# Energy-momentum relation in SR and GR

There is the famous energy-momentum relation

$$E^2 - p^2 c^2 = m^2 c^4.$$

I thought it is always valid. "Always" means it is valid in general, so in GR and SR. It is something like an elemantary construction. But now I am not sure and if I understand right it is valid only in SR. Right?

Below I write some more calculations and thoughts for a better understanding of my question.

In SR it is very simple to derive the formula. We start with the metric

$$g_{\mu\nu} = \eta_{\mu\nu} = {\rm diag}(1,-1,-1,-1)$$

and the product

$$g_{\mu\nu} p^\mu p^\nu = m^2 c^2$$

where $$p^\mu = m u^\mu = m \frac{{\rm d} x^\mu}{{\rm d} \tau}$$ is the four-momentum. This gives

$$g_{\mu\nu} p^\mu p^\nu = (p^0)^2 + \eta_{ij} p^i p^j = (E/c)^2 - p^2 = m^2 c^2$$

with the definitions $$p^0 = E/c$$ and $$\eta_{ij} p^i p^j = {\mathbf{p}}^2 = p^2$$. The last equation is equivialent to the above energy-momentum relation.

Now, when we turn to GR and consider the general metric in spherical coordinates

$$g_{\mu\nu} = {\rm diag}(g_{00}, g_{11}, -r^2, -r^2 \sin^2 \vartheta)$$

gives

$$g_{\mu\nu} p^\mu p^\nu = g_{00} (p^0)^2 + g_{ij} p^i p^j = g_{00} (p^0)^2 - p^2 = m^2 c^4$$

Here we have to be carefully now with the definition of $$p^0$$. In nearly all books about GR we find for the above spherical metric the following relation (or equivalent)

$$g_{00} m \frac{{\rm d} x^\mu}{{\rm d} \tau} = E/c = \rm const$$

which is interpreted as the conservation of energy. In SR we have $$g_{00} = \eta_{00} = 1$$ and therefore $$p^0 = E/c$$. But in GR we have the $$g_{00}$$ coefficent, e.g.

$$g_{00} p^0 = E/c$$

This would give the following energy-momentum relation

$$\frac{E^2}{g_{00}} - p^2 c^2 = m^2 c^4$$

Am I missing or do I mix something, e.g the definition of the time-component $$p^0$$ through an energy?

Or are the calcutions all right and in GR the energy-momentum relation is indeed different compared to the above famous and well known, which seems to be valid only in SR.

I am heavily confused.

My only idea how to resolve this is to redefine the energy in the general case, e.g.

$$\sqrt{g_{00}} p^0 = \mathcal{E}/c$$

Note the square root compared to the above definition. This would give an epxression analogue to the famous relation

$$\mathcal{E}^2 - p^2 c^2 = m^2 c^4$$

But $$\mathcal{E}$$ is not conserved here any more, $$\mathcal{E} \neq \rm const$$. Instead we have $$\sqrt{g_{00}} \mathcal{E} = E = \rm const$$.

Is this a problem?

In SR we have simply $$E = \mathcal{E}$$ and there is no need to worry. So, SR automatically implies that $$p^0$$ is conserved. But it is not necessary the case in GR, right? Or is this all just a question of definition/convention?

• – probably_someone Feb 14 at 14:11
• @probably_someone hmm.. I don't think I enter (or want to enter) into the discussion of mass definition here. In the above equations for me m is simply a constant scaling factor, usually interpreted as the rest mass. The question is rather about the definition of the energy... for a given mass definition. Thanks anyway! – user3356865 Feb 14 at 15:36

GR is locally the same as SR -- that's one way of stating the equivalence principle. So stuff like this is exactly the same in GR as in SR. There is actually no GR involved in your question. We can use arbitrary coordinate systems in SR. GR happens when there is curvature, which is not relevant here because we're talking about a single vector, which lives in its own neighborhood of spacetime.

A couple of aspects of your presentation are needlessly complicating things. Do yourself a favor and start working in natural units, with $$c=1$$, so you don't have to keep writing fatcors of $$c$$ all over the place. Also, there is no special reason to use spherical coordinates here. GR works with any coordinate system. If you want to reduce this to its essence, just do a 1+1-dimensional metric in a coordinate system where the metric is $$\operatorname{diag}(g_{00},g_{11})$$.

The basic issue here is that we don't just define energy as the 0th component of the energy-momentum vector. As a trivial example of why we wouldn't do this, consider the coordinate system $$(x,t)$$, where $$x$$ and $$t$$ are local Minkowski coordinates, i.e., we just swap the order. In these coordinates, energy would be the 1th component, not the 0th.

In general, energy is the timelike component of the energy-momentum. For an observer with normalized velocity vector $$u$$, the energy of an object is $$u^\mu p_\mu$$.

The identity $$E^2-p^2=m^2$$ is always valid. It equates the norm of a four-vector to a scalar, and equations of that form are valid in all coordinate systems.

• So, the energy-momentum relation is - simply speaking - a convention, e.g. m is defined as the rest mass, $p^2 = g_{ij} p^i p^j$ and $p^0 = E$ (with c=1) where E is some energy but not further described. In my example I have the additional condition that E is a constant. In that case I run into a conflict with the usual identity if the metric is not Minkowskian. I can resolve this conflict when I relax the definition of the energy E (e.g. not a constant). – user3356865 Feb 14 at 15:45
• You said $u^\mu p_\mu$ is the momentum. Isn’t it the energy? – G. Smith Feb 14 at 20:16
• @user3356865: So, the energy-momentum relation is - simply speaking - a convention No, I don't agree. and p0=E No, I gave a counterexample to this. In my example I have the additional condition that E is a constant. Not sure what you mean by this in this context. – Ben Crowell Feb 15 at 4:20
• uhh, I thought I gave enough details to understand my problem/questions. But it seems there are to many conventions between us for a proper communication. Please, let my try again in short: it is a convention to describe the zeroth component as the time component. This component then is usually associated to an energy, e.g. $p^0 = E$. For a flat metric (SR) we have ${\rm d}p^0/{\rm d}\tau = 0 \to p^0 = E = {\rm const}$, so conservation of energy. That's what I mean a constant. For a non-flat metric (GR) we have in my example $g_{00} p^0 = {\rm const}$. Here we are free to define an "energy". – user3356865 Feb 15 at 19:22
• @user3356865: The example of reordering the coordinates was just a simple example of a coordinate transformation that shows why it doesn't make sense in general to try to identify energy as one coordinate component of the energy-momentum vector. We don't even have to have any coordinate that's timelike. It's true that if we have a timelike Killing vector there is a conserved energy, but I don't see why you think that is relevant here. – Ben Crowell Feb 16 at 23:51