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. If when you're 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.
*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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