B Deriving Energy and Momentum in GR: 4-Component vs. Tensor Approach

[PeterDonis]

I would say as for contribution of an electron to geometry, even m is so small, it has Schwartzushild metric around itself which has tiny difference from Minkowsky's. Tiny but not zero stress energy tensor of electron shall be nececessary as RHS of Einstein equation.

Not quite. In a pure classical treatment, an electron would be represented as a Kerr-Newman BH because it has both spin and charge.

If you want to analyze an electron as a source of gravity, yes, this is what you would do. But again, as I said in post #28, if you do this, everything we have discussed up to now about the geodesic motion of an object in the Schwarzschild geometry is invalid. We would have to start from scratch and analyze things using the spacetime geometry including the effects of the electron (or the ball, or whatever object you want to use). Which is way, way beyond the scope of a "B" level thread or even an "I" level thread.

 

[sweet springs]

You're still missing the point. If the ball is a test object, it has no stress-energy tensor; it's stress-energy is zero. Anything with nonzero stress-energy changes the geometry of spacetime. That means the metric is no longer the Schwarzschild metric, so everything we have done up to now is no longer valid.

I do not catch your comment as for to my post#24 on energy-momentum tensor of a particle.
A particle cannot form energy-stress or energy-momentum ,I think they refer the same, tensor ? In
https://www.reed.edu/physics/courses/Physics411/html/page2/files/Lecture.19.pdf
I find the formula (19.5).

 

[PeterDonis]

I do not catch your comment as for to my post#24 on energy-momentum tensor of a particle.

The issue is not how you would define the stress-energy tensor of a point particle. The issue is that if you want to talk about "energy" in terms of the stress-energy of a point particle, or a baseball, or anything else, that is a different topic from the topic we have been discussing, which is the energy at infinity of a test object moving geodesically in Schwarzschild spacetime. As I've said before, you need to decide what specific topic you want to discuss; trying to mix them together is just increasing your confusion.

 

[sweet springs]

What I said in post #30 in the previous thread is that m√1−2M/(R+H)m1−2M/(R+H)m \sqrt{1 - 2M / (R + H)} is the energy at infinity of the ball. Then I used the Newtonian approximation to show where mgHmgHm g H comes from in that approximation.

So my question in other words is how Newtonian approximation, e.g. K.E=1/2 m v^2 v for what ? still to what ? observer should be staying still at the same radius or free-falling? coordinate time or proer time?, is introduced and authorized in GR.
K.E.=\frac{1}{2}mv^2 in Newtonian Mechanics
E=m\frac{1}{\sqrt{1-v^2}} in SR, But we are in GR. How do you allow or not allow them to use formula in GR as approximation suggesting such and such observation conditions.
If we have GR formula of mass+kinetic energy, Newton and SR formula will be given from it by approximation. Do we have such a formula ? If we do not have it, how we can admit Newton and SR formula as they are?

 

[PeterDonis]

If we have GR formula of mass+kinetic energy, Newton and SR formula will be given from it by approximation. Do we have such a formula ?

The simple "B" level answer? Yes, there is such a formula. (I thought that was already clear from my previous answers, but I guess not.) I've already told you, a number of times now, that you find out what it is by solving the geodesic equation. You can use the fact that the energy at infinity is a constant of the motion to simplify the calculation.

Now go do that, and start a new thread (probably at the "I" level at least; solving the geodesic equation is probably beyond "B" level) if you have questions about it. This thread is closed.