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In terms of gravity and electric fields, I'm not sure what the difference is between field strength and potential is and how they are related? Both using maths and not.

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The field strength is the negative first derivative of the potential.

For example, in Cartesian coordinates, with electric potential $V = V(x,y,z)$, the electric field is:

$$ \boldsymbol E =-\nabla V= -\frac{\partial V}{\partial x}\boldsymbol i - \frac{\partial V}{\partial y}\boldsymbol j - \frac{\partial V}{\partial z}\boldsymbol k $$

Note that you can add or subtract an arbitrary constant potential to $V$ and it will not affect the field strength, since its derivative will be 0.

For the field/potential, you can calculate the force/energy by just multiplying by a charge. In terms of gravitational field, use $\boldsymbol g$ instead of $\boldsymbol E$, and you multiply by mass, rather than charge, to get $\boldsymbol F$.

A "physical" way to think of the derivatives is that they show how rapidly the potential changes in space; big changes in potential over a short distance make for big fields in the direction of maximum change.

Note that the relation between electric field and potential above is true only for static fields. The more general formula is:

$$ \boldsymbol E=-\nabla V-\frac{\partial\boldsymbol A}{\partial t} $$

where $\boldsymbol A$ is the magnetic vector potential.

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Hi Will. Welcome to Physics.SE. Thanks for your answer. As a part of your contribution, please use TeX markups (MathJax) in equations of your answers. If you could spend some time, have a look over here or our FAQ. Feel free to respond to guys who comment (for instance, me) on your posts using @ before the commentator's name :-) – Waffle's Crazy Peanut Jan 14 '13 at 14:50
Thanks, will do! – Will Cross Jan 14 '13 at 14:56
@CrazyBuddy D'oh! How about this? At any rate, I'm researching TeX and the FAQ before I go too crazy on comments & answers. – Will Cross Jan 14 '13 at 15:22

The units of electric field should be a big hint here: Volts/meter. It is literaly the electric potential per unit length along the field vector.

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I don't understand what it means by a meter the field is an area (or really a volume), but a meter is one dimensional. And that would seem to suggest the field is constant? – Jonathan. Jan 14 '13 at 12:53
The meter here is measured along the direction of the field. So the field strength is a measure of the difference in potential along one meter of field line. – dmckee Jan 14 '13 at 12:57
So the field strength defines the potential at each point? (or the difference in potential between two point?) – Jonathan. Jan 14 '13 at 12:58
And what is the units for gravitational field? I can't find such a clear answer as there is for electric fields? – Jonathan. Jan 14 '13 at 12:59
You can add or substract an arbitray constant from the potential without affecting the physics (a simple gauge transform), so the field tells you about potential differences rather than absolute potential. – dmckee Jan 14 '13 at 12:59

In terms of electromagnetism, the potential is given by the gauge field $A_\mu$, whereas the field strength is given by $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu.$ The main difference is that the field itself is can be changed to a different value by a so-called gauge transformation (which is given by an element of the group $U(1)$), while the field strength itself is invariant (not changed) by it.

For gravity, one can find a similar structure: here the field strength corresponds to the Riemann tensor ${R^\mu}_{\nu\rho\sigma}$, from which one can construct a quantity that is invariant under a change of coordinates (which corresponds to the electromagnetic gauge transformation), i.e. the Ricci scalar $R$. The coordinates are represented by the metric tensor $g_{\mu\nu}$, and the relations between the aforementioned quantities are given by

\begin{equation}{R^\mu}_{\nu\rho\sigma}=\partial_\rho{\Gamma^\mu}_{\nu\sigma}-\partial_\sigma{\Gamma^\mu}_{\nu\rho}+{\Gamma^\mu}_{\lambda\rho}{\Gamma^\lambda}_{\nu\sigma}-{\Gamma^\mu}_{\lambda\sigma}{\Gamma^\lambda}_{\nu\rho},\end{equation} \begin{equation}{\Gamma^\mu}_{\nu\rho}=\frac12g^{\mu\sigma}(\partial_\rho g_{\nu\sigma}+\partial_\nu g_{\sigma\rho}-\partial_\sigma g_{\nu\rho}),\end{equation} \begin{equation}R_{\mu\nu}={R^\lambda}_{\mu\lambda\nu}\end{equation} and \begin{equation}R=R_{\mu\nu}g^{\mu\nu}.\end{equation}

$R_{\mu\nu}$ is the Ricci tensor and ${\Gamma^\mu}_{\nu\rho}$ is the Christoffel symbol. The latter can be interpreted as a gauge field for gravity. The analogy is even clearer if one considers the field strength of a non-abelian gauge symmetry. For example, the field strength of $SU(3)$ is given by \begin{equation}F_{\mu\nu}^{ab}=\partial_\mu A_\nu^{ab}-\partial_\nu A_\mu^{ab}+iA_\mu^{ac}A_\nu^{cb}-iA_\nu^{ac}A_\mu^{cb},\end{equation}

where $a$ and $b$ are indices that transform under SU(3).

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For someone who is clearly starting, and doesn't "see" the difference, this may not be the best explanation. – MyUserIsThis Jan 14 '13 at 17:39
While I agree that my answer might be a little over the top, I'm convinced that it is on topic and useful to some. – Frederic Brünner Jan 14 '13 at 17:41

Field strength determines the force in the Newton equation. It is responsible for the velocity changes.

The potential has no such a meaning - its absolute value is not important if kept constant throughout consideration. So its absolute value is a matter of convenience. It is involved in such a thing as the total energy. And the velocity absolute value will change only if the potential changes. A particle never feels the potential itself, but a force (a local potential difference in the direction of maximum potential change).

The local potential difference $\phi(x+\delta x)-\phi(x)\;$ divided by $\delta x\;$ with small $\delta x\;$ determines the field strength at $x$.

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You have to view the field as a vector field, that is, for every point, you will have a vector, that is $E$, that represents the force that a unit charge will feel at that point, and the direction of the force. If you don't feel comfortable with Volts/meter units, then use N/C (Newtons/Coulomb) That if: force per unit charge, that happens to be the same as volts/meter.

The potential talks about energies. The potential is now a scalar field, that is: for every point in the space we have a number, that number represents the energy a unit charge would have if it was located there (potential electric energy). By definition the potentials are defined so their gradient is the force: $E=-\nabla V$, as someone said before, in units, we have that $N/C=[V]/m$, being $N$=newtons, $C$=coulombs, $m$=meter, and $[V]$ the units of our new potential unit, and we decide to call that volts, as you see, the measure of the electric field is the same as volts/meter: $V/m$

See it as the slope of the potential, that is how it's defined: $-\nabla V$, the field goes down/up $x$ volts per meter in point $P$.

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In a region with electric potential $V$ that varies with position, the electric field $E$ points along the potential's direction of greatest decrease. A positive charge will naturally go towards points of lower electric potential, and so at any given position, it will feel a force proportional to the electric field at that point, for the $E$-field points downhill, so to speak, exactly where the particle is compelled to go.

You can imagine the $V$ field as determining a landscape with hills and valleys--the value of $V$ determines the heights of these hills and valleys. The $E$-field points along the unique direction that most quickly goes downhill.

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