**3.1.1 Lorentz transformation of vectors**

No, This particular subsection is supposed to be about theThroughout 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.

*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.

I said that transforming a spatial displacement vector obviously doesn't rely on transforming a velocity vector.You have it backwards, the velocity transformation is a CONSEQUENCE of the transformation of the displacement vector $$dr$$.

How is that backwards?

You keep talking about transforming dThis 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.

**r**. Why?

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

Why is differentiating necessary?

No, he is not differentiating at all.He actually is doing differentiation followed by setting $$dt=0$$

He is starting with a displacement vector

**r**, and deriving

**r'**.

Your document begins by presenting a general transformation for displacement vectors:I have already covered this subject in great extent. See the document I posted.

Why differentiate?Tach said:The Lorentz Transforms for Displacement Vectors

How does a vector transform in SR? Assume given a vector $$d\vec{r}$$ in frameSand a frameS'is moving with velocity $$\vec{V}$$ wrt frameS.

The Lorentz transformation betweenSandSis:

$$\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}$$

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:

So, $$d\vec{r}$$ is the differential displacement of pointTach said:$$\vec{v_p} = \frac{d\vec{r}}{dt}$$

**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**.

This is not consistent with the definition of $$\hat{P_t}$$.Tach said:In the above, $$d\vec{r}$$ represents $$\hat{P_t}(0)$$ and $$d\vec{r}'$$ represents $$\hat{P_t}'(0)$$.

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

And we of course must be clear about what it means for a vector to be tangent to the wheel:Pete said:$$\hat{P_t}(t)$$ 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'.

Pete said: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?