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Is Time a Scalar or a Vector?

  1. Aug 26, 2013 #1
    In the equation x = vt it is generally accepted that x and v are vectors and that they have a common eigenvector. Each vector is the product of a scalar and a unitary eigenvector. Dividing both sides by v works because in x/v = t the x and v vectors have identical and canceling eigenvectors. This would indicate then that t is the quotient of two scalar values, yielding a scalar value. If t is a vector then the multiplication of v and t doesn't work and the division of x/v doesn't work.

    Flat Minkowski space has a matrix of:

    | x 0 0 0 | - - - | 1 0 0 0 |
    | 0 y 0 0 | or - | 0 1 0 0 |
    | 0 0 z 0 | - - - | 0 0 1 0 |
    | 0 0 0 -ct | - - | 0 0 0 -c |

    This standard matrix is composed of vectors. Since x, y, z are vectors then -ct must also be a vector. The speed of light (c) is a scalar therefore t must be a vector for this matrix. Yet, the previous discussion indicates that t must be a scalar?

    Is Time (t) a vector or a scalar? If t is a scalar then there can be no time travel?
     
    Last edited: Aug 26, 2013
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  3. Aug 26, 2013 #2

    PeterDonis

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    Huh? Vectors don't have eigenvectors. Matrices have eigenvectors.

    This is all incorrect. Where are you getting this from?

    This is true, but it doesn't have anything to do with the rest of what you wrote above.

    I'm not sure that viewing t as the quotient of x and v is the best way to look at it, although it's not exactly wrong; the equation x = vt does say that the vector x is the product of the vector v and the scalar t, so mathematically t does work out as the quotient of x and v. But that way of looking at it leaves out the fact that t can vary; it's different for different events. See further comments below.

    The second "matrix" (the correct term is "metric tensor") is correct (though the -c is usually written as -1 since we usually use units in which c = 1 in relativity). The first one makes no sense at all; where are you getting it from?

    No, it isn't; a matrix is not "composed" of vectors, it's a matrix that you can multiply by a vector to get another vector (with some technicalities that I don't think we need to go into at this point).

    It's a coordinate; x, y, z, and t are all coordinates, at least as you're using those symbols now. When you wrote the equation x = vt above, you were implicitly using "x" to mean the coordinate triple (x, y, z), which can be thougt of as labeling a vector--but it's a 3-vector, not a 4-vector. In spacetime we use 4-vectors, so the 4-tuple (x, y, z, t) labels a 4-vector; it's the 4-vector that goes from the origin (0, 0, 0, 0) to the point labeled by the coordinates (x, y, z, t). t is therefore one component of this 4-vector, i.e., it's a coordinate.

    None of the above has anything to do with whether or not there can be time travel.
     
  4. Aug 27, 2013 #3

    Dale

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    Hi joenitwit, welcome to PF!

    t is neither a vector nor a scalar. It is a coordinate. Coordinates map positions to points in R4. Positions are not a vector space (what is the vector sum of the positions of New York and Paris). Coordinates are also not generally a vector space although they can be in specific cases, like Cartesian coordinates on an infinite flat manifold with simple topology.
     
  5. Aug 27, 2013 #4
    Consider a duration of time, say I took 2 minutes to type this. Then the duration is a scalar right?
     
  6. Aug 27, 2013 #5

    Dale

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    Proper time is a scalar. I believe that you are refering to proper time here, i.e. the time measured by your wristwatch as you typed.
     
  7. Aug 28, 2013 #6

    clem

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    Time is one component of the four-vector X=(ct,x,y,z).
     
  8. Sep 3, 2013 #7
    Coordinates are always vectors. Coordinates refer to a space-time location in reference to an origin. Without an origin and an axis system there can be no coordinates. Drawing a line from the origin to the coordinate always produces a vector. If it were not a vector then the sign, plus or minus, would not have any meaning since the coordinate would then just be a distance.

    My question concerned the flat-space Minkowski matrix since the first row is a multiplier for the x-vector, the second for the y-vector, the third for the z-vector and the fourth for the t-vector... yet t, time, is not and cannot be a vector? There seems to be a fundamental flaw here concerning the nature of time? Yes, time is definitely a coordinate, but time cannot be a vector.

    No need to worry about a tuple (it just confuses things but you can do it if you want). For a single coordinate you can always rotate the axis so the location is on the x-axis.

    A vector is always the multiplication of an eigenvector (vector of length one - unitary) and a scalar which expresses the length or magnitude of the vector. This is the only way to divide vectors. Mathematically, the only permissible quotient of two vectors is if they have identical eigenvectors. Eigenvectors are usually expressed as i-hat, or j-hat or k-hat, but I don't know how to make the symbols on this forum.

    My ultimate point is that there can never be negative time. Since x and v must always be in the same direction (common eigenvectors), t must always be a positive scalar.
     
    Last edited: Sep 3, 2013
  9. Sep 3, 2013 #8
    You could add these vectors. I don't know why you would want to but it is mathematically permissible.

    t is a scalar multiplier. If time can speed up or slow down then the multiplier might change in relation to a location in space (the depth of a g-field) or to the particular motion of the traveler. In any case, time is a multiplier, not a vector. Time has no direction, rather like entropy, it just expands.
     
  10. Sep 3, 2013 #9
    Right! This is where I want to go...

    What other time is there, other than proper time? Maybe this is the answer to my question?
     
  11. Sep 3, 2013 #10

    Nugatory

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    Coordinate time, which is the value of the t coordinate of a given point after we've made a particular choice of coordinate system.

    We usually choose our coordinate systems so that the coordinate time corresponds to the proper time on a particular worldline, but that correspondence doesn't make them the same thing.

    My wristwatch measures proper time along my worldline. When I say that the sweep hand went around the dial once so one minute has passed, that's one minute of proper time as experienced by me.

    When I use my wristwatch to attach times to events off my worldline ("The neighbor's dog barked at the same time that my wristwatch read 1:00 AM"), those times are coordinate times. I'm choosing to use a coordinate system in which the time coordinate is equal to the reading of my wristwatch (and I'm making another arbitrary choice as well, namely how I define "at the same time").

    Proper time is unambiguously and clearly a scalar; it's the distance along a worldline.

    Coordinate time is treated on an equal footing with the spatial coordinates; I disagree with your statement that "coordinates are vectors", but if it were correct, then internal consistency would require that it apply to the time coordinate as well as the spatial coordinates.
     
  12. Sep 3, 2013 #11

    D H

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    No, they are not. The coordinates (of a point) is *not* a vector. Affine spaces are not vector spaces. It doesn't make sense to add points.

    Nonsense. The point in time one chooses to represent t=0 is arbitrary.

    This mathematical discussion has absolutely nothing to do with whether time travel is possible.
     
  13. Sep 3, 2013 #12
    My comment was that coordinates mean nothing without an origin. The relation of coordinates with the associated origin does define a vector. But, that's not my question.


    t=0 is a reference, not a time. I'm trying to think of time in any context which is not actually Δt. Is there any equation where t is not actually Δt? When Δt is multiplied by the speed of light the result is a scalar distance, once again not associated with an axis since there is no direction information.

    My question concerned the fourth row of the Minkowski matrix. This row corresponds to a time vector/axis or perhaps it is better to talk about a vector multiplier (sometimes -ct is put in the first row). Each of the other "multipliers" correspond to an axis, but everyone here has agreed that time is a scalar. A scalar can't correspond to an axis. My point was that the row corresponding to time (-ct) is unlike the other three (x,y,z). Why?
     
  14. Sep 3, 2013 #13

    Dale

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    No, you cannot, at least not without imposing some additional structure.

    If you disagree then please answer the question: "What is the vector sum of the positions of New York and Paris?"

    Unless you define some additional structure beyond that inherent in positions there is no well-defined vector sum.
     
    Last edited: Sep 3, 2013
  15. Sep 3, 2013 #14

    PeterDonis

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    That's not how multiplying a matrix by a vector works. If I multiply a matrix M by a vector V to get a product P, then P is also a vector; and the first component of P (the x component) is obtained by multiplying the first row of M, component-wise, by the components of V, and adding those four numbers together. That is, ##P^x = M^x_x V^x + M^x_y V^y + M^x_z V^z + M^x_t V^t##, where the upper index on the matrix is the row and the lower index on the matrix is the column. Similarly for the other components of P.
     
  16. Sep 3, 2013 #15

    Dale

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    The other time is coordinate time, which I believe is what you are discussing here.
     
  17. Sep 3, 2013 #16

    Dale

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    Ahh! No! Coordinates are most defnitely NOT always vectors. For example, polar coordinates in a plane do not form a vector space. For polar coordinates 0<r and -π<θ<π, so any polar coordinate multiplied by -1 gives an r which is outside the space of polar coordinates, so scalar multiplication does not follow the axioms of vectors for polar coordinates. Similarly, for two vectors each with θ<π/2 the sum is outside the space, so addition of two polar coordinates also does not follow the axioms of vectors.
     
  18. Sep 4, 2013 #17
    This is exactly my point! You just imposed a Vector on Time, ##V^t##, which cannot be done! Time is not, cannot be, a vector!
     
  19. Sep 4, 2013 #18
    ??? Why must r be greater than zero ???

    Never mind... this is getting off on a tangent. My question was about Time being viewed as a vector in the Minkowski flat space matrix (flat space is when it is a diagonal matrix). So far everyone has agreed that Time must be a scalar, not a vector. Something seems amiss with the given matrix and thus with General Relativity.
     
  20. Sep 4, 2013 #19

    Nugatory

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    You are still confusing the scalar proper time with the coordinate time and with the t component of a four-vector. That "Minkowski matrix" (which is really the representation of the metric tensor for flat spacetime, using Cartesian coordinates) acts on the four-vector, and it treats all four components of the four-vector equivalently.
     
    Last edited: Sep 4, 2013
  21. Sep 4, 2013 #20

    PeterDonis

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    I don't understand. ##V^t## is just the 4th component of a 4-vector ##V## that is being multiplied by the matrix ##M##. Are you saying that 4-vectors can't have a 4th component? That makes no sense.

    Also, I only labeled the 4th component ##V^t## because when you wrote down the components of a 4-vector, you wrote them as (x, y, z, t); you'll note that I labeled the other three components of ##V## as ##V^x##, ##V^y##, and ##V^z##. In other words, my labels are taken from your labels. Are you saying your labels are wrong?
     
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