Debate: Lorentz invariance of certain zero angles

[Pete]

Updated tracking list

1.0 Scenario (Complete)

1.1 - Coordinate dependence vs. coordinate independence (Resolved)
1.2 - Definition of rods T1 and T2 (Resolved)
1.3 - Definition of points A and B (Obsolete)​

2.0 Methodology (Complete)

2.1 - Tach's proposed measurements (Complete)

2.1.1 - Transverse doppler effect (Complete)​

2.2 - Pete's proposed measurements (Complete)
2.3 - Measuring remote events using background Rods and Clocks (Complete)​

3.0 Calculations (Active)

3.1 Calculations for Pete's method (Active)

3.1.1 Lorentz transformation of vectors (Active)
3.1.2 Orientation of rod T1 (Pending)
3.1.3 Angle between surface and velocity in the low velocity limit (Galilean spacetime) (Pending)

3.1.3.1 Rindler's proof of angle invariance (Pending)​

3.2 Calculations for Tach's method (Pending)​

4. Summary and reflection (not started)​

 

[Tach]

Your post raises a number of issues, which we will address one at a time (see tracking list in the next post).
First, the most fundamental. The other issues might become obsolete when this is resolved.

3.1.1 Lorentz transformation of vectors
In your document you use three different transformations for vectors under a Lorentz boost.

The second transformation you use is this:

I agree that this is the correct lorentz transformation for a velocity vector.

On page 2, you have this:

I agree that this is the correct Lorentz transformation of a displacement 3-vector.

But I don't understand the first transformation you use:


Questions:
What is the source of this equation?
What does $$dt$$ represent in that equation?

See C. Moller "The Theory of Relativity". pages 41 and 47.

What is $$d\vec{r}$$? You seem to imply that it is a displacement vector, perhaps $$\hat{P_t}(0)$$. Is that correct?

Yes, more generally, it is $$\hat{P_t}(t)$$

I think you have also incorrectly applied the equation to the parallel and perpendicular components.
Plugging in $$d\vec{r_\perp}$$ gives:
$$d\vec{r_\perp} - \gamma\vec{V}dt$$
Plugging in $$d\vec{r_\parallel}$$ gives:
$$\gamma(d\vec{r_\parallel} - \vec{V}dt)$$
Adding them together results in:
$$d\vec{r}' = d\vec{r_\perp} + \gamma(d\vec{r_\parallel} - 2\vec{V}dt)$$

No, see Moller, page 47.

Since I just answered your challenge to my writeup, I would like you to answer my challenge in the previous post. Thank you.

 

[Pete]

3.1.1 Lorentz transformation of vectors

See C. Moller "The Theory of Relativity". pages 41 and 47.

Which edition? The 1952 edition is freely available online. My university has the 1974 edition at another campus, so it will take a few days to get. I don't have reasonable access to the 2011 edition. Google books has snippets of others.
Can you please explain the context of the equation?
What does dt represent?

Yes, more generally, it is $$\hat{P_t}(t)$$

Thanks. It would be useful if you could stick to notation we've used previously, or at least explain any new notation you decide to use.

Now, we agreed that a displacement 3-vector is transformed according to what you wrote on page 2:

$$\begin{align}
\vec{a} &= \vec{a_\parallel} + \vec{a_\perp} \\
\vec{a}' &= \vec{a_\parallel}' + \vec{a_\perp}' \\
&= \frac{\vec{a_\parallel}}{\gamma} + \vec{a_\perp }
\end{align}$$

So why are you transforming $$\hat{P_t}(t)$$ according to the Moller equation?
They are not equivalent.

No, see Moller, page 47.

Or, you could just explain your understanding directly.
Do you understand the equation you used?

Since I just answered your challenge to my writeup, I would like you to answer my challenge in the previous post. Thank you.

Yes, that will be done, but we must proceed one issue at a time.

 

[Tach]

3.1.1 Lorentz transformation of vectors


Which edition? The 1952 edition is freely available online. My university has the 1974 edition at another campus, so it will take a few days to get. I don't have reasonable access to the 2011 edition. Google books has snippets of others.
Can you please explain the context of the equation?
What does dt represent?

1952 is good.

Thanks. It would be useful if you could stick to notation we've used previously, or at least explain any new notation you decide to use.

Now, we agreed that a displacement 3-vector is transformed according to what you wrote on page 2:


So why are you transforming $$\hat{P_t}(t)$$ according to the Moller equation?

Because different approaches show the SAME result. This is important in establishing validity.

They are not equivalent.

Please look at the web document, I inserted all the explanations necessary (the equations stayed the same).

Or, you could just explain your understanding directly.
Do you understand the equation you used?

Don't patronize me. You took 5 weeks to write up an incorrect solution, I don't need your patronizing tone.

Yes, that will be done, but we must proceed one issue at a time.

No, please answer my question, I raised the issue with your solution, according to the rules you should answer it in the very next post, so please address it. Did you check the Rindler reference I gave you?

 

[Pete]

3.1.1 Lorentz transformation of vectors
Can you please explain the context of the equation?
What does dt represent?

1952 is good.

$$\begin{align}
\vec{r}' &= \vec{r} + \vec{v}(\gamma-1)\frac{\vec{r}.\vec{v}}{v^2} \\
\vec{r_\parallel}' & = \gamma \vec{r_\parallel} \\
\vec{r_\perp}' &= \vec{r_\perp}
\end{align}$$

This is the correct transformation for a displacement 3-vector that is stationary in S'.

Do you agree?

 

[Tach]

3.1.1 Lorentz transformation of vectors
Can you please explain the context of the equation?
What does dt represent?

$$\begin{align}
\vec{r}' &= \vec{r} + \vec{v}(\gamma-1)\frac{\vec{r}.\vec{v}}{v^2} \\
\vec{r_\parallel}' & = \gamma \vec{r_parallel} \\
\vec{r_\perp}' &= \vec{r_\perp}
\end{align}$$

This is the correct transformation for a displacement 3-vector.

Do you agree?

Nope, Moller uses the convention $$dt=0$$ (mark the ends of the 3 vectors simultaneously in frame S). If you mark the ends of the vectors simultaneously in frame S' ($$dt'=0$$) the solution I have shown you produces the exact transformations you have already seen. The answer is a function on the simultaneity condition. Even IF we change the condition, the answer to the problem of vector parallelism remains the SAME, you end up simply moving $$\gamma$$ from the RHS to the LHS. Either way, $$\gamma$$ is a non-factor in the derivation of the angle between vectors.

Now, please could you stop the diversions and answer the question posed at post 119. That was the FIRST question, it should be addressed FIRST. If you need some time to track down the exact pages in Rindler, take your time (I have provided you with the exact derivation in my document, so you should be able to address the issue right away).

 

[Pete]

3.1.1 Lorentz transformation of vectors
I think we've almost resolved this issue. Let's be sure, then we can move on to another:

Lorentz transformation of a velocity vector

$$\vec{v_p}' = \left(\frac{v_{p\parallel} - V}{1-v_{p\parallel}V/c^2} \ , \ \frac{v_{p\perp}}{\gamma(1-v_{p\parallel}V/c^2)}\right)$$

We're both satisfied that this is the correct lorentz transformation for a velocity vector. Right?

Lorentz transformation of a displacement vector
We're both satisfied that a displacement vector stationary in S transforms as shown on page 2 of your document:

$$\begin{align}
\vec{a} &= \vec{a_\parallel} + \vec{a_\perp} \\
\vec{a}' &= \vec{a_\parallel}' + \vec{a_\perp}' \\
&= \frac{\vec{a_\parallel}}{\gamma} + \vec{a_\perp }
\end{align}$$

A displacement vector stationary in S' transforms as described in Moller, 1952, p47:

Let $$\vec{x_1}' $$ and $$\vec{x_2}' $$ be the coordinate vectors of two fixed points $$P_1' $$ and $$P_2'$$ in the system of coordinates S'. The straight line connecting these points represents a fixed vector $$\vec{r}'=\vec{x_2}'-\vec{x_1}'$$ in S'.
...
$$\begin{align}
\vec{r}' &= \vec{r} + \vec{v}(\gamma-1)\frac{\vec{r}.\vec{v}}{v^2} \\
\vec{r_\parallel}' &= \gamma \vec{r_\parallel} \\
\vec{r_\perp}' &= \vec{r_\perp}
\end{align}$$

Do you agree?

 

[Tach]

3.1.1 Lorentz transformation of vectors
I think we've almost resolved this issue. Let's be sure, then we can move on to another:

Lorentz transformation of a velocity vector

We're both satisfied that this is the correct lorentz transformation for a velocity vector. Right?

Right.

Lorentz transformation of a displacement vector
We're both satisfied that a displacement vector stationary in S transforms as shown on page 2 of your document:

Good.

A displacement vector stationary in S' transforms as described in Moller, 1952, p47:

Do you agree?

I already explained to you how the answer depends on the simultaneity condition. It all depends in which frame you mark the endpoints simultaneously. I prefer the condition $$dt'=0$$ since it (re)produces the longitudinal length contraction. Moller's doesn't.
More importantly, even if we used the simultaneity condition used by Moller, NOTHING changes in my solution, the conclusion is the SAME because $$\gamma$$ cancels out in the condition of parallelism. Now, could you please stop the diversions and address the flaw in your derivation I flagged at post 119? Thank you.

 

[Pete]

3.1.1 Lorentz transformation of vectors

I already explained to you how the answer depends on the simultaneity condition. It all depends in which frame you mark the endpoints simultaneously. I prefer the condition $$dt'=0$$ since it (re)produces the longitudinal length contraction. Moller's doesn't.

I don't understand what you're objecting to.
Do you agree that the transformation in Moller is valid or not?
I don't see why you should disagree, since it is just the inverse of the previous equation.

More importantly, even if we used the simultaneity condition used by Moller, NOTHING changes in my solution, the conclusion is the SAME because $$\gamma$$ cancels out in the condition of parallelism.

I really want to get to that issue, but you seem to imply that I'm saying something wrong about the vector transformations, and I'm reluctant to move on until we're both clear on this basic stuff.
So am I saying something wrong? If so, what?

Now, could you please stop the diversions and address the flaw in your derivation I flagged at post 119? Thank you.

Please don't be impatient.
I really want to conduct this discussion as I proposed: not with the goal of determining a winner and a loser, but of first finding common foundations, then building on those foundations to a mutual conclusion.

 

[Tach]

I really want to get to that issue, but you seem to imply that I'm saying something wrong about the vector transformations, and I'm reluctant to move on until we're both clear on this basic stuff.
So am I saying something wrong? If so, what?

Post 119 tells you exactly what is wrong with your solution.

Please don't be impatient.
I really want to conduct this discussion as I proposed: not with the goal of determining a winner and a loser, but of first finding common foundations, then building on those foundations to a mutual conclusion.

I have been very patient, so let's work on your understanding what is wrong with your solution, let's go back to post 119 and work together (in earnest) trying to understand the issue.

 

[Pete]

I think we do agree regarding the Lorentz vector transformation equations.
Moving on.

Pay special attention to the general method in Euclidian space. Any correct method in Minkowski space must reduce in the limit to the Euclidian one at non-relativistic speeds.
If I look at your solution, it is quite clear that at non-relativistic speeds ($$\gamma=1$$) your solution disagrees with Rindler's because $$v'_p$$ is still not parallel (according to your solution) to the microfacet.

3.1.3.1 Rindler's proof of angle invariance
Rindler correctly proves that the angle between two displacement vectors is invariant in Euclidian space, by relying on the fact that the norm to a displacement vector is invariant under a Euclidian velocity boost.
Tach correctly extends that proof to displacement 3-vectors in Minkowski space.

However, Rindler's proof doesn't not apply to the angle between a displacement vector and a velocity vector.
The norm of a velocity vector is not invariant under a velocity boost.
In the low velocity limit, a velocity vector $$\vec{v}$$ boosted by velocity $$\vec{V}$$ transforms as:
$$\begin{align}\vec{v_\parallel}' &= \vec{v_\parallel} - \vec{V} \\
\vec{v_\perp}' &= \vec{v_\perp}
\end{align}$$

 

[Tach]

I think we do agree regarding the Lorentz vector transformation equations.

Yes.

3.1.3.1 Rindler's proof of angle invariance
Rindler correctly proves that the angle between two displacement vectors is invariant in Euclidian space, by relying on the fact that the norm to a displacement vector is invariant under a Euclidian velocity boost.
Tach correctly extends that proof to displacement 3-vectors in Minkowski space.

Thank you for finally addressing the issue. Rindler's proof is about arbitrary three-vectors.

 

[Pete]

3.1.3.1 Rindler's proof of angle invariance

Rindler's proof is clearly intended to apply to arbitrary displacement 3-vectors.
It clearly does not apply to velocity vectors.

Do you agree that the norm of a velocity vector is frame dependent in Euclidian space-time?

 

[Tach]

3.1.3.1 Rindler's proof of angle invariance

Rindler's proof is clearly intended to apply to arbitrary displacement 3-vectors.

Actually, Rindler, just above the proof, talks about speed, acceleration, momentum as examples of three-vectors.

Do you agree that the norm of a velocity vector is frame dependent in Euclidian space-time?

So is $$\Delta x$$, nevertheless Rindler clearly uses $$\Delta x, \Delta y, \Delta z$$ as a prototype of a three-vector in his proof. So, I do not see your point.

 

[Pete]

3.1.3.1 Rindler's proof of angle invariance

Actually, Rindler, just above the proof, talks about speed, acceleration, momentum as examples of three-vectors.

Regardless of what Rindler says (I don't have ready access to it, so I can't confirm for myself), The proof you quoted clearly doesn't apply to velocity vectors.

It's obvious.
In the low velocity limit, a velocity vector $$\vec{v}$$ boosted by velocity $$\vec{V}$$ transforms as:
$$\begin{align}\vec{v_\parallel}' &= \vec{v_\parallel} - \vec{V} \\
\vec{v_\perp}' &= \vec{v_\perp}
\end{align}$$

Do you agree?
Can you see from that result that the angle between velocity vectors is not frame invariant?

Do you agree that the norm of a velocity vector is frame dependent in Euclidian space-time?

So is $$\Delta x$$, nevertheless Rindler clearly uses $$\Delta x, \Delta y, \Delta z$$ as a prototype of a three-vector in his proof. So, I do not see your point.

Tach, when an explicit question is asked, an explicit answer should be given.

Do you agree that the norm of a velocity vector is frame dependent in Euclidian space-time?

The point is that your quote of Rindler's proof explicitly relies on the norm of a vector being invariant:

Proof, according to ref [2]:
Let $$\vec{a}$$ and $$\vec{b}$$ be two arbitrary vectors. Then:
$$\|\vec{a} + \vec{b}\|^2 = a^2 + b^2 + 2\vec{a}.\vec{b}$$
But, in Euclidian space, the norm of a vector is invariant, so $$\|\vec{a} + \vec{b}\|^2$$, $$a^2$$, $$b^2$$ are all invariant.

If $$\vec{a}$$ is a velocity vector, then obviously $$a^2$$ is not invariant, for example.

 

[Tach]

We need to go back to 3.1.1 Lorentz transformation of vectors because this issue is not resolved.

3.1.1 Lorentz transformation of vectors
I think we've almost resolved this issue. Let's be sure, then we can move on to another:

Lorentz transformation of a velocity vector

We're both satisfied that this is the correct lorentz transformation for a velocity vector. Right?

Correct. But you need to pay attention, the transformation of velocity vectors is based on the general transformation of displacement vectors, see formulas (1.1)-(1.2) here. In order to get the transformation for velocity, one needs to perform a couple of divisions, one by $$dt' \ne 0$$ followed by a second one, by $$dt \ne 0$$, see formula (1.3). Therefore, one cannot turn around and plug in $$dt=0$$ or $$dt' =0$$ into the general formula for transformation (1.1) as Moller does. While Moller formulas are correct for the particular case $$dt=0$$ , you are not allowed to apply it for the problem we are discussing because the derivation of the velocity transformation already precludes EITHER $$dt=0$$ OR $$dt'=0$$. You need to proceed in your calculations by using the general formula (1.2) for displacement vector transformation, you cannot use the particular formula derived by setting $$dt=0$$.

Lorentz transformation of a displacement vector
We're both satisfied that a displacement vector stationary in S transforms as shown on page 2 of your document:

Yes, the transformation I have shown you is correct. No, you cannot use its particular form derived for $$dt=0$$ or for $$dt'=0$$, for the reasons explained above.

 

[Pete]

3.1.1 Lorentz transformation of vectors

Correct. But you need to pay attention

We both need to pay careful attention to exactly what vector transformations we're describing.

the transformation of velocity vectors is based on the general transformation of displacement vectors, see formulas (1.1)-(1.2) here.

Equation (1.1) comes from Moller:
$$\begin{align}
\vec{r}' &= \vec{r} + \vec{V}(\frac{\gamma-1}{V^2}\vec{r}.\vec{V} - \gamma t) \\
t' &= \gamma(t - \vec{V}.\vec{x}/c^2)
\end{align}$$
This is the general transform of an arbitrary four-vector.

In order to get the transformation for velocity, one needs to perform a couple of divisions, one by $$dt' \ne 0$$ followed by a second one, by $$dt \ne 0$$, see formula (1.3). Therefore, one cannot turn around and plug in $$dt=0$$ or $$dt' =0$$ into the general formula for transformation (1.1) as Moller does.

You certainly can, if you're transforming space-like four-vectors, as Moller does.

While Moller formulas are correct for the particular case $$dt=0$$ , you are not allowed to apply it for the problem we are discussing because the derivation of the velocity transformation already precludes EITHER $$dt=0$$ OR $$dt'=0$$. You need to proceed in your calculations by using the general formula (1.2) for displacement vector transformation, you cannot use the particular formula derived by setting $$dt=0$$.

We already agreed on the velocity transformation.
You don't need differentiation to derive it, since a timelike four-vector easily translates to a velocity 3-vector, but that's irrelevant - we agree on the transformation.

Differentiation is irrelevant when transforming a spatial displacement vector.
On page 47 of Moller, (the specific citation you gave), Moller is explicitly talking about transforming the spatial displacement vector between two points at rest in S'.

To do so, he considers the four-vector between simultaneous events in S, one at each point. Obviously, this four vector has a t value of zero.

Transforming that four-vector to S' and considering the spatial component produces Moller's result:
$$\vec{r}' = \vec{r} + \vec{v}(\gamma-1)\frac{\vec{r}.\vec{v}}{v^2}$$
Note that the t' component of the transformed four-vector is non-zero, but is of no interest because the two points in question are at rest in S'.

So there's no problem with Moller, unless it's misapplied.

You used the inverse of this transformation yourself in your document, so we clearly agree on how to transform the spatial displacement between points which are both at rest in some frame.

We agree on the general transformation of a four-vector.
We agree on the transformation of a velocity vector.
We agree on the transformation of spatial vectors between points stationary in one frame.


So what's the issue?
It seems that the specific issue we're having is about how $$\hat{P_t}(t)$$ should be transformed.
You want to differentiate it and transform the resulting velocity vector.
I want to transform it as a simple spatial vector (actually not quite, but the result turns out to be the same).
I think a new subheading is in order:

3.1.1.1 Lorentz transformation of vectors

 

[Tach]

3.1.1 Lorentz transformation of vectors




You certainly can, if you're transforming space-like four-vectors, as Moller does.

No, you cannot do that when you already used $$dt' \ne 0$$ and $$dt \ne 0$$ in the derivation of the velocity transformation. You cannot have both $$dt=0$$ and $$dt \ne 0$$ at the same time. On the other hand, you can definitely use the general transformation (1.2) for the displacement vectors since there is no contradiction.

We already agreed on the velocity transformation.
You don't need differentiation to derive it,

Oh yes, you do, this is how the velocity transformation was derived in first place.

Differentiation is irrelevant when transforming a spatial displacement vector.

False, you cannot simply pick the form that suits you , you need to use consistent math throughout. You cannot use $$dt \ne 0$$ for the derivation of velocity transformation and then suffer sudden amnesia and set $$ dt =0$$ in the general transform of the displacement vectors because it serves your purpose.

On page 47 of Moller, (the specific citation you gave), Moller is explicitly talking about transforming the spatial displacement vector between two points at rest in S'.

To do so, he considers the four-vector between simultaneous events in S, one at each point. Obviously, this four vector has a t value of zero.

I understand very well what he's doing. As an aside, a much better approach is to set $$dt'=0$$ but this is beside the point, he's not mixing displacement vector transformation with velocity transformation.

So there's no problem with Moller, unless it's misapplied.

Yes, there is no problem. No, it does not apply in our debate since it leads to a mathematical contradiction.

You used the inverse of this transformation yourself in your document, so we clearly agree on how to transform the spatial displacement between points which are both at rest in some frame.

Yes, I did, because it reflects length contraction better but this should not be misconstrued as in a nod towards this transform in the case we are debating.

We agree on the general transformation of a four-vector.
We agree on the transformation of a velocity vector.
We agree on the transformation of spatial vectors between points stationary in one frame.


So what's the issue?

See above. You can't mix and match as it suits your PoV, you need to use the general transform (1.2), you can't make $$dt=0$$ in the general transform because you already assumed $$dt \ne 0$$ in the derivation of the velocity transform.

It seems that the specific issue we're having is about how $$\hat{P_t}(t)$$ should be transformed.
You want to differentiate it and transform the resulting velocity vector.
I want to transform it as a simple spatial vector (actually not quite, but the result turns out to be the same).

It is not what we want, it is about what is right. What is right is not to particularize the transform for a specific case ($$dt=0$$) . This is the crux of the disagreement between our approaches. In your approach $$\hat{P_t}(t)$$ is frame-invariant, in my approach, it transforms exactly the same way as velocity, hence, thy remain parallel to each other.

I think a new subheading is in order:

3.1.1.1 Lorentz transformation of vectors

All the cases can be read from the posted document . This includes the argument for the only acceptable choice (the general form).

 

[Pete]

3.1.1 Lorentz transformation of vectors

No, you cannot do that when you already used $$dt' \ne 0$$ and $$dt \ne 0$$ in the derivation of the velocity transformation. You cannot have both $$dt=0$$ and $$dt \ne 0$$ at the same time. On the other hand, you can definitely use the general transformation (1.2) for the displacement vectors since there is no contradiction.

False, you cannot simply pick the form that suits you , you need to use consistent math throughout. You cannot use $$dt \ne 0$$ for the derivation of velocity transformation and then suffer sudden amnesia and set $$ dt =0$$ in the general transform of the displacement vectors because it serves your purpose.

Are you talking about vector transformations in general, or about the specific case of $$\hat{P_t}$$?

In general, transforming a spatial displacement vector obviously doesn't rely on transforming a velocity vector. Right?

I have not been talking about $$\hat{P_t}$$ or any other vector in our particular problem in this section so far. I'm talking in general about the simple transformation of a arbitrary spatial displacement vector.

I understand very well what he's doing. As an aside, a much better approach is to set $$dt'=0$$ but this is beside the point, he's not mixing displacement vector transformation with velocity transformation.

Yes, I did, because it reflects length contraction better but this should not be misconstrued as in a nod towards this transform in the case we are debating.

Well, Moller doesn't do anything with derivatives in that chapter, so he's not setting dt or dt' to anything.
He is just starting from two points at rest in S', considering events at those points with the same value of t, transforming those events to S', and finding the spatial component of the vector between the transformed events.

You can get the inverse result by starting with points at rest in S.
Both results correctly reflect length contraction.
The result in Moller shows that a rod at rest in S' is contracted in S. The inverse shows that a rod at rest in S' is contracted in S.
Obviously, both are correct.

In the next post, I'd like to move on to the sub issue of transforming $$\hat{P_t}$$.

 

[Tach]

3.1.1 Lorentz transformation of vectors

Are you talking about vector transformations in general, or about the specific case of $$\hat{P_t}$$?

Throughout all my posts I have been talking about the specific case of $$\hat{P_t}$$. This is what the debate has been about all along.

In general, transforming a spatial displacement vector obviously doesn't rely on transforming a velocity vector. Right?

You have it backwards, the velocity transformation is a CONSEQUENCE of the transformation of the displacement vector $$dr$$. This is why you are not allowed to make $$dt=0$$ in order to derive the transformation of $$dr$$ , then to "forget" about making it zero by now making $$dt \ne 0$$ in order to derive the velocity transformation, as already explained several times.

I have not been talking about $$\hat{P_t}$$ or any other vector in our particular problem in this section so far. I'm talking in general about the simple transformation of a arbitrary spatial displacement vector.

The debate is not about transformations of displacement vectors in general, it is about how $$\hat{P_t}$$ transforms with regards to the way $$v$$ transforms.

Well, Moller doesn't do anything with derivatives in that chapter, so he's not setting dt or dt' to anything.

He actually is doing differentiation followed by setting $$dt=0$$ but this is totally irrelevant, we are debating another subject: how $$\hat{P_t}$$ transforms with regards to the way $$v$$ transforms. So, let's keep it on point.

In the next post, I'd like to move on to the sub issue of transforming $$\hat{P_t}$$.

I have already covered this subject in great extent. See the document I posted.