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## Newtonian force as a covariant or contravariant quantity

While searching for information relative to this thread, I stumbled across this on Google Books (Richard A Mould, Basic Relativity, p.258). It claims that, under some circumstances, "covariant energy" (time component of covariant 4-momentum) is globally conserved, but "contravariant energy" (time component of contravariant 4-momentum) is conserved only locally but not globally. Unfortunately the next page is missing from the preview, so it's not clear what the "some circumstances" actually are, or how you prove the claim.

Does anyone know what is the correct statement and its proof?

Well, in the case of Rindler coordinates
$$ds^2 = g^2 z^2 \, dt^2 - dx^2 - dy^2 -dz^2$$(c=1) the contravariant 4-momentum of a particle at rest is
$$P^\alpha = \begin{bmatrix} \frac{m}{gz} \\ 0 \\ 0 \\ 0 \end{bmatrix}$$whereas the covariant 4-momentum is
$$P_\alpha = \begin{bmatrix} mgz && 0 && 0 && 0 \end{bmatrix}$$Of course $mgz$ is the correct formula for "gravitational" potential energy in Rindler coordinates, whereas $m/(gz)$ has no significance that I know of.

Another non-relativistic example I can think of is cylindrical polar coordinates in Euclidean 3-space
$$ds^2 = dr^2 + r^2 \, d\theta^2 + dz^2$$For an arbitrary particle
$$x^i = \left( r(t), \theta(t), z(t) \right)$$we have a contravariant 3-momentum
$$p^i = \begin{bmatrix} m \dot{r} \\ m \dot{\theta} \\ m \dot{z} \end{bmatrix}$$whereas the covariant 3-momentum is
$$p_i = \begin{bmatrix} m \dot{r} && m r^2 \dot{\theta} && m \dot{z} \end{bmatrix}$$Here again we see that the covariant component $m r^2 \dot{\theta}$ is the conserved angular momentum, whereas the contravariant component $m \dot{\theta}$ is not conserved.

So all of the above seems to be more evidence to suggest that momentum and its time-derivative, force, are naturally covariant rather than contravariant.
 Recognitions: Gold Member Science Advisor Is the statement that the 0 component of contravariant 4 - momentum is NEVER globally conserved and can only ever be locally conserved whereas the 0 component of covariant 4 - momentum CAN be globally conserved under the appropriate conditions?

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 Quote by DrGreg While searching for information relative to this thread, I stumbled across this on Google Books (Richard A Mould, Basic Relativity, p.258). It claims that, under some circumstances, "covariant energy" (time component of covariant 4-momentum) is globally conserved, but "contravariant energy" (time component of contravariant 4-momentum) is conserved only locally but not globally. Unfortunately the next page is missing from the preview, so it's not clear what the "some circumstances" actually are, or how you prove the claim. Does anyone know what is the correct statement and its proof?
I see. Since the metric can depend on position, we can have $p'_a=p_a$ (final=initial), but $p'^a \ne p^a$, because the particle can be in different places at the initial and final times.

 Quote by DrGreg we have a contravariant 3-momentum $$p^i = \begin{bmatrix} m \dot{r} \\ m \dot{\theta} \\ m \dot{z} \end{bmatrix}$$whereas the covariant 3-momentum is $$p_i = \begin{bmatrix} m \dot{r} && m r^2 \dot{\theta} && m \dot{z} \end{bmatrix}$$Here again we see that the covariant component $m r^2 \dot{\theta}$ is the conserved angular momentum, whereas the contravariant component $m \dot{\theta}$ is not conserved.
I guess the Killing vectors $\partial/\partial \theta$ and $\partial/\partial z$ lead to the 2 conserved components of the lower-index momentum.

Here's another thing I have to try to wrap my head around: I think Killing vectors are naturally lower-index vectors, and even though you do have a metric, it doesn't make sense to try to raise their indices. The reason is that the Killing vector is a field, not just a vector defined at a point, and the metric is varying...Does that make sense!?

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 Quote by bcrowell When I say there's no metric, I mean that there's no metric on the two-dimensional space of $(\phi,\theta)$.
I see. And regarding the definition of parallel - is it defined by a connection?

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 Quote by bcrowell Here's another thing I have to try to wrap my head around: I think Killing vectors are naturally lower-index vectors, and even though you do have a metric, it doesn't make sense to try to raise their indices. The reason is that the Killing vector is a field, not just a vector defined at a point, and the metric is varying...Does that make sense!?
I'm not sure I understand this. A killing vector field is defined as a vector field such that the lie derivative of the metric tensor along this vector field is zero. We can talk of killing vector fields in terms of their flows and so on. The lie derivative measures the rate of change of a tensor field along the flow of a vector field so it makes sense to take a killing vector field as, at each point, naturally an element of the tangent space. Of course you can use the metric to raise and lower indices as usual but the way a killing field is defined it uses the notion of a vector field as per the lie derivative and the notion of a vector field as a derivation. I don't see immediately why they would instead be naturally co -vector fields a priori. Note that we raise and lower indices of vector fields all the time as it is a point wise operation done at each point in space - time. This isn't an issue.

 Quote by WannabeNewton I'm not sure I understand this. A killing vector field is defined as a vector field such that the lie derivative of the metric tensor along this vector field is zero.
In the Wikipedia article on Killing vectors, it is said that $X^\mu$ is a Killing vector field if for all vectors $Y$ and $Z$,

$g(\nabla_Y (X),Z) + g(Y , \nabla_Z (X)) = 0$

which is an equation on vector fields $X^\mu$. However, it also says that in "local coordinates", this is equivalent to

$\nabla_\mu X_\nu + \nabla_\nu X_\mu = 0$

which is an equation on covector fields $X_\mu$. I don't quite understand this, because it seems that the latter equation doesn't even involve the metric (except indirectly, through the covariant derivative). It would seem to me that the latter could be defined even for a manifold without a metric, with just a connection, as a definition of a Killing co-vector field $X_\mu$.

 Quote by stevendaryl In the Wikipedia article on Killing vectors, it is said that $X^\mu$ is a Killing vector field if for all vectors $Y$ and $Z$, $g(\nabla_Y (X),Z) + g(Y , \nabla_Z (X)) = 0$ which is an equation on vector fields $X^\mu$. However, it also says that in "local coordinates", this is equivalent to $\nabla_\mu X_\nu + \nabla_\nu X_\mu = 0$ which is an equation on covector fields $X_\mu$. I don't quite understand this, because it seems that the latter equation doesn't even involve the metric (except indirectly, through the covariant derivative). It would seem to me that the latter could be defined even for a manifold without a metric, with just a connection, as a definition of a Killing co-vector field $X_\mu$.
Huh?, Killing (co)vectors are those that preserve the metric tensor, how can you define them if there is no metric?

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 Quote by stevendaryl In the Wikipedia article on Killing vectors, it is said that $X^\mu$ is a Killing vector field if for all vectors $Y$ and $Z$, $g(\nabla_Y (X),Z) + g(Y , \nabla_Z (X)) = 0$ which is an equation on vector fields $X^\mu$. However, it also says that in "local coordinates", this is equivalent to $\nabla_\mu X_\nu + \nabla_\nu X_\mu = 0$ which is an equation on covector fields $X_\mu$.
Yeah you can express them that way in local coordinates as you prolly already know because when you compute the first expression in coordinates the metric tensor ends up lowering the indices so I don't know if that would count as making the co - vector expression any more natural.

 Quote by bcrowell Here's another thing I have to try to wrap my head around: I think Killing vectors are naturally lower-index vectors, and even though you do have a metric, it doesn't make sense to try to raise their indices. The reason is that the Killing vector is a field, not just a vector defined at a point, and the metric is varying...Does that make sense!?
I don't understand this. What varying metric are you referring to? why doesn't make sense to raise or lower indices for fields if there is a metric?, fields assign a vector or covector to each point in the manifold.
Why the insistence on what is natural for a certain field when examples keep popping up that contradict any abstract "naturalness" in the presence of a metric of the covariance or contravariance of (co)vector fields?

 Quote by TrickyDicky Huh?, Killing (co)vectors are those that preserve the metric tensor, how can you define them if there is no metric?
I'm just saying that the equation for a Killing co-vector doesn't mention the metric tensor, and so the equation would make sense even in the absence of a metric, it would seem.
 Recognitions: Gold Member Science Advisor Staff Emeritus Trying to clear up my foggy thinking re Killing vectors... The Killing equation $\nabla_a \xi_b+\nabla_b \xi_a=0$ does use the metric, because the covariant derivative is defined in terms of the metric. Likewise if you define Killing vectors in terms of preserving distances, you're still appealing to a metric. On a bare manifold without a connection, there is no way to define a Killing vector. For instance, if I take the spacetime in and around the planet earth, it has Killing vectors such as rotation and time translation. If I take away the metrical information, then all I have is a topological space that's topologically isomorphic to R^3, and there's no way to say what its Killing vectors are. You can raise the indices like this $\nabla^a \xi^b+\nabla^b \xi^a=0$. Note how the structure of the equation forces you to raise the indices on the derivatives as well as the Killing vector. If we're thinking of derivatives as being "naturally" lower-index quantities, then this does provide some motivation for saying that Killing vectos are naturally lower-index. But the fact that this is a *covariant* derivative means that you must have a metric defined, and therefore there's nothing to stop you from raising its index. This is different from the case of a plain old partial derivative. People normally express Killing vectors using partial derivatives as a basis, e.g., $\partial_t$ for a metric that doesn't change over time. So if there is a "natural" way to write them, it would definitely have to be lower-index. On the other hand, it would also seem extremely natural to me to express the same symmetry as a translation of the time coordinate, $x^t \rightarrow x^t+dt$. When you have a Killing vector $\xi_a$, test particles follow trajectories that conserve $\xi_a v^a$. This seems odd to me because we normally think of momentum as being what's conserved, not velocity. We can multiply by m to get a momentum, but then it would be an upper-index momentum, which is not the natural way to express momentum. I would like to connect this to DrGreg's #103, which wasn't explicitly written in terms of a discussion of Killing vectors.

 Quote by bcrowell Trying to clear up my foggy thinking re Killing vectors... The Killing equation $\nabla_a \xi_b+\nabla_b \xi_a=0$ does use the metric, because the covariant derivative is defined in terms of the metric.
But covariant derivative only requires a connection, not a metric. A metric is sufficient for a connection, but not necessary.

 Quote by stevendaryl But covariant derivative only requires a connection, not a metric. A metric is sufficient for a connection, but not necessary.
The covariant dervative used with Killing fields is not a general connection but the Levi-Civita unique metric connection.

 Quote by stevendaryl I'm just saying that the equation for a Killing co-vector doesn't mention the metric tensor, and so the equation would make sense even in the absence of a metric, it would seem.
The equation doesn't make sense without a metric, the metric is implicit in the Christoffel coefficients of the covariant derivatives of the equation, you need the metric to obtain them.

 Quote by TrickyDicky The equation doesn't make sense without a metric, the metric is implicit in the Christoffel coefficients of the covariant derivatives of the equation, you need the metric to obtain them.
No, connections are more basic than metrics. It's possible to have a connection even when you don't have a metric.

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