Ray said:
Pete, many thanks for your reply.
Yes, you're right the web page arrives at a result in section 1.6 of the book by a different method, but that's precisely the problem. I do not understand how the diagram in Schutz is constructed nor how it proves the un-numbered result nor Schutz's use of the principle of relativity to prove figh(v) =figh(mod v). Any ideas?
Okay. We'll start from scratch. First off its not "figh(v)" its phi(v). It took me a while to recognize your spelling of it so let's use phi(v) (i.e. \Phi(v)) so that those who are following won't be as confused as I was. Okay?
Let's start from the beginning. We are trying to completely comprehend Section 1.6 in Schutz's text
A first course in general relativity. This section is covered from page 10 to 15. This section shows that the spacetime interval is a Lorentz invariant.
Schutz starts out by describing two events on the world line of the same light beam. He calls these events
E and
P. Inframe O the events themselves are labeled (t, x, y, z). The difference between two events is labeled (\Delta t, \Delta x, \Delta y, \Delta z). The two events
Eand
P satisfy the relation (Choosing units so that c = 1)
-(\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 = 0
Since the speed of light is a Lorentz invariant we must also have
-(\Delta t')^2 + (\Delta x')^2 + (\Delta y')^2 + (\Delta z')^2 = 0
Where events in O' are labeled (t', x', y', z'). This motivates Schutz to define the
spacetime interval, \Delta s^2 (which is the square of \Delta s) as
\Delta s^2 = -(\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 = 0
Note: Please read footnote on page 10 carefully.
It is readily seen that for the spacetime interval between the two events
E and
P] is \Delta s^2 = 0. This is true wheher you choose coordinates in O or in O'. In O' this reads \Delta s'^2 = 0. Therefore this is a
geometric relationship between the events, i.e. one that is not dependant on coordinates.
The steps from Eq. (1.2) to 1.5 are difficult. I recommend that you follow this with pen and paper and convince yourself that these steps are correct. If you have a problem in doing so then come back and let me know and I'll help you through it. However it appears that you've already done this so on we go. What we end up with is Eq. (1.5)
(1.5) \Delta s'^2 = \Phi(v)\Delta s^2
The next step is to show that \Phi(v)[/tex] = 1. Now consider the rod which is lying on the y axis. This is shown in Fig. (1.6). This diagram is supposed to show that a rod is at rest in O and perpendicular to the x-axis. The length of the rod will equal the square root of the spacetime interval between the events <i>A</i> and <i>B</i> as shown shown in Fig (1.6). How do we know this? Consider the difference in time between those events as measured in O. The diagram shows that they are simultaneous events as observed in O. Therefore the spacetime interval for these two events, given that<br />
<br />
\Delta t = \Delta x = \Delta z= 0<br />
<br />
\Delta s^2 = \Delta y^2<br />
<br />
\Delta y is the <i>proper length</i> of the rod. Why? Because in O the rod is at rest and since \Delta t = 0 the magnitude of the spacetime interval is the magnitude of the length of the rod which is at rest in frame O. Thus the term "rest length" is used for the intrinsic length of the rod. I prefer to use the term <i>proper length</i>. So if you see that term in the future you'll know what it means. Now take a look at the clock's worldline as shown in Fig. (1.7). The clock starts at the midpoint between events <i>A</i> and <i>B</i>. If you carefully study this diagram you'll see that the clock is at rest in O' and is thus moving with respect to O. The worldline is parallel to the xt-plane. However the distance spatial distance between events <i>A</i> and <i>B</i> are equal and this means that if at these events a flash of light is emitted from each end of the rod the flashes will reach the clock at the same time which means \Delta t&#039; = 0 as well. If we substitute \Delta t&#039; = \Delta x = \Delta z = 0 into the spacetime interval for \Delta s&#039;^2 we get the following relationship<br />
<br />
\Delta y&#039;^2 = \Phi(v)\Delta y^2<br />
<br />
which is exactly what the equation above Eq. (1.6 on page 13 reads). Now switch the direction that O' is moving and you will obtain<br />
<br />
\Delta y&#039;^2 = \Phi(-v)(\Delta y)^2<br />
<br />
This is interpreted to mean that \Phi(v) = \Phi(|v|). Follow the next paragraph to where it shows that \Phi(v) = +-1. This results in <br />
<br />
\Delta s&#039;^2 = \Delta s^2<br />
<br />
This means that the spacetime interval is a Lorentz invariant. This can also be shown rather easily by taking the expression for \Delta s^2 and substitute the variables for system O' by using the Lorentz transformation. This will yield the same result.<br />
<br />
Let me know if you need more help.<br />
<br />
Best wishes<br />
<br />
Pete
