0
$\begingroup$

A rotating mass will twist spacetime and cause orbiting bodies to precess. My question is whether the mass of the rotating body itself plays a bigger role in the dragging effect than energy and how that relates to E=MC^2.

Hypothetically, if the Sun and Earth had the exact same mass, radius, rotation, etc. would the frame dragging effect be the same around both objects or different?

$\endgroup$
0
$\begingroup$

It is the density and flow of energy and momentum that cause spacetime curvature. Mass is irrelevant. You could have a gravitating cloud of photons, which are massless.

Note: In modern physics, “mass” usually means the Lorentz-invariant mass, $m=\sqrt{E^2-p^2}$ (in units where $c=1$).

Two spheres with the same radius, mass density profile (say, as a function of the distance from the center), and angular velocity would produce the same frame-dragging, because they would have the same energy-momentum tensor $T^{\mu\nu}$. I’m assuming that they are rotating rigidly, which neither the Earth nor the Sun does.

I’m also assuming that they are at the same temperature. Perhaps since you asked about the Sun vs. the Earth, your question is whether if two spheres are the same except one is hotter, would it cause more curvature and more frame-dragging.

The answer to that is yes, because the hotter object would have more density and flow of energy and momentum at each point.

$\endgroup$
3
  • $\begingroup$ so, if we have two particles causing gravity next to each other, but their velocities are either the same or oppossite. The flow of momentum is less in the second case (and energy and energy flow are the same, correct)? so is its their curvature less? $\endgroup$ Nov 17 '18 at 23:20
  • $\begingroup$ The energy-momentum-stress tensor is a tensor field $T^{\mu\nu}(x, y, y, t)$, and for two point particles each component of the tensor has two 3D delta functions at the positions of the particles, each multiplied by appropriate factors of mass and 4-velocity. So for two oppositely moving particles the momentum density and momentum flow only cancel if you average the tensor over a region. In this average sense, the energy density would add but the energy flow would cancel. So, if you looked at the distant field, I think the curvature would be less if the particles are oppositely directed. $\endgroup$
    – G. Smith
    Nov 18 '18 at 3:29
  • $\begingroup$ For a point mass, $T^{\mu\nu}(\vec{r},t)=\frac{1}{\gamma} m u^\mu u^\nu \delta^{(3)}(\vec{r}-\vec{r}(t))$. $\endgroup$
    – G. Smith
    Nov 18 '18 at 3:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.