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What is a straight world-line?

  1. May 23, 2016 #1

    I am puzzled by a question about the concept of action. I have:

    [tex]S=\int_{\tau_{1}}^{\tau_{2}}-mc^2d\tau = -mc^2\int_{a}^{b}ds[/tex]
    [tex]ds = \sqrt{c^2dt^2-dx^2-dy^2-dz^2}[/tex]

    The textbook says: "By the principle of least action
    [tex]\delta S =0[/tex]
    and so
    The integral takes its maximum value along a straight world-line and so this implies reassuringly that free particles move along straight lines."

    I don't understand the last sentence. What is a straight world-line? Is it a path along the time axis? and how do we show the result mathematically?

    Thank you for any help!
  2. jcsd
  3. May 23, 2016 #2
    A world-line is a path through space-time (generally speaking, this is a coordinate system with one time dimension and three space dimensions). In one-dimensional space, a world-line is a simple position-time graph. The last equation in your question simply states that a free particle will always trace a straight line when traveling between two points in such a coordinate system. This means a geometrically straight path and constant velocity. Mathematically this is follows from the fact that the interval covered between the two points has to be an extremum (which in this case is a minimum) and a straight line is the shortest distance between two points.
  4. May 24, 2016 #3
    Thank you, I see better what a world-line is now. But this is still unclear how the math works here. If I start with:

    How do I show that the maximum value of S is obtained for the minimal path (straight-line) between a and b? Which argument am I supposed to use in 4 dimensions? Does that minimize space coordinates, like dx=0? Or do you simply imagine that ds is analogous to a 1D path dx?
  5. May 24, 2016 #4
    When integral(ds) is minimized, S is also minimized since the two are proportional. Note that S has to be a local minimum (not maximum) according to the principle of least action.

    The expression sqrt( c ^ 2 * dt ^ 2 - dx ^ 2 - dy ^ 2 - dz ^ 2) represents the length of a small path taken in Minkowski space (which is a 4D coordinate system consisting of three real spatial dimensions and one imaginary time dimension). More information on Minkowski space can be found at en.wikipedia.org/wiki/Minkowski_space.

    This is similar to how if you have a curve in x-y coordinates, the arc length of a small part of the curve is given by sqrt (dx ^ 2 + dy ^ 2), as found in standard calculus books. In 1D, yes, ds = dx simply. The only difference is that now we are adding time as a dimension, and making it imaginary (so that its square is negative). This explains the shift between signs in the square root (in this case the whole expression under the square root is multiplied by -1 but this is really just a matter of convention).

    Anyway, imagine that I say delta{ integral { sqrt ( dx ^ 2 + dy ^ 2) }} = 0. This is saying that the integral is an extremum (i.e., a minimum). So, integral{ sqrt( dx ^ 2 + dy ^ 2) } is a minimum. This simply says that the total arc length of the curve between the arbitrary points a and b is minimized. So, the curve can only be a straight line since this is the path that minimizes arc length between two points. The space-time equation is saying exactly the same thing, just now we have an extra dimension of space and also a dimension of time.

    Note that it is not correct to say dx = 0. All this equation is saying is that the arc length is minimized (basically the derivative of the arc length is what is equal to zero). The particle must follow a path given by z = Ax + By + Ct (a straight line in the above-mentioned coordinate system).
  6. May 25, 2016 #5
    My book disagrees with you on one point. It states that for the straight line the integral takes its maximum value and not minimum. It does not say much more about that but there is a little graph in the corner in 2D (x,t) with two points A and B linked by a line parallel to the t axis and another random path which is not a straight line. Below the graph, the authors say: "The straight-line path from A to B has a larger [tex]\int_{a}^{b}ds[/tex] than the wiggly one (because [tex]cdt > \sqrt{c^2dt^2-dx^2}[/tex]), even though the wiggly one looks longer."

    Actually this is why I asked if we could set dx=0, because for the path along the t axis, x does not change and the authors also make this comment about cdt being larger than the square root including dx. I like your comment and the analogy with other dimensions but it is still not clear to me why my book says that the minimal path maximizes the action if we cannot set dx=0 for this particular path. Is the integral too much general to develop some mathematical derivation further instead of using only intuition? For example can we derive a straight line equation like the one you propose (z = Ax + By + Ct) starting from the derivative of the action? So far I don't see it.
  7. May 25, 2016 #6
    Yes, I made a mistake on that point. The action is minimized but the negative sign in front of the integral means that the integral must actually take a maximum. It is probably not as intuitive to prove this as I first thought (due to the differing signs in the square root), but the solution is still a straight line. The mathematical proof relies on showing that the Euler-Lagrange equations give a solution with zero curvature. Details can be found at http://eagle.phys.utk.edu/guidry/astro421/lectures/lecture490_ch5.pdf (section 5.0.5).
  8. May 25, 2016 #7
    Thank you very much! One last question: in this case what is the process to go from the Lagrangian they define to the Euler-Lagrange equations? I tried to consider S as a functional and take its derivative to obtain the equation but I am stuck with this method. Then with the equation I think I understand well how we get the straight line.
  9. May 25, 2016 #8
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