0
$\begingroup$

When you draw a space-time diagram and the associated world line for an object moving through space-time, the diagram will be dependent upon the reference frame you are observing from, right? Like for instance the world line for a moving object when taking a reference frame riding along with that object will always be straight up since there does not appear to be any relative motion.

  1. If this the case, then how is it possible to compare two space-time diagrams — wouldn't the fact that they are from different reference frames discredit the validity of making comparisons between them?

  2. Second, when you observe a moving object from your inertial reference frame, that moving object will have a slanted path through space-time, with the planes of simultaneity also tilted but still orthogonal to the worldline of the object, right? Why must these planes be tilted?

  3. Finally, why are the worldlines for light drawn at a 45º angle? Doesn't this imply that the light travels a same distance in space as it does in time? Shouldn't the line be close to 0 slope since it travels a very great distance in a short amount of time?

  4. Finally, since 45º is the line for light, does this mean that you can not have any object with a world line of lower angle, since this would imply that its speed is faster than that of light?

$\endgroup$
3
  • 2
    $\begingroup$ When you say that an object's lines of simultaneity are orthogonal to the worldline of that object, be careful --- they are Lorentz orthogonal, not Euclidean orthogonal. $\endgroup$
    – WillO
    Commented Nov 2, 2015 at 0:20
  • 1
    $\begingroup$ "Finally, why are the worldlines for light drawn at a 45º angle? " Examine the time axis closely. They are usually labeled $ct$ (which has units of length and is scaled so that light has slope of one) or you've been told in some other context that they are drawn in natural units (which is equivalent). $\endgroup$ Commented Nov 2, 2015 at 0:29
  • $\begingroup$ This isn't a complete answer, but just something to add in the meantime. You're right that space-time diagrams are drawn for a particular Lorentz frame, but one doesn't directly "compare" space-time diagrams. You draw the world-line of the particle of interest, choose whatever point you're interesting in looking at, identify whatever stuff you're interested in in your S-T diagram (simultaneous points, etc.), perform a Lorentz transformation on your entire S-T diagram to get into the particle's Lorentz frame at that point, and then make comparisons. $\endgroup$ Commented Nov 2, 2015 at 1:00

1 Answer 1

1
$\begingroup$

The world lines exists independent of the frame you choose. That is, Minkowski space-time is an affine space (like the euclidean space $\mathbb E^n$, not to be confused with $\mathbb R^n$) where there are no frames. Here you can "draw" world lines, and doesn't matter that there is none inertial frames yet. Then, when you select the frame you are actually selecting an inertial frame and some "special" point in the Minkowski space-time to be chosen as the origin (nothing special with this selection).

As Einstein state, simultaneity is a relative concept... so, as long as you stay in this frame, time can be "absolute" for you. So, when you choose a frame, you are choosing a way time flows and a way to measure distances in the Minkowski space-time.

If there was another dimension, we could just jump on it and admire the way you select your frame in the space time (something like admiring a painting) and we probably see that this frame is all twisted and crumpled. But you are not in an extra dimension, you are in the inertial frame, and due to the way we construct this frame, for you this looks kinda like $\mathbb R^4$.

1.- Both frames need to be inertial in first place. Accelerated paths can't be chosen to be inertial paths. Then, a Lorentz transformation is just the way to compare how two inertial frames see something draw in the Minkowski space-time.

2.- Well, what you said is a plane "orthogonal" (remember you aren't in $\mathbb R^n$, here orthogonal means other thing, but I got your point) to the path, is how, respect to your frame, the body that is moving sees simultaneity. This, of course, can't be the simultaneity of your frame. Simultaneity to you are planes defined by the equations $t=0, \,t=1,\,t=\sqrt{2},\,\ldots$.

3.- This is a re-scale of the time. It is of common use in particle physics, which I hate by the way, but it has its advantages. This is called "Natural units", you can found a lot about it online.

4.- Yes, that's what it means. A path that is causal (remember, paths are draw in the Minkowski space-time) is one that its tangent vector, in every point of it, lies inside the null cone (or light cone) with center in this point (with respect to your inertial frame). Causal paths are for particles with mass. Would that mean that there aren't paths that aren't causal?.. well, remember, what is causal for one inertial frame is no necessarily causal for other inertial frame.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.