On Tue, Dec 24, 2024 at 12:12 AM Alan Grayson <[email protected]>
wrote:

>
>
> On Monday, December 23, 2024 at 3:04:58 PM UTC-7 Jesse Mazer wrote:
>
> 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?
>
>
> *In the car's rest frame, the back end of garage is at x' = 16, but in the
> garage's rest frame, front of car is x = 25 (not x'), so you can't subtract
> apples from oranges. AG *
>

No, the car's rest length is 25, so that's its length in the *car's* rest
frame, by definition. Since the back of the car is at rest at x' = 0 in
this frame, the front must be at rest at x' = 25 in this frame.

Meanwhile the car is moving at 0.6c in the garage's rest frame so its
length is Lorentz-contracted down to 25/1.25 = 20, meaning at t = 0 the
front of the car is at x = 20 in the garage's rest frame.

Jesse

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