The inertial mass of an object is defined as its resistance to acceleration by $\vec{F}_{net} = m_I\vec{a}$. The gravitational mass of an object is defined as the scaling of the gravitational force an object experiences by $\vec{F}_g = m_g\vec{g}$ where $\vec{g}$ is assumed to be uniform in the following. The equivalence principle states the following: A frame at rest relative to a uniform gravitational field $\vec{g}$ is equivalent to a frame accelerating at a constant rate $\vec{g}$ relative to a completely isolated inertial frame. It is my understanding that we may demonstrate that $m_I = m_g$ by the equivalence principle. We should not have to assume that all objects fall with the same acceleration in a uniform gravitational field, rather this is what we would like to demonstrate. To this end, I produced the following.

If we let S and S' be inertial and non-inertial reference frames with $\vec{A}$ the acceleration of S' relative to S then $m_I\vec{a'} = -m_I\vec{A} + \vec{F}_1 + \vec{F}_2 \,+ ...$ where $\vec{F}_i$ are real forces. Now, according to the equivalence principle, the gravitational force $\vec{F}_g = m_g\vec{g}$ is fictitious. If we consider a frame of reference accelerating at $\vec{g}$ relative to a completely isolated inertial reference frame then an isolated object of inertial mass $m_I$ and gravitational mass $m_g$ will have acceleration $\vec{a'}$ = -$\vec{g}$ by the formula above. Now, let's see what the equivalence principle says about this if we analyze the situation from a frame of reference at rest relative to the gravitational field. We must have the following for the freely falling object. $-m_I\vec{g} = m_I\vec{a'} = -m_g\vec{g}$ since the fictitious term is the gravitational force. We immediately see that $m_I = m_g$.

My question concerns my statement of the equivalence principle. Typically it is stated as follows: A freely falling frame in a uniform gravitational field $\vec{g}$ is equivalent to an isolated inertial reference frame. The problem is that if I start from there, I have to assume that the frame falls at a constant rate $\vec{g}$ in the field which is what we prove if we show that $m_I = m_g$. This exercise to show $m_I = m_g$ follows from the equivalence principle came from a book I'm reading. I'm curious if there's any way to approach the problem from the more traditional way of stating the equivalence principle or if I've essentially captured the essence of the problem.


1 Answer 1


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 systems in the absence of gravitation.

(Weak form would be just for the case of freely falling body,only under influence of gravity)

So, coming to your proof it's clear that from Newton's 2nd law you can write: $$\vec{F_{net}}=m_{I}{\vec{a}_{S' wrt \,S}={\vec{F}_{gravity}(\neq 0)}+\vec{F}_{other}}$$ The observer who is in S frame and looks at the S' frame(where the particle is stationary) writes the above equation. Now, to make $\vec{F}_{gravity}$ disappear you must add another force term on RHS.But you cannot usher in new forces, the only tool you have is coordinate transformations or in other words

Is it that the coordinates you have chosen are playing tricks with you,making you believe that gravity is a "real" force(as opposed to some pseudo force that comes because of ill choice of coordinates)?

And so as it turns out, that's exactly the case.To switch off gravity locally, just grab a seat on a frame S''(acceleration ${\vec{a}_{S'' wrt \,S}}$ wrt S). Now, the equation looks like:

$$\vec{F_{net}}=m_{I}{\vec{a}_{S' wrt \,S''}}=({\vec{F}_{gravity}(\neq 0)}-m_{I}{\vec{a}_{S'' wrt \,S})+\vec{F}_{other}}$$

To kill $\vec{F}_{gravity}$ what's the suitable ${\vec{a}_{S'' wrt \,S}}$?

Newtonian theory of gravitation for small bodies(otherwise gravitational field $\vec{g}$ will vary) allows you to write $$\vec{F}_{gravity}=m_g\vec{g}$$

Now because I have specifically mentioned transformation to locally inertial frames(unaccelerated small region in spacetime), you must require: $$ m_g\vec{g}=m_I{\vec{a}_{S'' wrt \,S}} \, for\,all\,m$$

The above equation is the essence of the principle of equivalence between gravitation and inertia. You can make several equivalent statements from it:

  1. If ${\vec{a}_{S'' wrt \,S}}=\vec{g}$, then $m_g=m_I$.You will here demand the existence of such a frame S''.
  2. If you take $m_g=m_I$ or even better just $\frac{m_g}{m_I}$ to be a constant independent of the the nature of the small mass, then we have that acceleration of small bodies is independent of their nature(shape,size etc). Note that here $\vec{g}$ is constant as the spacetime point is fixed and $\vec{g}$ is the field due to some other far away massive body.
  • $\begingroup$ No, you're wrong. $\vec{a'}$ is the acceleration of an object in a non-inertial frame S' and $\vec{A}$ is the acceleration of the non-inertial reference frame S' relative to S. The equation I have is not utterly wrong but exactly correct. In fact, it comes from a well established book I'm working with. Also, in 1. You're assuming all objects fall at the same rate and in 2. you assume the gravitational and inertial masses are equal. These two statements are equivalent and it's what we're trying to prove. I suggest you delete your answer or I'll down vote you. $\endgroup$
    – user7348
    Commented Jul 12, 2016 at 23:05
  • $\begingroup$ Hey, don't take it personal! I am not trying to attack your question and arguments. Sorry, if I sounded so. Now coming to the question, yes,I admit, I couldn't get the entire picture of the frames you were using in my quick read. So, okay $\vec{a'} $ and $\vec{A}$ are different and the equation stands well and good. However, it seems you have another problem. First tell me how is $m_I\vec{a'} = -m_g\vec{g}$ ? Here my concern is the minus sign. Once you let me know your logic, I can try to give you the exact line of my argument in another comment/edit the answer to make my point even clearer. $\endgroup$
    – user122637
    Commented Jul 13, 2016 at 5:42

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