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. With c multiplied by coordinate time along time axis for a particle spatially at rest, isn't this tantamount to a definition with the intended result that spacetime velocity is c? 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? 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/6c8d8b15-3a65-4bbb-b45f-e5580b0e33d8n%40googlegroups.com.
