Lorentz Transform Troubles: Seeking Help with Equations and Substitutions

[OneEye]

Can anyone help me with this?

I was tinkering with the Lorentz transform and ran into some trouble.

To begin with, I was looking at the idea that:

$${ x \over t } = v = { x^\prime \over t^\prime }\eqno 1$$

which I think is probably correct, but along the way I came up with:

$$\begin{equation*}<br /> \begin{split}<br /> {{x^\prime \over t^\prime} &= {(x-vt)\gamma \over (t-{v\over c^2}x)\gamma}}\quad\quad \eqno 2\\<br /> &= { x-vt \over t-{v\over c^2}x }\quad\quad \eqno 3<br /> \end{split}<br /> \end{equation*}$$

So far, so good. But in trying to simplify from equation 3 to equation 1, I substituted $v={x \over t}$ into equation 3 - and what happened then was not pretty:

$$\begin{equation*}<br /> \begin{split}<br /> {{x^\prime\over t^\prime} &= { x-{x\over t}t \over t-{v\over c^2}x }}\quad\quad \eqno 4\\<br /> &= { x-x \over t-{v\over c^2}x }\quad\quad \eqno OOPS!<br /> \end{split}<br /> \end{equation*}$$

No, I am not working for Starthrower.

Can anyone help me?

 

[arildno]

1 is wrong
(The observed velocities are in opposite directions..)

 

[turin]

1 is wrong
(The observed velocities are in opposite directions..)

Actually, there is absolutely no a priori relationship whatsoever between the coordinate velocity of a particle in two arbitrary frames. For instance, if the unprimed frame is the rest frame of the particle and the primed frame is some other, then v = 0 whereas v' = "not zero."

EDIT:
I should point out that I am assuming we are talking about some geodesic of some point particle in flat space-time.

 

[OneEye]

So, let's assume that the two frames are observationally connected and, rather than using particles, let's use (say) a fellow departing the sun using a solar sail, northward, out of the ecliptic (v=1/400th c). Thus, the two frames are not arbitrary, but are related.

One of the points here has to be that, relativistically, there is no positive or negative v. Right?

I still am missing some simple, obvious fact. The math is not all that complex - just a little algebra - but if my subsitution in step 3 is valid, then the Lorentz transform will always result in 0. This is certainly not right.

Can anyone help me here? I must be making some dumb mistake.

 

[turin]

OneEye,
I feel that your mistake is in interpretation, as evidenced in equation 1. Can you please define for us what are x, t, x', t', and v? Are you using equation 1 to do this? If so then the other equations are not correct (because in that case v would not be the velocity that shows up in the Lorentz transformation). Are we dealing with a free particle in a 1+1 D space-time?

 

[OneEye]

x and t are the proper measurements; x' and t' are the transformed measurements. v is the relative velocity.

It is clear that Einstein uses at least $v={x\over t}$ in the derivation of the Lorentz transform.

Use the train and embankment if that helps. I am quite surprised at what a tangle I have arrived at in such a few steps.

 

[turin]

x and t are the proper measurements; x' and t' are the transformed measurements. v is the relative velocity.

I find this ambiguous. Conventionally, unprimed coordinates are taken from initial observation (or "proper" observation, if you will), and then the primed coordinates are taken from a transformation. I suppose (as usual) I didn't make myself very clear. I'll give you two examples:

x is a function of t, x(t), such that x/t = v for all x,t. Therefore
x(t) = vt. This can be abreviated for clarity as
x = vt.

x' is a function of t', x'(t'), such that x'/t' = v for all x',t'. Therefore
x'(t') = vt'. This can be abreviated for clarity as
x' = vt'

If these coordinates describe the trajectory of the same particle, then x' = x and t' = t. Therefore the velocity that appears in the Lorentz transformation that takes you from (x,t) to (x',t') is zero (or, more explicitly, &beta; = 0, &gamma; = 1).

If these two sets of coordinates describe two different particles, then I disagree with your equations 2 and 3. The Lorentz transformations do not take you from the coordinates of one particle to the coordinates of the other, they merely "rotate" the space-time that is used to describe some trajectory.

It is clear that Einstein uses ...

Please don't take offense to this: I am not going to discuss with you what Einstein says.

 

[OneEye]

Okay, this is not really helping. I am looking for a basic math or conceptual error which is causing an unexpected result as I study the Lorentz transform.

First off, I am sure that equation 2 is correct. It is a trivial relation between the textbook expressions of the Lorentz transform.

For the moment, I consider equation 1 to be frangible. Intuitively, I think it is correct, but at the same time, this is what is in question, so I will simply drop it for now.

Equation 3 is clearly the obvious elimination of $\gamma$ from numerator and denominator. Totally unremarkable.

The question of equation 4 is whether $v=x/t$ is really correct.

I think that it is, but am more than willing to be corrected.

Here, v is the relative velocity of two RFs - the train, say, and the embankment. x is the proper measurement of distance at (say) the embankment frame. t is the proper measurement of time at the embankment frame.

So, nothing could seem more appropriate than to calculate the relative velocity, v by dividing the proper distance traversed by (say) a marker light on the train car - x - by the proper time which it takes for the marker light to travel the measured distance - t.

This all seems so simple. Yet it cannot be right.

Anyone, anywhere: Please help me with this? This ought to be textbook freshman stuff. I honestly am confused about this, and sure could use an expert's insight.

Thanks in advance.

P.S.

Please don't take offense to this: I am not going to discuss with you what Einstein says.

Whatever floats your boat. Far be it from me to tread on anyone's crotchet.

 

[Doc Al]

I was tinkering with the Lorentz transform and ran into some trouble.

I'm not sure what you are trying to do. The purpose of the LT is to relate measurements made in one frame (unprimed frame) to measurements of the same event made in another frame (primed frame) when the relative speed of those frames is v. (The speed of the primed frame is +v as measured by the unprimed frame, and the speed of the unprimed frame is -v as measured by the primed frame.)

I assume that's what you mean by "v" and x and x'.

To begin with, I was looking at the idea that:

$${ x \over t } = v = { x^\prime \over t^\prime }\eqno 1$$

I'm not sure what you mean by "v" here. If you mean the speed of something as measured in both frames, then you'd better use a different symbol for it so you don't confuse it with "v". Then the speed of the object in the primed frame is Δx'/Δt'; in the unprimed frame the speed is Δx/Δt. These speeds are, of course, not equal.

If, instead, you are using "v" to represent the relative speed of the frames again, then you must be careful. To measure the speed of the primed frame (from the unprimed frame) you must measure the displacement of a point on the primed frame (say x' = 0) in a certain amount of time. In this case v = Δx/Δt, but it's trivial: we measure the position of the x' = 0 (origin of primed frame) to be x = 0 at t = 0. Then it moves to position x = vt at time t. So... v = vt/t = v. But this should be no surprise. To measure the speed of the unprimed frame (from the primed frame), do the same thing, get the same answer (-v).

Just setting x/t = x'/t' assumes that both frames are somehow measuring the same thing, just transformed by a LT. But measurements of relative speed of the frames are NOT measurements of the same thing! One frame measures the movement of the x' = 0 point, the other measures the movement of the x = 0 point. Not the same at all.

Does this help at all?

 

[Doc Al]

one more point...

One more thing I'd like to point out...

...but along the way I came up with:

$$\begin{equation*}<br /> \begin{split}<br /> {{x^\prime \over t^\prime} &= {(x-vt)\gamma \over (t-{v\over c^2}x)\gamma}}\quad\quad \eqno 2\\<br /> &= { x-vt \over t-{v\over c^2}x }\quad\quad \eqno 3<br /> \end{split}<br /> \end{equation*}$$

You can use this to find the speed of the unprimed frame as measured in the primed frame, but only if you are measuring the movement of the origin (x = x' = 0). That's the only point for which Δx = x, Δt = t, Δx' = x', and Δt' = t'. But if you do that, then x always equals zero, so your x'/t' reduces to -v, as it must.

 

[OneEye]

First off, let me say this: I know that people often come on this board to peddle some sort of crackpot theory. I haven't surveyed the board, but I think that this particular forum is probably pretty prone to that sort of thing. And, while I admit that I am not above peddling a crackpot theory, that is not what I am up to here. I accept the Lorentz transformations de fidei as being the appropriate way to handle light-based measurements. So what I have here is a silly math or conceptual error, and I just need someone to spot my error, and get me over the hump.

Second, thanks for helping me wrestle with this. I am not trying to be a chore, and I really do appreciate the help. That's why I come here.

Now: I have two equations which I think to be rock-steady.

First, given two frames of reference, the relative velocity v between those two frames of reference is calculated as:

$$v = { x \over t } \eqno A1$$

(Consider, for instance, the train-and-embankment scenario. From either reference frame, we would certainly calculate the relative velocity by marking out a distance in our rest frame, x, and measuring how long it took for a moving point in the other reference frame to traverse the distance - hence, t.)

I'm pretty jolly sure that you can't express this relation any other way.

Then, we have the Lorentz equation to determine observed lengths from proper measurements and relative velocity:

$${x^\prime = {(x-vt)\gamma}\quad \quad \quad \eqno A2$$

A simple substitution is what gets us in trouble:

$$\begin{equation*} \begin{split} x^\prime &= (x-{x\over t}t)\gamma \quad \quad \quad \eqno A3 \\<br /> &= (x-x)\gamma \quad \quad \quad \eqno A4\\<br /> &= {0\gamma} \quad \quad \quad \quad \eqno A5\\<br /> \end{split} \end{equation*}$$

One thing that I noticed when I was modeling the Lorentz transforms in Excel was that you can't vary x and t at the same time. When you do, you end up with zeroes and negative numbers for your x' and t'.

Is that the clue as to what's going wrong here?

Surely, after over 90 years, this relationship between these equations is well-remarked. Someone must have covered this ground before. Probably lots of someones.

Help me! I've stumbled over basic algebra and I can't get up!

 

[Doc Al]

Now: I have two equations which I think to be rock-steady.

First, given two frames of reference, the relative velocity v between those two frames of reference is calculated as:

$$v = { x \over t } \eqno A1$$

In general, that's not true. First off, speed is not x-coordinate over t-coordinate, it is Δx over Δt. Second, this will only give you the relative velocity of the two frames if you are measure Δx and Δt of a single point in the moving frame.

(Consider, for instance, the train-and-embankment scenario. From either reference frame, we would certainly calculate the relative velocity by marking out a distance in our rest frame, x, and measuring how long it took for a moving point in the other reference frame to traverse the distance - hence, t.)

Exactly right! But don't mix up Δx (a distance) with x (a position).

I'm pretty jolly sure that you can't express this relation any other way.

See my comment above as to why this is incorrect (in general).

Then, we have the Lorentz equation to determine observed lengths from proper measurements and relative velocity:

$${x^\prime = {(x-vt)\gamma}\quad \quad \quad \eqno A2$$

This equation relates coordinates in one frame to coordinates in another. x and x' are general coordinates, not distances. But, lucky for you, in the special case in which you are measuring the motion of the primed origin starting from x = 0 at t =0, then Δx just happens to equal x.

A simple substitution is what gets us in trouble:

$$\begin{equation*} \begin{split} x^\prime &= (x-{x\over t}t)\gamma \quad \quad \quad \eqno A3 \\<br /> &= (x-x)\gamma \quad \quad \quad \eqno A4\\<br /> &= {0\gamma} \quad \quad \quad \quad \eqno A5\\<br /> \end{split} \end{equation*}$$

Actually there is no problem at all here. Again, in this special case Δx' happens to equal x' (assuming you start measuring at the point x=x'=0 and t=t'=0; the two frames share the same space-time coordinates for this one event). What you've determined is how far the primed frame says its origin has moved during your measurement of relative speed. Of course, the primed frame views its origin as not moving, so Δx' = x' = 0.

 

[OneEye]

Okay, this is kind of making sense to me, but then this causes four other questions:

1) In a relativistic coordinate system, what is the difference between x and Δx?

2) Einstein defines v as the motion of the origin of K' with respect to K, and substitutes this definition of v for x/t when deriving the Lorentz transform. I confess that I am probably not enough of a mathematician to say whether this helps or hinders your explanation, but it looked like it hindered.

3) Your explanation invalidates the derivation of the relativistic formula for the addition of velocities given me by another board member (Need Help with Derivation), in which he supplied the definition of W (the summary velocity) as W=x/t. Yet his derivation results in the right answer, and appears cogent - thus bolstering support for the idea that v=x/t in the Lorentz tranforms.

4) Assuming your position to be correct, what then is the formula which transforms Δx into Δx' ? It must not be the Lorentz transform per se, but rather something derived from it.

Please help me out here. I am not out of the woods yet. Any clarity you can offer will be most useful.

Thanks again for all your help.

 

[Doc Al]

1) In a relativistic coordinate system, what is the difference between x and Δx?

The same as in any coordinate system. x is a position; Δx is a difference of position. If Δx is the change in the coordinate of an object, then it's the distance traveled by the object--measured in the unprimed frame. If you really think that the general formula for speed is v = x/t, then if I'm at x=10m and t=2sec, what's my speed? Of course, that makes no sense, does it? Speed is change in position divided by change in time.

But for the special case that interests you here, where we start measuring at x=0 and t=0, then you can get away with using v = x/t as long as you realize what you are doing. x is the position of the x'=0 point at time t.

2) Einstein defines v as the motion of the origin of K' with respect to K, and substitutes this definition of v for x/t when deriving the Lorentz transform. I confess that I am probably not enough of a mathematician to say whether this helps or hinders your explanation, but it looked like it hindered.

I have no problem with that. I'll say it again: for the SPECIAL CASE that you are interested in, using v = x/t (where x and t measure the coordinates of x'=0 as it moves from the common origin) is OK.

3) Your explanation invalidates the derivation of the relativistic formula for the addition of velocities given me by another board member (Need Help with Derivation), in which he supplied the definition of W (the summary velocity) as W=x/t. Yet his derivation results in the right answer, and appears cogent - thus bolstering support for the idea that v=x/t in the Lorentz tranforms.

What makes you say that my explanation (which is just an explanation of what the Lorentz Transformations mean) "invalidates" a derivation of the addition of velocity formula?

Remember, the LT are general coordinate transforms. IN GENERAL, taking the x-coordinate of some event and dividing by t-coordinate of that event will NOT give you any kind of speed.

4) Assuming your position to be correct, what then is the formula which transforms Δx into Δx' ? It must not be the Lorentz transform per se, but rather something derived from it.

I have no "position". This is just Lorentz Transforms 101. There is no problem using v = x/t in the SPECIFIC situation that we are discussing. In that SPECIAL CASE (detailed above) it will give the speed of the primed frame with respect to the unprimed frame.

But, to answer your question, if you had a distance measurement in the unprimed frame (Δx = x2 - x1; and Δt = 0) and you wished to know what the primed frame would say is the separation between those two events: just apply the LT. In general, the primed frame will see those events at different times and positions.

That's essentially what you were doing when you found that a non-zero Δx turned out to equal Δx' = 0 in the primed frame in your earlier post. That bothered you for some reason, but it's correct.

I'm not sure we're making any progress.

 

[Doc Al]

one more try...

Let's start over and maybe I can make myself clearer. Early in this thread you made the statement:

$${ x \over t } = v = { x^\prime \over t^\prime }\eqno 1$$

Let's analyze this carefully.

First we'll look at the motion of the primed frame from the unprimed frame:

(a1) If x & t refer to the position of the origin of the primed frame (x'=0) as measured in the unprimed frame, then and only then:

(b1) vpf = x/t, where "vpf" means "velocity of primed frame as measured by unprimed frame". The magnitude of vpf is just v.

Note that v = x/t is ONLY true when (a1) is true; that was my point in saying that v = x/t is not true in general.

Now we'll look at the motion of the unprimed frame from the primed frame:

(a2) If x' & t' refer to the position of the origin of the unprimed frame (x=0) as measured in the primed frame, then and only then:

(b2) vuf = x'/t', where "vuf" means "velocity of unprimed frame as measured by primed frame". The magnitude of vuf is just v.

Again, b2 is true ONLY when a2 is true.

Now let's make some general statements:

(1) vpf = -vuf

And as long as you remember what x & t and x' & t' refer to (see a1 and a2), then I suppose you can write (but you are begging for trouble if you do):

(2) x/t = -x'/t'

But here (finally) is the point I want to make (and I believe the source of your confusion):

(3) x & t and x' & t' (as used here) DO NOT REFER TO THE SAME EVENT! One describes the motion of the origin of the PRIMED frame (x' = 0), the other that of the UNPRIMED frame (x = 0). SO: x, t and x', t' are NOT RELATED BY A LORENTZ TRANSFORMATION! If you start applying a LT to x and x' in equation (2) you will get nonsense!

The LT connects different measurements of the same thing. But x,t and x',t' (as used here) represent measurements of different things.

I hope that makes sense. Your turn.

 

[OneEye]

Dear Doc Al,

Please, again, be aware that I am not trying to substantiate some quixotic pseudo-science here. I am honestly just looking for answers to a silly little math problem I had. So don't think I am trying to "win" anything. Rather, I am just looking to understand what's going on here. I had a highly unexpected result from what should have been a simple bit of algebra, and am trying to get to the bottom of the cause of it. Thank you for your willingness to present a "101" level introductions. I appreciate your persistence, but, as you say, we are "not there yet."

If we're going to start over, I would much rather start over at my equations A1-A4. I am only too happy to discard my equation 1, since I was only speculating when I proposed that anyway. I don't think it's the cause of the trouble here, since I never took it as axiomatic or fundamental or even true, but rather only as putative. It is not a part of the mathematical problem I encountered, but only a bit of "local color" for the problem.

I'm going to put aside much of what you've said for brevity's sake (though I particularly think that the use of W=x/t in DW's derivation is critically important, and I am not at all convinced of the difference between x and Δx in special relativity), and just go straight to the point I think important:

4) Assuming your position to be correct, what then is the formula which transforms Δx into Δx' ? It must not be the Lorentz transform per se, but rather something derived from it.

But, to answer your question, if you had a distance measurement in the unprimed frame (Δx = x2 - x1; and Δt = 0) and you wished to know what the primed frame would say is the separation between those two events: just apply the LT. In general, the primed frame will see those events at different times and positions.

So, then, it's okay to say:

$$\Delta x^\prime=(\Delta x-vt)\gamma$$

Right?

 

[Doc Al]

Please, again, be aware that I am not trying to substantiate some quixotic pseudo-science here.

Why do you feel the need to preface your posts with such a strange disclaimer? You're not just pulling my leg, are you OneEye?

I am honestly just looking for answers to a silly little math problem I had. So don't think I am trying to "win" anything. Rather, I am just looking to understand what's going on here. I had a highly unexpected result from what should have been a simple bit of algebra, and am trying to get to the bottom of the cause of it.

The problem is not one of algebra, but of conceptual understanding of the formula you are combining. It's not a "silly little math problem". Until you understand:

(1) the meaning of velocity and under what conditions one can say that v = x/t
(2) the meaning and use of the Lorentz transformations​

you will not be able to combine them algebraically in any meaningful way.

If we're going to start over, I would much rather start over at my equations A1-A4. I am only too happy to discard my equation 1, since I was only speculating when I proposed that anyway. I don't think it's the cause of the trouble here, since I never took it as axiomatic or fundamental or even true, but rather only as putative. It is not a part of the mathematical problem I encountered, but only a bit of "local color" for the problem.

Again, despite the title of this thread, the problem is not a mathematical one. We'll get to any equation you like, but it must be done systematically, starting with my previous post. Once we get through that, I think you will see the light. Those concepts are fundamental.

I'm going to put aside much of what you've said for brevity's sake (though I particularly think that the use of W=x/t in DW's derivation is critically important, and I am not at all convinced of the difference between x and Δx in special relativity), and just go straight to the point I think important:

You see the critical importantance of a formula (w = x/t), but don't seem to want to spend time understanding it? I find that quite odd. If you don't understand the difference between Δx and x, or the conditions under which v = x/t, then how can you intelligently make use of them? Again, let's start with my previous post.

So, then, it's okay to say:

$$\Delta x^\prime=(\Delta x-vt)\gamma$$

Right?

No, the LT is this:
$$\Delta x^\prime=(\Delta x-v\Delta t)\gamma$$
But this will be of no use to you until you understand what x, x', t, and t' mean in the context of the LT as well as the purpose of the LT.

 

[robphy]

... I am not at all convinced of the difference between x and Δx in special relativity)...

[snip]

So, then, it's okay to say:

$$\Delta x^\prime=(\Delta x-vt)\gamma$$

Right?

Maybe some labels are needed for clarity.

Let A and B be events (points in spacetime).
In a reference frame (coordinate system), A has coordinates $$(t_A,x_A)$$ and B has coordinates $$(t_B,x_B)$$.
The displacement vector from A to B is a vector with temporal component $$(\Delta t)_{A to B}=t_B-t_A$$ and spatial component $$(\Delta x)_{A to B}=x_B-x_A$$.

The [average] velocity from A to B is a [vector] slope $$v_{\mbox{avg from A to B}}=\frac{(\Delta x)_{A to B}}{(\Delta t)_{A to B}}$$.

Clearly, $$x$$ is a different mathematical object from $$\Delta x$$.

Note everything I've written so far works for both special relativity and Galilean relativity.

Additionally, it seems to me that you would run into the same kind of difficulty if you repeat your first post with the Galilean transformations.
So, it may be more fruitful to first resolve this problem in the Galilean case.

It may also be useful to draw a spacetime diagram. (Here's one for the special relativity case.)
$$\unitlength 1mm<br /> \begin{picture}(100.00,100.00)(0,0)<br /> \put(0.00,0.00){\line(1,0){50.00}}<br /> \put(60.00,0.00){t}<br /> \put(0.00,0.00){\line(0,1){50.00}}<br /> \put(0.00,60.00){x}<br /> <br /> \multiput(0.00,0.00)(0.72,0.12){83}{\line(1,0){0.72}}<br /> \multiput(0.00,0.00)(0.12,0.72){83}{\line(0,1){0.72}}<br /> \put(70.00,10.00){t'}<br /> \put(10.00,70.00){x'}<br /> <br /> \multiput(0.00,0.00)(10,10){6}{\line(1,1){7.5}}<br /> \put(40.00,16.00){\circle{1.00}B}<br /> \put(30.00,18.00){\circle{1.00}A}<br /> \end{picture}$$

Concerning your last post, with my labels,
$$x_A^\prime=(x_A-Vt_A)\gamma$$
$$x_B^\prime=(x_B-Vt_B)\gamma$$
then
$$\begin{align*}<br /> x_B^\prime-x_A^\prime&=(x_B-Vt_B)\gamma-(x_A-Vt_A)\gamma\\<br /> (\Delta x)_{A to B}^\prime&=\left( (\Delta x)_{A to B}-V(\Delta t)_{A to B} \right)\gamma<br /> \end{align*}$$
where $$V$$ is the "boost" that relates the coordinates (t,x) in one frame with the coordinates (t',x') in another frame. (Generally, $$V$$ is not the same as $$v_{\mbox{avg from A to B}}$$).

 

[OneEye]

Why do you feel the need to preface your posts with such a strange disclaimer? You're not just pulling my leg, are you OneEye?

No, it's just that I am all-too-aware of the number of pseudo-Einsteins who work this forum, and I am particularly concerned at not being categorized as a "StarThrower" who is trying to prove that the Lorentz tranformations are mathematically wrong. Maybe I'm just being oversensitive, but I feel that it would be easy to be categorized as a froot-loop who is just grinding some crackpot axe. (Man, now that's how to mix the metaphors, ain't it?)

Anyway, it's just that I am interested in actually figuring this out rather than getting into an ego war - and I want you to know that I am sincere - so I dsiclaim myself a lot. I hope that this has not been an excessive vexation to you, and I will now cease to do so.

No, the LT is this:
$$\Delta x^\prime=(\Delta x-v\Delta t)\gamma$$
But this will be of no use to you until you understand what x, x', t, and t' mean in the context of the LT as well as the purpose of the LT.

Actually, I think that this is not right, since:
$$\begin{equation*}\begin{split}<br /> v &= { \Delta x \over \Delta t }\\<br /> \Delta x^\prime &= (\Delta x-v\Delta t)\gamma\\<br /> &=(\Delta x-{ \Delta x \over \Delta t } \Delta t)\gamma\\<br /> &=(\Delta x-\Delta x)\gamma\\<br /> \end{split}\end{equation*}$$

...and there we are again.

Last night, I thought that perhaps...
$$\begin{equation*}\begin{split}<br /> \Delta x^\prime &= x_2^\prime - x_1^\prime\\<br /> &= (x_2 -vt )\gamma - (x_1-vt)\gamma\\<br /> &= ((x_2 -vt) - (x_1-vt))\gamma\\<br /> &= (x_2 - x_1)\gamma\\<br /> &= (\Delta x)\gamma<br /> \end{split}\end{equation*}$$

...Which supports what you are saying, and perhaps is more helpful toward illustrating your concept.

But this still leaves me with a few loose ends... which I will bring up if we are aligned this far.

 

[OneEye]

robphy - Your point about this consideration affecting Gallilean transformations is a good one. I will give that some thought.

Peripherally (and not to muddy the waters): Every x is also a Δx (with respect to the origin, of course), and every t is a Δt (with respect to t₀, of course), and for every x, t relationship there is of course a v which describes that relationship (with respect to the x, t origin, of course).

I think that my problem is that I am thinking of v as a formulaic relationship when, with regard to a particular pair of inertial reference frames, v is actually a constant.

Maybe.

 

[Doc Al]

Regarding my statement that the LT is this:
$$\Delta x^\prime=(\Delta x-v\Delta t)\gamma$$
You say:

Actually, I think that this is not right, since:
$$\begin{equation*}\begin{split}<br /> v &= { \Delta x \over \Delta t }\\<br /> \Delta x^\prime &= (\Delta x-v\Delta t)\gamma\\<br /> &=(\Delta x-{ \Delta x \over \Delta t } \Delta t)\gamma\\<br /> &=(\Delta x-\Delta x)\gamma\\<br /> \end{split}\end{equation*}$$
...and there we are again.

Assuming that Δx is the distance that the primed frame origin has moved in time Δt... Then this is perfectly correct! (As I've already stated.)
Δx' = 0... absolutely! What's the problem? (Actually, I know what the problem is... you don't understand the LT! )

What this equation tells us is the displacement (Δx') of the prime frame origin (x'=0) as measured in the primed frame... of course it's zero!

Last night, I thought that perhaps...
$$\begin{equation*}\begin{split}<br /> \Delta x^\prime &= x_2^\prime - x_1^\prime\\<br /> &= (x_2 -vt )\gamma - (x_1-vt)\gamma\\<br /> &= ((x_2 -vt) - (x_1-vt))\gamma\\<br /> &= (x_2 - x_1)\gamma\\<br /> &= (\Delta x)\gamma<br /> \end{split}\end{equation*}$$
...Which supports what you are saying, and perhaps is more helpful toward illustrating your concept.

This doesn't support what I'm saying at all... in fact it's plain wrong.

Your mistake is in keeping t constant (setting Δt = 0) while x varies from x1 to x2. The correct way to write the general LT (at least half of it) is as I gave it:
$$\Delta x^\prime=(\Delta x-v\Delta t)\gamma$$

Perhaps you're mixing this up with the so-called "length contraction" formula? In that case, you are using Δx' to represent the distance between the ends of a measuring stick (for example) that is at rest in the primed frame. The question is: what is the length of that stick as observed in the unprimed frame? Of course, we need to observe the positions of the stick ends at the same time--so Δt = 0. (Beware that the primed frame will disagree that we are measuring the ends at the same time--simultaneity is relative.) But, again, this is a special case--not the general Lorentz transformation. And it's not relevant to anything you've brought up so far. Don't get ahead of yourself.

 

[turin]

$$\begin{equation*}\begin{split}<br /> v &= { \Delta x \over \Delta t }\\<br /> \Delta x^\prime &= (\Delta x-v\Delta t)\gamma\\<br /> &=(\Delta x-{ \Delta x \over \Delta t } \Delta t)\gamma\\<br /> &=(\Delta x-\Delta x)\gamma\\<br /> \end{split}\end{equation*}$$

You seem to want to deal with two different (and, more importantly, absolutely unrelated) velocities by using the same symbol interchangibly. In the LT, the velocity that shows up is exactly what Doc Al has been saying. It is ALWAYS the speed that the primed origin moves wrt the unprimed frame. It may or may not be some other velocity of interest. When you have your "equation 1," particularly the part that says v is the same velocity in the unprimed and primed coordinates, then this v ABSOLUTELY CANNOT be the same v that shows up in the LT, UNLESS v = 0, otherwise you would get something like:

v = 0
v /= 0

which has no solution.

Note arildno's comment that the best you can do [for non-trivial v] is to have one of these velocities to be the negative of the other. But, from the other part of your "equation 1" that states the velocity in the unprimed coordinates is also v, things start to get really ambiguous.

 

[OneEye]

Yes, well, okay. Thanks, all. I think I've got this worked out now. See my response to robphy, above - which response turin echoes and expands.

Thanks for all your help.