Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've heard before that everything in physics can be thought of as either a field, or its excitation. Is there some intuitive explanation of how I can look at gravity, light, electromagnetism, etc as a field/excitation?

share|cite|improve this question
There's not much that's "intuitive" about quantum field theory; that's why it took physicists so long to figure out. A better way to phrase it would be: fields are the fundamental building blocks of the universe (instead of particles). Particles, and the forces between them, are simply (quantum) excitations of those fields. – Dmitry Brant Jun 7 '13 at 15:54
Related: and links therein. – Qmechanic Jun 7 '13 at 16:22
Just a note: I think the idea of fields is much more intuitive than the standard pre-quantum field theory era interpretation. – centralcharge Jun 8 '13 at 10:23
up vote 4 down vote accepted

A field is just something that has a value at every spatial coordinate (or close to that). It is easy to think of gravity and electric/magnetic fields. Consider gravity, no matter where you go, you can always state the value of gravity at that position (and it's direction but that is not a necessary condition). For example, at sea level, we can say the acceleration due to gravity is $9.81m/s^2$ and at other places, like far out in intergalactic space, we can pick any location and say the gravity is (somewhere around) $0m/s^2$ (it's not, but that's pretty close). Similarly, we can do the same thing with electromagnetism; we can point to any spacial coordinate and give a value for the electric and magnetic fields. Barring some radical physics, usually you will find the values of a field to be continuous across space. Thus, you "could" think of a field like a body of water; for a vector field (like the ones you mentioned), each point in space would give the speed and direction of flow of the water. For a scalar field (no direction), each point would give the temperature of the water. Note while each point in the water may have different values, there will still be a smoothness/evenness to its flow.

Light is slightly different. As you pointed out, it is an excitation of the EM fields. An excitation of a field is when the values of the field in some localized area are fluctuating/changing relatively rapidly. In my water analogy, it would be like waves/ripples in the water (depending on the amplitude of the excitation). The overall body is still flowing more or less the same, but in a localized area, there are relatively rapidly fluctuating values.

Hope that helps.

share|cite|improve this answer
that does help thanks for that. And similarly a "force" between these excitations can be thought of as how? – user79950 Jun 8 '13 at 17:51

A classical field is simply a space-time function $\Phi(x,y,z,t)$, whose parameters are space-time coordinates $x,y,z,t$, and $\Phi$ is a quantity depending of the considered problem.

For instance, the temperature $T$ could vary from time to time, and vary from one place to another place, so you can modelize this temperature as a field $T(x,y,z,t)$ depending on place and time.

In this example, temperature is a scalar quantity, but you can have also vectorial quantities, for instance, you can modelize the speed of the wind as a vector : $\vec v(x,y,z,t)$

Quantum fields are differents because real quantities becomes operators (you may think "operators" as a kind of "infinite" matrices). The difference with usual quantities, is that operators don't commute. (same thing as for matrices)

For instance, a real bosonic scalar quantum field $\phi$ must verify these non-commutation rules :

$$[\phi(x,y,z,t), \frac{\partial \phi(x',y',z',t)}{\partial t}] = i\delta(x - x')\delta(y - y')\delta(z - z')$$

You can make a decomposition of $\Phi$ on momentum space:

$$\Phi(x,y,z,t) = \int d^4\tilde k ~ [a(k) e^{-i k.x} + a^+(k) e^{+i k.x}]$$

Here the $a(k), a^+(k)$, which are operators too (they verify $[a(k), a^+(k')] = \delta^3(k - k')$), and could be seen as creation or anihilation of "particles" or "excitations of fields".

You can describe electromagnetism as quantum fields, using the same kind of above tools

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.