general relativity - How energy curves spacetime?

We know through General Relativity (GR) that matter curves spacetime (ST) like a "ball curves a trampoline" but then how energy curves spacetime? Is it just like matter curvature of ST?

Answer

Theoretical viewpoint:

Einstein field equations can be written in the form:

$$\color{blue}{G_{\mu\nu}}=\color{red}{\frac{8\pi G}{c^{4}}} \color{darkgreen}{T_{\mu\nu}}$$ We can write in simple terms: $$\rm \color{blue}{Space-time \,\,geometry}=\color{red}{const.}\,\,\color{darkgreen}{Material \,\,objects}.$$ And the $T_{\mu\nu}$ is a mathematical object (a tensor to be precise) which describes material bodies. In that mathematical object, there are some parameters such as the density, the momentum, mass-energy... etc. So it is those parameters that determine 'how much space-time curvature' is around a body. And one of the parameters is of course energy. Therefore, energy do bend space-time.

Experiments that confirm this point:

First, do photons have mass? The answer is an emphatic 'no'. The momentum of a photon is $p=\frac{hf}c$, and from special relativity:

$$\begin{align}E=\sqrt{(mc^2)^2+(pc)^2}&\iff E^2=(mc^2)^2+(pc)^2\\&\iff E^2-(pc)^2=(mc^2)^2\\ \end{align}.$$ The energy of a photon is: $E=hf$ which is an experimental fact. It can also be expressed as $E=pc$ since $E=hf=\frac{hf}{c}\cdot c=pc.$ Therefore, $E^2=(pc)^2$ and so $E^2-(pc)^2=0$. Putting this in our previous derivation we get: $E^2-(pc)^2=(mc^2)^2=0$. Since $c^2$ is a constant, then $m=0$. Therefore, photons have no rest mass.

Claim: Photons are not subject to gravitational attraction since they have no rest mass.

Experimental disproof: Gravitational lensing:

You could see light being bent due to the presence of a strong gravitational field.

Conclusion: Even if light has no rest mass, it has energy and momentum. And it is being attracted due to gravity, so the natural conclusion is that energy do curve space-time.