Assumption in the derivation of the Lorentz transformation

[Kyouran]

TL;DR

Assumption in the derivation of the lorentz transformation

In the special theory of relativity, it seems impossible to derive the lorentz transformation without assuming that the lorentz factor is independent of the sign of the relative velocity. For some reason, I can't get my head around why this assumption is so easily made, as if it's trivial. Can anyone explain why this assumption can be made?

 

[ergospherical]

Can you try and articulate why you think the sign of the relative velocity would affect the Lorentz factor?

 

[Kyouran]

It's not that I have a physical reason for it, but it is an assumption and assumptions must be justified. In particular, I am thinking about the following step:

$x' = (x-vt) \gamma(v)$
$t' = (t-\frac{vx}{c^2}) \gamma(v)$

Now, to obtain the reverse transformation, one would simply replace v by -v and swap the x and x' and t and t'. Then, substitution of the reverse transformation into this transformation (e.g. first equation) gives:

$x'= x' (1-\frac{v^2}{c^2}) \gamma(v) \gamma(-v)$

So here's where I'm stuck. I need one more justification to exclude the possibilities that $\gamma(v) = \frac{1}{1+\frac{v}{c}}$ or $\gamma = \frac{1}{1-\frac{v}{c}}$

 

[FactChecker]

It's not that I have a physical reason for it, but it is an assumption and assumptions must be justified.

Consider two cases:
The mathematical assumption as a postulate: In mathematics, one can make any assumption as a postulate and determine the logical consequences of that postulate. There is no need to justify it.
The physical "assumption" due to experimental evidence: There is plenty of experimental evidence that the direction of velocity is not important in reality. So the conditions of the Lorentz transformation are satisfied and the transformation can be applied.

 

[robphy]

Trigonometrically, your question amounts to the question of
why the dot-product is independent of the sign of the angle-between them.

So, maybe one should consider the Euclidean analogue to derive
the cosine function, and worry about any needed constraints on the "angle between".

 

[ergospherical]

Now, to obtain the reverse transformation, one would simply replace v by -v and swap the x and x' and t and t'. Then, substitution of the reverse transformation into this transformation (e.g. first equation) gives:

$x'= x' (1-\frac{v^2}{c^2}) \gamma(v) \gamma(-v)$

So here's where I'm stuck. I need one more justification to exclude the possibilities that $\gamma(v) = \frac{1}{1+\frac{v}{c}}$ or $\gamma = \frac{1}{1-\frac{v}{c}}$

The underlying assumption is that of the isotropy of space, meaning that a priori a relative velocity of $v$ is physically equivalent to a relative velocity of $-v$.

 

[PAllen]

Note, there are many different derivations of Lorentz transform, many with no room for such ambiguity. One variant considers only one direction of transform, with the requirement a spherical light pulse is spherical in the primed frame, with the same speed of light. You get gamma explicitly, as a result (actually, you get the complete forward Lorentz transform). Then, simply algebraically solving for the unprimed variables given the full forward transformation you have derived, shows the reverse transform is the same except for the sign of v.

For example, this is the derivation given in Peter Bergmann’s 1942 text on relativity.

 

[Dale]

Summary:: Assumption in the derivation of the lorentz transformation

In the special theory of relativity, it seems impossible to derive the lorentz transformation without assuming that the lorentz factor is independent of the sign of the relative velocity. For some reason, I can't get my head around why this assumption is so easily made, as if it's trivial. Can anyone explain why this assumption can be made?

Space is isotropic. So a transformation between inertial frames has to be the same under spatial rotation. That means that you can always rotate it to be in the positive direction. In other words, it can only depend on the speed because speed is the isotropic part of velocity.

 

[Herman Trivilino]

Now, to obtain the reverse transformation, one would simply replace v by -v and swap the x and x' and t and t'. Then, substitution of the reverse transformation into this transformation (e.g. first equation) gives:

$x'= x' (1-\frac{v^2}{c^2}) \gamma(v) \gamma(-v)$

So here's where I'm stuck. I need one more justification to exclude the possibilities that $\gamma(v) = \frac{1}{1+\frac{v}{c}}$ or $\gamma = \frac{1}{1-\frac{v}{c}}$

But $\gamma=\sqrt{\frac{1}{1-(v/c)^2}}$. When you reverse the sign of $v$ to form the inverse transformations, you don't reverse the sign of $v$ in the expression for $\gamma$.

 

[ergospherical]

@Mister T the op is following a particular derivation where the functional form of $\gamma(v)$ is yet to be determined

 

[PeroK]

It's not that I have a physical reason for it, but it is an assumption and assumptions must be justified. In particular, I am thinking about the following step:

$x' = (x-vt) \gamma(v)$
$t' = (t-\frac{vx}{c^2}) \gamma(v)$

Let's consider what the x-axis looks like in the primed frame at $t' = 0$. Which means that $t = \frac{vx}{c^2}$ and: $$x' = \gamma(v)(x - vt) = \gamma(v)x(1 - \frac{v^2}{c^2})$$Let's assume that $v > 0$ and choose a point $x > 0$.

Now, if we reverse the velocity so that the primed frame has velocity $-v$, then the point $-x$ is physically the same as $x$ was previously. Let's use double prime in this case. We know from physical symmetry that: $x'' = -x'$. It must be the same coordinate as before, with the sign reversed. In this case: $t'' = 0$, $t = \frac{vx}{c^2}$ and: $$x'' = \gamma(-v)(-x + vt) = -\gamma(-v)x(1 - \frac{v^2}{c^2})$$
And, for $x'' = -x'$ we must have $\gamma(-v) = \gamma(v)$.

 

[vanhees71]

That touches on the "reciprocity theorem", i.e., the apparently self-evident property of inertial reference frames that if $\Sigma'$ moves with constant velocity $\vec{v}$ wrt. $\Sigma$ then $\Sigma$ moves with velocity $-\vec{v}$ relative to $\Sigma'$.

It depends a bit on the assumptions you take as "postulates". One derivation uses only the special principle of relativity (existence of a global inertial frame of reference), the Euclidicity of space for any inertial observer (implying homogeneity and isotropy of space) and the homogeneity of time for any inertial observer. Additionally it is assumed that the symmetry transformations of the spacetime model form a group. With these assumptions you find out that the reciprocity theorem holds and that the only two spacetime models are either the Galilei-Newton spacetime model (with no "limiting speed") or the Einstein-Minkowski spacetime model of special relativity (with a "limiting speed" which empirically coincides with the speed of light in vacuo in Maxwell's equations).

https://aip.scitation.org/doi/10.1063/1.1665000

 

[Sagittarius A-Star]

$x' = (x-vt) \gamma(v)$
...
Now, to obtain the reverse transformation, one would simply replace v by -v and swap the x and x' and t and t'.

From that follows for the spatial distance between two events:
$\Delta x' = ( \Delta x-v \Delta t) \gamma(v) \ \ \ \ \ \ \ \ \ (1)$
$\Delta x = ( \Delta x'+v \Delta t') \gamma(-v) \ \ \ \ \ (2)$

If you define in equation (1) $L' := \Delta x'$ and set $\Delta t =0$, then you get the length contraction:
$\Delta x = \frac{1}{\gamma(v)} L'$

If you define in equation (2) $M := \Delta x$ and set $\Delta t' =0$, then you get the length contraction:
$\Delta x' = \frac{1}{\gamma(-v)} M$

For symmetry reasons, both length contraction factors must be equal.

Then, substitution of the reverse transformation into this transformation (e.g. first equation) gives:

$x'= x' (1-\frac{v^2}{c^2}) \gamma(v) \gamma(-v)$

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

 

[Kyouran]

Let's consider what the x-axis looks like in the primed frame at $t' = 0$. Which means that $t = \frac{vx}{c^2}$ and: $$x' = \gamma(v)(x - vt) = \gamma(v)x(1 - \frac{v^2}{c^2})$$Let's assume that $v > 0$ and choose a point $x > 0$.

Now, if we reverse the velocity so that the primed frame has velocity $-v$, then the point $-x$ is physically the same as $x$ was previously.

Ok, I have some problem with that last statement. Just because you change the relative velocity of a frame doesn't mean you change the orientation of its axes. Therefore, I don't get how you can say -x is physically the same as x, since both frames keep the same orientation?

 

[Dale]

Just because you change the relative velocity of a frame doesn't mean you change the orientation of its axes. Therefore, I don't get how you can say -x is physically the same as x, since both frames keep the same orientation?

All of the answers above boil down to using the fact that the laws of physics are isotropic. This means that if you have some experiment, and then you repeat the experiment with everything rotated by $\theta$ the two experiments are physically the same. Equivalently, you can take one experiment and rotate your axes by $\theta$ and use the same laws of physics.

So if you choose $\theta=\pi$ then you immediately have “-x is physically the same as x”

 

[PeroK]

Ok, I have some problem with that last statement. Just because you change the relative velocity of a frame doesn't mean you change the orientation of its axes. Therefore, I don't get how you can say -x is physically the same as x, since both frames keep the same orientation?

You could imagine an observer in S' reaching the origin and the point $x$ is a certain distance in front of it. Then, an observer in S'' reaches the origin and the point $-x$ must be the same distance in front of it. These must be physically identical scenarios.

I chose the $x''$ axis to be in the same direction as the others. You don't need to do this, but the orientation of the positive $x''$ axis cannot be physically relevant.

If you want the $x''$ axis in the opposite direction you are going to have to mess around with all the formulas. That can't change the physical equivalence or the validity of my analysis.

 

[Kyouran]

We know from physical symmetry that: $x'' = -x'$. It must be the same coordinate as before, with the sign reversed. In this case: $t'' = 0$, $t = \frac{vx}{c^2}$ and: $$x'' = \gamma(-v)(-x + vt) = -\gamma(-v)x(1 - \frac{v^2}{c^2})$$
And, for $x'' = -x'$ we must have $\gamma(-v) = \gamma(v)$.

Hmm...why is $t" = 0$ not $t = -\frac{vx}{c^2}$? Srry for being so confused guys :D

 

[Dale]

You are standing on the north embankment watching a train moving to the west (your right). I am on the south embankment watching the same train moving to the west (my left). If you and I each use coordinates where the +x direction is to our right then for you the train is moving at +v and for me it is moving at -v. Since we use the same laws of physics to determine the time dilation of the train it is therefore necessary that $\gamma(v)=\gamma(-v)$

 

[PeroK]

Hmm...why is $t" = 0$ not $t = -\frac{vx}{c^2}$? Srry for being so confused guys :D

It's not that I have a physical reason for it, but it is an assumption and assumptions must be justified. In particular, I am thinking about the following step:

$x' = (x-vt) \gamma(v)$
$t' = (t-\frac{vx}{c^2}) \gamma(v)$

It might be clearer if we reserve $v$ for the generic velocity there and use $v = u > 0$ and $v = -u < 0$ for our two frames. Then we have:

$x' = (x-ut) \gamma(u)$
$t' = (t-\frac{ux}{c^2}) \gamma(u)$

$x'' = (x+ut) \gamma(-u)$
$t'' = (t+\frac{ux}{c^2}) \gamma(-u)$

Now, we consider points $x = x_1 > 0$ and $x = x_2 = -x_1 < 0$:

$x'_1 = (x_1-ut_1) \gamma(u)$
$t'_1 = (t_1-\frac{ux_1}{c^2}) \gamma(u)$

$x''_2 = (x_2+ut_2) \gamma(-u)$
$t''_2 = (t_2+\frac{ux_2}{c^2}) \gamma(-u)$

At $t' =0$, we have a relationship between the $x_1$ and $t_1$ coordinates:

$t_1 = \frac{ux_1}{c^2}$

And, at $t'' = 0$, we have a relationship between the $x_2$ and $t_2$ coordinates.

$t_2 = -\frac{ux_2}{c^2} = +\frac{ux_1}{c^2}$

This gives us:

$x'_1 = (x_1-ut_1) \gamma(u) = x_1(1 - \frac{u^2}{c^2})\gamma(u)$

$x''_2 = (x_2+ut_2) \gamma(-u) = x_2(1 - \frac{u^2}{c^2})\gamma(-u) = -x_1(1 - \frac{u^2}{c^2})\gamma(-u)$

Now, we use the physical equivalence of $(t_1, x_1)$ in the S' frame and $(t_2, x_2)$ in the S'' frame and the chosen orientation of all axes to demand that:

$x''_2 = - x'_1$

And this requires $\gamma(u) = \gamma(-u)$

 

[PeroK]

At $t' =0$, we have a relationship between the $x_1$ and $t_1$ coordinates:

$t_1 = \frac{ux_1}{c^2}$

And, at $t'' = 0$, we have a relationship between the $x_2$ and $t_2$ coordinates.

$t_2 = -\frac{ux_2}{c^2} = +\frac{ux_1}{c^2}$

This is interesting as the time coordinates have sorted themselves out and are seen to be equal (as they must by a physical symmetry argument). In other words, a clock at $x_1$ measured in the S' frame is equivalent to a clock at $-x_1$ as measured in the S'' frame.

 

[Kyouran]

Thanks guys, I think I got it now

 

[swampwiz]

Summary:: Assumption in the derivation of the lorentz transformation

In the special theory of relativity, it seems impossible to derive the lorentz transformation without assuming that the lorentz factor is independent of the sign of the relative velocity. For some reason, I can't get my head around why this assumption is so easily made, as if it's trivial. Can anyone explain why this assumption can be made?

At random, point your finger in some direction. Now, is that direction any better of a random direction than any other? Of course not. Similarly, it does not matter what direction a photon is observed moving relative to the observer - and with that postulate and the one that a photon is always observed moving at the constant speed of light, the Lorentz transformation (and the transformations for General Relativity as well) must be location & orientation independent.

 

[ergospherical]

At random, point your finger in some direction. Now, is that direction any better of a random direction than any other?

I pointed my finger towards my lunch and I can confirm that this direction is indeed better than many other directions.

 

[FactChecker]

All of the answers above boil down to using the fact that the laws of physics are isotropic. This means that if you have some experiment, and then you repeat the experiment with everything rotated by $\theta$ the two experiments are physically the same. Equivalently, you can take one experiment and rotate your axes by $\theta$ and use the same laws of physics.

So if you choose $\theta=\pi$ then you immediately have “-x is physically the same as x”

Is the assumption of isotropic supported at relativistic speeds before the experiments like the Michelson-Morley experiment?

 

[Dale]

Is the assumption of isotropic supported at relativistic speeds before the experiments like the Michelson-Morley experiment?

No previous experiment detected any relativistic anisotropy, but the MMX was the first specifically designed to detect it. However, the isotropy assumption was pretty widely held since it was part of non-relativistic mechanics and Galilean invariance.

 

[PAllen]

No previous experiment detected any relativistic anisotropy, but the MMX was the first specifically designed to detect it. However, the isotropy assumption was pretty widely held since it was part of non-relativistic mechanics and Galilean invariance.

Actually, I don't think the aether theory was considered to imply anisotropy. It was considered to be a physically preferred 'frame' in the same way as air on earth. What its frame was would be considered an accident of initial condition (in modern parlance, a symmetry breaking in the origin of the universe). But there would be no difference between moving left, right, north, south etc. relative to the aether, nor would homogeneity be violated - different positions in the aether would be equivalent.

 

[Dale]

Actually, I don't think the aether theory was considered to imply anisotropy. It was considered to be a physically preferred 'frame' in the same way as air on earth. What its frame was would be considered an accident of initial condition (in modern parlance, a symmetry breaking in the origin of the universe). But there would be no difference between moving left, right, north, south etc. relative to the aether, nor would homogeneity be violated - different positions in the aether would be equivalent.

Yes, I agree. It would be considered a practical sort of anisotropy rather than a fundamental one.

 

[bhobba]

Space is isotropic.

That is why I always fall back on the derivation where symmetries are explicitly assumed (I know I have posted it many times before):
http://www2.physics.umd.edu/~yakovenk/teaching/Lorentz.pdf

Thanks
Bill

 

[bhobba]

Yes, I agree. It would be considered a practical sort of anisotropy rather than a fundamental one.

Yes. But in any other frame moving relative to it at a constant velocity, you have an aether wind that breaks isotropy. That's why if you do classical mechanics using the beautiful treatment in Landau Mechanics, while not disproving an aether, you feel like Einstein did when told of Eddingtons results. The best version of that story I know was his assistant was given the results first. Excitedly they showed it to Einstein, who glanced at it then carried on with what he was doing. The assistant was shocked and asked him why he was so calm. Of course, the reply was famous. 'I would feel sorry for the good Lord; the theory is correct.

Thanks
Bill

 

[vanhees71]

That is why I always fall back on the derivation where symmetries are explicitly assumed (I know I have posted it many times before):
http://www2.physics.umd.edu/~yakovenk/teaching/Lorentz.pdf

Thanks
Bill

One has only to add, why the "Euclidean case", $a>0$ is excluded. That's, because we want to establish a causality structure with our spacetime model, i.e., there must be some temporal order for causally connected "events". That implies that the physical symmetry group should be restricted to an "orthochronous subgroup". Now for the Euclidean case for the here considered (1+1)-dimensional spacetime the transformation group is O(2), i.e., the transformation matrix is of the form
$$\hat{T}=\begin{pmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix}.$$
The restriction to orthochronous transformations, i.e., $T_{00}>0$ restricts $\alpha$ to the interval $[-\pi/2,\pi/2]$, but with this restriction the transformation matrices do not form a group, which is why this possibility is ruled out, while $a<0$ leads to the Lorentz group and special-relativistic spacetime (Einstein-Minkowski spacetime) while $a \rightarrow \infty$ leads to the Galilei group and Galilei-Newton spacetime.