I thought about gravitational field. In Newtonian mechanics, gravitational energy between two matter is $U=-G\frac{M_1 M_2}{R^2}$ when mass of each matter is M1 and M2, having a distance R. With this equation, I was able to obtain the energy density of the gravitational field. But since Newtonian mechanics are good approximation of relativity, it should be used only in non-relativistic situation. Anyway, I obtained the formula that if the strength of gravitational field(which also means gravitational acceleration) is g, the energy density of field was $u_g=-\frac{1}{8\pi G}g^2$. Now here is my question.  Does gravitational field really has negative energy density? Or is my miscalculation due to an error between Newtonian mechanics and relativity? If so, can a system have a negative net energy?  edit) In the link from comment, there was solution using Schwarzschild raidus. Then what is the net energy difference between black hole with mass $M$ and system with same mass $M$, but nearly no potential energy due to large distane between particles. (Lets assume that black hole is Schwarzschild Black Hole for simple calculation :)
122502
With respect to both gravity and electromagnetism, to the best of my understanding potential energy is added or subtracted from a system based on the distance between two objects such as charged or massive objects.  Would this then imply that when oppositely charged objects or two massive objects are brought together within a small enough radius that the drop in potential energy would be greater than the combined mass of the two objects making the net mass/energy of the system negative?
77931
In a gravitational field, the gravitational force acting on a body (of mass m) at a point x metres away from the attracting body (of mass M) is $\frac{GMm}{x^2}$. Integrating this force from a point at infinity to x gives $-\frac{GMm}{x}$, or the work done by the gravitational field on the mass in moving the mass from a point at infinity to x. Then, the gravitational potential at point x is $-\frac{GM}{x}$.  However, the definition that my school gives is : "Gravitational potential at a point in a gravitational field is the amount of work done by an external agent in moving one unit of mass from a point at infinity to the point in the gravitational field, without any acceleration". The lecturers explained that as a body moves closer to the attracting body, it accelerates and gains kinetic energy. In order to maintain a constant velocity, negative work must be done by an external agent to remove this extra kinetic energy.  Now I am very confused, because both seem valid to me. What does "gravitational potential" actually mean?
73561
On John Baez's website, , he discusses the problem of how entropy increases when a cloud of ideal gas collapses gravitationally (no black holes - keeping it simple).  But he doesn't give away the answer - instead he implies that, I presume, either gravitons or photons are released during collapse and these increase the overall system entropy.  Can someone please answer which it is, gravitons or photons?  If gravitons (or gravity waves), then does this mean that pre-20th century physicists, with no notion of gravity waves, would have concluded that gravitational attraction violates the 2nd law?  And if photons, then what about if the gas particles were totally neutral - wouldn't they not emit any EM radiation in that case?  Or am I wrong that gravitons or photons are the answer? Thanks!
42793
I am confused by how to deal with the negative sign in the equation $U=-GMm/r^2$ in the following problem:  If the distance between two masses is tripled, then the magnitude of the gravitational potential energy is: A) Increased by a factor of 3 D) Decreased by a factor of 3 -- correct answer  The cause of my confusion is that if r = 1 and you assume U = -1, then when r = 3, U = -1/3. I am thrown off by the negative sign. Isn't -1/3 actually a larger value than -1? So why is this considered a decrease? Is it because they specify factor of 3, and an increase by factor of 3 would actually be -1*3 = -3? Furthermore, if potential energy decreases as r gets larger, then why does a pen gain potential energy as you lift it further away from a table and why does the potential energy decrease/convert to kinetic energy when you drop the pen? I have only taken intro physics, so some of the other potential energy sign convention questions went over my head.
113343
Almost every book on physics that I read have some weird and non-clear explanations regarding the potential energy. Ok, I do understand that if we integrate a force over some path, we'll get a difference in some origin-function values ($\int_{A}^{B} Fdx = U(B) - U(A)$). This function is the potential energy. Of course, whether we can define this term or not depends on the force.  Now, here's an example of explanation (to be more precise - lack of explanation) regarding the GPE from one of the books:     "...When a body moves from some point A to point B, gravity is doing work: $U_A-U_B=W_{A \to B}$. The magnitude can be calculated using an integral:   $W_{A \to B}=\int_{r_A}^{r_B} F(r) dr = \int_{r_A}^{r_B} \left(-\frac{GMm}{r^2} \right)dr=(-\frac{GMm}{r_A})-(-\frac{GMm}{r_B})=U_A-U_B$      ...      Thus,  when $r_A&gt;r_B$, the magnitude is positive and therefore $U_A&gt;U_B$. In other words: when the distance between the bodies is being increased - the gravitational potential energy of the system is also being increased.   Giving absolutely no explanation on why all of a sudden they put a minus sign into the integral.  From another book:     Work done by Coulomb's force: $W_{el}=\int_{r_1}^{r_2}\frac{q_1q_2dr}{4 \pi \epsilon_0 r^2}=\frac{q_1q_2}{4 \pi \epsilon_0 r_1}-\frac{q_1q_2}{4 \pi \epsilon_0 r_2}$      ... Calculating the work gravity is doing is no different from the calculation of the work done by an electric field, with two exceptions - instead of $q_1q_2/4 \pi \epsilon_0$ we should plug $G M m$, and we also should change the sign, because the gravitational force is always a force of attraction.   Now, this is not satisfying at all. So what if it is an attraction force? How this should influence our calculations, if the work is defined as $|F| |\Delta x| \cos \theta$, so the sign only depends on the angle between the path vector and the force vector? Why they put a minus sign? Is is just a convention or a must thing to do?  Some say the sign is important, others say the opposite. Some explain this as a consequence of that we bring the body from infinity to some point, while others say it is a consequence of an attractive nature of the gravitational force. All of that is really confusing me.  Also, in some of the questions like "what work is required to bring something from point $A$ to point $B$ in the field of gravitational/electric force", the books sometimes confuse $U_A-U_B$ and $U_B-U_A$ - as I understand it - the work that I must do is always $U_B-U_A$. However the work that the force that is being created by the field do is always $U_A-U_B$, am I correct?
112897
I've been wondering, in Coulomb's Law, $k_e = \dfrac{1}{4\pi\epsilon_0}$. Therefore, why do we use $G$ in Newton's Law of Gravitation? What if the constant is more like Coulomb's Law, e.g. $G = \dfrac{1}{4\pi G_0}$ where $G_0$ is some constant.  This would make Newton's Law of Gravitation look like the following: $$\bf{F}_{12} = -\dfrac{m_1 m_2}{4\pi G_0 |\bf{r}_{12}|^2} \bf{\hat{r}}_{12}$$  $GM = \mu$ is used for calculations but so could $\dfrac{M}{4\pi G_0} = \mu$.  If this is not the case, what is the significance of this definition?
37865
Recently, there was a rapid communication published in Phys.Rev.D (), titled "Gravity is not an entropic force", that claimed that , disproves Verlinde's entropic approach to gravity.  The neutrons experiment gave results consistent with the predictions of Newtonian gravity for the lowest energy state.  As I understand it, the author claims that the fact that Verlinde's entropic force comes from a thermodynamic process that is irreversible (or approximately reversible), leads to non-unitarity in the evolution of quantum systems. The non-unitarity then exponentially suppresses the eigenfunctions, predicting results very much different than the Newtonian. Thus, that experiment is in contradiction to what is expected if Verlinde's approach is correct.  My questions are,   First of all, is there anything else essential that I am missing? Is there any response to that argument? Is that a fatal problem with Verlinde's entropic approach? Is that a fatal problem for any entropic approach?   Updates on the discussion:   There is also this recent comment  Once more: gravity is not an entropic force
122606
Ok,this is a silly question. But for quite long I have been thinking about this. What is gravitation all about?? My book directly writes since force of gravitation is directly proportional to the masses and inversly proportional to the square of the distance ie. $$F = G\frac{M_1 . M_2}{r^2}$$ . So is this gravitation all about? What is the cause of it? The book then says     We are not in the scope in classical approach so as to discuss the cause of this. However,as a matter of fact that if there are more than one mass,then there is gravitation as it is their fundamental property.( Cause of gravitation can be attributed to exchange of non-classical particles between the masses) .    Then it ended the talkings. What are those non-classical particles? Why and how are they exchanged? So far I have studied physics , I never found such weird phenomenon like this. I want to know what is the cause of gravitation and what is the cause of exchange of those so-called non-classical particles? And how do they do it?
4313
In a perfectly symmetrical spherical hollow shell, there is a null net gravitational force according to Newton, since in his theory the force is exactly inversely proportional to the square of the distance.  What is the result of general theory of relativity? Is the spacetime flat inside (given the fact that orbit of Mercury rotates I don't think so)? How is signal from the cavity redshifted to an observer at infinity?
22605
Sorry to ask the long way, but trying to make the question clear so that I can get a clear answer.   Why does it make any physical sense for a body to have negative potential energy?   If a body A is at infinite distance from another body B, it has zero gravitational potential energy, but if it is at "zero" distance from another body B, it has negative infinite potential energy, and acquires infinite kinetic energy in the process of getting there.  Please don't tell me that the zero point of potential energy is arbitrary, and it could just as well have zero potential energy when it is at zero distance from another body B, and positive infinite potential energy when it is at infinite distance from the other body B. How can it have this? When body A is at infinite distance from the other body B, then it is not influenced by the other body B at all, it is as if the other body B does not exist at all, so then the potential energy in body A must be zero, as there is no gravitational field of the other body B to cause any potential energy in body A.  But then the question is if the potential energy contained in the body A is zero at infinite distance from the other body B, then how is it possible for this zero potential energy to be reduced further and get changed to kinetic energy as the body A accelerates towards the other body B, so that the potential energy becomes negative? As I see it, if the body A never had any potential energy to begin with, then this not present energy cannot change into kinetic energy, otherwise we would be creating kinetic energy out of nothing.   Why does it make any physical sense to say that the body is in "potential energy debt", ie the potential energy is reduced to negative when it was zero to begin with?
42574
On this site, the Phys.SE question  has been asked. See also  Phys.SE question. The 'answer' provided on this forum has been that the curvature of spacetime explains both. The answer is still cryptic for me as I am more a concrete thinker.  Newton said $F=ma$. I can use this formula to measure inertial mass. Experimentally I can measure the motion of an object while applying a constant force to it. Newton also said $F=\frac{GMm}{r^2}$. In this case, what simple experiment will allow the measurement of gravitational mass?
3100
So I have been trying to understand why there is a negative sign in the following$\Delta U= -\int\vec{F}\cdot d\vec{l}$  I wanted to try to understand it by trying to derive the gravitational potential energy $E_p=-\dfrac{GMm}{r}$ I start by defining potential energy to be $0$ at $\infty$ and see what I get if I let gravity do work on the mass until a distance r away from a planet  $W_G=\int_{\infty}^{r} \vec{G}\cdot{d\vec{l}} = \int_{\infty}^{r} G\cos{\theta}dl = \int_{\infty}^{r} G dx$ G is the gravitational force so:  $W_G= \int_{\infty}^{r} -\dfrac{GMm}{x^2}dx = \Bigg[\dfrac{GMm}{x}\Bigg]$ applying limits of integration gives $W_G = \dfrac{GMm}{r}$  Thus I am confused why the minus sign, from the derivation I wrote it seems to have to be $\Delta U= \int\vec{F}\cdot d\vec{l}$ at least for gravity. I also did the same derivation for a mass on earth ($W_G = -mgh$ Where $\vec{G}$ is considered to be constant) and the result indeed is the correct one $\Delta U= -\int\vec{F}\cdot d\vec{l}$ so my confusion arises, why don't i get the same result for both cases.   I am also wondering if I carry out the same derivation but for elastic potential energy, what would I get... Thank you for any help!
9129
In General Relativity Einstein's equation implies that stress-energy tensor on its RHS is conserved (has vanishing divergence), due to the Bianchi identity. Considering variational principles leading to Einstein's equation leads to conclusion that this stress tensor is equal to the variational derivative of full action with respect to the metric tensor. However, on several occasions I heard people stating that quite generally one can define stress tensor for a field theory in this way and it is automatically conserved. In flat spacetime and without any coupling to gravity! I wonder if this is true. I don't see a reason why it should be.
63475
Starting  with the Einstein-Hilbert Lagrangian  $$ L_{EH} = -\frac{1}{2}(R + 2\Lambda)$$  one can formally calculate a gravitational energy-momentum tensor   $$ T_{EH}^{\mu\nu} = -2 \frac{\delta L_{EH}}{\delta g_{\mu\nu}}$$  leading to  $$ T_{EH}^{\mu\nu} = -G_{\mu\nu} + \Lambda g_{\mu\nu} = -(R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R) + \Lambda g_{\mu\nu}. $$  But then, in the paragraph below Eq. (228) on page 62 of  paper, it is said that this quantity is not a physical quantity and that it is well known that for the gravitational field no (physical) energy-momentum tensor exists.  To me personally, this fact is rather surprising than well known. So can somebody explain to me (mathematically and/or "intuitively") why there is no energy-momentum tensor for the gravitational field?
7035
For the sake of simplicity, let's imagine that the entire universe is empty except for a single lump of (classical) matter with mass $m$. In its center of momentum frame, it is clear that the total energy is simply $E_\text{CoM}=mc^2$. However, in a frame moving relative to it with speed $v$, we have $E_\text{moving}=\sqrt{m^2c^4+p^2c^2}&gt; mc^2$ for all nonzero $v$.  Are we to infer that the energy content of a system can be different relative to the observer? Does this not violate the first law of thermodynamics?
23525
I was just  that any type of energy contributes to mass of the object. This must indeed include potential energy in gravitational field. But here, things cease to make sense, have a look:   I have object at some distance $r$ from radial source of gravitational field. The potentional energy is calculated like this: $E_p = m*a*r$ where $a$ is gravitational acceleration and $m$ is mass of your object. But that means, that the object is a bit heavier - because of the potentional energy of itself - $m = m_0 + \frac{E_p}{c^2}$ ($m_0$ here is the mass without the potentional energy) That would mean that the gravitational force is a bit stronger at higher distances.   Now, I do understand that the rules of physics are not recursive and the mass and force will be finite. But what is the correct approach to this situation? What is the correct equation for potentional energy?
40808
Following the measurement of gravitational waves, many sources described them and explained they carry energy away. What I don't get is how this energy will get transfered back to anything else.  If the fabric of space-time itself is vibrating, it would seem to be in impossible for any physical object to gain this energy.    What am I missing?  How would one hypothetically get energy out of gravitational waves?  If impossible, does the universe end up with nothing but GW?