What are the units of vectors and dual vectors?

[Phrak]

What are the units of vectors and dual vectors?

And where do the the units in the metric need to be placed so that $V^{\mu}V_{\mu}$ is a scalar?

Taking charge density and current density that combine as a 4-vector, $J^{\mu} \partial_{\mu}$, and it's dual vector $J_{\mu} dx^{\mu}$, which units go where on the these vector coefficients and which on their vector basis?

 

[Pengwuino]

Current is current and derivatives are inverse-length. The units still work out in 4-vector notation. With natural units there might be some confusion though I'd suspect.

 

[arkajad]

If you want $$\langle \omega,v\rangle$$ (where $$\omega$$ is a dual vector) to be dimensionless scalar that the units of dual vectors should be the inverses of the units of vectors.

There are several different conventions possible. For instance you may consider coordinates $$x^\mu$$ as dimensionless and you may want to have $$ds^2$$ having the dimension $$\mbox{length}^2$$. Then the metric should have this dimension. I do not know which is your convention.

As for the last question, you need to take into account the units of the metric when rising or lowering the indices of J.

 

[bcrowell]

Different people have different ways of looking at this. In particular, some people are happy to have a vector whose components have different units from one another, while that makes other people queasy. IMO it's more in the spirit of relativity to say that coordinates have no physical significance in and of themselves, and therefore there is no logical requirement we can impose on the units of the components of a vector. We can only impose requirements on things like the scalar product of two vectors. As a specific example, if you write the Schwarzschild metric in terms of $(t,r,\theta,\phi)$ coordinates, then clearly the metric's components will not all have the same units, nor will the components of the Christoffel symbol or the curvature tensors.

And where do the the units in the metric need to be placed so that $V^{\mu}V_{\mu}$ is a scalar?

Once you compute the components of $V^{\mu}$ and $V_{\mu}$, you will see that they have certain units. E.g., if you start from $V^{\mu}$, then lowering the indices will result in different units for $V_{\mu}$, because the components of the metric have their own units. When you take the inner product, you will find that all the terms have the same units, and they are the units you expect for the squared magnitude of V.

 

[Phrak]

All good replies. Thank you.

It seems rather natural to have the components of vectors hold units. For a displacement vector in two dimensional Cartesian coordinates,

$D = D^x \partial_x + D^y \partial_y \ .$

$D = D^k$ has units of length [L].
$D = \partial_k$ has units of $[L^{-1}]$.

However, this has the drawback that $D$ is unitless.

If we want D to have units of length, and keep the units of $[L^{-1}]$ on the bases then $D = D^k$ has units of $[L^2]$. Very unintuitive.

The only requirement, it seems, is that all $D^k \partial_k$, for each k, have the same units.

At this point I'm rather drawn to the idea of having the one-forms of displacement assigned units of length, and the basis carrying all the units on the right hand side. Has anyone heard of this convention, or does it have it's own drawbacks?

 

[bcrowell]

$D = D^x \partial_x + D^y \partial_y \ .$

$D = D^k$ has units of length [L].
$D = \partial_k$ has units of $[L^{-1}]$.

$\partial_k$ has nothing to do with D; it's simply a basis vector. The thing you need to talk about, and that you never talk about above, is $D_k$.

 

[Phrak]

All good replies. Thank you.

It seems rather natural to have the components of vectors hold units. For a displacement vector in two dimensional Cartesian coordinates,

$D = D^x \partial_x + D^y \partial_y \ .$

$D = D^k$ has units of length [L].
$D = \partial_k$ has units of $[L^{-1}]$.

$\partial_k$ has nothing to do with D; it's simply a basis vector. The thing you need to talk about, and that you never talk about above, is $D_k$.

I should have proof-read better. This was intended to say.

$D = D^x \partial_x + D^y \partial_y \ .$​

$D^k$ has units of length [L].
$\partial_k$ has units of $[L^{-1}]$.​

Do you have any objections to this or what follows?

 

[bcrowell]

Different people may have different ways of thinking about this, which can make it confusing (see arkajad's #3 and my #4).

Let's say p is a momentum vector, and let's say we're working in Cartesian coordinates, so that the metric is simply a unitless (1,-1,-1,-1). Then $p^\mu=g^{\mu\nu}p_\nu$ seems to give $p^\mu$ and $p_\nu$ the same units, and I would think these would just be units of momentum. Then when you do $p^\mu p_\mu$, you get a scalar with units of momentum squared, which makes sense.

Now suppose you do the same things as in the preceding paragraph, but in cylindrical coordinates $(t,z,r,\theta)$. Then the metric is $(1,-1,-1,-r^2)$; it's got different units in its different components. A momentum vector would also have different units in its components, but when you computed $p^\mu p_\mu$, you would still get something with units of momentum squared.

I dunno, maybe you have a different way of looking at it that works equally well, but I'd have to see it spelled out in more detail. You've only discussed upper-index vectors, not lower-index ones. It seems to me that the one absolute requirement is that when you take a norm like $p^\mu p_\mu$, you ought to get a scalar with the right units. I don't really see how that can happen in the system you're talking about.

 

[Phrak]

All very challenging thoughts, bcrowell. I have an early night tonight, and other constraints, so I'll make this brief for now. I have to confess, I thought everyone else had a good handle on this and I was behind, so I've been reluctant ask.

We spend so much time with "natural units", or no units at all, this gets swept under the rug. The key word you brought in is consistency. I'm trying to see how this is really true, and if so, hows any system of units can be made consistent.

One constraint on consistency, I gave above. Another seems to be consistency with multiple indices. If a dual vector has a particular set of units, the metric with lower indices should have these units squared. And more, where units, [A] are assigned to a vector, and units to its bases, then a (k,0) tensor should have A^k units for it's components and B^k for its bases.

 

[arkajad]

D
I dunno, maybe you have a different way of looking at it that works equally well, but I'd have to see it spelled out in more detail.

I would take the metric $$g_{\mu\nu}$$ to have the dimension [length^2], coordinates dimensionless, covariant momentum has dimension [mass x length], contravariant momentum [mass/length]. Then $$p_\mu p^\mu$$ has dimension [mass^2]. With this convention there are no problems with radial or whatever coordinates.

 

[bcrowell]

So far we have two different self-consistent systems, my #8 and arkajad's #10.

One constraint on consistency, I gave above. Another seems to be consistency with multiple indices. If a dual vector has a particular set of units, the metric with lower indices should have these units squared.

I don't think this is required for consistency. The system I gave in #8 has a metric that is typically unitless (in Cartesian coordinates), so it doesn't satisfy this constraint. The system arkajad gave in #10 has upper-index(=dual=contravariant) vectors that are unitless, but a lower-index metric that has units of length squared, so it also doesn't satisfy this constraint.

 

[arkajad]

The system arkajad gave in #10 has upper-index(=dual=contravariant) vectors that are unitless, but a lower-index metric that has units of length squared, so it also doesn't satisfy this constraint.

My contravariant vectors are not necessarily unitless. If a covariant vector has some physical dimension, then physical dimensions of its contravariant version is the product of the covariant dimension and of the dimension of the contravariant metric, which is = area^-1.

 

[bcrowell]

My contravariant vectors are not necessarily unitless. If a covariant vector has some physical dimension, then physical dimensions of its contravariant version is the product of the covariant dimension and of the dimension of the contravariant metric, which is = area^-1.

I think Phrak was referring to contravariant displacement vectors, $dx^\mu$. Those are unitless in your system, right?

 

[Phrak]

I don't think this is required for consistency.

Good catch. I missed that.

I dunno, maybe you have a different way of looking at it that works equally well, but I'd have to see it spelled out in more detail. You've only discussed upper-index vectors, not lower-index ones. It seems to me that the one absolute requirement is that when you take a norm like $p^\mu p_\mu$, you ought to get a scalar with the right units. I don't really see how that can happen in the system you're talking about.

Here's what I had in mind. I'm very much drawn to the idea of attaching the units of space and time displacements to the bases and units, that are not associated with the manifold, to the components. If you could consider the charge density three-form, $\rho$ with units of charge per unit volume,
$$\rho = \rho_{ijk} dx^i \wedge dx^j \wedge dx^k$$

then $\rho_{ijk}$ has units of charge and
$dx^i \wedge dx^j \wedge dx^k$ has units of volume.

This convention would have a metric with lower indices having units of length². A dual vector would have units of length, and a vector have units of length^-1.

The vector units may initially sound odd, but the directional derivative,

$$\frac{d}{d\lambda} =\frac{dx^i}{d\lambda}\partial_i$$

would have units length^-1 when the parameter $\lambda$ on the spacetime manifold is c$\tau$. Vectors are scaled directional derivatives of the proper time.

This may not hold up will, considering

1) time
2) angular displacement, as you've both touched upon
3) bases (vierbein?) that are not coordinate bases.​

A. Replacing dt with d(ct) would fix the first. I don't know about the others yet.

B. Other than mathematical constraints, are there physical constraints that we've ignored?

C. Also, my concern, all along has been a transitioning of units between spaces and subspaces. What happens to the assigned units when we look at a submanifold of spacetime, or consider spacetime as a submanifold.

D. What happens to the momentum 4-vector in a particular system when it is partitioned into time and space components? (My concern has been charge and current density--but it's similar). There are at least two ways to look at 4-momentum--as mass times velocity, or as inverse length and time scaled by Planck's constant.

There so many interesting ways to go with all this.

By the way, thanks to the both of you I resolved a long standing problem I've had with the 4-current. I made an error! The combination of charge density and current density is not a vector but a three-form. Applying a units convention resolved it.

 

[arkajad]

I think Phrak was referring to contravariant displacement vectors, $dx^\mu$. Those are unitless in your system, right?

Yes. But I should add that in fact I would not unite space and time dimensions assuming c dimensionless. I would keep Time and Length as separate, c with the dimension $$LT^{-1}$$, so that

$$\frac{ds}{dt}=\frac{c}\sqrt{1-\beta^2}$$

Here t is the "observers time". Planck constant has the dimension $$ML^2 T^{-1}$$, electric charge $$L^{3/2}M^{1/2}T^{-1}$$

In short, I would take the conventions from http://www.dma.unifi.it/~modugno/2-ricerca/papers/CovariantQuantumMechanics/EiQuFu.pdf"

 

[Phrak]

I'm perplexed. Given to the notion that tensors inherit their units from the underlying manifold, then vectors should have units of some independent quantities multiplied by L^-1. Where, by definition of a covector, the inner product of a vector with it's dual covector, invoking this inheritance idea, the units of the covector would contain L. Whether the vector is a complex probability amplitudes or not, the units inherited from the manifold will cancel.

The perplexing part involves this notion of 'inheritance'. I have not idea whether it is meaningful, or if it's pure imagination on my part.

 

[arkajad]

I'm perplexed. Given to the notion that tensors inherit their units from the underlying manifold.

Not necessarily. You can have tensors with values in whatever space you want. Your choice.

 

[Phrak]

I'm trying to codify this notion of inheritance. It may be rubbish. I don't know yet.

I understand that there could be additional useful values, independent of the underlying manifold, or we couldn't have things like charge density where the manifold does not have a dimension identified with electric charge.

The directional derivative upon a manifold maps R to Rⁿ. The set of directional derivatives comprises a vector space.

[text]A \frac{d}{d\lambda} = A \frac {d x^{\mu}} {d\lambda} \frac{\partial}{\partial x^{\mu}}[/tex]

Textbook stuff.

 

[Rasalhague]

I think Phrak was referring to contravariant displacement vectors, $dx^\mu$. Those are unitless in your system, right?

Hmm... Aren't differentials (exterior derivatives), including those of coordinate functions, classified as covector fields (synonyms: covariant vector fields, dual vector fields, cotangent vector fields, tangent covectors fields; rather than contravariant ones), since their values are scalar-valued linear functions of tangent vectors? My current strategy for reconciling the "a differential is an infinitesimal displacement" with "a differential is an exterior derivative function" is to guess that when I see expressions like $dx^{\mu_0}$ called infinitesimal displacements, what is being referred to is actually the scalar value of these function after they've acted on a notional tangent vector. So, for example, while some authors regard $ds^2 = g_{\mu\nu} dx^\mu \, dx^\nu$ as a "jargon notation" for $\textbf{g} = g_{\mu\nu} \, dx^\mu \otimes dx^\nu$, the more traditional meaning is, I'm guessing,

$$\left \| \textbf{s} \right \|^2=\left \| s^\mu \, \partial_\mu \right \|^2$$

$$= \textbf{g}(\textbf{s},\textbf{s})$$

$$= g_{\mu\nu} \, dx^\mu \otimes dx^\nu (s^\mu \, \partial_\mu, s^\mu \, \partial_\mu)$$

But perhaps I'm still adrift here.

I'd have expected a displacement vector to look like this: $s^\mu \, \partial_\mu$ (i.e. a contravariant). If we choose bigger units along one coordinate curve $x^{\mu_0}$, widening the gaps between unit marks along the $x_{\mu_0}$ curve, the coefficient of a vector representing length should get smaller. Say we have the coordinate transformation $y^\mu = x^\mu /2$. How does this affect the representation of a contravariant vector in a coordinate basis?

$$s^\nu = \frac{\mathrm{d} y^\nu}{\mathrm{d} x^\mu} s^\mu = \frac{1}{2} s^\mu.$$

Its coefficients are cut by half. Seems pretty length-like/displacementish behaviour to me ;-) On the other hand, the coefficients of a covariant vector get bigger in a dual coordinate basis, in this case,

$$s_\nu = \frac{\mathrm{d} x^\mu}{\mathrm{d} y^\nu} s_\mu = 2 s_\mu.$$

Just how we'd expect a 1d density to behave: whatever is the inverse of displacement.

It seems bizarre to reverse this and think of geometric objects that behave like length as "having units of inverse length" and vice-versa. But maybe "having units of length" is a different matter to "having the nature of length". If covariant and contravariant objects are all infinitesimal things, derivatives and things that can be integrated to give macroscopic properties, then I suppose they all have a density-like quality in that sense.

If it's just a matter of bookkeeping, and there are many, equally valid ways of chosing units, maybe it's best to regard them as belonging to coordinate systems, rather than to geometric objects such as tensors themselves.

 

[arkajad]

While $$s^\mu\partial_\mu$$ is a displacement vector, $$dx^0$$ is a "gauge" of the displacements that is being derived form the coordinate function $$x^0$$. Vectors have no values by themselves. Forms give values to vectors - they define a "grid" by which displacements can be converted into pure numbers. So each coordinate system defines such a grid. Densities are objects that are multiplied by an appropriate power of the Jacobian when we replace one such coordinate grid by another grid.

 

[Phrak]

It all makes sense, but I still have some nagging doubts. So I'm trying to take this to the seemingly absurd to see if it still holds-up: Would there be anything wrong, mathematically or physically, with assigning units of Celsius, for instance, to a displacement vector on the spacetime manifold?

To accomplish this, I assign units of Celsius to displacement vectors as follows.

A spacetime vector is a map from a parameter, λ in R to spacetime, R⁴. λ has units of Celsius^-1. I define a displacement vector, d/dλ. d/dλ has units of Celsius.

 

[arkajad]

Well, You can do it, but with some care. First of all you want $\lambda$ to have the dimension of the "inverse of the temparature". With this you still cannot differentiate with respect to $\lambda$. Now you choose a basis in the $\lambda$ space and you call it 1/Celsius. Now you can differentiate with respect to $\lambda$ and you can assign to your displacement the unit of Celsius.

 

[Phrak]

Yes, that's what I had in mind. In like fashion, given freedom of choice,vector coefficients have cents squared. Continuing on, the covectors have unitless coefficients and bases in Celcius, and the length of a yard is measured in cents. So something seems wrong.

 

[arkajad]

Nothing is wrong. You can have vectors in cents and other vectors in yards. They live in different spaces. There is no way to add them.

You can have vectors with values in matrices 2x2 and other vectors with values in matrices 10x10. You can't add them. They live in different spaces. Yet they all come from tensoring your original vector space by an auxiliary vector space. This tensoring operation gives your original vectors a "dimension".