Is there a name for the integral of 4-force with respect to spacetime distance?

Question

I understand that the integral of force with respect to distance is known as Energy. As I understand it even in special relativity the force is the same as the derivative of momentum with respect to time. In Special Relativity the four force is the derivative of spacetime coordinates with respect to proper time. I understand that the spacetime distance is given by the equation

$${\Delta}s^2={\Delta}x^2+{\Delta}y^2+{\Delta}z^2-c^2{\Delta}t^2$$

with ${\Delta}s$ being the spacetime distance. (I understand that the term "spacetime interval" is generally used, but I get confused as to whether it refers to ${\Delta}s$ or ${\Delta}s^2$, which is why I'm using the term "spacetime distance" instead).

Is there a name for the integral of the Four Force of an object with respect to spacetime distance in Special Relativity?

Answer 1

There is a simple name for this quantity: it's zero for ordinary forces.

To see this, suppose it's the change in some quantity $I$, $$\Delta I = \int F^\mu \, dx_\mu.$$ Then the rate of change of $I$, with respect to proper time, is $$\frac{dI}{d\tau} = F^\mu \frac{dx_\mu}{d\tau} = F^\mu u_\mu$$ where $u^\mu$ is the four-velocity. Now, the definition of the four-force is the rate of change of four-momentum, so $$\frac{dI}{d\tau} = \frac{d}{d\tau} (m u^\mu) u_\mu = m \left(u_\mu \frac{du^\mu}{d\tau} \right) = \frac12 m \frac{d}{d\tau} (u^\mu u_\mu).$$ Note that this bears a strong resemblance to the nonrelativistic expression for power. But the difference is that in relativity, the norm of the four-velocity is constant, $$\frac{d}{d\tau} (u^\mu u_\mu) = 0$$ which means that $I$ is a constant. In other words, your integral always vanishes.

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There's an exception, which is that I assumed that $m$ was independent of $\tau$ when performing the derivative above. If you have a force that is capable of changing rest mass, then we instead get $$\frac{dI}{d\tau} = \frac{dm}{d\tau} (u^\mu u_\mu) = \frac{dm}{d\tau}$$ which tells us that $\Delta I$ is just the change in rest mass. This is a perfectly sensible result, though we usually don't describe interactions that change rest mass as "forces". For example, if you have a stationary object and warm it up, then $dx_\mu$ is purely temporal, and so is $F^\mu$, so $F^\mu dx_\mu$ is nonzero. The rest mass of the object increases because it picks up more thermal energy, and $E=mc^2$.

Answer 2

Is there a name for the integral of the Four Force of an object with respect to spacetime distance in Special Relativity?

It is called the change in the four-momentum. You probably could call it the four-impulse and be understood, but as far as I know that is not standard terminology.