Derivation of the Lorentz Transformations from first principles

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Derivation of the Lorentz Transformations from first principles

10/03/2013 by RhEvans

Riding on a beam of light

In this previous blog, I discussed how an experiment involving electrodynamics was not invariant under a Galilean transformation. Or, to put it another way, the laws of electrodynamics as stated would allow someone to determine whether they were at rest or moving, something which deeply troubled a young Albert Einstein. It is said that one of Einstein’s first “thought experiments” was to imagine himself travelling along on a beam of light. Light is the ultimate “free lunch”, the changing magnetic field produces a changing electric field which produces a changing magnetic field. It self-propogates at a speed of $3 \times 10^{8}$ metres per second in a vacuum.

Einstein realised that if he were travelling with the beam of light then, relative to him, the light would disappear as the electric and magnetic fields would be stationary relative to him. This worried him, as it suggested that one would be able to tell whether one was travelling or at rest, just by measuring the properties of light. Einstein realised, in an insight which possibly no one else was capable of, that the speed of light was fundamental to physics, and needed to always be constant. This led him to develop what we now call the special theory of relativity, most of which is expressed in a paper he published in 1905 called “On The Electrodynamics of moving bodies“.

Einstein’s special theory of relativity

Einstein’s Special Theory of Relativity is based on two very simple but far reaching principles

No experiment, mechanical or electrodynamical, can distinguish between being at rest or moving at a constant velocity.
That the speed of light in a vacuum, c, is constant to any observer, no matter how quickly the observer is moving.

From the second of these principles, with a simple thought experiment, we can derive the Lorentz transformations from first principles. These are the equations which allow us to translate from one frame of reference to another so that all the laws of Physics are invariant.

An expanding sphere of light

The thought experiment we will use to derive the Lorentz transformations from first principles is one of a flash of light originating at the origin of two frames of reference S and S’ which are moving relative to each other with a velocity $v$ . We set up our experiment so that at time $t=0$ the origins of the two frames of reference are in the same place.

Two frames of reference S and S' moving relative to each other have a flash of light originate at their respective origins at time t=0

Two frames of reference S and S’ moving relative to each other with a velocity v have a flash of light originate at their respective origins at time t=0

The flash of light will expand as a sphere, moving with a velocity $c$ in both frames of reference, in accordance with Einstein’s 2nd principle of relativity. For reference frame S we can write that the square of the radius $r^{2}$ of the sphere is $x^{2} + y^{2} + z^{2} = c^{2}t^{2}$ so

$\boxed{ x^{2} + y^{2} + z^{2} - c^{2}t^{2}=0 } \qquad(1)$

For the reference frame S’ we can write that

$\boxed{ x^{\prime 2} + y^{\prime 2} + z^{\prime 2} - c^{2}t^{\prime 2} = 0 } \qquad(2)$

These two equations must be equal, as it is the same sphere of light and therefore the sphere must have the same radius in the two reference frames. Let us see if we can transform from one to the other using the Galilean transforms, which are

$\boxed {\begin{array}{lcl} x^{\prime} & = & x - vt \\ y^{\prime} & = & y \\ z^{\prime} & = & z \\ t^{\prime} & = & t \end{array} }$

$x^{\prime 2} + y^{\prime 2} + z^{\prime 2} -c^{2}t^{2} = (x-vt)^{2} + y^{2} + z^{2} - c^{2}t^{2}$

Expanding the brackets of the right hand side gives

$x^{2} - 2vtx + v^{2}t^{2} + y^{2} + z^{2} - c^{2}t^{2} \neq x^{2} + y^{2} + z^{2} - c^{2}t^{2}$

The the equation should be equal, but the terms highlighted do not exist on the right hand side of the equation.

The left side of the equation should be equal to the right side, but the terms highlighted do not exist on the right hand side of the equation.

As we can see, the two expressions are not equal as the left hand side has the extra terms $-2vtx + v^{2}t^{2}$ . This means that a Galilean transformations does not work. The extra terms involve a combination of $x$ and $t$ , which suggests that both the equations linking $x$ and $x^{\prime}$ and $t$ and $t^{\prime}$ need to be modified, not just the equation for $x$ as is the case in the Galilean transformations.

Modifying the Galilean transformations

Let us assume that the transformations can be written as

$\boxed {\begin{array}{lcl} x^{\prime} & = & a_{1}x + a_{2}t \qquad(3) \\ y^{\prime} & = & y \\ z^{\prime} & = & z \\ t^{\prime} & = & b_{1}x + b_{2}t \qquad(4) \end{array} }$

We need to find the values of $a_{1}, a_{2}, b_{1}$ and $b_{2}$ which correctly transform the equations for the expanding sphere of light. We do this by substituting equations (3) and (4) into equation (2). Before we do this, we note that the origin of the primed frame $x^{\prime}=0$ is a point that moves with speed $v$ as seen in the unprimed frame S. Therefore its location in the unprimed frame S at time $t$ is just $x=vt$ . So we can write equation (3) as

$x^{\prime} = 0 = a_{1}x + a_{2}t \rightarrow x = -\frac{a_{2}}{a_{1}} t = vt$

$\therefore \frac{ a_{2} }{ a_{1} } = -v$

Re-writing equation (3)

$x^{\prime} = a_{1}x + a_{2}t = a_{1}(x+\frac{ a_{2} }{ a_{1} } t) = a_{1}(x-vt)$

Now we substitute this expression and equation (4) into equation (2)

$a_{1}^{2}(x-vt)^{2} + y^{\prime 2} + z^{\prime 2} -c^{2}(b_{1}x+b_{2}t)^{2} = x^{2} + y^{2} + z^{2} -c^{2}t^{2}$

$a_{1}^{2} x^{2} -2a_{1}^{2} xvt + a_{1}^{2} v^{2} t^{2} - c^{2} b_{1}^{2} x^{2} - 2c^{2} b_{1} b_{2} xt -c^{2} b_{2}^{2} t^{2} = x^{2} - c^{2} t^{2}$

Equating coefficients:

$( a_{1}^{2} - c^{2}b_{1}^{2} ) x^{2} = x^{2} \rm{\;\; or \;\;} a_{1}^{2} - c^{2}b_{1}^{2} = 1 \qquad(5)$

$( a_{1}^{2} v^{2} - c^{2} b_{2}^{2} ) t^{2} = -c^{2} t^{2} \rm{\;\; or \;\;} c^{2} b_{2}^{2} -a_{1}^{2} v^{2} = c^{2} \qquad(6)$

$(2a_{1}^{2} v + 2b_{1} b_{2} c^{2} ) xt = 0 \rm{\;\; or \;\;} b_{1} b_{2} c^{2} = -a_{1}^{2}v \qquad(7)$

From equations (5) and (6) we can write

$b_{1}^{2} c^{2} = a_{1}^{2} - 1 \qquad(8)$

and

$b_{2}^{2} c^{2} = c^{2} + a_{1}^{2} v^{2} \qquad (9)$

Multiplying equations (8) and (9) and squaring equation (7) we get

$b_{1}^{2} b_{2}^{2} c^{4} = ( a_{1}^{2} - 1 )( c^{2} + a_{1}^{2} v^{2} ) = a_{1}^{4} v^{2}$

$a_{1}^{2} c^{2} - c^{2} + a^{4} v^{2} - a_{1}^{2} v^{2} = a_{1}^{4} v^{2}$

$a_{1}^{2} c^{2} - a_{1}^{2} v^{2} = c^{2}$

$a_{1}^{2} ( c^{2} - v^{2} ) = c^{2}$

$a_{1}^{2} = \frac{ c^{2} }{ c^{2} - v^{2} } = \frac{ 1 }{ 1 - v^{2}/c^{2} }$

$\boxed{ a_{1} = \frac{ 1 }{ \sqrt{ (1 - v^{2}/c^{2} ) } } }$

Thus we can write

$\boxed{ a_{2} = -v \cdot \frac{ 1 }{ \sqrt{ ( 1 - v^{2}/c^{2} ) } } }$

Using equation (8) we can write

$b_{1}^{2} c^{2} = \frac{ 1 }{ (1 - v^{2}/c^{2} ) } - 1$

$b_{1}^{2} c^{2} = \frac{ 1 - ( 1 - v^{2}/c^{2} ) }{ (1 - v^{2}/c^{2} ) } = \frac{ v^{2}/c^{2} }{ (1 - v^{2}/c^{2} ) } = \frac{ v^{2} }{ c^{2} } \cdot \frac { 1 }{ (1 - v^{2}/c^{2} ) }$

so $b_{1}^{2} = \frac{ v^{2} }{ c^{4} } \cdot \frac{ 1 }{ ( 1 - v^{2}/c^{2} ) }$

Taking the negative square root we can write

$\boxed{ b_{1} = - \frac{ v }{ c^{2} } \cdot \frac{ 1 }{\sqrt{ (1 - v^{2}/c^{2} ) }} }$

From equation (9) we can write

$b_{2}^{2} c^{2} = c^{2} + v^{2} \cdot \frac{ 1 }{ ( 1 - v^{2}/c^{2} ) } = \frac{ c^{2}( 1 - v^{2}/c^{2} ) + v^{2} }{ ( 1 - v^{2}/c^{2} ) } = \frac{ c^{2} - v^{2} + v^{2} }{ (1 - v^{2}/c^{2} ) } = \frac{ c^{2} }{ ( 1 - v^{2}/c^{2} ) }$

which leads to

$b_{2}^{2} = \frac{ 1 }{ ( 1 - v^{2}/c^{2} ) }$

and so

$\boxed{ b_{2} = \frac{ 1 }{ \sqrt{ ( 1 - v^{2}/c^{2} ) } } }$

which is the same as $a_{1}$ .

If we define

$\gamma = \frac{ 1 }{ \sqrt{ ( 1 - v^{2}/c^{2} ) } }$

we can write

$a_{1} = \gamma, \;\;\; a_{2} = -\gamma v, \;\;\; b_{1} = -\frac{ v }{ c^{2} } \cdot \gamma \rm{\;\;\ and \;\;\;} b_{2} = \gamma$

Thus we can finally write our transformations as

$\boxed {\begin{array}{lcl} x^{\prime} & = & \gamma (x - vt) \\ y^{\prime} & = & y \\ z^{\prime} & = & z \\ t^{\prime} & = & \gamma ( t - \frac{ v }{ c^{2} }x ) \end{array} }$

These are known as the Lorentz transformations.

The Lorentz factor

The term $\gamma$ is know as the Lorentz factor.

$The Lorentz factor gamma plotted against speed as a fraction of the speed of light.$

The Lorentz factor $\gamma$ plotted against speed as a fraction of the speed of light.

As this plot shows, the Lorentz factor is essentially unity until the ratio $v/c$ (the ratio of the speed to the speed of light) becomes about half of the speed of light, or about $1.5 \times 10^{8}$ m/s. Given that even our fastest space ships only travel at a tiny fraction of the speed of light, it is not surprising that we have no direct experience of the weird effects that a Lorentz factor deviating significantly from one produce. Of course we see these effects in particle accelerators and cosmic ray showers, but human beings are a long way from attaining speeds where the Lorentz factor will deviate from unity.

In a future blog I will discuss some of these weird effects. They include time passing more slowly and distances shrinking. Very very weird; but very very real, they are shown to happen every day in our particle accelerators.

Posted in History, mathematics, Physics | Tagged Einstein, Lorentz, Special Theory of Relativity, Speed of light | 44 Comments

44 Responses

on 11/03/2013 at 12:22 | Reply Phillip Helbig

Here’s a question involving GR. The equivalence principle says that one cannot distinguished between being at rest in a gravitational field (say, in a lab on the surface of the Earth) or experiencing constant acceleration in the absence of a gravitational field (say, in a lab in an accelerating rocket in deep space). On the other hand, it is well established that accelerating charges radiate. So, can I determine if my windowless lab is on the surface of the Earth or in an accelerating rocket by observing whether a charge within it radiates?

If the answer is that it radiates in the case of the rocket but this is only visible to someone outside the rocket, then nevertheless photons must travel from the charge to the observer. If I fill the inside of the rocket with reflecting walls, will I see these photons collect up inside the lab in the rocket?
- on 11/03/2013 at 13:17 | Reply RhEvans
  
  That is a very interesting question! I’m no expert on GR, but I am familiar with the equivalence principle. I’m going to have to think about your question, the answer doesn’t seem obvious so I’m assuming it’s not trivial.
  - on 11/03/2013 at 15:17 Phillip Helbig
    
    Of course, I can distinguish the two cases via tidal effects; the equivalence principle holds only in the limit of an infinitesimally small lab. However, I don’t think that that has anything to do with the problem. (Also, one can derive the formula for the deflection of light due to gravity via the equivalence principle, though this seems to depend on the lab not being infinitesimally small. I’m sure this isn’t relevant to the radiating-charge problem either, but interesting nonetheless.)
    
    There is some literature on this problem. However, I think it would be better for you and other readers here to have a stab without consulting the literature first; this might lead to some fresh insights.
  - on 12/03/2013 at 07:05 RhEvans
    
    I have derived the light deflection formula using the equivalence principle for a class I taught 5-6 years ago. As you say, for this to work your accelerating lab has to have finite width, to get the ray of light to move in a parabola and hence be bent.
    
    But I have not thought about your particular problem before. I have a busy 48 hours finishing up something, but after that I will give it some thought.
  - on 29/04/2013 at 14:24 Phillip Helbig
    
    Solved it yet? And I don’t mean solving it like this.
on 27/04/2013 at 16:52 | Reply Alexey Vasin

Your derivation seems incomplete. The Euclidean solution also obeys the constancy of the involved characteristic speed (commonly referred to as “the speed of light”). This solution is rulled out by the additional assumption of casualty that you do not make. In fact, even the Gililean solution imply the constant (infinite) characteristic speed. This solution in turn is ruled out by the experiment (not necessarily with light) showing that the characteristic speed is not infinite.

Furthermore, there is no theoretical relation between coordinate transformations and light as a physical phenomenon or its specific speed. The Lorentz transformations only imply that the maximum observed speed is the same in any inertial frame. With what specific speed photons, gluons, or neutrinos actually move is a matter of experimental measurement unrelated to the derivation of the coordinate transformations.

For example, if at the time Instead of light it was the speed of neutrinos measured, it would also have been found constant within the margin of error. Obviously, calling the Lorentz characteristic speed “the speed of neutrinos” would have been wrong, as we know now that neutrinos move slower than light. The historical way a physical theory is first formulated is often different from its true logical structure.
on 30/05/2013 at 05:59 | Reply Einstein and time travel | thecuriousastronomer

[…] this blog I derived, from first principles, the Lorentz transformations which are used in Einstein’s […]
on 23/08/2013 at 10:41 | Reply LUKKA KARTHIK SUBHASH CHANDRA

THE FUNDAMENTAL IS THAT WE CAN SAY WHETHER A FRAME IS ACCELERATING OR NOT BY USING A MASS OR A CHARGE.

THE LOGIC FOLLOWS AS ” YOU ARE IN DARK SPACE! AND ARE MAINTAINING CONSTANT ACCELERATION .YOU CANNOT TELL THIS PHENOMENON IS DUE TO AN ENGINE OR DUE TO GRAVITY UNLESS AN UNTIL YOU SEE AN ENGINE OR A STAR.

SUPPOSE M (INERTIAL) = M SQUARE (GRAVITATIONAL). THEN BY DOUBLING GRAVITATIONAL MASS U CAN CALCULATE THE FORCE ACTING WAS 4 TIMES OR 2 TIMES THE ORIGINAL. ACCORDINGLY U COULD DETERMINE WHETHER IT WAS GRAVITY OR NOT,
- on 23/08/2013 at 10:43 | Reply RhEvans
  
  Um, I think you are confusing inertial and non-inertial frames. The special theory of relativity only holds for inertial frames.
- on 23/08/2013 at 10:45 | Reply RhEvans
  
  What you have stated above is essentially Einstein’s “principle of equivalence” which he came up with in 1907 and which started him on his path to his theory of general relativity.
on 08/12/2013 at 08:59 | Reply str

well you didn’t prove galileo wrong there. there is a big hole in your consideration regarding v^2t^2 and -2xvt being extra terms. Either you put t=0, then these 2 terms vanishes or if you don’t take t=0 then don’t equate your equation to ct because that is distance from center not where s’ is after time t.
- on 08/12/2013 at 20:14 | Reply RhEvans
  
  The centres are coincident at time t=t’=0. Only considering the equations at time t=0 is not going to tell you how things are at any other time in the two reference frames, so doesn’t get you anywhere.
on 28/08/2014 at 14:18 | Reply Usman Ch

Its Alternative method……? 🙂
on 16/12/2014 at 07:31 | Reply Einstein’s General Relativity – part 1 – the Principle of Equivalence | thecuriousastronomer

[…] have already blogged about Einstein’s ground-breaking Special theory of Relativity here. Just to recap, based on two […]
on 11/02/2015 at 11:00 | Reply Time dilation in General Relativity | thecuriousastronomer

[…] have already derived the Lorentz Transformations from first principles in this blog, and these equations are at the heart of SR, and show why time dilation occurs when one travels […]
on 12/02/2015 at 12:36 | Reply salchon

The time dilation effect predicted by SR is illogical and mathematically inconsistent. See my view on it at salchon.wordpress.com.
- on 12/02/2015 at 12:48 | Reply RhEvans
  
  You wouldn’t be the first to think that, nor the first to be shown to be wrong.
  - on 12/02/2015 at 13:49 salchon
    
    Can you point out the mistake logically and mathematically without appealing to authority or ad hominem arguments? I agree that the argument I present might be defeated, but I don’t understand why SR is a priori considered infaillable.
  - on 12/02/2015 at 13:59 RhEvans
    
    Of course it’s not infallible, no theory is. But, if you read through this derivation step by step, it makes logical sense as long as you adhere to the two premises of SR – the constancy of the speed of light and that all inertial frames are equivalent. If you do that, and follow the derivation in this blog, the Lorentz transformations come out naturally.
  - on 12/02/2015 at 14:09 salchon
    
    Likewise, step by step, I show on the link I listed above that time dilation is mathematically inconsistent within the context of SR hypotheses (constancy of the speed of light and invariance of physical laws in inertial systems).
  - on 12/02/2015 at 14:10 RhEvans
    
    Well then either you are wrong, or the remaining 99.99% of physicists are wrong. I suspect the former.
  - on 12/02/2015 at 14:12 salchon
    
    Now you are appealing to authority…
  - on 12/02/2015 at 14:19 RhEvans
    
    No, I’m appealing to my own understanding of the subject and my own ability to follow the logic.
on 01/09/2015 at 19:55 | Reply Steven DeRyke

He is not appealing to authority, he is appealing to the success of the theory. Science is not a search for truth (see philosophical argument about the white crow experiment), it is in search of the theory with the best predictive power. Special Relativity has that in spades. If your theory can do a better job of predicting the behavior of muons as they fall through the atmosphere from space and do not decay as quickly as they are expected to (and a host of other experimental results as well) then the 99.99% of the physicist would consider it and abandon SR.
- on 01/09/2015 at 20:10 | Reply RhEvans
  
  Yes, I agree.
on 10/09/2015 at 21:38 | Reply tycoparri

This was very well done and straightforward,thanks. You’re only missing the square root in the denominator of the coefficient b1 when you first show it. Keep up with the good work!
- on 18/03/2016 at 11:43 | Reply iillyyaa
  
  I need help understanding one step in this derivation. The place, where you say: “Equating coefficients” and go on to drive (5) , (6) and (7). I’m missing the reason why we can equate the coefficients independently, whereas before the weighted sums were equated.
  
  Probably something simple that isn’t clicking without a nudge. Would appreciate said nudge.
  - on 20/03/2016 at 17:49 RhEvans
    
    If you have (a_1 + a_2)x^2 + (b_1 + b_2)x + c_1 + c_2 = dx^2 + ex + f then you can say that (a_1 + a_2) = d (equating the x^2 coefficients), (b_1 + b_2) = e (equating the x coefficients) and (c_1 + c_2) = f (equating the constant terms). That is all I have done here. Does that now make sense?
  - on 21/03/2016 at 23:46 iillyyaa
    
    Sorry, I was not familiar specifically with the technique of “equating coefficients”, though wikipedia helped in that regard.
    
    It does make sense now. I think I omitted to notice that the equality must hold for all x (as opposed to there existing some x for which it could be solved.)
    
    Thank you for the answer, and thanks for the walkthrough of the derivation.
  - on 22/03/2016 at 02:24 RhEvans
    
    You’re welcome 🙂
on 11/12/2016 at 02:26 | Reply Lazer

I have a doubt in how you arrived at your expression for b1. While I understand that the root of something can be positive or negative (like the root of 4 is either 2 or -2), could you please explain to me the logic of choosing the negative root for b2?
- on 11/12/2016 at 05:49 | Reply RhEvans
  
  I am flying back from Australia so will respond to this tomorrow (Monday).
on 28/06/2017 at 21:55 | Reply The Lorentz Transformation – physics derivations

[…] have seen derivations of this transformation from the axioms of relativity alone, but this approach leads to quite a mess. I will take the approach I just read in Taylor’s […]
on 04/09/2017 at 06:30 | Reply How to add velocities in special relativity | thecuriousastronomer

[…] I mentioned in this blogpost, in special relativity any observer will measure the speed of light in a vacuum to be , […]
on 15/09/2017 at 03:01 | Reply Paul

Why is the negative square root of b1 taken as the solution for b1 ?
I am unable to figure out the reason. Would like some help.
Thanks
- on 26/11/2017 at 15:18 | Reply Michael Bradnick
  
  I would also like to know the answer to this if possible 🙂
- on 28/11/2017 at 00:57 | Reply RhEvans
  
  Because a2 is negative, to retain symmetry b1 has to be negative too.
on 10/11/2017 at 16:14 | Reply Serious Physicist

I m a huge fan of yours. The way you explain things from the scratch is amazing. I always wanted to see some heuristic derivation of Lorentz transformation for space and time and eventually you answered my queries. Thanks a lot man!
- on 10/11/2017 at 21:04 | Reply RhEvans
  
  Thank you for your kind words.
on 02/06/2018 at 09:53 | Reply Uri Halpern

Well done . I am speecless
Good writing.
on 28/08/2018 at 13:33 | Reply Lev

SRT is completely erroneous since it is based on the wrong kind of transformations: they have lost the scale factor characterizing the Doppler effect. First, Lorentz considered a more general form of transformations (with a scale factor), but then he, and also Poincare and Einstein equated it 1 without proper grounds. Their form was artificially narrowed, the formulas became incorrect. This led to a logical contradiction of the theory, to unsolvable paradoxes. For more details, see my brochure “Memoir on the Theory of Relativity and Unified Field Theory” (2000):
http://vixra.org/abs/1802.0136
on 11/02/2019 at 16:00 | Reply Hamza

hi. in this derivation the velocity of S’ frame is only in x-direction what if the velocity vector was in all three directions. could you do the lorrentz transformation for that case. or could you point me in directions of general symmetries that we establish in this example that we could use in for the more general case
on 18/01/2020 at 14:11 | Reply Vivek Karunakaran

That’s quite intriguing and easy to understand. But why is the a1 you get after taking the square root turns out to positive. It could have been +/- right?
- on 29/01/2020 at 13:03 | Reply RhEvans
  
  You can reject the negative solution based on the physics