Inverse Lorentz transformation confusion I've been tripped up for a very long time by this question. I hope that someone can explain it for me once and for all. My question is that when does one use the Lorentz transformation and when does one use the Inverse Lorentz transformation? Can someone give an example of when it is right to use one and when it is right to use the other? 
 A: Let us say you have two frames of reference; frame $F$ and frame $F'$ such that $F'$ is moving at velocity $v$ in the positive $x$ direction of $F$. Given a space time event that occurs at $(ct,x,y,z)$ in frame $F$ the Lorentz transform helps us to find the space-time coordinates $(ct',x',y',z')$ of that event in frame $F'$. If, however, you know the event occurs at  $(ct',x',y',z')$  the inverse Lorentz transform  helps us find the space time coordinates $(ct,x,y,z)$  of that event in frame $F$. 
The Lorentz transform for the $x$ coordinate is given by:
$$x'=\gamma (x-vt)$$
Everything on the RHS of this equation is measured in the frame $F$ and every thing on the LHS is measured in frame $F'$. From the first postulate of special relativity the laws of physics in frame $F'$ must be the same as those in frame $F$ so to find $x$ we can use:
$$x=\gamma(x'-v't')$$
Where $v'$ is the velocity of frame $F$ in the frame $F'$ which is $-v$, thus: 
$$x=\gamma(x'+vt')$$
This is the inverse Lorentz transform and again notice that everything on the LHS is measured in frame $F$ and on the RHS in frame $F'$. 
So the Lorentz transform and inverse Lorentz transform  are the same thing just between different frames. The Lorentz transform is used when going from frame $F$ to $F'$ and the inverse transform is used when going from frame $F'$ to frame $F$. 
A: 1.0 Introduction.
The Lorentz Transformation pertains to two inertial observers. The primary form of the transformation is generally used to transform the first observer's spacetime coordinates of an event into the second observer's  spacetime coordinates for the same event. The inverse form of the transformation is generally used in the opposite way, to transform the second observer's spacetime coordinates of an event into the first observer's spacetime coordinates for the same event. However, the reader should understand that the primary and inverse forms of the transformation are mathematically indistinguishable. To be clear, they are mathematically identical. They simply express the relationship of the first and second observer's spacetime representation of the same event in two different ways. It has been over a hundred years since the transformations were placed in the limelight, and yet even today, many people still believe there is a fundamental difference between the primary and inverse form of the transformation. I think I understand why since I was one of those people. I became interested in physics two years ago, and one of the first things I did was derive the primary and inverse form of the transformation which are presented below in Sections 1 and 2. The two sets of transformation equations looked considerably different to me, and I erroneously concluded they were different. But, they're not! You can use whichever set you want to use; you'll get the same answer with either set. However the amount of mathematics involved can be different using one set versus the other set. I'm pretty sure that there are some math gurus who can intuitively grasp the equivalence of the two forms of the transformation, but for the rest of us we need step by step proof to believe it. I provide that proof in Section 4. I derive the INVERSE form of the transformation from the PRIMARY form thereby proving the two forms are identical. In all of the derivations that follow  I apply a $(x,\,y,\,z,\,t)$ spacetime coordinate system to the first observer who I call Jane, and a $(x',\,y',\,z',\,t')$ spacetime coordinate system to the second observer who I call Dick. For more information about Dick and Jane consult my book |1| on relativity which is available on Amazon.
2.0 The Primary Transformation.
The primary form of the Lorentz Transformation is shown below. The equations were derived assuming Jane to be at rest in space and Dick to be in motion relative to Jane at velocity v in the positive xx' direction.
\begin{align}
  2.1\qquad x'  \quad&=\quad  \gamma\,(x - vt)\\
  2.2\qquad y'  \quad&=\quad  y\\
  2.3\qquad z'  \quad&=\quad  z\\
  2.4\qquad\, t'  \quad&=\quad  \gamma ( t - x v/c^{2})\\
  2.5\qquad\, \gamma\quad&=\quad \dfrac{1}{\sqrt{1- v^{2}/c^{2}}}
 \end{align}
3.0 The Inverse Transformation.
The inverse form of the Lorentz Transformation is shown below. The equations were derived assuming Dick to be at rest in space, and Jane to be in motion relative to Dick at velocity v in the negative xx' direction.
\begin{align}
  3.1\qquad x  \quad&=\quad  \gamma\,(x' + vt')\\
  3.2\qquad y  \quad&=\quad  y'\\
  3.3\qquad z  \quad&=\quad  z'\\
  3.4\qquad\, t  \quad&=\quad  \gamma (t'+ x' v/c^{2})\\
  3.5\qquad \gamma\quad&=\quad \dfrac{1}{\sqrt{1- v^{2}/c^{2}}}
 \end{align}
4.0 Deriving The Inverse From The Primary.
The primary form of the transformation shown in Section 2, expresses each of Dick's four spacetime coordinates (primed variables) in terms of combinations of Jane's four spacetime coordinates (un-primed variables). The inverse form of the transformation shown in Section 3 is formatted in the opposite way. Each of Jane's four spacetime coordinates (un-primed variables) are expressed in terms of combinations of Dick's four spacetime coordinates (primed variables). The following steps show that the inverse form can be derived from the primary form by simple algebra. The step by step derivation to show this equivalence begins with the first equation in Section 2 as shown below.
Expand right side of equation (2.1) then rearrange and solve for x as shown below.
\begin{align}
4.1\qquad x'\quad&=\quad  \gamma\,x - \gamma\,v\,t\\
4.2\quad\,\,\,\, \gamma x  \quad&=\quad  x' + \gamma v t\\
4.3\qquad\, x \quad&=\quad  x'/ \gamma + vt
 \end{align}
Expand right side of equation (2.4) and solve for t as shown below.
\begin{align} 
\,\,\,\,\,4.4\qquad t'\quad&=\quad \gamma t- \gamma x v/c^{2}\\
4.5\qquad t\quad\,&=\quad  t'/\gamma + x v /c^{2}
 \end{align}
Substitute t from equation (4.5) into equation (4.3) as shown below.
\begin{align}
\qquad\qquad\quad\,4.6\qquad x \quad&=\quad  x'/ \gamma + v(t'/\gamma + x v / c^{2})\\
  \quad&=\quad     x'/ \gamma + v\,t'/\gamma + x v\,^{2} / c^{2}
 \end{align}
Rearrange equation (4.6) so primes are on the right.
\begin{align}
 \qquad\qquad4.7\qquad x(1-v\,^{2} / c^{2})\,=\,x'/ \gamma + v\,t'/\gamma
 \end{align}
Apply identity $(1-v\,^{2}/ c^{2})\,=\, 1/\gamma^{2}$ to equation (4.7) then solve for x.
\begin{align}
  \,\,4.8\qquad x/\gamma^{2}\,&=\,x'/ \gamma + v\,t'/\gamma\\
  4.9\qquad\quad\,\,\, x\,&=\quad  \gamma (x' + v\,t')  
 \end{align}
The inverse of equations (2.2 and 2.3) is a simple left-right reversal as shown below.
\begin{align}
        4.10\qquad y \quad&=\quad  y'\qquad\qquad.\\
  4.11\qquad z \quad&=\quad  z'\qquad\qquad.
 \end{align}
Substitute x from equation (4.9) into equation (4.5) and then solve the expression for t. The identity $1/ \gamma^{2}=1-v^{2}/c^{2}$ is applied midway through the derivation as shown below.
\begin{align}
  \quad\qquad\qquad\qquad\qquad 4.12\qquad t \quad&=\quad  t'/\gamma + x v /c^{2}\\
         &=\quad  t'/\gamma + \gamma (x' + v\,t') v/c^{2}\\
      &=\quad  t'/\gamma + \gamma x'v/c^{2} + \gamma \,t'v^{2}/c^{2}\\
      &=\quad  \gamma t'( 1/\gamma^{2}+ v^{2}/c^{2}) + \gamma \,x'v/c^{2}\\
      &=\quad  \gamma t'( 1-v^{2}/c^{2}+v^{2}/c^{2}) + \gamma \,x'v/c^{2}\\
      &=\quad  \gamma t'+ \gamma \,x'v/c^{2}\\
      &=\quad  \gamma (t'+ x'v/c^{2})
 \end{align}
The $\gamma$ factor in eqquation (2.5) is not a function of Dick and Jane's spacetime coordinates, and is the same in both forms of the transformation.
\begin{align}
  4.13\qquad \gamma\quad&=\quad \dfrac{1}{\sqrt{1- v^{2}/c^{2}}}
 \end{align}
5.0 Summary.
The concordance of the derivation in Section 4 with the inverse form of the transformation in Section 3 is shown below.
\begin{align}
Equation (3.1)\quad = \quad Equation (4.9\,\,)\,\\
Equation (3.2)\quad = \quad Equation (4.10)\\
Equation (3.3)\quad = \quad Equation (4.11)\\
Equation (3.4)\quad = \quad Equation (4.12)\\
Equation (3.5)\quad = \quad Equation (4.13)
\end{align}
The above concordance proves that the two forms of the transformation are identical, and yet I erroneously concluded the two forms were different when I first derived them. My error in logic stemmed from the assumptions I made in deriving the two forms of the transformation. I was trying to mimic the way I envisioned Lorentz would have derived his equations. The fact that I assumed Jane to be at rest and Dick in motion for the first derivation, and the opposite assumption for the second derivation caused me to subconsciously infer that this distinction in the different assumptions would impart a corresponding distinction in the two forms of the transformation. This is of course an illogical inference. The correct inference is that the transformations must be compatible with the conditions in the assumptions used to derive them, not that they must mandate the conditions in the assumptions. I'm pretty sure I'm not the only on who has made an error in logic like that. For those of you interested in the logic leading to the Lorentz Transformation check out my book |1| about Einstein's throry of relativity.
Bibliography
|1| Richard Alan, Everything You Ever Wanted To Know About Dick & Jane And Mary.  "The derivation of Einstein's Special Theory of Relativity in nauseating detail".  Available on Amazon.
A: It has been more than a hundred years since The Lorentz Transformation and the so called Inverse Lorentz Transformation were placed in the limelight. You would think that by now everyone would realize that the two transformations are absolutely identical which means the inverse transformation does not really exist. It is a separate transformation in name only; an alias for the Lorentz Transformation disguised by a different format that makes it appear to be a different transformation. You can use whichever transformation you want. You'll get the same answer no matter which one you choose.
