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. 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* 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 * *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 -- 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/94deb10a-65fd-4ee7-84a2-2a2267bd7705n%40googlegroups.com.
