move the heaviest object. So linear momentum is some kind of a weighed velocity.

Another useful concept is conservation of linear momentum. Consider two objects, one that has no momentum (the objects stands still), and one that moves. Assume that when the collide, they'll stick together (this is an inelastic collision). The total momentum of the two objects together, will be the same as the original momentum of the single moving object. So now both objects will be moving. Since you want $p=mv$ to be constant (according to the law), $v$ of the two objects together will be lower, since $m$ of the two objects is larger than $m$ for only one object. I think this should be clear intuitively too.

As you stated, a photon has a momentum too, although it has no (rest) mass. The definition here is $p=h/\lambda$, with $h$ the constant of Planck, and $\lambda$ the wavelength of the photon. To explain the correlation between these two definitions, is intuitively a little more difficult, since one would need theory of relativity. But that's why I mentioned an inelastic collision: when photons collide inelastically on an object in space (which means the object absorbs the photons, so the object and photons will "stick together" after collision"), they'll give extra momentum to the object. So the object will accelerate, which proves experimentally that photons have a momentum.

To conclude, linear momentum has actually nothing to do with spin. The reason why you're probably confused, is there's also something called angular momentum. It describes how much a particle rotates (like the linear momentum describes how much a particle moves). And there exists a conservation law of angular momentum too. It's because of this law, that eventually one came up with the spin of a particle, also called spin angular momentum. It's an intrinsic characteristic of a particle; if it wouldn't exist, e.g. conservation of angular momentum wouldn't apply for the electrons in an atom. You can imagine spin of an electron as if it was rotating around it's own axis, although that's not a real accurate description.

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Very good explanation! And what does that have to do with if you know the position, you may not know the momentum (uncertainty principle)? If h is constant, then that means the wave length is unknown, once I know the exactly position of a photon? Am I right? – user2489690 Jun 18 '13 at 16:08
Actually that's a whole different thing. It's Heisenberg's uncertainty principle which states you cannot know the exact position and momentum of a particle at the same time. The uncertainties on the different quantities, are related by $\Delta x \Delta p \ge {h\over 4 \pi }$, where h *is* a constant, **Planck's constant**.$x$ is the position of the particle, $p$ the impulse. The $\Delta$ in front of the quantity means the interval in which the quantity is located. If $x$ is exactly known, $\Delta x$ is infinitely small. $\Delta p$ should be infinitely large then, which means $p$ is unknown. – BNJMNDDNN Jun 18 '13 at 16:22

The momentum is defined in non-relativistic mechanics as: $$p=mv=m\frac{\mbox d x}{\mbox d t}$$

It can intuitively (but very hand-wavily) be thought of has "how much a mass moves".

The integral of momentum with respect to the velocity is the Kinetic Energy.

$$\int p\mbox{ d}v=\frac{mv^2}{2}=K\mbox{ or } T, \mbox{ whatever notation you use.}$$

The derivative of the momentum with respect to time is the force:

$$F=\frac{\mbox{d}p}{\mbox{d} t}=ma$$

I would say it is best for you to learn some Classical (Newtonian, Lagrangian, Hamiltonian) Mechanics first . And also Old Quantum Theory.

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v = distance/time, then what is d? And I already know classical mechanics. But thanks for the explanation. – user2489690 Jun 18 '13 at 16:03
"d" means differential, you will learn it if you try to learn calculus (which is very important to learn even something as basic Newtonian mechanics) – centralcharge Jun 19 '13 at 2:53
And v is the rate of change of distance over time, which is the differential in x divided by by the differential in t, not x divided by t. – centralcharge Jun 19 '13 at 4:48