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Lorentz invariance of four-volume element [itex] d^4x [/itex]

  1. Sep 4, 2012 #1
    I am slightly confused with the invariance of four-volume element. The orthodox way to show it is to prove that Jacobian is one, that I did, however in many textbooks I find a reasoning that because we have Lorentz contraction on one hand and time dilation on the other hand, the product is invariant then. I can not agree with that:

    The problem can be stated like this: one has a cube of sides [itex] \Delta x [/itex] which lives only for time [itex] \Delta t [/itex]. What are the sides and life of the cube in moving reference frame?

    I think that solution is like this: in moving reference frame there will be a length contraction but time also will seem to shorter [itex] ds = dt \sqrt{1-v^2} [/itex] so the product is not invariant.

    Am i wrong? would appreciate any answer, spent lots of time on thinking about it.
    Last edited: Sep 4, 2012
  2. jcsd
  3. Sep 4, 2012 #2


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    Write down the Lorentz transformation. Differentiate it, getting dx' and dt' in terms of dx and dt. Multiply them together, showing that dx' ∧ dt' = dx ∧ dt. This is basically what you already did if you calculated the Jacobian, but it also shows that there's more to the Lorentz transformation than just a length and time contraction -- e.g. dx' depends on both dx and dt.
  4. Sep 4, 2012 #3
    Thank you, Bill for your reply.

    So you think that it is wrong to use the argument of cancellation of time dilation and length contraction? And my way of showing it is wrong, is correct? I used a textbook which served good for me for like 5 years and learned a lot from it, so just want to make absolutely sure that it is wrong.
  5. Sep 4, 2012 #4
    And may i ask what the symbol ∧ stands for?
  6. Sep 4, 2012 #5


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    LayMuon, My advice is to keep reading the book, you still have a few things to learn from it! :wink:

    The Lorentz transformation is a hyperbolic rotation in the x-t plane, more general than a simple length contraction or time contraction.

    The symbol ∧ stands for the wedge product, and is the easiest way to calculate a Jacobian determinant. See the Wikipedia page on "exterior algebra", where they use it to calculate a 2 x 2 determinant.
  7. Sep 4, 2012 #6
    Thank you, Bill.

    yes , I know that one can introduce rapidity which is basically a parameter for Lie Group representation of Lorentz (boost) transformation and cast Lorentz transformation in a form containing hyperbolic sine and cosine thereby ensuring the additive nature of rapidity.

    I knew about Jacobian and how to use its determinant to calculate volume element transformation, but didn't hear about wedge product. I will study, thanks.
  8. Sep 4, 2012 #7
    In the moving reference frame, the spatial extent of the volume is smaller. But the time extent is *longer*. The volume lasts for a time dt, measured in its own *proper* time. A moving observer observes a stationary clock to be *slow*, so someone in a moving reference frame observes the volume's clock to take longer than dt seconds to count off dt seconds. So the volume lasts for a longer time than dt, from the perspective of a moving observer. This cancels the length contraction.
  9. Sep 4, 2012 #8
    The Duck, just to clarify:

    Here are Lorentz transformations into primed (moving) reference frame:

    [tex] \Delta x' = (\Delta x - v \Delta t) \gamma [/tex]
    [tex] \Delta t' = (\Delta t - v \Delta x) \gamma [/tex]

    in the moving reference frame observer is at rest, right? so [itex] \Delta x'=0 [/itex] which gives [itex] \Delta t' = \Delta t \sqrt{1-v^2} [/itex]. Am i wrong?
  10. Sep 4, 2012 #9
    But apparently on the other hand cube is fixed at the non-primed system so [itex] \Delta x = 0 [/itex] too?? :confused:

    How all this fits into equations and reason?
  11. Sep 4, 2012 #10
    I think i got it: [itex] \Delta x' [/itex] refers to an observation of the clock in the primed system so naturally it is not the one that should vanish but [itex] \Delta x [/itex] which is a statement that the cube is fixed in the non-primed system.
  12. Sep 4, 2012 #11


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    The ^ operator is the "wedge product". The wedge product of two vectors generates an oriented area, the wedge product of three vectors generates an oriented volume, etc.

    Thus the wedge product of 4 vectors would be a 4-volume. Which is what your original question was about, and demonstrates why we'd even talk about wedge products.

    However, we can take advantage of the fact that the vectors perpendicular to the boost are not changed by the Lorentz transform, so we really only need to worry about the "area" in the x-t plane.

    This suggests the alternate approach to your original problem. though - transform a square volume element "at rest" via a Lorentz transform. Draw a little square on a space-time diagram, then transform it via the Lorentz transform and look at the resulting figure.

    You should find that the resulting figure is a trapezoid, and not a square.

    Calculate the area of the resulting trapezoid. Note that the area of a trapezoid is NOT the product of the length of the two side. Which is what you were calculating, I assume you were thinking that the square transformed into another square, rather than realizing it transformed into a trapezoid.

    Mathematically, the wedge product is the anti-symmetrized tensor product. Thus the wedge product of two vectors is NOT a vector rather it's a rank 2 tensor. Because it's anti-symmetric, it's also known as a "bivector".

    You'll see the wedge product discussed in "Clifford Algebras", which will also talk about bivectors and such. See for instance http://www.mrao.cam.ac.uk/~clifford/introduction/intro/intro.html. It's advanced but very useful.
  13. Sep 4, 2012 #12
    In other words, the Lorentz transformation does not dilate the plane of the boost. Hence, it keeps the volume element invariant.

    This is an easy proof: take the differential element to be a 4-vector. [itex]dV \equiv e_t \wedge e_x \wedge e_y \wedge e_z |dt \; dx \; dy \; dz| = e_{txyz} |dV|[/itex]

    A boost in, say, the tx plane, only alters those two vectors.

    [tex]\underline L(e_t) = \gamma(e_t + \beta e_x) \\
    \underline L(e_x) = \gamma(e_x + \beta e_t)[/tex]

    The bivector [itex]e_t \wedge e_x = e_{tx}[/itex] then goes to

    [tex]\underline L(e_{tx}) = \gamma^2 (e_t + \beta e_x) \wedge (e_x + \beta e_t) = \gamma^2 (1-\beta^2) e_{tx} = e_{tx}[/tex]

    It should be clear at this point that [itex]dV' = dV[/itex] in every way. The pseudoscalar of the space has been kept invariant. This is the same as saying [itex] \det \underline L = 1[/itex], as any rotation (of which Lorentz boosts are a subset) must do.
  14. Sep 4, 2012 #13
    Actually I understood that it's not going to be a square after rotation, that's why I used jacobian which basically takes into account the angle because it's in general not diagonal.I was rather confused with conceptual questions of "what the measurement is actually?" and "What is the differential element under boosts?" I think I now understand but would need to read about wedge product.

    And thank you for the link, I am reading it now.
    Last edited: Sep 4, 2012
  15. Sep 5, 2012 #14


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    A minor point: it's a rhombus, isn't it? (Both sets of opposite sides are parallel, and all sides are still the same length; only the interior angles change.)
  16. Sep 5, 2012 #15


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    I suppose it will be, at least if you use sensible units.
  17. Sep 5, 2012 #16


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    Hm, yes, I was assuming units in which c = 1.
  18. Sep 5, 2012 #17


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    This is not just a property of the Lorentz transformation. It's a property that has to be satisfied by any transformation that's linear and treats the velocity parameter v in a manner that's consistent with isotropy of space. It applies to Galilean boosts, and it also applies to unphysical examples like a rotation in the x-t plane through an angle proportional to v. Proof: http://www.lightandmatter.com/html_books/0sn/ch07/ch07.html#Section7.2 [Broken]

    Since it's not specifically a property of the Lorentz transformation, it seems backwards to me to prove it by arguments that are specific to the Lorentz transformation. (It's also much more complicated to do it that way.)
    Last edited by a moderator: May 6, 2017
  19. Sep 5, 2012 #18
    So isotropy necessarily leads to a transformation with unit Jacobian? your link is so long, can you bring a shorter version of the proof?
  20. Sep 5, 2012 #19
    Really, it suffices to say that there is a whole class of linear transformations with determinant 1, and any of those will necessarily maintain the invariance of the 4-volume element. The Lorentz transformations are just some of such transformations.
  21. Sep 5, 2012 #20


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    The proof is extremely short. It's in the figure at the end of section 7.2.1.

    But you have to prove that it has determinant 1.
  22. Sep 7, 2012 #21
    Yes, the statement that isotropy is enough is correct indeed. Thanks
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