# General physics questions (experts)

1. Aug 7, 2008

### jspstorm

1. Classical momentum is derived using this formula: p=mv. The mass of a photon is zero, so then, how can it have momentum? Does it have some other kind of momentum that differs from classical momentum?

2. Photons are particles, why then do they travel in a wave? I've heard it's because space-time is distorted in a higher spatial dimension, and light being a vibration from this higher dimension, follows a straight line in the higher dimension, but from our limited perspective, appears to be taking a path, which is not the shortest route between two points.
But this doesn't make sense to me, why would a vibration, a form of harmonic motion in a higher dimension cause particles(photons) to appear?

3. Does anti-matter only explode upon contact of its matter counterpart, or upon contact with any form of ordinary matter? And why does it explode at all? For example, why would a neutron, and its anti-matter equivalent explode when they came into contact? They both carry no charge. And what exactly is the difference between a matter neutron, and an anti-matter neutron? Are they composed of different point particles, or what?

4. Why is our universe all matter, and no anti-matter? Is there some special property, or was it simply luck which caused our universe to be composed of matter? Obviously both cannot coexist, but surely it was just as probably for the universe to form from anti-matter.

5. Could someone please explain the 'spin' of subatomic particles to me? Apparently it isn't actually rotation of a particle around it's center of mass, but something different from classical rotational momentum.

2. Aug 7, 2008

### JesseM

In relativity the momentum for a massive object with rest mass m and velocity v is not p=mv, but $$p = \frac{mv}{\sqrt{1 - v^2/c^2}}$$. If you make m=0 and v=c, this formula breaks down, giving 0/0, so it doesn't really tell you anything about the photon's momentum. But in relativity we also have the formula for the total energy E of an object with nonzero momentum p, which is:

$$E^2 = m^2 c^4 + p^2 c^2$$

Rearranging, this gives:

$$p^2 = \frac{E^2 - m^2 c^4}{c^2}$$

And with a rest mass of m=0 for the photon this reduces to p = E/c. We know from quantum mechanics that the energy of a photon is given by E=hf, where f is the frequency and h is Planck's constant, so this means the momentum of a photon must be p = hf/c.
Photons don't travel on wavy spatial paths, if that's what you mean. If you see a photon depicted as a sine wave, the horizontal dimension represents space but the vertical dimension would represent electromagnetic field strength at each point along the line in space, at least in classical electromagnetism; in QM I think it might represent the quantum-mechanical "phase" of the photon's wavefunction at different points in space, though I'm not really sure (someone correct me if I'm wrong). "Higher dimensions" appear in string theory but they are not part of the normal theory of relativity, which just has 3 space dimensions and 1 time dimension.
This is still an active area of research among physicists; the basic idea seems to be that very shortly after the Big Bang the universe originally had an almost exact balance of matter and antimatter, but with a very slight bias towards matter, so that when the vast majority of the matter and antimatter annihilated there was still a little matter left over. There are theoretical ways to get this sort of small imbalance, but they involve some new physics beyond the "Standard Model" of particle physics. See here for a good introductory discussion.
In terms of experiments, I think it's basically a property that determines how particles will move in magnetic fields (at least in the case of charged particles, for neutral particles like photons spin is measured differently, see http://electron6.phys.utk.edu/cat/electron_an_photon_spin.htm [Broken] for a good discussion of spin measurements for both kinds of particles), but the rules governing it are strange and don't match what we'd expect if the particles were little classical spinning charged balls.

Last edited by a moderator: May 3, 2017
3. Aug 7, 2008

### Fredrik

Staff Emeritus
Welcome to PF jspstorm. I see that this is your first post, so you are probably not aware that all of these questions have been asked and answered many times here. I hope you don't think I'm rude if I suggest that you try to search for the answers in older posts first.

I don't have time for a long answer right now, so I'll just address the detail that's definitely wrong. The explanation you have heard about the wave properties of light is definitely incorrect. Some of it sounds like a part of the explanation of why the path of a photon is deflected when it passes a heavy object like a star, so I'm guessing that you have heard the answer to another question and got the two mixed up somehow.

Photons are not particles in the sense of classical physics. If they were, the state of a photon could be represented by six numbers (3 for position, 3 for velocity). It can't. Photons are much more complicated. In particular, they don't even have a well-defined position. So they are neither classical particles nor classical waves. They are quantum particles which is something completely different. Quantum particles have both particle-like and wave-like properties. An example of the first is that they can make a detector click each time it gets hit by one. An example of the second is that the probability of detection by a certain detector at a certain time depends on all the paths through spacetime from the emission event to the possible detection event.

I see JesseM has answered some of your questions now. Maybe you'll have a complete set of answers before you have time to do a search.

4. Aug 7, 2008

### rbj

actually, classical mechanics defines momentum as:

$$\vec{p} = m \vec{v}$$

and has empirically determined that the net composite sum of this other quantity called "force" follows this behavior:

$$\sum_n \vec{F_n} = \frac{d\vec{p}}{dt}$$

so closely, that they call it a "law". now, if the mass is constant, the above comes out as

$$\sum_n \vec{F_n} = \frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt} = m \frac{d\vec{v}}{dt} = m \vec{a}$$

now there is a dispute between folks about the value of differentiating the concepts of "rest mass" (a.k.a. "invariant mass") and "relativistic mass" (which i have seen at least one other person here refer to as "inertial mass"). if you belong to the minority group, you like to define this inertial mass as whatever quantity necessary to multiply the velocity vector to get the resulting momentum vector as in the top equation (thus choosing to keep the p=mv definition). it turns out that that relativistic mass is

$$m = m_0 \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$

where m0 is the rest mass, what inertial mass the body would apparently have at low velocities compared to c.

i personally like this minority way of looking at it because in the oft-repeated

$$E = m c^2$$

if the mass m[/m] means this relativistic mass above, then the E means the total energy of a body (as seen by some observer that the body is moving at a velocity of v relative to), the sum of the "rest energy" (what the majority means by "E") and the kinetic energy.

5. Aug 7, 2008

### jspstorm

"Photons are not particles in the sense of classical physics. If they were, the state of a photon could be represented by six numbers (3 for position, 3 for velocity). It can't. Photons are much more complicated. In particular, they don't even have a well-defined position. So they are neither classical particles nor classical waves. They are quantum particles which is something completely different. Quantum particles have both particle-like and wave-like properties. An example of the first is that they can make a detector click each time it gets hit by one. An example of the second is that the probability of detection by a certain detector at a certain time depends on all the paths through spacetime from the emission event to the possible detection event."

Surely any particle, even a Quantum particle must have a definate position in space-time? If a photon is striking a detector, then it must be occupying a certain part of space-time, one directly adjacent to the detector.

When you say that photons can't be represented using six numbers, are you refering to the uncertainty principle? I thought that you could definitively locate the position of the particle, but only while not simultaneously knowing the momentum.

6. Aug 7, 2008

### Fredrik

Staff Emeritus
Nope. Quantum mechanics is pretty cool, isn't it?

Yes, this is correct. The particle has a reasonably well-defined position immediately after the measurement, but not before the measurement and not when some time has passed after the measurement. The position measurement changes the state of the particle into a state of (almost) well-defined position. (If we had chosen to measure its energy instead, the measurement would have put the system into a state of well-defined energy and less well-defined position).

I'm talking about stuff that you can think of as the underlying causes of the uncertainty principle. The uncertainty in position is a property of the quantum state of the particle, and so is the uncertainty in momentum. The uncertainty principle says that the product of the two is $\geq\hbar/2$, but it's also the case that both of them are non-zero at all times. A position measurement will make $\Delta x$ small and $\Delta p$ large, and a momentum measurement will accomplish the opposite.

There is however no lower bound on how accurately you can measure the position or the momentum, so it does make some sense to say that you can locate the particle.

By the way, there are lots of threads about the uncertainty principle, and most of them have "uncertainty" somewhere in the title, so they shouldn't be hard to find. There are some good explanations of the finer details in several of them.

7. Aug 23, 2008

### jspstorm

I have one more method of determing a quantum particles location I was wondering about.
Assume first, that time travel is possible. Now, a light bulb, floating in vacuum, is given enough electricity to produce just one photon. A scientist places one molecule of h2o2 (photosensitive) one meter away from the bulb. When this particle fails to react, he travels backward in time to the instant after the photon was released, and relocates the detector particle. He does this again and again, variating in all three dimensions until he gets a reaction. He then notes the position (relative to the bulb), and travels back once again, then does the same, 1 centimeter closer to the bulb. After doing this for nearly an eternity, he plots the collisions on a three dimensional graph, and discovers the trajectory of the particle. Now he can calculate its position at any given time.

Where am I going wrong?

8. Aug 23, 2008

### Fredrik

Staff Emeritus