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abhinavjeet
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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
abhinavjeet said: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
abhinavjeet said: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...
abhinavjeet said:I don't have any reason to prefer >v or <v
Yes, exactly. This is called a symmetry principle.abhinavjeet said:I don't have any reason to prefer >v or <v
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.bcrowell said: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."
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),
strangerep said: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.
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.)bcrowell said: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.
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.strangerep said: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.)
strangerep said:I do not see this. What do you think I "claim to prove".
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.bcrowell said:@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.
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:Erland said:Symmetry, spatial isotropy, how are these principles formulated in a mathematically stringent way?
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.)Erland said: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?
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.strangerep said: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.
The presence of the object breaks the symmetry. This is not because of asymmetry in the laws, but asymmetry in the boundary conditions.Erland said: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?
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...?DaleSpam said:The presence of the object breaks the symmetry. This is not because of asymmetry in the laws, but asymmetry in the boundary conditions.
Erland said: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}##?
Yes, I think so too.PeroK said: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 to formulate it stringently and economically. I have been considering something like this:But, if you are trying to justify why space is isotropic, then what do you take as your axioms?
In a stringent way, no. In a rough handwaving way:Erland said: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...?
Erland said:How does "the differential geometry of spacetime" imply the isotropy w.r.t the Lorentz transformation?
Erland said:Isotropy here means, I suppose, that the Lorentz transformation is somehow invariant w.r.t. rotations.
Erland said:isn't it much simpler in the SR case, where we have linearity and flatness?
Erland said: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?
Erland said: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.
Wow, excellent point.PeterDonis said: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.
DaleSpam said: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?
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.PeterDonis said:No. The Killing vector fields I described are for the SR case, flat Minkowski spacetime.
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?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 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.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.
It must apply to every equation in the theory under consideration. The question then is how to specify a "theory".Erland said:Very good, but this then leads to the question: Which are the equations in the theory to which this applies?
Here, you're talking about the more general 1+3D case, and what you say cannot be proven. Instead, one assumes that a rotation of the transformed axes around the boost direction has been performed (if necessary) to ensure that they're aligned with the original axes. (Strictly speaking, one might also have to perform a parity reversal as well.)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.)
Well, that requires a fair bit more work, applying the assumptions I outlined above to derive further constraints on the unknown functions. I'm happy to work through that detail if you wish -- provided we treat it like a homework exercise: i.e., you must do at least as much of the work as I do, and show it here on PF.And how is it used to prove that ##v_{12}=-v_{21}##?
Erland said:what do you mean by "commutation relations"?
Erland said: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.
This is wrong, both in Euclidean and Minkowski space. As a counterexample in 2d Minkowski space, with ##c=1##, take ##x=(3,0)## and ##y=(5,4)##. Their Minkowski lengths are ##l(x)=3^2-0^2=9## and ##l(y)=5^2-4^2=9##. If what you wrote is true, we would have ##l(x+x)=l(x+y)##, since ##x+x## is a linear combination of ##x## and ##x## with the same coefficients as in the linear combination ##x+y## of ##x## and ##y##, and ##x## and ##y## have the same Minkowski lengths. But ##l(x+x)=6^2-0^2=36## and ##l(x+y)=8^2-4^2=48##.PeterDonis said:A linear combination of the intervals x and y is another interval, and which interval it is is independent of the choice of coordinates. (This is just the Minkowski spacetime analogue of vector addition in ordinary Euclidean space.) So the Minkowski length of the linear combination can only depend on the lengths of x and y, and the coefficients of the linear combination.
Erland said:take ##x=(3,0)## and ##y=(5,4)##.
Erland said:If what you wrote is true, we would have ##l(x+x)=l(x+y)##
Erland said:##x+x## is a linear combination of ##x## and ##x## with the same coefficients as in the linear combination ##x+y## of ##x## and ##y##
Erland said:in 2d Minkowski space, and I am quite sure in 3d and 4d also, we cannot find two lightlike, linearly indepent vectors which are orthogonal to each other.
##l(x)## here means the Minkowski length of ##x##. Let ##x=(1,1)## and ##y=(1,-1)##. These are both null vectors. Yet, ##l(x+x)=l(2,2)=0## while ##l(x+y)=l(2,0)=4\neq 0##.PeterDonis said:Of course this is trivially true for any pair of null intervals ##x## and ##y##, since their lengths are always zero. So what my statement really amounts to is saying that, for the case under discussion (##x## and ##y## null intervals), the length of the linear combination depends only on the coefficients of the linear combination.
Linear combinations are defined for all finite sequences of vectors. It doesn't matter if some of them happens to be equal or not. So ##x+x## can be considered as a linear combination of the pair ##(x,x)##, with coefficients ##(1,1)## (another possibility is ##(0,2)##, and there are infinitely many more, since the pair ##(x,x)## is linearly dependent). Likewise ##x+y## can be considered as a linear combination of ##(x,y)## with coefficients ##(1,1)## (and infinitely many more possibilities, if ##(x,y)## is linearly dependent). So, the coefficients are the same in both cases.Um, what? That doesn't even make sense. The coefficients of a linear combination of ##x## and ##y## are the numbers multiplying ##x## and ##y## in the linear combination, in order; i.e., for the linear combination ##ax + by##, the coefficients are ##a, b##. So ##x + x## has coefficients ##2, 0##, while ##x + y## has coefficients ##1, 1##. So these are two different linear combinations.
It is not clear to me what you actually claim. Do you claim that every linear transformation ##T## which maps (Minkowski) null vectors to null vectors, preserve Minkowski length? If so, you are wrong.Basis vectors don't have to be orthogonal, they just have to be linearly independent. Finding a basis of four null vectors just means finding four null vectors that are linearly independent. (You are correct that we can't find two lightlike, linearly independent null vectors that are orthogonal; two null vectors can only be orthogonal if they are collinear.) The basis vectors in the case you're used to, an inertial frame, are all orthogonal, but that does not mean basis vectors always have to be orthogonal; they just happen to be in that particular case.