What assumptions underly the Lorentz transformation?

[Ad VanderVen]

TL;DR Summary

The Lorentz transform for velocities is as follows:
$$u=\frac{v+w}{1+\frac{vw}{c^{2}}}$$
But which assumption exactly underlies this so that you get exactly this formula and no other formulas with approximately the same properties?

The Lorentz transform for velocities is as follows:
$$u=\frac{v+w}{1+\frac{vw}{c^{2}}}$$
But which assumption exactly underlies this so that you get exactly this formula and not any other formula with approximately the same properties?

 

[Orodruin]

I would not call that “Lorentz transformation” but “relativistic velocity addition”.

The assumption is that the velocities are colinear (and of course that special relativity holds).

 

[PeroK]

But which assumption exactly underlies this so that you get exactly this formula and no other formulas with approximately the same properties?

It follows directly from the Lorentz (spacetime) Transformation. And that follows from, for example:

a) The postulates of SR; and, specifically, the invariance of the speed of light.

b) Some basic assumptions about the isotropy and homogeneity of spacetime - which leads to two possible options: the Galilean Transformation or the Lorentz Transformation.

 

[Ad VanderVen]

I would not call that “Lorentz transformation” but “relativistic velocity addition”.

The assumption is that the velocities are colinear (and of course that special relativity holds).

What exactly are the assumptions of special relativity (preferably in formulas)?

 

[Ad VanderVen]

assumptions about the isotropy and homogeneity of spacetime

What exactly are the assumptions about the isotropy and homogeneity of spacetime?

 

[PeroK]

What exactly are the assumptions about the isotropy and homogeneity of spacetime?

http://www2.physics.umd.edu/~yakovenk/teaching/Lorentz.pdf

 

[Orodruin]

What exactly are the assumptions of special relativity (preferably in formulas)?

I recommend reading a basic textbook on special relativity. There is no point in us repeatIng the stuff that can be found there.

 

[Sagittarius A-Star]

The Lorentz transform for velocities is as follows:
$$u=\frac{v+w}{1+\frac{vw}{c^{2}}}$$
But which assumption exactly underlies this so that you get exactly this formula and not any other formula with approximately the same properties?

As others wrote, the formula is usually called "relativistic velocity addition". But in SR it is not really an addition, therefore:

The addition law is also called a composition law for velocities.

Source:
https://en.wikipedia.org/wiki/Velocity-addition_formula#Special_relativity

I think you are correct calling it a transformation of a velocity, although the formula is usually not called so. A derivation of the formula for "relativistic velocity addition" from inverse Lorentz transformation:

$\require{color}x = \gamma (\color{red}x'\color{black} + vt'), \ \ \ \ \ t = \gamma (t' + \frac{v}{c^2}\color{red}x'\color{black})$

with

$\color{red}x'\color{black} := w \ t'$​

$u = x/t$​

=>
$$u = \frac{v + w} {1+ vw/c^2}$$
Here I transformed $w$ from the primed frame to the unprimed frame.

 

[robphy]

In my opinion,
to give an answer about what assumptions are needed,
some other assumptions (the ground rules, the starting points) must be given first…. especially if “exact”ness is required.

 

[Ad VanderVen]

As others wrote, the formula is usually called "relativistic velocity addition". But in SR it is not really an addition, therefore:

Source:
https://en.wikipedia.org/wiki/Velocity-addition_formula#Special_relativity

I think you are correct calling it a transformation of a velocity, although the formula is usually not called so. A derivation of the formula for "relativistic velocity addition" from inverse Lorentz transformation:

$\require{color}x = \gamma (\color{red}x'\color{black} + vt'), \ \ \ \ \ t = \gamma (t' + \frac{v}{c^2}\color{red}x'\color{black})$

with

$\color{red}x'\color{black} := w \ t'$​

$u = x/t$​

=>
$$u = \frac{v + w} {1+ vw/c^2}$$
Here I transformed $w$ from the primed frame to the unprimed frame.

If we use the more general formulas

$x' = a x + b y$ and $y' = c x + d y$

where $y = c t$ and $y' = c t'$ and require

$x^{2} - y^{2} = x'^{2} - y'^{2}$

then a possible solution is

$x'= cosh(\omega) \, x - sinh(\omega) \, y$ and $y'= - sinh(\omega) \, x + cosh(\omega) \, y$

with $-\infty < \omega < \infty$.

The question however is whether this is the only non-trivial solution.

 

[Orodruin]

Up to time and space inversion (i.e., proper orthochronous transformations), yes it is the only solution.

Inserting into the requirement:
$$
x^2 - y^2 = x'^2 - y'^2 = x^2(a^2 -c^2) + y^2 (b^2 - d^2) + 2xy (ab - cd)
$$
Since this should hold for any $x$ and $y$, we can identify
$$
a^2 - c^2 = 1, \quad d^2 - b^2 = 1, \quad ab = cd.
$$
The first two equations may be parametrised by two hyperbolic angles $\varphi$ and $\theta$ such that
$$
a = \pm\cosh(\varphi), \quad c = \sinh(\varphi), \quad d = \pm \cosh(\theta), \quad b = \sinh(\theta).
$$
The $\pm$ signs are both positive for the proper orthochronous transformations. Insertion into the last identity results in
$$
ab - cd = \cosh(\varphi)\sinh(\theta) - \cosh(\theta)\sinh(\varphi) = 2\sinh(\theta - \varphi) = 0
$$
which implies $\theta = \varphi$ (switching the signs will result in different relations between $\theta$ and $\varphi$ but not be proper and orthochronous transformations).

 

[Ad VanderVen]

Up to time and space inversion (i.e., proper orthochronous transformations), yes it is the only solution.

Inserting into the requirement:
$$
x^2 - y^2 = x'^2 - y'^2 = x^2(a^2 -c^2) + y^2 (b^2 - d^2) + 2xy (ab - cd)
$$
Since this should hold for any $x$ and $y$, we can identify
$$
a^2 - c^2 = 1, \quad d^2 - b^2 = 1, \quad ab = cd.
$$
The first two equations may be parametrised by two hyperbolic angles $\varphi$ and $\theta$ such that
$$
a = \pm\cosh(\varphi), \quad c = \sinh(\varphi), \quad d = \pm \cosh(\theta), \quad b = \sinh(\theta).
$$
The $\pm$ signs are both positive for the proper orthochronous transformations. Insertion into the last identity results in
$$
ab - cd = \cosh(\varphi)\sinh(\theta) - \cosh(\theta)\sinh(\varphi) = 2\sinh(\theta - \varphi) = 0
$$
which implies $\theta = \varphi$ (switching the signs will result in different relations between $\theta$ and $\varphi$ but not be proper and orthochronous transformations).

Thanks a lot!

 

[Ad VanderVen]

Up to time and space inversion (i.e., proper orthochronous transformations), yes it is the only solution.

Inserting into the requirement:
$$
x^2 - y^2 = x'^2 - y'^2 = x^2(a^2 -c^2) + y^2 (b^2 - d^2) + 2xy (ab - cd)
$$
Since this should hold for any $x$ and $y$, we can identify
$$
a^2 - c^2 = 1, \quad d^2 - b^2 = 1, \quad ab = cd.
$$
The first two equations may be parametrised by two hyperbolic angles $\varphi$ and $\theta$ such that
$$
a = \pm\cosh(\varphi), \quad c = \sinh(\varphi), \quad d = \pm \cosh(\theta), \quad b = \sinh(\theta).
$$
The $\pm$ signs are both positive for the proper orthochronous transformations. Insertion into the last identity results in
$$
ab - cd = \cosh(\varphi)\sinh(\theta) - \cosh(\theta)\sinh(\varphi) = 2\sinh(\theta - \varphi) = 0
$$
which implies $\theta = \varphi$ (switching the signs will result in different relations between $\theta$ and $\varphi$ but not be proper and orthochronous transformations).

My original question was:

But which assumption exactly underlies this so that you get exactly this formula and no other formulas with approximately the same properties?

So the answer seems to be that there is indeed one assumption, and that assumption is that there exists an absolute velocity, i.e. a velocity that is not relative.

 

[Orodruin]

So the answer seems to be that there is indeed one assumption, and that assumption is that there exists an absolute velocity, i.e. a velocity that is not relative.

No, that is not an assumption. What makes you think there is?

 

[Ad VanderVen]

No, that is not an assumption. What makes you think there is?

What makes you think it is not an assumption?

 

[Orodruin]

What makes you think it is not an assumption?

That’s not really answering the question. There is nothing anywhere that requires an absolute velocity, quite the contrary. We cannot help you understand unless you tell us what your reasoning is.

 

[Ad VanderVen]

That’s not really answering the question. There is nothing anywhere that requires an absolute velocity, quite the contrary. We cannot help you understand unless you tell us what your reasoning is.

The requirement
$$
x^2 - y^2 = x'^2 - y'^2
$$
is simply the expression of the idea that there is an absolute velocity. So that's the assumption. You could say it's an observation given the Michelson & Morley experiments. But to be able to explain the results of those experiments, you just assume that there is an absolute speed.

 

[vanhees71]

I'm not sure, what you mean by "absolute velocity", but you can indeed ask, what are the symmetry transformations for a spacetime model, in which the special principle of relativity holds in addition to the other usual symmetries, i.e., homogeneity of time and space as well as Euclidicity of space for any inertial observer. With these assumptions you indeed get only two possible spacetime models: Galilei-Newton spacetime without any additional fundamental parameter or Einstein-Minkowski spacetime, which introduces a "limiting speed" as a fundamtental parameter, which empirically is given by the speed of light in vacuo. A nice paper deriving this is

V. Berzi and V. Gorini, Reciprocity Principle and the Lorentz Transformations, Jour. Math. Phys. 10, 1518 (1969),
https://doi.org/10.1063/1.1665000

 

[Vanadium 50]

Invariant is not the same as absolute.
Speed is not the same as velocity.
A bunch of words is not the same as a theory.
Personal theories do not belong on PF.

 

[Orodruin]

is simply the expression of the idea that there is an absolute velocity

No, it is not, it is expressing the idea that the speed of light is invariant. You are mixing up the ideas of absolute velocity and invariant speed.

Invariant speed means that an object moving at that speed will move at that speed regardless of the inertial frame.

Absolute velocity is a velocity of an object relative to some absolute rest frame. Such an absolute velocity does not exist in special relativity. Nor does it exist in classical mechanics.

 

[pervect]

Summary: The Lorentz transform for velocities is as follows:
$$u=\frac{v+w}{1+\frac{vw}{c^{2}}}$$
But which assumption exactly underlies this so that you get exactly this formula and no other formulas with approximately the same properties?

The Lorentz transform for velocities is as follows:
$$u=\frac{v+w}{1+\frac{vw}{c^{2}}}$$
But which assumption exactly underlies this so that you get exactly this formula and not any other formula with approximately the same properties?

There are multiple answers to this. Einstein chose one set of assumptions in his 1905 paper. There are other possible assumptions that will lead to the same theory. Unfortunately, I don't have a good summary of references to the various possible assumptions that lead to special relativity.

Certainly there are simple arguments that can narrow down the range of possibilites. I'll suggest a set that comes to mind:

1) The velocity "addition" rule (perhaps better called the velocity composition rule) must be symmetric on interchange of the two velocites, which you call v and w.

2) For low velocities, when v<<c and w<<c, the velocity "addition" rule must approximate the Gallilean velocity addition rule, v+w.

3) We must have $v \oplus c$ = c. By rule 1 this implies that $c \oplus w = c$.

I believe that if we take these basic requirements, and also specify that there are terms no higher than quadratic, we'll wind up the standard relativistic velocity addition formula you cite. (I don't have a formal proof or a reference). Basically, we want functions that involve v, w, v+w, and v*w only, plus the constant c, with the above properties. We can eliminate the constant c by normalizing our velocites $\beta = v/c$ so that all velocites fall in the range zero to one.

I believe it is possible through more sophisticated arguments to rule out the existence of terms of higher order than quadratic, which is the v*w term. But I don't recall any specific demonstration. More commonly, the focus is not so narrow as to be only on the velocity composition rule, but involves treating coordinate transformations as having the basic properties of groups, namely closure, invertibility, the existence of an identity, and associativity. This group theoretical formulation is very powerful, but as I noted, it's not the only approach to special relativity. And usually this sort of approach is not the first way one learns special relativity. I would suggest Bondi's approach, (my favorite), in "relativity and common sense" as a good way to first learn special relativity. Then other approaches can be learned and studied later.

 

[robphy]

With the question of what assumptions underlie relativity,
I think it useful to appreciate that there are many ways, which depend on what one starts with.

Following up on my earlier post (#9) in this thread is a link to a very old post of mine
https://www.physicsforums.com/threa...-challenge-for-experts-only.83373/post-694535
which featured
a chart from "Spacetime and Electromagnetism" by J.R. Lucas, P.E. Hodgson
where they try to diagram the various approaches to obtain the Lorentz Transformations.
[Certainly, the chart is not complete... but merely representative.]

(Sorry about the small size... that was what I had at the time.)

So, the point is... where does one wish to start?

 

[Orodruin]

With the question of what assumptions underly relativity,
I think it useful to appreciate that there are many ways, which depend on what one starts with.

Following up on my earlier post (#9) in this thread is a link to a very old post of mine
https://www.physicsforums.com/threa...-challenge-for-experts-only.83373/post-694535
which featured
a chart from "Spacetime and Electromagnetism" by J.R. Lucas, P.E. Hodgson
where they try to diagram the various approaches to obtain the Lorentz Transformations.
[Certainly, the chart is not complete... but merely representative.]

View attachment 302781
(Sorry about the small size... that was what I had at the time.)

So, the point is... where does one wish to start?

I think it should also be noted that just because one road can be taken or was taken in the historical development of the theory that does not mean that that road is in any way privileged. The theory is what it is and it can be tested experimentally. One could just as well start by defining Minkowski spacetime and introduce Lorentz transformations as coordinate transformations between orthonormal affine coordinate systems. Then there is the question of what experiments say.

 

[vanhees71]

Indeed, and I think that's the most straight-forward way. Often also the apparent "kinematical paradoxes" are overstressed. These are just not familiar to us, because in our everyday-life we don't experience them since we don't move with relative velocities close to the speed of light nor are strong gravitational fields around us. That's why "relativistic effects" are non familiar to us, but they are not paradoxical in any way.

 

[Orodruin]

"Paradoxes" arise when people try to squeeze a round peg into a square hole demanding that it is a consistent way of doing things and then complaining about the shapes not fitting together rather than making a proper analysis within the framework of the theory itself.

 

[robphy]

Often also the apparent "kinematical paradoxes" are overstressed.

Having students deal with misconceptions is helpful
when they can be resolved properly to become "gotchas"... otherwise, they become (or remain) "paradoxes".

Paradoxes in relativity appear to be part of its historical folklore,
which seems like an obstacle-course along the beaten path taken by typical textbooks.

To oppose the typical textbook storyline, I really like this quote:

J.L. Synge in Relativity: The Special Theory (1956), p. vii

Therefore, I apologise, if apology is necessary, for departing from certain traditional approaches which seemed to me unclear, and for insisting that the time has come in relativity to abandon an historical order and to present the subject as a completed whole, completed, that is, in its essentials. In this age of specialisation, history is best left to the historians.

Maybe this quote fragment (suitably re-interpreted) also applies
to textbook presentations

Spock in “Star Trek II: The Wrath of Khan”

[Khan] “exhibits 2-dimensional thinking”

that is, non-spacetime thinking.

 

[Meir Achuz]

One point to note is that it is circular reasoning to say that the transformation of velocities formula is derived from the Lorentz transformation, because the velocity transformation formula is often a step in the derivation of the Lorentz transformation.

 

[weirdoguy]

because the velocity transformation formula is often a step in the derivation of the Lorentz transformation.

Can you show us this step?

 

[Meir Achuz]

Can you show us this step?

It's in Sec. 14.2 of "Classical Electromagnetism" by Jerrold Franklin. I thought that other texts had it too, so "often" is not appropriate. But some texts have unstated (are they 'obvious') assumptions. For instant, Jackson goes from Eq. (11.15) to (11.16) with "it is straightforward to show...".

 

[Sagittarius A-Star]

It's in Sec. 14.2 of "Classical Electromagnetism" by Jerrold Franklin.

In Sec. 14.2, Jerrold Franklin does not derive the LT from the "relativistic velocity addition" formula. He only mentions above his derivation of the LT the Galilean velocity addition formula in eq. (14.5).

For deriving the LT, he makes use of the principle of relativity (SR postulate 1) in step 4, and of the invariance of the speed of light (SR postulate 2) in step 6.

The table contents says:

14.2 The Lorentz Transformation ... 411
14.3 Consequences of the Lorentz Transformation ... 414
14.3.1 Relativistic addition of velocities ... 415

Source (Google books preview viewed with Chrome):
https://books.google.de/books?id=33...old Franklin 11.15&pg=PR8#v=onepage&q&f=false

In his book "Solved Problems in Classical Electromagnetism" you can find in chapter 5.1 a derivation of the "relativistic velocity addition" formula from LT:
https://books.google.de/books?id=pt...rrold Franklin lorentz transformation&f=false

 

[Meir Achuz]

Step 5 is the transformation law for velocities, and step 6 uses the transformation law and the invariance of c to identify beta as v/c.

 

[Sagittarius A-Star]

Step 5 is the transformation law for velocities, and step 6 uses the transformation law and the invariance of c to identify beta as v/c.

Step 5 is not yet the relativistic transformation law for velocities, although it looks so. The text in step 5 says "We can derive a transformation law for velocities ... $u' = ... = \frac{u-v}{(1-\beta u/c)} \ \ \ \ \text{(14.10)}$". Up to that point, he has for example not excluded the possibility, that $\beta$ (which he had formally introduced in the transformation formula for time, in step 3) might be zero and he gets the Galileo transformation.

First in step 6, $\beta = v/c$ is derived by demanding, that, if $u$ is the speed of light $c$, then $u'$ must be also $c$.

 

[pervect]

One point to note is that it is circular reasoning to say that the transformation of velocities formula is derived from the Lorentz transformation, because the velocity transformation formula is often a step in the derivation of the Lorentz transformation.

I'm not familiar with such a derivation. The derivation I favor requires the following three assumptions. One: if we have two observers in relative motion at the same location, light emitted by one will experience a doppler shift factor k, when received by the other, the ratio k being the ratio between the interval of emission and the interval of reception. Secondly, we need the assumption that the speed of light is constant for all inertial observers. Thirdly, to handle two observers in relative motion who are not at the same location, we add another assumption, which states that light emitted from an inertial observer and received by another inertial observer at rest with respect to the first at a different location will experience no doppler shift, that the k factor will be unity.

This is basically Bondi's argument, though Bondi did not, to my knowledge, codify the third and last assumption. I feel the third assumption is a useful approach, though, and shores up some things I don't like about Bondi's approach. I should add, I've never seen the third assuption in a published paper, but I still like it.

We don't need any arguments about velocity addition to use the above assumptions, which are simple and straightforards, and only require high school algebra to analyze. The analysis can be found in Bondi's book, "Relativity and Common Sense", and basically uses a simple radar setup to determine the relationship between the factor k and relative velocity. Looking up "k-calculus" on the internet will probably also work for those who don't want to track down Bondi's book.

 

[hartmutneff]

Everything follows from the invariance of Maxwell's wave equation. The velocity of waves is not derived from any kind of dx/dt. The velocity of a wave is more like a geometric speed. If you take a mass-spring system, with overall mass M and overall spring constant K, where the length of the system is L, the velocity of a wave is given by (KL^2/M). It is kind of a different type of velocity, as I said, not dx/dt.
Now, if you put the mass-spring system on a train, then it goes at the wrong speed, for an observer on a platform, since, classically, the velocity is KL^2/M+v, where v is the speed of the train. Therefore, the wave equation is not valid for a spring on a train.

For a mass-spring system, you can sweep this under the rug, as you can see the masses and the springs on the train. One can say, there is a preferred reference frame for a spring-mass system, the one where the springs and the masses rest.

For light, and Maxwell's wave equation, the sweeping under the rug doesn't work, as light doesn't require a medium. There is no preferred reference frame in which the medium rests, as there is no medium.

Then you have two options, you give up one the wave equation, or you fix the Galilean invariance, such that the speed of light is always the same, for every observer. If you assume a general space and time transformation, which leaves the wave equation invariant, you end up with the Lorentz transformation. This is pretty straightforward.

Once you got the Lorentz transformation, the rest, like relativistic addition of velocities, follows quite easily.

By the way, if you assumed for a minute, that light was a mass spring system, then, with the Lorentz contraction of L, the relativistic mass follows directly from the constance of KL^2/M.

 

[Sagittarius A-Star]

Can you show us this step?

I found a derivation of the LT from "relativistic velocity addition" and time dilation.

Relativistic velocity addition law derived from a machine gun analogy and time dilation only
...
1. Deriving the relativistic velocity addition law without using the Lorentz transformations
...
2. Lorentz transformations from the relativistic addition law of velocities.

Source (PDF):
https://arxiv.org/pdf/physics/0703157
via (Abstract):
https://arxiv.org/abs/physics/0703157