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$\textbf{I have two questions.}$

On this lecture note: http://www.damtp.cam.ac.uk/user/hsr1000/part3_gr_lectures_2016.pdf on page 2, we face a discussion about Weak Equivalence Principle (WEP).

First of all, we all know that $\textbf{F}=m\textbf{a}$ is our "cannon" definition of a force and under this notion, we have the concept of inertial mass $m := m_I$. On the other hand we have the weight of a body $\textbf{W} = m'\textbf{g}$ that gives us the definition of gravitational mass $m' := m_G$. Both notions arise from experimental facts. But, sometimes, I think that the weight of a body and the force (under the general form of $\textbf{F}=m\textbf{a}$) are both equivalent definitions of force: one "using" gravity and the other using "inertia".

But if we think like an freshman on physics, the weight is just another kind of force that enters under the general form of $\textbf{F}=m\textbf{a}$ (and this is a correct notion, I know).

$\textbf{My first question is}$: what is really the case (the physics) with $\textbf{F}=m\textbf{a}$ and $\textbf{W}=m'\textbf{g}$?

For my second question I will begin quoting the author:

In Newtonian theory, one can distinguish between the notions of inertial mass $m_I$ , which appears in Newton’s second law: $\textbf{F} = m_{I}\textbf{a}$, and gravitational mass, which governs how a body interacts with a gravitational field: $\textbf{F} = m_{G}\textbf{g}$. Note that this equation defines both $m_G$ and $\textbf{g}$ hence there is a scaling ambiguity $g\to \lambda g$ and $m_G\to \lambda ^{-1} m_{G}$ (for all bodies). We fix this by defining $\displaystyle \frac{m_I}{m_G}=1$ for a particular test mass, e.g., one made of platinum. Experimentally it is found that other bodies made of other materials have $\displaystyle \frac{m_I}{m_G}-1= 0 \times 10^{−12}$ (Eötvös experiment).

$\textbf{My second question is}$: What suppose to mean the ambiguity $g\to \lambda g$ and $m_G\to \lambda ^{-1} m_{G}$ and ow can we relate the ambiguity $g\to \lambda g$ and $m_G\to \lambda ^{-1} m_{G}$ with my first question?

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You have to combine both masses:

$$m_I a = m_G g.$$

Then with rulers and clocks you can measure $a$, which gives you an experimental value of

$$\frac{m_Gg}{m_I}.$$

But you have 3 values in there. So to proceed further, you do the measurement for another material and get an experimental value of

$$\frac{m'_Gg'}{m'_I}.$$

Why $g'$? A priori, $g$ could depend on the material, couldn't it? But actually this is physically not sound because the first measurement could be that of

$$\frac{\frac{m_G}{2}2g}{m_I}$$

instead and that of the second one

$$\frac{\frac{m'_G}{3}3g'}{m'_I}$$

instead. Actually, this is even worse because the same little game could be played with $m_I$ and $m'_I$. So if we proceed as naively as above, we do not learn anything.

Thus let' reboot. First, we settle for measuring the ratio

$$r = \frac{m_G}{m_I}.$$

So now we can measure the acceleration, leading to a measurement of $rg$ but that is also a measurement of $\frac{r}{\lambda}\lambda g$ for any real number $\lambda$. So we need to get rid of that ambiguity. The simplest way to do so is to state that $r=1$ for some compound, which then fixes $g$. Then any measurement of $rg$ for other compounds will be a measurement of $r$ since we have now $g$.

A different but eventually equivalent point of view is to decide that we absorb the ambiguity $\lambda$ entirely into the ration $r$, i.e. we decide that $g$ is constant. Then we can measure $rg$, $r'g$, and $r''g$ for respectively Platinum, Copper, and Iron, let's say, and therefore get the ratios $r'/r$ and $r''/r$, thus using Platinum as the reference.

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  • $\begingroup$ Unfortunatelly this answer didn't solve my doubts. $\endgroup$ – M.N.Raia Oct 19 '17 at 22:49

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