Talk:Special relativity/Archive 21

Archive 15

←

Archive 19

Archive 20

Archive 21

Archive 22

Archive 23

Duplicate material: Consequences of special relativity

The article Consequences of special relativity is a massive duplication of this article special relativity. Suggest it be deleted. Comments on that talk page give a similar sentiment. 78.148.9.158 (talk) 14:52, 5 October 2010 (UTC)

Question?

In classical Physics, as shown by Newton in his Principia, The physical principles involved in causing motion in one direction can be be correctly calculated without consideration of the degree of motion in an orthogonal direction. So 2 parallel travelling particles may be caused to orthogonally deviate from their parallel paths by the same amounts regardless of their joint travelling velocity in the parallel direction. The question is: Is this still true in accordance with the theory of special relativity?WFPM (talk) 13:41, 21 October 2010 (UTC)

Yes. In each case, i.e. under Newtonian laws and under special relativity, they have the same inertial frame of reference, so each to the other looks stationary. If you were somehow unable to see the rest of the universe, and were moving with one of these particles looking at the other, there would be no way to tell you are moving. But according to special relativity the 'travelling' frame is no different from the stationary frame. You might as well say the particles are stationary and everything else is moving relative to them, and any 'fixed' frame you choose is moving relative to others: the earth is moving relative to the sun, the sun is moving within the galaxy, the galaxy is moving relative to others, and so on. There is no fixed frame and all frames are relative to all others.--JohnBlackburne^words_deeds 14:38, 21 October 2010 (UTC)

I think that John Blackburne misunderstood WFPM's question.

To WFPM: No. For example, motion in the x direction increases the relativistic mass which reduces the ability to move in the y and z directions. If this were not so, then an object could travel at anything less than the speed of light in each of the x, y, and z directions simultaneously. This would mean that the object could travel at up to sqrt(3) c diagonally which is not possible. JRSpriggs (talk) 15:48, 21 October 2010 (UTC)

Hm, yes, if you put it that way. If you consider two moving particles, i.e. moving relative to your current frame, then relativistic effects would make a difference for them moving at relativistic i.e. near-light speeds. One reason is as things go faster and gain energy they gain mass, so take a larger force to move them. They could never reach √3c as it would take an infinite amount of energy to get them to and past lightspeed. In their own frame of reference they can be pushed apart as quickly as at rest, but because of time dilation they will appear to move apart more slowly, again limiting their speed in any frame. So, yes, there would be a relativistic effect, but it depends on your frame of reference.--JohnBlackburne^words_deeds 16:02, 21 October 2010 (UTC)

Right! But we have this expanding universe concept, which involves galaxies moving away from a presumably center of mass, and we're on a gravitational slope somewhere, seeing a reddening of light in all directions, due to the universe expansion factor. But within this 3 dimensional space/1 dimensional time continuum, there must be a plane that separates the greater expansion galaxies on one side and the lesser expansion on the other, and is more or less perpendicular and thus points in the local direction of the expansion vector. In other words. is the Red shift value a directional value or just an expansion value? And shouldn't that be a factor in astronomical observations? And is that a factor of consideration in Special Relativity subject matter?WFPM (talk) 21:32, 21 October 2010 (UTC)

The usual model for the expanding universe is a 3D space that wraps back on itself in 4D, i.e. so there is no 'centre' in the 3D space. Instead that centre is in the 4D space and the universe is in a sense expanding away from that. But we can't detect this as we only exist in 3D, we can only infer it from the things we can observe such as that space is expanding but with no centre and no boundary. If we could go far enough we could in theory also measure the curvature, but the distances are so large this is not a practical prospect. A historical parallel is the surface of the Earth: primitive peoples often thought the world flat, bounded and with a centre, which they often thought significant so associated it with e.g. a religious centre like Jerusalem. But we know the earth is round, and the 2D surface is curved, unbounded and has no centre, or at least not one with a latitude and longitude.--JohnBlackburne^words_deeds 22:06, 21 October 2010 (UTC)

That's not strictly true; even if you assume that the shape of the universe is closed, that doesn't mean it's embedded in any higher-dimensional space. The whole point of Riemann geometry is that we only have to define the local topology of spacetime (at all points in spacetime) to get a description of it. The idea of it being "within" some higher-dimensional space stems from the teaching examples used (the expanding balloon analogy), not anything in the math itself.

With regards to the original poster's question about expanding from a central point, that concept only makes sense if one assumes that the universe is finite and has a boundary. As you point out, if it wraps back on itself, any given choice for "central" point looks as good as any other. The usual default assumption is that it's infinite and approximately "flat" (not curved). In this case too, you can pick any point and make it look like a center of expansion (this is just an arbitrary choice of origin for the coordinate system you choose to use). The mathematical description of the model is at FLRW metric, and of the expanding coordinate system that results is at metric expansion of space. --Christopher Thomas (talk) 22:26, 21 October 2010 (UTC)

I'm not defining a boundary. I'm merely listening to a current comment that they have found evidence of a galaxy that is 10+ billion light years from ours and must only be a few hundred million years old. And if I can point to something that old in a certain direction, why can't I draw a conclusion about a spacial geometry concept? Do you say that that distance is still local geometry?WFPM (talk) 03:02, 22 October 2010 (UTC)

I guess that my trouble is that I consider the shape of the universe to be a sphere that fills and thus is .5236 of a cube. And the only question is the size of the cube. And before I got into universe expansion I estimated the size of the cube to be 10 billion Megaparsecs, and thus have a volume of 523+ billion cubic Megaparsecs, which is about the right number of Cubic Megaparsecs needed for 100 billion Galaxies. But now we have "Expansion" and I cant have a cubical volume of space concept for the universe? Well then what shape is it?WFPM (talk) 03:15, 22 October 2010 (UTC)

The shape of the part of the universe we can see (the observable universe) is a sphere, but that's just because the age of the universe limits how far light that reaches us can have travelled. Anything farther away is invisible to us; that doesn't mean that distance marks the end of the universe. Similarly, in an expanding universe there might be parts that always remain hidden to us (because in the coordinate system we're looking at them from they're moving away from us faster than light). There is still no edge at that point, so saying it's a sphere of radius 13.x billion light-years or 4x.x light-years or what-have-you (depending on coordinates chosen) is incorrect. As far as we can tell, the universe continues on indefinitely in all directions. That is why your question about shape puzzles me. --Christopher Thomas (talk) 05:49, 22 October 2010 (UTC)

So here we are, lost in space!! And we can't see an edge to our volume. And what we can see is reddening as if it is moving away in all directions. And we calculate from that movement that it was all in one location 13+ Billion years ago. And somebody comes up with the 11-11 theory re the base 10 log values of stars-galaxies in our universe, and I have a book called "The pictorial Atlas of the Universe" by Kevin Krisciunas and Bill Yenne (Brompton Books Corp 1989) that shows Galaxies in spacial volumes kind of floating around like cumulous clouds with a concentration of 1 Galaxy per Cubic Megaparsec. So what is an Engineer supposed to do other than imagine a volume of space that more or less contains this bevy of concepts? And now comes this news of a galaxy that considered to be a "few" hundred million years old, as it was when the light left it 10+ billion years ago, which sould point a space-time arrow through this volume from our position in the direction of the original position of the universe. Isn't that what I should think?WFPM (talk) 01:10, 24 October 2010 (UTC)

The key point is that from any other location within the universe, it also looks like it's expanding in all directions away from that location. Any location will do; making this a formal assumption is what the "cosmological principle" is all about. There are solutions to the equations of general relativity that give you a universe that looks like it's expanding in all directions from any point within the universe, so the usual assumption is that this is what's happening with our universe. --Christopher Thomas (talk) 06:51, 24 October 2010 (UTC)

If I have a 3 dimensional volume of space with contained objects, and wish to expand the volume, without distorting the shape, I can't just do that from any point. And if I chose the volumetric center as the point of expansion, (which retains the shape), I'm not accomplishing anything but changing the space unit dimension factor, which nobody knows the basics of in the first place. It's kind of like analyzing the system of hexagonal closest packing, where all objects are all spherical of the same size and are in contact with their 12 neighbors, and if you want to expand the volume you just make all the objects larger and the distance of expansion (away from the center) becomes accumulative. How does that differ?WFPM (talk) 08:45, 24 October 2010 (UTC)

Sure you can - scale it. Scaling it with one choice of origin gives you the same result as scaling with any other choice of origin, if you change the observer's position and velocity such that they're on top of and at rest with respect to the new origin. The whole point of relativity is that both observer positions/velocities are equally valid (they're both inertial frames). The whole point of the cosmological principle is that we wouldn't expect one location to be any more special than another. In this case, you get a solution that is consistent with both. For detailed math, please see FLRW metric, per my previous comment. --Christopher Thomas (talk) 21:39, 24 October 2010 (UTC)

I would think that in order to agree with Hubble's law of universal expansion, there would have to be evidence of accumulative expansion values related to the comparative Galactic distances. How else are you going to arrive at a value of 75 km/second per Megaparsec or whatever.WFPM (talk) 09:06, 24 October 2010 (UTC)

But hey now! If I give you the Spherical concept of the universe plus the spacetime arrow pointing in a certain direction, and then I look into space in the opposite direction, What am I supposed to see? Should I expect to see an equivalently spacetime removed Galaxy for the sake of Symmetry? And should I more closely check the Expansion factor in that direction for uniformity? Its kind of like wondering how fast 2 beams of light are separating when they're pointing in opposite directions, and particularly when you think that those beams consist of matter particles.WFPM (talk) 14:56, 24 October 2010 (UTC)

Time dilation and length contraction

THE BEGINNING OF THIS SECTION IS NOT CLEAR. PLEASE REVIEW WITH THE CAPITAL CHARACTERS.

See also: Twin paradox

Writing the Lorentz transformation and its inverse in terms of coordinate differences we get △t’=γ(△t-(v△x/c^2))

DOES {in terms of coordinate differences} MEANS

(T2’-T1’)=γ((T2-T1)-(V(X2-X1)/C^2))?

△x’=γ(△x-v△t) and △t=γ(△t’+(v△x’/c^2)), △x=γ(△x’+v△t’)

Suppose we have a clock at rest in the unprimed system S. Two consecutive ticks of this clock are then characterized by Δx = 0. If we want to know the relation between the times between these ticks as measured in both systems, we can use the first equation and find: △t’=γ△t (for events satisfying Δx = 0)

DOES {the relation between the times between these ticks} MEANS ABOVE EQUATION, △t’=γ△t, IS FOR TICKS T2’=M+1 AND T1’=M FOR △t’, THEN, T2=N+1 AND T1=N FOR △t?

IF THAT IS WHAT IT MEANS, THEN △t’=γ△t MEANS 1=γ*1, SO THAT γ=1 AND △t’=△t, NO TIME DILATION.

IF THAT IS NOT WHAT IT MEANS, THEN COULD SOMEONE TELL ME WHAT DOES IT MEAN?

This shows that the time Δt' between the two ticks as seen in the 'moving' frame S' is larger than the time Δt between these ticks as measured in the rest frame of the clock. This phenomenon is called time dilation.

COULD SOMEONE REWRITE IT?

Regards, John Huang Jh17710 (talk) 05:37, 24 October 2010 (UTC)

The way that section Special relativity#Time dilation and length contraction is written, the observer would have to be in the primed reference system and the clocks or measuring rods being distorted would have to be in the unprimed reference system. This is the reverse of way that I would have chosen to do it, if it were up to me. But either way should work if the text explained it well, which it does not. JRSpriggs (talk) 15:21, 24 October 2010 (UTC)

I think the text explains it very well. Starting from the standard transformations, it indeed takes the "moving" clock/rod in the unprimed sysem (thus with "proper" time/length in unprimed coordinates) and calculates the coordinates in some other system, using some set of "primed" coordinates. DVdm (talk) 18:02, 24 October 2010 (UTC)

Thanks for your note, JRSpriggs. It is really hard to stay posted in Wikipedia, but, I will try to follow their regulations.Jh17710 (talk) 15:43, 24 October 2010 (UTC)

John, you cannot combine the equations for time dilation and length contration the way you tried to do it. When you do that, you effectively combine two events satisfying both Δt’=0 and Δx = 0. When you look at the transformation equations, you will notice that this can be only be valid for two events satisfying Δt’ = Δt = Δx' = Δx = 0, in other words for equations of the form 0=0. You deduced "... THEN Δt’=γΔt MEANS 1=γ*1, SO THAT γ=1 AND Δt’=Δt, NO TIME DILATION", but you forgot the case Δt’=Δt=0, from which 1=γ*1 does not follow. You "divided by zero", so to speak. I think this was explained to you a few times before somewhere on Usenet. DVdm (talk) 18:02, 24 October 2010 (UTC)

Thanks a lot, DVdm. You are the person I need to talk to. I thought that some thing wrong in this short paragraph, from "Writing..." to "...is called time dilation." because I was unable to explain the "△t’" mathematically to fit in the "Two consecutive ticks of this clock". I asked for help.

Your answer to JRSpriggs did not mention about any detail of the CALCULATES THE COORDINATES like using regular concepts of {a time period △t’}, which is normally represented by {from time t1' to another time t2', so that △t’=(t2'-t1')}. What is the mathematical value of the first tick of that clock and what is the value for the next tick of that clock? If there is no such value, then, don't use "△t’", use something else like Ut' to represent the time period between two consecutive ticks of the moving clock. However, if you do so, you must figure out how you adjust the equations of LT to fit your Ut' and Ut. We know that △x’=γ(△x-vUt) and Ut’=γ(Ut-(v△x/c^2)) is not going to fit for every event location with x-coordinate x in the stationary system and x'-coordinate x' in the moving system. Could you provide more detail regarding your statement "I think the text explains it very well."? Thanks.

Regards, JohnJh17710 (talk) 19:54, 24 October 2010 (UTC)

John, if you have problems with the special relativity theory, please ask on the wp:reference desk/science. Please read the wp:talk page guidelines: this article talk page is for discussing the article, and not for discussing the subject. It is also not for educating those who have problems understanding the subject. This was explained to you on your old talk page User talk:John C. Huang many times. I have left a new warning on your new talk page User_talk:Jh17710. Good luck (and look) at the reference desk. DVdm (talk) 20:33, 24 October 2010 (UTC)

Thanks DVdm. As you can tell, I just try to make this section more reader friendly. My main point is that, the usage of a popular symbol △t in this section is very different from popular definition. As I know, △t is for the difference of two times, △t=(t2-t1), so that I hope someone can bridge the gap.

The evidence is in the statement {the time Δt' between the two ticks (of the clock in S) as seen in the 'moving' frame S'}. There is no t2' and t1' so that it looks like Δt' represents the time interval for one unit of time in S'. That means it will be miss recognized as Δt'=1, and the same for Δt=1. I mean, physically, {the one unit of time} in S' is larger than {the one unit of time} in S, that is clearly described but mathematically it is not presented properly. Hope you can improve it.

Especially, when you think about the popular time equation of SR, t'=t/γ, which is also true for △t'=△t/γ with regular definition as (t2'-t1')=(t2-t1)/γ; don't you think this section needs some modification?

Regards, JohnJh17710 (talk) 00:32, 25 October 2010 (UTC)

I don't think it is necesserary to explicitly mathematically present the definitions of the Deltas, since everything is well explained in words, but I did it anyway. Hope this helps. DVdm (talk) 07:07, 25 October 2010 (UTC)

Thanks DVdm. Now, we have △t for the difference of two times, △t=(t2-t1). In this paragraph, t1 will be the time of the first tick in S and t2 is for the time of the second tick in S. If we arrange the clock to stay in S, we will derive △t'=γ△t like in the current paragraph; if we let the clock to stay in S', like what Einstein did in his paper dated 6-30-1905, then we will derive △t'=△t/γ, like what Einstein had derived before. That is why JRSpriggs said {This is the reverse of way that I would have chosen to do it, if it were up to me.}

The reason original paragraph did make sense physically is, if △t' means the length of {one unit of time} in S' and △t means the length of {one unit of time} in S, then △t'=γ△t did clearly mean time dilation. To avoid 1=γ*1, it can be written like 1s'=γ*1s, or Ut'=γUt.

However, if we use 1s' to represent the length of <time unit> in S', and 1s to represent that in S, then the definition of 1s' and 1s will make them not compatible with Lorentz equation. Then, we need to figure out some other way to use the symbols 1s' and 1s to explain time dilation.

I think, probably, if we let that reference clock stay in S' for now, we will have more time to find a proper solution for 1s' and 1s, or Ut' and Ut. It will be nice if we can see one unit of time in S' is longer than one unit of time in S mathematically, because it makes more sense, physically.

Regards, JohnJh17710 (talk) 04:09, 26 October 2010 (UTC)

You've got it all wrong and you clearly do not understand the setup. The text clearly explains that the clock is at rest in S, so, when we have two consecutive tick events on that clock, then Δt can indeed mean the " length of one unit of time in S ", but then Δt' does not mean the " length of one unit of time in S' ". Δt' would then mean the " length as measured in S' of one unit of time in S ", and the equation Δt' = γ Δt expresses time dilation. I don't think there is more clear way to explain this than is done in the text. If you really do not understand this, then I strongly suggest that you go elsewhere to have it explained to you. This is not the place for that. DVdm (talk) 08:29, 26 October 2010 (UTC)

You are right about the setup, {Δt' would then mean the "length (t2'-t1')Ut' as measured in S' for one unit of time 1*(Ut) in S}. But, in that setup the equation Δt' = γ Δt expresses time speeding, not dilation, because for all |v|<c, we have γ>1 and Δt'>Δt. Assume that γ=5, we will have Δt=1 and Δt'=5, that means {the event of two ticks in S} is measured as 1*Ut in S and measured as 5*Ut' in S'. That means Δt'=5*Ut'=1*Ut=Δt, Ut'=Ut/5; the clock in S' is running faster. But, Einstein wanted it the other way. Isn't it?Jh17710 (talk) 02:21, 27 October 2010 (UTC)

I think you might find someone to help you understand some basics on the wp:Reference desk. This is not the place. I have put a 3rd level warning on your talk page (User talk:Jh17710) DVdm (talk) 12:24, 27 October 2010 (UTC)

4th dimension contraction mathematics

In the Geometry of space time section section it is desired to treat time as just a 4th dimension, but with the explicit assumption that it is either orthogonal or maybe just equally related to the other 3 dimensions. However, since we don't have an orthogonal 4th dimensional space direction, how about settling for an equally related 4th spacial dimension, which is the diagonal of a cube dimension, and which is equally related to the orthogonal 3 dimensions of the sides of the cube? Then, in space time considerations, the 3 side dimensions of the cube could be the 3 spacial dimensions, and the diagonal dimension could be the time dimension. Then we could plan a 3 dimensional Michelson-morley light interference test and see if we get the same results in 3 orthogonal directions as we get in only 2 dimensions. And it allows the possible spacial locations of a point in 3 dimensional space (such as the path of the light beams) to be reduced to 2 dimensional direction versus time location system on a flat surface in the same manner as is done for the Lorentz time-distance correction calculations, but the mathematics of the correction are not the same because of the lack of orthogonality of motion of the light beams, (on the 2 dimensional chart), and I haven't been able to work it out.WFPM (talk) 21:24, 24 October 2010 (UTC)

Google "hyperbolic Minkowski". You'll find that the natural geometry of these rotations in flat or Minkowski space, involves entirely tanh, cosh, and sinh functions of angles, and the inverse hyberbolic functions to get angles from ratios of quantities. For example: [1]. SBHarris 03:21, 25 October 2010 (UTC)

The diagonal direction is NOT orthogonal to the three principal directions of a cube. JRSpriggs (talk) 05:57, 25 October 2010 (UTC)

Didn't say it was! Said it was "equally related" which is about the best you can do in a 4th spacial dimension. And I've figured out that the point of return of a 3 dimensional signal to the cubic diagonal line is at the 2/3rds point on the line which has a total length within the cube of 1.732 (square root of 3). but that's when there is no relative velocity of movement in any direction. but if there's a transition in any of the 3 orthogonal directions, the math gets beyond me. And I appreciate the hyperbolic geometry reference but I don't know how to use it and shame on me! But it's not the same math as the orthogonal directions correction of the Lorentz correction, because in this case when one signal travels up and downstream in the river, the other is traveling at a not 45 degree angle and not all the way across, and I think the correction might be different. And I'm sure that some mathematician can work that right out and tell me how wrong I am. And anyway I think that the results of a 3 dimensional interference test would be better than a 2 dimensional as a means of disproving the ether theory.WFPM (talk) 07:31, 25 October 2010 (UTC)

The experiment would probably involve a light transmission to three mirrors at say a 1000 ft hypotenuse distance and at a 120 degree azimuth angle from each other, but the mirrors would have to reflect the light back from an elevation of 577 ft above the ground plane. In this instance the diagonal of the tested 1000 foot cube stands in a vertical direction above the light emission location.WFPM (talk) 05:10, 29 October 2010 (UTC)

It doesn't matter, however you measure it, i.e. in whatever directions, the speed of light is the same. That is what Michelson and Morley found. As long as you are in an inertial frame all directions are the same and the speed of light is the same constant in all directions, or at least that's what has been confirmed to a extremely high degree of accuracy many times since M & M.--JohnBlackburne^words_deeds 15:34, 29 October 2010 (UTC)

The only way I can see for that to happen is for all light emission to be at right angles to whatever direction the emitting particle is moving.WFPM (talk) 22:08, 29 October 2010 (UTC)

The whole point of the Lorentz transform, and of special relativity itself, is that each observer has a (different) deformed coordinate system such that they all agree on the speed of light even if they disagree about what direction it's travelling, and about the speeds, masses, and rate of time flow for each of the observers. For further information, please consult a forum, tutorial, or textbook outside of Wikipedia; the purpose of these talk pages is to discuss changes to the article, not to answer questions about the subject itself. --Christopher Thomas (talk) 22:13, 29 October 2010 (UTC)

It's how it is: light speed is constant in any inertial frame. The discrepancy which I think you're seeing is accounted for in two ways. First any light transmitted by a moving source is redshifted or blueshifted, so it stays the same speed but loses or gains frequency. This relates to another way of relating the frames, that their clocks run at different rates, a phenomena called time dilation. This means two observers can both measure the same light beam going at c, despite being in different frames of reference so one is moving relative to the other. Again this is all experimentally tested physics: time dilation in particular has been demonstrated in experiments, despite perhaps being counter-intuitive.--JohnBlackburne^words_deeds 22:18, 29 October 2010 (UTC)

Thanks for the comment. And I understand about the Doppler effect. But consider the emission of light by a moving entity. If he is moving in the direction of the light beam, and the beam cant advance faster than the speed of light, then he is loosing his ability to get rid of kinetic energy of translation in that direction (which is proportional to M times delta V squared). So even if he has (time dilation) time enough to get rid of the same number of light particles, they wont be carrying off the same amount of kinetic energy. And since they cant move faster than c, they can't individually carry off a greater amount of energy. And the closer he gets to c velocity, the less ability he has to give off energy in that direction. And I can't think of anything else the light emission process does other than carry off released energy as the result of some physical process.WFPM (talk) 01:57, 30 October 2010 (UTC)

The light is blue-shifted in the frame of the observer you're using, thus carrying off more energy (and momentum). This is a special case addition of velocities at relativistic speeds (which doesn't work the way you probably expect it to). For further details, once again, please ask at a forum that's actually intended for teaching people about relativitiy. --Christopher Thomas (talk) 07:57, 30 October 2010 (UTC)

Well thanks for bearing with me, and maybe some mathematician will calculate the Lorentz correction that should occur in my 3 dimensional MM test, and then we'll see if it is the same as for the conventional 2 direction MM test. And with regard to the blue-shift property, you might consider the Doppler effect analogy re the energy rate of reception of the sound energy of an approaching airplane and remember what happens when his rate of approach becomes that of the velocity rate of sound transmission.WFPM (talk) 12:46, 30 October 2010 (UTC)

The analog for electromagnetic radiation of a sonic boom is Cherenkov radiation. JRSpriggs (talk) 14:59, 31 October 2010 (UTC)

Well Thank you, and that's how you learn. And I finally have a reason to think about Cherenkov radiation. But I never thought that the velocity of light slowed down to c/n anyway, since I was never able to explain the return of the exit light velocity to c at the "last atom" exit point. And thanks for rthe interest and information.WFPM (talk) 16:45, 31 October 2010 (UTC)

Space travel limits

I added this section:

As we can not travel faster than light one might conclude that a human can never travel further from the earth than 40 light years, if the pilot is active between the age of 20 and 60. So human space pilots would never be able to reach more than the very few solar systems which exist within the limit of 20-40 light years from the earth. But that is wrong. Because of the time dilation he can travel thousands of light years during his 40 years as a space pilot. If he has a spaceship which accelerates with a suitable 1G he will after 10 years reach speeds close to the speed of light, and the time dilation will increase his lifetime to thousands of years, seen from the reference system outside his spaceship. See the sentences about muon decay time above. If he returns to the earth he will land thousands of years into the future. His speed will not be seen as higher than the speed of light by observers on earth, and he will not measure his speed as being higher than the speed of light, but he will see a length contraction of the universe in his traveling direction. So, although no observer can observe speeds higher than the speed of light humans can travel a lot faster than the speed of light if we use a bastard unit for speed measurement. It is a bastard unit because we mix measurements from different reference systems. We use the time measurement from the space pilot's reference system inside his spaceship, and we use the length measurement from the reference system outside the spaceship. But although it is a bastard unit it is very useful for practical space travel.

We can compare this with the bastard unit which is created when we measure how far a muon can travel during its lifetime. It should only be able to travel 660 meter before it is dissolved but because of the time dilation at the speed it has it can travel for many kilometers before it is dissolved. In this calculation too, we are creating a bastard unit because we mix measurements from two different reference systems. In the reference system of the muon it can only travel 660 meters in the earth atmosphere, but the earth atmosphere is very thin because of length dilation, so the muon can be detected at the earth surface.

I can add the mathematical formula which describes how far you can travel, you can insert the acceleration and travel time and the result is the length of the travel, I have that formula somewhere in my archives. Do you want it? Roger491127 (talk) 13:27, 15 March 2011 (UTC)

Searching for a source we can cite as reference for this section

I found a scientific paper on internet 10 years ago which says the same thing, but I can not find it now.

At http://wiki.riteme.site/wiki/Faster-than-light I found a sentence which partially supports it:

"Proper speeds

If a spaceship travels to a planet one light year (as measured in the Earth's rest frame) away from Earth at high speed, the time taken to reach that planet could be less than one year as measured by the traveller's clock..."

I found a reference, about muons but, of course, also applicable for spaceships:

http://library.thinkquest.org/C0116043/specialtheorytext.htm

I will continue to search. Roger491127 (talk) 12:26, 21 March 2011 (UTC)

Dear Roger491127 (talk), I have added a comment on your talk page, about the physics issue and the necessity for certain restrictions in our local Wikipedia context. Wwheaton (talk) 15:21, 21 March 2011 (UTC)

Copy of a comment on my user talk page:

As it happens your general point, that special relativity actually enables extremely high speed travel (far beyond the speed of light) as defined by "ship measure" vship = (apparent distance traveled measured by stationary observer)/(apparent time elapsed by ship clocks), is quite correct, and has been known for decades. Arthur C Clarke described it well in one of his early books (? I think it must have been The Exploration of Space, ca 1951?), referencing a paper published in the 1940s I think it was. It is also laid out mathematically in Misner, Thorne, and Wheeler's standard text Gravitation on relativity, around page 105 I think it is. The "gotcha" is of course that "you can't go home again" (alas), together with the cruel fact that no currently conceivable rocket or vehicle can ever approach the performance needed to take advantage of this loophole. You might want to look also at Minkowski diagram, which has a section on the fundamental impossibility of exceeding c, if current notions of time and causality are valid. Cheers, Wwheaton (talk) 15:08, 21 March 2011 (UTC)

Who says you can't go home again? You can travel back to earth, but, of course, thousands of years have passed on earth.

Other problems which can be overcome: You need a motor which can deliver a lot of power for a long time, like a nuclear submarine motor which can work for 20 years without re-fueling. Add some extra capacity or extra fuel and you have the motor you need.

You need food and water for the pilot for 40 years. By recycling the water and compress the food into small tablets or grow some of the food in the ship this is not an impossible feat.

At extremely high speeds all kinds of space dust and even the two hydrogen atoms per cubic meter we find in "absolute empty space" may become a strong resistance to the ship, like thin air can burn a spaceship returning from space at high velocity.

We also need some mass to accelerate to increase the speed of the spaceship.

We can solve both of these problems by traveling out of this galaxy before we go up to really high speeds. The space outside the galaxy is free from dust and we can gather up the hydrogen atoms and accelerate them to increase the speed, after we have used up the store of mass we need at the beginning of the travel.

Anyhow, this is a fact of physics which we may, or may not, be able to use to really high effect while using the earth as the starting point, but we can probably use it after moving the starting point to a place where the space is very clean. As our technology progress and the population of the earth become bigger we will probably start producing very big cylindrical spaceships with a bright lamp along the middle axis, shielding half of it with a reflector. So when the ship rotates to give the millions of passengers a 1G environment they will experience a day, a sundown, a night and a sunrise.

Using such ships we can search for other inhabitable planets, or live on them forever, or travel to very empty and clean places in space, where this kind of travel faster than light, in simple terms, is relatively safe. Roger491127 (talk) 15:49, 21 March 2011 (UTC)

Building such very big ships will become necessary to spread out the human race, for safety reasons. To have all humans on one single planet is very risky, a single big comet can kill all of us. So, following the old saying "Do not put all of your eggs in one basket." we simply need to spread out the human race. Another reason is that our sun will explode in around 4 billion years, but comets is a much more urgent reason. Roger491127 (talk) 16:00, 21 March 2011 (UTC)

With the best will in the world, (a) This is largely nonsense technically, and (b) has little place here on the WP special relativity discussion page, for the reasons you can find via the links I have put on your talk page. If you want to talk about Interstellar travel, you can try that article's discussion page, but even there guidelines for article talk pages need to be observed. Conversations on your own talk page are of course almost unrestricted, as long as minimally courteous. Wwheaton (talk) 16:34, 21 March 2011 (UTC)

This section is completely wrong so please remove it. You can't travel faster than light no matter in what frame of reference you're in. So 1ly of distance could be covered in no less than 1 year, NO MATTER how fast you're moving. To be clear if you travel with near light speed your time would slow down, but you'd be getting shorter, so in total you'd travel from point A to point B for the same amount of time, as one would observe while standing still. Remember that the space-time interval is constant, so if your time dilates the length you'd have to travel increases! —Preceding unsigned comment added by 78.90.43.45 (talk) 16:24, 19 May 2011 (UTC)

Tough question

Note:- This needs a bit of imagination. I am an astronomer and have also studied about light. Last month I and my colleagues made an astonishing discovery…. Aliens, who can change their shape, colour etc!!! 2 of them, Zap and Zest Al came to live on Earth. Yesterday, Zap mentioned that Al Nazeer (their planet) had great leaps of technology. Then my love for knowledge, especially in the field of light, rekindled. He took the shape of a machine and I stepped into it. We were transported into a dark room. A ray of light came out of the machine. Then we accelerated to 3 * 10^8 m/s. Then the acceleration reduced to 0. So we were in an inertial frame. Yet according to the 2nd postulate, the speed of the ray = 3*10^8m/s w.r.t us. Thus the ray is travelling faster than us. But we are travelling at the speed of light. Thus the ray is travelling faster than the speed of light!!!??? How is this possible???

thanks Richu1996 13:23, 27 March 2011 (UTC)

It's not possible for material bodies to reach the speed of light (otherwise relativistic energy and momentum would become infinite). So in Earth's frame, light travels at the speed of light, and the "dark room" with a speed less than that of light. And in the rest frame of the "dark room", light travels at the speed of light while the room is at rest. The velocities in the different frames are mutually connected by the relativistic Velocity-addition formula. --D.H (talk) 08:00, 28 March 2011 (UTC)

To Richukuttan: It would be better to avoid sci-fi or fantasy in your questions since that just obfuscates the issues.

A entity traveling at the speed of light (photon, graviton, neutrino) experiences no passage of time, that is, it is annihilated at the same subjective moment that it was created. Thus it could not 'measure' the speed of any other entity. JRSpriggs (talk) 10:03, 28 March 2011 (UTC)

sorry about the sci fi Richu1996 08:16, 9 April 2011 (UTC) A entity traveling at the speed of light (photon, graviton, neutrino) experiences no passage of time, that is, it is annihilated at the same subjective moment that it was created. Then how do we know it exists if we cannot do any experiments on it (it would annihilate before the experiment nears completion)? Richu1996 08:12, 9 April 2011 (UTC) — Preceding unsigned comment added by Richukuttan (talk • contribs)

Both correct, ofcourse. I'd like to add that photons are mass-less particles so relativistic energy and momentum don't become infinite for them, and in fact, every photon travels at the speed of light. The difference in photons is their relative energy, which manifests as the wavelength aka frequency of light. this could be interpreted as it's "momentum" so to speak - replacing kinetic energy w/an electric/potential energy of sorts.

and as far as gravity is concerned (according to relativity, at least), it's actually mass+energy that causes gravity so it's the same-difference to gravity. (it's just that since e=m^c^2+...; since m has the c^2 term on it, it tends to be the dominate form of energy so it tends to do the most bending of space in massive particles.

in truth, however, a universe guided purely by the laws of special relativity would be hamiltonian; that is, the future would be exactly predictable from the past and vice-versa, so there'd really be no meaningful concept of "time", just different ways to look at what is ultimately the same frozen state. that's where quantumn physics and gravity come into play... they both (individually) introduce statistical irreversibility (better known as "entropy"). and with it the conventional notion of time as novelty. however this version of "time" is incompatible with special relativity's version. (or at least the relation is very non-trivial.) that is one thing that makes it difficult to combine SR and quantumn physics w/gravity. oh, and on that note, SR and quantumn physics were combined decades ago by Paul Dirac. The resulting equation predicted the existence of antimatter. gravity being such a weak effect, however, it's doubtful -- in my opinion -- that such a synthesis would really be all that useful, as a practical matter. but i digress. look at lorentz transform and proper velocity to understand the relations of time and speed proposed by SR. Kevin Baas^talk 00:23, 30 March 2011 (UTC)

All the points here have been discussed and resolved many years ago and the conclusions published in reliable sources. There is no need for us to speculate here, special relativity has been well understood for over a century. Martin Hogbin (talk) 08:38, 9 April 2011 (UTC)

Incorrect statement about massless particles

The article states: "Furthermore, he assumed that the energy/momentum of light transforms like the energy/momentum of massless particles, which was known to be true from Maxwell's equations." This isn't true. There is no mention of massless particles in the two 1905 Einstein papers ("On the electrodynamics of moving bodies" and "Does the inertia of a body depend upon its energy content?"). In the first paper, he proves based on Maxwell's equations that the energy of a light wave transforms in a certain way (identically to the scaling factor of the frequency given by the relativistic Doppler shift). In the second paper he refers back to the same result. Neither paper says anything about massless particles. I'll correct the article.--75.83.69.196 (talk) 00:04, 21 May 2011 (UTC)

Out of date historical information on the influence of the Michelson-Morley experiment

The article contradicted itself on this issue. In one place, it stated that the MM experiment had no influence, citing a 1974 book by Polanyi in which Einstein is quoted. In another place, it referred to a 2009 paper by van Dongen that shows that Einstein's statements could not have been completely accurate. Van Dongen's scholarship is more recent and contains information not available to Polanyi. I've changed the text appropriately.--75.83.69.196 (talk) 00:30, 21 May 2011 (UTC)

Out of date definition of SR

The lead states: 'The theory is termed "special" because it applies the principle of relativity only to the special case of inertial reference frames, i.e. frames of reference in uniform relative motion with respect to each other.' This is an extremely old-fashioned definition. Essentially all relativists today define the SR/GR distinction as one between flat spacetime and curved spacetime. I've edited the lead to bring it more up to date.--75.83.69.196 (talk) 02:13, 21 May 2011 (UTC)

I reverted this change. The "old-fashioned" definition is indeed the historical reason for the name "special". Also general relativity has to do with general covariance, i.e. the invariance of the laws of physics under arbitrary curvilinear coordinate transformations. Gravity was just the icing on the cake. Also see my comments at Talk:Special relativity/Archive 20#Special relativity in accelerating frames. Also see Coordinate conditions#Lorentz covariant coordinate conditions. JRSpriggs (talk) 06:00, 21 May 2011 (UTC)

I think you are wrong to revert. The article is about SR not the history of SR and it should therefore reflect current thinking on the subject. I will edit the article to try to present both fact appropriately. Martin Hogbin (talk) 08:41, 21 May 2011 (UTC)

As I said in the comments to which I linked, the theory actually described in the bulk of the article is the old-fashioned special relativity limited to inertial frames; it is not the general relativity minus gravity to which you are referring. This is necessary for pedagogical reasons (you have to walk before you can run). JRSpriggs (talk) 02:26, 23 May 2011 (UTC)

I see your point and have read the link you have provided, however, I tend to agree with the statement by Michael C. Price: Then the solution is not to remove the statement(s) that SR is applicable to non-gravitational situations, but to improve the article to make it clear that SR does apply to non-gravitational setups. i.e. that it applies to more than just inertial frames or locally around geodesics. We cannot change the generally accepted current definition of SR just because the article is limited in scope. Martin Hogbin (talk) 15:37, 23 May 2011 (UTC)

Addition

Is it possible to add to text about "force" the next:

Another expression for force is: (Fedosin S.G. Fizicheskie teorii i beskonechnaia vlozhennost’ materii. – Perm, 2009-2011, 858 pages, Tabl. 21, Pic. 41, Ref. 293. ISBN 978-5-9901951-1-0. (in Russian).)

$~{\mathbf {F} }=m\gamma ^{3}\left(\mathbf {a} +{\frac {\mathbf {v} \times [\mathbf {v} \times \mathbf {a} ]}{c^{2}}}\right)$.

Also an addition to "Force in 4D" :

With the help of operator of proper-time-derivative expression for four-force may be rewritten so:

$~F_{\nu }=U^{\mu }\partial _{\mu }p_{\nu }$.

Fedosin (talk) 05:23, 21 August 2011 (UTC)

Both of those expressions are unnecessary since there are superior expressions already in the text for the same quantities. Furthermore they are potentially confusing since it may be difficult for some readers to determine what they mean. In other words, they just clutter up the article. So I reverted the second and the first was already reverted by another editor. JRSpriggs (talk) 13:13, 21 August 2011 (UTC)

Application to motion in cyclotrons

A "main article" link to cyclotron at the head of this section should take care of the concern about "straying". Cyclotrons cover a lot of territory, and this little corner of this big subject provides such a neat example of relativity as to more than adequately justify a section on it in this article. Does anyone disagree? --Vaughan Pratt (talk) 23:02, 12 April 2011 (UTC)

I agree with whoever marked it as off topic in January. The section is far too long in relation to the article. I deleted it.--75.83.69.196 (talk) 02:47, 29 October 2011 (UTC)

Housecleaning of external links

The external links section had accumulated a very large number of links that were of low value or even spammish. I've moved a couple to the textbooks section and deleted a bunch more. Of course there is no guarantee that other wikipedians will agree with all of my value judgments, but I ask that anyone looking over these edits consider them carefully on a case by case basis rather than doing a wholesale revert.--75.83.69.196 (talk) 03:27, 17 October 2011 (UTC)

Science News resource, regarding Neutrinos and the speed of light and OPERA experiment

Neutrinos seen to fly faster than light ; Experiment hints at possible violation of Einstein’s speed limit by Devin Powell October 22nd, 2011; Vol.180 #9 (p. 18) 97.87.29.188 (talk) 23:13, 1 November 2011 (UTC)

Notice that there is already a paragraph on this subject at the bottom of the section Special relativity#Causality and prohibition of motion faster than light. At this time, this paragraph may or may not be justified, but certainly no greater coverage is justified since this is merely one unconfirmed observation. JRSpriggs (talk) 12:10, 2 November 2011 (UTC)

How does it work?

In the section about reference frames, one can read the following: "For example, the explosion of a firecracker may be considered to be an "event". We can completely specify an event by its four space-time coordinates: The time of occurrence and its 3-dimensional spatial location define a reference point. Let's call this reference frame S. In relativity theory we often want to calculate the position of a point from a different reference point. Suppose we have a second reference frame S′, whose spatial axes and clock exactly coincide with that of S at time zero, but it is moving at a constant velocity v with respect to S along the x-axis."

Physicists claim that two observers considering the same event (the explosion of a firecracker) will assign different values (x,t) and (x',t') to the coordinates of this event. How can special relativity decide whether x is larger or smaller than x' ? How can special relativity decide whether t is larger or smaller than t' ? On which ground?

S' is moving in respect to S and S is moving in respect to S'. But the principle of relativity forbids stating that one of them moves objectively faster than the other. So which physical criterion are they going to invoke to break the symmetry?

Is it the assumption that both observers move away from each other? What would happen if they were moving toward each other?

Is it the fact that their relative motion actually reflects that one observer has objectively a larger velocity in respect to the firecracker than the other observer? … But which one?

The Lorentz transformation cannot be derived from fully symmetrical hypotheses, so what does the special relativity theory actually say on this topic? Sugdub (talk) 20:12, 16 November 2011 (UTC)

This page is for discussing the article Special relativity and how to improve it, not for questions about the topic. You may find Introduction to special relativity to be a more accesible introduction to the topic, or one of the references and external links especially at that topic. For particular questions there's the Science reference desk, though be warned that if your question is too vague you may just be referred to articles like this for further reading.--JohnBlackburne^words_deeds 20:41, 16 November 2011 (UTC)

Collapsed off-topic discussion. Please take it elsewhere

Well, this sounds like an automatic message triggered by my "?". Indeed it is the content of the article which is at stake. In sketchy terms, special relativity is presented as follows: I) Observer-1 receives the light of the explosion of a firecracker and assesses that the event occurred at a distance x from him/her, at time t on his/her clock. The event is assigned the coordinates (x,t) in the reference frame of observer-1. II) Observer-2 …. the same firecracker ... x' … t' …. (x',t') … reference frame of observer-2. III) Observer-1 and observer-2 are in relative motion, moving away from each other. IV) Special relativity concludes that x' is smaller than x. What is the trigger for this conclusion? I'm not seeking any personal views, but if special relativity has any relevance to science, physicists must have a good reason for reaching this conclusion better than the opposite one. Or do they play it heads and tails? With this explanation the article will improve a lot.Sugdub (talk) 21:15, 17 November 2011 (UTC) To Sugdub: 'Special relativity' does not decide anything; it is just a theory not a person. Ask yourself the same questions about two observers in classical physics, say with one on a train and one on the ground. The asymmetry is not in the theory; it is in the factual situation in which the observers find themselves. Notice that observers who are first approaching one another will after a time be moving away from one another, assuming that neither accelerates. JRSpriggs (talk) 01:53, 17 November 2011 (UTC) Of course the theory must build-up on a difference in the respective experimental conditions of the observers in order to conclude that x' is smaller than x. So for the example of the firecracker proposed in the article, which parameter of the experimental conditions for observer-1 is assumed having a different value in the experimental conditions for observer-2 ? If you don't know the answer, just let physicists explain how their theory works. This is precisely the purpose of this article.Sugdub (talk) 21:15, 17 November 2011 (UTC) In special relativity, the observers determine the time and position of an event in just the same way that they would in classical physics, using measuring rods and clocks, geometry, photographs, tape measures or chains, theodolites, etc.. (Except if their equipment moves relative to the spatial origin of their system, then they must compensate for the time dilation or Lorentz contraction of the equipment, just as they have to compensate for effects due to temperature and humidity of the air.) JRSpriggs (talk) 06:42, 18 November 2011 (UTC) It seems you didn't catch the issue at stake. We are dealing with a section in the article which suggests that the relative motion between two observers is the key explanation of their difference of appreciation of their respective distance to a material target (e.g. a firecracker). This does not make sense since: i) it implies that the measurement performed by one observer gets impacted by the measurement performed by another observer; ii) it does not provide any criterion for deciding which observer will find the larger value; iii) one can easily produce a counter example showing that it contradicts the principle of isotropy of space (e.g. two observers moving away from the firecracker in opposite directions at the same time and the same constant speed v/2 will receive the light of the firecracker at the same time after it has exploded and will therefore assess the same distance for the firecracker, being exposed to fully identical experimental conditions thanks to the principle of isotropy of space; however, since both observers are in relative motion to each other, special relativity will - according to this article - predict that they will find different values for their distance to the firecracker, although it is unable to predict which observer will find the larger value). This can be expressed in a very synthetic way: if the relative motion between two observers could be seen as a proof that they are exposed to different experimental conditions, then the absolute velocity of each observer would be an objective physical quantity of its own, in full contradiction with the principle of relativity set by Einstein. It is difficult to accept the idea that the sense of logic by physicists is so ill-fated as it would look like if the presentation displayed in this article actually reflected their theory. So let's see if physicists have something consistent to offer concerning the genuine cause of the phenomenon, in order to counteract this disastrous image. If not, every reader of this page will be free to draw his/her own conclusions ...Sugdub (talk) 20:33, 20 November 2011 (UTC) I've found that the mentally easiest way to see all this is something like Einstein's thought experiment with trains, lightning and embankments, except that we update it for things we're familiar with, like strobe-flashes and atom bombs and spaceships. We begin with an "infernal device" which is simply two strobes at the end of a 600 meter rigid rod, floating in space. Exactly midway on the rod (300 m from each strobe) we put a electric eye looking at both strobes, and an atom bomb (and an astronaut in a spacesuit, or a cat in a Shroedinger box or whatever). We synch the strobes so that they flash each second, in a way that their signals travel 300 m to reach the center at exactly the same time, which is (of course) 1 microsecond = 1000 ns after they flash. After we have all this rigged up, we set a trigger so that if either signal arrives even 1 ns early or late, the bomb explodes. Then, we just let the whole thing run and watch how everybody has to view it. Now, if we take the atom bomb and rod as being the "rest frame" we see that the strobes must flash very closely at the same time (simultaneously) if the signals are to reach the middle simultaneously, and the bomb not go off. Thus, the strobe events/flashes are simultaneous in this frame. But what about other frames? For these, you can either have the rod move through space, or you can have somebody fly by the rod. The first is where we find out if there is an ether, for if we did this experiment with sound and air, as soon as the thing started to move through the air, one signal would arrive ahead of the other and the bomb would blow up. So we'd have a motion detector. Also an aether detector, since such a device would also show if the wind was blowing (some windgauges actually work like that). Let us suppose that all such experiments fail with light. Thus, we have no way to detect absolute motion, and no way to detect aether winds or aether-drift. For the problem with simultaneity, it's easy to see that even for the person riding the rod-ship that the strobes mush flash at the same time, even if the rods contract in both directions, since the speed of light is constant, so the distances (even if short) must be the same, and the arrival-time cannot change. However, just as clearly, a person flying by the rod in such a way as to arrive at the mid-point just as the signals arrive, CANNOT see the bomb go off either (it cannot both go off and not go off). So if the arrival "event" (the bomb not blowing up) is seen in one frame, it must be seen in all fames AT THAT SPOT (where the bomb is). You can call it a non-event (since nothing happens) but actually it's an event because the signals have to arrive at the middle at the same time (or within 1 ns) in all frames, and that simultaneous arrival is our "event". However, that causes a problem for moving frames. Suppose an astronaut flies by the bomb in the direction of the rod, at a velocity of 3000 m/sec. He sees the pulses arrive at the same time, but he knows that in the millisecond the pulses are traveling, the rod has moved 3 mm and so the light from the trailing flash has 3 mm extra to move, and the light from the forward flash has 3 mm less distance to cover. Thus, the two flashes cannot be emitted at the same time if the flashes arrive simultaneously, although the time difference is only that required for light to cross 3 mm, which is 3e-3/3e8 = 1e-11 sec = 1/100 ns. But now you see the problem, as already our arrivals are not QUITE simultaneous, and moving the observer to a frame 100 times faster (300 km/sec = 3e5 m/sec = c/1000) gets us into real trouble with the (arbitrary) way we set the sensitivity of our photoelectric detonation device. Now, at the faster speed of c/1000 the rod moves 0.3 m while the pulses are in motion, and one arrives 1 ns early and the other 1 ns late if they are emitted at the same time. This is enough to blow up the bomb and we can't have that, since all we did was move the observer. To preserve the observer's observation that the pulses arrive at the same time (and qualitatively, that the bomb doesn't explode just as he flys by it), the trailing strobe must emit 2 nanosec before the forward one! So these distant events are simultaneous for the observer riding on the rod along with the bomb, but they CANNOT be for the observer in the same spot (flying by the bomb very close to it-- infinitely close to it), but moving at 300 km/sec with respect to it. That's special relativity. The moving observer figures that for what he sees to be true, the forward clock and strobe are running behind his time, by an offset amount equal to t-t' = Δt = vd/c^2 = 3e5*3e2 / (3e8)^2 = 1 ns. A similar thing happens to the trailing clock which must be running fast by 1 ns and emits 1 ns early, because of this. Note also that time-dilation and length-contraction are not important in this problem, as we're only going c/1000. So forget getting involved with calculating the clock RATES and the rod LENGTHS. They change according to the usual Lorentz formula and get multiplied in, but not enough to affect the principal result, which is the clock offset (not clock rate) which requires that non-co-local events (the two strobe flashes) not be simultaneous in a "moving" frame if they are simultaneous in a "rest" frame. Only events at the same physical spot in both frames (like the arrival of the pulses at the electric eye and the bomb not exploding) can happen at the same time and place in both frames. All other events at distance, have to be "offset" by vd/c^2 to make it all come out. So you can see easily where that factor comes from, even if we don't have the gamma factor to contend with (which we don't in this example, where the speeds are too slow to need it). Note also the odd affect that if the observer goes past the bomb in the OTHER direction, he sees the clock times synch reverse, and the clock that was slow now runs fast, and vice-versa. He has no way to decide which is the "truth" since the clocks snap back into synch every time he matches velocities with them. SBHarris 21:54, 20 November 2011 (UTC)

Can Sbharris and Sugdub please take this to their talk pages now? - DVdm (talk) 22:18, 20 November 2011 (UTC)

Equivalence of mass and energy: is it advisable to perpetuate this phrasing?

The section Special_relativity#Consequences contains a sentence beginning "Equivalence of mass and energy". Although this is a common description, would it not be more accurate to say "Relation between mass and energy" or some other wording? After all, mass and energy are not two separate but somehow "equivalent" properties, but are related to a four vector [p₀, p₁, p₂, p₃] with p₀=mc²γ. The idea in using this description probably is that it is "popular" (and was used even by Einstein in several different ways that have proven confusing). Even so, a better description should be introduced immediately following this popular header. This recommendation applies also to the subsection titled Mass-energy equivalence and to the article Mass–energy equivalence that appear to perpetuate this misleading description as well. Brews ohare (talk) 19:21, 19 December 2011 (UTC)

I don't see why you think it misleading. Give something energy and you give it mass. E.g. heat something or accelerate something to give it thermal or kinetic energy and it gains mass. Extract energy from something and it loses mass: burn it, or extract energy through fission or fusion, and the energy you extract exactly equals the mass lost. Annihilate an electron and positron and 100% of their mass becomes energy. Whether we can measure the change in mass is another question but always if energy is gained then mass is lost, with the quantities related by the formula E = mc².--JohnBlackburne^words_deeds 06:53, 20 December 2011 (UTC)

The relativistic mass is the energy divided by c². The rest mass is the relativistic mass in the co-moving frame of reference. Inertia is a side effect of the conservation of energy and momentum together with

$E={\sqrt {p^{2}c^{2}+m^{2}c^{4}}}\,$

$\mathbf {p} c^{2}=E\mathbf {v} \,.$

So it seems to me that saying that mass and energy are equivalent is quite justified. JRSpriggs (talk) 18:32, 20 December 2011 (UTC)

Indeed, energy and mass are both scalars, just with different units. However, they are interconverable with a c^2. The fact that invariant mass is the length of a 4-vector and relativistic mass is not, only means they are two different kinds of mass. Invariant mass is equivalent of the rest energy (or system energy in the COM frame). Relativistic mass is the equivalent of relativistic energy (total energy). It's only when we try to cross relativistic THIS with invariant THAT, that we have problems with these 4 quantities. They are not all equivalent, except in the COM frame. But there's a pair of E,m that are equivalent for whatever situation you like. $E={\sqrt {p^{2}c^{2}+m^{2}c^{4}}}\,$ is just the way of converting relativistic E to rest mass, and so on. These aren't equivalent. It's only the m and E with the proper TAGS that are equivalent (except in the special COM case, again, where all four E(rel), m(rel), E(rest) and m(rest/invariant) are equivalent, save for units). SBHarris 07:43, 5 June 2012 (UTC)

Reference frames, coordinates and the Lorentz transformation

Diagram 1. which shows "rapidly accelerated" motion seems out of place, and potentially confusing, in this article about Special Relativity, since Special Relativity deals only with uniform motion (no acceleration). — Preceding unsigned comment added by Gregg Lee (talk • contribs)

I replaced the (````) with a signature. That should be tildes (~~~~). I hope you don't mind.

That is a common misunderstandig. It is true that inertial reference frames play a special role in SR because they make some calculations easier, but SR deals with any kind of motion. See, for instance Time dilation#Time dilation at constant acceleration and Twin paradox#Difference in elapsed times: how to calculate it from the ship. Special stands for "with exclusion of gravitation". - DVdm (talk) 09:23, 13 February 2012 (UTC)

Very Good Wikipedia Entry

Hey everyone, I am just saying that the banner that is at the top of this page should be reinstated as a "good article" since I recently went to a lecture on spacetime and Einstein and this article reinforces what a university professor was saying. The article has plenty of equations that are explained and I think someone should nominate the page to a "good article" status again. Thank you 67eldorado (talk) 22:41, 1 March 2012 (UTC)

How long does it take...?

I changed the section. It had one egregiously confusing error in calling traveler's proper time 'subjective'. Technically it is correct, but for the lay-person it implies "not objective" rather than "as experienced". You need to know your audience. I added some examples of round-trip durations and distances using the equations garnered from here:http://www.astrophysicsspectator.com/topics/specialrelativity/TravelDilation.html, but available in numerous on-line and text book sources. Please clean-up, I tend to be too verbose. I did NOT remove, although I feel it SHOULD be removed, the sentence on 4-velocity. I think it completely confusing and irrelevant: divide the trip's distance as seen from Earth by time as seen by him? Apples and Oranges. It does not advance the article one iota, IMHO.71.31.149.105 (talk) 08:05, 28 April 2012 (UTC)

transfer/removal of content

Any content removed is in my sandbox - find anything there quicker than the edit history.

The subsection Special relativity#Electromagnetism in 4D will be cut and paste/merge it into Covariant formulation of classical electromagnetism. It's out of place and this article is FAR too long... F = q(E+v×B) ⇄ ∑_ic_i

Also completley remove the stuff on 4-vectors (except for metric and lorentz transformations) in the section Physics in spacetime - the subject of another article "4-vector" (!). F = q(E+v×B) ⇄ ∑_ic_i 05:44, 4 June 2012 (UTC)

At least for now - move the subsection Relativistic mechanics overfilling this article into the underfilled article with the identical name Relativistic mechanics, and adjust the surrounding text of this article. F = q(E+v×B) ⇄ ∑_ic_i 06:35, 4 June 2012 (UTC)

In the section Reference frames, coordinates and the Lorentz transformation, the more general transformation was removed - there is no point cramming it there since half the explaination is what the symbols are and the x-boost is what people learn and pick up easier first, just link to main article. Also switched the order of the images/animations and pulled the text out of the animation into the mainspace of the section. F = q(E+v×B) ⇄ ∑_ic_i 07:22, 4 June 2012 (UTC)

Looks good. I made a slight restore of those two restrictions next to their equations. - DVdm (talk) 07:47, 4 June 2012 (UTC)

Thanks for finding those capital Deltas! I knew I double clicked the capital deltas before positioning the cursor, when I tried to find them I flicked the page up by mistake, then couldn't find them anywhere!

Also, your restore of the text next to the equations for time dilation and length contraction is fine, but the reason for the change was the odd formatting:

<math>\Delta t' = \gamma\,\Delta t \qquad ( \,</math> for events satisfying <math>\Delta x = 0 ).\,</math>

For consistency maybe it's better to use a table like this:

$\Delta t'=\gamma \,\Delta t$

(for events satisfying Δx = 0)

which has a more followable code:

{|
|-
| <math>\Delta t' = \gamma\,\Delta t</math>  ||      (for events satisfying Δ''x'' = 0)
|}

(using 4 space characters & nbsp; before the text, which will not show for some reason)? Just my suggestion - its not essential of course. Thanks again! =P F = q(E+v×B) ⇄ ∑_ic_i 19:18, 4 June 2012 (UTC)

I really would prefer rendering the deltas and variables in the same way, just to emphasise the importance of the condition.

Perhaps we can do something like this:

$\Delta t'=\gamma \,\Delta t\qquad \left({\text{for events satisfying }}\Delta x=0\right).\,$

(source: <math>\Delta t' = \gamma\,\Delta t \qquad \left( \text{for events satisfying } \Delta x = 0 \right).\,</math>)

That way we have consistent math rendering, but visually I find the current style much more pleasing:

$\Delta t'=\gamma \,\Delta t\qquad (\,$ for events satisfying $\Delta x=0).\,$

- DVdm (talk) 19:31, 4 June 2012 (UTC)

I agree. Lets not use the first one (with LaTeX all the way through - it looks horrible). It will be left alone as it is in the current article.F = q(E+v×B) ⇄ ∑_ic_i 19:50, 4 June 2012 (UTC)

"well-known" transformation law...

In this equation:

$T_{\left[j_{1}',j_{2}',\dots ,j_{q}'\right]}^{\left[i_{1}',i_{2}',\dots ,i_{p}'\right]}=\Lambda ^{i_{1}'}{}_{i_{1}}\Lambda ^{i_{2}'}{}_{i_{2}}\cdots \Lambda ^{i_{p}'}{}_{i_{p}}\Lambda _{j_{1}'}{}^{j_{1}}\Lambda _{j_{2}'}{}^{j_{2}}\cdots \Lambda _{j_{q}'}{}^{j_{q}}T_{\left[j_{1},j_{2},\dots ,j_{q}\right]}^{\left[i_{1},i_{2},\dots ,i_{p}\right]}$

in section Physics in spacetime, are there actually supposed to be square brackets around the indices, which mean tensor antisymmetrization? I doubt it (though not sure)... And if this is well known, why is there no source? I can't find this in any source I have... F = q(E+v×B) ⇄ ∑_ic_i 13:53, 5 June 2012 (UTC)

Indeed, it's in Carroll's book without antisym brackets. See equation 1.51 on p. 14. - DVdm (talk) 14:13, 5 June 2012 (UTC)

You are correct that the square brackets should not be there. Good catch. I removed them. JRSpriggs (talk) 14:23, 5 June 2012 (UTC)

Thanks the two of you for clarifying. =P F = q(E+v×B) ⇄ ∑_ic_i 14:25, 5 June 2012 (UTC)

It's equation (1.63) on page 22 in my hardcover copy of the book. I have added the source to the text. - DVdm (talk) 15:28, 5 June 2012 (UTC)

Thanks for the reference. If its also ok I'll remove the commas between the indices where no partial differentiation is required, since a reader who stumbles on any tensor calculus article may think that may be implied. F = q(E+v×B) ⇄ ∑_ic_i 22:10, 5 June 2012 (UTC)

Relativistic particle as a Nonholonomic system

Since the square of the 4-momentum vector $P_{\alpha }$ is a constant:

$P_{\alpha }P^{\alpha }-m^{2}c^{2}=0,$

a relativistic particle can be considered as mechanical system with nonholonomic connection in the 4-dimensional Pseudo-Euclidean space

• O. Krupková and J. Musilová, «The relativistic particle as a mechanical system with non-holonomic constraints», J. Phys. A: Math. Gen. 34 (2001) 3859-3876.
• O. Krupkova, J. Musilova, «The relativistic mechanics in a nonholonomic setting: A unified approach to particles with non-zero mass and massless particles» arXiv:0904.2933.
• V.E. Tarasov «Relativistic non-Hamiltonian mechanics» Annals of Physics. Vol.325. No.10.(2010) p.2103-2119.

This information is important to understand the special relativity. It should not be removed from the article. Baz.77.243.99.32 —Preceding undated comment added 16:45, 6 September 2012 (UTC)

Please sign your talk page messages with four tildes (~~~~). Thanks.

It seems that —at this moment— there is no wp:consensus for adding this content. Perhaps other contributors can comment. - DVdm (talk) 18:18, 6 September 2012 (UTC)

This information is not appropriate for this article — it is way above the level of the people who would be reading this article. I do not even understand what you are saying; and I am good at General Relativity, not merely Special Relativity. At the very least, you need to explain here why you think it is important to this article. JRSpriggs (talk) 06:53, 7 September 2012 (UTC)

α = 1/γ ?

In the composition of velocities, why introduce another factor

$\alpha (\mathbf {v} )={\frac {1}{\gamma (\mathbf {v} )}}={\sqrt {1-{\frac {|\mathbf {v} |^{2}}{c^{2}}}}}\ ?$

What is the point? Why not just divide by γ(v)? Maschen (talk) 11:20, 21 September 2012 (UTC)

Forget to ask, if this is used by the referenced author (ref 17: Y. Friedman, Physical Applications of Homogeneous Balls), why not take the statement "some authors use α = 1/γ" with that reference to the main article Lorentz factor? This article needs every possible trim of inessential details... It is very long and wordy to read...Maschen (talk) 11:27, 21 September 2012 (UTC)

I think that alpha can be directly replaced with 1/gamma per wp:CALC, but I don't think it does much harm here. - DVdm (talk) 11:35, 21 September 2012 (UTC)

True, not much harm, but it's one less thing to define. I'll proceed by transferring this formula (with reference) to the Lorentz factor article (thereby removing it from this article). Maschen (talk) 11:44, 21 September 2012 (UTC)

No problem. Good call. - DVdm (talk) 12:01, 21 September 2012 (UTC)

Thanks, although after making that fuss I decided to just delete and point to the main article on velocity addition formula. The x-axis formulae are enough for this article. Maschen (talk) 12:30, 21 September 2012 (UTC)

"a pair of diagrams, with x-t and x'-t' coordinates would help here"

In section Causality and prohibition of motion faster than light, this quote is commented out. Would these (from Lorentz transformation) help?

Particle travelling at constant velocity (streight worldline coincident with time t′ axis).

Particle with varying acceleration (curved worldline).

Lorentz transformations on the Minkowski light cone spacetime diagram, for one space and one time dimension.

Or is the comment out of date? I wonder who wrote it... Maschen (talk) 16:58, 22 September 2012 (UTC)

It appears to me that the accelerating particle is changing its acceleration. This is undesirable without a caption comment to that effect. JRSpriggs (talk) 06:42, 24 September 2012 (UTC)

Fixed, better? Maschen (talk) 06:47, 24 September 2012 (UTC)

Question

Based on my incredibly limited exposure, study and knowledge of simple physics, I humbly submit questions related to the concept that atoms are 99.9999 percent empty space. Is it possible that the incredible gravitational forces found in exotic matter such as a black hole could compress an atom of any given element to near 0 percent empty space? If possible, could this account for any of the 'missing' visible universe? Based upon my exposure to quantum theory, I must challenge some of the paradoxes associated with Relativity and Quantum Theory. Firstly, they cannot be unified as Electricity and Magnetism. They are incompatible. Where Relativity precisely describes celestial mechanics, its formulas fail utterly when applied to the subatomic world of quantum mechanics. The converse is true for quantum theory. However precisely, beautifully and accurately the formulas of quantum theory describe the very strange world of microscopic matter they too utterly fail if applied to larger matter. Where they meet and essentially leave humanity in the 'dark' is where both theories must be applied to attempt ANY understanding of a black holes exotic state and properties. I theorize (foolishly) that when a black hole is formed in the implosion of a supermassive star, elements are created that have not been observed by any scientific discipline. That is to say, yes, I believe fusion ends when a star produces iron but, without this implosion, we may not even have black holes in our universe as we understand it. Ok, my point: Could the 'infinate' gravity of a black hole drastically and in fact completely squeeze the empty space from each and every atom consumed by the black hole? If so, wouldn't that then mean that the potential energy of each atom is released as energy in it's purest form, electrons? Even tho light isn't supposed to have the capability to escape the gravitational force of this celestial body, Hawking radiation does escape. It is this readers humble opinion that the potential energy of each atom consumed by a black hole is released as pure energy as postulated by Albert Einstien and described by Steven Hawking (hawking radiation)? This leads me to a troubling question: Example:

Consider the atomic weights of each element and its' corrisponding atomic number. If a black hole compresses and overcomes the other 3 fundemental laws of physics: electromagnetism, the weak and strong nuclear forces, could one atom distinguish itself from another? Practical excersise: if one iron atom contains 63 protons and 64 neutrons totalling 107 (for arguments sake, I will plug the actual numbers in later) wouldn't 64 hydrogen atoms then be equally as heavy as the single iron atom? Each hydrogen atom contains one each proton and neutron. Should the space of these atoms be compressed to the forces applied by the 'infinate' state of a black hole, would they then weigh the same. If any given proton or any given neutron weighs the same dispite the atom it composes, and those weak and strong nuclear forces have been overwhelmed, it seems to make sense that the gravitational force is therefore NOT infinate but, in DIRECT proportion with the number of protons and neutrons which have been consumed by the black hole that trapped it.

Having scratched this surface, I find myself now contemplating Potential Energy and its counterpart, (precise name unknown) 'actual energy - heat and light' and certainly not limited to the frequency densities of the spectrum that human eyes can detect or human skin to feel. This energy released by the black hole could account for the missing Mass of the universe simply by applying Ensteins formula E=MC2 he states that energy is matter. The IMMENSE amount of energy released by a black hole should therefore eventually cool and spontaneously produce matter as we know it in the periodic table. This 'Dark energy' would then eliminate some fundamental paradoxes as predicted by both relativity and quantum theory. The energy we cannot detect could manifest and become the matter that in turn we cannot detect because it exists as pure energy. Leading to a unification of Relativity and Quantum mechanics? David E. Eaves. Age 43. Texas — Preceding unsigned comment added by Deighvid69 (talk • contribs) 04:36, 1 November 2012 (UTC)

Hi, please note that, per wp:talk page guidelines, this is an article talk page for discussions related to improving the article, not for general discussions or questions about the topic. For such questions you can visit our our science reference desk. Good luck. - DVdm (talk) 16:54, 1 November 2012 (UTC)

19 december edit?!

The 19 december edit looks like a mess: refs are (improperly) introduced, and vast portions of txt deleted. Why all this??Super48paul (talk) 17:23, 19 December 2012 (UTC)

It was vandalism by an IP user. Thanks are owed to Hhhippo (talk · contribs) for reverting it. JRSpriggs (talk) 09:43, 20 December 2012 (UTC)

Time Dilation & Length Contraction - Correct Expressions

Hi all,

I'm surprised to find some, lets say subtle, but critical to this article... I was unable to leave this that way.

Time dilation and Length contraction expressions are incorrect. Im based on the same topic, but taken from specialized books on the particle acceleration physics topic, which as review always present some physics summary. In particular, taken from [1, p.91]

In the section 'Time Dilation', it is customary to derive the formula: t=\gamma t', not the expression indicated in the article (t' = \gamma t) which is incorrect and lead to misinterpretations.

Several specialized books used the expression t = \gamma t', where is also customary to refer S' as the "moving" box, and S' as the "static" space. In this case, you use the expression: ct = \gamma ( ct' + \beta x' ) and not the inverse expression ct' = \gamma ( ct - \beta x ).

The mistake arises by considering x = 0 for proving time dilation. The correct assumption is to consider x'=0, that is, there is no displacement inside the moving S' box. This leads to the correct expression.

Here are the correction.

Correction on Time Dilation Expressions

The Time Dilation can be proved by taking the S' moving box transformation: ct = \gamma ( ct' + \beta x' ). Consider you 'lived' an interval t' inside the S' moving box. For simplicity, take two instants 'lived' at the same position under the S' moving box frame, that is, x'=0.

Hence, the interval at the S static space frame is: ct = \gamma ( ct' + \beta 0) t=\gamma t'.

Being \gamma > 1 for high speeds, for example lets say \gamma=3, means: 1 year living in the moving box is equivalent to 3 years living in the static space frame. This is the "twin paradox".

Correction on Length Contraction Expressions

Similarly, the Length Contraction can be proved by taking the S static space transformation: x' = \gamma ( -\beta ct + x ). Again, consider you 'measured' a distance x' inside the S' moving box. In this case, for simplicity, take two positions 'measured' at a same time under the S static space frame, that is, t=0.

Hence, the distance at the S static space frame is: x' = \gamma ( -\beta 0 + x ). x = 1/\gamma x'

Again, for example lets say \gamma=3, means: a 1 m rod measured in the moving box is equivalent to a 1/3 m rod measured in the static space frame.

Thanks to all, Juan F. Araya hypfco

[1] David Griffiths, Introduction to Elementary Particles, Second Revised Edition, Wiley-VCH, 6th Reprint 2011 — Preceding unsigned comment added by Juan Fco. Araya (talk • contribs)

It all depends on in which frame of reference the clock and ruler are taken at rest.

In our article we have the clock and rod at rest in the unprimed frame and the effects are measured in the primed frame, such that the clock is at same place in the unprimed frame, and the rod's ends are measured simultaneously in the primed frame:

$\Delta t'=\gamma \,\Delta t$    for events satisfying    $\Delta x=0\$

$\Delta x'={\frac {\Delta x}{\gamma }}$    for events satisfying    $\Delta t'=0,\$

whereas in Grifith's book the clock and rod are at rest in the primed frame and the effects are measured in the unprimed frame, such that the clock is at same place in the primed frame, and the rod's ends are measured simultaneously in the unprimed frame:

$\Delta t=\gamma \,\Delta t'$    for events satisfying    $\Delta x'=0\$

$\Delta x={\frac {\Delta x'}{\gamma }}$    for events satisfying    $\Delta t=0.\$

So there are no errors here and all the equations match the equations in the corresponding articles where they are properly sourced. - DVdm (talk) 17:54, 6 January 2013 (UTC)

Thanks for your answer, now i can see the corrected expression. They are now shown correct. As note of standardization, i would suggest to use primed coordinates for the 'moving box' frame, proper coordinates, and let the unprimed coordinates for the 'static space' frame, laboratory coordinates. This is also consistent with four-vectors and tensorial notation.

Please, also note the Time Dilation and Lorentz Contraction expressions cannot be equal!. they should be symmetrical, as shown. I've found several other -even specialized- references where it is no differences, and put x=\gamma x' and t=\gamma t'.

I've note also some inconsistencies on the 'Lorentz Contraction' article. Here both the sub-zero notation for proper coordinates and the primed notation are introduced, at some places they are even mixed. If you are also in charge of these pages, please make the proper corrections ( i could make them too, but could be interfering your work).

Thanks in advance. User:Juan Fco. Araya

Hi, note that this article is not only my work, but the work of many others -- see the edit history and the many talk page archives. The form that it has now has been pretty stable for a long time, and apparently found to be correct and appropriate by many others. It is of course true that some authors have other preferences, so the way it has been done here makes sure that no mistake can be made, and the physical meanings of all the variables are carefully explained. So I don't see any reason to swap primed and unprimed here so to speak. - DVdm (talk) 18:39, 6 January 2013 (UTC)

Special relativity (alternative formulations)

I would like to draw attention to the article Special relativity (alternative formulations). It's of very poor quality and is not linked to from any other article. I posted in its talk page my reasons for thinking that it should be deleted. I haven't started an AFD process for it because I'm not active enough on WP to want to invest the time in that process.--75.83.64.6 (talk) 01:35, 23 February 2013 (UTC)

Journal articles

This section is goofy. Unless I'm missing something, the article doesn't seem to explicitly state any connection between these papers and the contents of the article. Some of the papers are good and well known, e.g., the one by Alvager, which is a test of SR -- but it's unclear why these particular papers are referenced. Others appear to be fringe papers published in low-quality journals. I'm going to delete several of the most obviously low-value papers. IMO the whole section serves no purpose and should be deleted as well.--75.83.76.23 (talk) 02:11, 14 April 2013 (UTC)

I agree with the general plan. Sources supporting specific material in the article should be connected to that material via an inline citation. In cases where the topic of the source is not described in this article moving the reference to a more specialized article might be appropriate. Remaining general references could go to a "Further reading" section. At the moment it looks like each of the general sources supports the whole article, which is surely not the case.

@Maschen: A source being outdated is by itself indeed no reason for removal. But in this case the reason given was that there are already better sources supporting the same material, with better being defined as less outdated and more on-topic. To me that seems to justify removal. That said, since we're dealing with general references here, one should check if there's anything in the article that's not covered by the other sources and requires an inline citation to MTW. — HHHIPPO 09:56, 14 April 2013 (UTC)

Well it never hurts to keep as further reading... Anyway the section Comparison between flat Euclidean space and Minkowski space already mentions MTW (the authors, not just the book) in the context of the metric - why replace that? There is no reference for the section Invariance (OK plenty of refs could be found for this). But they also emphasize that the Lorentz force law is the "predictor of motions and definer of fields", leading to the definition of the field strength tensor by inspection, and again there is no reference for Relativity and unifying electromagnetism, which could include this detail.

In any case I dislike the idea of removing a classic book written by three of the greatest 20th century theoretical physicists, more generally of those who make/made considerable contributions to knowledge academically (even just by writing well-written texts) because this makes them and their work less known in a main, central, and popular article like this one. Shall we replace the Landau and Lifshitz course of Theoretical Physics and Nicolas Bourbaki books replaced by other modern refs in the relevant articles too? M∧Ŝc²ħεИ_τlk 11:32, 14 April 2013 (UTC)

I think there's two separate points here: the use of sources to establish verifiability of an article and the notability of sources by themselves. You actually found an inline citation to MTW, so it should indeed stay in the reference section. Maybe we should improve the formatting and clarify the connection by using a cite template. The same for the other references: some of them might be unlinked inline citations rather than general references.

Regarding notability of sources: that book is definitely notable by itself, it even deserves and has it's own article (which should be linked here btw). I wasn't sure how far it is relevant for an article on Special relativity (don't have the book here right now), but the passage you mentioned is clarifying that as well.

So yes, keep MTW. While the quoted reason for removal would in principle be valid, the situation it describes is not given here. — HHHIPPO 14:00, 14 April 2013 (UTC)

Maschen reverted my entire edit, which included both the removal of MTW and the removal of three journal articles. AFAICT the only controversial issue is MTW, so I'm undoing the revert of the journal articles while leaving MTW in. Maschen, please look more carefully at an edit before reverting it, and please pay attention to discussion on the talk page before reverting.--75.83.76.23 (talk) 14:04, 14 April 2013 (UTC)