5.8.3 First Example: Back to SR (General Relativity)
Description
This article is from the Relativity and FTL Travel FAQ, by Jason W. Hinson jason@physicsguy.com with numerous contributions by others.
5.8.3 First Example: Back to SR (General Relativity)
The most simple application of the ideas expressed in Section 5.8.2 is
one which we have already looked at (though without using the concepts
discussed in that section). It is the situation where there is no
gravitational field. That is exactly the situation we were considering when
we discussed special relativity. In special relativity, there is no
gravitational field. All the components of the stress-energy tensor are
identically zero.
Now, we will figure out the metric of space-time in such a case by
examining what we already know about special relativity. So, let's go back
to our space-time diagrams. (By the way, our diagrams only considered one of
the spatial dimensions, but we will incorporate the other two in this
section.) Consider two observers who start out moving parallel to one
another on the diagram. This would mean that they start out with the same
velocity in any inertial frame. Well, in special relativity (with no
gravitational field) the two observers will continue to remain on parallel
paths on the space-time diagram. This is the property of a flat manifold, so
in SR, space-time is "flat".
Before we go on, it will be helpful for us to redefine the time
variable in our space-time coordinates. Instead of "t", consider the
combination "c*t" (where c is the speed of light). For convenience, we will
simply define a new variable, w, where
(Eq 5:30)
w = c*t
Then we can use w in place of t in our coordinates. This is actually a
fairly natural substitution in a couple of ways: First, note that w has the
units of length, just like x, y, and z do. Second, using w on our space-time
diagrams makes them a little more general. Why? Well, remember how we
defined the units of length and time to be the light-second and the second?
We did this so that a light ray would make a line at a 45 degree angle on
our diagram. Well, with a w-x coordinate system, this will automatically be
the case, regardless of what units you use. To see this, note that the value
of t at a certain value of w is just the time it takes for light to travel
that length, w (because t = w/c). For example, the point x = 1 light-second
and t = 1 second corresponds to the point x = 1 light-second and w = 1
light-second. So, on both an x-t diagram and on an x-w diagram, a light beam
would make a 45 degree angle with the x axis by going through the point
(1,1). However, if we wanted to, we could now use a meter as our unit of
length. Then, when w = 1 meter, t would just be the time it takes for light
to travel 1 meter. So, the point x = 1 meter, w = 1 meter also lies on the
light path, and again, that light path would automatically make a 45 degree
angle with the x axis by going through the point (1,1). For consistency, we
will continue to use units of seconds and light-seconds, but we will now use
"w" in units of light-seconds to indicate time in our discussions and
diagram (remember, the length "w" just represents the time it takes light to
travel that length).
Now, let's look at a change in coordinates on the flat space-time of
SR. In space-time, a change in coordinates can represent a change in an
observer's frame of reference. So, when we discussed two observers who were
moving with respect to one another, we were looking at two different
coordinate systems (x-t and x'-t', or now, x-w and x'-w') which both
correctly described space-time in SR. This leads us to consider the
invariant interval, because we know it must be the same for each of these
two coordinate systems. So, let's take a closer look at these coordinate
systems on our diagrams and see if we can't define the invariant interval
(which, remember, is just another way of writing the metric).
We will specifically want to consider infinitesimal lengths like dx.
So, let's look at a small line segment which lies on a particular
geodesic--a geodesic we know a little about. That geodesic is the path which
light follows. Like anything else being acted on only by gravity, light must
follow a geodesic on the space-time manifold. So, for the particular case of
a light path, a small segment on that path would have an x component (dx)
and a t component (dt); however, we now want to begin thinking of w as the
unit which represents time, so we note that a small change in t (dt)
represents a change in w of dw = c*dt. Now, since the small distance light
travels (dx) divided by the time (dt) it took it to travel that distance is
defined as the speed of light, then we have the following:
(Eq 5:31) dx -- = c (where c is the speed of light) dt
which can be rewritten as
(Eq 5:32) dx -- = 1 dw
That means that dx = dw (for light). Now, since we always define the
invariant interval in terms of the infinitesimal lengths squared, we will
actually want to square both sides of that equation and then bring
everything to one side so as to get the following:
(Eq 5:33) dx^2 - dw^2 = 0 (For light)
Now, because the speed of light is the same for all inertial observers, the
above equation must be true for all frames of reference. Thus, we might
consider the idea that the invariant interval for any small line segment
(not just for light) is given in SR by
(Eq 5:34) ds^2 = dx^2 - dw^2,
and this turns out to be the case. The light path, then, is just the case
where ds^2 = 0.
Now, let's note a few things about this interval. First, it is
independent of where you are in space-time. All that matters is the lengths
dx and dw, regardless of what actual x and w position you have. This means
that the distances (like dx) don't have to be infinitesimal, because the
equation remains true regardless of how far you extend dx and dw. Thus,
let's consider the case where one side of the line segment is at x = w = 0
(the origin). Then dx will be the x distance from the origin to the end of
the line segment (which in this case can be as far away as we like), and dw
will be the w distance to that point. In other words, for SR, dx and dw can
be replaced with x and w when we consider one side of the line segment to be
at the origin. Further, consider a point in space-time with coordinates
(x,w) in the o observer's coordinates and (x',w') in the o' observer's
coordinates. Since the value of the invariant interval is the same for any
frame of reference, the following must be true:
(Eq 5:35) x^2 - w^2 = x'^2 - w'^2
Let's see that this is the case on our space-time diagrams. Diagram 5-9
shows a space-time diagram with two coordinate systems indicated, one for an
observer o, and a second for an observer (o') moving with velocity 0.6 c
with respect to o. (Note that now we use w = ct for the time axes.) There is
also a point marked "*" on the diagram. The x'-w' coordinates for that point
are clearly shown to be x'= 1 light-second and w'= 2 light-seconds (i.e. t'=
2 second, remember?). The x-w coordinates are x = 2.75 light-seconds and w =
3.25 light-seconds, and I tried to show this as best I could with an ASCII
diagram.
Diagram 5-9
w w'
| /
| /
| /
w=3.25 |-> / *
+ / ' '
| / ' '
| w'=2+' '
| / '
| / '
+ / ' x'
| / ' '
| / ' '
| / ' '
| + ' +'
| / ' '
+ / ' '
| / + '
| / 'x' = 1
| / '
|/ '
--+-----------o---------+----------+---------+--->x
' /| ^
' / | x=2.75
We therefore find the following:
(Eq 5:36)
ds^2 = x^2 - w^2 = (2.75)^2 - (3.25)^2
= -3 light-seconds^2
and
ds'^2 = x'^2 - w'^2 = (1)^2 - (2)^2
= -3 light-seconds^2
There are a couple notes to make about this outcome. First, of course,
we note that ds^2 = ds'^2, as it must be. In fact, it is the form of the
invariant interval and the fact that it must be invariant from one
coordinate system to another that causes the transformation from x-w to
x'-w' to look as it does. If the x' and w' axes didn't look the way they do
relative to the x and w axes in our diagrams, then the interval would not be
invariant. Note that if the "-" sign in the invariant interval were a "+"
sign, then the invariant interval would look just like the one for a normal,
space-only x-y coordinate system where ds^2 = dx^2 + dy^2. Then, the
coordinate transformation to x'-w' would be just like a rotation of
coordinates (see Diagram 5-10). The "-" sign in the SR interval causes one
of the axes to rotate in the opposite direction from the other when we do
our space-time coordinate transformation.
Second, note that the interval squared is, in fact, negative. This is
not too distressing, because we know that _physical_ lengths on our diagram
do not represent the space-time "lengths" which the invariant interval gives
us. If they did, then the invariant interval for special relativity would be
just like the x-y form of the invariant interval (since the physical lengths
on our diagrams are just normal lengths on the flat paper/screen we draw
them on). Now, the actual length of an infinitesimal interval on a manifold
is usually defined to be the square root of the absolute value of ds^2.
Thus, we can still make sense of lengths, even when the invariant interval
squared is negative.
Diagram 5-10
x'-y' is rotated from x-y, and the line segment
in the two diagrams are identical
y y'
| /
| /
| / / /
| / . / / '
| ds / . / ds / '
| / . dy / / '
| / . / / 'dy'
| /.......... / / '
| dx / ' . '
--+------------------ x + dx' '
| \
\
\
Note: the length of the line segment \
doesn't change just because you rotated x'
the coordinate system, so
dx^2 + dy^2 = dx'^2 + dy'^2
The reader may have noted that thus far in our look back at special
relativity we have still only included two of the four dimensions of
space-time. The other two (y and z) could actually replace x in any of our
discussions, and so they play the same roll in the invariant interval as x
does. Therefore, the total four dimensional invariant interval for special
relativity is given by
(Eq 5:37) ds^2 = dx^2 + dy^2 + dz^2 - dw^2
Finally, let's talk about some physics in this space-time using the
concepts discussed in the previous section. First, consider the proper time
between two time-like separated events. Recall that we defined this time
such that:
(Eq 5:38)
ds^2 = g (of SR)*dT^2
tt
We now know that g_ww = -1 for SR from the above, so g_tt = -c^2 for SR.
This is how we got Equation 5:28, which is duplicated here:
(Eq 5:39--Copy of Eq 5:28) dT^2 = -ds^2/c^2.
in the previous section. However, since we are now working with w for our
time coordinate, we should define dW = c*dT, and rewrite Equation 5:39 as
(Eq 5:40) dW^2 = -ds^2
Now, let's consider the observer which followed the t' axes in Diagram 5-9
such that his velocity was 0.6 c. Consider the O observer's frame of
reference, and note that if it takes O' a certain time (dw) to travel a
certain distance (dx) in the O observer's coordinates, then it must be the
case that dx/dt = 0.6 c. So dx/dw = 0.6, or
(Eq 5:41) dx = 0.6*dw
This, then, is true all along the w' axes (the line that O' follows through
the O observer's coordinate system). So, the invariant interval (considering
only two dimensions once again) at any point along the w' axes must be given
by the following (using Equation 5:37 with only x and w coordinates and
substituting Equation 5:41):
(Eq 5:42)
ds^2 = dx^2 - dw^2
= [0.6]^2*dw^2 - dw^2 = -[1 - 0.6^2]*dw^2
plugging this into Equation 5:40 we find that
(Eq 5:43) dW^2 = [1 - 0.6^2] * dw^2
so,
(Eq 5:44)
1
dw = --------------- * dW = gamma*dW
SQRT[1 - 0.6^2]
Since dW just represents an infinitesimal time as measured on our "moving"
observer's clock, and dw an infinitesimal time measured on our clock,
Equation 5:44 is just the equation which shows time-dilation effects in SR,
and it was quickly derived using our new knowledge.
For another physics consideration, look at the momentum four-vector. We
defined this earlier (Equation 5:29) and it is duplicated here:
(Eq 5:45--Copy of Eq 5:29) a p = m*da/dT
Again, we want to use dW = c*dT, and we thus find
(Eq 5:46) a p = m*c*da/dW
For us, we consider the situation where "a" is the x dimension. Then, p^x'
for the "moving" observer himself is zero (because all along the w' axes we
have dx' = 0 by definition, i.e. he is not moving relative to himself).
However, for the O observer (for whom the "moving" observer moves a distance
dx in a time dw) we find the following from Equation 5:46 by substituting x
for a(Note that from Equation 5:44 we can write dW = dw/gamma, and we are
substituting that here. We also use dw = c*dt and v = dx/dt in this
equation.):
(Eq 5:47) x p = m*c*dx/[dw/gamma] = gamma*m*c*dx/dw = gamma*m*dx/dt = gamma*m*v.
This is exactly the definition of the momentum we saw in our discussions of
special relativity.
However, now we can also look at the time component of the momentum
four-vector and figure out what it represents. Again we use Equation 5:46,
but here we substitute w for x:
(Eq 5:48) w p = m*c*dw/[dw/gamma] = gamma*m*c
But this is just the energy we had defined in SR (E = gamma*m*c^2) divided
by c:
(Eq 5:49) w p = E/c.
And so, we now know all about the components of the momentum four-vector of
a particle: three are the spatial components of the momentum of the
particle, and the time component represents the energy of the particle
divided by c.
As a final bit of physics, consider the dot product (as defined in
Equation 5:26) of the momentum four-vector with itself:
(Eq 5:50)
w w x x
p (dot) p = g *p *p + g *p * p
ww xx
= -[E/c]^2 + p^2
(Note that the total momentum of this observer is p^x, and so we write p^2
in the last line to mean the total momentum squared). Now, recall that the
dot product is invariant, so that if any observer measures the energy and
momentum of a particle and calculates the above equation in his frame of
reference, he must find the same number that any other observer would find
in any other frame of reference. This shouldn't come as too much of a
surprise if we look back for a moment. Back when we discussed energy and
momentum in special relativity, we found in Equation 1:7 that E^2 = m^2*c^4
+ p^2*c^2. Thus, we find that the dot product in Equation 5:50 is simply
equal to -m^2*c^2. Since m and c are invariant (remember, m is the rest
mass), we could have already known that the formula in Equation 5:50 would
be invariant.
We have therefore been able to find all the major physics equations we
saw in special relativity by simply apply some tensor analyses using the
metric of flat space-time.
So, to sum up, we have found the following: For SR, where there is no
gravitational field, space-time has the properties of a flat manifold. The
invariant interval of a flat space-time manifold is given by the following:
(Eq 5:51--Copy of Eq 5:37) ds^2 = dx^2 + dy^2 + dz^2 - dw^2
That interval tells us all about the nature of space-time in SR. The fact
that the contribution of the time component (dw) is negative where as the
spatial components have positive contributions is what gives the coordinate
transformation between different frames of reference its unique form. Thus,
it is the negative sign which essentially causes time dilation and length
contraction effects, and it is the fact that the speed of light is invariant
which causes that sign to be negative.
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