Most fundamental reason why boosts have to have L(-v)L(v)=I?

In summary: This is a pretty complicated example, but it seems to me that you would still need the principle of relativity to ensure that the coordinates you assigned to events would match the coordinates that I would assign.
  • #1
bcrowell
Staff Emeritus
Science Advisor
Insights Author
Gold Member
6,724
429
This question may be one of those questions that is hard to talk about because it seems so obvious that a certain thing has to be a certain way that it becomes impossible to figure out how you know that it has to be that way.

When you derive the Lorentz transformations from a certain set of axioms (possibly Einstein's 1905 axioms, or possibly something less archaic), a typical piece of input to the calculation is that when we do a boost by v, followed by a boost by -v, we have to get an identity transformation: L(-v)L(v)=I.

What is the most fundamental reason why this has to be this way, and how is this derivable strictly from your favorite axiomatization without having to bring in some other ad hoc principle?

You can certainly get some silly results if you don't require this. For example, suppose that I let [itex]L(-v)L(v)=(1+v^2)I[/itex], i.e., a dilation of both time and space by the same factor. This leaves the speed of light (which equals 1) invariant. I could oscillate a meter-stick to the left and right, and as time went on, the meter-stick's length would grow exponentially. This violates time-reversal invariance, but other than that it's not obviously logically impossible. For example, in the Weyl gauge theory, the rate of a clock depends on its past history of motion. It's nutty, but it's not obviously logically impossible.

-Ben
 
Physics news on Phys.org
  • #2
Isn't it necessary that there is one and only one (oriented) inertial frame for each state of motion?
If so, isotropy dictates that L(-v) transforms back to the initial state of motion and thus to the initial frame.
 
  • #3
Consider 1+1 dimensions first. If your velocity relative to me is v, and our x axes point in the same direction, then my velocity relative to you must be -v, "by the principle of relativity". If I can find out what coordinates you assign to events by applying a function to the coordinates I would assign, you must be able to find out what coordinates I would assign, by applying a function to the coordinates you would assign, again "by the principle of relativity". If we call those functions Λ and Λ', then for consistency, we must have Λ'=Λ-1.

The words "by the principle of relativity" are in quotes because the principle of relativity isn't a mathematical statement, but our conclusions are. Since it isn't possible to prove mathematical theorems from non-mathematical axioms, I would say that we didn't really derive our result from the principle of relativity. I would say that the intermediate results are the axioms in the derivation, because they are mathematical statements that make certain aspects of the principle of relativity precise.

To go from Λ'=Λ-1 to Λ(-v)=Λ(v)-1, we need to know that the coordinate change function is completely determined by the velocity v. We need additional assumptions for that. See e.g. what I did here. (Start reading at "The explicit formula...").

In 3+1 dimensions, it's not true in general that Λ(-v)=Λ(v)-1 (since a rotation can be involved), but since you asked specifically about boosts, you have reduced the problem to the 1+1-dimensional one.
 
  • #4
I believe that if that were not the case then the Poincare group would not actually be a group, which I think is essentially the mathematical statement of the first postulate. Of course, then you can ask what is the fundamental reason for the Poincare group, but symmetries are considered the most fundamental answers.
 
  • #5
I would just go with a physical intent argument:

Given physics in frame x, L(x) describes physics in frame x'. The purpose of L-inv is to map from physics in x' to physics in x. Therefore one requires that L-inv(L(x)) = I

Similarly, given physics in x', you want L-inv(x') to be physics in x. Thne you want L(L-inv(x'))=I to be physics in x'.

So, to me, it is a physically motivated additional axiom.
 
  • #6
Thanks, all, for your interesting replies.

Ich said:
Isn't it necessary that there is one and only one (oriented) inertial frame for each state of motion?
This seems like a reasonable requirement, but I don't think I've seen it explicitly stated as an assumption or axiom in derivations -- maybe because it seems too obvious to give explicitly? The Weyl gauge theory does violate this requirement.

Fredrik said:
If I can find out what coordinates you assign to events by applying a function to the coordinates I would assign, you must be able to find out what coordinates I would assign, by applying a function to the coordinates you would assign, again "by the principle of relativity".

It seems to me that this doesn't follow just from the principle of relativity. I think you need the principle of relativity plus Ich's principle that a frame is uniquely determined by one's state of motion. In the Weyl gauge theory, two rulers, A and B, that were initially calibrated side by side against one another could be moved around in different ways, and disagree when reunited. Suppose observer Alice has traveled around with ruler A while Betty has traveled with ruler B. Alice and Betty are then reunited with the same state of motion, but their rulers disagree. We then want transformations that go between Alice and Betty's coordinates and the coordinates of some third observer Charlie, who is in motion relative to Alice and Betty with velocity v. The transformation from A to C is different from the transformation from B to C, so the transformation does not just depend on v. It also depends on the history of motion of each of the three observers.

DaleSpam said:
I believe that if that were not the case then the Poincare group would not actually be a group, which I think is essentially the mathematical statement of the first postulate. Of course, then you can ask what is the fundamental reason for the Poincare group, but symmetries are considered the most fundamental answers.
It could be a group, but just a bigger group, labeled by more parameters than just v. Its elements might also need labels that would describe something about the observer's history of motion. In the Weyl gauge theory, I believe that what's relevant about the history of motion is a single real number [itex]f=\int A_b dx^b[/itex], where A is the electromagnetic potential. So then we would have not L(v) but L(v,f).

To be a little more concrete, here's my own favorite axiomatization of SR:

A1. No point in time or space has properties that make it different from any other point.
A2. Likewise, all directions in space have the same properties.
A3. Motion is relative, i.e., all inertial frames of reference are equally valid.
A4. Causality holds.
A5. Time depends on the state of motion of the observer.

These are logically equivalent to Einstein's 1905 postulates.

It seems to me that we can't prove L(-v)L(v)=I without assuming a sixth axiom. Two possible choices would be:

A6. "There is one and only one (oriented) inertial frame for each state of motion." -- Ich
A6'. Time-reversal symmetry holds.

To me, A6' actually seems somewhat more appealing than A6, since it restores the symmetrical treatment of space and time that is otherwise broken by A2. On the other hand, A6' is stronger than A6, and may rule out interesting physical theories that would be allowed by A6 (e.g., ones in which the Copenhagen model's collapse of wavefunctions is really time-asymmetric as proposed by Penrose, or ones in which black hole evaporation breaks unitarity).

-Ben
 

Related to Most fundamental reason why boosts have to have L(-v)L(v)=I?

1. What is the concept of boosts in physics?

Boosts, also known as Lorentz transformations, are mathematical operations that describe how an observer's measurements of space and time change when transitioning between different inertial frames of reference. In simpler terms, they explain how an object appears to move differently when observed from different perspectives.

2. Why is it necessary for boosts to follow the equation L(-v)L(v)=I?

This equation is a fundamental property of boosts and is known as the composition law. It states that a boost in one direction followed by a boost in the opposite direction will result in no overall change, or the identity transformation (I). This is necessary because it ensures that the laws of physics remain consistent regardless of which inertial frame of reference they are observed from.

3. What does the equation L(-v)L(v)=I represent?

The equation L(-v)L(v)=I is a mathematical representation of the symmetry of space and time. It shows that there is no preferred direction or reference frame in the universe, and that all observers will measure the same physical laws and phenomena regardless of their relative motion.

4. How does the equation L(-v)L(v)=I relate to Einstein's theory of relativity?

Einstein's theory of relativity is based on the principle of relativity, which states that the laws of physics are the same for all inertial observers. The equation L(-v)L(v)=I is a consequence of this principle and is a fundamental aspect of special relativity. It also plays a crucial role in Einstein's famous equation, E=mc^2, which describes the relationship between energy, mass, and the speed of light.

5. Are there any exceptions to the equation L(-v)L(v)=I?

No, the equation L(-v)L(v)=I holds true for all boosts and is a universal principle in physics. However, it is important to note that it only applies to inertial frames of reference, where there is no acceleration or gravitational forces acting on the objects being observed. In non-inertial frames, other equations and principles may apply.

Similar threads

  • Special and General Relativity
Replies
33
Views
2K
  • Special and General Relativity
2
Replies
57
Views
4K
  • Special and General Relativity
Replies
5
Views
984
  • Electrical Engineering
Replies
4
Views
2K
  • Special and General Relativity
2
Replies
55
Views
3K
  • Special and General Relativity
Replies
1
Views
1K
  • Special and General Relativity
3
Replies
75
Views
3K
  • Special and General Relativity
Replies
13
Views
4K
  • Special and General Relativity
Replies
1
Views
987
  • Special and General Relativity
Replies
2
Views
1K
Back
Top