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This question is about one of the assumptions involved when one is deriving the Lorentz transformation, for example under "Principle of Relativity" in Wikipedia's page on the derivations. Let me elaborate:

Consider two inertial frames $S$ and $S'$ in the stantard configuration, i.e., $S'$ moves with velocity $v$ along the x-axis of $S$. Let us forget about the directions $y$ and $z$. The coordinates $(x,t)$ of $S$ and $(x',t')$ of $S'$ are related by (due to the Inertia law and first postulate of special relativity):

$$\begin{align} x' &= \alpha_1 x+\alpha_2 t \tag{1} \\ t' &= \alpha_3 x+\alpha_4 t\tag{2} \end{align}$$

and we want to fix $\alpha_i$.

The first step is to consider that a particle at rest in $S'$ is moving with velocity $v$ in $S$. Then,

$$x=vt\quad \text{if}\quad x'=0$$

which fixes

$$\alpha_2=-\alpha_1 v\tag{3}$$

We use now the second postulate, i.e.,

$$\frac{x}{t}=\frac{x'}{t'}=c$$

which, together with (1) and (2), allows us to fix more one $\alpha$, let us say $\alpha_4$:

$$\alpha_4=\left(1-\frac{v}{c}\right)\alpha_1-c\alpha_3\tag{4}$$

We now claim the first postulate: we can consider that the reference frame $S$ is moving with velocity $-v$ with respect to $S'$. Now, the coordinates would be related by,

$$\begin{align} x &= \tilde\alpha_1 x'+\tilde\alpha_2 t' \tag{5} \\ t &= \tilde\alpha_3 x'+\tilde\alpha_4 t' \tag{6} \end{align}$$

Now, we have the condition

$$x=0\quad \text{if} \quad x'=-vt'$$

which fixes

$$\tilde\alpha_2=\tilde\alpha_1 v\tag{7}$$

Using again the second postulate, we obtain now,

$$\tilde\alpha_4=\left(1+\frac{v}{c}\right)\tilde\alpha_1-c\tilde\alpha_3\tag{8}$$

Finally, we assume that the transformations (1),(2) and (5),(6) are inverses of each other. This would fix all the missing $\alpha_i$ and $\tilde\alpha_i$, except for one of them. This means, we can write all $\alpha_i$ and $\tilde\alpha_i$ in terms of, let us say, $\alpha_1$.

Now comes the part I cannot understand: one says that by the first postulate we actually have:

$$x=\alpha_1 (x'+ v t')\tag{9}$$

i.e., we have $\tilde\alpha_1=\alpha_1$, which would fix then all the constants and we would obtain the Lorentz transformation.

However, for me it looks like that the first postulate only leads to (5), not to (9). Can someone elaborate on this? One argument seems to be just change "prime" to "not-prime", but this is not convincing me; I can not see why the first postulate implies $\tilde\alpha_1=\alpha_1$.

Another way to look at the problem

The coefficients $\alpha_i$ can be fixed by means of the following assumptions (no need to use the transformation related to $\tilde\alpha_i$):

Assumption 1: The following holds:

$$ x^2-c^2t^2=x'^2-c^2t'^2 $$

Assumption 2: the coefficients $\alpha_i$ do not depend on $x$ and $t$.

Then, the question can be reformulated as: how Assumption 1 and Assumption 2 follows from the Postulates of special relativity?

An old book

I have found an old reference: The Theory Of Relativity (1914) by L. Silberstein (the book can be read online). In this book, the author seems to claim that indeed $\alpha_1=\tilde\alpha_1$ is kind of arbitrary, see the discussion culminating on page 110. I would need to read it more carefully to confirm that this is indeed what he is saying, but I am mentioning the reference in case someone already did it.

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  • $\begingroup$ Couple of things you might use to improve your math markup. First delimiting equations in double-dollar-signs will get block type setting $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ will get set on it's own line (and centered, which is nicer than what you have done). Second you can use \tag{} to get equation numbering in block-set equation (and they'll be right justified in the usual form). $\endgroup$ – dmckee Jan 18 '16 at 18:14
  • $\begingroup$ I put in a better title (and a few formatting tweaks) because this is a really good question and I want it to be in the best shape possible so we can show it off :-P more or less. I'm not sure if I correctly edited the part where it says that the transformations (1),(2) and (5),(6) are inverses of each other - or in the previous revision, "have inverse". Could you check? $\endgroup$ – David Z Jan 18 '16 at 18:40
  • $\begingroup$ @DavidZ Thank you for your changes. Yes, that was what I meant, that (1),(2) and (5),(6) are inverses of each other (much better in this way!). $\endgroup$ – aprendiz Jan 18 '16 at 19:08
  • $\begingroup$ @DavidZ About the title, I think it should contain ''Derivation of the Lorentz transformations'', since my question is really about this point. Every book/notes I looked in, eq.(9) is assumed without a convincing argument (at least for me)! (including the appendix I of A. Einstein, "Relativity: The Special and the General Theory" !!) But it is ok, a better idea is not pumping up ;) $\endgroup$ – aprendiz Jan 18 '16 at 19:14
  • $\begingroup$ @aprendiz well, feel free to change the title if you want. But "derivation of the Lorentz transformations" is a pretty vague phrase, and I'm not sure it would fit into a good title. $\endgroup$ – David Z Jan 18 '16 at 19:40
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No need to force $\tilde\alpha_1 = \alpha_1$.

After taking advantage of the fact that (1), (2) and (5), (6) must be inverses of each other, your transformation reads, in matrix form, $$ \left(\begin{array}{c} x'\\ t' \end{array}\right) = \alpha_1(v)\left(\begin{array}{cc} 1 & -v\\-\frac{v}{c^2} & 1 \end{array}\right)\left(\begin{array}{c} x\\ t \end{array}\right) $$ where the dependence of $\alpha_1$ on the relative velocity $v$ is shown explicitly, and if we identify $\tilde\alpha_1 = \alpha_1(-v)$ we also have $$ \alpha_1(v) \alpha_1(-v) \left( 1 - \frac{v^2}{c^2}\right) = 1 $$ Consider now 3 distinct inertial frames, $S$, $S'$, and $S"$. Let $S"$ move at velocity $w$ relative to $S$, and denote $W(w)$ the corresponding coordinate transformation. Similarly, let the relative velocities of $S'$ and $S$, and of $S'$ and $S"$ be $u$ and $v$ respectively, and denote by $V(v)$ and $U(u)$ the corresponding coordinate transformations from $S$ to $S'$ and from $S'$ to $S"$. By the principle of relativity transformations $U(u)$, $V(v)$, and $W(w)$ must have the same form and the same dependence on the respective relative velocities. In addition, by the same principle of relativity, transforming coordinates from $S$ to $S"$ by means of $W(w)$ must produce the same result as transforming coordinates first from $S$ to $S'$ by means of $V(v)$ and then from $S'$ to $S"$ by means of $U(u)$. Otherwise we could have 2 different transformations from $S$ to $S"$, generating different results for at least some events, and we could easily use such a feature to distinguish between frames. It follows that necessarily $$ W(w) = U(u) V(v) $$ The right hand side compound transformation amounts to
$$ \left(\begin{array}{c} x"\\ t" \end{array}\right) = \alpha_1(u)\alpha_1(v)\left(\begin{array}{cc} 1 & -u\\-\frac{u}{c^2} & 1 \end{array}\right)\left(\begin{array}{cc} 1 & -v\\-\frac{v}{c^2} & 1 \end{array}\right)\left(\begin{array}{c} x\\ t \end{array}\right) $$ and doing the algebra gives eventually $$ \left(\begin{array}{c} x"\\ t" \end{array}\right) = \alpha_1(u)\alpha_1(v)\left(1+\frac{uv}{c^2}\right)\left(\begin{array}{cc} 1 & -\bar w\\-\frac{\bar w}{c^2} & 1 \end{array}\right)\left(\begin{array}{c} x\\ t \end{array}\right) $$ where $$ \bar w = - \frac{u+v}{1 + \frac{uv}{c^2}} $$ But this must be the same as $$ \left(\begin{array}{c} x"\\ t" \end{array}\right) = \alpha_1(w)\left(\begin{array}{cc} 1 & -w\\-\frac{w}{c^2} & 1 \end{array}\right)\left(\begin{array}{c} x\\ t \end{array}\right) $$ Equating the previous result element by element with this direct transformation shows that $$ \alpha_1(w) = \alpha_1(u)\alpha_1(v)\left(1+\frac{uv}{c^2}\right) $$ $$ - w \alpha_1(w) = - \bar w \alpha_1(u)\alpha_1(v)\left(1+\frac{uv}{c^2}\right) $$ The first identity above gives a composition law for $\alpha_1$. Substituting this in the 2nd identity gives a form of the velocity addition rule, even though we don't yet know the explicit expression of $\alpha_1$: $$ w = \bar w = - \frac{u+v}{1 + \frac{uv}{c^2}} $$ Now notice that $$ w = v \;\;\Rightarrow \;\; u = - \frac{2v}{1+\frac{v^2}{c^2}} $$ and substitute into the composition rule for $\alpha_1$: $$ \alpha_1(v) = \alpha_1(u) \alpha_1(v) \frac{1 - \frac{v^2}{c^2}}{1+\frac{v^2}{c^2}}\;\;\Rightarrow\;\;\alpha_1(u) = \frac{1 + \frac{v^2}{c^2}}{1-\frac{v^2}{c^2}} $$ for $u = - \frac{2v}{1+\frac{v^2}{c^2}}$. But let us notice further that $$ 1 \pm \frac{u}{c} = 1 \mp \frac{2v}{1+\frac{v^2}{c^2}} = \frac{\left(1 \mp \frac{v}{c} \right)^2}{1+\frac{v^2}{c^2}} $$ Then we must also have $$ 1 - \frac{u^2}{c^2} = \left(1 - \frac{u}{c}\right)\left(1 + \frac{u}{c}\right) = \left(\frac{1- \frac{v^2}{c^2}}{1+\frac{v^2}{c^2}} \right)^2\;\;\Leftrightarrow\;\; \frac{1 + \frac{v^2}{c^2}}{1-\frac{v^2}{c^2}} = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} $$ and therefore $$ \alpha_1(u) = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} $$ Obviously $\alpha_1(-u) = \alpha_1(u)$ and the original condition $\alpha_1(u) \alpha_1(-u) \left( 1 - \frac{u^2}{c^2}\right) = 1$ is satisfied.

Note: The light speed condition isn't actually necessary. Homogeneity and isotropy of space also lead to Lorentz transformations through a couple of extra algebraic steps ($c$ enters as parameter and must be identified as light speed a posteriori), with a side alternative of Galilei transformations (corresponding to $c \rightarrow \infty$).

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z Jan 23 '16 at 13:41
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The first postulate of Special Relativity - which is also the Principle of Relativity is "Laws of Physics are invariant in all the inertial frames."

Now consider the following scenario:

Suppose there are two events which are happening simultaneously in frame $O$ and are spatially separated by distance $l$. From your Equation $1$, the spatial interval between the same events is $\alpha_1$$l$ in a frame $O'$ which is moving uniformly with respect to the frame $O$.

Now consider the reverse situation. There are another two events which are happening simultaneously in frame $O'$ and have spatial interval $l$ between them. Then from your Equation $5$, it is clear that the spatial interval between the same events is $\tilde\alpha_1l$ in frame $O$.

But from the principle of relativity, the laws of Physics must be same in all the inertial frames. So the factor by which the spatial interval between the simultaneous events of one frame gets multiplied when we see them from a different frame should be same for a given relative speed between the two frames. It is just the symmetry argument based on the symmetry already postulated in the principle of relativity. Thus, $\alpha_1$ must be equal to $\tilde\alpha_1$.

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  • $\begingroup$ I think you do not answer the question. Your last paragraph is precisely what is claimed in a lot of textbooks. However, I think this is not correct. The Principle of Relativity only says that $x$ is a linear combination of $x'$ and $t'$, as well as that $x'$ is a linear combination of $x$ and $t$. It does not says anything about the relation between their coefficients. $\endgroup$ – aprendiz Jan 20 '16 at 18:20
  • $\begingroup$ @aprendiz Linearity of transformation equations is not confirmed by the Principle of Relativity. They are the manifestation of the Homogeneity and Isotropic ness of Space and Time. On the other hand, what Principle of Relativity suggests is precisely (without any aid of another assumption - provided once the linearity is assumed) $\tilde\alpha_1 = \alpha_1$. Try to think in the form of symmetry of the inertial frames and the relation between the measurements of two frames. $\endgroup$ – Dvij Mankad Jan 20 '16 at 19:28
  • $\begingroup$ @Dvji, I think you are wrong. Consider two inertial frames $S$ and $S'$, with some relative velocity. In general, an event $(x,t)$ of $S$ and an event $(x',t')$ are related by $x'=f_1(x,t)$ and $t'=f_2(x,t)$. Consider now a particle with constant velocity in $S$, i.e., its trajectory is a straight line. By the law of inertia (a consequence of the Principle of Relativity), the trajectory on $S'$ will be also a straight line. By definition, a map that maps straight lines to straight lines is a linear map. There is absolutely no need to bring up homogeneity and isotropy of space . $\endgroup$ – aprendiz Jan 20 '16 at 20:19
  • $\begingroup$ @Dvji Complementing. By the same line of reasoning, one would have $x=\tilde f_1(x',t')$ and $t=\tilde f_2(x',t')$, with $\tilde f_1$ and $\tilde f_2$ linear functions. This is precisely what the Principle of Relativity implies. Nothing can be said about the coefficients of the linear functions, except by some "initial" conditions. $\endgroup$ – aprendiz Jan 20 '16 at 20:24
  • $\begingroup$ @aprendiz Can you prove "Law of Inertia is nothing but Principle of Relativity"? As far as I know, Einstein said that if the laws of Physics hold in a frame in the simplest form then they hold in all the uniformly translating frames in the same way. Now the simplest law that is followed at least in some frames is the law of inertia and that is an experimental fact that there exist such frames in which law of inertial holds and there has never been detected any frame in which any simpler law can work. So Relativity and Law of Inertia - they are different things. $\endgroup$ – Dvij Mankad Jan 20 '16 at 21:18

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