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mass of the body. And so we have
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The Einstein Field Equations
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F1 = m I a1 = m 1 1 F2 = m I a2 = m 2 2 Using the second equation, we solve for the acceleration of mass 2: m I a2 = m 2 2 a2 = m2 mI 2
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However, the experimental results obtained by Galileo tell us that all bodies in a gravitational eld fall with the same acceleration, which we denote by g. This means that a1 = a2 = g, and we have a2 = g = m2 mI 2
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Since a1 = a2 = g, we can rewrite F1 = m I a1 = m 1 in the following way: 1 F1 = m I a1 = m I g = m 1 1 1 g= m1 mI 1
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Equating both expressions that we have obtained for g, we nd that m m1 = 2 I m1 mI 2 Canceling from both sides, we get m1 m = 2 I m1 mI 2 Masses m 1 and m 2 used in this experiment are completely arbitrary, and we can substitute any body we like for mass m 2 and the result will be the same. Therefore, we conclude that the ratio of passive gravitational mass to inertial
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The Einstein Field Equations
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mass is a constant for any body. We can choose this constant to be unity and conclude that mI = m1 1 That is, inertial mass and passive gravitational mass are equivalent (this result has been veri ed to high precision experimentally by the famous E tv s exo o periment). We now show that active gravitational mass is equivalent to passive gravitational mass. Consider two masses again labeled m 1 and m 2 . We place mass m 1 at the origin and m 2 is initially located at some distance r from m 1 along a radial line. The gravitational potential due to mass m 1 at a distance r is given by 1 = G mA 1 r
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where G is Newton s gravitational constant. The force on mass m 2 due to mass m 1 is given by F2 = m 2 1 Since we are working with the radial coordinate only, the gradient can be written as F2 = m 2 1 = m 2
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mA m Am 1 r = G 1 2 2 r r r
Similarly, the force on mass m 1 due to the gravitational eld produced by mass m 2 is F1 = r G m Am 1 2 r2
To understand the difference in sign, note that in this case we have since r the force points in the opposite direction. Now, Newton s third law tells us that F1 = F2 ; therefore, we must have m Am m Am G 22 1 = G 12 2 r r
The Einstein Field Equations
We cancel the common terms G and r 2 , which give m Am 1 = m Am 2 2 1 This leads to m2 m1 = A A m1 m2 Again, we could choose any masses we like for this experiment. Therefore, this ratio must be a constant that we take to be unity, and we conclude that mA = mp that is, the active and passive gravitational mass for a body are equivalent. We have already found that the passive gravitational mass is equivalent to inertial mass, and so we have shown that m = mI = mp = mA where we have used the single quantity m to represent the mass of the body.
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Test Particles
Imagine that we are studying a region of spacetime where some distribution of matter and energy acts as a source of gravitational eld that we call the background eld. A test particle is one such that the gravitational eld it produces is negligible as compared to the background eld. In other words, the presence of the test particle will in no way change or alter the background eld.
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