1

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.

  1. Does gravitational field really has negative energy density? Or is my miscalculation due to an error between Newtonian mechanics and relativity?
  2. 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 :)

Qmechanic
  • 201,751
littlegiant
  • 254
  • 2
    Does this answer your question? Is it possible for a system to have negative potential energy? – DavidH Jun 26 '20 at 14:54
  • 1
    Well, it is also a good answer, but it doesn't solve my curiosity. What if we 'can' make a system that satisfy the condition for negative energy? Like, if we use newtonian approximation, the self potential energy of the sphere with uniform density and has mass $M$ and radius $R$, would be $-\frac{3GM^2}{5R}$. And the total energy of the system would be $Mc^2 -\frac{3GM^2}{5R}$. Lets limit the condition to use this equation so that the maximum field strength must be same or less than $1000m/s^2$. – littlegiant Jun 26 '20 at 15:07
  • Then if the radius $R=7.510^{13}m$ and mass $M=8.4410^{40}kg$, my calculation gave me that the net energy is exactly zero. So what is wrong in this calculation? – littlegiant Jun 26 '20 at 15:11
  • May I ask one more question? I will edit my question. – littlegiant Jun 26 '20 at 15:13
  • The radius you quote is smaller than the Schwarzschild radius for that mass, so the Newtonian approximation is invalid. I think it would be best to open a new question if you have a different one. – DavidH Jun 26 '20 at 15:23
  • Understood. Should I delete this question? – littlegiant Jun 26 '20 at 15:28
  • @littlegiant: Better edit this question by removing the newtonian-mechanics/gravity tags and refocusing the question on general-relativity, then the question could be reopened. – A.V.S. Jun 26 '20 at 16:19
  • What if I already posted a new question? – littlegiant Jun 26 '20 at 16:22
  • If you were to try to come up with a gravitational system with zero or negative net energy, then its gravitational energy would have to exceed its rest mass $mc^2$ (actually be less than $-mc^2$). If you do the math with the expression for the gravitational energy, you obtain that the point masses have to be at a distance within an $O(1)$ factor of the Schrwarzschild radius. Hence, at least in the case of simple gravitational point masses, it seems that nature abhors so much negative energy systems that before it can happen, an event horizon will occur around it – lurscher Jun 26 '20 at 18:57

0 Answers0