What is spacetime made of?

Discussion in 'Physics & Math' started by Xmo1, Apr 29, 2017.

  1. hansda Valued Senior Member

    Consider spacetime is space + time. We can ask: What is space made of? We can also ask: What is time made of? If we can know the physicality of space and time; probably then only we can know about the physicality of spacetime.
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  3. Schmelzer Valued Senior Member

    Thanks for the link to the overview about the anisotropy of the universe. I have to admit that I would like it a lot, because my ether model contains an anisotropy on the level of the fundamental fields.

    Take an electroweak pair. If you consider it not as two particles, but two components of the same, more complex particle, then you can add the spin and the isospin and obtain a representation of the usual rotation group SO(3). This rotation group you can associate with rotations in space. But then the isospin fixed a preferred direction. So, this geometric interpretation of spin and isospin together requires a preferred spatial direction.
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  5. QuarkHead Remedial Math Student Valued Senior Member

    I am not sure this makes much sense.

    Intrinsic angular momentum (i. e. spin) has, by definition, a symmetry given by the group SU(2). I am pretty sure that isospin has the same symmetry. Can you prove that "adding" these gives the group SO(3)?

    To me me this seems to defy principal group theoretic axiom of closure.

    Please give proofs.
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  7. rpenner Fully Wired Staff Member

    SU(2) and SO(3) both have dimension 3 and are locally isomorphic, but are not globally isomorphic.

    By definition, SU(2) is Unitary which means its inverse is conjugate transpose. And it's determinant is 1, which is "Special."
    So if \(M = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \) then \(| M | \equiv \alpha \delta - \beta \gamma = 1\) (which simplies finding the inverse) and
    \(M^{-1} \equiv \begin{pmatrix} \delta & -\beta \\ -\gamma & \alpha \end{pmatrix} = \begin{pmatrix} \bar{\alpha} & \bar{\gamma} \\ \bar{\beta} & \bar{\delta} \end{pmatrix} \equiv M^{*}\)
    so \(\delta = \bar{\alpha} , \gamma = - \bar{\beta}, \alpha * \delta - \beta * \gamma = | \alpha |^2 + | \beta |^2 = 1\) . Thus we have two complex variables and one remaining algebraic condition for 4 - 1 = 3 dimensions of freedom.

    Alternatively, a SU(2) matrix is the matrix exponential of i times a traceless (sum of diagonal is zero) Hermitian matrix (its conjugate transpose is equal to itself). Such a matrix has only three dimensions of freedom.

    Thus all SU(2) matrices can be written as
    \( W = \cos \sqrt{x^2 +y^2 + z^2} ; \quad X = \frac{x}{\sqrt{x^2+y^2+z^2}} \sin \sqrt{x^2 +y^2 + z^2} \\ Y = \frac{y}{\sqrt{x^2+y^2+z^2}} \sin \sqrt{x^2 +y^2 + z^2} ; \quad Z = \frac{z}{\sqrt{x^2+y^2+z^2}} \sin \sqrt{x^2 +y^2 + z^2} \\ e^{ i \begin{pmatrix} z & x - i y \\ x + i y & -z \end{pmatrix} } = \cos \sqrt{x^2 +y^2 + z^2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + i \frac{ \sin \sqrt{x^2 +y^2 + z^2} }{\sqrt{x^2 +y^2 + z^2} } \begin{pmatrix} z & x - i y \\ x + i y & -z \end{pmatrix} = \begin{pmatrix} W + i Z & Y + i X \\ - Y + i X & W - i Z \end{pmatrix} \)
    W = Cos[Sqrt[x^2+y^2+z^2]]
    X = x Sin[Sqrt[x^2+y^2+z^2]] / Sqrt[x^2+y^2+z^2]
    Y= y Sin[Sqrt[x^2+y^2+z^2]] / Sqrt[x^2+y^2+z^2]
    Z = z Sin[Sqrt[x^2+y^2+z^2]] / Sqrt[x^2+y^2+z^2]
    MatrixExp[I {{z, x - I y}, {x + I y, -z}}] =Cos[Sqrt[x^2+y^2+z^2]] IdentityMatrix[2] - Sin[Sqrt[x^2+y^2+z^2]] / Sqrt[x^2+y^2+z^2] I {{z, x - I y}, {x + I y, -z}} = {{ W + I Z, Y+I X }, {-Y+I X, W - I Z}}
    Det[{{ W + I Z, Y+I X }, {-Y+I X, W - I Z}} ]  = W^2 + X^2 + Y^2 + Z^2 = 1
    \( \textrm{det} \, \begin{pmatrix} W + i Z & Y + i X \\ - Y + i X & W - i Z \end{pmatrix} = W^2 + X^2 + Y^2 + Z^2 = 1 = e^0 = e^{i \, \textrm{trace} \, \begin{pmatrix} z & x - i y \\ x + i y & -z \end{pmatrix} } \)

    This collection of matrices is isomorphic with the set of unit quaternions W + i X + j Y + k Z which is isomorphic to the surface of a unit hypersphere in four dimensions.

    Since W + i X + j Y + k Z and -W - i X - j Y - k Z lead to the same rotation (via conjugation) in 3-space of a purely imaginary vector, it follows that this is a double-cover of the rotation group SO(3).

    R(i B + j C + k D) = (±W ± i X ± j Y ± k Z) ( i B + j C + k D ) (±W ∓ i X ∓ j Y ∓ k Z) = ( ±1 ) (W + i X + j Y + k Z) ( i B + j C + k D ) ( W - i X - j Y - k Z) ( ±1 ) = (W + i X + j Y + k Z) ( i B + j C + k D ) ( W - i X - j Y - k Z)

    Last edited: May 2, 2017
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  8. Schmelzer Valued Senior Member

    Of course, you do not add groups. The group SU(2) acts on the doublet of fermions, in two ways, and you can combine them. On the group level, multiplication would be the better word. The spin operator is part of the corresponding Lie algebra $\mathfrak{su}(2)$, roughly the tangential space of the group at 1. Now, if you have two representations of the same Lie algebra, and they commute, then you can add them and get another one. The corresponding representation of the group is obtained by an exponential map, for a spin operator, say, $\sigma_3$, the corresponding rotations around this direction would be defined by $e^{it\sigma_3}$. So, the combination in terms of the group would be the multiplication of the two actions.

    Given that SO(3) is an image of SU(2) where the kernel (all what is mapped into 1) is $\pm 1$ (the "double cover" explained by rpenner), some actions of SU(2) define actions of SO(3) too. All what one needs for this is that the action of $-1$ is trivial. For the basic representation of SU(2), which is spin $\frac12$, it is nontrivial, $-1$ is mapped to $-1$. A $360^o$ rotation does not give the same state, but the same state multiplied by $-1$. But if you combine two of them, you get a factor $-1$ from above, and this gives $1$, so that this combined representation of the group SU(2) is one of those representations which are also representations of SO(3), with integer spin.
  9. Equinox Registered Member

    Space is the nothing that exists between stars.

    It's a 'nothing' that can be warped by certain effects - (in essence making it a 'something').

    We refer to space as space because that is exactly what it is - a void between objects. However we also call it 'the fabric of space' - this fabric made of nothing, is something that we understand in an abstract and partially in a scientific manner (equations in this case are far less clumsy than semantics and words) but it is something that we do not yet fully understand.
  10. QuarkHead Remedial Math Student Valued Senior Member

    I think there may have been some misunderstanding; I absolutely do not need a tutorial on the Lie groups (thanks all the same), I was merely questioning whether SO(3) was an appropriate symmetry to use for spin.

    Schmelzer gave a justification for which I am willing to accept, not being a physicist.
  11. Confused2 Registered Senior Member

    Misunderstanding? I was under the impression rpenner's post was intended for readers of the the thread who (like myself) absolutely need a tutorial on Lie groups. I found it a stunning post - a lure - a model of clarity. Thank you rpenner.

    Edit - belatedly, sorry about the off-topic post.
    Last edited: May 2, 2017
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  12. river Valued Senior Member

    Any mathematical symbols in the equation(s) of physics , represent a physical something . Which is what makes physics so complicated .

    Space-time , is made of the movement in that space , of an object , within a defined space ,nothing more . And this movement to Einstein is based on gravity .
  13. rpenner Fully Wired Staff Member

    Depends on choice of notation.

    Unclear. I believe that what Schmelzer is saying is that the Dirac equation is written to operate on the four components of a Dirac spinor. The first two components and the last two components are related by a SU(2)-derived term with opposite signs, famously "flipping the sign" of energy and momentum in the lower pair.

    Where I wrote a generic element of SU(2) as :
    \( e^{ i \begin{pmatrix} z & x - i y \\ x + i y & -z \end{pmatrix} } = e^{i x \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} + i y \begin{pmatrix} 0 & - i \\ + i & 0 \end{pmatrix} + i z \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} } = e^{i x \sigma_x + i y \sigma_y + i z \sigma_z } \),
    The Dirac equation is written in terms of the momentum operator ( the derivative ) and we have
    \(\begin{pmatrix} I_2 & 0 \\ 0 & - I_2 \end{pmatrix} i \frac{\partial \; }{\partial t } \psi + \begin{pmatrix} 0 & \sigma_x \\ - \sigma_x \end{pmatrix} i \frac{\partial \; }{\partial x } \psi + \begin{pmatrix} 0 & \sigma_y \\ - \sigma_y \end{pmatrix} i \frac{\partial \; }{\partial y } + \begin{pmatrix} 0 & \sigma_z \\ - \sigma_z \end{pmatrix} i \frac{\partial \; }{\partial z } \psi - m \psi = 0 \)
    Or more colorfully, using the metaphor of a vector of sigma matrices:
    \( i \begin{pmatrix} \frac{\partial \; }{\partial t } & \vec{\sigma} \cdot \vec{\nabla} \\ - \vec{\sigma} \cdot \vec{\nabla} & - \frac{\partial \; }{\partial t } \end{pmatrix} \psi - m \psi = 0 \)

    Lost me. How do representations of this Lie algebra commute when any faithful representation of a single group has 3 generators which do not commute?

    Are we talking about a block diagonal matrix where each block is a representation of the same Lie group? Sure, but that's not what's going on about. A matrix representation has to map the identity element to the identity matrix, so you can't just have the direct sum of \(\begin{pmatrix} \alpha & \beta \\ - \bar{\beta} & \bar{\alpha} \end{pmatrix} \) and \(- \begin{pmatrix} \alpha & \beta \\ - \bar{\beta} & \bar{\alpha} \end{pmatrix} \) and still have a representation.

    Last edited: May 3, 2017
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