# Postulate implies existence Lorentz transformation?

My textbook about Special Relativity says that the existence of Lorentz Transformation is guaranteed by the postulates of Special Relativity.

So, I'm assuming it's the first postulate we're talking about; laws of physics remain unchanged in different inertial frames.

However, how does this guarantee we can find the Lorentz Transformation between two inertial frames?

I've never doubted their existence, but I never thought of it as an implication of the first postulate.

• Maybe a wording question: the existence of Lorentz Transformation doesn't need to be guaranteed, it is a mathematical object that is self consistent (its definition itself says it exists). But the fact that these transformations correspond to a physical reality is a consequence of the first postulate.
– user130529
Jan 5 '17 at 17:02
• Are you perhaps looking for the proof that the interval-preserving transformations are the Lorentz transformations? Jan 6 '17 at 15:02
• I have come to the conclusion that this is an ill formed question (I mean that in a formal sense, not as a criticism, because it's a good conceptual question that I have been thinking quite a bit about to reach this conclusion). That is, to answer a question like this, you would need to encode the relativity principle as a set of axioms - no one axiom will capture such an informal statement - and it is then arguable how one does this. In my opinion, there would still be other axioms one would need after you have encoded the principle in this way - see .... Jan 11 '17 at 3:31
• .... my answer here. Jan 11 '17 at 3:32

The historical formulation of the relativity principle: the Lorentz transformation derivation a-la Einstein

It seems that the fact which confuses You is the applicability of the statement "laws of physics remain unchanged in different inertial frames", which says something about the dynamics of the system, for derivation of the kinematical transformations relating the coordinate 4-vector $x^{\mu}$ (and hence the velocity 4-vector $v^{\mu}$) in different inertial frames. The reason why the relativity principle in the above form allowed Einstein to derive the Lorentz transformations is another postulate stating that the speed of light is constant. This postulate imposes obvious restrictions on kinematic law relating coordinates (and hence velocities) relative to two frames.

The "speed of light" postulate historically appeared as the result of heuristically derived Maxwell equations, due to which the speed of light is invariant on the EM waves source velocity (at least in one frame). Since the Maxwell equations are dynamical equations, then due to the relativity principle this statement must be true in any frame.

The relativity principle: modern formulation

But as we know now, the fact that the speed of light isn't truly fundamental quantity. Indeed, the gluon and the graviton (and any massless particle) also propagate with the speed equal to speed of light. The constant $c$ has in fact more general meaning - it is maximally possible propagation velocity, related rather to the space-time properties; the fact that the speed of light is equal to $c$ is just connected to the masslessness of the photon. Therefore, from the modern field theory perspective, the postulate about of the speed of light based on the Maxwell equations is very special and should be changed on a more general one, which is independent on details of particular dynamical theory; instead it must depend on properties of the space-time.

This will, of course, lead to the fact that the relativity principle in Einstein's formulation is not enough for the derivation of the Lorentz transformations. Without the "speed of light" postulate it just says that f the given kinematic transformations the laws of dynamics must be covariant under kinematic transformations. Instead, let's formulate it in slightly another way: heuristically, all inertial frames have equal rights. This implies two statements:

1) Descriptions of the same system relative to two different inertial frames have one-to-one correspondence, and the rule determining of this correspondence is the same for any two pairs of inertial frames;

2) The evolution law for the given system has the one form independently on the inertial frame in which we consider it.

I will call the first statement as the "kinematical" part of the relativity principle, while the second statement as the "dynamical" part of the relativity principle.

The first statement is actually what You need to require the existence of Lorentz transformations as well as their unique form. Let's discuss its consequence for two examples: the first one is the derivation of the vector-like Lorentz transformations (in which You're interested in), and the second one is the derivation of the properties of the quantum theory Hilbert space which respects relativity. The second example will also demonstrate the difference of consequences of the first and the second statements.

Derivation of the Lorentz transformations

One way to derive the Lorentz transformations, i.e., the transformations which relate $x^{\mu}$ in different inertial frames with relative velocity $v$, namely $g(x^{\mu}, v)$, is to require the set of axioms. One of possible choice is (see, for example, this paper):

• The transformation $g$ is smooth and inversed;
• The space-time is homogeneous; in particular this means that if the relative velocity of two objects is equal to zero in one frame, then it will be equal to zero in another frame;
• The relativity principle;
• The space is isotropic;

It can be shown that the first two axioms leads to the following form of the function $g$: $$\tag 1 \begin{cases} x' = \gamma(v)(x - vt), \\ t' = \gamma(v)(t - \sigma(v)x) \end{cases}$$ The third postulate (the relativity principle) can fix the form of the functions $\gamma(v), \sigma(v)$.

First, suppose three frames $S_{1}, S_{2}, S_{3}$, and perform both step-to-step transformations between the first, the second and the third frame, and the transformation between the first and the third. By using $(1)$ we have $$\tag 2 \begin{cases} x_{3} = \gamma_{2}\gamma_{1}((1+v_{2}\sigma_{1})x_{1}-(v_{1}+v_{2})t_{1}), \\ t_{3} = \gamma_{2}\gamma_{1}((1+v_{1}\sigma_{2})t_{1} - (\sigma_{1}+\sigma_{2})x_{1}) \end{cases},$$ From the other side, the "kinematical" statement of the relativity principle implies existence of the transformation $x_{3}^{\mu} = g(x_{1},v_{3})$ $$\tag 3 \begin{cases} x_{3} = \gamma_{3}(x_{1} - v_{3}t_{1}), \\ t_{3} = \gamma_{3}(t_{1} - \sigma_{3}x_{1})\end{cases}$$ The relativity principle says us that $(3)$ must be equal to $(2)$, from which follows $$\tag 4 \frac{\sigma_{i}}{v_{i}} \equiv \frac{\sigma (v_{i})}{v_{i}} = \alpha = \text{const}$$ Second, the relativity principle says that the the form of the direct transformation $x \to x'$ must coincide with the form of inversed transformation $x' \to x$ up to the sign of $v$. By using this requirement, we have $$\gamma (v)\gamma(-v) = \frac{1}{1 - \alpha v^{2}}$$ I.e., the relativity principle almost uniquely fix the Lorentz transformation! The fourth axiom fixes $\gamma(-v) = \gamma(v)$, and only the value (the sign) of $\alpha$ remains undetermined! Adding the physical requirement that the energy of the particle must increase with increasing of its velocity fixes also the sign of $\alpha$ as positive. It can be then easily shown that the $\frac{1}{\sqrt{\alpha}} \equiv c$ has the sense of maximal propagation velocity.

P.S. One may add the axiom of the independence of time intervals on the choice of the inertial frame, which will reduce the Lorentz transformation to the Galilei transformation.

Relativistic quantum mechanics

Suppose now the consequences of the relativity principle in the Hilbert space of rays $|\Psi\rangle$, defining the quantum-mechanical state. We don't know about any Maxwell equations and want to formulate the general properties of the realization of the Lorentz symmetry in quantum mechanics. This means that we need to use the above mentioned definition of the relativity principle containing both "kinematical" (the first) "dynamical" (the second) statements.

The first statement requires that states $|\Psi{'}\rangle$ and $|\Psi\rangle$ defining the one quantum-mechanical state measured in different inertial frames must be related by the uniquely fixed transformation $$\tag 5 |\Psi'\rangle = U(\Lambda , a)|\Psi\rangle,$$ where $U$ is the operator of the inhomogeneous Lorentz group (called the Poincare group) transformation, $\Lambda$ is the Lorentz group transformation, while $a$ is translation 4-vector. Repeating the idea of $(3)$-$(4)$ one sees that $$U(\Lambda_{2},a_{2})U(\Lambda_{1},a_{1}) = e^{i\omega\big((\Lambda_{2},a_{2}),(\Lambda_{1},a_{1})\big)}U(\Lambda_{2}\Lambda_{1},\Lambda_{2}a_{1} + a_{2}),$$ where $\omega$ is the fixed phase which in fact can be $e^{i\omega} = \pm 1$. I.e., the first statement of the relativity principle ("kinematical") says us that on the space of quantum-mechanical rays must be realized the so-called projective representation of the Poincare group by operators $U(\Lambda, a)$.

Let's turn on the second statement ("dynamical"). The quantum theory dynamics describes us how by having the initial state $$|\Psi(t = 0)\rangle = |\Psi_{0}\rangle$$ to calculate the state $|\Psi (t > 0)\rangle$, which, being expanded on the full basis $\{ |\Psi_{j}\rangle\}$ of the Hilbert space, gives the set of probabilities to find the system in different states: $$\tag 6 P_{|\Psi_{0}\rangle \to |\Psi_{j}\rangle} = |\langle \Psi(t > 0)|\Psi_{j}\rangle|^{2}$$ The second statement of the relativity principle implies us that $(6)$ must be invariant under the transformation $(5)$. By the Wigner theorem, this means that the $U(\Lambda,a)$ must be linear unitary or anti-linear anti-unitary operator!

To conclude: You see that the "kinematical" and "dynamical" parts of the relativity principle imposes important consequences on the realization of the relativistic symmetry separately from each other.

• Good answer, but I think the statement "The first statement is actually what You need to require the existence of Lorentz transformations as well as their unique form. " could be a little misleading: it kind of implies (or could be read to mean) that the Lorentz transformation is contained in this statement (i.e. that it is sufficient to derive the transformation), whereas what you mean is that it is a necessary statement: as you show, you also need to add the spacetime homogeneity postulate, smoothness and so forth to get there. Jan 9 '17 at 13:43
• Actually, just continuity, rather than $C^\infty$ smoothness, will also do fine. $g(x,\,v)$ continuous in $x$ along with spacetime homogeneity (see my variation of JoshPhysics's answer here proves that $g$ is linear and so can be represented by a matrix. Thus a matrix group is the group of all Lorentz transformations, and then a continuity (and monotonicity) assumption forces the Lie group structure (see here). BTW, before I forget, +1 :) Jan 9 '17 at 13:52
• @WetSavannaAnimalakaRodVance : Thank You for the comments! I need to think a little. Jan 9 '17 at 22:54

Postulate I: The laws of physics behave the same in all inertial frames.

Postulate II: The speed of light, $c$, is constant to all inertial observers.

To be clear, both of these postulates are necessary to guarantee the existence and validity of the Lorentz transformations. As for why it's the Lorentz transformations and not some other transformation, the answer is simple—it's the only transformation that is mathematically consistent with the postulates.

Formally, the Lorentz transformations can be proven to exist (satisfy the postulates) and to be unique (the only such transformation). A very good derivation that illustrates these properties can be found in Resnick's Introduction to Special Relativity.

• I don't see what the postulates have to do with the existence of the Lorentz transformation. This transformation exists as a mathematical object, it is one family of linear coordinate transformations among infinitely many linear coordinate transformations, and can be defined without knowing anything about physics.
– user130529
Jan 5 '17 at 20:03
• @claude chuber : Lorentz had derived the homonymous transformation before Special Relativity from his intuition that Maxwell Equations of Electromagnetism must valid in exactly the same form (later term covariance) in all inertial frames. Now, I don't understand how the Lorentz transformation is a standing alone 'mathematical object defined without knowing anything about physics'. Is this a modern scientific opinion? Would this transformation exist before Maxwell's equations, Lorentz's Physics and Einstein's Special Relativity ? Jan 6 '17 at 0:23
• @Frobenius : nothing such sophisticated, it is a particular family of linear transformations belonging to the linear space of transformations from $\mathbb R^4$ into itself (as far as we are talking about existence).
– user130529
Jan 6 '17 at 8:05

The constancy of the speed of light in vacuum is a parameter/consideration in many physical laws.

c is further related to permeability and permitivity of free space, and if it changed there would be many physical consequences.

The Lorentz transform allows this to be upheld.

• Actually, Lorentz transformations can be derived from the constancy of $c$ Jan 5 '17 at 17:22
• "And further" you can derrive lorentz transforms from this. Jan 5 '17 at 17:30

"My textbook about Special Relativity says that the existence of Lorentz Transformation is guaranteed by the postulates of Special Relativity."

Summary

Depending on what one means by Lorentz transformation, this statement is incorrect. It's a necessary assumption, but not sufficient. The following is the flow of logic:

1. First, one must postulate that a manifold structure / co-ordinates on spacetime are even meaningful, and then that motion can be described by transformations on these co-ordinates.

2. The relativity postulate then ensures that these transformations depend only on the relative velocity between inertial frames and also completes the group structure for the set of transformations kitted with composition (through enforcing associativity). But you need much else besides (steps (3) through (7)) to get to the Lorentz transformation! That is, unless of course, you take the postulates in steps (3) to (7) as "obvious" and assume them tacitly;

3. Assumptions of homogeneity of spacetime together with continuity of the transformations in their dependence on the spacetime co-ordinates then show that the group of transformations is a linear, matrix group;

4. Assumption of continuity of the transformations in their dependence on the relative velocity between frames then shows that this matrix group is a Lie group of $4\times4$ matrices;

5. Isotropy of space then shows that the Lie group is the identity connected component of one of the orthogonal groups $O^+(4)$ or $O^+(1,\,3)$ (or the Galilee group, in the special case where the free parameter $c$ is infinite);

6. Causality then rules out the rotations $O^+(4)$;

7. Experiment shows us that the free parameter $c$ is finite and our group is $O^+(1,\,3)$, not the Galilee group.

Details

It depends on what "level" of proof your looking for, but let's for this answer assume your seeking high rigor with minimal clutch of axioms.

Certainly Galileo's principle (i.e. that physical law is unaffected by inertial motion, à la Salviati's Ship Allegory) is necessary to existence of the LT, but you do need quite a good many other assumptions to make it work.

First of all, one needs to assume that one's neighborhood in the World can be described by a co-ordinate system. This is wholly an experimental result, and a very deep and basic one at that. It's justified by countless everyday observations - if measure the spot on the wall where I think I want the builder to put my window, and I describe it to the builder by Cartesian co-ordinates, the end result (if the builder is competent) is in accordance with my expectations. Or, that people can use maps to meet at an agreed location. All of these seemingly trivial observations justify a postulate like:

Spacetime is modelled by a pseudo-Riemannian manifold and the events of spacetime can be uniquely specified by co-ordinates in a suitable chart of that manifold

We need "pseudo-Riemannian" because this asserts that the notion of distance given by an inner product is meaningful.

Now, this postulate has enabled our use of co-ordinates to describe events. This is what justifies any meaningful notion at all of "transformation" between relatively moving observers.

Next, the relativity postulate enters. We have a laboratory frame $A$, and two frames $B$ and $C$ observable within it. We wish to investigate the transformation of co-ordinates between $B$ and $C$ when they are in relative motion with each other. Suppose first we impart the same inertial motion on both $B$ and $C$ relative to $A$, so that $B$ and $C$ are still at rest relative to one another. Then we impart a second motion on $C$ alone so that $C$ moves relative to $B$. The relativity postulate tells us that the transformation between $B$ and $C$ must depend on this second motion alone. In general, we must allow for the transformation to depend on the common relative motion to $A$ as well, but if there were such a dependence, then $B$ and $C$ could detect that common motion from within their frames by observing that the co-ordinate transformation between them would change, even though their motion relative to each other stayed the same. This would tell against Galileo's principle, and so:

The co-ordinate transformation between two inertial observers depends only on their motion relative to each other, and is independent of any motion relative to any third reference frame

and this is the crucial property one derives from the relativity principle, i.e. that co-ordinate transformations between inertial frames is wholly defined by the relative velocity vector between those frames.

This conclusion also contains another important fact. It implies inverses to any transformation, and, with a bit of work, you can also work out from the complete characterization by relative velocity that the compositions of transformations is associative (by considering successive transformations and noting that the above implies that the order of their application must give the same result, because that result is characterized by the endpoint inertial frames alone). That is,

the transformations between inertial frames form a group

From here, one still needs several assumptions to get to the Lorentz transformation:

1. Homogeneity of spacetime together with an assumption that the co-ordinate transformation between inertial frames is continuous then tells us that the transformations are linear, as I discuss in my answer here; see also Mark H's answer (although this assumes differentiability as well: this is stronger than needed). So our group of transformations is a matrix group;

2. An assumption that the transformation is a continuous function of the relative velocity together with a monotonicity assumption (that two relative motions in the same direction compose to give a faster relative motion) then shows that the linear matrix group of transformations is a Lie group, with the transformation matrix given by $\exp(\eta\,K)$ where the $4\times4$ matrix characterizes the direction of motion, and the rapidity parameter $\eta$ is a continuous, monotonic function of relative speed and characterizes the relative speed. This fact I prove in the addendum to my answer here;

3. An assumption of isotropy of space (independence of transformation from boost direction) then means that we can choose any direction for the relative motion and if we align our co-ordinate system with it so that the $x$ direction points along the relative motion, then our we'll always get the same matrix $K$, independent of direction;

So now we are left with two possibilities for the transformation that mixes the $t$ and $x$ co-ordinate:

$$K = \left(\begin{array}{cc}0&\pm c^{-2}\\1&0\end{array}\right)$$

where $c$ is a free parameter (strictly speaking, there are possibilities of diagonal elements of $K$ as well, but these lead to transformations that are wildly at variance with reality! - they are what Sean Carroll calls the "Alice in Wonderland" transformations). Note that Galilean relativity is the special case where $c\to\infty$.

So, if $c$ is finite, we're left with essentially boosts and rotations that mix the time and space co-ordinates. But rotations are untenable in a causal universe: there would be relative motion that would reverse the direction of any timelike interval, and you could boost to some inertial frame moving relative to me to see me eating my boiled eggs for breakfast before I cooked them!

So, at last the form of the Lorentz transformation must be:

$$\exp(\eta\,K) = \left(\begin{array}{cc}\cosh\eta& \frac{1}{c}\sinh\eta\\c\,\sinh\eta&\cosh\eta\end{array}\right)$$

whence we can read off the relationship between speed and our rapidity parameter $v = c\,\tanh\,\eta$.

The last step is experimental: we must check whether $c$ is finite, and what its value is. The form of the Lorentz transformation shows us that, if $c$ is finite, it is a speed that would be invariant - the same in all frames. So we must look for something whose speed has this property - if we find it then this proves that $c$ is finite and moreover gives us its value experimentally.

Of course you know of something whose speed behaves in this way!