Is the “Force” of Gravity Simply Hamilton's Principle on a Curved Spacetime?
Answers
Answered by
0
Hey mate ^_^
It's my understanding that General Relativity abstracts away the concept of gravity as a force, and instead describes it as a feature of spacetime by which massive objects cause curvature.....
Then it follows that what we experience as a force is simply the difference between a geodesic on this curved surface and our perceived Euclidean space.....
#Be Brainly❤️
It's my understanding that General Relativity abstracts away the concept of gravity as a force, and instead describes it as a feature of spacetime by which massive objects cause curvature.....
Then it follows that what we experience as a force is simply the difference between a geodesic on this curved surface and our perceived Euclidean space.....
#Be Brainly❤️
Answered by
3
Hello mate here is your answer.
It follows that what we experience as a force is simply the difference between a geodesic on this curved surface and our perceived Euclidean space. What I am unsure of, exactly, is the implication of this.
S[q]≡∫L(q(t),δqδt(t),t)dtS[q]≡∫L(q(t),δqδt(t),t)dt
and Hamilton's Principle states that
δSδq(t)=0.δSδq(t)=0.
F(q(t))=−∇U(q(t))=∇(T(δqδt(t))−U(q(t)))=∇L,F(q(t))=−∇U(q(t))=∇(T(δqδt(t))−U(q(t)))=∇L,
which, as I understand is true for a conservative field like gravitation.
Hope it helps you.
It follows that what we experience as a force is simply the difference between a geodesic on this curved surface and our perceived Euclidean space. What I am unsure of, exactly, is the implication of this.
S[q]≡∫L(q(t),δqδt(t),t)dtS[q]≡∫L(q(t),δqδt(t),t)dt
and Hamilton's Principle states that
δSδq(t)=0.δSδq(t)=0.
F(q(t))=−∇U(q(t))=∇(T(δqδt(t))−U(q(t)))=∇L,F(q(t))=−∇U(q(t))=∇(T(δqδt(t))−U(q(t)))=∇L,
which, as I understand is true for a conservative field like gravitation.
Hope it helps you.
Similar questions
Biology,
7 months ago
Science,
7 months ago
Social Sciences,
7 months ago
Physics,
1 year ago
Physics,
1 year ago
Computer Science,
1 year ago