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Do all four-vectors live in a tangent space?

  1. Aug 3, 2012 #1
    Working through intro GR at the moment and I'm a little unclear on how tangent spaces are used to carry four-vectors over from SR to GR.

    So, at every point in spacetime we construct a tangent space. We can form a basis for this space with the tangent vectors (i.e. the four-velocities) of one time-like and three space-like worldlines passing through this point (am I doing OK so far?) Thus, we can construct various vectors and one-forms in this tangent space and, if a metric is provided, turn one into the other. Extend it to tensors, and we're done!

    Except...what about four-vectors that (as far as I can tell) can't be expressed as a tangent vector times some scalar? Where do they "live"? For instance: the four-potential. It's a four-vector because its components transform like the spacetime coordinates, not because they are spacetime coordinates. With something like four-velocity and four-momentum, I can imagine a vector in a tangent space. Since the components of four-momentum are just [itex]p^\alpha=mU^\alpha[/itex], I can actually think of the four-momentum as [itex]p=p^\alpha e_{(\alpha)}[/itex], where the [itex]e_{(\alpha)}[/itex] are the tangents to four worldlines. Can I do that with something like the four-potential?

    I guess I can rephrase my question like this, if that's not clear: if the components of four-potential at some point in spacetime are [itex](\phi,\vec{A})[/itex], what is the basis that those components refer to? Is it a basis of tangent vectors at that point? If that's not the case, and not every four-vector lives in a tangent space, then how do such non-tangent vectors carry over into GR?
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  3. Aug 3, 2012 #2
    I wouldn't say you construct one-forms in the tangent space. Rather, you construct a basis of cotangent vectors in the cotangent space, and this is where the electromagnetic four-potential naturally lives.

    Otherwise, yes, the components refer to the same "worldlines" you used to determine a basis for momentum. It's unusual to call them worldlines; I would just say axes. The basis vectors in the tangent space are tangent to their respective axes.
  4. Aug 4, 2012 #3
    Thanks for the reply!
  5. Aug 4, 2012 #4
    Hm, OK: one thing I'm still trying to wrap my head around with respect to this tangent business. If the four-potential at a point does in fact live in the tangent space at that point, then it must, by definition, be expressible as the tangent to some worldline through that point (in this case I really do mean worldline in the usual sense). But a worldline is a coordinate trajectory parametrized by proper time, so its tangent vector has units of [distance]/[time]. But those aren't units of the four-potential. How does that work out, then? If the four-potential lives in tangent space, what curve is it the tangent vector to?
  6. Aug 4, 2012 #5
    Again, the four-potential lives in cotangent space, not tangent space. Units are not relevant because you can multiply any suitable cotangent vector by scalars with different units.

    Edit: note that while it's common to use four-velocity as [itex]dx(\tau)/d\tau[/itex], you could also consider the slightly different quantity [itex]dx(c\tau)/dc\tau[/itex], which is a closer analogue of arclength parameterization and spits out dimensionless basis vectors. This distinction is academic when [itex]c = 1[/itex], and hence, it's usually ignored.
    Last edited: Aug 4, 2012
  7. Aug 4, 2012 #6
    OK, forget about about the four-potential, since the distinction between tangent space and the co-tangent space isn't really relevant for what I'm trying to understand. Let's think about the four-current instead, which, as far as I know, is canonically a vector rather than a co-vector. What I mean is, if the four-current lives in tangent space, then by definition there must be a spacetime trajectory—a worldline—whose tangent vector is the four-current. But I don't see how that's possible since the tangent vector of a worldline is always a velocity (and four-current is not a velocity, its a scalar with units of charge times a velocity). So what's wrong in my understanding?

    As I understand it, "tangent space at a point" means "the vector space of all tangent vectors to every possible worldline through the point". Is that incorrect?
  8. Aug 4, 2012 #7
    That's fine, you're just assuming that the tangent vectors to a worldline are unique when they're not. They can be arbitrarily rescaled with scalars and can have their units changed by scalar multiplication just fine. Their directions are unique, yes, and all that means is that the four-current is some unit timelike vector multiplied by a magnitude.
  9. Aug 4, 2012 #8
    But a worldline is parametrized by proper time. That means its tangent vector is unique: its the four-velocity of a particle traveling along that worldline. If you rescaled it with a non-unitless scalar, it would no longer be a tangent vector to a worldline. I'm not arguing with you, it's just that I know at least one of the definitions I'm using must be wrong and I'm trying to figure out which one. First I worried that my definition of a four-vector was wrong; you said it wasn't. Then I thought it might be my definition of a tangent space; apparently not. I can only think that my definition of a worldline is wrong. To summarize:

    - All four-vectors at a point live in tangent space (and all four-covectors live in cotangent space)
    - A tangent space at a point is the vector space of tangent vectors to all worldlines through that point
    - A worldline is a function that maps proper time to each of the spacetime coordinates
    - The tangent vector to a worldline is thus [itex]dx^i/d\tau[/itex], which has units of velocity
    - Thus, all four-vectors have units of velocity

    Something in that chain of reasoning is wrong since the conclusion is wrong. I just don't see which part it is.
  10. Aug 4, 2012 #9
    A worldline is a path through spacetime. A worldline has a unique four-velocity at a given point, but any scalar multiple of the four-velocity is tangent to the worldline.
  11. Aug 4, 2012 #10
    Got it. Thank you for your help and patience!
  12. Aug 4, 2012 #11
    Worldlines in general are not uniquely parametrized by proper time, they can have different parametrizations so they can have multiple tangent vectors.
  13. Aug 5, 2012 #12


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    Every tangent vector can be defined as a kind of "velocity vector". If you have a parametrized path, a function [itex]P(s)[/itex] that gives a point in spacetime for every value of the parameter s, then there is a corresponding 4-vector [itex]\dfrac{dP}{ds}[/itex]. The parameter [itex]s[/itex] does not need to be time, or proper time; the only requirement is that it increase monotonically along the path.

    To go in reverse, given a 4-vector [itex]V^\mu[/itex], at a point [itex]P[/itex], we can construct a corresponding parametrized path as follows: Define [itex]x^{mu}(s) = [itex]P^\mu + V^\mu s[/itex]. (There are infinitely many parametrized paths corresponding to the vector [itex]V^\mu[/itex] if it is only defined at a single point.)

    A current 4-current [itex]J^\mu[/itex] is actually not a 4-vector, but a 4-vector field. That is, it assigns a 4-vector to each point in spacetime. A vector field [itex]J^\mu[/itex] is associated with a family of parametrized paths [itex]P(s,\lambda)[/itex], one path for each value of [itex]\lambda[/itex]. To find [itex]P(s,\lambda)[/itex] given [itex]J^\mu[/itex] is a matter of integration.
  14. Aug 5, 2012 #13


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    That line is not correct. A worldline can use any monotonically increasing real-valued parameter; it doesn't have to be proper time. Proper time is a convenient choice for massive particles, but a massless particle, such as a photon, doesn't have a proper time. It can still have a worldline, though.
  15. Aug 5, 2012 #14


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    I like this answer, and I think it directly addresses the question. I would just point out that there are multiple ways of formalizing this kind of thing, and formalizing it in terms of tangent spaces, etc., is only one way of doing it. It is not, for example, how Einstein would have done it in the early years of GR -- in fact, he probably didn't worry about formalizing it at all.
  16. Aug 5, 2012 #15
    Fair enough, though I have a difficult time understanding things like this when they aren't treated rigorously enough. I used Schutz the first time I went through GR and found it so casual that I wasn't convinced by many of the derivations.

    As another follow up question—maybe this should go in a different thread, but it seems to follow smoothly from my earlier questions—is it proper to talk to about a set of tangent spaces at a point? Four-momentum and four-current are both four-vectors (yes, ok, four-vector fields but I just care about one point of spacetime and its associated tangents) but they clearly can't live in the same vector space since they can't be added. At a particular point, every different kind of four-vector inhabits a distinct vector space from every other different kind (even if they are just related by dimensionful scaling factors). So, what then is meant by the tangent space at a point? There can't just be one.
  17. Aug 5, 2012 #16
    The only reason they can't be added is because we don't define addition for numbers of different units. This would be like saying there are multiple vector spaces even in regular old 3d space. I mean, it's not objectionable in and of itself, but I think you're getting hung up on stuff out of unfamiliarity.

    Picture a 2d surface embedded in 3d space. At any point on the surface, there is a tangent plane, and that is the tangent space at that point. How vectors of different units interact is a separate question. Even if you want to say there are multiple spaces because of units, I would say the tangent space has no units involved. It is dimensionless, and all vectors within it are dimensionless, and all vectors that do have units are built from that.
  18. Aug 5, 2012 #17


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    Note that there is nothing in this question that requires more than one dimension, so this is sort of equivalent to asking whether there is only one real line or many different ones for describing scalars having various units. I think people are unlikely to stop talking about "the" real line.

    People think of a vector space as being the space spanned by some basis. They would probably be annoyed to have to talk about different spaces having the same basis vectors.

    You can formalize all this if you like, but it would just get klutzy. A unit like m/s2 can be modeled as an element like (1,0,-2) taken from the group [itex]\mathbb{Z}^n[/itex] or [itex]\mathbb{Q}^n[/itex] under addition of the exponents, where, e.g., n=3 for the SI, which has three base units. You then take the direct product of this group with the reals, and you get a number system which models the way physicists think about unitful quantities. But physicists usually don't care about formalizing this kind of thing, and mathematicians usually don't care about units. Another reason not to bother is that implementing this kind of thing on a computer is inefficient, and it gets messy when you have to figure out rules for operations like exponentiation and logs. (It makes sense to want to take the log of a unitful quantity, e.g., you can say that the log of 37 kg is log 37 + log kg.)
  19. Aug 6, 2012 #18
    Awesome, thanks everyone!
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