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Basic math operations on 4-vectors

  1. Mar 29, 2008 #1

    Dale

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    If a and b are four-vectors then are ka and a+b also four-vectors?

    My question arises because of the four-velocity, which always has magnitude c. So the sum or difference of two four-velocities will not generally be a four-velocity, but will it be a Lorentz invariant four-vector? If so, does it have any physical interpretation?
     
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  3. Mar 29, 2008 #2

    Jonathan Scott

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    Yes, they are four-vectors, but not unit-magnitude four-vectors. An example of a meaningful four-vector of this type is a four-momentum vector (where the timelike part is energy).
     
  4. Mar 29, 2008 #3

    tiny-tim

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    Actually, in a sense, they are unit-magnitude four-vectors …

    As DaleSpam says, they always have magnitude c. :smile:
     
  5. Mar 29, 2008 #4

    pam

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    The sum of two four-velocity vectors is also a four-velocity vector with magnitude c.
     
  6. Mar 30, 2008 #5

    Jonathan Scott

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    No. This isn't correct. What you are probably thinking of is the way in which relative velocities combine via Lorentz transformations, which for small speeds is similar to the way in which Newtonian velocities add together.

    It is true that the velocity of any frame relative to any other is also a four-velocity vector with magnitude c. However, the method of calculating the velocity of one frame relative to another moving frame is not to add them as four-vectors.
     
  7. Mar 30, 2008 #6

    George Jones

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    Yes.

    The set of all 4-vectors is a vector space, so any linear combinations of 4-vectors (4-velocities or not) is also a 4-vector.

    If a and b are 4-velocities, then:

    a+b is a future-directed timelike 4-vector, but, as Jonathan Scott said, it is not a 4-velocity, since its Lorentz invariant magnitude is strictly greater than c;

    a-b is (for a not equal to b) a spacelike 4-vector, since its Lorentz invariant magnitude is strictly negative.
     
  8. Mar 30, 2008 #7

    robphy

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    In fact, neither the sum nor the difference of two 4-velocities [i.e. future-timelike unit vectors] is ever a 4-velocity. [Upon first "Preview", I saw George's post.]

    Using natural units...
    [tex]\hat a\cdot \hat a =1[/tex],
    [tex]\hat b\cdot \hat b =1[/tex]
    but
    [tex]

    1\quad \stackrel{?}{=}\quad
    (\hat a+\hat b)\cdot (\hat a+\hat b)
    = \hat a\cdot \hat a + \hat b\cdot \hat b +2 \hat a\cdot \hat b
    = (1)+(1) +2(1)(1) (\cosh \theta_{a,b})
    [/tex]
    where we used Minkowskian geometry explicitly in the last step.

    Since [itex]\gamma = (\cosh \theta_{a,b}) \geq 1[/itex], the right-hand side can never be equal to 1.
    You can do a similar calculation for the difference and for a scaling.

    Geometrically, you are adding two vectors [using the parallelogram rule] whose tips lie on the future unit-hyperbola. The resulting sum is never on that hyperbola. Another interpretation is that there are NO EQUILATERAL TRIANGLES with timelike-legs.

    Can you see the Euclidean analogue of the question [which has a slightly-different answer]?


    A 4-velocity can be thought of a single tick of that observer's clock.
    [Strictly speaking, a 4-velocity refers to the tangent space whereas proper-time intervals refer to the spacetime.]

    So, a sequence of two 4-velocities [to be later added tip-to-tail] can be thought of as two legs of a trip...each one-tick at its corresponding velocity. The sum of those 4-velocities [which is not a 4-velocity] can be thought of as the proper-time for an inertial observer who would have made the trip from start event to end event.
     
    Last edited: Mar 30, 2008
  9. Mar 30, 2008 #8
    That's very interesting - I've never run across it before. I'm wondering about your final sentence, however. When you speak of the "start event" and "end event", am I right to understand that you mean the tail of the first 4-velocity vector and the tip of the second, respectively (where I've drawn them tip-to-tail, as you suggest)? I guess I'm having a little trouble since we're drawing velocity vectors on a space spanned by space-time vectors - that's not strictly correct is it? Is that what it means that "a 4-velocity refers to the tangent space"? Would the 3-space analog be the difference between x-y-z position space and the space of velocity vectors, like phase space?

    This seems like simple enough stuff, but I've honestly never thought about it this way, so I'm not sure I'm thinking about it correctly ...
     
  10. Mar 30, 2008 #9

    pam

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    You are right. I wrote too glibly. The sum will be a four-vector, but not a four-velocity.
     
  11. Mar 30, 2008 #10

    robphy

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    A more accurate description is this:

    Start at event O.
    From that event O, consider a segment of an inertial worldline with [4-velocity] future-unit-tangent-vector a and proper-time 1-tick, ending up at event P. Picture that P is on a unit-hyperbola centered at O. (...the analogue of sweeping an arc of a unit-circle with a compass centered at O.)

    From that event P, consider a segment of an inertial worldline with [4-velocity] future-unit-tangent-vector b and proper-time 1-tick, ending up at event Q. Picture that Q is on a unit-hyperbola centered at P.

    The inertial segment OQ is "along 1a+1b".
    That is, a+b is a tangent vector to OQ.
    (a+b)/|a+b| is the [4-velocity] future-unit-tangent-vector of the inertial observer along OQ, and |a+b| is his elapsed proper-time from O to Q (i.e. Q is on a hyperbola centered at O with radius |a+b|).
     
  12. Mar 30, 2008 #11
    Thanks - I think I get it. The one thing that's still hanging me up is the meaning of the sum of the two 4-velocities. I know that you can't simply add them and get another 4-velocity, but what then is it? It's tangent to the world line from O to Q, I get that, so it points in that 4-space direction, but how do you equate its magnitude to the proper time from O to Q? That must be true only because you picked your events to be separated by inertial world-lines of 1 unit proper time, right?

    If I picked my first displacement to be, say, three ticks of proper time, and the second to be two ticks, the 4-velocities a and b don't change, of course, but now does a + b tell me anything about the world-line from the starting event to the final event? Or, to put it another way, if I add these two displacement vectors to get that world line, can the 4-velocity along that path be related in any simple way to a and b?

    This seems very simple - I'm not sure why I feel unsure about it. I guess it seems that the 4-velocity is nothing more than a vector of unit length tangent to a given world-path. Actually, now that I say it aloud, that makes more sense. I was having trouble since the word "velocity" kept making me think of it as some kind of time derivative of a position vector, but if I think of it as the tangent vector to a curve parametrized by proper time, that makes more sense. I think.
     
  13. Mar 30, 2008 #12

    robphy

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    I'm technically not equating them... but more of a "similarity" ["scaling"] in the geometrical sense.
    Yes, I am taking advantage of the fact that the 4-velocities are unit-vectors and that, in my example, the proper-times of each leg are unit-intervals.

    There is nothing generally fundamental about adding two 4-velocities.
    The OP asked for an interpretation and I gave a specific one.
    One could probably play a similar game with unit-rest-masses and find an interpretation in terms of 4-momenta (instead of 4-displacements).

    The tangent-vector interpretation is probably the best to start out with... then derive subsequent interpretations.
     
    Last edited: Mar 30, 2008
  14. Mar 30, 2008 #13

    Dale

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    Hi Everyone,

    Thanks for all of the useful comments!
     
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