If the matter in the Universe formed from the energy of the Big Bang, and matter has weight, does this mean that energy has weight?

Moreover, as the Universe expands over time, does its overall weight stay the same?


2 Answers 2


What would energy having "weight" mean?

Relativity teaches us that $E = \sqrt{(m_0 c^2)^2 + (p c)^2}$ where $m_0$ is rest mass, $c$ is speed of light and $p$ is momentum. This is the mass-energy equivalent you are referring to, and it allows particles without rest-mass (e.g. photons) to convert into particles with rest-mass (e.g. electrons) by "investing" extra energy into "generating" a rest mass.

We can consider having "weight" as having interaction with a gravitational field. Then we can rephrase your question in a more well-defined way: Can particles without rest mass interact with gravitational fields? The answer is yes, and this can be seen for example in gravitational lenses, where the path of light is bent by the gravity of a massive object (e.g. a star). A more dramatic example is a black hole, which is so massive that light cannot escape its gravity.

Be careful, though. What we in our daily lives call weight refers to the rest mass, which is constant. So in that sense no. For example, the rest mass of a football does not change if you invest energy into it by kicking it.

For your second question: The total energy of the universe is constant even under inflation (to the best of our current knowledge). The "weight" of the universe is not a well-defined concept, since there is no outside gravitational field to interact with.

  • $\begingroup$ If you charge a capacitor, the total number of particles remains the same, but the energy increases, so the weight increases too, correct? $\endgroup$
    – endolith
    Jun 8, 2020 at 17:43
  • $\begingroup$ Hm, good question. I would say no, since it is potential energy, and so corresponds to kicking a football up a hill. But on the other hand, nucleons can gather in atomic nuclei where they release potential energy to lower the overall rest mass. Maybe you should pose it as a separate question? $\endgroup$ Jun 9, 2020 at 16:23
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    $\begingroup$ Ok, I think the answer is yes, but I'll ask a separate question: physics.stackexchange.com/questions/558479/… $\endgroup$
    – endolith
    Jun 10, 2020 at 14:46

When discussing about energy in Physics, one of the most important things to have clear is that energy is not a physical system in itself. Energy is a property of a physical system. There is no "pure mass", as well as there is no "pure energy". Instead, there are physical systems which have energy, have mass.

The first consequence of such preliminary statement is that, in a strict sense, there is no "matter formed from energy". There are physical systems which have mass, and other which do not have mass. Both have energy. Moreover, in the case of a system with mass, increasing the energy of the system, in the frame where the system is at rest, implies an increase of mass. That's the content of the celebrated relation $E=mc^2$. So, provided that what you call "weight" is actually the mass, we arrive to the correct statement: the fact that matter has mass implies it has energy. However, mass and energy remain properties of the physical system and do not have any meaning as standalone entities, without the physical system.

  • $\begingroup$ A sufficiently dense system of gravitationally interacting EM waves would act as if it had a rest mass. $\endgroup$
    – S. McGrew
    Mar 13, 2019 at 22:37
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    $\begingroup$ @S.McGrew I know. But, which part of my answer is in contrast with this fact? Take a two photon system with opposite momentum. It is a system with mass and its mass is proportional to the energy of the system. Once again, mass and energy are properties of the physical system of two photons. $\endgroup$ Mar 13, 2019 at 22:44
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    $\begingroup$ Such a system would be a "geon", which would be basically indistinguishable from a black hole. It would be difficult to argue that a black hole does not have rest mass and is not "matter". $\endgroup$
    – S. McGrew
    Mar 14, 2019 at 0:15
  • $\begingroup$ @S.McGrew Still I do not see any contradiction with my answer. I did not enter in the definition of what would be matter. $\endgroup$ Mar 14, 2019 at 4:54

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