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Lorentz transformations for spacetime

  1. Dec 30, 2009 #1
    I've tried several hours to understand Lorentz transformations(for space and for time)...it simply dosn't make any sense...I've posted here,on math section,because I need a better mathematical view over it... whitout this I can not understand much out of the restricted theory of relativity,thus the whole modern physics...
  2. jcsd
  3. Dec 30, 2009 #2


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    Welcome to PF!

    Hi physicsXS! Welcome to PF! :smile:

    What stage are you at?

    Have you done calculus? Linear algebra? Geometry? Coordinates?
  4. Dec 30, 2009 #3
    You are familiar with rotations, yes? A rotation matrix looks something like this:

    [tex]\begin{bmatrix} cos x & sin x & 0 \\ -sin x & cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}[/tex]

    Rotations are amazing in math and physics because they have nice properties. In particular, they preserve angles and lengths.

    Well, there is a special kind of function that is very similar to a rotation called a hyperbolic trig function (http://en.wikipedia.org/wiki/Hyperbolic_function). We call them hyperbolic sine (sinh) and hyperbolic cosine (cosh). Their definitions are very similar to the definitions of sine and cosine. They obey a number of similar looking rules. But they are slightly different.

    The most important, if you're learning about relativity, is that instead of preserving length, they preserve spacetime intervals.

    A hyperbolic "rotation" has a matrix which looks like this:

    [tex]\begin{bmatrix} cosh x & -sinh x & 0 \\ -sinh x & cosh x & 0 \\ 0 & 0 & 1 \end{bmatrix}[/tex]

    (Note the sign flip on one of the sinh's).

    So what is a Lorentz transformation? Simply put, it's any combination of (regular) rotations between two space dimensions and hyperbolic rotations between one space dimension and the time dimension.

    So, simply put, a Lorentz transformation is just a generalized kind of rotation. The details are straightforward after you get the general idea.
  5. Jan 2, 2010 #4
    thank you,everyone,for the warm welcome:)
    @tiny-tim : thanks,I'm in 12 grade,and I've done a bit of everything,but not very deep.
    @Tac-Tics : I like your sugestion...I've studied matrixes,but not rotations...I would like to study them also,master them in a week,then I will ask for future asistence.thanks a lot!
  6. Apr 12, 2010 #5
    I think rotations and matrix multiplication of vectors is a little bit above the pay grade for PhysicsXS.

    PhysicsXS - the mathematical derivation of Lorentz is available in Einstein's Book "Relativity"

    If you can accept the equations as is on page 34 of his book you at least have the mathematical mechanics of calculating the coordinates in the moving frame of reference if you know the coordinates in the original frame of reference and the velocity the moving frame of reference with respect to the original F.O.R.

    You do NOT need calculus to understand this. Also, you do not need to understand matrix operations to calculate this. Staying with the simple algebra that Einstein presents is actually more intuitive and more "hands on."

    You have to do a bunch of problems to get the feel of what you are doing.
  7. Apr 13, 2010 #6

    In the above matrix, what does the x represent?

    Also, it is a matrix. What would be the vector you would apply it to so that it would work?

    I presume when you say cosh x you mean [tex]\frac{1}{2}[/tex](ex + e-x) and for sinh x you mean [tex]\frac{1}{2}[/tex](ex - e-x)

    How is that matrix applied? Again, what is the vector to which it multiplies Where is time in that calculation?

    This is CLEARLY above the pay grade of a high school senior. It has been my experience that even those good math students in high school haven't a damn clue as to what they are doing but learn how to manipulate symbols, so the application of this matrix must have some basis in logic and intuition as to why it would work. I didn't understand the true meaning of calculus to years later whenI really understood the concept of the Riemann sum and the anti-derivative. You can always calculate, by numerical methods, a Riemann sum, but not all functions have an anti-derivative yet are still integrable,

    Maybe a simple illustration with a very simple problem may assist in that endeavor.

    Steve G
  8. Apr 13, 2010 #7
    It's a hyperbolic angle. It's analogous to the rotation angle in a rotation matrix. It's a frame-changing operation, so "the vector you apply it to" is actually every vector in the world.

    And yes, Cosh and sihn above are the "hyperbolic trig" functions hyperbolic cosine and hyperbolic sine. They generally aren't taught in high school or even undergraduate math classes (I got a minor in mathematics in college and I never had one teacher mention them in any class). However, their basic definitions and properties are simple to understand for anyone familiar with trigonometry and euler's identity.

    I will point out I made no promise that my answer subscribe to any "pay grades" :P Physics is hard. Math is hard. Some gifted inquisitive personalities can handle calculus at age 12. Some will never get it. But there is no harm in learning the names of important concepts at any age. Even if you have no idea what a hyperbolic rotation is, that's fine. You can pick out the details you can grok and leave the rest as an exercise for an older, wiser you. (Hell, even I don't understand the geometric significance of hyperbolic trig functions. I just know their definitions. I don't even know what cosh(0) is without looking it up.)

    My point wasn't to explain the details of the math. Rather to illustrate the major idea:

    Dilation is just a cousin of rotation.
  9. Apr 13, 2010 #8
    "Pay Grade" is a figure-of-speech. It means, here, that this stuff is above a 12th grade level.

    My other "real" question is, accepting the matrix as is, how would you use it? I presume you would have one vector in -ct, x, y, z and twist and turn it to another vector -ct', x', y', z'. So what I need is an example of a simple vector and a simple translation-rotation to another vector. I don't see where that matrix applies.

    One can use the Lorentz transformations one by one on each component of the primary vector to arrive at the new vector. But with the matrix - I am clueless and pulseless.

    Give me a kickstart.
  10. Apr 15, 2010 #9
    Check out this wiki page for a simple 2D example (1 space dimension, 1 time dimension). You can see it uses the same matrix (just tossing out the extra dimension).

  11. Apr 15, 2010 #10


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    Every event is characterized by the place where it happens and the time when it happens. In special relativity due to the constancy of the speed of light this interval is invariant.
    We can write the infinitesimal interval between two events in one frame as [itex]ds^2=c^2dt^2-dx^2[/itex] and in another frame as [itex]ds'^2=c^2dt'^2-dx'^2[/itex]. The requirement that [itex]ds^2=ds'^2[/itex]
    gives us the equation [itex]c^2dt^2-dx^2=c^2dt'^2-dx'^2[/itex] or for a finite interval [itex]c^2t^2-x^2=c^2t'^2-x'^2[/itex]. A solution to this equation is [itex]x=x' \cosh \phi+ct' \sinh \phi[/itex] and [itex]ct=x' \sinh \phi + ct' \cosh \phi[/itex] (check it). This solution can also be written in matrix form by using the hyperbolic matrix from Tac-Tics post.
    If we substitute [itex]x'=0[/itex] we get [itex] x=ct'\sinh \phi[/itex] and [itex]ct=ct'\cosh \phi[/itex]. From this follows that [itex]\tanh \phi=\beta[/itex], [itex] \cosh \phi=\gamma[/itex] and [itex]\sinh \phi = \beta \gamma[/itex], with [itex]\beta=v/c[/itex] and [itex] \gamma=1/\sqrt{1-\beta^2}[/itex].

    Plug this into the general solution of the equation [itex]s=s'[/itex] and we find the Lorentz Transformations (check it).
  12. Apr 15, 2010 #11


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    Hi PhysicsXS. Have you looked at the transforms in just one spatial dimension as per :

    [tex] \begin{bmatrix} x \\ ct \end{bmatrix} = \begin{bmatrix} \gamma & \gamma v/c \\ \gamma v / c & \gamma \end{bmatrix} \begin{bmatrix} x' \\ ct' \end{bmatrix} [/tex]
  13. Apr 15, 2010 #12
    My working knowledge of the hyperbolic functions is between zero and nothing other than knowing the formula as I indicated in post #6 above.

    uart in post 11 presents a matrix that I understand although it is backwards to me. I think of the Lorentz transformation as Einstein presented them on page 34 in which he expresses x' and t' given x and t. The matrix above will give x and t if x' and t' are given. I suppose if I find the inverse matrix of the above, I would have a matrix that will give me the form that I am used to.

    Call the above matrix presented by uart [A]. Then [A]^(-1)[A] =

    So, multiply what he has above on both sides by [A]^(-1) we get

    [A]^(-1)[x ct]^t = [x' ct']^t or [x' ct']^t = [A]^(-1)[x ct]^t which would be the "correct" format for me.

    Now, I have to remember how to invert a matrix. Also, my skills at using the advanced screen to post an answer are totally lacking. I don't know how to write a matrix or to write the SQRT[1 - v^2/c^2] in proper form.
  14. Apr 15, 2010 #13
    Inverting matrices is pain.

    Inverting rotations and boosts is easy.

    You have some Lorentz transformation L.

    Because it's lorentz, you know it is the product of a finite number of rotations and boosts:

    L = t1 t2 t3 ... tn

    Where ti is the i-th transformation.

    If ti represents a rotation of x degrees around some axis, ti^-1 is just the rotation of -x degrees around the same axis.

    If ti rerpesents a boost of x units (the hyperbolic angle) around some axis, ti^-1 is similarly just a boost of -x units around the same axis.

    Once you know t1^-1 through tn^-1, you can find L^-1 by

    L^-1 = tn^-1 ... t3^-1 t2^-1 t1^-1

    (Notice the order is flipped).

    If you only have L without knowing how to factor it into ti's, you have to use the standard matrix inverse algorithm (which I don't remember).
  15. Apr 15, 2010 #14


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    My post was intended to show you where those hyperbolic functions come from and how they relate to the Lorentz transformations you know. With the identity [itex]\cosh^2x-\sinh^2x=1[/itex] you should be able to verify the solutions.
  16. Apr 15, 2010 #15


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    Hi Stevmg. An essential requirement of those transforms (required by Einstein) was precisely that the inverse transform be almost identical to the forward transform. This is required by the hypothesis that there is no preferred inertial reference frame. In other words, relative to the (x,t) reference frame the (x',t') frame is moving away to the right at speed "v". While relative to the (x',t') reference frame the (x,t) frame is moving away to the left at speed "v". So apart from exchanging "v" with "-v" the two situations are absolutely identical for each reference frame! This is one of the most fundamental physical concepts associated with the transform.

    So what that means is that the transform must be invertible by simply replacing "v" with "-v", how cool is that.

    [tex] \begin{bmatrix} x \\ ct \end{bmatrix} = \begin{bmatrix} \gamma & \gamma v/c \\ \gamma v / c & \gamma \end{bmatrix} \begin{bmatrix} x' \\ ct' \end{bmatrix} [/tex]

    [tex] \begin{bmatrix} x' \\ ct' \end{bmatrix} = \begin{bmatrix} \gamma & -\gamma v/c \\ -\gamma v / c & \gamma \end{bmatrix} \begin{bmatrix} x \\ ct \end{bmatrix} [/tex]

    It's just personal preference, but like to remember the form where all the matrix elements are positive and then use the v to -v exchange trick to get the other one.
    Last edited: Apr 15, 2010
  17. Apr 16, 2010 #16


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    I have always thought that it's really weird that people always try to teach SR without using matrices. Matrices and matrix multiplication are orders of magnitude easier to learn and understand than most things in special relativity, and once you are familiar with them, they make SR significantly easier. You could probably learn matrices and SR in less time than you can learn just SR.

    For physicsXS and stevemg. A matrix is just a bunch of numbers arranged in rows and columns. If we write the number on row i, column j of an arbitrary matrix X as Xij, multiplication of matrices can be defined by

    [tex](AB)_{ij}=\sum_k A_{ik}B_{kj}[/tex]

    Note that the number of columns of A and the number of rows of B must be the same for this to work. That issue doesn't come up much in SR, since we're mostly working with 2×2 or 4×4 matrices. We also define addition, and multiplication by a number:



    I recommend this post for people who want to understand Lorentz transformations better. I also recommend this post for people who want to understand the relationship between linear operators and matrices. (Look up the definition of a vector space first).
  18. Apr 16, 2010 #17
    Tic-Tacs -

    Actually, inverting a matrix is pretty brutal.

    First you find the minors of each term of the matrix which is the value of the matrix which is created by omitting the row and column the term itself is in. You do that for each term. Then you multiply each of those minors by (-1)^(i+j) where i is the row and j is the column of that entry. This new matrix is xcalled the matrix of cofactors. Then you transpose this matrix and this is called the adjoint of the original matrix. You obtain the value of the determinant of the original matrix and divide it into each term of the adjoint of the original matrix which gives you the inverse of the original matrix.

    Am I right? Is that NOT brutal?

    But, to the rescue comes uart with his simple explanation with the negative of the v and voila! We got it.

    I still don't know how to write matrices using this screen with all its latex symbols. I think I am in a surgical suite using latex gloves.

    By the way, Tic-Tacs, using the hyperbolic functions as you pointed out to poor PhysicsXS what would be the correct form of the Lorentz transformations using the x, x', ct and ct'?


    I PAINFULLY know what a matrix is. I can't even print them on this blog and I've tried using the advanced screen. Relearning inverting matrices this afternoon was pure torture!


    You are trully a genius and saved me tons of algebra. We will elect you to the Matrix Hall of Fame. I am in the Matrix Hall of Shame. I see if you do the matrix multiplication you do wind up with Lorentz's original equations as posited by Einstein.

    Now, to all, how the hell did the hyperbolic functions ever get involved with this? Does it go back to Hugo (Lorentz) himself? I didn't notice this notation in Einstein's "Relativity" book.

    Also, to all, do you think we have PhysicsXS (the topic originator) completely turned around, clueless and pulseless? If so, we have done our job.
  19. Apr 16, 2010 #18


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    Hendrik Lorentz (1853-1928)

    Hendrik … see http://en.wikipedia.org/wiki/Hendrik_Lorentz" [Broken] :wink:
    Last edited by a moderator: May 4, 2017
  20. Apr 16, 2010 #19


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    One way to get hyperbolic functions into relativity is as I described in my previous post (#10).
  21. Apr 16, 2010 #20
    Cyosis -

    Get me from point A to B.

    [itex]x=x' \cosh \phi+ct' \sinh \phi[/itex] and [itex]ct=x' \sinh \phi + ct' \cosh \phi[/itex]

    Show me... I can't even get started.

    Where does [tex]\phi[/tex] come from?
  22. Apr 16, 2010 #21
    Tiny Tim

    Sorry - Hendrik Lorentz - Dutch Physicist. I don't know who "Hugo" was.
  23. Apr 17, 2010 #22


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    Stevmg, I agree that the whole thing with the matrix of cofactors is pretty brutal, but you only need it when you need an explicit formula for the inverse of an arbitrary n×n matrix. That formula isn't useful very often. I have only had to use it once. It was during my fourth year at the university. I didn't have to use it, but it was the only way I could prove a claim that was made in a book we we're studying. I talked to some of the others in my class about it, and none of them had any memory of ever even seeing that formula. If it was covered in our linear algebra class three years earlier, we had all forgotten it.

    You wouldn't use that formula to invert an arbitrary 2×2 matrix. Just do this: Let A be an arbitrary matrix. Call its components a,b,c,d. Call the components of the inverse e,f,g,h. Multiply the two matrices together and use the fact that the product is the identity matrix. From this you can determine e,f,g,h in terms of a,b,c,d (if, and only if, ac-bd≠0).

    In SR, we really only need to be able to invert this matrix:

    [tex]\Lambda=\gamma\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}[/tex]



    And it's very easy to verify that all you need to do is to change the sign of v. In one of the posts I linked to in #16, I took another approach. I let [itex]\gamma[/itex] be an unspecified function of v, and used the fact that the principle of relativity suggests that we should assume that the inverse can be obtained simply by changing the sign of v. Then I used that assumption to determine [itex]\gamma[/itex].

    It's not hard to post matrices that look good in this forum. Use tex tags. Start with \begin{pmatrix} and end with \end{pmatrix}. (Use bmatrix instead of pmatrix if you want them to look like the ones uart posted). Put a & between components on the same row, and start new lines with line with \\. Make sure there's a space after each \\. Click the quote button below this line to see how I did the matrix in this post.
    Last edited: Apr 17, 2010
  24. Apr 17, 2010 #23


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    The equation that we need to solve is given in post #10. The solution to that equation is also given in post 10 which you quoted in post 20. Plug this solution into the equation and check that it indeed is a solution.

    The angle [itex]\phi[/itex] is a hyperbolic angle. It doesn't really matter what it is right now. In the same way as [itex]x=\cos \phi[/itex] and [itex]y=\sin \phi[/itex] is a solution/parametrisation to the equation of a unit circle [itex]x^2+y^2=1[/itex] so is [itex]x=\cosh \phi[/itex] and [itex]y=\sinh \phi[/itex] a solution/parametrisation to the equation of the unit hyperbola [itex]x^2-y^2=1[/itex]. Furthermore you can check by using the definition of the hyperbolic functions that most of the identities you know from trigonometry hold for the hyperbolic functions.


    If you look at the Minkowski metric [itex]ds^2=c^2dt^2-dx^2[/itex] you can immediately see that this is the equation of a hyperbola with x=cdt and y=dx for a fixed value of ds.

    Anyhow grab pen and paper plug the solutions in, use the identities and see that it works out.
    Last edited: Apr 17, 2010
  25. Apr 17, 2010 #24
    Cyosis -

    I did work through the inversion the matrix (not the hyperbolic one but the algebraic one) and got it to work...
    I will now go back and work out the cosh and sinh bit.

    Do you think we really now have PhysicsXS completely confused???
  26. Apr 17, 2010 #25


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    The last time PhysicsXS replied to this thread is months ago. I think it is safe to say it is your thread now.
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