0
$\begingroup$

It is a central tenet of relativistic mechanics that there can be no such thing as an absolute frame of reference, and as such, it appears to me that it is physically impossible to measure the total energy of the universe, as calculations of energy depend on the velocity of objects. From this, I can consider a number of possible conclusions, which I wish to pose to you:

Does the universe have a theoretical total energy, which just happens to be incalculable?

Does the universe have a theoretical total energy, which happens to be calculable? For instance, if the calculated energy in every single reference frame would be the same?

Is there something incorrect about my logic earlier?

Or, is conservation of energy on a universal scale nonsensical? If this is the case, why would this not have significant ramifications on modern theory?

$\endgroup$
2
  • $\begingroup$ What do you mean by energy? Can you define it in such a way that would allow you measuring or calculating its value for the entire universe? Is it a tensor? A pseudo-tensor? Anything else? There is no unique definition. Energy of the universe can be defined as conserving, non-conserving, or zero. Thus your question is not well defined. Check out this post by the founder of Vixra who used to be a member here in the past: google.com/amp/s/vixra.wordpress.com/2010/08/06/… - but keep in mind that not everyone agrees with him. $\endgroup$
    – safesphere
    Apr 30, 2020 at 4:20
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/2838/2451 , physics.stackexchange.com/q/296/2451 and links therein. $\endgroup$
    – Qmechanic
    Apr 30, 2020 at 4:43

1 Answer 1

0
$\begingroup$

To answer your question, the conservation of energy on the scale of the entire universe does not work as for conservation of energy, the conserved system has to be time-shift symmetric. As the universe is exponentially growing, it is obviously not, and hence, while conservation of energy may work in smaller systems, it does not make sense on such large scales.

$\endgroup$

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