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Is time scalar?

  1. May 13, 2010 #1
    The speed of light in a gravitational field is not a scalar quantity because at any given coordinate its speed is a function of its direction.
    Does general relativity treat time in a gravitational field as a scalar quantity?
     
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  3. May 13, 2010 #2

    Dale

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    The speed of anything is a scalar, including the speed of light.

    You can say that time is a scalar, but I prefer to think of it as one component of a four-vector.
     
  4. May 13, 2010 #3

    atyy

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    There are several definitions of time.

    Time is a direction in spacetime for each person (time-like direction). Different people have different directions.

    If that person sets up a coordinate system which divides spacetime into "space" and "time", then "coordinate time" is one scalar component of the coordinate system.

    If that oerson measures the elapsed time on a clock he carries, that is the integral of the "proper time" on his wordline, which is an scalar that is the same regardless of anyone's choice of coordinate system.
     
  5. May 13, 2010 #4
    By definition speed is a scalar as it is the magnitude of the velocity, independent of direction. You can say the coordinate velocity of light in a gravitational field is not isotropic ( but locally it is).

    I guess you could construct a vertical light clock and a horizontal light clock and compare the tick rates by using an inteferometer (basically a Michelson-Moreley type apparatus with one arm vertical and one arm horizontal) and they should always tick at the same rate for a small and ideal apparatus. Generally speaking, as far as I know, time is treated as a scalar in GR, but as Dalespam points out it can be thought of as one of four vectors at right angles to the three spatial vectors if you can picture that (I can't). You can't really say for example, that a single clock has a tick rate in the x direction and a different take rate in the y or z directions.
     
    Last edited: May 13, 2010
  6. May 13, 2010 #5

    Stingray

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    Speed is not a scalar in any natural sense. It's frame-dependent. Time is also not a component of a 4-vector. It is common to represent spacetime points in special relativity as if they were 4-vectors, but this is misleading. It only works in globally inertial (i.e. Minkowski) coordinates. Edit: Actually, this fails even for a uniform translation in Minkowski coordinates.

    Time can be represented as a scalar in general or special relativity. Mathematically, it is a map that takes a spacetime point and returns a real number. This map is not unique, however. There are many possible times that one may define, so this interpretation might not be very helpful.
     
    Last edited: May 13, 2010
  7. May 13, 2010 #6

    Cyosis

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    Different people seem to be talking about different definitions of what a scalar is. Some define a scalar as just a number and a vector as something with a magnitude which is just a number and a direction. This however is not how scalars and vectors are usually defined in physics or relativity.

    A scalar is an object that transforms as follows [itex]v=\overline{v} [/itex], a vector [itex]v^i[/itex] is an object that transforms as [tex]\overline{v}^i=\frac{\partial \overline{x}^i}{\partial x^j} v^j[/tex], a tensor is an object that transforms as [tex]\overline{v}^{ij} =\frac{\partial \overline{x}^i}{\partial x^k}\frac{\partial \overline{x}^j}{\partial x^l}v^{kl}[/tex] etc. In this sense speed is definitely not a scalar.
     
    Last edited: May 13, 2010
  8. May 13, 2010 #7
    yes it is.
    It may be sometimes a component of a vector, but that doesn't mean it's a vector.
     
  9. May 13, 2010 #8
    You are right that the confusion stems from different definitions of the meaning of scalar. This wikipedia article http://en.wikipedia.org/wiki/Scalar_(physics [Broken]) illustrates the problem. It defines the term "scalar" as used in physics as:
    By this definition speed is not a scalar because speed is changed by coordinate transformations while the "relative speed" of two particles is a scalar.

    Lower down, the very same short article says:
    Clearly a contradiction to its first definition.

    Further meanings are given to the term "scalar" in mathematics. For example, if the result of a mathematical operation on matrices results in a 1x1 matrix then the result is a scalar. See http://en.wikipedia.org/wiki/Scalar_(mathematics [Broken])
     
    Last edited by a moderator: May 4, 2017
  10. May 13, 2010 #9

    Stingray

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    I think one other confusion is that people can use the words scalar, vector, etc. to mean things that transform in the usual way under some particular class of coordinate transformations that may not be completely general.

    So speed is a scalar in Newtonian mechanics with respect to any (time-independent) spatial coordinate transformation. This excludes the "Galilean boosts," and is natural in that theory.

    In relativity, you specify time and space coordinates equally, so it would make a lot less sense to make a restriction like this.
     
  11. May 13, 2010 #10

    DrGreg

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    As Cyosis correctly says it depends what you mean by "scalar". It depends what you mean by the "speed of light". And it depends what you mean by "time"! :smile:

    If "scalar" just means one-dimensional and "vector" means multi-dimensional, then the speed of light is scalar and time is scalar.

    If "scalar" means "invariant" (i.e. all observers agree on its value) then proper time is a scalar, but coordinate time is not. Proper time is the time measured by a single clock that is at zero distance from the events being measured. In the standard Minkowski coordinates of special relativity, coordinate time is one component of a vector. In general, it's not even that, it's just a timelike component of a manifold's coordinate chart.

    The speed of light measured by a local observer using local proper distance and local proper time is a scalar in this second sense. But in general the coordinate speed of light in an arbitrary coordinate system in GR is not a scalar.
     
  12. May 13, 2010 #11
    Good points Dr Greg :smile:
     
  13. May 13, 2010 #12
    If the "scalar" you are talking about means simply a "one-component" tensor, then yes it is in some sense treated in such a way that one can easily put it anywhere in the equations. But you can't do this with a vector or tensor: they always appear to be related to each other by a transformation law that Cyosis showed earlier. And since this reduces to an obvious equality in two different coordinates for scalars, so we are free to choose how to have time appear in the equations. Nonetheless, time in GR is only a component of the coordinate tuple i.e. [tex]x^{\alpha}:=(x^0,x^1,x^2,x^3)=(t,x,y,z),[/tex] in the relativistic units, but yet it is a scalar.

    AB
     
  14. May 13, 2010 #13

    Fredrik

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    With the definitions I'm using (i.e. the ones people should be using...and not just because I'm using them), it's the other way round. A scalar (field) is a function from an open subset of spacetime into the real numbers. A coordinate system is a function from an open subset of spacetime into [tex]\mathbb R^4[/tex], so each of its components is a scalar field. Proper time on the other hand, is a number assigned to a curve, not to a point in spacetime. So it's definitely not a scalar field. (Of course, if you choose one specific point p to be the starting point of all curves, and then supply additional information to narrow down the choice of curves to an arbitrary point q to one unique curve, then you can define a scalar field whose value at q is the proper time of the unique curve from p to q that meets the requirements).

    What the "transformation law" for scalars [itex]\phi'(x')=\phi(x)[/itex] really means is this: If [itex]\psi:U\rightarrow\mathbb R[/itex] is a scalar field, and y and z are coordinate systems such that U is a subset of the intersection of their domains, we have

    [tex]\psi(p)=\psi\circ y^{-1}(y(p))=\psi\circ z^{-1}(z(p))[/tex]

    for all p in U. Now define

    [tex]\phi'=\psi\circ y^{-1},\quad \phi=\psi\circ z^{-1},\quad x'=y(p),\quad x=z(p)[/tex]

    and we get [itex]\phi'(x')=\phi(x)[/itex].
     
    Last edited: May 13, 2010
  15. May 14, 2010 #14
    Yes this is the matheamatical procedure to get the transformation law for scalar fields. Speaking of which, sometimes people make this mistake that they assume "coordinate tuples" as four vectors so in this case time can be contained as a scalar (field) in the componential set of a four vector, though sounding believable in some sense, but it makes possibly this confusion that how the transfomration law for scalars can be read from the transformation of the four vector. So the best stance for me is to agree that we better use time as a scalar field so this field maps a (not necessarily unique) number to any point in the spacetime.

    AB
     
  16. May 14, 2010 #15
    I've often thought that misunderstanding the nature of time is the central obstacle to understanding both special and general relativity. I really like this discussion as it approaches the subject in a unique way. It is natural to underestimate the complexity involved---we wear it on our wrists, after all. But this is MY time (proper time). Relativity explains how my time is related to yours. In this way time is a scalar as DrGreg said earler.

    When I start to talk about what I think of your time (i.e., separated in space) we have to deal with relativity, space-time events, and such. In this context it is best to think of time as the fourth dimension of a particular frame of reference.

    BTW: Speed is definitely a scalar. It is the square root of [itex] g_{ij} v^i v^j [/itex].
     
  17. May 14, 2010 #16

    Cyosis

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    This is the norm of the four-velocity, which isn't exactly 'speed'.
     
  18. May 14, 2010 #17

    Ich

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    It's neither speed nor 1, it's http://en.wikipedia.org/wiki/Proper_velocity" [Broken]. At least in a "standard" coordinate system, and with "standard" index convention (latin=space only).
     
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  19. May 14, 2010 #18

    DrGreg

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    Having had a chance to think about this, I have to agree with almost all of this. Assuming we are talking about "scalars" in the context of tensors, yes, "scalar" should be interpreted as "scalar field", and I agree with all you said about proper time.

    In the case of coordinate time, I think it depends on how you interpret the words. "Scalar" implies "invariant", something all observers can agree on. So if we are talking about one specific coordinate system, that system's coordinate time is indeed a scalar field. However if each observer is referring to their own coordinate time, that's not a scalar. This is a matter of the "physical interpretation", or interpretation of the words, rather than any mathematical issue.

    (Of course, none of this applies when we use "scalar" in the wider sense of any real number, i.e. member of the (algebraic) field associated with an abstract vector space.)
     
  20. May 14, 2010 #19

    Ich

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    That's definitely ok from a mathematical point of view, if you identify a "scalar" with the corresponding map.
    However, my understanding was rather that a scalar quantity is independent of the chosen map.
    For example, spatial distance from the origin is a scalar according to this definition. Every number that can be deterministically defined is a scalar. That somehow makes it a synonym for "number". (Ok, except for the physically hardly motivateable (is there such a word in English?) constraint of "defined on an open subset". You know, you're talking about mathematics here, not necessarily physics.)
    The interval between two events is independent of the map. How do you call such a quantity? Does it deserve a distinct name?
    (These are not rethorical questions)
     
    Last edited: May 14, 2010
  21. May 14, 2010 #20

    DrGreg

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    The rest of my last post essentially agrees with the point you are making. I would say "spatial distance from the origin" isn't a scalar, but "spatial distance from the origin measured in frame I" (where we have already defined what I is) is a scalar field.
     
  22. May 14, 2010 #21

    Ich

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    Yes, something along these lines. If you fail to specify a map, and it makes a difference, a number is not a scalar.
    So "speed" or "time", without some qualifiying remarks, are not scalars. Right?
     
  23. May 14, 2010 #22

    DrGreg

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    In my view, yes.
     
  24. May 14, 2010 #23

    Dale

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    Of course, that means that the scalar potential also requires some qualifying remarks before calling it a scalar, right? :smile:
     
  25. May 14, 2010 #24

    Ich

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    Why?
     
  26. May 14, 2010 #25

    Dale

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    The scalar potential is frame-dependent. It transforms as the timelike component of the four-potential.
     
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