Talk:Special Relativity/Spacetime

Finally, some clarity!
I finally figured out a way to look at it that makes sense. Let T be the time interval measured by each observer. This will be the dimension that's perpendicular to x. We can make a dimension perpendicular to another by proclaiming it, provided we don't arrange the math to be incompatible with the claim.

Then t is the pythagorean distance from (0,0,0,0) and (T,x,0,0). It is sqrt(T^2 + x^2). It is not normal to x unless v=0 or x=0. But that's OK.

When we change v, we don't rotate in a euclidean circle or a minkowski hyperbola. We merely extend or contract x. t= sqrt((T^2 + x^2)) extends or contracts to fit. The triangle gets longer or shorter, and one short side stays the same. This would be a hyperbolic rotation if t was perpendicular to x, but it's easier to see this way.

I can do lorentz transformations by using my function from T to t, and then continuing as usual with t. I don't need to compute t' because T^2 = t'^2 - x'^2. My Lorentz transforms all stay on a single T plane which I find easier to visualize. T is not changed by a Lorentz transform. No matter what you decide the other guys times and distances should have been based on your relative velocity, the time his clock measures will still be the same.

The light-cone turns into two different mappings. Sometimes you care about when light arrives at (_,0,0,0). If light from (T,x,y,z) arrives precisely at (0,0,0,0) then T=0. If it arrived 2 time-units earlier then T=2. Two time-units later, T=-2. The total distance is the same from (2,x,y,z) to (0,0,0,0) as from (-2,x,y,z). You can't be at (_,x,y,z) and have the distance be less than sqrt(x^2 + y^2 + z^2). So far, I have had no need to consider a distinction between time-like or space-like intervals. Sometimes, light can leave one location and arrive at (0,0,0,0) before time 0. Sometimes, light can leave (0,0,0,0) and arrive at (T,x,y,z) before time T. So far each problem I've worked may require me to know one or the other of those but not both. It is not important to either map or to calculations made from either map. It only matters when you must consider both maps at once.

So far I haven't seen that time t is directly connected to time measured at (_,0,0,0). It is calculated to be the time that would make sense for something else, somewhere else. t makes sense as the total euclidean distance. So v= space-distance/total-distance, rather than space-distance/time. At slow speeds these are almost the same, so Newtonian stuff works.

I am not the first to do this, I have found a number of links using the keywords "euclidean relativity". They tend to start out the same way I did, and then add extra complications. One added a fifth dimension. Another decided that we have four four-dimensional spaces that interact in some ways but not others. But the essentials here are not a crackpot idea, or at least they get a degree of respect.

I think you will need to keep your book on traditional lines even though they are probably harder to comprehend, because Wikibooks must follow the cultural traditions. Thank you for helping me see the problems and how to fix them.

I have removed some comments that I think don't help revision of the book. They are in the history and I won't object if you think any of them should be restored. Jethomas5 (discuss • contribs) 19:06, 8 November 2013 (UTC)

Minkowski metric

 * The more I consider your comments the more it seems that the book needs to be amended to emphasise that relativistic geometry is different from Euclidean geometry. It is not good enough to just say that the metrics are different and then calculate using the Minkowski metric.  The approach taken in the book is simple but can be misleading.


 * Agreed.

contribs) 18:18, 5 November 2013 (UTC) RobinH (discuss • contribs) 10:55, 5 November 2013 (UTC)

Jethomas5 (discuss • contribs) 00:11, 5 November 2013 (UTC)

another confusion
Here is my reasoning -- first, if it works with all the c's explicitly set out, then it will also work when we set c=1 and express velocities in fractions of lightspeed. So I will set c=1 and remove it from all the equations.

Earlier there was the claim that when s^2=0, -T^2 = (vt)^2 - t^2.


 * Strictly when $$s^2 = 0$$ then $$0 = (vt)^2 - t^2$$. The metric is $$s^2 = x^2 + y^2 + z^2 - t^2$$. ( Although we should really be using $$\Delta$$ notation to indicate that the equations apply to intervals).


 * Delta notation would help, and yet it would make it look more cluttered and harder for someone who's new to it to follow. But if they think they follow it easily but they don't understand that it's intervals, that isn't good either.


 * I say "apparent" because the capitalised symbols are being used for proper lengths and times, not for lengths and times in the moving or stationary reference frames.  Each section is complete in itself.  The section on coordinate times and lengths is correct and the symbols are consistent.  The section on phase is correct and the symbols are consistent within that section.  At first sight it seems that  the symbols have been switched in the phase example but in fact the capitalised X and T apply to proper lengths and times so are being used in the same way as in the previous section. However, as you are pointing out, this is not really good enough and misled us both.


 * Specifically, in the section on proper lengths and times its says:


 * A: John will also observe measuring rods at rest on Bill's planet to be shorter than his own measuring rods, in the direction of motion. This is a prediction known as "relativistic length contraction of a moving rod". If the length of a rod at rest on Bill's planet is $$X$$, then we call this quantity the proper length of the rod. The length $$x$$ of that same rod as measured from John's planet, is called coordinate length, and given by


 * $$x = X \sqrt{1 - v^2/c^2}$$.


 * And in the section on phase it says:


 * B: But Bill's distance, x, is the length that he would obtain for things that John believes to be X metres in length. For Bill it is John who has rods that contract in the direction of motion so Bill's determination "x" of John's distance "X" is given from:


 * $$x = X \sqrt{1 - v^2/c^2}$$.


 * Notice that X in "A" refers to a proper lenth in Bill's inertial frame of reference whereas in "B" X refers to a proper length in John's inerial frame of reference. This is indeed misleading, as you say.


 * I don't think that's the problem, but I'm having trouble saying what the problem is instead. I'm sure the issue is not whether we consider Bill's inertial frame versus John's inertial frame. I got a hint from you saying that proper lengths and times are not lengths and times in stationary reference frames. I had thought that I got a proper time when I considered the frame stationary.


 * I get T^2 = s^2 which for typical homework problems gives me a real s instead of imaginary, which happens with the stationary frame. That means, when Bill's clock measures 4 seconds it will measure 4 seconds regardless of what any other observers think their coordinate time t should be, so T is a relativistic invariant.   X is a proper length regardless whether it's John's or Bill's. John and Bill have symmetry anyway, and given similar circumstances will measure the same proper times and proper lengths. Won't they? (Is the issue that we mustn't compare one of their coordinate times or lengths to their own proper time or length, but only to the other's proper time and length?)


 * I have been pondering how to overcome this problem. We could use a lot of subscripts to clarify which observer is making which measurement but I am not sure that this will improve either the readability of the text or make it clear to the reader what is happening.  Perhaps the simplest change is to explain that the capitalised symbols refer to proper lengths and times.RobinH (discuss •

contribs) 17:55, 3 November 2013 (UTC)


 * Maybe so. I had thought that it might be enough to just use a different symbol, like perhaps the tau that I used, with a note that it didn't refer to exactly the same thing as the T before. But now I don't in fact see why it isn't the same T.


 * Sticking with the old symbols $$X = x \sqrt{1-v^2}$$. This is the classical length contraction result where moving lengths are measured by the conventionally stationary observer to be less than stationary lengths. With your symbols when v=c:
 * $$c = X/T$$ and $$c = \frac{x/0}{t*0}$$ which is infinity over zero.
 * with the correct symbols:
 * $$c = X/T$$ and $$c = \frac{x*0}{t*0}$$ which is 1.

This is two sides of a triangle congruent with the first. X is the long side, and x is one of the short sides, and the third side is vX. It is argued that vX=T. But (sqrt(1-v^2)/v)x = (1-v^2)/v)X = T. vX is not T. vX is the phase change of T, not T.


 * This problem does not occur if the coordinate symbols are not mixed. Notice in the symbol conventions of the article:
 * $$x = X \sqrt{1-v^2}$$ and $$t = T \sqrt{1-v^2}$$
 * not:
 * $$X = x \sqrt{1-v^2}$$ and $$t = T \sqrt{1-v^2}$$

''I don't understand. The way the article is actually written, it says''
 * $$t = T / \sqrt{1 - v^2/c^2}$$.
 * $$x = X \sqrt{1 - v^2/c^2}$$.


 * mea culpa. I rushed this and got hold of the wrong end of the stick. Having jumped to the conclusion that the problem lay with wrong IFRs I hastily got the wrong ones! The actual problem lies with the symbols used in the section on phase in the article - see above.  I feel slightly chastened by missing the actual error and look forward to your comment on the changed text.  RobinH (discuss • contribs) 17:55, 3 November 2013 (UTC)

In the section on relativistic phase, we start with x^2 = X^2 - T^2. X^2 = x^2 + T^2. But we already have t^2 = x^2 + T^2. Somehow, the meaning of T has been changed.

As a minor matter, in the figure "How clocks become out of phase along the line of travel" you make T perpendicular to X. Probably people mostly won't look at it closely enough for this to matter, but it makes more sense to make T perpendicular to x, because X^2 = x^2 + T^2.

Jethomas5 (discuss • contribs) 10:15, 19 October 2013 (UTC)


 * The phase $$xv/c^2$$ is the term that appears in the Lorentz Transformation prior to adjustment by $$\gamma$$.
 * $$T = \frac {t - vx/c^2}{\sqrt{1 - v^2/c^2}}$$
 * It is a time interval in the frame of reference of the conventionally stationary observer and so orthogonal to their x-axis.RobinH (discuss • contribs) 23:30, 1 November 2013 (UTC)
 * Incidentally, there is a phase term that is perpendicular to the conventionally moving observer's x-axis, re-writing the transformation for the time of an event:
 * $$T = \frac {t}{\sqrt{1 - v^2/c^2}} - \frac {vx/c^2}{\sqrt{1 - v^2/c^2}}$$
 * The phase term $$vx/c^2$$ in the conventionally stationary observer's IFR is $$\frac {vx/c^2}{\sqrt{1 - v^2/c^2}}$$ in the conventionally moving observer's IFR. This term is indeed orthogonal to the moving observer's x-axis.RobinH (discuss • contribs) 18:18, 3 November 2013 (UTC)

Here is my reasoning:

$$w = \sqrt{1 - v^2}$$

$$T = \frac {t - vx}{w}$$

--comment by RobinH (discuss • contribs) 17:49, 4 November 2013 (UTC) Consider the Lorentz Transformation for time:



Let $$\tau = t - vx/c^2$$, then, given that the moving observer sees that stationary observer move vT metres in T secs, from the Minkowski metric:

$$(vT)^2 -(cT)^2 = - (c\tau)^2 $$

$$T^2 = \frac {{\tau}^2}{(1-v^2/c^2)}$$

So:

$$T = \frac {\tau}{\sqrt {1-v^2/c^2}}$$

or, substituting back for $$\tau = t - vx/c^2$$:

$$T = \frac {t - vx/c^2}{\sqrt {1-v^2/c^2}}$$

Notice that the Minkowski metric for 4D spacetime is used rather than the Pythagorean metric for 3D space. These metrics are just the relationship between the coordinate axes: Pythagorean 3 space: $$r^2 = x^2 + y^2 + z^2$$, Minkowskian spacetime: $$s^2 = x^2 + y^2 + z^2 - (ct)^2$$.

When the Minkowski Metric is used then:

1. Time must always be at right angles to space in a given inertial frame of reference (IFR) because it is a coordinate axis like x,y and z.

2. $$s^2 = r^2 - (ct)^2$$ not the Pythagorean "square on hypoteneuse = sum of squares on other two sides" ie: not $$s^2 = r^2 + (ct)^2$$

In Special Relativity it is usually valid to use the Pythagorean form $$s^2 = r^2 + (ict)^2$$ where $$i = \sqrt{-1}$$. Feynman says somewhere that he used ict to simplify calculations in developing QED before expressing them again in terms of real time. (Note however that imaginary time comes unstuck in General Relativity and the real form of the metric is used).

If we use the Pythagorean metric for 3D space to analyse the diagram above then:

$$(vT)^2 +(cT)^2 = (c\tau)^2 $$

$$T^2 (v^2/c^2 + 1) = {\tau}^2$$

$$ T = \frac {t - vx/c^2}{\sqrt {1+v^2/c^2}}$$

Which is not the Lorentz transformation for time. However, if we resort to imaginary time then:

$$(vT)^2 +(icT)^2 = (ic\tau)^2 $$

gives us:

$$(vT)^2 -(cT)^2 = - (c\tau)^2 $$

$$T^2 = \frac {{\tau}^2}{(1-v^2/c^2)}$$

$$T = \frac {t - vx/c^2}{\sqrt {1-v^2/c^2}}$$

which is the correct result, at least as far as SR goes... ---

$$wT + vx = t$$

So when T, x, and t are all real, we have an angle α so that v=sin(α) and w = cos(α). We can arrange T, x, and t in a right triangle so that $$t = sin(\alpha)*x + cos(\alpha)*T = v*x + w*T. $$


 * If we are going to treat time as just another spatial dimension - using a Pythagorean rather than a Minkowskian form of the metric - the complex form is needed. Whichever convention is chosen for time (real or imaginary), time must always be at right angles to space in a given IFR, this means that the right angle should go between x and t.

$$x = v*t$$

$$T=w*t.$$

$$T=\frac{vx}{w} = $$ $$\frac {vx/c^2}{\sqrt{1 - v^2/c^2}}$$

which makes T perpendicular to x.



Similarly, $$X=\gamma*(x-v*t)$$

$$w*X + v*t = x$$ And by the same reasoning $$x = sin(\alpha)*t + cos(\alpha)*X.$$

$$t = v*x,   X = w*x.$$

If these are both true at the same time, we have t = v*x, x=v*t, t=v*v*t. Somewhere I have jumped to a conclusion.

You can surely see how eager I was to accept the idea that $$x = X \sqrt{1 - v^2}$$ as stated, rather than the other way round. That way it made sense.


 * I think you have made an important contribution to the book. You are obviously very capable of understanding the algebra but the book has failed to make the distinction between Pythagorean and Minkowskian metrics sufficiently clear and so is misleading in its current form.  The question is how to make the use of the different metrics sufficiently clear without losing the reader...RobinH (discuss • contribs) 17:49, 4 November 2013 (UTC)

Jethomas5 (discuss • contribs) 07:29, 4 November 2013 (UTC)

confusion
In the example right after Figure 5, it says that phase difference is proportional to v (which is clearly true) while time dilation is related to the square of velocity (which is not clearly true, time dilation is proportional to 1/sqrt(1-v^2)). It then compares (v/c)^2 to (v/c) to show that loss of simultaneity can be a dominant effect compared to time dilation.


 * The binomial expansion of 1/sqrt(1-v^2)) leads to the expression being proportional to (v/c)^2 when v is less than c - of course, v is usually much less than c. This means that phase can become become the dominant effect if x is large.  But this is an unimportant point in the example and only weakly relates phase to the Andromeda paradox in the next section. On balance I would still leave the comment out because it detracts from the worked example.  RobinH (discuss • contribs) 09:48, 2 November 2013 (UTC)


 * I think I see! When v is small the first term of the binomial expansion dominates, and that's why you used it. Is this an example where classical physics is approximately correct? The larger v gets, the bigger the error in the classical approximation and the more necessary it is to use the relativistic form? It might deserve a longer explanation, somewhere else.

But aren't we actually trying to compare something like T/sqrt(1-v^2) to something like X*v/c^2? It might be simpler to compare (v/c)^2 to (v/c), but it isn't immediately obvious that this is what you want to compare. Jethomas5 (discuss • contribs) 07:47, 9 October 2013 (UTC)
 * Removed offending item pending further consideration. RobinH (discuss • contribs) 09:28, 9 October 2013 (UTC)

Inaccurate remark: “more properly, the time of the object as a function of position”
In this sentence,
 * A world line is a plot of the position of some object as a function of time (more properly, the time of the object as a function of position) on a spacetime diagram.

I believe the remark “more properly, the time of the object as a function of position” is inaccurate. Things can only be in one place at a given time, but they may be in a given place multiple times. Hence, one should give the position as a function of time, not the other way around. I’d be bold and change this if I had a tad more self-confidence (relativity is weird!).

Thanks for the great introduction to special relativity, by the way (I’ve only yet read this chapter). This text has done more for me in fifteen minutes than my high school teacher could in months. — Daniel Brockman 10:43, 29 March 2006 (UTC)


 * This is an interesting point. In mathematical terminology a graph with time as the ordinate should be describing change in time with x (ie: y is a function of x). A thing can be at any point on its worldline. The issue of whether a thing can travel up and down its own worldline is a different problem. The following discussions are about the hypothetical events due to space-time if it were actually possible to move around in the 4D manifold:


 * 1. In a 4D manifold such as spacetime each point is unique. This means that if you travelled back down your worldline by 10 secs you would be the person who was there 10 secs ago (you would not be the person 20 secs in the future). If time exists then it would not be impossible that we are always jumping around our worldline - we would never know.


 * 3. Travelling back through space-time by 10 secs but off your own past worldline is a bizarre idea. Your brain would reset to its state 10 secs ago but when you looked around the world would be entirely different. Your sudden materialisation would surprise the natives and surprise you because you would have no memory of 10 secs in the future. If such travel were possible you would certainly have travelled to a parallel universe because any description of space and time would need to contain both the place where you arrived without you in it and with you in it.


 * 4. Travel "off world line" has other problems, in SR material objects would become infinitely massive as they negotiated the space-time path from now to the past so could not make the trip without consuming a universe of energy, even if the trip were to a parallel universe. However, this is not impossible in general relativity where, for instance, light cones can lean over in areas of curved space-time to provide circular paths through space-time back in time. Even this form of time travel would probably require the concept of parallel universes because the same volume of space-time, the volume where you arrive, could have two states.


 * So, although visiting the same time coordinate many times is not forbidden by the nature of space-time other considerations make such visits either unknowable, require new theories such as parallel universes or energetically impossible.


 * Notice that I introduced this discussion with the caveat "if it were actually possible to move around in the 4D manifold". Relativity, whether Galilean, Special or General cannot explain change - QM cannot either.

RobinH 09:59, 30 March 2006 (UTC)

Intelligibility
Why is this explanation so much more difficult to follow than Einstein's original 1905 paper? (I refer to the 1923 translation to English). I can understand that modern understanding of the theory would be more complete, but modern explanations are much inferior when it comes to the ability of a newcomer to comprehend.

PAS 02:50, 24 July 2006 (UTC)


 * This book is about Special Relativity. Special Relativity is a theory that evolved between 1905 and 1915. It is the theory that the manifold of events that constitute the universe can be described using a metric with signature (---+). Physical laws and the constant "c" are consequences of the invariants in this metrical structure. Einstein's 1905 paper is not the sole, or even definitive, text in the evolution of Special Relativity, Minkowski's 1908 presentation is probably more definitive. Einstein would probably have been given equal precedence with Poincare and Lorentz had he not produced GR in 1915.


 * Einstein's 1905 paper is not an explanation of SR, it is the mathematical application of a particular assumption about a constant velocity for all observers. The assumption was immediately problematical because "velocity" is a relationship between length and time intervals (esp. in Galilean Relativity). Within 4 years of Einstein's 1905 paper the origin of the assumption about a constant velocity had become understood in terms of metrical geometry and Einstein's proposal was placed fully in a theoretical context.


 * In my experience students fail to understand relativity because the assumption of a constant velocity without any explanation appears absurd to them. If they are told that by making an apparently absurd assumption they can derive the Lorentz transformation many of them will feel that the whole exercise is apparently absurd. On the other hand I have not found any students who find the geometrical approach to be absurd and 100% understand SR having learnt it. RobinH 08:57, 24 July 2006 (UTC)

87.160.202.247 (talk) 07:25, 7 August 2010 (UTC)
 * Einstein stated himself: "After my theory of relativity has been dissected by mathematicians I don't understand it anymore". This seems to be a general problem of modern physics - mathematics and particularly mathematical models become more more primary over observation, experiment and conclusive predictions based on them. SR describes a set of laws of nature and as such it can and must be understood in terms of natural effects predicted by the theory, geometrical approach is secondary and cannot balance out lack of imagination or poor logic skills (both are ridiculously often demonstrated by students or even respectable full-size physicists).