Refer to:
http://www.pma.caltech.edu/Courses/p...p01/0201.2.pdf - pg 16.

The link is to Blanchard and Thorne's Applications of Classical Physics pdf
texts. In ch. 1 - sect 1.4 they state that "The squared length of the
particle's 4-velocity is easily seen to be -1:" - then show that by eq.
1.27.

eq. 1.27: u^2 = g(u,u) = dx/dt dot dx/t - (dx dot dx)/(dt)^2 = -1

In Eq. 1.27 - why does (dx dot dx)/(dt)^2 equal -1?

As the text states, u is some particle's 4-velocity, t is time as measured
by an ideal clock that the particle carries (the particle's proper time),
and x is the location of the particle in spacetime when its clock reads t
(or, equivalently, the vector from the arbitrary origin to that location).

Thanks,

Perion

The "d" in the original subject was a typo.

Perion

"Perion" wrote in message ...
Refer to:
http://www.pma.caltech.edu/Courses/p...p01/0201.2.pdf - pg 16.

The link is to Blanchard and Thorne's Applications of Classical Physics pdf
texts. In ch. 1 - sect 1.4 they state that "The squared length of the
particle's 4-velocity is easily seen to be -1:" - then show that by eq.
1.27.

eq. 1.27: u^2 = g(u,u) = dx/dt dot dx/t - (dx dot dx)/(dt)^2 = -1

In Eq. 1.27 - why does (dx dot dx)/(dt)^2 equal -1?

The next two lines in the text explain:
| "The last equality follows from the fact that dx dot dx
| is the squared length of dx which equals the invariant
| interval (Ds)^2 along it, and (d tau)^2 is minus that
| invariant interval."
See also equation (1.11) page 10 and equation (1.7) on page 9.

Dirk Vdm

"Dirk Van de moortel" wrote in message ...

Refer to:
http://www.pma.caltech.edu/Courses/p...p01/0201.2.pdf - pg 16.

The link is to Blanchard and Thorne's Applications of Classical Physics pdf
texts. In ch. 1 - sect 1.4 they state that "The squared length of the
particle's 4-velocity is easily seen to be -1:" - then show that by eq.
1.27.

eq. 1.27: u^2 = g(u,u) = dx/dt dot dx/t - (dx dot dx)/(dt)^2 = -1

In Eq. 1.27 - why does (dx dot dx)/(dt)^2 equal -1?

As the text states, u is some particle's 4-velocity, t is time as measured
by an ideal clock that the particle carries (the particle's proper time),
and x is the location of the particle in spacetime when its clock reads t
(or, equivalently, the vector from the arbitrary origin to that location).

Thanks,

Perion

Post a follow-up to this message

From: Dirk Van de moortel )
Subject: Why does the 4-vel squared = -1d?
Newsgroups: sci.physics.relativity
Date: 2003-10-10 07:03:57 PST


"Perion" wrote in message ...
Refer to:
http://www.pma.caltech.edu/Courses/p...p01/0201.2.pdf - pg 16.

The link is to Blanchard and Thorne's Applications of Classical Physics pdf
texts. In ch. 1 - sect 1.4 they state that "The squared length of the
particle's 4-velocity is easily seen to be -1:" - then show that by eq.
1.27.

eq. 1.27: u^2 = g(u,u) = dx/dt dot dx/t - (dx dot dx)/(dt)^2 = -1

In Eq. 1.27 - why does (dx dot dx)/(dt)^2 equal -1?

The next two lines in the text explain:
| "The last equality follows from the fact that dx dot dx
| is the squared length of dx which equals the invariant
| interval (Ds)^2 along it, and (d tau)^2 is minus that
| invariant interval."
See also equation (1.11) page 10 and equation (1.7) on page 9.

Dirk Vdm

So far as I know, there is no mention of sqrt(-1) in Einstein's
original Electrodynamics of 1905. This concept was introduced
by Minkowski in 1908, and he died before completing his work
to it's logical conclusion. Too maintain some conventional
orthogonality, but with a caveat he said, (see P of R,
pg 77, "Now what has the orthogonality...perfect freedom
of the time axis".)
In his caveat, he is perfectly aware non-orthogonal axes
do not need the sqrt(-1), and I have confirmed these ideas
in a previous thread, and I agree.
In summary, any introduction of -1 or sqrt(-1) is pure
convention.
Regards
Ken S. Tucker

"Dirk Van de moortel" wrote
in message ...

The next two lines in the text explain:
| "The last equality follows from the fact that dx dot dx
| is the squared length of dx which equals the invariant
| interval (Ds)^2 along it, and (d tau)^2 is minus that
| invariant interval."
See also equation (1.11) page 10 and equation (1.7) on page 9.
Yes - thanks - I did go back and study those two eqs but I guess my real
confusion was in eq 1.7 where it seems just to be an ad hoc manipulation
just to eliminate an imaginary number for a timelike interval. Is this so?

Perion

"Perion" wrote in message ...

"Dirk Van de moortel" wrote
in message ...

The next two lines in the text explain:
| "The last equality follows from the fact that dx dot dx
| is the squared length of dx which equals the invariant
| interval (Ds)^2 along it, and (d tau)^2 is minus that
| invariant interval."
See also equation (1.11) page 10 and equation (1.7) on page 9.

Yes - thanks - I did go back and study those two eqs but I guess my real
confusion was in eq 1.7 where it seems just to be an ad hoc manipulation
just to eliminate an imaginary number for a timelike interval. Is this so?

Not so much an ad hoc manipulation, but rather a neat way
of abbreviating the meaningful (and apparently important)
physical quantity
- Dt^2 + Dx^2 + Dy^2 + Dz^2
to something that has physical meaning and that involves real
numbers only.

For two arbitrary events we could just as well have defined
S = - Dt^2 + Dx^2 + Dy^2 + Dz^2
and then considered 3 separate cases:
1) if S 0
- define Ds = sqrt(S)
- DON'T define D\tau
2) if S = 0
- define Ds = 0
- define D\tau = 0
3) if S 0
- define D\tau = sqrt(-S)
- DON'T define Ds

This way there would be no (potentially) imaginary numbers,
but it would require *every* subsequent calculation, proof or
problem to separately consider these 3 separate cases for any
possible pair of events in the problem at hand. That would be
extremely tedious and very silly since, through the well known
properties of the real and complex numbers, we perfectly know
that we would get the same results anyway.

Dirk Vdm

"Dirk Van de moortel" wrote
in message ...

Not so much an ad hoc manipulation, but rather a neat way
of abbreviating the meaningful (and apparently important)
physical quantity
- Dt^2 + Dx^2 + Dy^2 + Dz^2
to something that has physical meaning and that involves real
numbers only.

For two arbitrary events we could just as well have defined
S = - Dt^2 + Dx^2 + Dy^2 + Dz^2
and then considered 3 separate cases:
1) if S 0
- define Ds = sqrt(S)
- DON'T define D\tau
2) if S = 0
- define Ds = 0
- define D\tau = 0
3) if S 0
- define D\tau = sqrt(-S)
- DON'T define Ds

This way there would be no (potentially) imaginary numbers,
but it would require *every* subsequent calculation, proof or
problem to separately consider these 3 separate cases for any
possible pair of events in the problem at hand. That would be
extremely tedious and very silly since, through the well known
properties of the real and complex numbers, we perfectly know
that we would get the same results anyway.

I understand. It does simplify the situation. I'm curious why many authors
prefer the -1,1,1,1 signiture for the metric rather than 1,-1,-1,-1. Since
any possible observable event for is either timelike or null (lightlike) it
seems that it would be more natural for the time component in the interval
to be positive. Then ds^2 = dt^2 - dx^2 - dy^2 - dz^2 which would always be
greater than or equal to zero.. I know it makes no difference because then
they'd just define x_0 = ct, x_1 = -x, x_2 = -y, x_3 = -z, etc. Hmmm....
Never mind - I see why someone might prefer the former signiture.
[Sorry... I think I'm rambling]

Perion

"Perion" wrote in message ...

"Dirk Van de moortel" wrote
in message ...

Not so much an ad hoc manipulation, but rather a neat way
of abbreviating the meaningful (and apparently important)
physical quantity
- Dt^2 + Dx^2 + Dy^2 + Dz^2
to something that has physical meaning and that involves real
numbers only.

For two arbitrary events we could just as well have defined
S = - Dt^2 + Dx^2 + Dy^2 + Dz^2
and then considered 3 separate cases:
1) if S 0
- define Ds = sqrt(S)
- DON'T define D\tau
2) if S = 0
- define Ds = 0
- define D\tau = 0
3) if S 0
- define D\tau = sqrt(-S)
- DON'T define Ds

This way there would be no (potentially) imaginary numbers,
but it would require *every* subsequent calculation, proof or
problem to separately consider these 3 separate cases for any
possible pair of events in the problem at hand. That would be
extremely tedious and very silly since, through the well known
properties of the real and complex numbers, we perfectly know
that we would get the same results anyway.


I understand. It does simplify the situation. I'm curious why many authors
prefer the -1,1,1,1 signiture for the metric rather than 1,-1,-1,-1. Since
any possible observable event for is either timelike or null (lightlike) it
seems that it would be more natural for the time component in the interval
to be positive.

But the time component can be either positive or negative,
in both cases giving a real square.
I guess you mean that for observable events "it would be
more natural for the time component in the interval to be
real" and thus taking the (+---) signature. But that doesn't
hold either since with the (-+++) signature the definitions
are made the other way anyway ;-)

Then ds^2 = dt^2 - dx^2 - dy^2 - dz^2 which would always be
greater than or equal to zero..

Only for observable events as seen by a specific observer
who is present at one particuar event.
But there are other events as well. If you are an observer
then you can think about two observable events
(t,x,y,z) and (t+dt,x+dx,y+dy,z+dz)
while the second events can be observable or not by the
observer present at the first event. And even then you will
have to deal with unobservable events as well.

I know it makes no difference because then
they'd just define x_0 = ct, x_1 = -x, x_2 = -y, x_3 = -z, etc.

certainly not, since that would make no difference for the
squares.

Hmmm....
Never mind - I see why someone might prefer the former signiture.
[Sorry... I think I'm rambling]

perhaps a bit confused :-)
What really matters is that the invariant quantity is given a
name and a letter and it is convenient that physical quantities
are represented by real numbers. Whether one author uses
a quantity and another uses 'minus that quantity' or even
'10 times that quantity' does not matter.

Dirk Vdm

This is basically units: 4-velocity is the tangent vector to an object's
trajectory. As such one can use any affine parameter along the curve,
and the norm of the 4-vector will depend on which affine parameter one
selects. By requiring its norm to be -1 one is choosing to use the same
units for proper time along the trajectory and the coordinates onto
which one projects the 4-velocity.

Naturally for this to make sense one must select units with c=1, but
that was implicit in your original post.


Tom Roberts

Perion wrote:

I'm curious why many authors
prefer the -1,1,1,1 signiture for the metric rather than 1,-1,-1,-1. Since
any possible observable event for is either timelike or null (lightlike) it
seems that it would be more natural for the time component in the interval
to be positive.

If you're looking at a canonical formulation of gravity, in which
a spacelike hypersurface evolves in time, it's natural to want the
induced spatial metric to be positive. If you're doing Euclidean
quantum gravity, it's easiest to have the ``imaginary time'' metric
positive definite rather than negative definite. On the other hand,
as you say, if you're looking at motion of observers in a spacetime,
it's easiest to have ds be proper time, and if you want to talk to
particle physicists, who usually use a +--- signature, it's easiest
to not have to keep explaining signs.

In the end it's an arbitrary choice, and the textbooks are split
almost evenly. The only place it makes a difference is if you aren't
careful about matching your definition of spinors to your choice
of metric signature; then the sign of the metric can matter. (See
S. Carlip and C. DeWitt-Morette, ``Where the Sign of the Metric
Makes a Difference,'' Phys. Rev. Lett.60 (1988) 1599.)

Steve Carlip