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I Understanding Lorentz-transformability, invariant operations

  1. Mar 26, 2016 #1
    An operation is frame-invariant if it maintains the Lorentz-transformability of an input, yes? So for example, the coordinate ##x## transforms according to ##x^\prime = \gamma(x-vt)##, and multiplying the unprimed space and time coordinates by ##c## would give ##cx^\prime = c\gamma(x-vt)##. In other words, we multiplied the unprimed coordinates by ##c##, and we can still convert them into their primed-frame equivalents using the Lorentz transformation.

    But we could multiply both sides of the equation by any old number, couldn't we? Doesn't have to be an invariant. If it weren't an invariant, though, then the inverse transformation (to go from primed frame to unprimed) would have to be multiplied by a different number, I think.

    So would it be accurate to say that the key to an invariant operation is that it leaves the Lorentz transformation and its inverse "reversible" or "symmetrical"? (Word choice, sorry, but is it clear what I'm getting at? That, if the operation is multiplication, the multiplier should be the same in the two equations?)

    I have another question, which is related. I'm trying to understand what happens when we take the ##d\tau##-derivative of ##c\, dt## and ##dx##. I know we end up with ##\gamma c## and ##\gamma v_x##, and I understand that these quantities are Lorentz-transformable, but I'm not sure that I understand why. Is it just that ##d\tau## is invariant, and so the transformation and inverse transformations remain "reversible" in the sense I described above? So for ##dx##:

    ##dx^\prime = \gamma(dx-v \, dt) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, ⇔ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, dx = \gamma(dx^\prime+v \, dt^\prime)##

    ##(\gamma v_x)^\prime = \dfrac{d}{d\tau}\gamma(dx-v \, dt) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,⇔\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \gamma v_x = \dfrac{d}{d\tau}\gamma(dx^\prime+v \, dt^\prime)##

    (where the ⇔ indicates the "reversibility" I was speaking of).

    Hope that isn't complete gibberish.
     
  2. jcsd
  3. Mar 26, 2016 #2

    andrewkirk

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    The term 'Lorentz Invariant' is only used for scalars. Those are things that are the same across all inertial frames such as ##c## and the length of an interval. They do not include components of a representation of a vector in a particular frame. So for instance coordinate-dependent distance and time are not scalars - although proper time is.

    One does not - in my experience - talk about Lorentz invariance of an operation. I don't know what that would mean.

    As well as Lorentz Invariance of scalars, we have Lorentz Covariance of equations. That applies to an equation that remains true when every component is transformed by the applying the appropriate Lorentz transformation for the relationship between two inertial frames. In cases like your above example there is an additional requirement that the coordinate axes coincide at time ##t=0##.

    Your second equation above would still be Lorentz Covariant if you multiplied both sides by a quantity that is not a Lorentz-Invariant scalar, such as ##x, y, x'## or ##y'##. That's because the equation would remain true, since the new factor cancels.

    I don't understand your last para. It refers to a quantity ##\tau## but doesn't say what it is.
     
  4. Mar 26, 2016 #3
    Thanks, andrewkirk.

    I'll try to rephrase some of this.

    In flat Minkowski spacetime, if we start with a set of coordinates (##t,x,y,z##) in one inertial frame, we can convert them to the coordinates (##t^\prime,x^\prime,y^\prime,z^\prime##) in another frame by using the Lorentz transformation, and we can convert them back by using the inverse Lorentz transformation (given all the standard assumptions about coinciding origins at ##t=t^\prime=0## and whatnot).

    If both the unprimed frame and the primed frame multiply their coordinates by the same invariant (or divide them by it, or differentiate them with respect to it), then the Lorentz transformation and its symmetrical inverse will still hold for the resulting products (or quotients, or derivatives). Such an operation is what I was loosely referring to as a "frame-invariant" operation. I wonder if there's a word for that?

    ##\tau## was proper time.
     
  5. Mar 27, 2016 #4

    Dale

    Staff: Mentor

    It sounds like a change of units to me. The Lorentz transform is a linear operation, so if you multiply it by a scalar then you get another linear operation.
     
  6. Mar 27, 2016 #5
    Indeed, but it's not just multiplication, right? One can also, say, differentiate the coordinates with respect to an invariant (like proper time), and that too maintains Lorentz-transformability. So is the rule simply that if you subject the coordinates to the same linear operation (anything that effectively scales them all the same way), then the results still transform Lorentzianly?
     
  7. Mar 27, 2016 #6

    Dale

    Staff: Mentor

    Hmm, I am not sure what you mean here. Since the proper time is path dependent I don't know how you would do this in a way that would leave you with a 1-to-1 mapping between events in spacetime and points in R4. I.e. you would not have valid coordinates.
     
  8. Mar 27, 2016 #7
    Isn't that how we go from the displacement four-vector to the velocity four-vector? ##dt/d\tau## and ##d\mathbf x/d\tau##?
     
  9. Mar 27, 2016 #8

    Dale

    Staff: Mentor

    The velocity four-vector is only defined along a specific worldline, not on general coordinates or all coordinates.

    So are you asking what operations can be done an a four-vector (or more generally a tensor) such that the result is another four-vector?
     
  10. Mar 27, 2016 #9
    Yes, that's what I'm asking. (Isn't it the same thing, since a four-vector's coordinates must Lorentz-transform?)
     
  11. Mar 27, 2016 #10

    Dale

    Staff: Mentor

    I think the word "coordinates" was throwing me off. I think you mean "components" instead.

    So if you do tensor addition, tensor multiplication, contraction, and covariant differentiation, then a tensor is still "manifestly covariant". I think that there are probably a lot of other operations that still leave it covariant, but not "manifestly". For those I don't know of a shortcut besides checking that it transforms correctly.
     
  12. Mar 27, 2016 #11
    Thanks, Dale.
     
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