## 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 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 . We set up our experiment so that at time the origins of the two frames of reference are in the same place.

The flash of light will expand as a sphere, moving with a velocity 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 of the sphere is so

For the reference frame S’ we can write that

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

Expanding the brackets of the right hand side gives

As we can see, the two expressions are not equal as the left hand side has the extra terms . This means that a Galilean transformations does not work. The extra terms involve a combination of and , which suggests that both the equations linking and *and* and need to be modified, not just the equation for as is the case in the Galilean transformations.

## Modifying the Galilean transformations

Let us assume that the transformations can be written as

We need to find the values of and 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 is a point that moves with speed as seen in the unprimed frame S. Therefore its location in the unprimed frame S at time is just . So we can write equation (3) as

Re-writing equation (3)

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

Equating coefficients:

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

and

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

so

so

Thus we can write

Using equation (8) we can write

so

Taking the negative square root we can write

From equation (9) we can write

which leads to

and so

which is the same as .

If we define

we can write

Thus we can finally write our transformations as

These are known as the Lorentz transformations.

## The Lorentz factor

The term is know as the *Lorentz factor*.

As this plot shows, the Lorentz factor is essentially unity until the ratio (the ratio of the speed to the speed of light) becomes about half of the speed of light, or about 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.

on 11/03/2013 at 12:22 |Phillip HelbigHere’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 |RhEvansThat is a

veryinteresting 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 15:17Phillip HelbigOf course, I

candistinguish 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 labnotbeing 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

withoutconsulting the literature first; this might lead to some fresh insights.on 12/03/2013 at 07:05RhEvansI 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 14:24Phillip HelbigSolved it yet? And I don’t mean solving it like this.

on 27/04/2013 at 16:52 |Alexey VasinYour 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 |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 |LUKKA KARTHIK SUBHASH CHANDRATHE 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 |RhEvansUm, 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 |RhEvansWhat 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 |strwell 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 |RhEvansThe 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 |Usman ChIts Alternative method……? ๐

on 16/12/2014 at 07:31 |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 |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 |salchonThe 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 |RhEvansYou wouldn’t be the first to think that, nor the first to be shown to be wrong.

on 12/02/2015 at 13:49salchonCan 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 13:59RhEvansOf 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 14:09salchonLikewise, 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 14:10RhEvansWell then either you are wrong, or the remaining 99.99% of physicists are wrong. I suspect the former.

on 12/02/2015 at 14:12salchonNow you are appealing to authority…

on 12/02/2015 at 14:19RhEvansNo, 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 |Steven DeRykeHe 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 |RhEvansYes, I agree.

on 10/09/2015 at 21:38 |tycoparriThis 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 |iillyyaaI 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 17:49RhEvansIf 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 23:46iillyyaaSorry, 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 02:24RhEvansYou’re welcome ๐

on 11/12/2016 at 02:26 |LazerI 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 |RhEvansI am flying back from Australia so will respond to this tomorrow (Monday).

on 28/06/2017 at 21:55 |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 |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 |PaulWhy 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 |Michael BradnickI would also like to know the answer to this if possible ๐

on 28/11/2017 at 00:57 |RhEvansBecause a2 is negative, to retain symmetry b1 has to be negative too.

on 10/11/2017 at 16:14 |Serious PhysicistI 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 |RhEvansThank you for your kind words.

on 02/06/2018 at 09:53 |Uri HalpernWell done . I am speecless

Good writing.