On Sunday, February 2, 2025 at 9:39:04 PM UTC-7 Jesse Mazer wrote:

On Sun, Feb 2, 2025 at 11:04 PM Alan Grayson <[email protected]> wrote:

*In mathematics, functions have domains and ranges, standard terminology. A 
function maps domain sets to range sets. The image of a function is the set 
containing its range. Again, standard mathematical terminology. The 
contraction formula, derived for the LT, is a function. With me so far? AG*


In math the domain and range of a function have no physical interpretation, 
they are just sets of mathematical objects with no comment about what they 
"mean". For example, in purely mathematical terms, the length contraction 
formula l = L*sqrt(1 - v^2/c^2) has as its domain values of the variables L 
and v where L can be any member of the set of real numbers > 0, and v is 
any member of the set of real numbers larger than or equal to 0 but smaller 
than the value of c (in whatever units you're using like 299792458 
meter/second), and the range is the value of l which can likewise be any 
member of the set of real numbers > 0.

If you are speaking more metaphorically, basically just saying that any 
equation like length contraction can be thought of as a sort of machine 
that takes two values (a proper length L and a speed v) as input/"domain" 
and spits out another value (a length l in whatever frame measured the 
object to be moving at v) as output/"range", then I'm fine with that.


*Now about the substance. If coordinate frame O2 is the domain of the 
formula function, the x values are elemments in its domain, and the x' 
values are elements in its range.*


If you are talking about the LT function when applied to a problem where we 
start with coordinates in O2 as input and get coordinates in O1 as output, 
sure. If you're talking about the length contraction formula, no, the only 
inputs to that are a proper length and a velocity in a single frame, and 
the output is the length in that same frame.


*Here's where we disagree. ISTM, there's a convention, that the image frame 
of the formula, is moving wrt the frame applying the contraction formula, 
and L' is the contracted length in the frame in relative motion.  If you 
claim the contraction occurs in the same frame from which the formula is 
applied, then won't we get no contraction? AG *


Also note that there is nothing in the LT formula that restricts which 
frame you take as input and which you get as output. You can just as easily 
start with the coordinates in O1 as input and use the formula to find the 
coordinates in O2 as output. I just said this in the post above and even 
offered to give you a numerical example of how this would work.


*Of course, but why do that? AG * 

 

* The contraction formula is a mapping or correspondence from coordinate 
sets O2 to coordinate sets O1, the moving frame in relative motion wrt O2.*


The contraction formula doesn't do that, the LT *can* be used to do that if 
you start with coordinates in O2 as input and then get coordinates in O1 as 
output, but as I say above it can just as easily map coordinates in O1 
taken as input to coordinates in O2 given as output.
 

* That is, the contraction formula maps O2, the frame with no rod, to O1, 
the frame with the rod.*


I asked you several times in the last comment (and in a number before that) 
to clarify if by "no rod" you just mean the rod is not *at rest* in O2, or 
if you mean that O2 literally doesn't "see" the rod to assign it 
coordinates, or something else.


*There is one rod is at rest in O1, and its frame in moving relatively wrt 
O2. AG*

 

* You say, and I now agree, that there's no contraction of the one and only 
rod in O1.*


Yes.
 

* So what happened to contraction?*


The rod is contracted in other frames like O2, and as I said you're free to 
use the LT to start with the coordinates in O1 used as input and then use 
the LT formula to get the coordinates in O2 as output (once you know the 
coordinates of the front and back of the rod in O2 it's easy to get the 
length of the rod in O2 from that). And if you use the length contraction 
formula rather than the LT it's an even simpler matter to derive the rod's 
contraction in O2, you just need the rod's proper length L as well as its 
velocity v in O2, you enter that L and v into the length contraction 
formula as input and get the contracted length l in O2 as output.


*When we map from O2 to O1, you agreed, no contraction, so if we map from 
O1 to O2, won't there also be no contraction? And if we map from O2 to O2, 
using the rod's coordinates in O2 as input, won't there also be no 
contraction? AG *


Jesse
 

* So, AFAICT, contrary to what relativity claims, contraction doesn't 
exist! Note also what happens to the Parking Pardox. No contraction of any 
object in any frame. Paradox solved! Are we having fun yet? AG*


On Sunday, February 2, 2025 at 7:52:55 PM UTC-7 Jesse Mazer wrote:

On Sun, Feb 2, 2025 at 8:17 PM Alan Grayson <[email protected]> wrote:

On Sunday, February 2, 2025 at 4:31:48 PM UTC-7 Jesse Mazer wrote:

On Sun, Feb 2, 2025 at 5:08 PM Alan Grayson <[email protected]> wrote:

On Sunday, February 2, 2025 at 2:43:09 PM UTC-7 Jesse Mazer wrote:

On Sun, Feb 2, 2025 at 12:44 PM Alan Grayson <[email protected]> wrote:

I will study your post and respond later. For now, let me say that the GPS 
situation is irrelevant. It just shows that time dilation is real. Nothing 
to do with length contraction. Also, after reading some of your earlier 
comments, I agree that in the frame containing the rod, its length is not 
contracted. This is the rest frame with the rod at the origin. The frame 
from which the LT is applied has an observer at the origin, but no rod, and 
is in relative motion compared to the frame with the rod. I hope you have 
no objections to this comment. If you have any objections, please let me 
know. AG


Mostly sounds fine but the only thing I'd want to double check is that when 
you say "The frame from which the LT is applied has an observer at the 
origin, but no rod", do you just mean that the rod is not at rest in this 
observer's frame? The rod is still measurable and can be assigned 
coordinates with changing position as a function of time in this observer's 
frame (the observer you called O2 in your earlier post), agreed? 


I want two frames with the rod in one, which I thought was your initial 
model. The rod is situated and fixed at the origin, and there is no rod in 
the frame using the LT;


But as I asked you repeatedly, when you say no rod "in" the O2 frame do you 
just mean there is no rod that's *at rest relative to* the O2 frame, or are 
you somehow denying that any given physical object like a rod is assigned 
coordinates by *all* frames including the O2 frame in which the rod is 
moving?


You still haven't answered this question, and it seems like it might be 
important given some of your other phrases below...
 

 

or if you prefer we can model the situation with a rod in each frame, at 
rest, both at origin, and their rest lengths are unimportant.


No need for two rods, provided you agree above that the O2 frame still 
assigns coordinates to the rod even though the rod is not at rest in that 
frame.


*OK. One rod, and frame with rod is given coordinates in both frames. For 
me, x ---> x' means a LT from frame with no rod, *


Does "frame with no rod" just mean "frame with no rod at rest in it", or 
are you somehow claiming there's a frame that doesn't "see" the rod at all 
in terms of being able to measure it and assign coordinates to it?

 

*to frame with rod, and from this there's the implication of contraction in 
x' frame, with rod. Do you agree or not? AG*

 

Drawing on the GPS situation, from any rod/frame applying the LT, the 
formula IMO predicts the measured length in the frame it is observing,


GPS is distinct because the clocks don't just tick at their natural rate


*They tick naturally and are then reset to presumably synchronize them with 
orbiting clock. AG*

but are artificially adjusted, as I said. If the rod is at rest in O1 and 
moving relative to O2, can we assume we are initially given the coordinates 
of the rod as measured in O2, then then O2 frame is the one "applying" the 
LT to predict the coordinates in the frame O1, so that O1 would be "the 
frame it is observing" in your statement above?


*Yes, except we don't have to assume the moving rod has coordinates in O2. 
AG *


Do you just mean it doesn't have *fixed* coordinates in O2, or do you mean 
it isn't assigned coordinates at all in O2? If the latter, are you 
imagining it's somehow invisible to the O2 observer? If so that's not how 
things work in relativity, the rod is just an ordinary physical object, of 
course the O2 observer is going to be able to measure it as it passes by 
his own system of rulers and clocks, and say things like "when the clock 
attached to the 3-light-second mark on my ruler showed a time of 5 seconds, 
the back of the rod was passing right next to it (as seen in a photo taken 
at that location at that moment, for example), therefore the worldline of 
the back of the rod passes through the coordinates x=3 light seconds, t=5 
seconds in my coordinate system"

 

 

similar to the Earth bound clocks in GPS which predict the time delays in 
the orbiting clocks. For this reason, in the contraction case, the 
frame/observer applying the LT, doesn't predict the contraction in 
observer's own frame (which doesn't exist if there's one rod in the model), 
but in the frame with the rod.


No, as my numerical example shows, if we start with the coordinates of the 
rod in O2 and use the LT to predict its coordinates in O1, we get a 
prediction of NO contraction of the rod in O1; the prediction will be that 
the rod has its "proper length" in the O1 frame.
 

However, and this is where I get my prediction which you object to; in this 
frame, the frame with the rod, the only prediction possible is zero 
contraction.


If you are talking about the type of calculation I describe above, I 
*agree* the prediction would be zero contraction in the O1 frame, which 
matches the fact that no contraction is MEASURED in the O1 frame. 


*Yes, this is what I've been saying. AG*
 

It was you who claimed that there was some prediction using the LT that 
would conflict with the fact we both agree on that no contraction is 
measured in the frame where the rod is at rest.


*I changed my pov when reading one of your previous posts. But since 
there's no contraction measured in frame where the rod exists*


Are you saying the rod literally does not "exist" in other frames in the 
sense of not being measured at all, or are you just saying the rod is not 
at rest in other frames? If you're somehow saying the rod is not assigned 
coordinates at all in the O2 frame, that doesn't make sense, see above.

 

* and is at rest (even though the frame is in relative motion), the LT has 
no other possible predictions, so it seems that length contraction never 
occurs!*


Sure length contraction occurs, in the example it occurs for the O2 
observer who sees the rod in motion. If the rod has a proper length of 10 
light-seconds, and the O2 observer says the BACK of the rod passed by x=3 
and t=5 in his coordinate system, and the FRONT of the rod passed by x=11 
and t=5 in his system, then the distance between the front and back at the 
single moment t=5 in this frame must be 11 - 3 = 8, so that's what the O2 
judges the rod's length to be, a length which is shorter than the rod's 
proper length measured by the O1 observer.

If you want to start from coordinates in one frame and then use the LT to 
predict coordinates in another, and you're asking about how this could ever 
lead to a prediction of contraction, one option would be to *start* from 
the coordinates in O1's frame where the rod is at rest (unlike in my 
numerical example where I started with the rod's coordinates in the frame 
where the rod was moving), then apply the LT to *predict* the coordinates 
in O2's frame where the rod is moving. If we did that instead of the 
opposite, in that case we *would* get a prediction of a contracted length 
in the predicted frame O2. I could give a different numerical example 
illustrating this, if you would actually be interested in reading through 
it.

 

* This is where the rubber hits the road in our disagreement about what the 
LT predicts, and what is measured. If contraction is never measured because 
it never occurs, the "predictions" of the LT are worthless to the point of 
not existing. I hope you're not going to tell me now, that x' in x --> x' 
refers to the frame applying the LT. AG* 

 

Hence, the LT doesn't do as you claim, and it doesn't function like the GPS 
situation. Moreover, I recall you used spacetime diagrams to show length 
*expansion 
*in your frame at relative speed, but never before have I heard or read of 
such a claim, which raises the proverbial red flag. AG  


As I stated repeatedly, by "expansion" I just meant the length would be 
predicted as larger in the second frame (O1 above) compared to the first 
frame where the rod was moving (O2 above), 


*How could that be if the result of the LT is x', not x?*


In my numerical example I treated the primed frame as the one where the rod 
is at rest, i.e. O1 in your terminology. So, going from x-->x' takes you 
from O2 to O1.

Also note however that the complete set of LT equations include both x-->x' 
and x'-->x, so you're free to go in either direction.

 

* The rod is not contracted in O1 even though its moving relative to O2, 
and the length of the rod is not in the image of anything in O2?*


If you want to do the LT from O2 to O1 you have to start with the 
coordinates of the rod in O2, I don't know if that's what you mean by "the 
image of anything" or if you just mean that we don't obtain the O2 
coordinates as a result of a LT.

 

* Remember; the LT prediction in O1 is no contraction. I think this is 
exactly where we disagree about what the formula predicts.*


What do you think we are disagreeing about in terms of what the formula 
predicts? I've said that if you start with the coordinates in O2 and use 
the LT to predict O1, the prediction is no contraction in O1.

Jesse

 

* Maybe in your reply we can finally resolve this issue. AG*

I wasn't talking about expansion relative to the rod's proper length. The 
prediction will be that the rod has its proper length in frame O1, i.e. "no 
contraction" relative to the proper length.

Jesse

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