On Mon, Dec 23, 2024 at 4:10 PM Alan Grayson <[email protected]> wrote:

> On Sunday, December 22, 2024 at 10:05:54 PM UTC-7 Jesse Mazer wrote:
>
> BTW, since you seem to be interested in a scenario where the car and
> garage are exactly matched in length in the garage frame, something which
> isn't true in Brent's scenario, here's a different scenario you could look
> at, where I'm again using units where c=1, let's say nanoseconds for time
> and light-nanoseconds (i.e. distance light travels in one nanosecond) for
> distance.
>
> --Car's rest length is 25, garage's rest length is 20, car and garage have
> a relative velocity of 0.6c, so gamma factor is 1/sqrt(1 - 0.6^2) = 1.25
>
>
> *OK. *
>
>
> --In garage rest frame, garage has length 20 and car has length 25/1.25 =
> 20. In the car rest frame, the garage has length 20/1.25 = 16 and the car
> has length 25.
>
>
> *OK, assuming car is moving, but I wouldn't call that "in the car rest
> frame" since you have garage length as contracted. AG *
>
>   |  - In both frames, set the origin of our coordinate system to be the
> point where the back of the car passes the front of the garage--then that
> point will have coordinates x = 0 and t = 0 in the garage frame, x' = 0 and
> t' = 0 in the car frame.
>
> *OK.*
>
> --In the garage frame, at t = 0 the front of the car is at the same
> position as the back of the garage, at position x = 20, so that's the
> position and time of the event of the front of the car passing the back of
> the garage in the garage frame.
>
>
> *OK. *
>
>      --In the car frame, at t' = 0 the back of the garage must be at x' =
> 16 (since we know the front of the garage is at position x' = 0 at time
>  t'=0, and using Lorentz contraction in the car frame we know the garage
> has length 16 in this frame), and the front of the car is at rest at x' =
> 25, so a distance of 25-16 = 9 from the back of the garage, which in this
> frame has already passed the front of the car at that moment.
>
> ?
>

You agreed above that in the car frame, the front of the garage was at
position x' = 0 at time t' = 0, yes? And you also agreed that in the car
frame, the garage has length 16, yes? So why would you have any doubt that
if the front end of the garage is at position x' = 0 at time t' = 0 in the
car frame, then the back end must be at position x' = 16 at the same time
t' = 0 in the car frame? That's just what "length" in a given frame means,
the distance between the two ends of an object at a single moment in time
in that frame. To put it another way, if this was just a classical 1D
problem and I told you a rod had length 16 and at t' = 0 the front end was
at position x' = 0, and it was moving in the -x direction, would you have
any doubt the back end would be at x' = 16 at the same moment?

Or do you agree that this is straightforward, but have questions about why
the front of the car would be at rest at x' = 25 (this also seems
straightforward since you agreed its back end is at x' = 0 and its length
is 25)? Or why, granted the back end of the garage is at x' = 16 and the
front end of the car is at x' = 25 at this moment, the distance between
them at this moment must be 9?

Please clarify your confusion on this sentence, and if we can straighten it
out we can move on to your next "?"


> About ambiguities in your defintion of local events, I was referring to
> the comparison of a spacetime event which is transformed to another frame
> using the LT.  Is the transformed event also local? AG
>

If you already know the local physical facts at a given point in spacetime
(like a physical clock reading in the neighborhood of some other physical
event like the front of the car reaching the back of the garage), you don't
transform them at all when switching to a different frame, you only
transform the coordinate labels assigned to these facts.

If you *don't* already know the local physical facts at a given point in
spacetime, but are given some boundary conditions in that frame (like a set
of 'initial conditions', though you can also work backwards from boundary
conditions rather than forward, which is why I think it's better to just
call them 'boundary conditions'), then you can use various equations
derived from LT (like length contraction and time dilation) to *predict*
the local physical facts at a point in spacetime that occurs later or
earlier than the boundary conditions. This is the sort of thing I was doing
in my example when I calculated that the clock at the front of the car
would read -15 at the moment it passed the back of the garage, working
backwards from the boundary conditions at t' = 0 in the car frame, and
making use of the Lorentz contraction equation to figure out the length of
the garage in this frame. If we were given the corresponding boundary
conditions in a different frame and used them to predict what the clock at
the front of the car would read when it passes the garage, we'd get the
exact same answer of -15.

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 [email protected].
> To view this discussion visit
> https://groups.google.com/d/msgid/everything-list/1211e981-797c-40fa-8926-12d6d45b6697n%40googlegroups.com
> <https://groups.google.com/d/msgid/everything-list/1211e981-797c-40fa-8926-12d6d45b6697n%40googlegroups.com?utm_medium=email&utm_source=footer>
> .
>

-- 
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 [email protected].
To view this discussion visit 
https://groups.google.com/d/msgid/everything-list/CAPCWU3JDSmug5qdyhe%3D0C2T_Xa84dQbd5tymggzn8rE9Q%3DGvhg%40mail.gmail.com.

Reply via email to