Positive and negative spacetime considerations (Hypothesis) in general relativity

Hi all,

Once again I am here to discuss my hypothesized theory concerning mass of a particle and its spacetime curvature. The main essence of this article is the spacetime curvatures for different types of mass .But because of lack of knowledge and an excess of enthusiasm and obsession for these theories and hypothesis, there will be many assumptions made by me in the Einstein Field equations (EFE) in their physical meaning. So let's begin........

  1. Mass having particles :- Consider the equation.\[{R_{\mu \nu }} - \frac{1}{2}R{g_{\mu \nu }} + \Lambda {g_{\mu \nu }} = \frac{{8\pi G}}{{{c^4}}}T\mu \nu \] Also consider that the L.H.S. represents the spacetime curvature by a body (please do educate me about the about its actual meaning in the comments section below). As the equation for a mass having particle is already given here, hence we'll not talk much about it.

  2. Mass-less particles :- Again using, \[{R_{\mu \nu }} - \frac{1}{2}R{g_{\mu \nu }} + \Lambda {g_{\mu \nu }} = \frac{{8\pi G}}{{{c^4}}}T\mu \nu \] And , \[\begin{gathered} E = m{c^2} \\ \frac{E}{m} = {c^2} \\ \frac{{{E^2}}}{{{m^2}}} = {c^4} \\ \end{gathered} \]Putting in the field equation we get,\[{R_{\mu \nu }} - \frac{1}{2}R{g_{\mu \nu }} + \Lambda {g_{\mu \nu }} = \frac{{8\pi G{m^2}}}{{{E^2}}}T\mu \nu \]But for massless particles like photon we have \[m = 0\]Hence the modified equation yields,\[{R_{\mu \nu }} - \frac{1}{2}R{g_{\mu \nu }} + \Lambda {g_{\mu \nu }} = 0\]Therefore for a particle having zero rest mass the spacetime curvature must be zero. We know the according to QED (Quantum Electrodynamics) a photon is considered as an excitation in the electromagnetic field. Hence, we can conclude that the presence of the E.m field has no effect on the spacetime curvature. Q: Does this mean that elemagnetic force and gravity cannot be unified into a single framework ?

  3. Negative mass particles (Exotic matter) :- Sounds weird ? But various hypothesis do predict the existence of this type of matter exhibiting a special property of negative mass. In theories, this matter is required to produce an anti-gravity effect to stabilize a traverse-able wormhole because of the tendency of gravity to close it. Let's get back to our topic. Now appy the same field equation to the negative mass particles, we get\[\begin{gathered} E = m{c^2} \\ \frac{E}{{{c^2}}} = - m \\ \frac{{{E^2}}}{{{c^4}}} = {m^2} \\ \end{gathered} \] And,\[{R_{\mu \nu }} - \frac{1}{2}R{g_{\mu \nu }} + \Lambda {g_{\mu \nu }} = \frac{{8\pi G{{\left( { - \frac{E}{{{c^2}}}} \right)}^2}}}{{{E^2}}}{T_{\mu \nu }}\]\[{R_{\mu \nu }} - \frac{1}{2}R{g_{\mu \nu }} + \Lambda {g_{\mu \nu }} = \frac{{8\pi G}}{{{c^4}}}{T_{\mu \nu }}\]This was not expected, according to the field equation it must have a negative spacetime curvature but there is not negative sign. Q: Can we conclude that G.R. has left our hand when it comes to anti-gravity. Or maybe my hypothesis of considering the left hand side of the field equation to represent the spacetime curvature of a particle. For those who have enough knowledge about G.R., to them I ask a simple question. Can really anti-gravity be employed to G.R. to obtain negative spacetime curvatures ? Or if the hypothesis is correct, does it mean that negative mass particles curve the spacetime positively ?

  4. Infinite spacetime curvature ? :- Most of us who do physics know about black holes and also that they curve the spacetime infinitely. I this article we'll also peek into this phenomenon. Again consider,\[{R_{\mu \nu }} - \frac{1}{2}R{g_{\mu \nu }} + \Lambda {g_{\mu \nu }} = \frac{{8\pi G}}{{\frac{{{E^2}}}{{{m^2}}}}}{T_{\mu \nu }}\] where if term \[{R_{\mu \nu }} - \frac{1}{2}R{g_{\mu \nu }} + \Lambda {g_{\mu \nu }}\] MAY represent the spacetime curvature. When the right hand side of equation approaches infinity, then the left side must represent infinite spacetime curvature. This implies, for\[\frac{{8\pi G}}{{{c^4}}}{T_{\mu \nu }} \to \infty \] implies the term\[\frac{{{E^2}}}{{{m^2}}} \to 0\] this also implies\[c \to 0\]Does it means deceleration of photon travelling at lightspeed to zero ? Does it means the removal of all the energy of the electromagnetic wave consisting of electric and magnetic field vectors ? This result can be interpreted as follows:

A photon entering a boundary after which it encounter infinite spacetime curvature be slowed down by a force in infinitesimal time interval resulting in the complete loss of energy of an E.M. field. After the photon enters this region its velocity becomes zero, hence it cannot move on its own to get out of there. During the preparation of the hardcopy of this article I realized that the inability of photon to move out of that region (let's give this region a special name , 'special region') is the behaviour as showed by the particle after it crosses the event horizon of a black hole.Thus, an actual result a-kind-of 'agreed' to the result of my hypothesis. That's why I was motivated to go further.

Now consider an electron and positron pair annihilate each other to form two photons. As for the photons, they will be entangled because they are created at the same instant. Both the photons will be able to determine each other's properties no matter how large the distance of separation between them is. This is known as the principle of quantum entanglement. The change in the properties of one will enable us to determine the properties of the other particle. Now consider that they move at an angle of 45 degrees to each other.

Let us designate the particles as 'A' and 'B'. Consider that particle 'A' moves normally in space while particle 'B' encounters that special region of the universe, about which we talked above. Let after 5 seconds particle 'B' crosses the boundary after which the special region in universe begins and comes to rest in the 'special region'. Will both the particles be entangled now ? Will the particle 'B' will pass through the black hole and travel in a wormhole known as Einstein-Rosen bridges (in this case) but it will take he photon an infinite amount of time to reach the other end. Will the particle 'A' also come to rest ? What about other properties and quantum numbers ?

  1. Quantum retrocausality (The party pooper) :- The phenomenon by virtue of which the future of a particle can influence its past is known as quantum retrocausality. Though it has not been experimentally verified, some in the community believe in existence of this phenomenon. Currently I am working on how to incorporate this idea into my article and will let you know after I am done.

Finally, I would like to say that it's you, readers, who have enough knowledge about the topic and will decide the fate of this article. So please highlight flaws in my assumptions (if any) and do educate me what's right, your valuable suggestions will boost my confidence so that I may continue coming up with such hypothesis and theories in future.

Once again, It's Amit panghal, signing out. Keep learning !!!

Note by Amit Panghal
3 months, 3 weeks ago

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Very interesting. Great work!

Krishna Venkatraman - 1 month, 3 weeks ago

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Thank you so much @Krishna Venkatraman for appreciation.

Amit Panghal - 1 month, 3 weeks ago

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