Talk:Galilean transformation

Invariance

I have a serious question. Can anyone work out for me how Maxwell's equations (differential forms, of course) are not Galilean invariant, but are Lorentz invariant? I attempted the problem before but couldn't see anything conclusive in the math. Every book I've run across says "it's easy to show...." ub3rm4th 19:49, 30 Dec 2004 (UTC)

The fact that all electromagnetic waves propagate at the speed of light is enough to show they're not Galilean covariant. As for Lorentz covariance, see Faraday tensor. Phys 17:23, 18 Jan 2005 (UTC)

in the language of differential forms, Maxwell's equations are dF=0 and *d*F=j. The first is invariant under all diffeomorphisms of spacetime. The second is only invariant under transformations which preserve the metric. Lorentz transformations preserve the metric but Galilean transformations do not. Just check what happens to the Hodge star in a Galilean boosted frame. -Lethe | Talk 17:07, Apr 24, 2005 (UTC)

Actually, the latter is invariant under all conformal transformations. Phys 01:08, 13 November 2005 (UTC)[reply]

Maybe any transformations which preserve the volume form as well? I forget, but it sounds plausible... -Lethe | Talk 01:21, 13 November 2005 (UTC)

The "macroscopic" Maxwell equations (div D = rho, curl H - dD/dt = J, div B = 0, curl E + dB/dt = 0) have an invariance group containing GL(4) (as well as a superset of the conformal group), since they can be expressed in the language of differential forms. Their invariance, therefore, is independent of the distinction between Galilean versus Poincare (versus Euclidean, even!)

The distinction arises from the constitutive laws. The Lorentz relations (D = epsilon_0 E, B = mu_0 H) are Lorentz invariant. That's directly tied to the duality operation F -> *F, which both requires a metric to be defined, and effectively defines a metric. When the Maxwell field is regarded also as a special case of a gauge field, the duality also involves the metric associated with the gauge group. Here, the role of the gauge group metric and its dual is played respectively by (epsilon_0 c) and (mu_0 c).

In contrast, the older Maxwell relations, which are equivalent to (D = epsilon_0 (E + G x B), B = mu_0 (H - G x D)), would be Galilean covariant, with G transforming as G -> G + v under a boost by velocity v. Maxwell's G was removed by the time of Lorentz once its non-observation became apparent. This is why you see a gap in Maxwell's alphabet soup A, B, C, D, E, F, H, I, J.

The invariance of the macroscopic equations is a "superset" of the conformal group because it also includes a residue of the complexion transformation: D -> D + kB, H -> H - kE, E -> E, B -> B. Consequently, it's best to distinguish the field (D,H) from the dual *F. The fields that come out of the Heisenberg-Euler Lagrangian in effective field theory, for instance, embody a transformation by a non-zero k, as well as having a variable epsilon.

This article is terribly written

The English in this article is terrible - example:

"From the point of view of group theory, Einstein's special theory of relativity the Lorentz transformation group events, it is essentially a particle do uniform motion of the Lorentz transformation group - This is the core of Einstein's special theory of relativity all."

I'm not an expert in this field so don't feel comfortable editing it, but that sentence is embarrassing.

What is worse: the article contains absolute nonsense. JocK (talk) 21:31, 4 January 2008 (UTC)[reply]

Have made a start with a thorough clean-up. JocK (talk) 21:57, 4 January 2008 (UTC)[reply]

Do we have t' = t when v = 0?

Let me arrange two inertial systems first. One with observer Mr. A holding his clock A at the origin point O of the stationary system S and the other one with observer Mr. B holding his clock B at the origin point O’ of the moving system S’ and x axes is on the same line of x’ axes while system S’ is moving at a constant speed v toward the positive direction of x axes. The y axes is parallel to the y’ axes and so does z axes to z’ axes. Clocks A and B are identical and set to 0 when O’ = O.

1. Postulate

The main purpose of Galilean Transformation, GT, is to study the result of measurements for location and time when an event is measured by Mr. A and Mr. B. The time formula t' = t is a postulate when GT started.

2. Believed

However, if we let v = 0, then the time formulas in Lorentz Transformation, LT, and Special Theory of Relativity, STR, all become t’ = t. Even scientists cannot let LT and STR coexist, most of them believe that one of LT and STR must be correct. Since both LT and STR indicate if v = 0 then t' = t it is fair to say that most scientists believe in that "If v = 0, then GT is correct." But how do we prove that if v = 0 then t’ = t?

3. Investigation

Let me describe an experiment just finished this morning for your reference. The system S’ stopped when Mr. B was about 900 meters away from Mr. A this morning and Mr. B threw an orange into the air then caught it 2 seconds later. Mr. B could use his clock B to record the beginning and ending time of that event. At that time, how did Mr. A measure the t’ for that same event? Do we have t’ = t for that event? My answer is no.

4. t' = t is not correct

Since that event was 900 meters away from Mr. A when it started, we know that the picture of the moment the orange was leaving Mr. B’s hand would take some time to arrive Mr. A’s eyes. (900 meters) / (300000000 (meters/second)) = 3000 nanoseconds. That means if Mr. B recorded the beginning of that event as 5 minutes according to the clock B then Mr. A would record the beginning of that same event as 5 minutes and 3000 nanoseconds according to the clock A. If we assume that both clocks are able to read nanoseconds then we already proved that t’ = t is not true when v = 0.

5. Nanosecond Formula

To an observer at point o, an event starts ta at point a is recorded as it is started t’a = ta+(ao/c) so that t’a is always later than ta by the difference of the time that photons spend in traveling the distance ao from point a to point o. If the event ends tb at point b, then the time period of that event tab = tb-ta is recorded as t’ab = t’b-t’a = (tb-ta)+((bo-ao)/c). The formula t’ab = tab+((bo-ao)/c) is the nanosecond formula.

John C. Huang (talk) 02:29, 19 May 2008 (UTC)[reply]

An "experiment just finished this morning" belongs in the realm of Original Research. Wikipedia does not deal with Original Research.

I have reverted your most recent edits to the pages talk:Time and talk:Lorentz transformation and left another warning on your talk page. How many times and in how many way do we have to explain this to you? DVdm (talk) 14:45, 19 May 2008 (UTC)[reply]

"Under the Erlangen program, the space-time (no longer spacetime)" what does that mean? ("no longer spacetime"). 217.132.109.1 (talk) 07:21, 5 November 2008 (UTC)[reply]

Please, explain the "operators"

My problem when reading this article was the lack of introduction of the "operators", E, C etc. Which transformations are they associated with? Why is there a hbar in this. I believe Galilei was unaware of quantum mechanics. It would be much better to use operators without hbars. Anyone ready to contemplate the Lie-algebra aspect of GT, will not be particularly disturbed by this deviation from the standard qm-textbook variants. —Preceding unsigned comment added by 212.37.14.186 (talk) 14:06, 28 April 2009 (UTC)[reply]

x' ≠ x - vt

I am a first time writer to Wikipedia and would appreciate some comments. Please note, that this is a mathematical challenge to this equation. That is, by letting vt = distance (d) my challenge is to x' = x - d. In no manner then, is this related to any Physics concerns, such as time, motion, observers, light, time dilation, length contraction, wave-fronts, velocity, etc.

My math challenge of x' ≠ x - vt since x' ≠ x - d

Given x = +2 and d = +3

Then according to x' = x - d, x' = -1.

However, Woldemar Voigt's 1887 equation, and consequently Relativity's x' = x - d should instead be

x' = x + d, where x' = 5.

The length of 3 is not a coordinate ( in the unmoved frame ) since it goes from -1 to + 2. By definition, all coordinate lengths must start from 0,0,0 in their own frame.

Please excuse my jumping the gun and editing the on-line article. I was under the impression it was to be screened first. Hopefully, this page/place is the correct procedure to follow. I do hope for some feedback/discussion/criticism/critique.

Steve waterman (talk) 17:49, 16 September 2009 (UTC)[reply]

Remarks:

"'... 'by letting vt = distance (d) ..." => Here you go wrong already. Distance is a positive quantity, whereas the time coordinate of an event can be positive or negative. A time coordinate is zero for all events simultaneous with the null-marking of a clock. An event taking place before the null-mark has a negative time coordinate. An event taking place after the null-mark has a positive time coordinate. Likewise, a velocity in the positive x-direction is modelled with a positive value for v, whereas a velocity in the negative x-direction is modelled with a positive value for v. So the product vt can be positive or negative. Distances cannot be negative, by definition. The x-coordinate and the x'-coordinate of an event can be negative or positive.
"Please note, that this is a mathematical challenge to this equation." => Please note that is an article about a topic in physics. Unlike in mathematics, in physics the variables in an equation have a physical meaning. Physics is not just another application of algebra.

DVdm (talk) 18:47, 16 September 2009 (UTC)[reply]

"Here you go wrong already." So vt ≠ d ? http://www.1728.com/velocity.htm Please show your reference that vt ≠ d.

I believe every textbook will allow velocity times time = distance.

Time is not relevant as I mentioned. The problem exists mathematically, AFTER the frame has moved 3 and stopped.

"Please note that is an article about a topic in physics." I am saying that the foundation for that Physics is based upon a math error... and thus has implications directly related to Relativity.

—Preceding unsigned comment added by Steve waterman (talk • contribs) 20:02, 16 September 2009 (UTC)[reply]

Remarks:

"So vt ≠ d ?" => Quite indeed, not with the definitions of v (velocity) and t (time coordinate) as they are used in the article, combined with the definition of d (distance) which you seem to have in mind. I tried to explain that the product of a time coordinate with a velocity is a location coordinate and not a distance. As opposed to the former, the latter is positive by definition. Please re-read my previous comment until you understand. Again, physics is not just another application of algebra.
"Time is not relevant as I mentioned" => Since the symbol t represents a time coordinate in the article, time is highly relevant. If you say it is not relevant, you are working outside the context of the article. Please note that an article's talk page is not meant to help readers understand the subject, get rid of their misconceptions, or entertain them about unrelated subjects, but to discuss the content and format of the article in question. I'm afraid you really have come to the wrong place with this.

DVdm (talk) 20:32, 16 September 2009 (UTC)[reply]

I thought using vt = d was going to be easier, it has not been so in this particular discussion. So I will just stick to the original equation.

My math challenge of x' ≠ x - vt

Given x = +2,0 and vt = +3

Then according to x' = x - vt, or x - vt, then x' = -1 after x' having been transformed to the unmoved frame.

However, Woldemar Voigt's 1887 equation, and consequently Relativity's x' = x - vt should instead be x' = x + vt, where x' = 5.

The length of 3 is not a coordinate ( in the unmoved frame ) after being transformed to the unmoved frame, since it extends between -1 to + 2.

question - Do you agree that the length of 3 is not a coordinate in the unmoved frame after being transformed to the unmoved frame ?

Noting that by definition, all coordinate lengths must start from 0,0,0 in their OWN frame.

Of course time is used in the Wiki article, you missed my point, I believe. Time is not relevant to my math only challenge discussion.

"I'm afraid you really have come to the wrong place with this."

I do appreciate your responding in the discussion. I will drop this thread immediately, as soon as you can answer my one question above. Fair enough ? If you could supply a brief logic for your yes or no choice, I would much appreciate it. I will let you have the last word, which quite likely, I will not be in agreement with. I do understand the nature of your observation "wrong place". I admittedly am as far from aiding in the construction of the current Galilean page as perhaps one can get.

Steve waterman (talk) 17:19, 26 September 2009 (UTC) —Preceding unsigned comment added by Steve waterman (talk • contribs) 22:22, 16 September 2009 (UTC)[reply]

Remarks:

"Given x = +2,0 and vt = +3. Then according to x' = x - vt, or x - vt, then x' = -1". => In the article's context where "x' = x - vt" this means:

The (x',y',z',t') system moves with velocity v in the positive x-direction of the (x,y,z,t) system. At time t=t'=0 the origins of both systems coincide.
Some event takes place at some time t at location x=2 as measured in the (x,y,z,t) system, i.e. at a location at 2 distance units from the origin of the (x,y,z,t) system in the positive x-direction, i.o.w. in front of someone sitting at this origin and looking in the +x-direction.
As measured in the (x',y',z',t') system this same event takes place at location x'=-1, i.e. at a location at 1 distance unit from the origin of the (x',y',z',t') system in the negative x'-direction, i.o.w. behind someone sitting at this origin and looking in the +x'-direction.

"Time is not relevant to my math only challenge discussion." => As you are discussing an article equation in which time is modelled as a variable, tIme is extremely relevant.
"I admittedly am as far from aiding in the construction of the current Galilean page as perhaps one can get." => Then your remarks and your questions are not relevent to this article's talk page. <== Please click this link and aquaint yourself with the talk page guidelines. For starters, use indentation and sign your comments. Thanks.

DVdm (talk) 07:03, 17 September 2009 (UTC)[reply]

As measured in the (x',y',z',t') system this same event takes place at location x'=-1, i.e. at a location at 1 distance unit from the origin of the (x',y',z',t') system in the negative x'-direction, i.o.w. behind someone sitting at this origin and looking in the +x'-direction.

Thank you for your nicely detailed response and that your made a numerical choice; being -1.

Quoting you from just above- As measured in the (x',y',z',t') system this same event takes place at location x'=-1

We likely agree that at t = 0 that x = 2 and x' = 2, being coincident frames at t = 0, therefore if x is present at t= 0, so too must be x'.

The transformation is to determine what the value of x' [ which is now at 2 at t = 0, ] is in the x,y,z,t system AFTER the (x',y',z',t') system has gone +3. You will disagree surely, believing it to be what x is transformed to AFTER the (x',y',z',t') system has gone +3.

Thus you would get the result ...What is x (2) in the (x',y',z',t') system ? x = -1 um, since x = 2, then in reference to the (right-moved x',y',z',t') system - the (x,y,z,t) system is itself left moved, relatively speaking. Obviously, if two systems are coincident, then moving one right moves the other left relative to each another.

What is x' (2) in the (x',y',z',t') system transformed value in the (x,y,z,t) system ? x' = +5

According to the equation, we what to know what x' is in the (x,y,z,t) system as would seem evident by the fact the equation was solved for x'. x' [the (x',y',z',t') system ] = x - vt the (x',y',z',t'] system.

I no longer need you to answer my previous question that was not addressed. I will keep my word then, and will make this my last post here. I really do thank you for playing along with my given values and representing your position. I will look for a response to this assessment above, hopefully.

Recap - are we not trying to transform a point x' from the (x',y',z',t') system to the (x,y,z,t) system ?

Think about it please....you have a point x' at 2 and a point x' at 2 at t = 0 and you move the (x',y',z',t') system three places to the right. Where is x' now in relationship to the (x,y,z,t) system ? I know it is most impossible to believe that we all got this wrong.

So we agree that x was transformed to -1 in the (x',y',z',t') system.

x = 2 vt = 3

If we solve x' = x - vt for x, we get x = x' + vt = 2 + 3 = +5.

relativity x = -1 x' = 5 reality x = -1 x' = 5

We agree where the transformed points end up; at -1 and +5, but not the direction.

x = -1 in the (x',y',z',t') system x' = 5 in the (x,y,z,t) system

FINAL SYNOPSIS given x = 2 vt = 3. Relativity says x' = x - vt = -1 and x = x' + vt = +5. However, x to x'y'z't = -1 and x' to xyzt = +5

and we disagree about directions. While in fact x' = x + vt = 5 and x = x' - vt = -1. x ≠ x - vt ≠ -1 and x ≠ x' + vt ≠ 5.

.............................................

You may only be considering point x = 2 at t = 0 ???

However, at t = 0, if x = 2 in the (x,y,z,t) system then the (x',y',z',t') system MUST also have a point x' at 2.

After all, by definition they are coincident systems. This is absolutely critical to agree upon this above singular issue or not.

As the (x',y',z',t') system moves to the right, that point X' at 2 in the (x',y',z',t') system ALSO moves to the right. It goes three and ends up at 5.

I believe relativity would just say x = 2 vt = 3. So x' = -1 and even if we solve for x, being x' + vt...we still get x' to be -1. Again likely you see 5, that is x' in the x,y,z,t) system as not involved at all. It certainly depends upon there also being at point x' at 2, at t = 0, should x be 2. Coincident frames have coincident points.

Perhaps worse even yet, one can be further be mislead by the sheer fact that EVERY x point in the (x,y,z,t) system is now vt greater than EVERY x' point in the (x',y',z',t') system. Which would also make it seem that x = x' + vt is true. whereas, X' in the (x',y',z',t') system, is 2 at t = 0, and 5 after vt is completed. So, in fact X' = X + vt. Steve waterman (talk) 17:19, 26 September 2009 (UTC)[reply]

Please note that you are disrupting this talk page. Do have a look at your user talk page. Thank you. DVdm (talk) 08:25, 19 September 2009 (UTC)[reply]

If you wish to remove my thread from the discussion, I am okay with that. I would be pleased if you do allow it to remain. In conclusion, I have finished a video of 2 and a half minutes that portraits my contention much clearer than I could do here. I would like to thank you Dr. Rick for your feedback upon this issue. I hope you will at least enjoy the accompanying music. http://watermanpolyhedron.com/videoforweb17-show0.html This video replaces the url that was here before as I was informed that some could not view it. Steve waterman (talk) 17:19, 26 September 2009 (UTC)[reply]

Spacetime video?

Surely that video of spacetime distorting with velocity changes in accordance with Einstein's relativity shouldn't be on a page about Galilean relativity? 165.124.205.18 (talk) 23:27, 8 June 2010 (UTC)[reply]

Huh? That video uses the Galilean transformation, not the Lorentz transformation. mike4ty4 (talk) 10:27, 31 December 2010 (UTC)[reply]

About Translation

The last sentence of Tanslation stated {Note that the last equation expresses the assumption of a universal time independent of the relative motion of different observers.} I think that is just one of the explanations for {t'=t}. There are at least two other ways to explain {t'=t}:

1. If the speed of light is infinitely fast, then, the picture of event-happening will be sent to all observers instantly so that all observers will record same event time t'=t.

2. If we assign observers at the location where the event is happening to record event time for their systems, then, since observers are all at the same place with zero distance from the event-location, they should record same event time t'=t.

Actually, the assumption of a universal time is not as clearly defined as above two explanations for the possible relation of t'=t. Because the Galilean Transformation (GT) is dealing with two sets of records for the same event eA, happens at a time tA and a location A. If in the stationary system (t,x,y,z), named S, the recorded time is tA=t and coordinate of A is A(x,y,z) while in the moving system (t',x',y',z'), named S', the recorded time is tA=t' and the coordinate of A is A(x',y',z'), then, GT states (t',x',y',z')=(t,x-vt,y,z). As you can tell, the assumption of a universal time does not care about that {an actual observation} is related to LIGHT, hence, the distance away from event or the speed of light should be handled. That assumption does not care about the distance away from event, nor the speed of light, so that it is not a detailed expalnation for t'=t, I think.

John Huang Jh17710 (talk) 05:14, 23 October 2010 (UTC)[reply]

transformed axis graph at t=0

Hi. I think it wouldn't hurt to have a graph similar to this one: http://www.phys.vt.edu/~takeuchi/relativity/notes/Lorentz-STD.gif — Preceding unsigned comment added by 78.179.87.40 (talk) 11:01, 9 September 2011 (UTC)[reply]

Perhaps a different approach

My 2c worth: The approach that this article currently takes, namely that the Galilean transformation is something only used in Newtonian physics helps perpetuate the confusion that existed in the early 20th century and which still persists today regarding relativity and continues to make relativity so difficult to understand or even accept. The Galilean transformation is a logically - indeed trivially logical - proveable transformation between ordinary Cartesian coordinates and a set of convenient generalized coordinates used in the analysis of a systrem moving at a constant velocity v relative to the observer. Standard Cartesian coordinates are projections of the positions of objects in the system onto the stationary axes of the observers reference frame. The Galilean transformation transforms these standard coordinates into a set of generalized coordinates consisting of projections of the positions of the objects onto a set of axes which according to the observer are moving with the same velocity as the system being analyzed. This really isn't any different to using any other generalized coordinates to simplify the analysis of a system and it is valid regardless of special reletivity. Saying that the Galilean transformation doesn't hold for relativity is what confuses people and creates misunderstanding and makes relativity sound like hocus pocus. The real issue leading to special relativity and the Lorentz transformation comes in when we introduce a second obsever and wish to convert from the standard Cartesian coordinates of the first observer to those of the second. That is where the Lorentz transformation is needed and where the Galilean transformation is the wrong tool, because that is not what the Galilean transformation is doing. The reason its not doing that is that all observers measure time and space against their own perception of electromnagnetic signals traveling between them and the objects being observed and thus in particular they all measure the speed of light to be same - its like if everyone used their own hand to measure length and then marvelled at the fact that everyone found their own hand to be exactly one hand long :) Because every observer measures with his "own hand", the Galilean transformation is not a conversion from one observers frame to anothers - we need the Lorentz transformation which scales "hand length" so to speak. However the Galilean transformation is still valid as a transformation used by a _single_ observer going from standard to convenient general coordinates. Kuratowski's Ghost (talk) 14:36, 10 December 2012 (UTC)[reply]

This article is missing so much information.

I would expect this article to have the following sections:

1. Some discussion of the relation between Galilean Transformation and Newton's First Law. Note that the standard physics version of the Galilean transformation does not include transformations to rotating or accelerating frames, since those are non-inertial frames.

2. Galilean Transformations in Dynamics; Transformation of displacement, velocity(1D, 2D, 3D), momentum, kinetic energy.

3. Galilean Transformations in systems of particles; Center of Mass Transformtions; Galilean Transformations in Fluids

4. Invariance of Classical Lagrangians under Galilean Transformations

5. Galilean Transformations in non-relativistic Quantum Mechanics; Relation to Unitary Operator, Momentum operator

6. An independent section discussing the limitations/extensions of Galilean transformation to Relativistic Phenomena. It is tiresome when the article mentions special relativity every other paragraph.

7. The stuff on group theory really seems to pertain to the Galilean Group (which is Galilean boost, translation, rotation, time translation) and not specifically to the Galilean transformation. The equations should really be limited to the article on the Galilean Group (which has several other issues of its own). And maybe just a paragraph in this article about how the galilean transformation is one part of a larger class of transformations known as the Galilean Group with a link.

70.124.92.52 (talk) 16:36, 26 October 2013 (UTC)[reply]

Go ahead and fix it. - DVdm (talk) 20:18, 26 October 2013 (UTC)[reply]

I'll get to work on it 70.124.92.52 (talk) 14:23, 1 November 2013 (UTC)[reply]

What is i ?

Some of the commutation relations like $[C_{i},H]=iP_{i}$ use a multiplicative i. But isn't the Lie algebra of the Galilean group a real algebra? Isn't there a confusion in the article with Representation_theory_of_the_Galilean_group and Quantum mechanics considerations? 94.245.87.72 (talk) 09:30, 5 November 2013 (UTC)[reply]

"moving relative ..." doesn't read right

Seems it should be "moving relativistically". 97.113.0.146 (talk) 17:42, 5 October 2024 (UTC)[reply]