On 10/25/2024 2:49 PM, Alan Grayson wrote:
On Friday, October 25, 2024 at 11:34:13 AM UTC-6 Jesse Mazer wrote:
On Fri, Oct 25, 2024 at 5:44 AM Alan Grayson <[email protected]>
wrote:
On Friday, October 25, 2024 at 2:44:06 AM UTC-6 Brent Meeker
wrote:
On 10/25/2024 1:36 AM, Alan Grayson wrote:
On Thursday, October 24, 2024 at 11:07:18 PM UTC-6 Brent
Meeker wrote:
On 10/24/2024 5:46 PM, Alan Grayson wrote:
On Thursday, October 24, 2024 at 1:30:32 PM UTC-6
Brent Meeker wrote:
Here's how a light-clock ticks in when in
motion. A light-clock is just two perfect
mirrors a fixed distance apart with a photon
bouncing back an forth between them. It's a
hypothetical ideal clock for which the effect of
motion is easily visualized.
These are the spacetime diagrams of three
identical light-clocks moving at _+_c relative
to the blue one.
*Three clocks? Black diagram? If only this was as
clear as you claim. TY, AG*
*You can't handle more than two? The left clock is
black with a red photon. Is that hard to comprehend?
Didn't they teach spacetime diagrams at your
kindergarten?
Brent
*
*What makes you think you can teach? *
*That I have taught and my students came back for more.*
*I can handle dozens of clocks. I know what a spacetime
diagram. It was taught in pre-school. Why did you
introduce a red photon? A joke perhaps? How can a clock
move at light speed? *
*None of the clocks in the diagram are moving at light
speed. The black one and the red one are moving at 0.5c
as the label says. What is it you don't understand about
this diagram?
Brent
*
*One thing among several that I don't understand is how the LT
is applied. For example, if we transform from one frame to
another, say in E&M, IIUC we get what the fields will actually
be measured by an observer in the target or primed frame. (I
assume we're transferring from frame S to frame S'). But when
we use it to establish time dilation say, we don't get what's
actually measured in the target frame, but rather how it
appears from the pov of the source or unprimed frame.
Presumably, that's why you say that after a LT, the internal
situation in each transformed frame remains unchanged (or
something to that effect). AG*
Can you give a concrete example? If you some coordinate-based
facts in frame S (source frame) and use the Lorentz transformation
to get to frame S' (target frame), the result should be exactly
what is measured in the target frame S' using their own system of
rulers and clocks at rest relative to themselves (with their own
clocks synchronized by the Einstein synchronization convention).
Jesse
*Glad you asked that question. Yes, this is what I expect when we use
the LT. We measure some observable in S, use the LT to calculate its
value in S', and this what an observer in S' will measure. But notice
this, say for length contraction. Whereas from the pov of S, a moving
rod shrinks as calculated and viewed from S, the observer in S'
doesn't measure the rod as shortened! This is why I claim that the LT
sometimes just tells how things appear in the source frame S, but not
what an observer in S' actually measures. AG*
*Yes, although "appear" can be misleading when you consider things
moving near light speed. More accurate is "measure", using the
invariant speed of light.*
*
*
*On another point concerning time dilation; I demonstrated that given
two inertial frames with relative velocity v < c, it's easy to
synchronize clocks in both frames provided we know the distance of
clocks from the location of juxtaposition, but I was mistaken in
concluding this alone shows time dilation doesn't exist. It does,
because we insist on using the LT as the only transformation between
these frames, and the reason we do this is because the LT is
presumably the only transformation that guarantees the invariance of
the velocity of light. So time dilation is, so to speak, the price we
pay for imposing the invariance of the velocity of light on our frame
transformation. But I remain unclear how a breakdown in simultaneity
resolves the apparent paradox of two frames viewing a passing clock in
another frame, as running slower than its own clock. AG*
*Look at the diagram I provided. At the bottom (t=0) the three clocks
are passing by one another. The blue clock sees the other two as
running slower.*
*
*
*Finally, for Brent, a word about "snarky". _You_ get snarky when I
don't understand something, like your "kindergarten" reference in one
of your recent replies. And occasionally I am correct in my
criticisms. Moreover, if you have typos in your explanation of your
graph, you shouldn't be surprised if they make it hard to understand
your graphical explanation of time dilation. AG*
*So that one typo, which was correct elsewhere made it muddled for you?
Brent*
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