# Where is potential energy in relativistic formula of energy?

1. ### ndung200790

520
The relativistic energy E=m.square(c)/squareroot(1-square(v)/square(c)) is determined by momentum p(because of ''square''(4-p)=square(m)).Then what is the role of potential in relativistic energy?When we consider the interaction between particles,how can we express the kinetic energy plus potential energy in the same formula?
Thank you very much in advance.

2. ### bcrowell

5,883
Staff Emeritus
You may be more likely to get helpful responses to your posts if you make them easier to read by marking the math up in LaTeX. Here is an example: $\sqrt{x^2+y^2}=1$. To see how I did that, click the QUOTE button on my post.

3. ### mathfeel

181
The question has to be more specific: what kind of force? Because if it is gravity, you need to do some GR and it's not exactly a potential anymore. If it's EM, the potential is a 4-vector: $(\phi, \mathbf{A})$, where $\phi$ is the scalar potential and $\mathbf{A}$ the vector potential. The Hamiltonian in this case is:
$$H = \sqrt{m^2 c^4 + \left(\mathbf{p} - e\mathbf{A}\right)^2 c^2} + e\phi$$

4. ### inottoe

25
If it's gravity, then both potential and kinetic energy are generalised as (letting c=1):
$-m\frac{d\tau}{dt}$​
where
$d\tau^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$​
We can see this clearly when we assume spherical symmetry and consider the Newtonian limit:
$-m\frac{d\tau}{dt}=-m\sqrt{\frac{g_{tt}dt^2+g_{mm}(dx^m)^2}{dt^2}}$​
$=-m\sqrt{g_{tt}-\dot{x}^2}$​
$=-m\sqrt{1-\frac{2GM}{r}-\dot{x}^2}$​
$\approx-m+\frac{GMm}{r}+\frac{1}{2}m\dot{x}^2$​
And given that the action in a gravity well is
$-m\int{d\tau}=0$​
then we recover the non-relativistic
$E=\Delta{U}+KE$​

Last edited: Jul 24, 2011