Physical reason for Lorentz Transformation

Seeing the mathematical derivation of the Lorentz Transformation for time coordinates of an event for two observers we get the term.

Now how to make sense physically of the $t-\frac{vx}{c^2}$ factor.

I am looking for an argument along the lines of the following. When relating spatial coordinates,

one observer measures the length separation between an event and the second observer in his frame and tells the other observer that this should be your length, which the second observer denies due to relativity of simultaneity and multiplies by the gamma factor to get the correct length.

• Relativity does not relate absolute time values to each other. It only relates time differentials. The only way you are getting an absolute time transformation from one coordinate system to another is by fixing the "start times" of your two clocks and the histories of both coordinate systems, which is a completely arbitrary choice that has nothing to do with physics. But maybe I am missing something about this specific scenario? – CuriousOne Aug 16 '14 at 1:17
• @CuriousOne I am sorry, I am assuming that there was a place when the observers met at a particular time and place and their clocks were set to 0 at that instant. – Isomorphic Aug 16 '14 at 9:38
• Thanks for clarifying. You are basically integrating the time transformation for a special case. Somebody may have a better idea, but I would say, what you are looking for is basically just an integration constant that happens to depend on the location where the two observers are synchronizing their clocks to a common value (in this case 0). I would not assign any specific physical meaning to it. Having said that, real world applications of relativity, like the GPS system, do have to keep track of these "unphysical" integration constants very carefully, though! – CuriousOne Aug 16 '14 at 20:10
• @CuriousOne Interpreting the terms as differentials. I mean take 't' to be 'dt' and so on. Can you explain what is the physical reason behind appearance of the '-$vdx/c^2$'factor. – Isomorphic Aug 17 '14 at 6:44
• Special Relativity basically states that there is no way to find the "absolute" physical reasoning that you are looking for. However, if one "absolutely" understands Special Relativity, then all the physical reasons behind all of the equations become instantaneously clear, since "absolute" extends beyond the boundaries of relativity. Thus, in turn, those that "absolutely" understand Special Relativity, should be able to provide you with the answer. – Sean Aug 17 '14 at 15:30

3 Answers

Consider two observers $O$ and $O'$ moving with velocity $v$ with respect to each other. Both of them will use a photon bouncing off a mirror to define any kind of duration, so it is easy to show that their duration will satisfy the relation $$\Delta t' = \gamma \Delta t \,,$$

and a similar argument with measuring any relative distance $\Delta l$ between points in the direction of relative movement gives

$$\Delta l' = \gamma \Delta l \, .$$

The stated relations will hold under any coordinate convention. Now suppose observer $O$ decides to establish a coordinate convention by setting $t=t'=0$ when her coordinate origin $x=0$ met with the coordinate origin $x'=0$ of $O'$. She then asserts that the relative distance between their origins grows as $-vt$, so they have after transforming any relative distances $$x'= \gamma (x - vt) \, .$$ By the same argument, she concludes that $O'$ must see the relative distances as changing in the opposite way with $+v t'$ and after using the same transform (due to the relativity principle), $O$ concludes that $$x = \gamma (x' + v t'\!) \,.$$ But, as we know, the relative growth $\Delta t$ is not the same as $\Delta t'$. By substituting for $x'$ and turning the last equation into an expression for $t'$ we have. $$t'= \gamma(t - \frac{vx}{c^2})$$ I.e., it is a question of consistence of the coordinate convention picked out by $O$ that we need this correction $vx/c^2$. If we chose $O'$ to fix her origin on the one of $O$, we would get

$$x'=\gamma x,\, t'=\gamma t \,.$$

The term in the transformation is thus a consequence of the fact that we choose to mix a $vt$ factor into the coordinate convention which in turn requires a skewing of time through space. It is very important during the discussion of special relativity to remember that this is a conventional (non-physical) coordinate transform fix.

EDIT: As requested, I will present a direct derivation which gives first a conventional meaning to simultaneity (the $t=\rm const.$) before requesting the consistency of the $x$ coordinate and without any explicit statement of length contraction. We only have to presume we already know the standard derivation of the relation $$\Delta t' = \frac{\Delta t}{\sqrt{1- v^2/c^2}}$$

Consider the following simultaneity convention:

Consider an array of clocks in rest with respect to $O$, this is a fixed observable feature. A light signal is sent from the origin of $O$ (this is the conventional point, the referential signal point could be anywhere in the coordinate system, even moving with respect to it) denoting the $t=0$ moment.

All the clocks set their time to $t=0$ upon receiving the signal even though there is a certain delay $\delta t_C$ for every individual clock $C$ and immediately send a light signal with their identification back to $O$. $O$ receives the signal from a clock $C$ at a time $t_C$.

She naturally concludes that $t_C = \Delta l/c + \Delta l/c$ and from that computes that when she sends a signal to $C$, the signal will be there at a delay $\delta t_C =t_C/2$. She then sends a signal to $C$ at her own time $t$ to set itself to $t + \delta t_C$. Once this process is finished for every clock $C$, a notion of the time coordinate $t$ or simultaneity is established globally for all points seen by $O$.

But consider now that $O$ watches $O'$ while establishing a similar convention with respect to a similar set of clocks, i.e., a set of clocks which is moving with a velocity $v$ in the $x$ direction with respect to $O$ but is in rest with respect to $O'$. $O'$ sends out the signal at $t'=0$, but the clocks in front of her (in the direction of motion) are "rushing away from the signal" with the speed of light constant, so the delay will be longer. On the contrary, the clocks behind $O'$ are "rushing towards the signal", so the delay is shorter. For clocks strictly in front of and behind $O'$, we have a delay $$\delta t_{b/f} = \frac{\Delta l}{c \pm v}$$ Once the clocks send their signal back, an opposite effect takes place for the clocks in front/ behind. The resulting time the second clocks send back their signals is thus identical for both cases (as measured by the clock system established by $O$): $$t_C = \frac{\Delta l}{c + v} + \frac{\Delta l}{c - v} = \frac{2 \Delta l}{c}\frac{1}{1 - v^2/c^2}$$ $O$ now uses the knowledge that $O'$ actually measured this time delay with a dilation, so she knows $O'$ will be sending out a correction time $$\delta t'_C = \frac{ \Delta l}{c}\frac{1}{\sqrt{1 - v^2/c^2}}$$ $O$ is perfectly clear on the fact that the duration from the point of view of $O'$ $t'$ since the time they met will be rescaled by the factor $1/\sqrt{1-v^2/c^2}$. However, $O$ concludes that also the sent out time/correction should be different to conform to her view of simultaneity. $O$ sees the duration of flight of the signal to be $\Delta l/(c \pm v)$ from her own point of view, but also realizes the clocks co-moving with $O'$ run slower, so she has to multiply the correction by $\sqrt{1-v^2/c^2}$ to compensate for the fact. She then sees a total discrepancy between her simultaneity and the $O'$ simultaneity of the magnitude $$\delta t'_{b,f} - \delta t'_C = \frac{\Delta l}{c}\left(\frac{\sqrt{1- v^2/c^2}}{1 \pm v/c} - \frac{1}{\sqrt{1-v^2/c^2}} \right)$$ Which gives after a bit of algebra $$\delta t'_{b,f} - \delta t'_C = \mp \frac{v \Delta l}{c^2} \frac{1}{\sqrt{1- v^2/c^2}}$$ Using a coordinate $x$ which is 0 at $O$, and the positive sign means "in front" and negative "behind", we can conclude that $O$ will summarize the "deformation" of $O'$ global notion of time as $$t' = \frac{t}{\sqrt{1-v^2/c^2}} - \frac{ v x}{c^2} \frac{1}{\sqrt{1- v^2/c^2}} = \gamma (t - vx/c^2)$$

$\blacksquare$

• Thank you for your answer. I would be grateful if you could derive the time dilation factor independely using t=$\gamma t'$ and not using the lorentz transformation for distance. – Isomorphic Aug 24 '14 at 6:11
• Yeah, I kinda remembered why I did not want to do it this way. Enjoy the behemoth full derivation, I think this should be exactly the stuff. – Void Aug 24 '14 at 13:24

The physical reason IS the constancy of the velocity of light... since I'm writing in a tablet the answer won't be complete, but expect to get you to the mathematical cross-road.

Constancy of velocity of light implies that $$\frac{d|\vec{x}|}{dt} = c, \quad\Rightarrow\quad d|\vec{x}| = c\,dt.$$ Since $d|\vec{x}| = \sqrt{dx^2 + dy^2 + dz^2}$, it follows that $$dx^2 + dy^2 + dz^2 = c^2\,dt^2 \quad\Rightarrow\quad dx^2 + dy^2 + dz^2 - c^2\,dt^2 = 0.$$

From here it is straightforward to see that the set (or group) of transformations preserving this quantity are those known as Lorentz transformations.

Now I leave you to analize the "generalization" to nonvanishing intervals, for massive particles. Hint: define a four-dimensional metric!

(continuation... after a few days)

The interval

As exposed previously, the physical condition of constancy of the speed of light leads to the conclusion that

All equivalent observers are connected through a transformation which keep the quantity $$dx^2 + dy^2 + dz^2 - c^2\,dt^2 = 0.$$

This can be generalized to the preservation of the quantity $$I = dx^2 + dy^2 + dz^2 - c^2\,dt^2,$$ called interval.

Notice that the interval can be written as $$I = X^t\, \eta\, X = \begin{pmatrix} ct & x & y & z \end{pmatrix} \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} ct \\ x \\ y \\ z \end{pmatrix}$$

Invariance of the interval

In order for the interval to be invariant under a transformation $X' = M\, X$, one needs to \begin{align} I &= I' \notag \\ X^t\, \eta\, X &= (M\, X)^t\, \eta\, M\,X \notag \\ &= X^t\, M^t\, \eta\, M\,X \notag \\ \Rightarrow\quad \eta &= M^t\, \eta\, M. \tag{*} \end{align} Therefore the problem is to find a set of transformations $M$ satisfying Eq. (*).

Two-dimensional case

Finding a general 4 by 4 matrix $M$ preserving the Minkowskii metric ($\eta$) requires a lot of algebra, but one can easily find the transformation preserving the 2 by 2 restriction to the $(ct,x)$-plane.

Propose a matrix $$M = \begin{pmatrix} a & b \\ c & d \end{pmatrix},$$, and solve the equation $$\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} a & c \\ b & d \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} -a^2+c^2 & -ab+cd \\ -ab+cd & -b^2+d^2 \end{pmatrix},$$ which is simple if one uses the identities $\cosh^2\theta - \sinh^2\theta = 1$, and the condition $M(\theta\to 0) = \mathbf{1}$.

Thus, $$M = \begin{pmatrix} \cosh\theta & -\sinh\theta \\ -\sinh\theta & \cosh\theta \end{pmatrix}.$$

Relation with the velocity

In a similar fashion of Euclidean geometry, in which $$\frac{y}{x} = \tan\theta,$$ one uses the transformation $M$ above to relate the $ct$ coordinate with the $x$ coordinate $$\frac{v}{c} \equiv \frac{x}{ct} = \mathop{\mathrm{tanh}}\theta.$$

Now, \begin{align} \mathop{\mathrm{tanh}^2}\theta &= 1 - \mathop{\mathrm{sech}^2}\theta \notag \\ &= 1 - \tfrac{1}{\cosh^2\theta} \notag \\ \Rightarrow\quad \cosh\theta &= \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \notag \\ \sinh\theta &= \frac{\frac{v}{c}}{\sqrt{1 - \left(\frac{v}{c}\right)^2}}. \notag \end{align}

Finally, from the relation $X' = M\, X$, one obtain the usual relations \begin{align} x' &= -\sinh\theta\cdot t +\cosh\theta\cdot x \notag\\ &= \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}}\left( x - vt \right) \notag \\ t' &= \cosh\theta \cdot t -\sinh\theta\cdot \tfrac{x}{c} \notag\\ &= \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}}\left( t - \tfrac{v}{c^2}t \right). \notag \end{align}

• Your answer shows the reason for the transform of the tangent bundle, i.e. the space of velocities, but this is very far from a direct demonstration/derivation of the $vx/c^2$ term in the actual coordinate transform as asked for in the OP. – Void Aug 19 '14 at 17:43
• You're only looking at a light-like interval; you need to show how invariance of this implies invariance of space-like and time-like intervals also. – Physiks lover Aug 24 '14 at 14:15

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. The geometric representation is then quickly used to produce the full set of equations related to SR. This of course includes the Lorentz Transformation equations and a complete understanding of them. This analysis is to lengthy to be exposed directly here, therefore only mathematical descriptions will be accepted, therefore this post will receive negative results. But the actual exposure is available at http://goo.gl/fz4R0I if you have 1 1/2 hours available. (Drawing below is used within the videos to create all of the SR equations.)

Via motion, we encounter spatial "length contractions" and "time dilation". Thus ones rulers contract in length, the clocks slow down, but also, clocks that are displaced from each other in the same frame of reference, are no longer in sync. These clock offsets are thus included in the overall change of our measurement instruments. The $-(v/c^2)x$ equation concerns the clock offsets which occur. The clock offsets are due to the occurrence of rotation with the 4 dimensional environment of space-time.

So yes, one can create the $-(v/c^2)x$ equation via a step by step mathematical analysis, but at the same time, the full understanding is not being revealed. If SR is fully understood, all of the equations can be created at the same time rather than in a sequential manner.

So yes, you can start from the top, aka the speed of light, and then work your way backwards via math, or you can understand motion via a logical analysis of motion, and understand it all, and then convert the understanding into equations.

• The author confuses Euclidean and pseudo-Euclidean geometry. This is a widespread mistake among those who did not master SR and its maths. – firtree Aug 26 '14 at 15:10
• Perhaps you have jumped to a conclusion. In the above diagram I have stacked on top of each other, both motion vectors and Space-Time distances. This is the key to revealing the link between ones direction of travel in Space-Time, Time Dilation, Length Contraction, Clock Offsets ( Thus the overall change of the measurement instruments.), thus in turn producing a flawless understanding of SR. – Sean Aug 26 '14 at 16:14
• I know how correct spacetime diagrams look like, either for distances or motions or even velocities. Yours look like nothing close to them. The clearest sign of your confusion is the dached circle (and the segment tangent to it), which can never appear in the correct diagram with this shape. I'm sure you could not write down the Lorentz transformations for your $R$ and $F$ points. – firtree Aug 26 '14 at 18:29
• As I have said, all of the SR equations are derived using the above geometric representation, including the the Lorentz transformations for the (R)ear and (F)ront points, and all these derivations are of course shown within the videos. – Sean Aug 26 '14 at 19:32