One definition of mass is 'a measure of the quantity of matter in an object at rest relative to the observer'. What do 'at rest' and 'relative to the observer' mean here? I know it has to do with mass resisting motion, but I cannot get what these mean.

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5The principal problem with that definition is that "quantity of matter" does not mean anything. That definition would need another definition to explain the notion of quantity if matter. This is not good physics in my view. – Valter Moretti Aug 10 '22 at 09:41
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1@ValterMoretti Well, it is not so bad as a long-standing tradition going back to Mach thinks. I agree that there are better ways to introduce the concept of mass, but for pedagogical reasons and to provide a first idea, I would not be so negative. Remeber that quantity of matter is a well definite concept within the SI. – GiorgioP-DoomsdayClockIsAt-90 Aug 10 '22 at 09:53
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1In fact, it needs another definition to be understood. – Valter Moretti Aug 10 '22 at 09:54
6 Answers
The issue is a bit tangled because the notion of mass in relativity historically relies upon the notion of mass in classical physics.
There, in classical mechanics, the mass can be defined as follows, with an argument which can be traced back to E. Mach when replacing velocities with accelerations. I will explicitly avoid to introduce the notion of force because, in my view, it makes the discussion even more complicated and it is by no means necessary (the notion of force is quite subtle and its use would open a number of related issues really unnecessary). What it is really necessary is just the law of conservation if total momentum.
Let us refer to bodies whose interactions are local (essentially contact interactions with some generalisation).
We consider a body and we make it interact with other bodies, describing what happens with respect to an inertial reference frame.
Let us consider first the case of two-body interactions, assuming that the two bodies do not interact with anything except possibly each other.
It turns out that there is a positive constant $m$, called the mass of the body, associated to our body and a second positive constant $m'$ associated to the other body interacting with the former, such that the vector $$m \vec{v} + m'\vec{v'}$$ remains constant in time even if they interact (provided they interact with nothing else).
Furthermore
The constant $m$ does not depend on the choice of the second body and on the value of $m'$.
The constant $m$ does not depend on the inertial reference frame we are using to describe the system.
In case we consider interactions with many bodies, the sum $$\sum_i m_i \vec{v}_i$$ turns out to be constant in time, where we used the values of the constants $m_i$ obtained in the experiment with two-body interactions.
This is only valid in classical mechanics. If a number of points merge to give rise to a single body, or a single body breaks into many bodies, the mass of the single body is the sum of the masses of the bodies it is/was made of.
It is clear that, within an idealised procedure of two-body interactions, the mass $m$ of a body can be measured by choosing a reference body with mass $1$ by definition and velocity denoted by $\vec{u}$, and solving the equation $$m \vec{v}(t) + 1 \vec{u}(t) = m \vec{v}(t') + 1 \vec{u}(t')$$ in the unknown $m$, after having measured the velocities at different times. That equation has one solution if the velocities changes, i.e., if the bodies interact.
When we pass from classical to relativistic physics, everything but 4 remains valid (with some restrictions on the type of interactions which must occur in single events), provided the vectors $m\vec{v}$ in the formulas above be replaced by the more complicated expression: $$\frac{m\vec{v}}{\sqrt{1-\frac{v^2}{c^2}}}.\tag{1}$$ You see that, in the limit of small velocity (with respect to the speed of light $c$) everything collapses to the classical formulas, since the expression in (1) can be expanded as $$m \vec{v} + m \vec{v} \frac{v^2}{2c^2}+ \vec{v} O(v^4/c^4)\:.$$ So, staying at small velocities with a body, its constant $m$ plays the same role as in classical physics in the law used to define it.

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1+1, but one extremely minor quibble: I think "just because the two bodies interact" should be "assuming that the two bodies do not interact with anything except possibly each other". – WillO Aug 10 '22 at 12:20
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I should have been more precise. I was looking at the line that immediately precedes "furthermore". It is not correct to say that the sum of the momenta remains constant because the two particles interact; instead, it remains constant even if they interact (provided they interact with nothing else). My issue wasn't the failure to state that there are no other interactions (which I think was naturally implied and didn't need to be explicit). My issue was specifically with the word "because". – WillO Aug 10 '22 at 12:37
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I changed the text as you suggested. But I am not completely sure to agree with you: one may assert that the nature of the interaction is the cause of the conservation of the total momentum. However it is a subtlety...And the form of the text proposed by you is however correct. – Valter Moretti Aug 10 '22 at 12:45
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I would have said that the interaction cannot be the cause of the conservation of momentum, because momentum is conserved even if there is no interaction. But of course this is a very minor point. – WillO Aug 10 '22 at 15:31
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"When we pass from classical to relativistic physics, everything but $4$ remains valid, provided $mv$ is replaced by $\gamma mv$" - Regarding $3$: In classical mechanics, the equation $$\frac{\mathrm d}{\mathrm d t}\sum_i \vec{p_i}=0$$ is meaningful as it is invariant under Galilei transformations. Is this also the case for Poincaré transformations? – Filippo Aug 10 '22 at 16:51
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@WillO Even if I understand well your logical viewpoint (A is cause if B if removing A, B does not occur), I stress that the type of interactions plays an active role in the conservation of the particle momentum. I am here expliciting considering contact forces, satisying the third principle. If I considered the Lorentz force for instance, we would not have conservation of mv + m'v' because part of the momentum would be embodied in the field, in presennce of radiation fields. However from your logical perspective I agree with you. – Valter Moretti Aug 10 '22 at 18:08
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1@Filippo Yes it holds if the interactions are localized in events and using the time coordinate of the reference frame. – Valter Moretti Aug 10 '22 at 18:23
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...Obviously, in the relativistic case $\vec{p}_i = \frac{m\vec{v}_i}{\sqrt{1-\frac{v_i^2}{c^2}}}$. – Valter Moretti Aug 10 '22 at 19:58
The cited definition mixes in an uncontrolled way different concepts.
The concept of mass is firmly bound to the theory we use for the dynamics. It underwent a long sequence of modifications as classical mechanics evolved, but also to cope with the conceptual changes introduced by Special Relativity firstly and Quantum Mechanics after.
The definition of mass as a measurement of the quantity of matter goes back to Newton's Principia, which was enough for a long time. Later, this definition was criticized and substituted with others in the nineteenth century. Today, such a definition is still present in textbooks and, in my opinion, could be used at the introductory level, but it would require some words of caution about its meaning and limits.
Let me try to explain them. First, quantity of matter nowadays has a technical meaning. In the International System (SI) of measurements, the mole is the unit for the quantity amount of substance or quantity of matter and is a measure of how many elementary entities of a given substance are in an object or sample. As you see, this is not directly the mass. However, if by "substance" or "matter" one means the nucleons (protons and neutrons), it is possible to define the mass as a quantity approximately proportional to the number of nucleons.
Such a definition is meaningful but highly unsatisfactory because it is not directly related to dynamical properties. Moreover, we know that the mass of a nucleus is only approximately given by the sum of the masses of nucleons. Last but not least, measuring a mass according to such a definition would require determining the exact isotopic composition of the sample.
If we maintain such a definition of mass, the further specification "in an object at rest relative to the observer" is understandable as a signal that who wrote the sentence is using the old-fashioned concept of relativistic mass. I do not want to discuss it (on PSE, there are many related Q&A). Still, I notice that even if we would like to keep using the relativistic mass, the previous specification is redundant if the quantity of matter has the same meaning as in SI. Indeed, independently of the relativistic mass, the number of units is not modified by the status of motion of the sample.
In conclusion, I would not recommend the cited definition for the mass. I hope the previous discussion could help.
I would guess the confusion arises because of the (widespread but misguided) belief that in special relativity mass increases with velocity. For more on this see Why is there a controversy on whether mass increases with speed?
If we define the relativistic mass as:
$$ m_r = \gamma m = \frac{m}{\sqrt{1 - v^2/c^2}} $$
then $m_r = m$ only when the speed $v = 0$. That is, the object is at rest relative to the observer measuring the mass.
A better way of defining the mass is to use the expression:
$$ E^2 = p^2c^2 + m^2c^4 $$
where $E$ is the total energy of the mass and $p$ is the momentum. When you define the mass this way it does not change if the object is moving relative to the observer.

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From the comments to my answer, the OP seems to be struggling more with the concept of "at rest", than relativistic energy equations . . . – m4r35n357 Aug 13 '22 at 08:43
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My point is if the definition is just "a measure of the quantity of matter in an object", What meaning does adding "at rest relative to the observer" add to the definition? – Salma Abdelaziz Abdeldayem Aug 17 '22 at 17:45
The terms 'at rest' and 'relative to the observer' simply mean that the physical distance between the observer and the "quantity of matter" is not changing.
They are nothing to do with "mass resisting motion", or inertia.

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Thank you. Would you please elaborate more on why it matters to have the distance between the observer and the mass unchanged? what meaning does this add to the mass? – Salma Abdelaziz Abdeldayem Aug 12 '22 at 13:34
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There is no more to say, as I have answered both points in your OP. I do not understand your supplementary question; mass has no "meaning". – m4r35n357 Aug 13 '22 at 08:24
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I mean why is the distance between the mass and the observer mentioned in a mass definition? why does that matter? – Salma Abdelaziz Abdeldayem Aug 15 '22 at 13:13
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The definition in Newtonian physics has three steps.
First we develop definitions of distance and time, and with these we can define velocity and acceleration.
Next we introduce Newton's second law: force is equal to rate of change of momentum, and for a body which merely accelerates without any other change, the increase in momentum is owing to the change in velocity, so we can write $$ f = m a $$ where $a$ is acceleration. From this we can deduce $$ m = \frac{f}{a}. $$ This is almost enough to define mass, but not quite because we don't yet have a definition of the size of the force $f$.
The third step goes as follows. We simply pick some physical object, such as a bar of platinum alloy sitting in a vault somewhere, and announce that it has a mass which we agree to call 'one kilogram'. Or, for more precision, we define a constant called Planck's constant and then from that figure out how much platinum is needed to make a kilogram. The second method is the modern method, but it does not change the reasoning I am about to explain. Once we have one physical object whose mass is known (e.g. one kilogram) then we can use $f = m a$ to compare the mass of one object with another. All we need to do is subject them both to the same force and compare the accelerations. Note, for this part we don't need to know the size of the force! We are simply comparing: $$ f = m_1 a_1 = m_2 a_2. $$ The force here could be, for example, that provided by a given spring at some given temperature under a given amount of compression, so that we know it is the same force in each experiment. Then, by using the above equation, we can deduce $$ m_1 = m_2 \frac{a_2}{a_1}. $$ If, in this equation, $m_2$ is the standard mass, and $a_2$ and $a_1$ are measured, then we have managed to find out the value of $m_1$.
The above gives a complete definition of mass. Seen this way, mass is the propensity of a body to resist acceleration.
In practice measurements of the kind I have been discussing are often done using a balance and comparing weights, which amounts to comparing gravitational attractions between planet Earth and two different objects. This method works because one can check in separate experiments that weight is indeed proportional to mass, and this has been checked to very high precision (see equivalence principle for more information on that).
The above answer is equivalent to the one provided by Valter Moretti; you can take your pick which one you prefer. His approach focusses the attention on conservation of momentum, which is closely connected to Newton's third law; my approach focusses attention on Newton's second law. In Moretti's argument the third law is functioning as a way to guarantee that two forces are equal in size and opposite in direction, but, as he points out, you don't have to reason in terms of force and acceleration. You can just consider momentum if you prefer.

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body being at rest for a given instant means that the instantaneous relative velocity of a body with respect to other body is 0.
Like simply assume 2 persons standing, for one person(who may be in some sort motion with respect to a third person in universe) if the other person appears to him at the same distance and in the same direction then we can say both are at rest with respect to each other.
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Thank you. But what meaning does 'having the distance between the mass and the observer unchanged' add to the mass definition? like why does that matter? – Salma Abdelaziz Abdeldayem Aug 12 '22 at 13:36
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I am not very much familiar with relativity but what I can think is that, lets say you are in front of me and you are moving, now I would simply think that even after having same forces from the rest of universe why are you able to move and I am not (remember I may also be moving but for me I am at rest), now I would think that somehow you would generate that energy internally. then according to Einstein's equation that energy will come from mass(I don't know the derivation of this eq).This is all I can think, tell me if this helps I will add this to my answer then. – failedtitration Aug 13 '22 at 04:55