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Relativity Question

  1. Jun 29, 2010 #1
    1. The problem statement, all variables and given/known data

    Assume that the coordinates of O' are given as the following linear combination's of those of O:

    t' = alpha t +beta x

    x' = mu t + nu x

    y' = ay

    z' = bz

    where alpha, beta, mu, nu, a and b may be functions of the velocity v of O' relative to O, but they do not depend on coordinates. Find the numbers {M_nm n,m = 0,...,3} of equation 1.2 (the equation below) in terms of alpha, beta, mu, nu, a and b.


    2. Relevant equations

    delta S^2 = sum sum M_nm (delta X^n) (delta X^m)

    Both of the sums are from m = 0 -> 3 and n = 0 -> 3)

    3. The attempt at a solution

    I haven't tried a question like this before and was wondering if someone could help get me started.
     
  2. jcsd
  3. Jun 30, 2010 #2
    I only want to make clear the initial conditions.
    (1)
    [tex]ds^2=c^2dt^2-dr^2[/tex]

    You want to find
    (2)
    [tex]ds^2=\sum_{m,n}M_{mn}dX^mdX^n[/tex]

    where
    [tex]X=(ct,r)[/tex].

    Also this condition is true
    (3)
    [tex]ds^2=ds'^2[/tex]
     
    Last edited: Jul 1, 2010
  4. Jun 30, 2010 #3
    Yes those conditions are true, im just having trouble understanding how to use

    [tex]
    ds^2=\sum_{m,n}M_{mn}X^mX^n
    [/tex]

    in relation to the O' equations i was given.

    edit: wow this is anoying, i copied the second eqn you listed but the third one keeps appearing??
     
    Last edited: Jun 30, 2010
  5. Jul 1, 2010 #4
    Please find dt', dx', ... and insert into (3).
     
  6. Jul 2, 2010 #5
    Alright im starting to understand this.

    So far i have:

    [tex] M_{00} = \mu^{2} - \alpha^{2} [/tex]

    [tex] M_{11} = \nu^{2} -\beta^{2}[/tex]

    [tex] M_{22} = a^{2}[/tex]

    [tex] M_{33} = b^{2}[/tex]

    with the rest being zero.

    This is that same as the answer given except the answer has a term:

    [tex] M_{01} = \mu\nu - \alpha\beta[/tex]

    How did they get this term?
     
  7. Jul 4, 2010 #6
    At the risk of adding my confusion to the mix, I'm probably missing something stupidly obvious here, but...

    zzzoak, your condition (1) seems to me to say that the coordinate matrix in O of the metric tensor [itex]M=\eta[/itex] = diag(1,-1,-1,-1), in units such that c = 1, which means that O is an orthonormal coordinate system. Together with (1), your condition (3) says the transformation from O to O' is a Lorentz transformation, and that therefore O' is also orthonormal. Is that right? If so, why doesn't the problem become a trivial matter of expressing 1 and -1 in terms of any combination of those variables we like?

    fys iks, could you explain (for my curiosity!) how you got those answers?

    The transformation is, in matrix form:

    [tex]T = \begin{bmatrix}
    \alpha & \beta & 0 & 0\\
    \mu & \nu & 0 & 0\\
    0 & 0 & a & 0\\
    0 & 0 & 0 & b
    \end{bmatrix}[/tex]

    Where mu, nu, alpha and beta are functions of velocity, making T a boost or, synonymously, a hyperbolic rotation.

    A general boost along the x-axis is given by

    [tex]\begin{bmatrix}
    \frac{1}{\sqrt{1-v^2}} & \frac{-v}{\sqrt{1-v^2}} & 0 & 0\\
    \frac{-v}{\sqrt{1-v^2}} & \frac{1}{\sqrt{1-v^2}} & 0 & 0\\
    0 & 0 & 1 & 0\\
    0 & 0 & 0 & 1
    \end{bmatrix} = \begin{bmatrix}
    \cosh(\phi) & -\sinh(\phi) & 0 & 0\\
    -\sinh(\phi) & \cosh(\phi) & 0 & 0\\
    0 & 0 & 1 & 0\\
    0 & 0 & 0 & 1
    \end{bmatrix}[/tex]

    where [itex]\phi = \text{artanh}(v)[/itex], the inverse hyperbolic tangent of v. Then [itex]\mu\nu - \alpha\beta = 0[/itex], and [itex]a^2=b^2=1^2=1[/itex]. But [itex]\mu^2-\alpha^2=-\sinh^2(\phi)-\cosh^2(\phi)[/itex] is never equal to 1 for any real value of phi. And [itex]\nu^2-\beta^2=\cosh^2(\phi)+\sinh^2(\phi)[/itex] is never equal to -1 for any real value of phi.

    http://www.wolframalpha.com/input/?i=-sinh^2%28x%29-cosh^2%28x%29%3D1

    Whereas:

    [tex]\mu^2+\alpha^2=-\sinh^2(\phi)+\cosh^2(\phi)=1[/tex]

    [tex]-\nu^2-\beta^2=-\cosh^2(\phi)+\sinh^2(\phi)=-1[/tex]

    Even if conditions (1) and (3) weren't requirements of the original problem, shouldn't the answers be consistent with the case where (1) and (3) are true, at least?

    On the assumption that

    [tex]ds^2=x^TMx=(Tx)^TMTx=x^TT^TMTx=(ds')^2[/tex]

    [tex]\Rightarrow M=T^TMT[/tex]

    I tried working back from the given answers to M, but this resulted in a horrible mess! Zeros in all the right places, but I had

    [tex]M_{33}=a^4[/tex]

    [tex]M_{44}=b^4[/tex]

    and the top left 2x2 square full of stuff like this:

    [tex]M_{00}=\alpha^2\mu^2-\alpha^4+\alpha\mu^2\nu+\alpha^2\beta+\mu^2\nu^2-\mu^2\beta^2[/tex]

    Argh.

    (By the way, the LaTeX problem you mention in your edit is a bug in PF that appeared a few months ago. I get around it, in Firefox, by copying the message I'm working on, closing the tab I'm typing my message in, and selecting "Clear Recent History" from the Tools menu, then opening a new tab, going back to PF, starting a new message and pasting in where I'd got up to. Bit of a faff, but you get stoical after a while, except late at night, if you keep noticing mistakes to correct...)
     
  8. Jul 4, 2010 #7
    To get my answers all i did was take the O' equations and used eq'n 1.

    For M_11 i did (dt'/dt^2) + (dx'/dt)^2 + (dy'/dt)^2 + (dz/dt)^2

    then M_22 the same thing wrt to dx

    M_33 wrt dy

    M_44 wrt dz

    I not really sure were u were going with all the hyperbolic stuff...
     
  9. Jul 5, 2010 #8
    Thanks for the tip, fys iks! Don't worry about the hyperbolic stuff--that's just another way of writing the matrix on the other side of that equation. How about my reasoning in the second paragraph, can you see where I went wrong with that, or which of my assumptions was wrong? You agreed with zzoak's conditions (1) and (2); doesn't that make M equal to diag(1,-1,-1,-1)? Oh, wait on, I've just realised... In your notation, does X=x' rather than x? Maybe that's where I was going astray. I'll have to think about this...
     
  10. Jul 5, 2010 #9
    For the 'relevant equations' i posted then X can equal anything because it is just a general equation. In this case however i think you are right where X=x'. But I'm still having trouble understanding the equation in general.

    And i thought that for the proper time the metric is (1,-1,-1,-1). Maybe in this case you have to use the metric for length which is (-1,1,1,1)?
     
  11. Jul 5, 2010 #10
    I'm still struggling myself, but maybe getting closer, I hope... I supect at least part of the problem might be to do with translating between different notations.

    In this post, I'll use capitals for values wrt O' and lower case for values wrt O, and dispense with prime symbols except for the names of the reference frames.

    Does ds=dS? I think so, although the formula for ds in terms of dxi will be different from the formula for dS in terms of dXi unless the transformation, T, from O to O' is a Lorentz transformation. If it's a Lorentz transformation, then M = diag(1,-1,-1,-1) or diag(-1,1,1,1) depending on sign convention. The equations you give for T are more general than a Lorentz transformation; they needn't preserve the magnitude of a spacetime separation vector (a vector representing a displacement in spacetime), but they will preserve arc length, and hence the line element ds=dS. (I don't know the mathematical reason for this last condition, but since this is a quantity with physical consequences, I assume valid coordinate transformations are limited by some rule to those that preserve arc length.)

    Using the time positive convention:

    [tex]dX^i=\frac{\partial X^i}{\partial x^j} dx^j[/tex]

    [tex][dX]=T[dx][/tex]

    [tex]dS^2=M_{mn}dX^mdX^n=ds^2=\eta_{mn}dx^mdx^n[/tex]

    [tex]M_{mn}dX^mdX^n=M_{mn}\frac{\partial X^m}{\partial x^i}dx^i \frac{\partial X^n}{\partial x^j}dx^j[/tex]

    [tex][dx]^T \eta [dx]= [dX]^TM[dX][/tex]

    [tex]= (T[dx])^TMT[dx]=[dx]^TT^TMT[dx][/tex]

    [tex]M=(T^T)^{-1} \eta T^{-1}=(T \eta T^T)^{-1}[/tex]

    [tex]=\begin{bmatrix}
    \alpha^2-\beta^2 & \alpha\mu-\beta\nu & 0 & 0\\
    \alpha\mu-\beta\nu & \mu^2-\nu^2 & 0 & 0\\
    0 & 0 & -a^2 & 0\\
    0 & 0 & 0 & -b^2
    \end{bmatrix}^{-1}[/tex]

    That would give M22 = -a-2 and M33 = -b-2, and in the top 2x2 square:

    [tex]\frac{\begin{bmatrix}
    \mu^2-\nu^2 & -\alpha\mu+\beta\nu\\
    -\alpha\mu+\beta\nu & \alpha^2-\beta^2
    \end{bmatrix}}{-(\beta\mu-\alpha\nu)^2}[/tex]

    But if I switch notation and use capitals for O and lower case for O', then I get

    [tex]M_{mn}dx^mdx^n=\eta_{ij}dX^idX^j[/tex]

    [tex][dx]^T M [dx] = (T[dx])^T \eta T[dx]=[dx]^TT^T \eta T[dx][/tex]

    [tex]M=T^T \eta T[/tex]

    [tex]=\begin{bmatrix}
    \alpha^2-\mu^2 & \alpha\beta-\mu\nu & 0 & 0\\
    \alpha\beta-\mu\nu & \beta^2-\nu^2 & 0 & 0\\
    0 & 0 & -a^2 & 0\\
    0 & 0 & 0 & -b^2
    \end{bmatrix}[/tex]

    Which is the same as the given answers, except that they've used the other sign convention, and they have M10 = 0. (But shouldn't a metric tensor be symmetric?) Supposing this is what the answer was meant to be, everything would work fine as long as the question was using capital letters for values wrt O. And it since it doesn't say one way or the other, maybe we can assume it was.

    I'm sure I've seen these ideas presented somewhere using capitals and lower case letters to distinguishg between two coordinate systems, but I can't remember which way round it was: which were the old coordinates and which the new. Maybe there's a convention about that.
     
    Last edited: Jul 5, 2010
  12. Jul 5, 2010 #11
    Thanks a lot! that is correct, and i think the book made a typo because i re-checked the answer and they didnt even list a value for M_01. The metric tensor is always symmetric too so you second solution is correct.

    Could you just help me out a bit. I'm new to this Tensor stuff.

    What do the following equations mean:


    [tex]
    [dX]=T[dx]
    [/tex]

    and

    [tex]
    [dx]^T M [dx] = (T[dx])^T \eta T[dx]=[dx]^TT^T \eta T[dx]
    [/tex]

    I am confused with the [ ] notation.
     
  13. Jul 5, 2010 #12
    You're welcome, and thanks for the puzzling-practice! Those are matrix versions of the equations with index notation. I should have explained. I just put square brackets around the differential displacement vectors to show that, for example, [dx] was all one matrix, rather than a matrix d and a matrix x. By [itex][dX]=T[dx][/itex], I meant

    [tex]\begin{bmatrix}dX^0\\ dX^1\\ dX^2\\ dX^3\end{bmatrix}=\begin{bmatrix}\alpha & \beta & 0 & 0\\ \mu & \nu & 0 & 0\\ 0 & 0 & a & 0\\ 0 & 0 & 0 & b\end{bmatrix}\begin{bmatrix}dx^0\\ dx^1\\ dx^2\\ dx^3\end{bmatrix}[/tex]

    The transformation matrix is

    [tex]T=\begin{bmatrix}\alpha & \beta & 0 & 0\\ \mu & \nu & 0 & 0\\ 0 & 0 & a & 0\\ 0 & 0 & 0 & b\end{bmatrix}[/tex]

    and the capital superscript T means the transpose, flipping the matrix about the main diagonal; in this case beta (hello beta) and mu (hello mu) swap places.

    [tex]T^T=\begin{bmatrix}\alpha & \mu & 0 & 0\\ \beta & \nu & 0 & 0\\ 0 & 0 & a & 0\\ 0 & 0 & 0 & b\end{bmatrix}[/tex]

    In terms of indices, for a square matrix A,

    [tex](A^T)_{ij}=A_{ji}[/tex]

    I used round brackets to show the scope of the transpose operation. So by [itex](T[dx])^T[/itex], I meant the transpose of the matrix T[dx]. In rearranging, I used the following propery of transposes

    [tex](AB)^T=B^TA^T[/tex]

    I used the sign convention where

    [tex]\eta=\begin{bmatrix}1 & 0 & 0 & 0\\0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1\end{bmatrix}[/tex]

    but it turned out the question was using the other sign convention, which is just my eta times minus one. What's your book, by the way?
     
  14. Jul 5, 2010 #13
    O i see now, thanks a lot that last post cleared up a lot of confusion i had. Its not that i found your posts confusing but its just i am getting used to all of the mathematical notations and such. I'm self teaching myself and only have first year calculus/ linear algebra carrying me so getting used to the matrix notations is confusing but im starting to get the hang of it now.

    The book I'm working through is A First Course in General Relativity By: Bernard F.Schultz
     
  15. Jul 5, 2010 #14
    Thanks Rasalhague. Very clear and definite.

    May be shown that M is symmetric

    [tex]M=T^T{\eta}T[/tex]

    [tex]M^T=(T^T{\eta}T)^T=T^T(T^T{\eta})^T=T^T{\eta}T[/tex]
     
  16. Jul 6, 2010 #15
    Me too, only I started learning calculus and linear algebra from the Khan Academy videos on YouTube and MIT Open Courseware. Without the internet I'd never have known where to begin.

    Thanks, zzzoak. Good point.
     
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