I want to prove that ##E = -g_{\mu \nu}u^\mu p^\nu## is the energy measured by an observer with velocity ##u^\mu## of an object with momentum ##p^\mu##. My reasoning is that in special relativity we know that ##\gamma m = E##. We can transform to coordinates where ##u'^\mu = (1,\vec{0})##. Since the measurement is happening locally to the observer, the space appears flat, and ##g = \eta## and hence ##dt/d\tau = \gamma##.(adsbygoogle = window.adsbygoogle || []).push({});

$$

-g_{\mu \nu}u^\mu p^\nu = -\eta_{\mu \nu}u'^\mu p'^\nu =

-(1,\vec{0})\cdot(-\gamma m, \gamma \vec{v}) = \gamma m = E

$$

My question. Is this kind of reasoning legal? Is it enough to argue that since the measurement happens locally, we can take the space to be flat? The reason I ask is because it would surprise me since pretty much any kind of measurement I can think of is a local concept.

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# A General relativity -- Proof of energy measured by observer

Have something to add?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**