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I recently read that the mass that we deal with in the equation $F=M_{i}a$ is called the inertial mass and the mass that we deal with in $F=M_{g}g$ is called the gravitational mass.

I find a hard time understanding why we need these two different names. It's the stuff that makes up the ball that is responsible for pulling it towards the ground when dropped from a height and it is the same stuff responsible in determining how hard I should push it if I want to accelerate it on a level plane. So, why do we need to invent two different notions of mass in the first place? What is the fundamental difference between the gravitational mass and the inertial mass that one should not apriori ignore?

I have also read that the inertial mass and the gravitational mass are equivalent to each other and the general theory of relativity is based on this result. So, I feel that there must be physical meaning to the equality of these two masses--they shouldn't be trivially equal to each other. How can a fruitful physical theory (the general theory of relativity) be based on the equivalence between the two masses if they can be assumed equivalent apriori? In other words, I want to say that the equality between the two masses must be a physical result in order for a theory to stem out of it--it cannot be a trivial mathematical identity (otherwise, it would be physically unfruitful). So, what is it that makes these two distinct concepts necessary in the first place and what forces upon us their equivalence?

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marked as duplicate by Kyle Kanos, jinawee, user10851, Qmechanic Jan 30 '14 at 21:10

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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/8610/2451 and links therein. $\endgroup$ – Qmechanic Jan 30 '14 at 20:35
  • $\begingroup$ There similarity is complex relating to quantum entanglement and consciousness. Otherwise there is no automatic equivalence. Air, for example, has inertial mass, due to Newton 1st law, but no gravitational mass as there is no entanglement to earth. The equivalance of GR is somewhat dependent on the "hidden" reference frame of the container. Make it a glass cage and it might not be equivalent. Seasickness arises due to this non-equivalence. $\endgroup$ – TheDoctor Oct 4 '16 at 17:20
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Given a point of matter, its motion is obtained solving the following equation:

$$m\frac{d^2 {\bf x}}{dt^2} = {\bf F}\left(t, {\bf x}, \frac{d {\bf x}}{dt}\right)$$

The RHS describes the causes of the motion, the interactions on the point due to external objects and it is a given function, containing several constants associated to the point and to the other points interacting with it. ( See my answer to Is force a "real thing", or a tool for explaining changes in measurable phenomena? for more details ).

The LHS describes the effect of the intereaction, the acceleration. The scalar $m$ appearing therein is a constant associated to the point, it is constant in the sense that it does not depend on the particular form of the function in the RHS, namely $m$ does not depend on the possible various interactions acting on the point. $m$ is the inertial mass of the point.

Let us focus on the RHS. Considering two interacting points very far from the other bodies, the RHS can be written as, where ${\bf y}$ is the position of the other point: $${\bf F}\left({\bf x}, \frac{d {\bf x}}{dt},{\bf y}, \frac{d {\bf y}}{dt}\right)\:.$$ A well known example is Coulomb's force which only depends on positions $${\bf F}\left({\bf x},{\bf y}\right) = k_C\frac{qq'}{|{\bf x}-{\bf y}|^3}({\bf x}-{\bf y})\:.$$ The constant $k_C>0$ is universal, depends on the type of interaction and on the used system of units. Constants $q$ and $q'$ are constants associated with the two particles called, as is well known charges, They can have values with sign.

Another type of interaction has a similar form, but now, the charges $M$, $M'$ can assume only positive values and the resulting force is always attractive, $k_G>0$, as before, depends on the type of interaction and on the used system of units:

$${\bf F}\left({\bf x},{\bf y}\right) = -k_G\frac{MM'}{|{\bf x}-{\bf y}|^3}({\bf x}-{\bf y})\:.$$

The interesting physical fact is that, with a suitable, universal, choice of the constant $k_G$, it happens that $m=M$ for every point of matter.

In this picture $M$ is called gravitational mass of the point.

Notice that the two notions of mass play a completely different role in all this picture: One is independent form the interactions acting on the point and enters the general form relating interactions (causes) and motion (effects), the other is a coupling constant of a very particular type of interaction, the gravitational one.

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  • $\begingroup$ nice summary of this old and important topic! +1 $\endgroup$ – Zoltan Zimboras Jan 30 '14 at 21:29
  • $\begingroup$ Ah, so $\ k_{G}$ is the constant such that m equals inertial mass, is it? $\endgroup$ – SNB Jul 9 '18 at 5:37
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    $\begingroup$ $k_G$ is the constant such that $M=m$, $\endgroup$ – Valter Moretti Jul 9 '18 at 5:57
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Inertial mass describes an object's resistance to change in velocity. The more inertial mass something has, the harder it will be to change its velocity.

Gravitational mass describes an object's ability to attract other matter (and under GR, to curve spacetime). The more gravitational mass something has, the more attracted to it other things will be.

When the equivalence principle says the two are the same, it is really saying that you cannot distinguish between a constant acceleration and a constant gravitational field. This leads to many interesting consequences and the beginnings of general relativity.

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  • $\begingroup$ It is worth noting that within relativity “inertial mass” is a dubious or ambiguous concept: see physics.stackexchange.com/questions/8610/… $\endgroup$ – Incnis Mrsi Oct 29 '14 at 10:39
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    $\begingroup$ @IncnisMrsi Even though inertial mass may be difficult to attribute a value to in relativity, it still remains a description of an object's resistance to change in velocity. That resistance just becomes dependent on frame of reference, which makes it of little practical use in relativity but does not detriment the underlying idea of what it represents. Thanks for reminding everyone of its ambiguity though $\endgroup$ – Jim Oct 29 '14 at 13:38
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Mass is one of fundamental attributes of a particle. These fundamental attributes are defined based on their interactions we observe in nature. There's no other way for us to assign a valued attribute to a particle. For example, charge is defined based on electromagnetic interaction. We observe the motion of particles under electromagnetic interaction and assign charge value based on this observation.

Similarly, we can assign attributes based on Strong, Weak, Gravitational and inertial interactions. That's the thing.. Based on Gravitational interaction, we have got Gravitational Mass. And, based on inertial interactions, we have got Inertial Mass (Inertial interaction is fictional, but so is the Gravitational interaction.... Read on).

General Theory of Relativity says that Gravity is an inertial force. Meaning, gravity is similar to what we feel at the time of breaking car. In decelerating car, your head accelerates forward despite there's no actual force to accelerate it. This illusion occurs because your reference frame (the car) isn't inertial... It's in acceleration. In case of Gravity, your car is Spacetime. Earth is actually moving straight in uniform speed. You see it revolving around Sun due to curvature of Spacetime. Similarly, a freely falling body is actually in uniform speed (due to which a freely falling body is inertial reference frame in General Theory of Relativity).

As Gravitational interaction is actually Inertial interaction, the calculated mass by both means should be same. That's what mass equivalence is.

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