From the rotation curve of spiral galaxy it is found that for (>>r ) the total mass is not concentrated at the centre but varies as ~r for which the rotational velocities of the stars far from the centre remains nearly constt. and near or within the central core varies linearly. this extra mass for >>r is named as dark matter since it is not expected from the theoretical point of view . Is the energy possessed by this matter is called the dark energy?

  • 5
    $\begingroup$ Did you check the definitions on Wikipedia? $\endgroup$ – Qmechanic Feb 12 '19 at 19:06

this extra mass for >>r is named as dark matter since it is not expected from the theoretical point of view

$\let\Lam=\Lambda \def\tR{\tilde R} \def\half{{\textstyle {1 \over 2}}}$ The reason for that name is another. It was generally assumed that matter in galaxies was mainly composed of stars or luminous gas clusters. Then a definite relationship was expected between mass and luminosity. On the contrary, the so-called dark matter manifests itself by gravitational effects alone, with no light emission.

Dark energy (BTW, a name I don't like much) has no significant effects at galaxies scale. It's only relevant on a much wider scale, i.e. cosmological. It has a long history, starting from Einstein's paper itself, where he introduced a "cosmological term" in his equation in order to allow for a static universe. This was before Hubble and expansion. When it became known that universe isn't static Einstein disowned his correction, defining it (according to Gamow) "the greatest blunder of my life".

Let me detail better this point. I have to write Einstein's equation - don't worry if you can't understand it, I hope you may grasp the basic idea. In suitable units it is $$\tR_{ik} - \half \tR\,g_{ik} = 8 \pi T_{ik}.$$ This is the original form. The modified one, with cosmological term added, is $$\tR_{ik} - \half \tR\,g_{ik} - \Lam\,g_{ik} = 8 \pi T_{ik}.\tag1$$

In both cases Einstein's equation has the general structure of a field equation with source. A much simpler example is Poisson's equation for classical electrostatic field: $$\nabla^2 \Phi = -4\pi\rho$$ where $\Phi$ is electrostatic potential and $\rho$ is its source, the density of electric charge. In both Einstein's equation the left-hand side replaces $\nabla^2 \Phi$ and $8 \pi T_{ik}$ is the source, in this case the energy-momentum tensor. This means that curvature of spacetime is generated by energy and momentum content in every region of spacetime. Thus Einstein modified first member, leaving unaltered the source.

The history of dark energy begins when eq. (1) is written in a slightly different form: $$\tR_{ik} - \half \tR\,g_{ik} = 8 \pi T_{ik} + \Lam\,g_{ik}.\tag2$$ You can see that eqs. (1) and (2) are mathematically the same - there's only a term displaced by left to right-hand side. Elementary algebra says it's always allowed. (I find it a typical example of why physics can't be reduced to mathematics.)

Eq. (1), following Einstein, leaves unaltered the source side but tries a correction to the field side. On the contrary eq. (2) maintains the original Einstein's proposal, but assumes a novel contribution to source: it's the contribution which received the name of dark energy. Dark because it can't be seen - it only manifests through its cosmological effects. Energy because its energy-momentum tensor has a peculiar form, totally unlike the one of any other kind of matter previously known, be it the so-called "cold matter" (baryons free or bound in nuclei) or "hot matter" (improperly also named "radiation") i.e. highly relativistic particles like photons or neutrinos.

Since eqs. (1) and (2) are mathematically the same, there's no hope of telling one from the other by cosmological observations. On the theoretical side we had to find an explanation for the novel "energy". AFAIK such an explanation is lacking right now.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.