Debate: Lorentz invariance of certain zero angles

[Pete]

3.1.1 Lorentz transformation of vectors

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.

No, This particular subsection is supposed to be about the general case of transforming vectors.
You started your document by presenting a general case, not related to a particular vector, and that's what I want to get sorted before moving on.
You can't present a general derivation, and then claim that it's only about a specific case.

You have it backwards, the velocity transformation is a CONSEQUENCE of the transformation of the displacement vector $$dr$$.

I said that transforming a spatial displacement vector obviously doesn't rely on transforming a velocity vector.
How is that backwards?

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.

You keep talking about transforming dr. Why?
Transforming a displacement vector doesn't mean transforming a differential.
Why is differentiating necessary?

He actually is doing differentiation followed by setting $$dt=0$$

No, he is not differentiating at all.
He is starting with a displacement vector r, and deriving r'.

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

Your document begins by presenting a general transformation for displacement vectors:

The Lorentz Transforms for Displacement Vectors
How does a vector transform in SR? Assume given a vector $$d\vec{r}$$ in frame S and a frame S' is moving with velocity $$\vec{V}$$ wrt frame S.

The Lorentz transformation between S and S is:
$$\begin{align}
\vec{r}' &= \vec{r} + \vec{V}(\frac{\gamma-1}{V^2}\vec{r}.\vec{V} - \gamma dt) \\
t' &= \gamma(t - \vec{r}.\vec{V}/c^2)
\end{align}$$

Differentiating, we obtain:
$$\begin{align}
d\vec{r}' &= d\vec{r} + \vec{V}(\frac{\gamma-1}{V^2}d\vec{r}.\vec{V} - \gamma dt) \\
dt' &= \gamma(dt - d\vec{r}.\vec{V}/c^2)
\end{align}$$

Why differentiate?
Your heading suggests that you are presenting the lorentz transform for displacement vectors in general, but your approach only applies to infinitesimal displacements of some other vector over time.
Is that what you intended?

3.1.1 Lorentz transformation of $$\hat{P_t}$$
Further down in your approach, it is implied that $$\vec{r}=\vec{P}$$, ie it is the position vector of point P in S, since:

$$\vec{v_p} = \frac{d\vec{r}}{dt}$$

So, $$d\vec{r}$$ is the differential displacement of point P in frame S in infinitesimal time.
$$d\vec{r}'$$ is the differential displacement of point P in frame S' infinitesimal time.
Obviously, $$d\vec{r}'$$ is parallel to $$\frac{d\vec{r}'}{dt'}$$.
But that is not relevant to $$\hat{P_t}$$, which is the unit tangent vector to the wheel at point P.

In the above, $$d\vec{r}$$ represents $$\hat{P_t}(0)$$ and $$d\vec{r}'$$ represents $$\hat{P_t}'(0)$$.

This is not consistent with the definition of $$\hat{P_t}$$.

Recall how $$\hat{P_t}$$ and $$\hat{P_t}'$$ were defined:

$$\hat{P_t}(t)$$ is the unit displacement vector tangent to wheel element P at time t in S.
$$\hat{P_t}'(t')$$ is the unit displacement vector tangent to wheel element P at time t' in S'.

And we of course must be clear about what it means for a vector to be tangent to the wheel:

If necessary, whether a given displacement vector is tangent to the wheel can be tested in at least two ways:

• A displacement vector is tangent to the wheel at a given point at a given instant if it is a scalar multiple of the displacement between two points on a straight rod tangent to the wheel at that point at that instant.
• A displacement vector is tangent to the wheel at a given point at a given instant if and only if there is no other point on the wheel at that instant such that the displacement between the given point and the other point is a scalar multiple of the displacement vector.

We know that rod T1 is a straight rod tangent to the wheel at P at t=0 in S, and also at some time t' in S', so we can use that rod to define $$\hat{P_t}(0)$$ and $$\hat{P_t}'(t'_0)$$:
$$\hat{P_t}(0)$$ is defined as a unit vector between two points on rod T1 at t=0 in S.
$$\hat{P_t}'(t'_0)$$ is defined as a unit vector between two points on rod T1 at $$t'=t_0$$ in S.

Do you agree so far?

 

[Tach]

3.1.1 Lorentz transformation of vectors




I said that transforming a spatial displacement vector obviously doesn't rely on transforming a velocity vector.
How is that backwards?

The part that you got backwards is that the velocity transformation is the consequence of the vector displacement dr transformation, not the other way around is you claim.

You keep talking about transforming dr. Why?

Because the transformation for dr determines the transformation for $$v$$. Once you have these two transformations, the problem is solved.

Transforming a displacement vector doesn't mean transforming a differential.

They transform the same exact way.

No, he is not differentiating at all.
He is starting with a displacement vector r, and deriving r'.

A displacement vector is the difference between two positional vectors, same thing as taking the differential of a positional vector.

Why differentiate?
Your heading suggests that you are presenting the lorentz transform for displacement vectors in general, but your approach only applies to infinitesimal displacements of some other vector over time.
Is that what you intended?

They are the same exact thing. The transforms are exactly the same since there is no notion of "big" difference vs. "infinitesimal" difference, $$d$$ and $$\Delta$$ are fully interchangeable.

 

[Pete]

The part that you got backwards is that the velocity transformation is the consequence of the vector displacement dr transformation, not the other way around is you claim.

I made no such claim.

Because the transformation for dr determines the transformation for $$v$$.

That is irrelevant to the general transformation of displacement vectors.

They transform the same exact way.

A displacement vector is the difference between two positional vectors, same thing as taking the differential of a positional vector.

Which doesn't change what I said.
Moller is not doing differentiation, and differentiating is not needed to transform a displacement vector.

Your heading suggests that you are presenting the lorentz transform for displacement vectors in general, but your approach only applies to infinitesimal displacements of some other vector over time.

They are the same exact thing. The transforms are exactly the same since there is no notion of "big" difference vs. "infinitesimal" difference, $$d$$ and $$\Delta$$ are fully interchangeable.

The difference is that your approach only applies to some change in another displacement vector over time.
It only applies to the spatial component of timelike 4-vectors.
It does not apply to displacement vectors in general.
Is that what you intended?

 

[Tach]

I made no such claim.

See this quote:

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

It is the transformation of velocity that depends on the transformation of the displacement vector.

That is irrelevant to the general transformation of displacement vectors.

This is the crux of our disagreement, so it is very relevant. You aren't using the "general transformation of displacement vectors", you are using a "particular" form obtained by constraining $$dt=0$$. But this form is not the right one for our problem since you need to have $$dt \ne 0$$ in order to derive the transformation of velocity as a direct consequence of the transformation of displacement vectors. In other words:

$$dr'_{perpendicular}=dr_{perpendicular}$$
$$dr'_{parallel}=\gamma(dr_{parallel}-Vdt)$$

In the above , you cannot set $$dt=0$$ (nor can you set $$dt'=0$$) , so, you cannot use the "truncated" form :

$$dr'_{parallel}=\gamma(dr_{parallel}-0)$$

Which doesn't change what I said.
Moller is not doing differentiation, and differentiating is not needed to transform a displacement vector.

Both above claims are incorrect but this is irrelevant in the grander scheme.

The difference is that your approach only applies to some change in another displacement vector over time.
It only applies to the spatial component of timelike 4-vectors.
It does not apply to displacement vectors in general.
Is that what you intended?

I look at this the other way around, I have shown you the general formula for displacement vector transformation . By setting $$dt=0$$ one generates a particularization of the general formula as applied to stationary vectors. But, the tangent to the "microfacet" is not stationary because none of the microfacets is stationary , they move in both frame S and in S'.

 

[Pete]

This is the crux of our disagreement, so it is very relevant. You aren't using the "general transformation of displacement vectors", you are using a "particular" form obtained by constraining $$dt=0$$.
But this form is not the right one for our problem since you need to have $$dt \ne 0$$ in order to derive the transformation of velocity as a direct consequence of the transformation of displacement vectors.

...

But, the tangent to the "microfacet" is not stationary because none of the microfacets is stationary , they move in both frame S and in S'.

These comments specifically relate to the transformation of $$\hat{P_t}$$, not to the general case.

3.1.1 Lorentz transformation of displacement vectors

See this quote:

transforming a spatial displacement vector obviously doesn't rely on transforming a velocity vector.

It is the transformation of velocity that depends on the transformation of the displacement vector.

I really don't understand what you're objecting to in my statement, unless you're not seeing the word "doesn't".

$$
\begin{align}
dr'_{\perp} &= dr_{\perp} \\
dr'_{\parallel} &=\gamma(dr_{\parallel}-Vdt)
\end{align}
$$

In the above , you cannot set $$dt=0$$ (nor can you set $$dt'=0$$) , so, you cannot use the "truncated" form :

$$dr'_{\parallel}=\gamma(dr_{\parallel}-0)$$

Right, you can't set dt=0, because your approach only applies to the displacement of some point over time.

Both above claims are incorrect but this is irrelevant in the grander scheme.

Both claims are quite correct, as shown by a simple reading of Moller (1952) p.47 - the reference you introduced to the discussion, and on which you rely to introduce your approach.

I look at this the other way around, I have shown you the general formula for displacement vector transformation. By setting $$dt=0$$ one generates a particularization of the general formula as applied to stationary vectors.

Your differential approach explicitly excludes setting dt=0, because that would also imply dr=0, meaning you're just transforming the 4-vector (0,0,0,0).
The only general formula you presented was the original formula quoted from Moller, before the unnecessary differentiation.

We agree on that formula.

Everything that comes after is limited to displacement over time of a moving point.

If that's what you intended, that's OK. Just say so, and we can move on.

 

[Tach]

The only general formula you presented was the original formula quoted from Moller, before the unnecessary differentiation.

This is incorrect, since differentiation is the same exact thing as calculating differences. But this is not important, important is that we agree on the....

We agree on that formula.

Good.

Everything that comes after is limited to displacement over time of a moving point.

No, this is not true, it is the general transformation for displacement vectors.

 

[Pete]

No, this is not true, it is the general transformation for displacement vectors.

This is the general transformation for displacement vectors between events separated in time by t:
$$\begin{align}
\vec{r}' &= \vec{r} + \vec{V}(\frac{\gamma-1}{V^2}\vec{r}.\vec{V} - \gamma dt)
\end{align}$$

Or equivalently, when broken into components:
$$\begin{align}
\vec{r_{\perp}}' &= \vec{r_{\perp}} \\
\vec{r_{\parallel}}' &=\gamma(\vec{r_{\parallel}}-\vec{V}t)
\end{align}$$

Right?

 

[Tach]

This is the general transformation for displacement vectors between events separated in time by t:
$$\begin{align}
\vec{r}' &= \vec{r} + \vec{V}(\frac{\gamma-1}{V^2}\vec{r}.\vec{V} - \gamma dt)
\end{align}$$

Not exactly, the exact formula is the one I derived:

$$\begin{align}
d \vec{r}' &=d \vec{r} + \vec{V}(\frac{\gamma-1}{V^2}d \vec{r}.\vec{V} - \gamma dt)
\end{align}$$

Or equivalently, when broken into components:
$$\begin{align}
\vec{r_{\perp}}' &= \vec{r_{\perp}} \\
\vec{r_{\parallel}}' &=\gamma(\vec{r_{\parallel}}-\vec{V}t)
\end{align}$$

Right?

Not , quite, the correct formula is:

$$\begin{align}
d \vec{r_{\perp}}' &=d \vec{r_{\perp}} \\
d \vec{r_{\parallel}}' &=\gamma(d \vec{r_{\parallel}}-\vec{V}dt)
\end{align}$$

 

[Pete]

Small typo in post 147. In the first equation, dt should be t.

This is the general transformation for displacement vectors between events separated in time by t:
$$\begin{align}
\vec{r}' &= \vec{r} + \vec{V}(\frac{\gamma-1}{V^2}\vec{r}.\vec{V} - \gamma t)
\end{align}$$

Not exactly, the exact formula is the one I derived:

$$\begin{align}
d \vec{r}' &=d \vec{r} + \vec{V}(\frac{\gamma-1}{V^2}d \vec{r}.\vec{V} - \gamma dt)
\end{align}$$

That's not a new derivation, it's exactly the same formula.
The only difference is that you've applied it to the special case of (dr,dt), which is necessarily timelike.
So, it's less general than the original.

 

[Tach]

Small typo in post 147. In the first equation, dt should be t.

That's not a new derivation, it's exactly the same formula.
The only difference is that you've applied it to the special case of (dr,dt), which is necessarily timelike.
So, it's less general than the original.

What do you mean by "original"? The starting point is the generic transformation for position vectors:

$$\begin{align}
\vec{r}' &= \vec{r} + \vec{V}(\frac{\gamma-1}{V^2}\vec{r}.\vec{V} - \gamma t)
\end{align}$$

By differentiation, you get the general transformation for displacement vectors, whereby there is no condition of simultaneity imposed:

$$\begin{align}
d \vec{r}' &=d \vec{r} + \vec{V}(\frac{\gamma-1}{V^2}d \vec{r}.\vec{V} - \gamma dt)
\end{align}$$

 

[Pete]

What do you mean by "original"?

I mean the starting point quoted from Moller.

The starting point is the generic transformation for position vectors:

$$\begin{align}
\vec{r}' &= \vec{r} + \vec{V}(\frac{\gamma-1}{V^2}\vec{r}.\vec{V} - \gamma t)
\end{align}$$

This, together with
$$ t' &= \gamma(t - \vec{r}.\vec{V}/c^2)$$
is presented in Moller (1952, p41) as the general transformation of any 4-vector with one end at the origin. The other end is at an arbitrary time.

 

[Tach]

I mean the starting point quoted from Moller.

This, together with
$$ t' &= \gamma(t - \vec{r}.\vec{V}/c^2)$$
is presented in Moller (1952, p41) as the general transformation of any 4-vector with one end at the origin. The other end is at an arbitrary time.

...and , if you take the differential, as I have shown you, you get the general transform of an arbitrary displacement vector.

 

[Pete]

...and , if you take the differential, as I have shown you, you get the general transform of an arbitrary displacement vector.

No, you already have the general transform.
Taking the differential gives you the exact same formula you already had, just substituting dr for r, and dt for t.
You haven't derived a new formula, you just have the old formula with different variables.
What's the point?

 

[Tach]

No, you already have the general transform.

...for a position vector. In order to get the formula for a displacement vector you need to subtract two position vectors, which means that you need to differentiate.

 

[Pete]

3.1.1 Lorentz transformation of displacement vectors
OK, I see what you're saying. You say that Moller's equation is not general, because Moller defines r with one end at the origin; ie r is not an arbitrary displacement vector.
Fair enough.

I say that your equation is not general, because you define dr as the differential of a moving position; ie dr is not an arbitrary displacement vector.

So, let's agree on the transform of the displacement vector between two arbitrary events, then we can move on.

OK?

Try this (I apologize for the awkward exposition. I'm trying to be as concise as I can, without being ambiguous):

S and S' are defined in the usual way, with $$\vec{V}$$ the velocity of S' relative to S.

Consider two arbitrary events, A and B.
$$\vec{a}$$ and $$\vec{a}'$$ are the position vectors of A in S and S' respectively.
$$\vec{b}$$ and $$\vec{b}'$$ are the position vectors of B in S and S' respectively.
$$t_A$$ and $$t_A'$$ are the times of A in S and S' respectively.
$$t_B$$ and $$t_B'$$ are the times of B in S and S' respectively.

Define $$\vec{r}$$ and $$\vec{r}'$$ as the displacement vectors from A to B in S and S' respectively:
$$\begin{align}
\vec{r} &= \vec{b} - \vec{a} \\
\vec{r}' &= \vec{b}' - \vec{a}'
\end{align}$$

Define $$\Delta t$$ and $$\Delta t'$$ as the time difference between A and B in S and S' respectively:
$$\begin{align}
\Delta t &= t_B - t_A \\
\Delta t' &= t_B' - t_A'
\end{align}$$​

Questions:
Do you agree that $$\vec{r}$$ is an arbitrary displacement vector?
Do you agree that:
$$\begin{align}
\vec{r}' &= \vec{r} + \vec{V}(\frac{\gamma-1}{V^2}\vec{r}.\vec{V} - \gamma \Delta t) \\
\Delta t' &= \gamma(\Delta t - \vec{r}.\vec{V}/c^2)
\end{align}$$

?

 

[Tach]

3.1.1 Lorentz transformation of displacement vectors
OK, I see what you're saying. You say that Moller's equation is not general, because Moller defines r with one end at the origin; ie r is not an arbitrary displacement vector.
Fair enough.

Good.

I say that your equation is not general, because you define dr as the differential of a moving position; ie dr is not an arbitrary displacement vector.

Sure it is, as proof, you end up getting the same exact result, only your notation is slightly different. Let me show you:

So, let's agree on the transform of the displacement vector between two arbitrary events, then we can move on.

OK?

OK

Try this (I apologize for the awkward exposition. I'm trying to be as concise as I can, without being ambiguous):

S and S' are defined in the usual way, with $$\vec{V}$$ the velocity of S' relative to S.

Consider two arbitrary events, A and B.
$$\vec{a}$$ and $$\vec{a}'$$ are the position vectors of A in S and S' respectively.
$$\vec{b}$$ and $$\vec{b}'$$ are the position vectors of B in S and S' respectively.
$$t_A$$ and $$t_A'$$ are the times of A in S and S' respectively.
$$t_B$$ and $$t_B'$$ are the times of B in S and S' respectively.

Define $$\vec{r}$$ and $$\vec{r}'$$ as the displacement vectors from A to B in S and S' respectively:
$$\begin{align}
\vec{r} &= \vec{b} - \vec{a} \\
\vec{r}' &= \vec{b}' - \vec{a}'
\end{align}$$​

The only difference between what I have been showing you and the above is that I use the notation:

$$\begin{align}
d \vec{r} &= \vec{b} - \vec{a} \\
d \vec{r}' &= \vec{b}' - \vec{a}'
\end{align}$$

or:

$$\begin{align}
\Delta \vec{r} &= \vec{b} - \vec{a} \\
\Delta \vec{r}' &= \vec{b}' - \vec{a}'
\end{align}$$


That's all ....

Define $$\Delta t$$ and $$\Delta t'$$ as the time difference between A and B in S and S' respectively:
$$\begin{align}
\Delta t &= t_B - t_A \\
\Delta t' &= t_B' - t_A'
\end{align}$$

Questions:
Do you agree that $$\vec{r}$$ is an arbitrary displacement vector?
Do you agree that:
$$\begin{align}
\vec{r}' &= \vec{r} + \vec{V}(\frac{\gamma-1}{V^2}\vec{r}.\vec{V} - \gamma \Delta t) \\
\Delta t' &= \gamma(\Delta t - \vec{r}.\vec{V}/c^2)
\end{align}$$

?

I simply prefer a more uniform notation, as in:

$$\begin{align}
\Delta \vec{r}' &= \Delta \vec{r} + \vec{V}(\frac{\gamma-1}{V^2}\Delta \vec{r}.\vec{V} - \gamma \Delta t) \\
\Delta t' &= \gamma(\Delta t - \Delta \vec{r}.\vec{V}/c^2)
\end{align}$$

 

[Pete]

3.1.1 Lorentz transformation of displacement vectors

Questions:
Do you agree that $$\vec{r}$$ is an arbitrary displacement vector?
Do you agree that:
$$\begin{align}
\vec{r}' &= \vec{r} + \vec{V}(\frac{\gamma-1}{V^2}\vec{r}.\vec{V} - \gamma \Delta t) \\
\Delta t' &= \gamma(\Delta t - \vec{r}.\vec{V}/c^2)
\end{align}$$

?

I simply prefer a more uniform notation, as in:

$$\begin{align}
\Delta \vec{r}' &= \Delta \vec{r} + \vec{V}(\frac{\gamma-1}{V^2}\Delta \vec{r}.\vec{V} - \gamma \Delta t) \\
\Delta t' &= \gamma(\Delta t - \Delta \vec{r}.\vec{V}/c^2)
\end{align}$$

So, is that yes or no to those two questions?
It seems like you're implicitly agreeing, but I've been wrong before.

 

[Tach]

3.1.1 Lorentz transformation of displacement vectors


So, is that yes or no to those two questions?
It seems like you're implicitly agreeing, but I've been wrong before.

Yes, on both questions, with the caveats pointed out.

 

[Pete]

3.1.1 Lorentz transformation of $$\hat{P_t}$$
Further down in your approach:

$$vec{v_p} = \frac{d\vec{r}}{dt}$$

This implies that $$\vec{r}=\vec{P}(t)$$, ie it is the position vector of point P in S.

In the above, $$d\vec{r}$$ represents $$\hat{P_t}$$, and $$d\vec{r}'$$ represents $$\hat{P_t}'$$.

That doesn't follow.

$$d\vec{r}$$ is the differential displacement of point P in frame S in infinitesimal time
$$d\vec{r}'$$ is the differential displacement of point P in frame S' in infinitesimal time.

Recall how $$\hat{P_t}$$ and $$\hat{P_t}'$$ were defined:

$$\hat{P_t}$$ is the unit displacement vector tangent to wheel element P at time t in S.
$$\hat{P_t}'$$ is the unit displacement vector tangent to wheel element P at time t' in S'.

It was also made clear what it means for a vector to be tangent to the wheel:

If necessary, whether a given displacement vector is tangent to the wheel can be tested in at least two ways:

• A displacement vector is tangent to the wheel at a given point at a given instant if it is a scalar multiple of the displacement between two points on a straight rod tangent to the wheel at that point at that instant.
• A displacement vector is tangent to the wheel at a given point at a given instant if and only if there is no other point on the wheel at that instant such that the displacement between the given point and the other point is a scalar multiple of the displacement vector.

• 
  Note that $$\hat{P_t}$$ and $$\hat{P_t}'$$ do not come from the same 4-vector.
  $$\hat{P_t}$$ is between events simultaneous in S.
  $$\hat{P_t}'$$ is between events simultaneous in S'.
  
  This means that $$\hat{P_t}'$$ is not necessarily a simple transformation of $$\hat{P_t}$$, which is why rod T1 is useful.
  
  We know that rod T1 is a straight rod tangent to the wheel at P at t=0 in S, and also at some time t'=t'_0 in S', so we can use that rod to define $$\hat{P_t}(0)$$ and $$\hat{P_t}'(t'_0)$$:
  • $$\hat{P_t}(0)$$ is defined as a unit vector between two points on rod T1 at t=0 in S.
  • $$\hat{P_t}'(t'_0)$$ is defined as a unit vector between two points on rod T1 at $$t'=t'_0 \$$ in S.

 

[Tach]

$$\hat{P_t}$$ is between events simultaneous in S.
$$\hat{P_t}'$$ is between events simultaneous in S'.

Neither of the above is correct, we have been over this, several times. There is no requirement of events to be simultaneous in either definition. Neither tangent vector is stationary, so you have no reason to impose either $$dt=0$$ or $$dt'=0$$