Einstein's explanation of the Equivalence Principle

Question

At the top of the Wikipedia page on the Equivalence Principle is this quote attributed to Einstein:

A little reflection will show that the law of the equality of the inertial and gravitational mass is equivalent to the assertion that the acceleration imparted to a body by a gravitational field is independent of the nature of the body. For Newton's equation of motion in a gravitational field, written out in full, it is:

(Inertial mass) (Acceleration) = (Intensity of the gravitational field) (Gravitational mass)

It is only when there is numerical equality between the inertial and gravitational mass that the acceleration is independent of the nature of the body.

I read this statement to imply that two statements are equivalent:

• Acceleration due to gravity is independent of the mass of the body
• Inertial mass and gravitational mass are numerically equal.

I also logically understand "equivalent" to mean, "if and only if."

I understand one direction of this argument. If gravitational mass is numerically equal to inertial mass, then acceleration due to gravity will be the same for all bodies (because is the same for all things on Earth). But I can't understand how the independence of the mass of the body and acceleration due to gravity implies that inertial mass and gravitational mass are the numerically equal. For instance, that would be true if inertial mass were exactly twice gravitational mass--we could just choose a different value for G.

Am I misreading Einstein's statement? Does it not mean what I think it means?

Answer 1

(Answering my own question).

I read the quote wrong.

I read it as if Einstein were saying, "look at this equation: mathematically it must be true since acceleration due to gravity is independent of mass that gravitational mass and inertia mass are equal." And in fairness to myself, when you point to an equation and say what it means, that's usually the sense in which you mean it--mathematically. It is, after all, an equation.

But he isn't saying that. He's using a much deeper meaning of the notion of equality. He's saying, "look at this equation: the nature of a body under the influence of gravity is equal to the nature of that body experiencing inertia. Not just 'equal to' in the numerical sense, but 'equal to' in the 'these two things are the same thing' sense.'" Falling is equal to inertia. He is looking at the equation and realizing a much more profound meaning behind what it means for two things to be equal.

That's just incredible. Genius, man. It's so simple once you understand it.

Answer 2

The Newtonian version of this idea is that the second law of mechanics is that the force on a body is equal to the mass of that body times the acceleration $\vec F~=~m\vec a$. The force is attributed to gravitational attraction to a mass $M$ at a distance $r$ is $$ m\vec a~=~\frac{GMm\vec r}{r^3} $$ for $\vec r$ the vector in the direction of the mass $m$ from the center of the mass $M$. In this the mass $m$ on the left, the inertial mass, is equal to the mass $m$ on the right. However, the issue is whether this is really the case. And Eotvos did measurements of determine this, and further measurements have been made to determine to what degree $m_i~=~m_g$. Einstein's equivalence principle says

What does this have to do with spacetime? This requires a bit of knowledge of general relativity. We start with the Schwarzschild metric for a spherically symmetric mass centered at $r~=~0$ $$ ds^2~=~\left(1~-~\frac{GM}{rc^2}\right)c^2dt^2~-~\left(1~-~\frac{GM}{rc^2}\right)^{-1}dr^2~-~r^2d\Omega^2 $$ We consider a weak gravity field where the coefficient $\frac{GM}{rc^2}$ is small. The time component is $O(c^2)$ larger and so we approximate this with $$ ds^2~=~\left(1~-~\frac{GM}{r}\right)dt^2~-~dr^2~-~r^2d\Omega^2. $$ Now divide through by $ds^2$ to get $$ 1~=~\left(1~-~\frac{GM}{r}\right)\left(\frac{dt}{ds}\right)^2~-~\left(\frac{dr}{ds}\right)^2~-~r^2\left(\frac{d\Omega}{ds}\right)^2. $$ Now we differentiate this with respect to $ds$, employ $$ \frac{d}{ds}\frac{GM}{r}~=~\frac{\partial}{\partial r}\frac{GM}{r}\left(\frac{dr}{ds}\right), $$ drop the angular term and we find $$ \frac{d^2r}{ds^2}~+~\frac{GM}{r^2}\left(\frac{dt}{ds}\right)^2~=~0. $$ This is the geodesic equation in the weak limit that recovers Newton second law of motion and gravity. This connects the equivalence principle to spacetime physics.

Answer 3

This question seems to be about Newtonian physics, so I'll restrict to that.

tl;dr The ratio doesn't matter because we can't measure it and the gravitational constant independently. We have chosen that the ratio is $1$ because that gives us the simplest formulas.

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Let $\alpha$ be the ratio of gravitational mass to inertial mass. For two bodies with inertial masses $m_i$ and $M_i$ and gravitational masses $m_g$ and $M_g$, the gravitational force has magnitude

$$ F = \frac{H M_g m_g}{r^2} = \frac{\alpha^2 H M_i m_i}{r^2}, $$ where $H$ is a constant called the gravitational constant; I chose an uncommon symbol for that on purpose.

Now, note that to know $\alpha$ or $H$ we have to perform measurements to determine the number. But we have a problem here! In every equation we have that contains $\alpha$ or $H$, they occur as product $\alpha^2 H$. So $\alpha^2 H$, not the individual values of $\alpha$ or $H$ is the only thing that matters in how the universe works. So we can't even measure $\alpha$ and $H$ individually, only $\alpha^2 H$.

Therefore, we can as well define a constant $G=\alpha^2 H$, which gives the equation $$ F = \frac{G M_i m_i}{r^2}. $$

We may then stop caring about $\alpha$, $H$ and gravitational masses, because they say nothing about gravity that this new equation doesn't say. We might as well go on and say that $G$ is the gravitational constant.

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In particular, you could do every aspect of Newtonian mechanics by saying that inertial mass is $1/2$ times gravitational mass. Everything would work out, you'd just have an extra factor of $1/2^2$ in the formula for the gravitational force. You would for example measure the gravitational constant ($H$) to be $2.669632 × 10^{-10} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2}$ (four times the value we have now).