# I The nature of four-momentum in general relativity

1. Apr 22, 2016

### davidbenari

Conserved quantities in GR deal with $p_\mu$ not $p^\mu$ and while in Minkowski spacetime its easy to see what each of the components mean (since the metric is so simple) in general relativity I think its not and its starting to confuse me.

Why exactly is $-p_0$ the energy in general relativity? Is this because of the equivalence principle making the analogy to special relativity? What exactly does this energy component include? Only kinematic energy, what else can it include and how?

Also, many terms like $p_\phi$ arent exactly angular momentum even if they are referred to as such. Should I think of this sloppy language like we take generalized momenta in Lagrangian mechanics?

My main dilemma is with respect to the energy component of the four momentum (with covariant indices, that is, as a form)

Thanks.

2. Apr 22, 2016

### Staff: Mentor

As long as you have a metric, you can always raise and lower indexes, so this isn't really a meaningful distinction.

It isn't unless two conditions are met: first, the spacetime must have a timelike Killing vector field, and second, you must be using coordinates in which the "0" basis vector is tangent to the orbits of the Killing vector field. The first condition is necessary for there to be a meaningful notion of "energy" as a conserved quantity; the second is necessary for the "0" component of $p$ to be the only relevant one for the energy.

No. It doesn't have anything to do with the EP.

The more precise name for it is "energy at infinity", and it is only a constant of motion for geodesic motion (i.e., motion solely under gravity) in a stationary spacetime (i.e., one with a timelike Killing vector field, as above). Heuristically, it can be thought of as telling you how tightly an object is bound: an object in a bound orbit around a massive object will have an energy at infinity less than its rest mass, and the smaller the ratio of energy at infinity to rest mass, the more tightly it is bound. (Conversely, energy at infinity larger than rest mass indicates an object that is not bound at all--it is coming in from infinity and will escape to infinity.)

Why do you say "aren't exactly"?

3. Apr 23, 2016

### davidbenari

Well $p^\phi$ is $m\frac{d\phi}{d\tau}$, but if I have some complicated metric then $p_\phi$ wont "exactly" be $m\frac{d\phi}{d\tau}$ (rather it would be some complicated expression with the metric terms), so I don't know why people refer to it as the angular momentum.

Why do you think its not a meaningful distinction? The conservation of $p_u$ need not imply the conservation of $p^\mu$ so I think they are different in a meaningful way. For me $p^{\mu}$ seems more directly connected to equations of basic physics than $p_u$.

4. Apr 23, 2016

### Staff: Mentor

Because angular momentum is not $m d \phi / d\tau$ in general, it just happens to be equal to that in special cases (the same ones where the metric is simple enough that $p_\phi$ is equal to $p^\phi$).

First of all, there is the same issue here as with energy at infinity: $p_\phi$ is only a constant of geodesic motion if the spacetime has an axial Killing vector field and the coordinates are chosen such that the $\phi$ basis vector is tangent to orbits of that axial Killing vector field. The first condition is necessary for there to be an "angular" constant of motion at all; the second is necessary for the $\phi$ component of $p$ to be the only relevant one for that constant of motion. Given those conditions, the constant $p_\phi$ works like the usual constant $L$ for geodesic orbits in Newtonian mechanics, so it is called "angular momentum" (or more precisely "orbital angular momentum") for the same reasons that $L$ is in Newtonian mechanics.

Second is understanding the correct expression for the angular momentum constant of the motion (and indeed for energy at infinity as well). See further comments below.

We need to be more precise here about what "conservation" means. Strictly speaking, what is conserved is a scalar quantity, not a "component" of a vector or 1-form; that scalar only looks like a particular component of a vector or 1-form because we chose particular coordinates. But the conservation laws in question are independent of any choice of coordinates, so to really understand them we need to express them in a form that is independent of coordinates.

The way to do that is to use the fact that I've already referred to, that each conserved quantity corresponds to a Killing vector field. So if energy is conserved, that means there is a timelike Killing vector field, which we can call $T^\mu$. The conserved quantity "energy at infinity" is then simply the contraction $T^\mu p_\mu$. Similarly, if angular momentum is conserved, that means there is an axial Killing vector field $\Phi^\mu$, and the conserved quantity "angular momentum" is the contraction $\Phi^\mu p_\mu$. These expressions are scalar invariants (like any contraction of a vector and a 1-form), so they are true independently of any choice of coordinates.

If we choose coordinates such that $T^\mu = (1, 0, 0, 0)$ (i.e., the "0" basis vector is tangent to the KVF), then we find that $T^\mu p_\mu = - p_0$ (assuming we are using the -+++ metric sign convention). Similarly, if we choose coordinates such that $\Phi^\mu = (0, 0, 0, 1)$ (where the "3" component is the $\phi$ component), then we find that $\Phi^\mu p_\mu = p_3 = p_\phi$. But those expressions are only valid in the chosen coordinates; they aren't valid generally.

The above should also answer the question of why the conserved quantities are usually written with a lower index, i.e., as 1-form components: because in order to obtain the coordinate-independent scalar invariant, we have to contract the object's 4-momentum with the appropriate Killing vector field. The latter is a vector field (at least in its "natural" formulation), so we have to use the 4-momentum 1-form.

However, it is perfectly possible to swap the indexes using the metric, and write the conserved quantities as $T_\mu p^\mu$ and $\Phi_\mu p^\mu$. These must be the same conserved quantities--i.e., we must have $T_\mu p^\mu = T^\mu p_\mu$ and similarly for the other one--regardless of what the metric is. This is true of any scalar formed by contracting tensors: swapping the positions of contracted indexes leaves the scalar invariant. That is why I said it doesn't really matter which one you consider to be a vector and which you consider to be a 1-form. Indeed, with the metric, you can contract two vectors, like so: $g_{\mu \nu} T^\mu p^\nu$. This is still the same scalar as the other two expressions. (Similarly, you can use the inverse metric to contract two 1-forms.) All of these expressions represent the same physics, so there's no reason to worry too much about index placement.

5. Apr 23, 2016

### davidbenari

Thanks. I think I've got it now. The thing is the course I'm taking (based on Bernard Schutz's book) only marginally mentions Killing vectors. I guess I'll read more about them as they seem to be really important to this discussion.