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Sacroiliac
2003-Mar-19, 10:20 PM
Hello

From what I have read spacetime curvature is equal to the mass/energy density plus 3 times the pressure. Now if you and I are at the exact same location in spacetime but moving at different velocities we will measure different mass densities and pressures. Does this mean that gravity/curvature is velocity dependant and relative?

Zathras
2003-Mar-19, 10:38 PM
On 2003-03-19 17:20, Sacroiliac wrote:
Hello

From what I have read spacetime curvature is equal to the mass/energy density plus 3 times the pressure. Now if you and I are at the exact same location in spacetime but moving at different velocities we will measure different mass densities and pressures. Does this mean that gravity/curvature is velocity dependant and relative?

First, welcome to the BABB board!

Secon, in short, the answer to your question is yes. The stress energy tensor (which has measures of pressure and energy as components) will have different values in a different reference frame. I am not familiar with any equation of the form that you mention, but the basic form of the GR field equation states that the curvature tensor is directly proportional (by a scalar constant) to the the stress energy tensor. Because the stress energy tensor has a different form, the curvature tensor will have a different form.

Sacroiliac
2003-Mar-19, 11:59 PM
Thanks. Its not really a formula its just what the authors of Black Holes and Timewarps say on page 118.

The sum of the strengths of these 3 curvatures is proportional to the density of mass in the particles vicinity ..plus 3 times the pressure of matter

They then go on to say that if we move relative to one another we will experience different curvatures. Does this just mean we will follow different geodesics? Certainly they cant be talking about relativistic mass can they?

Also does the magnitude of the energy momentum four vector play any role in GR?