How to derive the $vx/c^2$ term from first principles? In Lorentz transforms, the formula for time transformation is
$$t' = \gamma \left( t - \frac{v x}{c^2} \right)$$
I understand that the term $\frac{v x}{c^2}$ represents "time delay" seen by a stationary observer but I don't understand how to derive it from first principles. I understand $v/c$ as speed and $x/c$ as distance. Why multiply speed with distance? I thought time is distance divided by speed?
 A: As JG mentioned in the comment section, the derivation can be obtained in many ways. However, if you want to see some derivation based on first principles without heuristics like assuming that $x'$ and $t'$ are linear functions of $x$ and $t$, this is the most physical formulation I can think of, with only the second postulate of the special relativity,

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*The speed of light is a constant $c$ regardless of which inertial frame the observer chooses.

Starting Point: Measure the Spacetime Coordinate of an Event
Suppose an observer, Alice, is standing at the origin of the $x$-axis with her spacetime coordinate $(0,0)$ at the instant, and there are infinitely many mirrors uniformly buried underground along the $x$-axis. Alice has a device which has been and will be emitting light continuously over time, and information is encoded into light at emission. This can be theoretically done through altering power or frequency of light, but the point is Alice can tell which light is emitted at which time.
Suppose at time $t$, a mirror at the position $x>0$ quickly pops out the ground, reflects some light, and resorts back underground. Alice therefore receives a signal at time $t+{x \over c}$, and since she knows this is the light she emitted at time $t-{x \over c}$ through encoded information, she can locate the event of the mirror popping out at
$$({1 \over 2}\big(\big(t-{x \over c}\big)+\big(t+{x \over c}\big)\big),{1 \over 2}c\big(\big(t+{x \over c}\big)-\big(t-{x \over c}\big)\big))=(t,x)$$
Now suppose another observer Bob joins the experiment. At time $0$, he is also at the origin of the $x$-axis, but he is moving at velocity $v$ along the $x$-axis. He has a device similar to Alice's, and he also tries to measure the spacetime coordinate of the mirror popping out, which is $(t',x')$ in his frame. To make this done, we define two important events:

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*Event 1: When Bob is at $(t_1,vt_1)$ in Alice's frame, he emits the light which will be intercepted and reflected by the mirror popping out at $(t,x)$.

*Event 2: When Bob is at $(t_2,vt_2)$ in Alice's frame, Bob receives the light he emitted.

Since Bob is not moving in his frame, the spacetime coordinate of Event 1 in Bob's frame is $(t'_1,0)$, and that of Event 2 in Bob's frame is $(t'_2,0)$ where the relationship between $t'_1$, $t'_2$ and $t'$, $x'$ are unknown currently. However, Bob now can schematically do the measurement procedure similar to Alice, and he has the spacetime coordinate of the mirror popping out at
$$({1 \over 2}\big(t'_1+t'_2\big),{1 \over 2}c\big(t'_2-t'_1\big))=(t',x')$$
where we have assumed $x-vt>0$ so it is the light emitted to the right by Bob will be reflected by the mirror.
Crucial Step: Resolve Spacetime Coordinates in Different Frames
In this part, we determine $t_1$, $t_2$, $t'_1$ and $t'_2$ with $t$, $x$, $v$ and $c$. The spacetime coordinates of Event 1 and Event 2 in Alice's frame is easy to solve. Since in Alice's frame, the light emitted by Bob at time $t_1$ travels over the distance $c(t-t_1)$ at time $t$,
$$x-vt_1=c(t-t_1) \Rightarrow t_1={ct-x \over c-v} \tag{1}$$
About Event 2, since in Alice's frame, the light travels over the distance $vt_2-x$ from time $t$ to time $t_2$, we have
$$vt_2-x=-c(t_2-t) \Rightarrow t_2={ct+x \over c+v} \tag{2}$$
To relate $t'_1$ with $t_1$ and $t'_2$ with $t_2$, we consider a  light clock traveling with Bob (see figure),

and from the figure, since the light of the light clock in Alice's frame has traveled over the length by a factor of $\gamma={1 \over \sqrt{1-{v^2 \over c^2}}}$ in comparison with that in Bob's frame, we get
$$t'_1={1 \over \gamma}t_1, \ t'_2={1 \over \gamma}t_2 \tag{3}$$
Therefore, with Eq. (1), (2), and (3)
\begin{align}
(t',x') & = {1 \over 2\gamma}\big({ct-x \over c-v}+{ct+x \over c+v},c\big({ct+x \over c+v}-{ct-x \over c-v}\big)\big) \\
& = {1 \over 2\gamma}\big({2(-xv+c^2t) \over c^2-v^2},{2c(xc-vct) \over c^2-v^2}\big) \\
& = (\gamma\big(t-{vx \over c^2}\big),\gamma\big(x-vt\big)) \tag{4}
\end{align}
Supplementary Note
In Eq. (1) and Eq. (2), we have assumed $x-vt>0$. To derive the complete result, we can change them into
\begin{align}
x-vt_1 & =\text{sgn}(x-vt) \cdot c(t-t_1) \\
vt_2-x & =-\text{sgn}(x-vt) \cdot c(t_2-t)
\end{align}
with the spacetime coordinate $(t',x')$
$$(t',x')=({1 \over 2}\big(t'_1+t'_2\big),{1 \over 2}\text{sgn}(x-vt) \cdot c\big(t'_2-t'_1\big))$$
And I leave the rest verification to you.

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*Image from https://commons.wikimedia.org/wiki/File:Light-clock.png.

A: I will try with the diagram below, we suppose that the container ijfg is filled with water, the light crosses this container of the face $f$ towards the face $g$ with a speed $v$ and put a time $t$ to cross it, we have $$l=vt$$
if there is no water, the light will travel a distance L =ct (during the same time t),we have : $$\frac{L}{c}=\frac{l}{v} $$
which gives $$L=\frac{c}{v}l=nl\;\;\;\;\;(1)$$
which is the optical path.
we can see that : $$l=vt=cT=c(t-t')$$
wiht:$$ct=vt+ct'$$ $$t'=\left(t-\frac{vt}{c}\right)\;\;\;\;\;(2)$$
from (1), we have :$$\frac{l}{c}=\frac{v}{c^{2}}L=T$$
and (2) becomes :$$t'=\left(t-\frac{vL}{c^{2}}\right)$$

A: One route (explained broadly) is as follows:
Once you've written down $x'(x,t)$ as a length contraction, (which itself follows from time dilation), then it follows from the first postulate there exists a mode of description for $x(x', t')$ in terms of $-v$. Writing both sides equal to $\gamma$ (which is with the square magnitude of $v$, irrespective of its direction) you can set equal and solve for $t'(x,t)$
A: I'm not sure there's a meaningful way to derive it apart from the full Lorentz Equations, but one thing reflected in the equation is that a time interval $t'$ for one observer will correspond to partly a time interval $t$ and partly a space interval $x$ for a different observer.
This happens in exactly the same way that in 2 dimensional Euclidean geometry, in a plane with two different sets of axes set at an angle, an interval that is fully in the $x$ direction in one axes will have components in both the $x'$ and $y'$ of the other axes.
A: If "first principles" means "The Minkowski Metric", then note that the spacetime interval (I use units where $c=1$ for this whole discussion.  Feel free to replace $v$'s with $v/c$'s if you like):
$$ds^{2} = -dt^{2} + d^3{\vec x}^{2}$$
is invariant under the transformation:
$$\begin{align}
T & = t\cosh\phi  + x\sinh \phi \\
X & = t\sinh\phi  + x\cosh\phi \\
y & \rightarrow y\\
z & \rightarrow z\\
\end{align}$$
First, let's note the relative velocities of the frames.  First, let's consider an object moving at constatnt spacetime velocity on the interval $(0,0), (t_{1}. 0)$, which is obviously stationary in the base frame.  Applying the above transform, we have, in the new frame, the object traveling the interval $(0,0), (t_{1}\cosh\phi, t_{1}\sinh\phi)$.  It is moving in both T and X, so its velocity must be: $v = \frac{\Delta X}{\Delta T} = \tanh\phi$, which means the velocity of the frame must be opposite this, $-\tanh\phi$
Now, note that $\cosh^{2} - \sinh^{2} = 1$ tells us that $1-\tan^{2} = 1/\cosh^{2}$.  So, we have $\cosh \phi = \frac{1}{\sqrt{1-v^{2}}})$ and $\sinh\phi = -\frac{v}{\sqrt{1-v^{2}}}$
Knowing this, go back to our transformation for time, and we have:
$$T = \frac{t - vx}{\sqrt{1-v^{2}}}$$,
which is precisely your relationship, which derives solely from Minkowski geometry.
