# Symmetry assumption in derivation of SR

Gold Member
I would like to ask if anybody knows about some analysis of this part in Einstein's derivation of SR where he gets rid of unknown scaling function φ(v):

"From reasons of symmetry it is now evident that the length of a given rod moving perpendicularly to its axis, measured in the stationary system, must depend only on the velocity and not on the direction and the sense of the motion. The length of the moving rod measured in the stationary system does not change, therefore, if v and −v are interchanged. Hence follows that l/φ(v) = l/φ(−v), or
φ(v) = φ(−v)."

As it seems to me these reasons of symmetry are basically assumption that there is no preferred reference frame and transformations between different reference frames should be completely symmetric without any reference to some absolute state of motion.

Matterwave
Gold Member
That should mean that the velocity's direction doesn't matter in the transformations. This is isotropy of space. Homogeneity of space would imply that terms like φ(x,v) cannot occur since every point in space is the same as any other.

Gold Member
That should mean that the velocity's direction doesn't matter in the transformations.
That's right.
This is isotropy of space. Homogeneity of space would imply that terms like φ(x,v) cannot occur since every point in space is the same as any other.
No this is not about isotropy of space. Scaling function is φ(v) not φ(x).
This is isotropy of spacetime i.e. no matter what space slice of spacetime we take as simultaneous we should observe isotropy in that slice. But this is not exactly what we observe, right?

Fredrik
Staff Emeritus
Gold Member
That's right.

No this is not about isotropy of space. Scaling function is φ(v) not φ(x).
This is isotropy of spacetime i.e. no matter what space slice of spacetime we take as simultaneous we should observe isotropy in that slice. But this is not exactly what we observe, right?
It's isotropy of space for all observers. That may or may not be equivalent to isotropy of spacetime, I'm not going to think about that now.

You may find this post useful. It's just about the 1+1-dimensional case, but it still contains some useful ideas about how to think about these "derivations" and the mathematical assumptions that go into them.

You're of course right that we don't observe exact isotropy, but the point of these "derivations" isn't to find the ultimate theory of everything, including my shoes. The goal is to learn how one can find out if there's a theory in which each non-accelerating observer is associated with a coordinate system such that the functions that represent a change from one of those coordinate systems to another is a group.

Matterwave
Gold Member
Time is not isotropic...not all processes are time-reversal invariant...and we always experience time going forward. But I don't believe you need time to be isotropic to derive that φ should not depend on the direction of travel, but only on the magnitude of the velocity.

samalkhaiat
I would like to ask if anybody knows about some analysis of this part in Einstein's derivation of SR where he gets rid of unknown scaling function φ(v):

"From reasons of symmetry it is now evident that the length of a given rod moving perpendicularly to its axis, measured in the stationary system, must depend only on the velocity and not on the direction and the sense of the motion. The length of the moving rod measured in the stationary system does not change, therefore, if v and −v are interchanged. Hence follows that l/φ(v) = l/φ(−v), or
φ(v) = φ(−v)."

As it seems to me these reasons of symmetry are basically assumption that there is no preferred reference frame and transformations between different reference frames should be completely symmetric without any reference to some absolute state of motion.
See posts # 9 & 14 in

Sam

Gold Member
It's isotropy of space for all observers. That may or may not be equivalent to isotropy of spacetime, I'm not going to think about that now.

You may find this post useful. It's just about the 1+1-dimensional case, but it still contains some useful ideas about how to think about these "derivations" and the mathematical assumptions that go into them.

You're of course right that we don't observe exact isotropy, but the point of these "derivations" isn't to find the ultimate theory of everything, including my shoes. The goal is to learn how one can find out if there's a theory in which each non-accelerating observer is associated with a coordinate system such that the functions that represent a change from one of those coordinate systems to another is a group.
I would say that adding scaling function (φ(v)≠1) to transformation will give a group just the same.

But I was thinking if it is worth discussing this assumption. Obviously transformations between inertial observers are more restrictive with this assumption of isotropy. So if after some time we will find out that SR predictions deviate from experimental results under some conditions we can relax this assumption and see if it helps. Until then there is not much point in fiddling with this assumption.

And yet when thinking where such generalized transformation can be used I started to wonder if gravitating objects can be described using it. And it looks interesting. Say if you accelerate gravitating object then it's expansion (after adding energy into the system) will reduce gravitational potential and effect of gravity inside the body.

Fredrik
Staff Emeritus
Gold Member
I would say that adding scaling function (φ(v)≠1) to transformation will give a group just the same.
Yes, it certainly does. I was a bit sloppy there. Even if we add the assumption that the coordinate change functions take straight lines to straight lines, we're not getting rid of that factor. Another assumption is needed. There are probably lots of different assumptions that would do the job. In my 1+1-dimensional "derivation", I assumed that γ was an even function of v, and said that this is motivated by the principle of relativity. If it's not even, then statements that two observers make about the ticking rates of each others clocks won't be symmetrical.

I think these statements are more accurate than the one you quoted:
One way to find the Lorentz transformations is to consider a family of global coordinate systems $\{x_i:\mathbb R^4\rightarrow\mathbb R^4|i\in I\}$ such that each "coordinate change" function $x_i\circ x_j^{-1}$ takes straight lines to straight lines, and the set $\big\{x_i\circ x_j^{-1}|i,j\in I\big\}$ is a group, with a subgroup that's isomorphic to the translation group, and another that's isomorphic to the rotation group.

To drop the requirement of isotropy or homogeneity would be to drop the requirement that the group of coordinate change functions must have the appropriate subgroups. I don't know what we get if you do. I suppose that we don't want to completely drop them, but rather weaken them somewhat, so that the group almost has the translation and rotation groups as subgroups. I don't even know how to make that statement precise.
You can't actually derive them from Einstein's postulates. You derive them from mathematical statements that can be thought of as expressing aspects of Einstein's postulates mathematically. Are homogeneity and isotropy such aspects, or are they separate assumptions? I think that's actually a matter of taste. Einstein's postulates aren't very precise, so you can interpret them in more than one way.

Gold Member
Another assumption is needed. There are probably lots of different assumptions that would do the job. In my 1+1-dimensional "derivation", I assumed that γ was an even function of v, and said that this is motivated by the principle of relativity. If it's not even, then statements that two observers make about the ticking rates of each others clocks won't be symmetrical.
I agree with you - assumption follows from principle of relativity. So I can't see any clear reason to talk about separate assumption.
Thanks for helping to clear that up for me.