Fusion, Fission, Quarks and such nonsense.

Discussion in 'Discussions' started by OmniaNigrum, Apr 2, 2012.

  1. DavidB1111

    DavidB1111 Member

    I simply misunderstood you, Omni.
    I thought when you said time would shift back, I thought you meant in a way we would notice.
     
  2. OmniaNigrum

    OmniaNigrum Member

    Ah. That explains it. These subjects are not easy to speak of without occasional miscommunications. Do not worry. :)
     
  3. DavidB1111

    DavidB1111 Member

    Also, I never joke, I am always serious. :)
    I kid, I kid.
     
  4. mining

    mining Member

  5. Lahalito

    Lahalito Member

    Ya! That's a good way to think of things! I've never calculated nonparallel velocity addition myself, actually, but parallel velocity addition is derived in a fairly straightforward way using calculus. Calc ain't actually that difficult once they teach you the simplified version of things. And for everything that looks like it will be difficult, there are well-known rules (for which Newton has surely written extensive proofs). Typically they teach the proof of derivatives first:
    Take some polynomial function, say, y = x^2 + x^3. Now we want to make a function that described the slope at every point. To do this, we choose to measure the average slope over some distance h. It's the same as every other slope we calculate in geometry: Rise/run. Here, the "run" part is h, and the "Rise" is the difference between the y(x) position and the y(x+h) position. dy/dx is a notation thing, it just means "the derivative of y with respect to the variable x."


    So here we begin by writing dy/dx, and then the final position along the curve, at x+h, minus the initial position, at x:
    dy/dx = [ (x+h)^2 + (x+h)^3 - x^2 - x^3 ] / h in the limit as h-->0 Taking the limit is important. We want the slope at that point, so we have to construct the equation such that, eventually, we don't "run" anywhere.

    We perform the multiplication of (x+h)^2 and (x+h)^3, and write it down. Notice some terms have 0 factors of h, some have 1, and some have more than one factor of h.
    =[x^2+2xh+h^2 + x^3+3(x^2)h + 3xh^2 +h^3 - x^2 - x^3] / h

    Match like terms and subtract. This eliminates x^2 and x^3
    =[2xh+h^2+3(x^2)h + 3xh^2 +h^3] / h

    Every terms that's left has a factor of h, so perform the division by h. That "divide by h" term came from the "run" part of the slope
    =[2x+h+3(x^2) + 3xh +h^2] in the limit as h-->0

    OK now take the limit as h-->0. This means that we shorten the "run" h until it's gone, and we have a pointlike description of the slope!
    = 2x+3x^2

    OK that's calculus. The easy way is this:
    Describe the polynomial by powers of x. For instance:
    y = 5 + x + 3x^2 + x^3 ---> y = 5x^0 + x^1 + 3x^2 + x^3
    Next, to perform a derivative, first multiply each term by the current power of x. Then subtract 1 from that power:
    dy/dx = 0*5x^-1 + 1*x^0 +2*3x^1 + 3*x^2
    = 0 + 1 + 6x + 3x^2

    Not that bad! "Multiply by a power, lower by a power." That's calculus. There are rules you can learn for more complex things like e^ax or ln(x), as well as (ax+b)^p where p is some constant, but whatevs.

    Anyway, the point is you are well on your way to deriving the expression for relativistic velocity addition.
     
  6. mining

    mining Member

    Doesn't velocity addition just rely on the 'proper time' derivative of time, which is the Lorentz factor? This is where I get fuzzy :).

    Also, I know calculus/integration over two variables in two dimensions for a fair few cases, but not some of the substitutions.
     
  7. Lorrelian

    Lorrelian Member

    Even with substitutions and multiple variables, derivation and integration aren't that bad. I always had trouble when I hit related rates and similar problems. I passed my first two semesters but I'll confess that the difficulties I had were a large contributing factor to my changing majors to Journalism.
     
  8. mining

    mining Member

    The only thing that really makes me rage is surface integration. Don't like that at all.
     
  9. lccorp2

    lccorp2 Member

    Me, I loathe Bessel Equations myself.

    Damned cylindrical diffusion problems...
     
  10. Lahalito

    Lahalito Member

    Yeah! >=(

    Hmm. Ya, I think that's a fair thing to say: Velocity addition depends on the difference between proper time (typically tau) and coordinate time t. For slow things there's not enough of a difference, and velocity addition is fairly straightforward.
    This is probably because tau is invariant under lorentz transformations. I think my personal understanding of tau is kind of wrong. Like, my intuition isn't actually what tau is supposed to be, but it gives you good numbers. So I'm kind of disinclined to try to describe it that way.

    Wiki there gives you the official definition: tau = Integral ( square root [ -g]) where you integrate over the path of the object in space-time (that is, in coordinate time t, and spatial coordinates). It's not easy for me to picture what the metric means though. It's a tensor that describes the space-time in a region, and it is also awesome. It lets you do all of general relativity. You might say that the metric is in fact a derived property of the curvature tensor of a space, so in that sense, it might be a "fake" thing where the curvature is the "real" thing.

    related rates problems? I don't think I know what you mean.=P
    I kind of skipped over some important parts of calculus so I could keep doing physics and have been playing catch-up ever since. Most difficult calculus things are intended to just be looked up in a table of integrals. Russians have already done the calculations for us.