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The original form of the equivalence principle(strong form) is: In a region of gravitational field, for every spacetime point there exists a non-inertial coordinate system where the effect of gravitational field can be nullified in the sense that the laws of nature(laws of special relativity) take the same form as in unaccelerated Cartesian coordinate ...

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They are, as Einstein pointed out, equivalent. So why distinguish between the two? Well the only real difference is that they are measured differently. To measure inertial mass, we exert a given force to something with an unknown mass. To measure gravitational mass we compare the force of gravity from an object with an unknown mass to the gravitational force ...

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Gravity causes acceleration, but acceleration can happen from a lot of other things as well, for example, on electromagnetic effects. In most cases, the acceleration depends on some charge-like quantity. For example, a body with a mass of 1kg and with a charge of 1C will accelerate faster in the same electric field, as a body with 2kg of mass and with the ...

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As it is said in the article referred to in my post we can take any theory and reformulate it so that it is covariant under any group of transformations we pick; the problem is not physical, it is merely a challenge to our mathematical ingenuity. As @Lewis Miller pointed out the Lagrangian formulation of Newtonian mechanics is general-covariant, ...

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Actually the result is even stronger: Given a timelike geodesic $\gamma$ and a point $p \in \gamma$, there is a neighborhood $U \ni p$ equipped with coordinates, $x^0,x^1,x^2,x^3$ such that in the portion of $\gamma$ included in $U$, exactly along $\gamma$, the derivatives of the metric vanish in the said coordinates. Equivalently the Christoffel symbols $\... 0 Although Block B rests on Block A, the motion of Block B does not affect the motion of Block A, since there is no friction between them. The only forces on Block A are two tension forces in the string. So you can imagine that Block B rests on the smooth table, instead of on Block A. Both Blocks A and B then slide on the smooth table. Block A is attached ... 2 The system can be reduced to one with only one pulley: Using Newton's second law you can get three equations for each of the masses but as is often the case in such problems you need a fourth as there are four unknowns. The fourth equation you find by looking at the geometry of the system assuming that the string is inextensible. A way of getting that ... 3 I think it is easier to consider to the string length. String length is constant, so we have: $$x_C+2x_A-x_B=\textrm{constant}$$ $$\Longrightarrow\;a_A=\large{\frac{a_B-a_C}2}\;\tag 1$$ On the other hand, we know: $$m_Cg-T=m_Ca_C\;\tag 2$$ $$T=m_Ba_B\;\tag 3$$ $$-2T=m_Aa_A\;\tag 4$$ (You can obtain last three equations by drawing the free body diagram for$...

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