Why Two Defs of Four-Momentum?

[LarryS]

Apparently there are 2 definitions of four-momentum: (E, p), the contravariant form and (E, -p), the covariant form. Why is that, and are these 2 different vectors physically equivalent? Thanks in advance.

 

[HallsofIvy]

It's not just four-momentum. Every vector has both a covariant and contravariant form. It is, strictly speaking, better to refer to the covariant and contravariant components of a vector rather than covariant and contravariant vectors.

 

[LarryS]

It's not just four-momentum. Every vector has both a covariant and contravariant form. It is, strictly speaking, better to refer to the covariant and contravariant components of a vector rather than covariant and contravariant vectors.

Ok, but are (E,p) and (E,-p) considered to be elements of the same 4-D vector space?

 

[bcrowell]

Ok, but are (E,p) and (E,-p) considered to be elements of the same 4-D vector space?

No, they belong to two different vector spaces. One easy way to see this is that one is represented by a symbol with an upper index, the other lower. Since $p^i+p_i$ is ungrammatical, they clearly can't belong to the same vector space.

 

[PAllen]

Actually, I have an old relativity book which defines two momentum vectors, one contravariant, one covariant, but they are not components of the same vector. The covariant version is used (in this book) in the treatment of Lagrangians and Hamiltonians. They are defined as :

contravariant: E/c^2, p
covariant: -E, p

The covariant form corresponding to the contravariant form would be, instead:

E/c^2, -p/c^2

so the covariant version is -c^2 times the covariant version of the contravariant momentum 4-vector. In the OP's case, asssuming c=1, the vectors are covariant/contravariant components of the same vector, otherwise, they are not.

Anyway, my question is anyone seen anything like the version from my old book, and does anyone still use these definitions?

 

[dextercioby]

The 4-momentum of a point-particle A in a point q of the spacetime manifold M is a covector, i.e. a linear functional from the tangent space to the manifold of all the possible configurations of the particle A to the reals. So there's really one definition. Since the spacetime manifold admits a global metric, the tangent space to a point of it is isomorphic to the contangent space of the same point of it (the second algebraic dual of the tangent space is obviously isomorphic to the tangent space), so that's why one can describe the 4-momentum as a 4-vector as well. It is actually a 1-form.

$$P^{4} (q) = p_{\mu} dX^{\mu}(q)$$

The $dX^{\mu}$ form a basis of 1-forms in the cotangent space at "q" of the configuations manifold.

 

[PAllen]

The 4-momentum of a point-particle A in a point q of the spacetime manifold M is a covector, i.e. a linear functional from the tangent space to the manifold of all the possible configurations of the particle A to the reals. So there's really one definition. Since the spacetime manifold admits a global metric, the tangent space to a point of it is isomorphic to the contangent space of the same point of it (the second algebraic dual of the tangent space is obviously isomorphic to the tangent space), so that's why one can describe the 4-momentum as a 4-vector as well. It is actually a 1-form.

$$P^{4} (q) = p_{\mu} dX^{\mu}(q)$$

The $dX^{\mu}$ form a basis of 1-forms in the cotangent space at "q" of the configuations manifold.

Could someone try to translate this a little bit to older concepts of tensor analysis? I get basically nothing out of this, even consulting my books.

 

[pervect]

I think he's saying that the component of momentum in a particular direction is a map from a vector (a velocity) to a scalar (some number). Under the right circumstances, saind number will represent a conserved quantity. Though he didn't actually mention the part about it being a conserved quantity, that would take a much longer post to do it rigorously :-).

 

[PAllen]

The 4-momentum of a point-particle A in a point q of the spacetime manifold M is a covector, i.e. a linear functional from the tangent space to the manifold of all the possible configurations of the particle A to the reals. So there's really one definition. Since the spacetime manifold admits a global metric, the tangent space to a point of it is isomorphic to the contangent space of the same point of it (the second algebraic dual of the tangent space is obviously isomorphic to the tangent space), so that's why one can describe the 4-momentum as a 4-vector as well. It is actually a 1-form.

$$P^{4} (q) = p_{\mu} dX^{\mu}(q)$$

The $dX^{\mu}$ form a basis of 1-forms in the cotangent space at "q" of the configuations manifold.

Anyway, I don't see how this answers my question at all. I did find references about 1-forms (just seems like a new way of looking at covariant vectors, to me). Phrased this way, this book is still using a momentum 1-form that does not correspond to the unique 1-form associated with the momentum 4-vector; it is close, but no cigar. My question is whether there is some good reason for this, and has anyone else seen this, or is just an obsolete idiosyncracy of the author?

(I can see exactly one 'value' to it; in the case when c is not taken to be 1, both forms contain p, rather then one needing p/c^2).

 

[pervect]

I've seen people get passionate about the topic. If you define momentum from a Lagrangian (which is the natural way to do it in generalized coordinates), it seems more natural to define it as a co-vector, given that you generally think of velocities as tangent vectors to curves and hence as being vectors.

I.e. $p_{x} = \partial L / \partial \dot{x}$ basically says you're associating a scalar with a vector, given that you think of $\dot{x}$ as a vector. The same argument works with polar coordinates - angular momentum is $\partial L / \partial \dot{\theta}$. Given that you can convert vectors to their duals freely, though, it doesn't make a lot of difference.

[add]You'll also see the complementary relation between momentum and position in quantum mechanics

 

[LarryS]

The 4-momentum of a point-particle A in a point q of the spacetime manifold M is a covector, i.e. a linear functional from the tangent space to the manifold of all the possible configurations of the particle A to the reals. So there's really one definition. Since the spacetime manifold admits a global metric, the tangent space to a point of it is isomorphic to the contangent space of the same point of it (the second algebraic dual of the tangent space is obviously isomorphic to the tangent space), so that's why one can describe the 4-momentum as a 4-vector as well. It is actually a 1-form.

$$P^{4} (q) = p_{\mu} dX^{\mu}(q)$$

The $dX^{\mu}$ form a basis of 1-forms in the cotangent space at "q" of the configuations manifold.

Someday I hope to be versed in differential geometry. Are you saying that (E,p) and (E,-p) are mathematically equivalent? As simple vectors, they have different components, so how can they be the same thing?

 

[atyy]

Someday I hope to be versed in differential geometry. Are you saying that (E,p) and (E,-p) are mathematically equivalent? As simple vectors, they have different components, so how can they be the same thing?

They are not the same thing.

Let v be a column vector
The metric g is something that gives the length of column vectors g(v,v).

g(v, ) is not a column vector, but a row vector V.

Since V is a row vector it can act on the column vector v to produce the length of v.

V and v are different, but related by the metric.