| Topic: |
Science > Physics |
| User: |
"Scott H" |
| Date: |
17 Mar 2007 07:35:35 PM |
| Object: |
Lorentz transformation leaves (x_m)^2 invariant? |
In Quantum Electrodynamics by Gupta, it says that a linear
transformation of space-time coordinates x_m that leaves (x_m)^2 is
called a Lorentz transformation. However, in another book, a Lorentz
transformation is
x' = (x - vt)/sqrt(1 - v^2/c^2)
t' = (t - vx/c^2)/sqrt(1 - v^2/c^2)
When I calculate x'^2 or x'^2 + t'^2, I don't find either to be
invariant under the transformation. What's going on here?
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| User: "Androcles" |
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| Title: Re: Lorentz transformation leaves (x_m)^2 invariant? |
17 Mar 2007 07:40:48 PM |
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"Scott H" <zinites_page@yahoo.com> wrote in message =
news:1174178135.170933.281730@b75g2000hsg.googlegroups.com...
In Quantum Electrodynamics by Gupta, it says that a linear
transformation of space-time coordinates x_m that leaves (x_m)^2 is
called a Lorentz transformation. However, in another book, a Lorentz
transformation is
=20
x' =3D (x - vt)/sqrt(1 - v^2/c^2)
t' =3D (t - vx/c^2)/sqrt(1 - v^2/c^2)
=20
When I calculate x'^2 or x'^2 + t'^2, I don't find either to be
invariant under the transformation. What's going on here?
This is going on here:
http://www.androcles01.pwp.blueyonder.co.uk/Smart/Smart.htm
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| User: "Dirk Van de moortel" |
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| Title: Re: Lorentz transformation leaves (x_m)^2 invariant? |
18 Mar 2007 05:02:16 AM |
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"Scott H" <zinites_page@yahoo.com> wrote in message news:1174178135.170933.281730@b75g2000hsg.googlegroups.com...
In Quantum Electrodynamics by Gupta, it says that a linear
transformation of space-time coordinates x_m that leaves (x_m)^2 is
called a Lorentz transformation. However, in another book, a Lorentz
transformation is
x' = (x - vt)/sqrt(1 - v^2/c^2)
t' = (t - vx/c^2)/sqrt(1 - v^2/c^2)
When I calculate x'^2 or x'^2 + t'^2, I don't find either to be
invariant under the transformation. What's going on here?
(x_m)^2 is an abbreviation for the squared norm of a vector.
This quantity is calculated with the metric as
Sum( i=0..3, j=0..3, g_ij x_i x_j )
where the g_mm are the components of the metric.
In flat Minkowski space-time, when you use the 'standard'
coordinates
( x_0, x_1, x_2, x_3 ) = ( c t, x, y, z ) ,
the (g_ij) can be seen as a diagonal matrix
(-1,1,1,1) or (1,-1,-1,-1) ,
so you will find that
- (c t')^2 + x'^2 + y'^2 + z'^2 = - (c t)^2 + x^2 + y^2 + z^2
Dirk Vdm
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| User: "Eric Gisse" |
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| Title: Re: Lorentz transformation leaves (x_m)^2 invariant? |
17 Mar 2007 08:46:00 PM |
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On Mar 17, 4:35 pm, "Scott H" <zinites_p...@yahoo.com> wrote:
In Quantum Electrodynamics by Gupta, it says that a linear
transformation of space-time coordinates x_m that leaves (x_m)^2 is
called a Lorentz transformation. However, in another book, a Lorentz
transformation is
x' = (x - vt)/sqrt(1 - v^2/c^2)
t' = (t - vx/c^2)/sqrt(1 - v^2/c^2)
When I calculate x'^2 or x'^2 + t'^2, I don't find either to be
invariant under the transformation. What's going on here?
You aren't checking the right quantity. The book might be vague or you
are misreading it slightly.
The quantity that is preserved under a Lorentz transformation is the
quadratic form, using your notation, -(ct)^2 + x^2.
You should find that -(ct)^2 + x^2 = -(ct')^2 + (x')^2.
By the way, Ignore Androcles. I normally would say ignore Shubee as
well, but Shubee has it right for a change. Androcles is a loudmouthed
crank with more mouth than clue, and will only insult you when you
question his absolute psychosis.
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| User: "Shubee" |
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| Title: Re: Lorentz transformation leaves (x_m)^2 invariant? |
17 Mar 2007 11:29:29 PM |
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On Mar 17, 6:46 pm, "Eric Gisse" <jowr...@gmail.com> wrote:
I normally would say ignore Shubee as
well, but Shubee has it right for a change.
"Get your stinking mouth off of me, you damn dirty ape!"
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| User: "Eric Gisse" |
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| Title: Re: Lorentz transformation leaves (x_m)^2 invariant? |
17 Mar 2007 11:42:11 PM |
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On Mar 17, 8:29 pm, "Shubee" <e.shu...@yahoo.com> wrote:
On Mar 17, 6:46 pm, "Eric Gisse" <jowr...@gmail.com> wrote:
I normally would say ignore Shubee as
well, but Shubee has it right for a change.
"Get your stinking mouth off of me, you damn dirty ape!"
Shooby, you might want to limit your presence on this newsgroup
because you seem to be going down the same path as Androcles.
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| User: "Shubee" |
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| Title: Re: Lorentz transformation leaves (x_m)^2 invariant? |
17 Mar 2007 07:44:17 PM |
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On Mar 17, 5:35 pm, "Scott H" <zinites_p...@yahoo.com> wrote:
In Quantum Electrodynamics by Gupta, it says that a linear
transformation of space-time coordinates x_m that leaves (x_m)^2 is
called a Lorentz transformation. However, in another book, a Lorentz
transformation is
x' = (x - vt)/sqrt(1 - v^2/c^2)
t' = (t - vx/c^2)/sqrt(1 - v^2/c^2)
When I calculate x'^2 or x'^2 + t'^2, I don't find either to be
invariant under the transformation. What's going on here?
The invariant you're looking for is x'^2 - (ct')^2 which equals x^2 -
(ct)^2 under the Lorentz transformation.
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| User: "Timo A. Nieminen" |
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| Title: Re: Lorentz transformation leaves (x_m)^2 invariant? |
18 Mar 2007 12:07:21 AM |
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On Sun, 17 Mar 2007, Scott H wrote:
In Quantum Electrodynamics by Gupta, it says that a linear
transformation of space-time coordinates x_m that leaves (x_m)^2 is
called a Lorentz transformation. However, in another book, a Lorentz
transformation is
x' = (x - vt)/sqrt(1 - v^2/c^2)
t' = (t - vx/c^2)/sqrt(1 - v^2/c^2)
When I calculate x'^2 or x'^2 + t'^2, I don't find either to be
invariant under the transformation. What's going on here?
Not having Gupta at hand, I can only guess. (x_m)^2 is probably meant to
be read as
x_0^2 + x_1^2 + x_2^2 + x_3^2 (that is, the sum of x_m^2 for m = 0 to 3,
basically the commonly used Einstein summation convention)
where x_0 = ict, x_1 = x, x_2 = y, x_3 = z. There are other possibilities,
as the notation used in different books is diverse, non-standardised, and
often poorly explained.
Basically, under a Lorentz transformation, the magnitude of 4-vectors is
invariant. More generally, the scalar product of 2 4-vectors is invariant.
For 2 4-vectors a and b, we usually have
a.b = a_0 * b_0 - a_1 * b_1 - a_2 * b_2 - a_3 * b_3
or
a.b = - a_0 * b_0 + a_1 * b_1 + a_2 * b_2 + a_3 * b_3.
Note that
x' = (x - vt)/sqrt(1 - v^2/c^2)
y' = y
z' = z
t' = (t - vx/c^2)/sqrt(1 - v^2/c^2)
is a special case of a Lorentz transformation, not a definition of Lorentz
transformation.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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