New answers tagged

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I am only going to talk about massive particles as in your example. There are two ingredients: Higgs mechanism Today we do know that we live in a universe that is permeated by fields all through. This includes the Higgs field. Now the Higgs mechanism is a true phenomenon in that it shows you that when certain particles (which themselves are excitation of ...


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It doesn't, or not really and certainly not in your example. In a closed inertial system forces may apply themselves, but then their effect can only be determined by also translating whatever it is the force is applied on into pure force only, since on the applying side there is nothing other than force. Its not a real world situation, so it doesn't really ...


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Maybe it is because an object with more mass has to bend spacetime more in order to move through it.


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In classical non-relativistic physics, the inertial mass $m_{\mathrm{inertial}}$ is by DEFINITION the quantity of an object that acts as the proportionality factor relating force $\vec{F}$ and acceleration $\vec{a}$ in $$\vec{F}=m_{\mathrm{inertial}}\vec{a}.$$ We observe that different objects react differently to the same force, i.e. having a different ...


6

We call mass, inertial mass because it is the efficient (main) cause of inertial motion. How could there be motion if there was no time or no space? Hence, inertial mass is not the only cause, space is required as well as time. Aristotle called this natural motion, it's the motion natural to an object in virtue of simply being itself. We find out by ...


49

I do not think it is mass that "CAUSES" inertia. Rather mass is DEFINED to be the property of an object which gives it inertia For example, imagine that you were a theoretical observer who did not know of the concept of mass. There were just 3 objects in front of you. You applied $10 \mathrm{N}$ to each object and observed that the different ...


0

The answer to your question is stress-energy. Both gravity and inertia are rooted in stress-energy. In the case of gravity, it is more obvious, because we have general relativity, and one of its building blocks is that stress-energy is the cause of gravity, and everything and anything we know of that does possess stress-energy, does bend spacetime. yes they ...


1

Can I reason in the following way? As you mentioned, the reasoning is flawed because it relies on two premises which are not only counterfactual but mutually contradictory. However, the math works out for a different reason: The equation that describes a projectile is $\vec r'' = \vec g$ where $\vec r$ is the position vector at any time, $t$, and $\vec g$ ...


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Either premise 1 or premise 2 will be true, but not at the same time. The projectile is either launched in a gravitational field or it is not. It cannot be both ways at once so premise 1 and 2 do not conflict with each other.


2

You are correct when you say that the velocity won't change when the body has been subjected to no force.But the second case will be little different : When the body is acted upon by gravity the force due to gravity acts vertically downward .So the vertical component of the velocity will change but horizontal component will not change since there is no force ...


1

Classical mechanics has a more or less axiomatic framework. Here, gravitational and inertial mass is represented by the same concept and the same symbol, $m$. If we decide to disambiguate the concepts by representing them by different symbols, say $m$ for inertial mass and $M$ for gravitational mass and also by different concepts of mass, the resulting ...


1

Truly excellent question. All "why would we expect" questions are inherently subjective, and there is no single correct answer to this question. But I'm going to give a very heterodox (and I'm sure unpopular) answer: I would argue that you are correct, and in fact there never was any reason to expect that every object would have one inertial mass ...


2

Gravitation is not a force. If you stand on the Earth then you are accelerated upward by the electromagnetic force. There is no force pulling you towards the Earth. This upward acceleration is what makes you see your weight when standing on a scale. On a heavier planet your weight increases. But by scaling the scale you always get the same value for your ...


4

In addition to other good answers, the equivalence of inertial and gravitational masses is equivalent to the experimental fact that all masses fall at the same speed. Science knows that fact to be true since Galileo, and it seems obvious to us because we learned elementary physics long ago, but without doing the experiment it's actually far from obvious that ...


31

Objects have a property called "electric charge". This electric charge decides how strong a force they feel when close to other electrically charged objects. The electric charge of an object is more or less independent of inertial mass. So given a large, fixed, electrically charged object, you can make a small electrically charged test object feel ...


3

Considering only the two equations: one is the Newton's second law $F = ma$, the other is the gravitational law $F = \frac{Gm_1 m_2}{r^2}$. The second become $F = mg$ close to the surface. These, in principle, are two different law and then, forgetting for a moment the names, we can use $d$ instead of $m$ in $F = dg$. We can say that $d$ is a property that ...


6

Perhaps one starting point for thinking about the equivalence principle between gravitational mass and inertial mass is the example of an object falling towards the Earth. Here, we know from Newtonian mechanics that $mg = ma$ implies that $a = g$, that is, the acceleration due to gravity is the same regardless of the mass of the object. Even though it may ...


60

"isn't there just one property called m and it just appears in different equations (e.g. Newton's second law and the law of gravitation)? In a similar way that (say) frequency appears in many different equations." There IS indeed just one property called m which appears in both the equations. The point is that there is no intuitive reason why this ...


4

Not sure if Einstein thought this way, but imagine you want to create a special relativity version of gravity, meaning you want to introduce "gravitational field" and construct evolution equations for the field. Newtonian gravity looks similar to electrostatics, forces are proportional to $1/r^2$, so you would think there should be similar stuff ...


1

About mass and inertia. The two are correlated, obviously, but they're not the same thing. Example: the mass of each of the elements of the periodic system. The mass of a Helium nucleus is close to two times the mass of a proton plus two times the mass of a neutron, but not quite. The inertial mass of a nucleus involves the energy state of that nucleus. A ...


1

The mass of the Higgs is not due to the Higgs mechanism. There simply is a mass-term in the Higgs field Lagrangian. The whole idea that Higgs is the originator of mass of otherwise massless particles is quite mistaken, at least as it stands in the collective psyche of the popular culture. ;) The correct statement is that coupling to Higgs gives mass to ...


0

The integral doesn't mean that $x = f(m)$. On the contrary, $dm = f(x)dx$ and only after knowing that function it is possible to solve the integral, with the differential as function of $x$. A good example for a unidimensional situation is any turned part, as for example a roll for a rolling mill, as shown below. The roll diameter is a function of $x$. So, ...


2

The concept of a small particle of mass ${\rm d}m$ existing at some location $x$ contributes to the mass torque by $x \, {\rm d}m$ $$ x_{\rm CM} \int {\rm d}m = \int x \,{\rm d}m$$ But how do we integrate over mass? Consider mass as a result of a density field $\rho(x)$ such that $$ m = \int \rho(x) A {\rm d}x $$ and consider only a small portion to see that ...


3

What is generally know is the mass density distribution $\rho(\vec{x})$, and you can write that mass differential in terms of $\rho$ and a volume differential $dm=\rho(\vec{x})dV$, so the center of mass is $$\vec{x}_{com}=\frac{1}{M}\int_V\vec{x}\rho(\vec{x})dV,$$ where $M$ is the total mass, i.e. $M=\int_V\rho(\vec{x})dV$


0

Instead of thinking about inertia it is clearer to think about momentum. When you throw a ball upwards on a moving train, it has two components of momentum: horizontal (due to the train's motion) and vertical (because you threw it). These two components are independent (ie motion in one direction has no impact on the motion in a perpendicular direction), the ...


0

I think actually that the light-clock experiment will work if light were confined and restricted to travel in a tube or cable stretched perpendicularly from the light to the detector. Otherwise, the light will miss the detector, for the reasons stated above. The problem might be in the way the experiment is usually described or illustrated.


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The big difference is that a hard boiled egg is close to the model of a rigid body. When we apply a torque with our hand, its angular velocity increases (while our hand is in contact) by the relation $\tau = I\frac{d\omega}{dt}$, where $\tau$ is the applied torque, $I$ is the moment of inertia, and $\omega$ is angular velocity. After we release it, $\omega$ ...


0

In a hard boiled egg the mass is centered closer to the center of the egg since the yolk is the densest and thus the moment of inertia $\int r^2dm$ is smaller compared to a raw egg, where the dense yolk moves outwards as the centripetal force $F_{cpt} = \frac{mv^2}{r}$ implies a denser material will move outwards when rotating. This means that for the hard ...


2

You have not used the correct definition of $I$. The moment of inertia is defined as $$I=\int r^2_\perp dm=\int r^2_\perp\rho dV.$$ Using spherical coordinates $$\vec r=\begin{pmatrix}x\\y\\z\end{pmatrix}=r\begin{pmatrix}\cos\varphi\sin\theta\\ \sin\varphi\sin\theta\\\cos\theta\end{pmatrix},$$ we get $$r_\perp^2=x^2+y^2=r^2\sin^2\theta\,\,\,\text{and}\,\,\,...


1

While your calculation is correct, your derivation starts from the wrong premise. A moment of inertia is calculated about a given axis of rotation, not about a point. I assume in your case the axis of rotation runs through the Centre of Mass of the uniform sphere. In that case, you need to slice up the sphere into infinitesimal cylinders (circular plates) ...


1

You need to evaluate the moment of inertia about an axis through the center, and that distance is not always $r$ for a mass element $dm$ from the axis.


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