On Mon, Jan 20, 2025 at 4:13 PM Jesse Mazer <laserma...@gmail.com> wrote:
> > > On Mon, Jan 20, 2025 at 3:30 PM Alan Grayson <agrayson2...@gmail.com> > wrote: > >> Below is what Brent wrote to describe his plots: notice that he uses the >> car moving at 0.8c. But respect to what? He doesn't say. >> > > I would actually quibble with his statement "In both diagrams the car is > moving to the right at 0.8c", since in the diagram for the car's frame the > car isn't moving at all, but I'm sure if asked to clarify he'd say he just > meant that both diagrams show a scenario where the car has a velocity of > 0.8c relative to the garage, so this doesn't actually lead me to be > confused about his scenario the way some of your statements do (in part > because you refuse to give a straightforward numerical example like Brent > did). And everything else in his quote does clearly specify whether he is > talking about what's true in "the car's reference frame" or in "the > garage's reference frame" (naming the frame of a specific object as I have > asked you to do), he uses variations of those phrases several times. > > > >> And then he contracts the car from the garage frame. Is garage frame >> moving or at rest? He doesn't say. So much presumably ambiguous non >> standard terminology and not a peep out of you. >> > > Here you seem confused about what I am saying is "non standard > terminology"--I think it's GOOD that Brent doesn't declare one frame to be > "moving" and the other to be "at rest", because that is precisely the sort > of confusing non-standard terminology YOU use that I'm objecting to! In the > standard terminology, one wouldn't say a given frame is "at rest" or > "moving" except as part of a longer phrase that specifies some other object > or frame those words are supposed to be relative to, like "the car is > moving relative to the garage frame" or "the garage is moving relative to > the car frame" or "the car is at rest relative to the car frame", with > these phrases just telling you something about the velocity of the named > object in the named frame. > > > >> Let's do this; since this discussion has reached the point of tedious >> worthlessness, let's terminate it. AG >> > > As I said, one easy way to avoid terminological confusion would be to > answer my simple request for a numerical example: "give me a specific > number for the rod's speed in the Earth's inertial rest frame, its > direction (in the +x or -x direction), and the initial position of each end > of the rod at t=0 in the Earth frame". > > Is there some reason you are unwilling to give me a few numbers for > velocity and initial position of the rod in the Earth frame to work with? > If you did, I could then show you with a little simple algebra what happens > when we use the LT to transform these numbers into the rod's frame, proving > that when we do this the length of the rod is predicted to be EXPANDED > rather than contracted, compared to its length in the Earth frame. > Since I gave a similar numerical example in my recent comment at https://groups.google.com/g/everything-list/c/QgVdhXi3Hdc/m/SG_JbWiYDQAJ I'm going to re-use it with a few modifications to prove my point above: if we start with the equations for the worldlines of the back and front end of a rod which is moving in the Earth frame, and then use the LT to calculate the equations for the worldlines of the ends of the rod in the rod's own frame, we will see the its length in the rod frame is EXPANDED compared to the Earth frame, not contracted. Say that relative to the Earth frame, the rod is moving at 0.6c in the +x direction and is 8 light-seconds long as measured in this frame, so the equation for the back of the rod could be x = 0.6c*t and the equation for the front of the rod could be x = 8 + 0.6c*t. Now if we want to know the equations in the rod's own frame, we can substitute those expressions for x into the Lorentz transformation's position transformation equation, x' = gamma*(x - v*t). Since the rod frame has v=0.6c as measured in the Earth frame, we have gamma=1/sqrt(1 - 0.6^2) = 1.25, so the LT equation can be written as x' = 1.25*(x - 0.6c*t). Now, if you take the equation for the back of the rod in the unprimed (Earth) frame, x=0.6c*t, and substitute that in for x in the LT equation x' = 1.25*(x - 0.6c*t), you get x' = 1.25*(0.6c*t - 0.6c*t) = 0, meaning in the primed (rod) frame the back end of the rod has a fixed position x' = 0 which doesn't change with time (the rod is at rest in the primed frame). And if you take the equation for the front of the rod in the unprimed (Earth) frame, x = 8 + 0.6c*t and similarly substitute it into x' = 1.25*(x - 0.6c*t), you get x' = 1.25*(8 + 0.6c*t - 0.6c*t) = 1.25*8 = 10, meaning in the primed (rod) frame the front end of the rod is fixed at x' = 10. So you can see that if we start with the coordinates for a rod that in the unprimed (Earth) frame has a length of 8 light-second and is moving at 0.6c, and then we apply the LT equations, we end up with the coordinates for a rod that's 10 light-seconds long and at rest in the primed (rod) frame. Jesse -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To view this discussion visit https://groups.google.com/d/msgid/everything-list/CAPCWU3LE9gQiXF31SnJuuhffdNJm-2rrBvje5Xd%3DBt%2BapvGiFw%40mail.gmail.com.
