Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lorentz boost help

  1. Apr 28, 2010 #1
    I was reading a section on lorentz boosts and i need some help understanding what they did:

    the book starts off by defining the line element dS where:

    (dS)^2 = -(CΔt)^2 + dx^2 + dy^2 + dz^2

    then they say: "consider the analogs of rotations in the (ct) plane. These transformations leave y and z unchanged but mix ct and x. The transformations with this character that leave the analogies of rotations of (3.9) but with trig functions replaced by hyperbolic functions because of the non euclidean nature of space time. Specifically

    ct= [cosh(theta)]*[ct] - [sinh(theta)]*x
    x = [sinh(theta)]*[ct] + {cosh(theta)]*x
    y= y
    z = z

    and 3.9 was

    x = cos(gamma)*x - sin(gamma)*y
    y = sin(gamma)*x + cos(gamma) * y

    So what i dont understand is why did they decide to use the hyperbolic functions in 4 dimensions?
  2. jcsd
  3. Apr 28, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If you use a transformation with an ordinary sin and cos in it, you get a rotation. If you keep on rotating the x-t plane, at some point you'll have rotated it so it's 180 degrees upside-down. At that point you've reversed the direction of time. This violates causality.
  4. Apr 28, 2010 #3
    Thanks! That helped out a lot.

    So by using hyperbolic sin/cos they were able to create the same shift without violating causality.

    Do you have any links that further explain this in more detail?
  5. Apr 28, 2010 #4


    User Avatar
    Gold Member

    You have not been paying attention, have you. An ordinary rotation in 3D space leaves the length of a vector unchanged, where the length element is [itex]ds^2=dx^2+dy^2+dz^2[/itex]. Rotating this by [itex]\theta[/itex] gives in the xy plane gives

    ds'^2= (cos(\theta)dx+sin(\theta)dy)^2+(sin(\theta)dx-cos(\theta)dy)^2+dz^2
    and because [itex]cos(\theta)^2+sin(\theta)^2=1[/itex] it follows that [itex]ds'=ds[/itex].

    In the ct-plane, we wish to preserve [itex]ds^2=dt^2-dx^2-dy^2-dz^2[/itex]. If you do the calculation as above with cosh and sinh instead, you find [itex]ds'=ds[/itex] because [itex]cosh(\theta)^2-sinh(\theta)^2=1[/itex]. The geometry of the cx plane is not Euclidean.

    [posted simultaneously with bcrowell]
  6. Apr 29, 2010 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

  7. Apr 29, 2010 #6
    In 4 dimensional spacetime intervals of constant length are hyperbolas. Also it is sometimes easier to work with hyperbolic functions

    tanh(\theta) = v/c

    The addition of velocities is reduced to adding hyperbolic functions.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook