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What does imaginary time mean

  1. Apr 18, 2005 #1
    When I was reading the Landau's The Classical Theory of Fields, I found that when distance of four dimentional space is negative, the time's square should be a negative as well.Then time is an imaginary number. it's SPACELIKE. But what is imaginary time mean on earth?
    They say that in that case(spacelike), an event could be BEFORE or AFTER or SIMULTANEOUS with another event in the book. As far as I know, two imaginary number can't be compared. Then how is the time compared?
  2. jcsd
  3. Apr 18, 2005 #2


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    Is this what Landau's saying? 'Cause I don't see it.

    [tex](\Delta s)^2 = (c\Delta t)^2-(\Delta x)^2-(\Delta y)^2-(\Delta z)^2[/tex]

    If [itex](\Delta s)^2 < 0[/itex], it simply mean that the distance separating the two events is bigger than the distance light travels in the time interval which separates them. So they cannot be linked by a relation of causality.
  4. Apr 18, 2005 #3


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    I also understand that a set of events are said "space-like" in a certain referential if [itex]\Delta t = 0[/itex] in that referential, and they're "time-like" if [itex]\Delta x^2 + \Delta y^2 + \Delta z^2 = 0 [/itex].
  5. Apr 18, 2005 #4
    imaginary time means nothing. It's just a trick we use to make the problem easier to solve mathematically, or to make the problem solvable.
  6. Apr 18, 2005 #5
    The imaginary number in question is the proper time, [tex]\Delta \tau[/tex], between the two events. If the spacetime interval (distance) [tex]\Delta s[/tex] is given by

    [tex](\Delta s)^2 = -(c \Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2[/tex] then the proper time is given by

    [tex](\Delta \tau)^2 = - \frac{(\Delta s)^2}{c^2}[/tex].

    So if [tex](\Delta s)^2 > 0[/tex] (spacelike separation), the proper time [tex]\Delta \tau[/tex] is imaginary. To repeat what quasar987 said, "spacelike separation" means that the space distance between the two events is too large for a light signal emitted at one event to reach the other event. So one event cannot cause the other. The proper time between two events can be thought of as the time interval measured by someone moving in a straight line with constant velocity from one event to the other (he is present at both events). When the proper time is imaginary, as it is for spacelike separation, it just means that you can't be present at both events because you would have to travel faster than light to do it.

    edit: so to answer the last part of the OP's question, [tex]\Delta t[/tex] is always a real number. For a given pair of spacelike separated events, you can always choose a reference frame such that [tex]\Delta t[/tex] is positive, negative, or zero. However, [tex]\Delta s[/tex] is the same in all inertial frames.
    Last edited: Apr 18, 2005
  7. Apr 18, 2005 #6
    So what does that mean; how do you compare time if you can't compare two imaginary numbers?
  8. Apr 18, 2005 #7


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    Two imaginary numbers can be compared... that is to say, the imaginary numbers can be ordered.

    Define (ai) > (bi) when a>b, where a and b are real numbers.
  9. Apr 18, 2005 #8
    Oh, I wish we had gone over that in my math class. So how do you compare time if they are imaginary numbers? Would you check to see if the a term os greater than the b term?
  10. Apr 18, 2005 #9
    yes It's what I mean.
  11. Apr 18, 2005 #10
    Sorry, I have made a mistake.What I wanted to ask is if[tex](\Delta s)^2+(\Delta x)^2+(\Delta y)^2+(\Delta z)^2= (c\Delta t)^2[/tex]is a negative.
  12. Apr 18, 2005 #11
    No, it is not negative. [tex]\Delta t[/tex] is the difference between the time reading of a clock at one event and the time reading of a clock at another event. So it must be a real number. For example, the clock at event A reads 4:59pm and the clock at event B reads 5:02pm, then [tex]\Delta t[/tex] is 3 minutes.
  13. Apr 19, 2005 #12
    But according to the theory. It CAN BE imaginary. It may contradict causality. I just wonder HOW that contradict it!
  14. Apr 19, 2005 #13
    Sometimes, in analysis, we deal with real valued functions in complex space. This is done either to make finding the solution very much easier or to avoid the presence of singularities (rather, more correctly, where on the real line there are but two directions to apporach the discontinuity from, in the complex plane we have infinitely many). Is it possible the use of imaginary time is simply an approach to deal with singularities? Or are you actually dealing with roots of negative real numbers?
  15. Apr 19, 2005 #14

    I'm sorry, when I said "it" in that sentence, I was referring to [tex]\Delta t[/tex]. I should be more careful with pronouns. [tex]\Delta t[/tex] is ALWAYS a real number for the reason I gave in my last post. This means that [tex](c\Delta t)^2[/tex] is always positive or zero. But the proper time [tex]\Delta \tau[/tex] can be real or pure imaginary, just like [tex]\Delta s[/tex]. If one is real, the other is pure imaginary. But the point is that proper time does not always correspond to a measurement that can be made by a real clock. If two events are spacelike separated, then there is no single clock that can be present at both events. The "proper time" between the two events is still defined--it is just an imaginary number in this case.

    edit: for reference, here are some useful definitions regarding the spacetime separation between two events A and B:

    "Spacelike separated" means [tex](\Delta s)^2 > 0[/tex].
    "Timelike separated" means [tex](\Delta s)^2 < 0[/tex].
    "Lightlike" or "Null separated" means [tex](\Delta s)^2 = 0[/tex].

    See my post #5 for the sign convention being used here for [tex](\Delta s)^2[/tex].
    Last edited: Apr 19, 2005
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