On Thu, Feb 6, 2025 at 5:48 PM Jesse Mazer <laserma...@gmail.com> wrote:

>
>
> On Thu, Feb 6, 2025 at 3:46 PM Alan Grayson <agrayson2...@gmail.com>
> wrote:
>
>>
>>
>> On Thursday, February 6, 2025 at 6:48:34 AM UTC-7 Jesse Mazer wrote:
>>
>> On Thu, Feb 6, 2025 at 4:48 AM Alan Grayson <agrays...@gmail.com> wrote:
>>
>> On Monday, February 3, 2025 at 6:44:29 AM UTC-7 Jesse Mazer wrote:
>>
>> On Mon, Feb 3, 2025 at 3:16 AM Alan Grayson <agrays...@gmail.com> wrote:
>>
>> On Sunday, February 2, 2025 at 11:37:48 PM UTC-7 Jesse Mazer wrote:
>>
>> On Mon, Feb 3, 2025 at 12:09 AM Alan Grayson <agrays...@gmail.com> wrote:
>>
>> 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 <agrays...@gmail.com> 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,*
>>
>>
>> The length contraction formula is not exactly translating between
>> different frames at all, at least not in the sense that the input
>> exclusively consists of variables from one frame and the output a variable
>> from a different frame, the way the LT does. In the length contraction
>> formula, both the speed v which is used as input and the contracted length
>> L' which is given as output are measured in the SAME frame, the frame of
>> the observer who sees the object moving at speed v relative to their own
>> frame. The proper length L, which is also used as input, can be thought of
>> as length in a different frame, namely the object's own rest frame.
>>
>> Since there is a symmetry of motion where the speed of B relative to A is
>> the same as speed A relative to B, I suppose the v in the length
>> contraction formula *could* instead be defined as the speed of the observer
>> as measured in the object's rest frame, rather than defining it as the
>> speed of the object in the observer's frame as textbooks normally do. If
>> you think of it in that alternate way where v is the speed of the observer
>> in the object's frame, then you could consider both input variables L and v
>> to be measured in one frame (the object's rest frame) and the output
>> variable L' in another (the observer's frame). Even though v is not
>> normally described this way, it'd make no difference mathematically if you
>> did. But if we do think of it this way, I presume you'd then say the
>> object's rest frame is the frame "applying the contraction formula", and
>> the observer's frame is the "image frame"?
>>
>>
>>
>> * 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 *
>>
>>
>> See above, the normal way of describing the length contraction formula
>> involves a mix of measurements in two different frames as input so there is
>> no single 'frame from which the formula is applied', but if you want, you
>> do have the option to think of the meaning of v in a different way such
>> that both L and v are defined in terms of measurements in a single input
>> frame (the rest frame of the object), in which case the contracted length
>> L' would be in a different output frame (the rest frame of the observer).
>>
>>
>>
>>
>> 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 *
>>
>>
>> Because you asked how the LT could be used to predict a contracted
>> length--in order to do that, the output of the LT formula has to be defined
>> in a frame where the object has non-zero velocity.
>>
>>
>>
>>
>> * 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*
>>
>>
>> But you agree that the observer O2 can measure and assign coordinates to
>> the rod in O2's own rest frame?
>>
>>
>>
>>
>> * 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?*
>>
>>
>> No, since the rod is at rest in O1 and moving in O2, if you map from O1
>> to O2 using the LT you get a contracted length for the output. As I said I
>> could give you a numerical example showing this if you want.
>>
>>
>> *If rod is at rest in 01 and in relative motion wrt O2 (because O1 is the
>> moving frame) and we map from 02 to 01 we get no contraction, but if we map
>> in opposite direction, from O1 to O2, we get contraction? Could you explain
>> how you reach these conclusions? AG *
>>
>>
>> The quickest way to get this conclusion is just to know that the
>> predictions of the LT about length will always match those of the length
>> contraction equation which was derived from it, and if the rod has some
>> known proper length like L=10 and some nonzero velocity like v = 0.6c
>> relative to O2, then in O2 frame its length is 10*sqrt(1 - 0.6^2) = 8, but
>> since it has a velocity of v = 0 relative to O1, then in the O1 frame it
>> must have length 10*sqrt(1 - 0^2) = 10. The LT will agree with this so if
>> you use the equations with O2 as input and O1 as output, you'll have a
>> length of 8 as input and a length of 10 as output (no contraction in output
>> frame). But if you use the LT equations with O1 as input and O2 as output,
>> you'll have a length of 10 as input and a length of 8 as output
>> (contraction in the output frame).
>>
>> If you don't want to just trust the principle of "the LT agrees with the
>> length contraction equation", you can verify this by direct calculation
>> using the LT equations. Let's call O2 the unprimed frame and O1 the primed
>> frame. In the O2 frame we have the following equations for the position as
>> a function of time of each end of the rod (i.e. the worldline of each end):
>>
>> Back of the rod: x = 0.6c*t
>> Front of the rod: x = 8 + 0.6c*t
>>
>> These equations tell you that at t=0 the back of the rod is at x=0 and
>> the front of the rod is at x=8, so the rod has a length of 8 in the O2
>> frame, and it's moving at 0.6c.
>>
>> Then in the primed O1 frame we have the following equations:
>>
>> Back of the rod: x' = 0
>> Front of the rod: x' = 10
>>
>> These equations tell you that at any choice of time coordinate t' the
>> back is always located at x'=0 and the front is always located at x'=10,
>> i.e. the rod is at rest in this frame and has a length of 10 (its proper
>> length).
>>
>> One way to use the LT is to directly plug the full equations of motion in
>> one frame into the LT equations as input, and after a little algebra get
>> out the equations of motion in the other frame as output. That's what I did
>> in that post at
>> https://groups.google.com/g/everything-list/c/ykkIYDL3mTg/m/giZVF9PpDQAJ
>> going from the equations of motion in the frame where the rod was moving
>> (here the O2 frame) to the equations of motion in the frame where the rod
>> was at rest (O1). This does involve a little algebra though. A simpler way
>> of just checking that the LT map between those equations of motion is just
>> to pick the coordinates of some individual points that are along a given
>> end's worldline in the coordinates of one frame, and see that they always
>> map to coordinates of individual points that are along the same end's
>> worldline in the coordinates of the other frame.
>>
>> For example in the O2 frame, the points (x=0, t=0) and (x=3, t=5) and
>> (x=6, t=10) and (x=9, t=15) would all lie along the worldline of the BACK
>> of the rod given by x = 0.6c*t in this frame (I'm assuming we're using
>> units like light-seconds and seconds where c=1). If you take any of those
>> points as input and use the x-->x' LT to find the corresponding coordinates
>> of the point in the primed O1 frame as output, you will get (x'=0, t'=0)
>> and (x'=0, t'=4) and (x'=0, t'=8) and (x'=0, t'=12) [note that the x-->x'
>> equation in this case is x' = 1.25*(x - 0.6c*t) and the t-->t' equation is
>> t' = 1.25*(t - 0.6*x)]. You can see that all of these points do lie along
>> the line x'=0, the equation for the BACK of the rod in the O1 frame.
>>
>> Similarly, in the O2 frame, the points (x=8, t=0) and (x=11, t=5) and
>> (x=14, t=10) and (x=17, t=15) all like along the worldline of the FRONT of
>> the rod given by x = 8 + 0.6c*t in this frame. If you take any of those
>> points as input and use the x-->x' LT to find the corresponding coordinates
>> in the primed O1 frame as output, you will get (x'=10, t'=-6) and (x'=10,
>> t'=-2) and (x'=10, t'=2) and (x'=10, t'=6). You can see all these points
>> lie along the line x'=10, the equation for the FRONT of the rod in the O1
>> frame.
>>
>> Then if you want to go in reverse, starting with O1 coordinates as input
>> and getting O2 coordinates as output, you can use the LT equations for
>> x'-->x which in this example is x = 1.25*(x' + 0.6c*t'), and the one for
>> t'-->t which is t = 1.25*(t' + 0.6*x'). You can verify that the reverse
>> mapping here works too, for example if you take (x'=10, t'=2) as input (a
>> point I listed above as being on the worldline of the front of the rod) you
>> get back (x=14, t=10) as output. So you can verify this way that if you are
>> given the equations x' = 0 and x' = 10 for the rod in the O1 frame
>> (corresponding to a rod at rest with length 10 in the O1 frame), if you
>> pick any point along those lines and map to the O2 frame with the LT, as
>> output you always get points along x = 0.6c*t and x = 8 + 0.6c*t
>> (corresponding to a rod of length of length 8 and velocity 0.6c in the O2
>> frame). So, here the input coordinates represent a rod with its proper
>> length of 10, and the output coordinates represent a rod with a contracted
>> length of 8.
>>
>> Jesse
>>
>>
>>
>> Let's start from the beginning to make sure we're on the same page. Using
>> the contraction formula L' = L * sqrt ( 1 - (v/c)^2), where L is the rest
>> length of rod in frame f1, moving at velocity v wrt frame f2, and L' is the
>> contracted length of the rod as calculated from the pov of f2. Do you agree
>> with my interpretations of these variables? Do you agree that the measured
>> length of rod in f1 is never L', but always its rest length L? These are
>> Yes or No questions. TY, AG
>>
>>
>> As long as we are considering a case where f2 is different from f1 (which
>> is normally the only situation where anyone would bother to use the
>> formula), I agree with everything you just wrote. But the one caveat I'd
>> add is that one is technically free to consider the special case where f2 =
>> f1, i.e. you are imagining that the frame f2 of the observer who is using
>> the equation is the same as f1, the rest frame of the rod, so in this case
>> v=0 and L' = L. In my first paragraph above (the one starting with 'The
>> quickest way to get this conclusion') I had L=10, and I considered both the
>> case of an observer O2 moving at 0.6c relative to the rod who used L=10 and
>> v=0.6c as input to the formula to output a length L' = 8 in his frame, and
>> also the case of observer O1 who was at rest relative to the rod and used
>> L=10 and v=0 as input to the formula to output a length of L' = 10 in his
>> frame. Of course the latter is a trivial case and it's easier to just
>> remember that rest length/proper length is what will be measured in the
>> frame where the rod is at rest, but I included it to illustrate the point
>> that either observer can apply the formula to get the correct length in
>> their frame as the output.
>>
>> Jesse
>>
>>
>> In the scenario posited, you agree that the observer in the moving frame,
>> f1, where the rod is at rest, always measures the rod as having length L,
>> whereas the observer applying the contraction formula, resident in frame
>> f2, always measures the rod as having a contracted length as L', which is
>> always less than L,
>>
>
> I would not call f1 "the moving frame" as that is the sort of
> non-comparative designation I don't like to use. And I also made clear it's
> not *always* true that the observer in f2 measures a length less than L,
> because you can consider the special case where f2 = f1, i.e. the observer
> applying the length contraction formula is at rest relative to the rod
> (observer O1 in the previous discussion)--in that case the rod has length
> L' = sqrt(1 -0^2)*L = L according to the formula. So it would be more
> accurate to say that the length L' which is the output of the length
> contraction formula is always *greater than or equal to* the proper length
> L in the rod's rest frame.
>

Sorry, above I should have written that the length L' which is the output
of the length contraction formula is always *smaller* than or equal to the
proper length L.


>
>
>> why do you deny my claim (so it seems) that the LT, via the contraction
>> formula, never predicts what the observer in f1 will actually measure? AG
>>
>
> I don't understand what "the LT, via the contraction formula" is supposed
> to mean here. If you are continuing to endorse the strange idea that the LT
> would predict a contracted length in f1 (rather than the proper length L
> which is actually measured), then it seems to be an obvious consequence of
> your idea that the LT would have to *differ* in its predictions from the
> length contraction formula. Please tell me if you agree/disagree with the
> following four statements:
>
> 1. If observer O2 has a velocity of v=0.6c relative to the rod, and uses
> the length contraction formula to predict the length in his own frame, then
> the output of the formula is that the rod has contracted length L' = 0.8*L.
>
> 2. If the observer O1 has a velocity of v=0 relative to the rod, and uses
> the length contraction formula to predict the length in her own frame, then
> the output of the formula is that the rod has its non-contracted length L'
> = L.
>
> 3. In order for the LT's predictions to agree with the predictions of the
> length contraction formula when converting between O1's frame and O2's
> frame in either direction, then regardless of which frame's coordinates you
> use as input and which you use as output, the numerical predictions should
> agree with statements #1 and #2 above.
>
> 4. So, if your input coordinates are those of O2 where the length is 0.8*L
> and the velocity is 0.6c (as per statement #1 above), then if the LT agrees
> with the length contraction formula, that means the output coordinates in
> O1 should say that the length is L and the velocity is 0 (as per statement
> #2 above). Similarly if your input coordinates are those of O1 where the
> length is L and the velocity is 0 (as per statement #2 above), then if the
> LT agrees with the length contraction formula, the output coordinates in O2
> should say the length is 0.8*L and the velocity is 0.6c (as per statement
> #1 above).
>
> Agree/disagree?
>
> Jesse
>

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