On 11/3/2025 9:20 AM, Alan Grayson wrote:
On Saturday, November 1, 2025 at 9:07:12 PM UTC-6 Alan Grayson wrote:
On Saturday, November 1, 2025 at 5:15:36 PM UTC-6 Brent Meeker wrote:
On 10/31/2025 10:36 PM, Alan Grayson wrote:
On Friday, October 31, 2025 at 4:15:29 PM UTC-6 Brent Meeker
wrote:
On 10/31/2025 6:17 AM, Alan Grayson wrote:
On Friday, October 31, 2025 at 2:40:07 AM UTC-6 Alan
Grayson wrote:
1) For a body at rest, we multiply clock time, aka
proper time, and/or coordinate time by some
velocity, so its units become spatial. But why
multiply by c? Is this procedure really a
*definition* to get a velocity of c in spacetime?
2) Proper time and coordinate time are not equal
along some arbitrary path in spacetime.
*Note that for a body at rest, coordinate and proper
time are identical. Hence, d(tau)/dt = 1, where t is
coordinate time and tau is proper time. But this is not
true for a body not at rest. How does a physical clock
"know" is it moving, making that derivative non-zero. AG *
You're muddling things. For a clock moving inertially in
flat spacetime, the coordinate times are arbitrary up to
a linear transformation. So d(tau)/dt=const. not
necessarily 1. And the constant depends on the speed
(time dilation). So the coordinate speed depends on the
choice of coordinate time, i.e. relativity of motion.
Brent
*In the video toward the end, he claims d(tau)/dt=1, so every
1 sec increment in coordinate time is set to 1 sec increment
in proper time. *
I don't understand that.
*It's pretty straightforward. If you're at rest in some frame in
spacetime, you're moving along the time axis only. Along that axis
are coordinate labels, but since you've multiplied these lables by
c, you're left with distances (as on spatial axis), and the
distance separation of two adjacent coordinate unit times, has a
distance which light traverses in one second of proper time. IOW,
along the time axis, proper and coordinate time are identical.
Thus, d(tau)/dt=1. *
OK, you've used proper (clock) time to mark the intervals of coordinate
time.
*When motion is not strictly along time axis, that is, when you're
not at rest, coordinate and proper time no longer coincide, no
longer have equal values. The non trivial existential question is
why a clock which measures only proper time, "knows" to adjust its
rate when moving along some arbitrary path in spacetime? AG*
It doesn't "adjust its rate". The clock continues to measure proper
time along the new spacetime direction. But because the new direction
is not parallel to the old one the intervals don't match the intervals
of the clock that remained on the stationary worldline. Motion is only
relative. So each clock sees the other as running slow because they
judge the other clock to not be going in the futureward direction.
Brent
If when you'r.e stationary in some coordinate system you can
adjust you clock to sync with the stationary clocks and so
long as you remain stationary you would have d(tau)/dt=1. But
once you started to move relative that coord system
one-second-per-second in your frame is not
one-second-per-second in the stationary frame. Proper time is
one-second-per-second /along your path thru spacetime/. In
general it will be shorter as measured from the stationary
frame, but remember /inertial/ motion is strictly relative so
you will see the stationary clocks as running slow.
*With c multiplied by coordinate time along time axis for a
particle spatially at rest*
That would define a forward light cone for that particle. Why
would you do that?
*, isn't this tantamount to a definition with the intended
result that spacetime velocity is c? *
Along a path at velocity c the proper time is zero.
*
*
*No. When at rest, proper time increments, and is
indistinguishable from coordinate time. If this weren't the case,
it would be impossible to make the unintelligible conclusion that
all motion in spacetime is c. AG *
*One can always plot distance versus time, but in this case we must
convert time to units of distance because we want to calculate (ds)^2
and have consistent units (of length). Toward the end of the video the
author does exactly that, using the Pythagorean formula, but I don't
see how he gets the negative sign preceding the spatial
differrentials. AG *
*You refer to time dilation, but this definition seems
unrelated to that concept. The key question is how a physical
clock measures something other than coordinate time when
moving along some arbitrary path? *
Every clock measures only its proper time. Coordinate time in
general is just labels. But for convenience it is usually
chosen to be the the time kept by some clock(s) that are taken
to define a stationary frame (so they're stationary relative
to one another). So any clock moving relative to that frame
will measure proper time different from that frame.
Brent
*AG *
How does a clock "know" it isn't reading coordinate
time, but something else called proper time?
Alternatively, what principle can we apply to put
proper time on a logically necessary footing?
3) When moving along some arbitrary path in
spacetime, the Pythagorean theorem holds; that is,
(ds)^2 = (ct)^2 + (dx)^2. So how do we get a
negative sign preceding the spatial differentials?
Here I'm referring to a YouTube video whose link I
will post later.
4) If (ds)^2 is an *invariant *under SR, does this
hold only for the LT, but is it true for any linear
transformation, as well as non-linear transformations?
AG
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