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I am reading Landau's Volume 2 of the course of theoretical physics. I have a doubt after reading the first few pages of it which I explain below.

Landau first defines intervals and on pages 5 and 6 shows that two events having time like interval between them can never occur simultaneously in any reference system. Then he goes on to construct a 2D space-time graph (for visualization) with an event O occurring at (0,0,0,0). Then he considers any event which occurs in future in that frame and is time-like w.r.t. origin and says on page no. 7,

But two events which are separated by a time-like interval cannot occur simultaneously in any reference system. Consequently, it is impossible to find a reference frame in which any of the events in region aOc occurred "before" the event O, i.e. at time t<0.

The argument above just proves that because interval square should be positive, i.e. the events can't be simultaneous. But, if I replace the difference in time in the original frame with its negative in my proposed frame and let the space distance between them to be same in both frames, then I get an in my proposed frame an interval which is time like but in it the order of events is changed. Am I making some gross error or Landau has missed some argument?

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    $\begingroup$ The problem is, in some sense, semantics. Just because we artificially chose a new time coordinate in the opposite time direction, the physical notions of actual past and future (light-cones) do not change. $\endgroup$
    – Qmechanic ♦
    Commented Jul 14, 2011 at 14:28
  • $\begingroup$ So, does it mean that if I mathematically construct such a frame, in reality it cannot exist? Isn't it true that there is a one-one correspondance between the mathematical and physical reference systems? Please elaborate. $\endgroup$
    – user4235
    Commented Jul 14, 2011 at 15:55
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    $\begingroup$ @Lakshya: No, there is no one-to-one correspondence between mathematical and physical systems. Only a subset of mathematical systems are physically realized; in other words, it is possible (and common) to develop some mathematics which does not correspond to anything physical. Improper or non-orthochronous Lorentz transformations are examples of that. (Note that the frame you're talking about does exist, but there's no way to boost yourself, the observer, into it. It's that boost, not the frame, that would correspond to an improper Lorentz transformation.) $\endgroup$
    – David Z
    Commented Jul 14, 2011 at 17:12
  • $\begingroup$ @David Aaah I see. Probably this frame results because physical laws are isotropic in time? But then, how will we set such a boundary that we can see its reverse in time picture only in case of time-like intervals and not space-like intervals? There's got to be something very specific property about time-like intervals. And that thing is that it is discontinuous at a point. So, I think that we can resolve this dilemma by defining and using continuity of functions in higher dimensions. What do you think about this? $\endgroup$
    – user4235
    Commented Jul 15, 2011 at 15:25

2 Answers 2

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If you believe that (a) timelike separated events cannot be simultaneous in any reference frame, and (b) the set of inertial frames is (in some appropriate sense) a continuous set, then L&L's conclusion follows. After all, if there were two frames in which the order of two timelike separated events differed, then by continuously transforming one frame into another, you could find one in which they were simultaneous.

But without some such additional assumption, you're right that the conclusion doesn't logically follow. There are coordinate systems that preserve the spacetime interval but flip the direction of time, such as the substitution $t\to -t$ that you mention. As BebopbutUnsteady observes in a comment to Karsus Ren's answer, we often use the term "orthochronous Lorentz transformation" to refer to a transformation that preserves the direction of time. The full group of Lorentz transformations (i.e., of all transformations that preserves the interval) includes both orthochronous and non-orthochronous components, which are not connected to each other. Physically, we usually only consider the orthochronous ones.

You do have to be careful with the terminology: sometimes people use "Lorentz transformation" to mean just the orthochronous ones; sometimes it's the full group.

By the way, pretty much the same thing applies to spatial reflections: is $x\to -x$ (leaving $y,z,t$ unchanged) a Lorentz transformation? After all, it preserves the spacetime interval. Often we refer to non-reflecting Lorentz transformations as "proper." So when people are being careful with their terminology they often refer to "proper orthochronous Lorentz transformations."

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  • $\begingroup$ Thank you very much for the answer which deserves to be called an 'answer'. This is undoubtedly the best answer I ever received and you duly told everything to me that should be in an answer. Although, I would like to ask you to recommend some book on special relativity in which I can find more rigorous stuff than Landau like this. Thank you. :) $\endgroup$
    – user4235
    Commented Jul 14, 2011 at 16:14
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    $\begingroup$ Glad you liked the answer! I don't know the best recommendations for things for you to read. If you want treatments of the Lorentz transformation with more mathematical rigor, your best bet is probably mathematical physics books, rather than special relativity books. For instance, I think Arfken's Mathematical Methods for Physicists might be a good place to start. Also, general relativity books tend to treat the mathematics more carefully than special relativity books. I like Wald's General Relativity myself, although at a quick glance it doesn't seem to treat this particular topic. $\endgroup$
    – Ted Bunn
    Commented Jul 14, 2011 at 17:09
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I don't know what Landau says, but space-time interval doesn't distinguish order. To derive the non-interchangeability of time-like intervals, Lorentz transformation must be explicitly used.

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    $\begingroup$ By which you mean orthochronous Lorentz transformations presumably. $\endgroup$
    – BebopButUnsteady
    Commented Jul 14, 2011 at 14:48
  • $\begingroup$ Thanks for your answer. As I am following Landau, I haven't yet reached to the Lorentz transformations. But see, a reputed physicist Landau is saying 'Consequently....' And he is writing a very serious book, so when he says consequently, there must be some line of reasoning. That's why I put this here. $\endgroup$
    – user4235
    Commented Jul 14, 2011 at 16:00

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