The cosmological constant is a fiction. The negative energy is the energy of localized matter: E=-mc2/sqrt(1+v2/c2) (mind the signs). Negative energy attracts (creates a negative force) and that's why we have gravitation.

Also note that if in the above formula you substitute v = c, you get E=-mc2/sqrt(2) = -mc2/1.414 ~ -0.71 = -71%, that is, about 71% of all energy in the Universe is negative. This is the amount they ascribe to "dark energy" which the cosmological constant is purported to represent.

The cosmological constant is not dimensionless. It has dimension 1/length2.

Lyudmil Antonov "Negative energy attracts (creates a negative force) and that's why we have gravitation."

No. Positive energy attracts. Two positive masses attract each other. And the correct relationship between energy and mass is E=(m2c4+p2c2)1/2, not your expression with a minus sign. A negative mass repels a positive mass and vice versa, but since a negative mass would normally also have negative inertia, the repelling force exerted by a positive mass on a negative one will lead to the negative mass moving in the direction of the positive one! Repulsion will look like attraction. In fact, if negative masses existed, you could build a perpetual motion machine.

The dimensionless cosmological constant is different from the cosmological constant and it is the cosmological constant rescaled by the (square of) Hubble constant. When multiplied by the dimensionless Hubble parameter known as "little h", the product is also called dimensionless cosmological constant and its value makes a big difference to the age of the universe.

Cosmologists and astrophysicists may call the dimensionless cosmological constant itself cosmological constant with a symbol lambda or omega standing for the density for vacuum.

In Planck units,the cosmological constant Λ in Einstein field equations

becomes dimensionless and thus Λ can be either dimensionful or dimensionless.

Dear Prof. K. Kassner,

I appreciate your interest in the question and the time that you took to reply. The answer that you wrote corresponds to the current knowledge and the generally accepted theory (special and general relativity). However, I would like to highlight several points that throw doubt on its validity, especially on the question of metric and energy (and its conservation).

1. I regard the definition of the metric in special relativity as a misconception promoted by Einstein. He calls it "pseudo-Euclidean" when in fact, it is hyperbolic, as correctly pointed out by Minkowski. Indeed, c2s2 = c2t2 - r2 with r2 = x2 + y2 + z2 is an equation of a hyperbola (hyperboloid) and not an ellipse (ellipsoid). Therefore, the geometry in this Minkowski 4d space is Lobachevski-Bolyai rather than an elliptic geometry. The Lorentz transformations confirm this as they use hyperbolic functions (sinh, cosh) instead of trigonometric (sin, cos). According to my theory based only on Galilean transformations and Pythagoras' theorem (Preprint Cyclic-time mechanics of isolated systems

2. The metric used in the Hamilton principle (the principle of least action). Following, for example, Landau & Lifshitz, The Classical Theory of Fields, pp. 25-28, or Goldstein. Classical Mechanics, Section 7.9, the Hamilton principle for a free-moving relativistic particle is S = constant * integral from a to b of ds. They show later by analogy that constant = m*c. s is the metric which in the relativistic case is cds = sqrt(c2dt2 - dr2 ). Differentiating both sides by dt, they obtain dS/dt = mc2 * integral from t1 to t2 of sqrt(1-v2/c2)dt. Here comes the glitch. For an action to be minimal, we should have v=c. Then, we have integral of 0.dt = 0 and so S = 0, i.e., light and particles who move with light speed (neutrinos) carry no action, and, therefore, do not possess energy. Furthermore, and even more pathologic, at zero S is minimal, i.e., while a particle moving with light speed has a zero action, a particle at rest has maximal action. To bypass this, somehow, they needed to modify the principle to state that it is a principle of extremal action (that's why in volume I Mechanics, Landau & Lifshitz seek the minimum, and in volume II, Field Theory, they seek the extremum). Anyway, to get some minimum, they put a minus sign before the integral, to become dS/dt = -mc2 * integral from t1 to t2 of sqrt(1-v2/c2)dt. So, it becomes "the principle of the maximal negative action", or, in other words: "In order to not do anything, you must walk the maximal negative path", which is nonsense. On the basis of the above action, they get a negative Lagrangian: L = -mc2*sqrt(1-v2/c2) and a positive Hamiltonian H = E = mc2/sqrt(1-v2/c2). With v=0, the latter is the famous Einstein equation. However with v not zero, this equation has big problems. At v=c, the energy of the particle becomes infinite! They usually justify this by saying that light with v=c has no mass and, therefore, has no infinite energy because the zero in the numerator somehow balances the zero in the denominator. But how about neutrino with m > 0 and v=c? And another one: light has no mass but has a momentum, and momentum by definition is p = mv = mc (for light). In the action integral, if you put m=0, you get S = 0, so light must not have action and energy. You can say, light is not a material particle. Granted, but then why does light have a momentum?

Probably, led by these considerations, Einstein stated that his formula for E is valid for speeds much smaller than c. He then Taylor decomposed the formula for the small quantity v/c and obtained the following Taylor series E = mc2 + mv2/2 + 3/8 mv4/c2 + ... and he truncated the third term as small (Einstein. The special and general theory of relativity, p. 43). For the generalized momentum p = dL/dv, later authors obtained p = mv/sqrt(1-v2/c2), and comparing the results for the approximated E and the generalized p, they obtained the equation for energy E2/c2 = p2 + m2c2 and the Hamiltonian H = c*sqrt(p2 + m2c2) that you cited above. However, apart from the flaws, which I'll describe below, I must note that the Hamiltonian has an indefinite sign because it is obtained by taking the square root of E2. It should be H = +/- c*sqrt(p2 + m2c2). Also, the generalized momentum becomes p = 0 at m = 0 so light shouldn't have a momentum (and, in reality, it has).

Everything becomes different and sensible when I used the same approach with the metric c2s2 = c2t2 + r2. The principle of least action gives the least action when v=0, i.e., the particle is at rest. Then the action comes only from time (t) which is never at rest. Time is always moving, so S is never zero. In order for S to be constant, dS/dt = const, and since dS/dt = -Ht by the Hamilton-Jacobi equation, S is a wave function, which is another line in my theory leading to quantum mechanics.

Returning to the principle of least action, the Lagrangian becomes positive L = mc2*sqrt(1+v2/c2) (btw, I prove that L is the angular momentum) and the Hamiltonian becomes negative: H=-mc2/sqrt(1+v2/c2). At v=c, H=-mc2/sqrt(2), and at v=0, H=-mc2. In fact, with this metric nothing is zero or infinity, i.e., the theory is completely devoid of singularities unlike relativity.

3. Now, why a mass has to have negative energy in order to attract. By the same reason by which they attach a negative term -U/r = -Gm/r to the total energy of each physical system (even in quantum mechanics) which they call a Newtonian potential. A mass has to create a potential well around it in order for another mass to fall in. The bigger and denser the mass, the deeper the well. Smaller masses with shallow wells fall to the bigger masses with deeper wells. This is what the general relativity says but in another terms. GR involves spacetime and curves it when, in fact, everything is connected with the mass and the negative potential that every mass creates.

Stated in term of forces, F=ma=m.d2r/dt2. A positive force pushes away from the body (a positive direction) while a negative force pulls towards the body (a negative direction).

For a more advanced treatment, see, e.g., Mark Levi, Classical Mechanics with calculus of variations and optimal control, pp. 16-17 from which the attached picture is taken. It means that the Newton equation can be represented in the form d2x/dt2 = F(x). If F(x) = -x, this is an oscillating system dx/dt = v, dv/dt = -x which has an attractive force (concave potential, first panel in the picture). If F(x) = x, we have a hyperbolic phase space and a repulsive force (convex potential, second panel in the picture).

More in depth, in Goldstein, chapter 10, Hamilton-Jacobi theory and action-angle variables, they show that at negative Hamiltonians, the orbits are closed elliptic curves, while at positive Hamiltonians we have repulsion (hyperbolic orbits).

4. One should not mix mass and charge, which are different concepts. There are positive and negative charges but there are no positive and negative masses. It is true that mass is equated to energy but energy comes in different forms. There is a potential energy, and there is a kinetic energy. Kinetic energy is created in motion when transitioning from higher to lower potential energy. The virial theorem states that, on the average, T = -V/2 where T is kinetic, and V is potential energy. To have a kinetic energy, the potential energy should be negative, otherwise things cannot go uphill. The Hamiltonian, which is the sum of the two energies, H = T + V, when negative, means that the potential energy (negative) dominates, and, hence, the system attracts (in other words, the system has a negative potential).

5. The above metric induces rotational (harmonic) motion, which is the default inertial motion of any isolated system (think of motion of electron around nucleus, Earth around the Sun). Inertial motion means that no energy is input or released. Yes, it is in a sense a perpetuum mobile but the difference is that the energy of the system stays constant in full abidance to the energy conservation principle and the first law of thermodynamics. This was the conundrum of early quantum physicists who wondered why the electron does not fall on the nucleus and why it does not release energy. By the same reason, I could ask why the earth does not fall on the Sun. The answer is that rotation is the default motion of an isolated system (a system that does not receive or release energy and matter). This follows from the equations of motion that I derived in the above preprint, and is also seen from the action which has a form of wave function. The canonical equations that use the above negative Hamiltonian lead to the same result.

6. I didn't understand why the cosmological constant has a dimension 1/lenght2. If the cosmological constant is some negative energy, it should have dimensions of energy, that is, J = N.m or, rather, -J = -N.m. If it is a force (energy) that creates negative pressure, it can be expressed in area (N.m2) or even volume (N.m3). If we take it to be mc2 in some form, the units could be kg.m2.s-2= J.

At the present time there is an ongoing research project called DESI involving more than 900 researches from all over the world investigating the validity of the cosmological constant. So far their conclusion is that there is nothing constant about it They are also finding that the Hubble constant is also not constant unless averaged over very great distances.

The bottom line will be that there is no cosmological constant, and a different explanation will replace dark energy as a mediator of the Hubble constant.

https://universemagazine.com/en/dark-energy-wanes-scientists-receive-contradictory-data-on-the-universes-most-mysterious-force/