# How is everything a field

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?

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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
–  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. –  Dimensio1n0 Jun 8 '13 at 10:23

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.

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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

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