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Which one do you think should be the greater?

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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...

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That wasn't the question. The question is:

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...

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

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

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Then it must be ##=v##.I don't have any reason to prefer >v or <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.

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Dale

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Yes, exactly. This is called a symmetry principle.I don't have any reason to prefer >v or <v

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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.

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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 $$V_{12}=-V_{21}$$.

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strangerep

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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.]

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.]

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This isn't a necessary assumption. This fact can be derived.3) An assumption that the axes transverse to the boost direction remain unchanged (as usual when deriving Lorentz transformations),

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

From (1)-(4) one canderivethat, for a boost transformation with parameter ##v##, the inverse transformationmusthave parameter ##-v##. Additional hand waving intuition is not necessary.

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

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- #13

strangerep

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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.)This isn't a necessary assumption. This fact can be derived.strangerep said:3) An assumption that the axes transverse to the boost direction remain unchanged (as usual when deriving Lorentz transformations),

I do not see this. What do you think I "claim to prove".By assuming a Lie group you are assuming what you claim to prove.

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

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

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.

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 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.)

Maybe I'm misunderstanding this as well. Is the complete argument written up somewhere?I do not see this. What do you think I "claim to prove".

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.

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- #15

strangerep

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

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

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

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.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.

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- #16

Erland

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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.

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strangerep

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

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

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.

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- #18

strangerep

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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.)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?

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##.

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- #19

Erland

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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.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. Ifeveryequation in the theory has this property, then the theory has the property of spatially isotropy.

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}##?

- #20

Dale

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

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Erland

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

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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.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}##?

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

- #23

Erland

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Yes, I think so too.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.

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

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

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

Given this, we now assume that the (squares of) the lengths (x

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?

- #24

Dale

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

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.

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

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

More info here:

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

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