A What assumptions underly the Lorentz transformation?

Click For Summary
The Lorentz transformation for velocities, expressed as u=(v+w)/(1+vw/c²), is derived from the principles of special relativity, particularly the invariance of the speed of light and the isotropy and homogeneity of spacetime. The discussion emphasizes that the formula is more accurately termed "relativistic velocity addition" rather than a transformation. Key assumptions include the collinearity of velocities and the requirement that the velocity addition rule be symmetric and approximate Galilean addition at low speeds. The conversation also clarifies that there is no concept of absolute velocity in special relativity, as all velocities are relative to inertial frames. Understanding these foundational assumptions is crucial for deriving the Lorentz transformation accurately.
Ad VanderVen
Messages
169
Reaction score
13
TL;DR
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?
 
Last edited:
Physics news on Phys.org
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).
 
  • Like
Likes Dragrath, pervect, David Lewis and 2 others
Ad VanderVen said:
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.
 
  • Like
  • Informative
Likes Dragrath, Delta2, FactChecker and 2 others
Orodruin said:
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)?
 
PeroK said:
assumptions about the isotropy and homogeneity of spacetime
What exactly are the assumptions about the isotropy and homogeneity of spacetime?
 
Ad VanderVen said:
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.
 
  • Like
Likes martinbn and Vanadium 50
Ad VanderVen said:
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:
Wikipedia said:
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.
 
Last edited:
  • Like
Likes vanhees71
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.
 
  • Like
Likes vanhees71
  • #10
Sagittarius A-Star said:
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.
 
Last edited:
  • #11
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).
 
  • Like
Likes vanhees71 and PeroK
  • #12
Orodruin said:
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!
 
  • #13
Orodruin said:
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.
 
  • #14
Ad VanderVen said:
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?
 
  • Like
Likes vanhees71
  • #15
Orodruin said:
No, that is not an assumption. What makes you think there is?
What makes you think it is not an assumption?
 
  • #16
Ad VanderVen said:
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.
 
  • #17
Orodruin said:
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.
 
  • Skeptical
Likes weirdoguy and Motore
  • #18
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
 
  • Like
Likes otennert and Orodruin
  • #19
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.
 
Last edited:
  • Like
Likes malawi_glenn, nasu, berkeman and 2 others
  • #20
Ad VanderVen said:
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.
 
  • Like
Likes Vanadium 50
  • #21
Ad VanderVen said:
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.
 
  • Like
Likes Dragrath, vanhees71, Vanadium 50 and 2 others
  • #22
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.]

1655161907053.gif

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

So, the point is... where does one wish to start?
 
Last edited:
  • Like
Likes Dragrath and martinbn
  • #23
robphy said:
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.
 
  • Like
Likes Dragrath, martinbn, robphy and 1 other person
  • #24
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.
 
  • Like
Likes robphy
  • #25
"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.
 
  • Haha
  • Like
Likes Dragrath, vanhees71 and robphy
  • #26
vanhees71 said:
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.
 
  • Like
Likes Dragrath and vanhees71
  • #27
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.
 
  • Skeptical
Likes strangerep, weirdoguy and PeroK
  • #28
Meir Achuz said:
because the velocity transformation formula is often a step in the derivation of the Lorentz transformation.

Can you show us this step?
 
  • Like
Likes strangerep, vanhees71 and Sagittarius A-Star
  • #29
weirdoguy said:
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...".
 
  • #30
Meir Achuz said:
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:
Classical Electromagnetism - Jerrold Franklin said:
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
 
  • Like
Likes Orodruin

Similar threads

  • · Replies 33 ·
2
Replies
33
Views
3K
Replies
3
Views
1K
  • · Replies 54 ·
2
Replies
54
Views
4K
Replies
0
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
10
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
5
Views
2K