Suppose we have a reference frame S with (position,time) variable as $(x,t)$ and in another frame$ S'$ we have another set of coordinates $(x',t')$ moving at a speed of $v$ with respect to $S$ then the Lorentz transform is the map which converts the first coordinates :

$$ x'=X(x,t,v)$$

Now the form of the equation takes:

$$ x'=A(v) x + B(v) t$$

So some function of velocity times the old x coordinate plus some other function of velocity times the old time coordinate. I found a way to show that it must neccesarily be that $x'$ is is straight line with respect to $x$ but how do I show that it should be with respect to $t$ as well? i.e: Manifestation of what physical property is it that $x'$ is a line with respect to $t$ for fixed $x$ and $v$?

Proof that $x'$ is linear function of $x$: Suppose in the frame with variables $(x,t)$ we have a rod with end points $x_1$ and $x_2$, then the length of rod is given as:

$$ x_2 - x_1 = \Delta x$$

Now suppose I displace the rod $x_2 \to x_2 +h$ and $x_1 \to x_1 +h$, then still length will be conserved.

In the other frame, the length is given as:

$$ x_2' -x_1'= \Delta x'$$

$$X(x_2,t)-X(x_1,t)=\Delta x'$$

Now, the length will be same when we transform the displaced rods coordinates:

$$X(x_2,t)- X(x_1,t) = X(x_2 +h,t) - X(x_1 +h,t)$$

Rearrange, divide by $h$ and take the limit:

$$ \frac{\partial X}{\partial x}|_{x_1} = \frac{\partial X}{\partial x}|_{x_2}$$

This can be argued for any points $x_1$ , hence the slope must be constant and the $X$ must be a straight line with $x$.

Now, how do I show a similar argument for time?

  • $\begingroup$ You say that there is a velocity $v$ in one reference frame and another $v'$ in another reference frame. What is the relationship between the two frames ? Usually SR has $(x,t)$ in $S$ and $(x',t')$ in $S'$ and there is one constant relative velocity between $S$ and $S'$. I also think that without any assumptions on what your transformation is supposed to preserve there is no hope to do any derivations. In SR the Lorentz-transformation preserves the Lorentz metric. $\endgroup$
    – Kurt G.
    Jul 13, 2021 at 11:08
  • $\begingroup$ Hi, the assumption here I used for proving that it is line with respect to x is that length of rod measured at any place in old frame is same. I want to figure out what would be the physical assumption for the $t$ $\endgroup$
    – Babu
    Jul 13, 2021 at 11:49
  • $\begingroup$ I see. You are trying to show that the transformation must be linear when there is some homogeneity in space and time. Space looks ok. Time should not be any different. If you measure a time interval $\Delta t$ in $S$ it does not matter what time your clock shows at $t_1\,.$ Same for $S'\,.$ In other words: time differences are time homogeneous in every Lorentz frame. $\endgroup$
    – Kurt G.
    Jul 13, 2021 at 12:10
  • $\begingroup$ This should have been written as an answer @KurtG. xD $\endgroup$
    – Babu
    Jul 13, 2021 at 14:15
  • $\begingroup$ Probably. Meanwhile I am seeing nice answers here already. Have you thought about the question what conditions imply not only the linearity but also the form of the Lorentz transformations as we know them ? $\endgroup$
    – Kurt G.
    Jul 13, 2021 at 16:39

2 Answers 2


Now, how do I show a similar argument for time?

It is actually easier to argue both together. We are trying to find the form of a transformation between inertial frames. An inertial frame is one where Newton’s 1st law holds: an isolated object moves in a straight line at a constant speed. Such a path is a straight line in spacetime. Therefore, we are looking for spacetime transformations that map straight lines to straight lines.

Such transformations are called affine transformations. An affine transformation can be written as $\mathbf{x’}=\mathbf{A \ x} + \mathbf{b}$. The $\mathbf{b}$ is just a translation of the origin, which we can defer for now and add on later. So we focus on $\mathbf{A \ x}$ which is a linear transformation and has the form you wanted above for both space and time


You can just displace the origin of the chosen co-ordinate axes along the time axis. When you displace the origin, the time co-ordinates of all the events change a bit.

Now consider the fact that the choice of origin is arbitrary. By changing the origin, you've merely changed the time-labels that you're attributing to the events. Physics doesn't care about the label you provide to the events. The labels are only for your book-keeping. Using this reasoning, you can conclude that this shift in origin doesn't have any impact on the time interval between any two events after Lorentz transformation.

Here's another way to put it: You and your friend are standing at the same location with stopwatches in your hands (no relative motion between you two). You both want to measure the interval between two events $A$ and $B$. Say, your stopwatch measures event $A's$ occurrence at $t=1s$ and $B$ at $t=2s$ Your friend's watch shows $A$ at $t=1+h$ and $B$ at $t=2+h$. Clearly, the only difference is that your friend started his stopwatch $h$ seconds earlier than you.

If you both independently calculate the time interval between the two events as observed by some observer moving relative to you, you two clearly should reach the same answer. You both belong to the same inertial frame. The fact that your friend started his stopwatch earlier to make time measurements should have no impact on his calculation.


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