Work in special relativity When computing work in special relativity, do you only use the equation
$$
E=\sqrt{p^2c^2+m_0^2c^4} 
$$
or can you also use
$$
W=\int_{\Gamma}F(x^\mu)ds
$$
(where $x^\mu$ means $(x,y,z,t)$. And if you can use the second equation, would you use apparent time and length or proper time and length
 A: Even without worrying about relativity, energy and work are different things. Energy is a function of the state of the system and is conserved. Work describes the mechanical transfer of energy, and it describes a process, not a state.
The relevant special-relativistic identity (in one dimension for simplicity) is $E=\int F dx$, where $F=dp/dt$, and $x$ and $t$ are the Minkowski coordinates. Since $t$ is not the proper time, $F$ is not a relativistic vector. (To get the relativistic force vector, you would need to divide the momentum by the proper time, which is a relativistic scalar.) The variable $x$ is not a proper length; to define a proper length, you need an extended object that sweeps out a ribbon in Minkowski space, and there is no such object here. To prove the identity:
$$\frac{dE}{dx}=\frac{dE}{dp}\frac{dp}{dt}\frac{dt}{dx}$$
The first factor is $dE/dp=v$, the second factor is the force, and the third factor equals $1/v$.
A: Good question, yes you can. Its not obvious that in special relativity that forces exist since the magnitude of velocities are not changing. One cannot use Newton's $\mathbf{F}=m\mathbf{a}$, but if one uses the relationship for force being
$$\mathbf{F}(x^\mu)=\frac{\text{d}\mathbf{p}}{\text{dt}}$$
with the relativistic momentum, then the work integral holds for conservative paths. In GR, one has to pay extra attention to the line element $ds$ and realize that energy conservation is not well defined as
$$\nabla_\mu T^{\mu\nu}=0$$
is a very very special case and works locally. See Sean Carroll
s post.
A: 
When computing work in special and general relativity, do you only use the equation
$$
E=\sqrt{p^2c^2+m_0^2c^4} 
$$

You can use this equation to determine the kinetic energy. Then, just like in non-relativistic kinematics, the work (the usual $\vec F \cdot \vec dx$ work) is the change in the kinetic energy.
You can see this is true because:
$$
\int \vec {dx}\cdot \vec F = \int \vec {dx}\cdot\frac{\vec {dp}}{dt}
= \int \vec {v}\cdot \vec {dp}
$$
which, by the canonical definition of velocity, is equal to:
$$
= \int \frac{\partial H(\vec p,\vec x)}{\partial \vec {p}}\cdot \vec {dp}
$$
and which, since $H = E(\vec p) + U(\vec x)$, where $E$ is your expresion for the kinetic energy, we have:
$$
=\int \frac{d E}{d\vec {p}}\cdot \vec {dp} = \Delta E
$$

or can you also use
$$
W=\int_{\Gamma}F(x^\mu)ds
$$
(where $x^\mu$ means $(x,y,z,t)$. And if you can use the second equation, would you use apparent time and length or proper time and length

No, you can't use this. We already saw above that it is $\int \vec F\cdot \vec{dx}$ that is equal to the change in kinetic energy.
To see more explicitly why $\int_{\Gamma}F(x^\mu)ds$ is wrong, note that:
$$
p = (E/c, \vec p)
$$
$$
F = \frac{dp}{d\tau} = (\vec v \cdot \frac{d\vec p}{d\tau}/c, \frac{d\vec p}{d\tau})
$$
and
$$
ds = (cdt, \vec {dx})
$$
So
$$
\int Fds = \int (\vec v \cdot \frac{d\vec p}{d\tau}dt - \vec {dx}\cdot\frac{d\vec p}{d\tau}) = 0
$$
See also, this answer.
