Friday, 17 August 2012

Special Relativity: From Galileo to Einstein


The intent of this article (adapted from an original paper I wrote about a decade ago) is to show how Special Relativity can be arrived at by examining Galilean boosts along with the inherent assumptions and then removing invalid assumptions. During the process it becomes clear that there is a potential source of confusion associated with the standard expression of time dilation. Note that this is the “methodical” version; there is also a simplified version which (I hope) will make more sense to those who are less gifted mathematically.

1. Introduction


Galilean Relativity relies on a number of assumptions. A number of these assumptions have been shown to be invalid in the real world and Einstein’s Special Relativity, which dispenses with those assumptions, is a far superior conceptual framework. Einstein arrived at Special Relativity via “mind experiments” and brilliant reasoning. I intend to demonstrate an alternate method of arriving at Special Relativity.

2. Galilean Boost

A Galilean boost is the equation which allows x’ to be calculated in Figure 1 where S is nominally stationary and M is nominally in motion with a velocity v. The equation relies on the following assumptions:

                                            when t = 0 and t’ = 0, S and M are coincident
                                            S and M observe the same event
                                            S and M measure time and space identically
                                            S and the observed event (X) share a preferred frame
                                            information is transmitted instantaneously

The boost is given by:
x’ = x – vt
t’ = t

3. Information is not transmitted instantaneously

We know that it actually takes time for information about the event X to reach each of the observers.

This results in a modification to an assumption, such that event X takes place when S and M are coincident, and t = 0 and t’ = 0.  The extant assumptions are, therefore:

                                            the observed event takes place when S and M are coincident, and t = 0 and t’ = 0
                                            S and M observe the same event
                                            S and M measure time and space identically
                                            S and the observed event (X) share a preferred frame

The modified Galilean boosts are:

x’ = x – vt
ct’ = x’ = x – vt = ct – v.x/c
t’ = t – v.x/c2

4. No preferred frame



We referred to S as stationary, but this merely means that S is stationary relative to a nominated “observer” (conceptually the reader). Relative to M, S has a velocity of -v. There is no valid reason to assume that either of the perspectives (or “frames”) is preferred.




Note that the unprimed frame pertains to S and the primed frame pertains to M.  That is to say:

·         in the unprimed frame, S is stationary and M has a velocity of vs according to S and

·         in the primed frame, M is stationary and S has a velocity of vm according to M.

The magnitude of vs and vm are such that:

vs = -vm = v

The extant assumptions are:

– the observed event takes place when S and M are coincident, and t = 0 and t’ = 0
– S and M observe the same event
– S and M measure time and space identically

The third assumption cannot stand without a preferred frame, since x’m > x’s and xm > xs where vs > 0.


5. S and M do not measure time and space identically


Figure 3 is modified slightly to reflect that vs = -vm = v.

The extant assumptions are:

                                            the observed event takes place when S and M are coincident, and t = 0 and t’ = 0
                                            S and M observe the same event.

According to each observer, the other utilises different, but consistent units of measurements – in the other’s own frame.  The unprimed frame pertains to S, so:

                                                                       xm = .xs        (1)
The primed frame pertains to M, so:

                                                                      x’s = .x’m        (2)

where is yet to be determined.  Taking first the perspective of S:
x’s = xs vt’s 
xs = x’s + vt’s

but since xm = .xs and t’s = x’s / c
                                          xm = .xs = .(x’s + vx’s / c) = . x’s (1 + v / c)         (3)

Then taking the perspective of M:

xm = x’m - (-vtm) = x’m + vtm
                                                                   x’m = xm - vtm         (4)

but since x’s = .x’m and tm = xm / c
                                          x’s = .x’m = .(xm - vxm / c) = . xm (1 - v / c)        (5)

Applying (5) to (3):

xm = . x’s (1 + v / c) = . [. xm (1 - v / c)].(1 + v / c)
xm =  2. xm (1 - v / c).(1 + v / c)
1 =  2. (1 - v2 / c2)
                                                               = 1 / (1 - v2 / c2)½          (6)

6. Lorentz boosts



Understanding a Galilean boost is relatively simple, since there is only one perspective. With a Lorentz boost, which we can derive from the information in the previous section, we have two perspectives to select from – that of S and that of M. Since we have already stated that S is stationary by virtue of being stationary relative to us, we will examine the situation from the perspective of S.
According to S, the distance between the event X and the point at which M observes that event is x’s = xs vt’s.  The Lorentz boost doesn’t seek to tell us this, it seeks to tell us what that distance is in terms of what M observes.  To arrive at the relevant equation, we make use of (2), (4) and (6):

                                                                      x’s = .x’m
                                            x’s = .(xm - vtm) = (xm - vtm) / (1 - v2 / c2)½          (7)

and since x’s = ct’s and xm = ctm so that tm  = xm / c:

                         x’s = ct’s = (xm - vtm) / (1 - v2 / c2)½ = (ctm - vxm / c) / (1 - v2 / c2)½
                                                    t’s = (tm - vxm / c2) / (1 - v2 / c2)½         (8)

It is trivial to show that a similar equation can be derived from the perspective of M, in terms of what S observes, so long as one keeps in mind that vs = -vm = v.

7. Finding x and t in the other frame

Rather confusingly, a different priming convention applies for denoting relativistic effects on space and time.  When considering relativistic effects, a primed value refers to a measurement within a frame in relative motion, as made by an observer from within that frame in relative motion – in terms of units which pertain to the nominally stationary frame.

Consider xs and xm:

                                            xs is a measurement made by observer S within the S frame which, according to observer M, is in motion
                                             what is xs in terms of xm?

This is equivalent to the question what is x’ in terms of x?  We apply (7) to (1):

                                                                       xm = .xs

                                                                        xs = xm / = xm . (1 - v2 / c2)½
                                
                                                            x’ = x . (1 - v2 / c2)½         (9)

 and since xs = cts and xm = ctm:

                                                       xs =  cts = ctm . (1 - v2 / c2)½                                        

                                                              ts = tm . (1 - v2 / c2)½                                              

                                                             → t = t . (1 - v2 / c2)½          (10)

The same logic can be applied to x’s and x’m and the same result achieved.

8. On the result achieved by standard derivations

Standard derivations achieve a different result, viz:

                                                              L' = L . (1 - v2 / c2)½          (11)

                                                               Δt' = Δt / (1 - v2 / c2)½         (12)

This is because, in the standard derivations, Δx’ corresponds with x’ as expressed in the boost while Δt’ refers to something quite different, something that is conceptually inverse to t’ as expressed in the boost.  Conceptually, the term Δt’ refers a period measured in a frame in motion (relative to the observer).  In other words, Δt’ is equivalent to Δtm rather than Δts:



It must, however, be noted that a key assumption in this derivation is that if x = ct then x’ = ct’, and by extension that if Δx = Δct then Δx’ = Δct’, so that the use of the prime notation is consistent.  If the use of the prime notation used in the length contraction and time dilation equations is intended to be consistent, then there is a problem with this assumption as used explicitly to transition from (8) to (9).  Therefore it behoves me to present a derivation which does not rely on this assumption.

9. Finding t in the other frame by other means


Recall that according to each observer, the other utilises different, but consistent units of measurements – in the other’s own frame.  The unprimed frame pertains to S, so:

                                                                       tm = .ts

                                                                       = tm / ts        (13)

The primed frame pertains to M, so:

                                                                      t’s = .t’m

                                                                      = t’s / t’m         (14)

where is yet to be determined.  Multiplying (13) by (14):

                                                            . = (t’s / t’m).(tm / ts)

Rearranging:

                                                             2 = (t’s / ts).(tm / t’m)        (15)

According to S, M moves towards the event at velocity v while information about the event moves towards both S and M at c, so:

                                                    ts / t’s = c / (c - v) = 1 / (1 - v / c)       (16)

According to M, s moves away from the event at velocity v while information about the event moves towards both S and M at c, so:

                                                   t’m / tm = c / (c + v) = 1 / (1 + v / c)          (17)

Applying (16) and (17) to (15):

                                    2 = (t’s / ts).(tm / t’m) = [1 / (1 - v / c)].[1 / (1 + v / c)]                    

                                                                2 = 1 / (1 - v2/ c2)

                                                            = 1 / (1 - v2/ c2)½ =         (18)

Applying (18) to (13):

                                                        ts = tm / = tm . (1 - v2/ c2)½                                        

Since ts is a measurement made by observer S within the S frame which, according to observer M, is in motion, then:

                                                              ts = tm . (1 - v2/ c2)½

                                                               t' = t . (1 - v2/ c2)½      (10)

As described in Section 8, this corresponds to the time dilation equation (where tm = Δt' and ts = Δt), so:

Δt = Δt' . (1 - v2/ c2)½
Δt' Δt / (1 - v2/ c2)½      (12)

10. Conclusion (as modified)

The equations for Special Relativity can be derived mathematically by means of systematically removing invalid assumptions associated with Galilean relativity.

No comments:

Post a Comment

Feel free to comment, but play nicely!

Sadly, the unremitting attention of a spambot means you may have to verify your humanity.