I have been trying to understand a more or less geometric derivation of the Lorentz transformation, and I'm getting stuck at one spot. The wikipedia article for the Lorentz transformation for frames in standard configuration lists the following equations:

$$x^{\prime} = \frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}}$$

$$y^{\prime} = y$$

$$z^{\prime} = z$$

$$t^{\prime} = \frac{t-(v/c^2)x}{\sqrt{1-\frac{v^2}{c^2}}}$$

I've been able to work everything out except for $-(v/c^2)x$ in the $t^{\prime}$ equation. I haven't seen any explanations for this, which makes me feel like I'm missing something simple. Where does this part of the equation come from? Shouldn't $t^{\prime} = \gamma \cdot t$?

EDIT: Ok, so I reviewed the idea I was using to derive the Lorentz factor and thus the transformation for $t^{\prime}$. Suppose you have the two frames I've described, and you have a light wave moving perpendicular to the X axis in the second ($\prime$) frame.

Light Path Diagram

Using basic trig with the diagram, you can derive:

$$t^{\prime}=t\cdot\sqrt{1 - \frac{v^2}{c^2}}$$

Obviously this would contradict the transformation provided by wikipedia. What step am I missing here? I don't really want a proof that I'm wrong or that the equation I've derived is incorrect - I'm already pretty convinced of that. What I would really like is an intuitive explanation as to why mine is invalid and how I would go about deriving the correct equation through similar means.

  • 1
    $\begingroup$ Marek's answer is fine, although possibly not at the right level for the OP. Note that if the time transformation were $t'=\gamma t$, then all observers would agree on simultaneity, but there are straightforward and well-known ways of showing that this can't happen, e.g., Einstein's famous thought experiment with the train and the two flashes of lightning: bartleby.com/173/9.html $\endgroup$
    – user4552
    Commented Jul 27, 2011 at 17:14
  • 1
    $\begingroup$ The other thing to understand here is that length contraction and time dilation are both different things from what the Lorentz transformations describe. Length contraction and time dilation describe the properties of clocks and rulers in frames where they're not at rest, compared to frames in which they are at rest. To recover time dilation as a special case of the Lorentz transformations, you have to pick two events, $(t_1,x)$ and $(t_2,x)$, and substitute them into the Lorentz transformations. Then the $\gamma v x/c^2$ terms cancel. $\endgroup$
    – user4552
    Commented Jul 27, 2011 at 17:17
  • $\begingroup$ Thanks @Ben-Crowell, that's really helpful. I'm thinking that I may have underestimated the complexity involved :) $\endgroup$
    – Jake
    Commented Jul 28, 2011 at 4:31

5 Answers 5


I'll not derive the transformation (that has been done in countless books and articles, I am sure you can find them yourself) but instead will try to explain why the formula you propose can't be correct.

For starters, observe that since you don't touch $y$ and $z$, we might as well work in 1+1 dimensions. Also, let $c=1$ so that we aren't bothered by unimportant constants (you can restore it in the end by requiring that formulas have the right units). Then it's useful to reparametrize the transformation in the following way $$x' = \gamma(x - vt) = \cosh \eta x - \sinh \eta t$$ $$t' = \gamma(t - vx) = -\sinh \eta x + \cosh \eta t$$ where we introduced rapidity $\eta$ by $\tanh \eta = v$ and this by standard (hyperbolic) trigonometric identities implies $\cosh \eta = \gamma = {1 \over \sqrt{1 - v^2}}$ and $v \gamma = \sinh \eta$, so that this reparametrization is indeed correct.

Now, hopefully this reminds you a little of something. In two-dimensional Euclidean plane we have that rotations around the origin have the form $$x' = \cos \phi x + \sin \phi y$$ $$y' = -\sin \phi y + \cos \phi x$$ and this is indeed no coincidence. Rotations preserve a length of vector in Euclidean plane $x'^2 + y'^2 = x^2 + y^2$ and similarly, Lorentz transformations preserve space-time interval (which is a notian of length in Minkowski space-time) $x'^2 - t'^2 = x^2 - t^2.$ You can check for yourself that only the stated transformation with hyperbolic sines and cosines can preserve it and consequently the change you introduced will spoil this important property. Also, if you are familiar with phenomena like relativity of simultaineity, one could also argue on physical grounds that your proposed change can't lead to physical results.

Incidently, there has recently been asked similar question to yours, namely how to derive that the transformation is linear purely because of the preservation of space-time interval. You might want to check it out too.


You should look at this answer, because it derives the term you want right away. Einstein's postulates $\leftrightarrow$ Minkowski space for a Layman

The reason it's $t'=(t-vx)/\sqrt{1-v^2}$ and not $t'=t/\sqrt{1-v^2}$ (you must set c=1 to follow anything in relativity) is simple--- it's failure of simultaneity at a distance. The coordinate lines t=constant can't stay horizontal in a space-time diagram--- they have to get tilted up by the same amount that the time axis is tilted right. The remaining factors can be understood by reproducing time-dilation and length-contraction arguments, but failure of simultaneity is the most important nonintuitive effect, and it is the first discussed by Einstein in his paper, for this reason.

The form of the Lorentz transformation should be constrasted with the form of a rotation of the x and y coordinates, so that the x coordinate gets a slope of m:

$$x' = { x+my \over\sqrt{1+m^2}}$$ $$y' = { y-mx \over\sqrt{1+m^2}}$$

or if you use different units for x and y, say x in inches and y in centimeters,

$$x' = { x + my \over \sqrt{1+{m^2\over c^2}} }$$ $$y' = {y - {mx\over c^2}\over \sqrt{1+{m^2\over c^2}} }$$

Where c is a universal constant of nature: the isoceles slope of right, which is the slope of an isoceles right triangle with legs along the x and y axis. It's magnitude is 2.54 cm/inch.


I also have struggled with this issue for sometime. So, here's the answer completely intuitive. It is obviously related with relativity of simultaneity. Now in the moving frame, if two events are simultaneous for the moving person, don't make it simultaneous for you. If you work out the calculations, you'll see that there is a time difference of $(vx/c^2)\gamma$, where the event farther in the direction of motion happens later by this much time difference in the rest frame. So you add this factor to the time of the clock of the observer in case of the fact that the person is moving along the positive side of your origin of spacetime

Now, why the negative sign in the reverse transformation, well it turns out that this is an interplay of choosing a right handed coordinate system or saying that all observers, even those moving in the opposite direction, must attribute the same side of origin as positive in their coordinate systems. This was implicitly done even in gallilean days when you solved simple kinematics problems.

Now what happens is in the negative velocity case, if you assign coordinates to events, the one which is more farther along your direction of motion now will be negative too. Hence, the 'x' coordinate will be negative, and velocity is also negative and two negatives make a positive. So, this again leads to addition of extra time for the event, since the clock tick happened earlier than the event you were recording.

All this can be rederived if you think carefully about two simulatenous events in a moving frame. I hope this clears it up.


That's interesting. You said you're able to work out everything, except the $-(v/c^2)x$ offset for $t$. You can derive this? $$x' = \frac{x - vt}{\sqrt{1 - (v/c)^2}}.$$ Ok. Then, what about the other way? Remember the inverse transform? $$x = \frac{x' + vt'}{\sqrt{1 - (v/c)^2}}.$$ Now, all you need to do is substitute for $x'$ and solve for $t'$.


Deriving the Lorentz Transformation equations clearly can be done via the math to math approach, but understanding can thus still be at a reach from you since math is an external tool, external from the mind.

A logical analysis of motion soon produces a geometric representation of motion which then soon produces the full set of equations related to SR. This of course includes the Lorentz Transformation equations and a complete understanding of them.


If you have time to watch the above playlist( 9 short videos, total of approx. 1 1/2 hours ), you will never see Einstein's Special Relativity the same again. The 9 videos playlist can also be found at youtube via "KSP Special Relativity" search.

You said "I've been able to work everything out except for $-(v/c^2)x$ in the $t^{\prime}$ equation.". This part of the equation is made crystal clear within the videos. In short, if a spaceship was at absolute rest in space, it would be extending across space only, but it is in motion across the dimension of time at the speed of light. Thus while within the 4 dimensional environment known as Space-Time, its current direction of travel is across time only.

However, if the spaceship now changes its direction of travel within Space-Time, rotation occurs, motion across space occurs, and less motion across the dimension of time occurs. Thus, time slows down, and due to the rotation, the spaceship begins to extend less across space now that it begins to partially extend across the dimension of time as well.

Thus we encounter spatial contraction and time dilation.

But what is to be noted is that the spaceship now extends partially across the dimension of time, thus a clock at the rear end of the spaceship is no longer in sync with a clock located at the front of the spaceship since both are no longer located at the same point in time. These clock offsets are thus included in the overall change of our measurement instruments.

Thus the rulers contract in length, both clocks have slowed down, and the two clocks are not in sync.

Thus the $-(v/c^2)x$ in the $t^{\prime}$ equation concerns the clock offsets which occur due to rotation.

  • $\begingroup$ If you give a down vote, please provide an explanation if you are able to. Thank you. $\endgroup$
    – Sean
    Commented Aug 15, 2014 at 13:26

Not the answer you're looking for? Browse other questions tagged or ask your own question.