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Before telling you why an observer in free fall does not feel any force acting on him, There is a couple of results that should be introduced to you. Newton's second law is only valid in inertial frames of reference: To measure quantities like position,velocity and acceleration of an object, you need a coordinate system $(x,y,z,t)$. Now the coordinates ...

13

It is incorrect to link the feeling of being accelerated to being accelerated itself. You can be under constant velocity or be continuously accelerated, yet you need not feel anything at all. Let me explain. The reason you feel compressed or stretched when you are accelerated in a lift is because of the presence of the normal force from the ground on you. ...

4

Well, everything depends on what you mean by "to experience a force". I suspect that you are thinking of some psycho-physical idea. Indeed both floating in space and freely falling we perceive similar sensations. The reason is simply due to the fact that, in both situations, all particles of our body moves with the same speed (due to a spatially uniform ...

3

Yes, well, sort of. Energy can be a bit tricky to keep track of in general relativity, and it's important to be precise about what we mean by energy. In this case the issue is whether the light is red shifted. The red shift does reduce the energy of individual photons, though overall the energy is not lost - it's just diluted. You probably know that the ...

3

This paper is interesting. It uses the method of calculating the number of nucleons in the neutron star, $N$, based on the radius, $r$, the number density as a function of radius, $n(r)$, and the metric function $\lambda$, which comes from the equations of general relativity: $$N=\int_0^R 4\pi r^2e^{\lambda/2}n(r)dr=\int_o^R4\pi r^2 ... 2 The gravitational mass of a neutron star is quite a lot less than its baryonic rest mass (plus the mass associated with the kinetic energy of its contents), because a bound neutron star, by definition, must have a total energy (the sum of its internal energy and gravitational potential energy) that is less than zero. In a “normal star” this is also true, ... 2 falling in a gravitational field is physically indistinguishable from floating in interstellar space Yes. Indeed, this is one of the founding principles of general relativity and is (one of the forms of) the equivalence principle. Your argument is that we can feel acceleration, and gravity makes you accelerate, so shouldn't you feel acceleration while ... 1 You need a coordinate system to decide a body’s position, velocity, acceleration, momentum or force on it. Assume the body is in free fall near the Earth. 1) First consider a coordinate frame (3 perpendicular rods and a clock) with its origin in free fall near the free falling body. By the equivalence principle we know the rods are falling in unison with ... 1 Gravitational field vectors conventionally are represented pointing toward the gravitating body whose gravitational field is being analyzed. If two gravitating bodies are mutually attracted to each other, and one is in orbit around the other, the center of mass of their system is called the barycenter, and both orbit around the barycenter. This results in ... 1 Based on the title of your question, there are only two objects. I'm going to infer that you are analyzing the motion behavior of one of those objects based on the influence of the other. I'm also going to assume that the influence is gravitational and not electrical. That's the context of your question. You will only show the gravitational force vector ... 1 Mass is a Lorentz invariant quantity! The relativistic mass is not the real mass, it is is just called relativistic "mass" for obvious reasons. This term is abandoned by most textbooks, as it often causes this confusion. 1 The equation E = m^2c^2 + p^2 c^2 is restricted to Special Relativity. However, in classical physics we have$$ \vec{F} = m \vec{a}, $$and$$\vec{F} = m \vec{g},$$whence$$ m \vec{a} = m \vec{g}. $$This can be written as$$ m \big( \vec{a} - \vec{g}\big) = \vec{0}. $$From a mathematical point of view we have$$ \big( m = 0 \big) \vee \big( \vec{a} - ...

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One of the basic principles in relativity is that spacetime always looks locally flat. By this I mean that if you restrict your observations to a small region surrounding you the curvature becomes negligable. You can make the effects of curvature arbitrarily small by making the region you consider arbitrarily small. The point of this is that for your clock ...

1

Classical Lagrangian field theory deals with fields $\phi: M \to N$, where $M$ is spacetime and $N$ is the target-space of the fields. We shall for convenience call $M$ and $N$ the horizontal and the vertical space, respectively. The metric $g$ can be viewed as a classical field of this kind. OP is asking about finding the Euler-Lagrange equations. In that ...

1

If you are above the water you will get accelerated down until the weight of the water you disperse is equal to your own weight (calling this level $x$). As soon as you are completely submerged the gravitational force downwards will be $\rho Vg$ and by Archimedes principle the force upwards will be $\rho_w V g$, where $\rho$ is your density, $V$ is your ...

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