Deriving Lorentz transformations

[abhinavjeet]

Why is relative speed taken to be symmetrical i.e. speed of one frame of reference from a second frame is equal to that of the second frame of frame refrence from the first frame

 

[PeroK]

Why is relative speed taken to be symmetrical i.e. speed of one frame of reference from a second frame is equal to that of the second frame of frame refrence from the first frame

Which one do you think should be the greater?

 

[abhinavjeet]

Well I think that if I observe an object moving with velocity v then how can I know how is the object observing me.
Also if the speeds are equal then can I say that in my refrence frame
X=vt
And in his frame X'=vt'
Please forgive me if I m mistaken
I m not so good at physics...

 

[PeroK]

Well I think that if I observe an object moving with velocity v then how can I know how is the object observing me.
Also if the speeds are equal then can I say that in my refrence frame
X=vt
And in his frame X'=vt'
Please forgive me if I m mistaken
I m not so good at physics...

That wasn't the question. The question is:

If you observe an object moving with velocity $v$ does that object measure your velocity as $> v$ or $< v$? (Assuming it's not $= v$)

 

[abhinavjeet]

I don't have any reason to prefer >v or <v

 

[PeroK]

I don't have any reason to prefer >v or <v

Then it must be $=v$.

Let's say you see an object moving towards you at speed $v$ and it sees you moving towards it at speed $v_1 > v$. Then, unless there is a lack of symmetry in the universe, you must see it moving towards you at $v_2 > v_1$. That can't be. You have the same argument if $v_1 < v$

The symmetry of the universe implies equality of relative velocities.

 

[Dale]

I don't have any reason to prefer >v or <v

Yes, exactly. This is called a symmetry principle.

 

[bcrowell]

I have a discussion of this in my SR book http://lightandmatter.com/sr/ , section 1.4, example 11, "Observers agree on their relative speeds."

 

[DrSirius]

I have a discussion of this in my SR book http://lightandmatter.com/sr/ , section 1.4, example 11, "Observers agree on their relative speeds."

The book seems excellent. I am glad to have found about that example, it is the first time that I see, explicitly stated, that the question is not completely obvious or trivial. A teacher of mine solved it by printing a paper and fliping it the other side to show the symmetry. IMHO, that is not completely trivial and the matter is a little more subtle than that.

 

[sweet springs]

Let velocity of IFR 2 measured in IFR 1 be $$V_{12}$$.
Let velocity of IFR 1 measured in IFR 2 be $$V_{21}$$.
Let X be axis along to the velocity. Changing the direction of X, V_12 also changes its sign.
Chanbing the direction of X, IFR 1 and IFR 2 exchange in the sense that one is along X and the other is reverse.
So we may have a reason to say $$V_{12}=-V_{21}$$.

 

[strangerep]

Afaik, one only needs the following assumptions:

1) The usual relativity principle,

2) An assumption of 3D spatial isotropy,

3) An assumption that the axes transverse to the boost direction remain unchanged (as usual when deriving Lorentz transformations), [Edit: meaning that the rotational freedom allowed by spatial isotropy is exploited to align the transverse axes of the original and transformed frames.]

4) An assumption that velocity boost transformations form a 1-parameter Lie semigroup (possibly a group).

From (1)-(4) one can derive that, for a boost transformation with parameter $v$, the inverse transformation must have parameter $-v$. Additional hand waving intuition is not necessary.

[Edit: I modified this post slightly in an attempt to soften my previous tone.]

 

[bcrowell]

3) An assumption that the axes transverse to the boost direction remain unchanged (as usual when deriving Lorentz transformations),

This isn't a necessary assumption. This fact can be derived.

4) An assumption that velocity boost transformations form a 1-parameter Lie semigroup (possibly a group).

From (1)-(4) one can derive that, for a boost transformation with parameter $v$, the inverse transformation must have parameter $-v$. Additional hand waving intuition is not necessary.

By assuming a Lie group you are assuming what you claim to prove.To justify this assumption, you will find yourself having to reason about physics. Physical reasoning does not equate to "hand waving intuition."

[Mentors note: This post has been edited as part of some overall thread moderation]

 

[strangerep]

3) An assumption that the axes transverse to the boost direction remain unchanged (as usual when deriving Lorentz transformations),

This isn't a necessary assumption. This fact can be derived.

I suspect we are talking at crossed purposes on this point. Please see my edit in post #11. If that doesn't put the discussion back on track, then please explain how this is "derived". (I did look in your book, but maybe I wasn't looking in the right place.)

By assuming a Lie group you are assuming what you claim to prove.

I do not see this. What do you think I "claim to prove".

To justify this assumption [Lie (semi)group], you will find yourself having to reason about physics.

Sure.

 

[bcrowell]

@strangerep: Sorry for my grumpy tone in my previous post.

I suspect we are talking at crossed purposes on this point. Please see my edit in post #11. If that doesn't put the discussion back on track, then please explain how this is "derived". (I did look in your book, but maybe I wasn't looking in the right place.)

I see. From your original version, I thought you were talking about an assumption that there is no transverse Lorentz contraction. From your edited version, I can see that you just mean there is no rotation between the two frames.

I do not see this. What do you think I "claim to prove".

Maybe I'm misunderstanding this as well. Is the complete argument written up somewhere?

In #4, you only assume a semigroup, but then at the end you refer to "the inverse transformation." Are you claiming that based on assumptions 1-4 you can prove that it's a group, and not just a semigroup?

My general comments on the outline you've presented are:

(1) I don't see the point of doing this in 3+1 dimensions rather than 1+1. Any successful argument is going to require an assumption of spatial isotropy, but that assumption can be expressed in one spatial dimension, where it's equivalent to parity symmetry.

(2) Proving some group-theoretical facts doesn't establish anything about the physics unless you provide some "glue" between the math and the physics, which you haven't done. The mathematical assumptions require physical justification, and the mathematical results require physical interpretation. Using fancy math may give the superficial impression of rigor, but fancy math without the glue is actually less rigorous than simple math with the glue. In any case, all you seem to be claiming to prove is that "for a boost transformation with parameter v, the inverse transformation must have parameter −v." This is easy to prove in even the simplest treatment of SR using high school algebra. The gee-whiz stuff about Lie semigroups comes off to me as obscurantism.

 

[strangerep]

@strangerep: Sorry for my grumpy tone in my previous post.

Thank you. Actually, I see now that (the original version of) my post could indeed be perceived as supercilious. I take your remarks on board for the future.

Maybe I'm misunderstanding this as well. Is the complete argument written up somewhere?

My writeup of the complete argument is still not in final form (and no one else has yet proofread the draft). Probably, I should keep my mouth shut until this occurs, but I don't always have the necessary self control.

In #4, you only assume a semigroup, but then at the end you refer to "the inverse transformation." Are you claiming that based on assumptions 1-4 you can prove that it's a group, and not just a semigroup?

I believe so, yes. (Strictly speaking, there's a couple of other inputs: i.e., that the original and transformed origins coincide, and some physical input to establish that the effect of the transformations does indeed correspond to a natural interpretation of what "velocity boost" means in terms of coordinates.)

(1) I don't see the point of doing this in 3+1 dimensions rather than 1+1. Any successful argument is going to require an assumption of spatial isotropy, but that assumption can be expressed in one spatial dimension, where it's equivalent to parity symmetry.

I wanted to investigate the most general case, to be sure I wasn't accidentally overlooking something.

(2) Proving some group-theoretical facts doesn't establish anything about the physics unless you provide some "glue" between the math and the physics, which you haven't done. The mathematical assumptions require physical justification, and the mathematical results require physical interpretation. Using fancy math may give the superficial impression of rigor, but fancy math without the glue is actually less rigorous than simple math with the glue.

I agree with all of the above. One of the reasons I haven't finished my writeup is that it's challenging to do this well, and without subtly introducing extra assumptions.

In any case, all you seem to be claiming to prove is that "for a boost transformation with parameter v, the inverse transformation must have parameter −v." This is easy to prove in even the simplest treatment of SR using high school algebra. The gee-whiz stuff about Lie semigroups comes off to me as obscurantism.

Well, I'm trying to write a treatise covering the most general (fractional-linear) case, with as few assumptions as possible. I decided to be more careful about the semigroup stuff when I investigated time displacement transformations in the same framework. The advanced investigations along related lines that I'm aware of always seem to assume time reversal symmetry. (E.g., Bacry & Levy-LeBlond, "Possible Kinematics", JMP vol 9, 1968, p1605.) Relaxing that assumption, one must investigate the physical domain (in velocity phase space) on which all the transformations are well-defined, and ask on what domain the inverse transformations are also well-defined. I found some interesting consequences (beyond the scope of this thread), and this motivated me to relax the group assumption for other transformations such as boosts and spatial displacements, and investigate how much could be derived starting from only a semigroup assumption. Whether this is perceived as "obscurantism" depends on one's interests, I suppose.

 

[Erland]

Symmetry, spatial isotropy, how are these principles formulated in a mathematically stringent way?

Early in this thread, it was claimed that the relative speeds $v_1$ of Frame 1 w.r.t. Frame 2 and $v_2$ of Frame 2 w.r.t. Frame 1 must be equal because of "symmetry". But then, suppose an object is at rest w.r.t Frame 1. Then, it has speed $v_1>0$ w.r.t Frame 2. Why doesn't this violate symmetry?

Also, I am reading bcrowells interesting book and looked up the example he mentioned, but I don't understand how "spatial isotropy" is applied in that case.

 

[strangerep]

Symmetry, spatial isotropy, how are these principles formulated in a mathematically stringent way?

Let's take spatial isotropy as an example. Heuristically, this means there is no distinguished (or "preferred") direction in 3-space. In mathematically more precise terms, I'd express it as follows:

A theory is spatially isotropic iff every equation of the theory remains invariant when all quantities in the equation are substituted by their (respective) rotated counterparts. To understand this in more detail, let's consider an equation in a theory, written as $$F(t,{\bf x}, {\bf v}, ...) ~=~ 0 ~~~~~~ (1) ~,$$ where the bold symbols denote 3-vectors. Now consider an arbitrary 3D rotation matrix $\bf M$, and substitute all 3-vectors in (1) by their rotated counterparts, i.e., ${\bf Mx}, {\bf Mv}$, etc. So we have $$F(t,{\bf Mx}, {\bf Mv}, ...) ~=~ 0 ~~~~~~ (2) ~.$$ If (2) is equivalent to (1), i.e., if by purely algebraic manipulations we can make (2) look exactly like (1), then (1) is called a spatially isotropic equation. If every equation in the theory has this property, then the theory has the property of spatially isotropy.

A similar idea applies to other symmetries: if we can take any equation in the theory, substitute the various quantities therein by their transformed counterparts, and by purely algebraic manipulations make the new equation look exactly like the original, then the theory has that symmetry.

Sometimes it's obvious that an equation is spatially isotropic, e.g., if it contains only scalar expressions explicitly, like (say) ${\bf x \cdot \bf p}$.

BTW, although I've only explained the case with 3-vectors explicitly, the same idea applies if there are higher rank tensors (or spinors) in the theory -- one simply has to use the correct rotation transformation formula for each type of quantity appearing.

 

[strangerep]

Early in this thread, it was claimed that the relative speeds $v_1$ of Frame 1 w.r.t. Frame 2 and $v_2$ of Frame 2 w.r.t. Frame 1 must be equal because of "symmetry". But then, suppose an object is at rest w.r.t Frame 1. Then, it has speed $v_1>0$ w.r.t Frame 2. Why doesn't this violate symmetry?

By introducing "an object" (which I'll denote by $O$), you must now consider 4 relative velocities: $v_{12}$, $v_{21}$, $v_{O1}$, $v_{O2}$. (Here, my notation $v_{AB}$ denotes velocity of "A" relative to "B". More precisely, $v_{12}$ denotes the velocity of the origin in Frame 1 relative to the origin of Frame 2, where we assume the origins coincide momentarily.)

We have $v_{21} = -v_{12}$ by spatial isotropy (or rather, parity reversal in the 1+1D case), but that (by itself) doesn't say anything about the speed of O relative to either frame. So even if $v_{O1} = 0$, it doesn't necessarily mean $v_{O2} = 0$, since $v_{12} \ne 0$.

 

[Erland]

Let's take spatial isotropy as an example. Heuristically, this means there is no distinguished (or "preferred") direction in 3-space. In mathematically more precise terms, I'd express it as follows:

A theory is spatially isotropic iff every equation of the theory remains invariant when all quantities in the equation are substituted by their (respective) rotated counterparts. To understand this in more detail, let's consider an equation in a theory, written as $$F(t,{\bf x}, {\bf v}, ...) ~=~ 0 ~~~~~~ (1) ~,$$ where the bold symbols denote 3-vectors. Now consider an arbitrary 3D rotation matrix $\bf M$, and substitute all 3-vectors in (1) by their rotated counterparts, i.e., ${\bf Mx}, {\bf Mv}$, etc. So we have $$F(t,{\bf Mx}, {\bf Mv}, ...) ~=~ 0 ~~~~~~ (2) ~.$$ If (2) is equivalent to (1), i.e., if by purely algebraic manipulations we can make (2) look exactly like (1), then (1) is called a spatially isotropic equation. If every equation in the theory has this property, then the theory has the property of spatially isotropy.

Very good, but this then leads to the question: Which are the equations in the theory to which this applies? This should be specified in a stringent exposition.

Also I wonder, how can this be used to prove that, for example, distances in directions perpendicular to the direction of motion between the frames are not changed by the transformation? (In this case, only rotations fixing the direction of motion should be used above.)

And how is it used to prove that $v_{12}=-v_{21}$?

 

[Dale]

Early in this thread, it was claimed that the relative speeds $v_1$ of Frame 1 w.r.t. Frame 2 and $v_2$ of Frame 2 w.r.t. Frame 1 must be equal because of "symmetry". But then, suppose an object is at rest w.r.t Frame 1. Then, it has speed $v_1>0$ w.r.t Frame 2. Why doesn't this violate symmetry?

The presence of the object breaks the symmetry. This is not because of asymmetry in the laws, but asymmetry in the boundary conditions.

 

[Erland]

The presence of the object breaks the symmetry. This is not because of asymmetry in the laws, but asymmetry in the boundary conditions.

Can you, in a stringent way, state the law(s) / symmetry principle(s) used here, from which it follows that the relative speed between the frames is the same w.r.t both frames...?

 

[PeroK]

Very good, but this then leads to the question: Which are the equations in the theory to which this applies? This should be specified in a stringent exposition.

Also I wonder, how can this be used to prove that, for example, distances in directions perpendicular to the direction of motion between the frames are not changed by the transformation? (In this case, only rotations fixing the direction of motion should be used above.)

And how is it used to prove that $v_{12}=-v_{21}$?

It seems to me that trying to answer his question requires a mixture of mathematical axioms and a physical explanation of why those axioms are adopted. For SR, for example, there is no problem is simply taking as an axiom the differential geometry of spacetime. Spatial symmetry and isotropy come along with the axiom.

But, if you are trying to justify why space is isotropic, then what do you take as your axioms?

 

[Erland]

It seems to me that trying to answer his question requires a mixture of mathematical axioms and a physical explanation of why those axioms are adopted.

Yes, I think so too.

For SR, for example, there is no problem is simply taking as an axiom the differential geometry of spacetime. Spatial symmetry and isotropy come along with the axiom.

I don't understand what you mean. How does "the differential geometry of spacetime" imply the isotropy w.r.t the Lorentz transformation?

But, if you are trying to justify why space is isotropic, then what do you take as your axioms?

Isotropy here means, I suppose, that the Lorentz transformation is somehow invariant w.r.t. rotations. I don't think it can be really justified by more "basic" physical principles, it just seems obvious. The problem is to formulate it stringently and economically. I have been considering something like this:

We assume that the two frames, called the "stationary" and the "moving" frame, move w.r.t each other along their x- and x'-axes, respectively, and that L(0,0,0,0)=(0,0,0,0). Consider the events 1, 2, and 3, with coordinates (0,1,0,0), (0,0,1,0), and (0,-1,0,0), respectively, in the stationary frame. The spatial vectors (x_i,y_i,z_i) of these events (i=1,2,3) in the stationary frame are perpendicular to the direction of relative motion of the frames, they have the same lengths (1) and the angles between these vectors of events 1 and 2, and 2 and 3, respectively, are the same (90 degrees). Also, the events are simultaneous in the stationary frame. Let (x_i',y_i',z_i',t_i'), i=1,2,3, be the coordinates of the events 1,2, and 3, respectively, in the moving frame.
Given this, we now assume that the (squares of) the lengths (x_i')²+(y_i')²+(z_i')² are equal, that the x_i':s are equal, and that the t_i':s are equal, (i=1,2,3), and that the angles between the spatial vectors (x_i',y_i',z_i'), for the pairs 1,2 and 2,3 respectively, are equal.
All this is justified by "isotropy". Using this and linearity of the Lorentz transform L, it is not hard prove that, after a suitable rotation of the spatial coordinates in the moving frame, there is a constant a such that if L(x,y,z,t)=(x',y',z',t'), then y'=ay and z'=az.

But I am not entirely satisfied with this. It feels a little bit too cumbersome. Can someone come up with something better, simpler, and more beautiful?

 

[Dale]

Can you, in a stringent way, state the law(s) / symmetry principle(s) used here, from which it follows that the relative speed between the frames is the same w.r.t both frames...?

In a stringent way, no. In a rough handwaving way:

Given inertial reference frames A and B with B moving at speed v wrt A then, by the principle of relativity there is nothing which distinguishes A from B, therefore by the principle of relativity we can exchange A and B and state that we have A moving at speed v wrt B.

 

[PeterDonis]

How does "the differential geometry of spacetime" imply the isotropy w.r.t the Lorentz transformation?

The "stringent" way of talking about isotropy is by means of Killing vector fields. Saying that a spacetime is "isotropic" means that, at every event in the spacetime, there is a 3-parameter group of spacelike Killing vector fields that have the commutation relations of the rotation group SO(3).

Isotropy here means, I suppose, that the Lorentz transformation is somehow invariant w.r.t. rotations.

The connection with Lorentz transformations, in the "stringent" way of talking that I describe above, is that the rotation group SO(3) is a subgroup of the Lorentz group SO(3, 1), which is a six-parameter group of Killing vector fields having three spacelike generators (generating the SO(3) subgroup) and three timelike generators (generating "boosts", which do not form a subgroup because the commutator of two boosts is a rotation--physically, this shows up as Thomas precession). So Lorentz transformations preserve the same invariants that are preserved by the spatial rotation group SO(3).

More info here:

https://en.wikipedia.org/wiki/Lorentz_group

 

[Erland]

PeterDonis, this seems quite advanced. I need to study this more to fully understand it. But isn't it much simpler in the SR case, where we have linearity and flatness?

Anyway, is it correct that in defining the Lorentz group, one assumes that the spacetime distance dx²+dy²+dz²-c²dt² is preserved by the Lorentz transformations?
I am not willing to take this as an axiom (for physics), except when this distance is 0, since this is just another way of stating the invariance of the light speed. Otherwise, it seems in no way obvious and no simple consequence of Einstein's postulates.

 

[PeterDonis]

isn't it much simpler in the SR case, where we have linearity and flatness?

No. The Killing vector fields I described are for the SR case, flat Minkowski spacetime.

is it correct that in defining the Lorentz group, one assumes that the spacetime distance dx²+dy²+dz²-c²dt² is preserved by the Lorentz transformations?

It is often presented that way, but IIRC you don't actually have to make that assumption. The Lorentz group can be defined by the Killing vector fields that generate it and their commutation relations. Showing that the group of transformations on spacetime that have those commutation relations preserves the spacetime interval can then, IIRC, be derived as a theorem.

As a warmup exercise, consider the simpler case of the rotation group SO(3). This group preserves ordinary Euclidean distances in Euclidean 3-space, i.e., it preserves $ds^2 = dx^2 + dy^2 + dz^2$. But I don't think you have to assume that in defining the group; I think you can define the group by its commutation relations, and then derive as a consequence the fact that the transformations in the group preserve Euclidean distances.

I am not willing to take this as an axiom (for physics), except when this distance is 0, since this is just another way of stating the invariance of the light speed.

You can construct a basis for Minkowski spacetime using only null vectors, i.e., any vector can be expressed as a linear combination of null vectors, so invariance of null intervals is sufficient to establish invariance of all intervals.

 

[Dale]

You can construct a basis for Minkowski spacetime using only null vectors, i.e., any vector can be expressed as a linear combination of null vectors, so invariance of null intervals is sufficient to establish invariance of all intervals.

Wow, excellent point.

How does that translate to curved spacetime where the basis vectors would only span the tangent space and not spacetime which is a manifold and not a vector space?

 

[PeterDonis]

How does that translate to curved spacetime where the basis vectors would only span the tangent space and not spacetime which is a manifold and not a vector space?

The tangent space is still Minkowski spacetime, so as long as we are just looking at infinitesimal line elements, everything works the same way.

Once we go beyond infinitesimal line elements, there is, as you know, no such thing as a global "Lorentz transformation" in curved spacetime. Any coordinate transformation must still preserve arc lengths along curves (and other geometric invariants), but there are no global coordinate charts that have all the properties of inertial charts on Minkowski spacetime. So the viewpoint that Lorentz transformations are defined as the ones that preserve spacetime intervals doesn't really have an analogue, globally, in curved spacetime.

This, btw, is another reason to learn the definition of isotropy (and other symmetries) in terms of Killing vector fields; those definitions carry over just fine to curved spacetime. An isotropic curved spacetime is simply one which has a 3-parameter group of KVFs at every event with the commutation relations of SO(3). That's all there is to it.

 

[Erland]

No. The Killing vector fields I described are for the SR case, flat Minkowski spacetime.

OK, sorry for my misunderstanding. I know I should study differential geometry more, but I have no good book about this avaliable for the moment. I want to thank you all for trying to enlighten me. I don't mean to be difficult, but I do question things I don't find obvious.

It is often presented that way, but IIRC you don't actually have to make that assumption. The Lorentz group can be defined by the Killing vector fields that generate it and their commutation relations. Showing that the group of transformations on spacetime that have those commutation relations preserves the spacetime interval can then, IIRC, be derived as a theorem.

As a warmup exercise, consider the simpler case of the rotation group SO(3). This group preserves ordinary Euclidean distances in Euclidean 3-space, i.e., it preserves $ds^2 = dx^2 + dy^2 + dz^2$. But I don't think you have to assume that in defining the group; I think you can define the group by its commutation relations, and then derive as a consequence the fact that the transformations in the group preserve Euclidean distances.

Maybe I should know this, but what do you mean by "commutation relations"? Do you simply mean relations of the type AB=BA, or A^-1B^-1AB=I, where A and B are rotations (in this case)? Or do you mean the set of all commutators: A^-1B^-1AB, or the subgroup generated by these (the commutator subgroup) given without explicitly displaying the A:s and B:s forming the commutators, or do you mean something else?

You can construct a basis for Minkowski spacetime using only null vectors, i.e., any vector can be expressed as a linear combination of null vectors, so invariance of null intervals is sufficient to establish invariance of all intervals.

I find it in no way obvious that just because the Minkowski lengths of x and y are preserved by L, the same is true for all linear combinations of x and y. It is of course true, but I know that just because I already know what L looks like.