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Peter Christensen
science forum Guru Wannabe
Joined: 02 Jan 2006
Posts: 130
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 Posted: Thu Jul 13, 2006 9:36 am Post subject:
P = m*V = (E/c,p) ?
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P [4-vector] = m*V [4-vector] = (E/c,p [3-vector] )
Does this always apply?
I just wonder, because I've seen the P (momentum 4-vector) defined in
both different ways. Maybe the relation m*V = (E/c,p) always holds, if
the momentum [3-vector] is defined as p = gamma*m*v [3-vector] and E =
gamma*m*c^2. Is this true?
Example: P = (E/c,p) = m*[gamma*c, gamma*v] = m*V
I know, that (E/c,p) makes sence for photons with m = 0, while m*V
really doesn't.
PC |
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tendon
science forum beginner
Joined: 13 Jul 2006
Posts: 18
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| Posted: Thu Jul 13, 2006 6:15 pm Post subject:
Re: P = m*V = (E/c,p) ?
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Peter Christensen wrote:
| Quote: | P [4-vector] = m*V [4-vector] = (E/c,p [3-vector] )
Does this always apply?
I just wonder, because I've seen the P (momentum 4-vector) defined in
both different ways. Maybe the relation m*V = (E/c,p) always holds, if
the momentum [3-vector] is defined as p = gamma*m*v [3-vector] and E =
gamma*m*c^2. Is this true?
Example: P = (E/c,p) = m*[gamma*c, gamma*v] = m*V
I know, that (E/c,p) makes sence for photons with m = 0, while m*V
really doesn't.
PC
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what is a 4 vector, a function a 4 parameters?
draw a 4 vector |
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Peter Christensen
science forum Guru Wannabe
Joined: 02 Jan 2006
Posts: 130
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| Posted: Thu Jul 13, 2006 10:38 pm Post subject:
Re: P = m*V = (E/c,p) ?
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my mother wrote:
| Quote: | Peter Christensen wrote:
P [4-vector] = m*V [4-vector] = (E/c,p [3-vector] )
Does this always apply?
I just wonder, because I've seen the P (momentum 4-vector) defined in
both different ways. Maybe the relation m*V = (E/c,p) always holds, if
the momentum [3-vector] is defined as p = gamma*m*v [3-vector] and E =
gamma*m*c^2. Is this true?
Example: P = (E/c,p) = m*[gamma*c, gamma*v] = m*V
I know, that (E/c,p) makes sence for photons with m = 0, while m*V
really doesn't.
PC
what is a 4 vector, a function a 4 parameters?
draw a 4 vector
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Something very fundamental to know in relativity. In the name of the
newsgroup, maybe you've noticed.
Please try to draw an ordinary 3-vector on a piece of paper...
PC |
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Tom Roberts
science forum Guru
Joined: 24 Mar 2005
Posts: 1399
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| Posted: Fri Jul 14, 2006 1:56 am Post subject:
Re: P = m*V = (E/c,p) ?
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Peter Christensen wrote:
| Quote: | P [4-vector] = m*V [4-vector] = (E/c,p [3-vector] )
Does this always apply?
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Yes, in SR. Note that your "p" is shorthand for 3 components, and there
is no gamma anywhere.
The 4-velocity V has components relative to coordinates {x^i} defined by:
V^i = (d/d\tau) x^i where \tau is the proper time of
the object.
The sometimes-seen equation p = m*gamma*v, for 3-vectors p and v, is not
an invariant equation and must be written in a specific inertial frame.
In particular, both p and v use _coordinate_time_ in their definition,
not the object's proper time (the change from coordinate to proper time
involves a factor of gamma in this case).
Tom Roberts |
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Peter Christensen
science forum Guru Wannabe
Joined: 02 Jan 2006
Posts: 130
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| Posted: Sat Jul 15, 2006 12:17 pm Post subject:
Re: P = m*V = (E/c,p) ?
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"Tom Roberts" <tjroberts137@sbcglobal.net> skrev i en meddelelse
news:IRCtg.64739$fb2.11731@newssvr27.news.prodigy.net...
| Quote: | Peter Christensen wrote:
P [4-vector] = m*V [4-vector] = (E/c,p [3-vector] )
Does this always apply?
Yes, in SR. Note that your "p" is shorthand for 3 components, and there is
no gamma anywhere.
The 4-velocity V has components relative to coordinates {x^i} defined by:
V^i = (d/d\tau) x^i where \tau is the proper time of
the object.
The sometimes-seen equation p = m*gamma*v, for 3-vectors p and v, is not
an invariant equation and must be written in a specific inertial frame. In
particular, both p and v use _coordinate_time_ in their definition, not
the object's proper time (the change from coordinate to proper time
involves a factor of gamma in this case).
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Yes I agree, on spacetime, it's known that:
c^2*(d\tau)^2 = c^2*(dt)^2 - (dx)^2
When using v=dx/dt, it's easy to show that dt = gamma*d\tau. -But it's hard
to understand the difference between t (coordinate time) and \tau (proper
time) intuitively, I think. The advantage with proper time (\tau) is that it
is an invariant. That's why the R 4-vector (position) and the P 4-vector
(momentum) can be differentiated with respect to \tau and still be
4-vectors: V (velocity), A (acceleration) and F (force) respectively.
Thanks for the reply.
PC |
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tendon
science forum beginner
Joined: 13 Jul 2006
Posts: 18
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| Posted: Sat Jul 15, 2006 3:52 pm Post subject:
Re: P = m*V = (E/c,p) ?
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Peter Christensen wrote:
| Quote: | my mother wrote:
Peter Christensen wrote:
P [4-vector] = m*V [4-vector] = (E/c,p [3-vector] )
Does this always apply?
I just wonder, because I've seen the P (momentum 4-vector) defined in
both different ways. Maybe the relation m*V = (E/c,p) always holds, if
the momentum [3-vector] is defined as p = gamma*m*v [3-vector] and E =
gamma*m*c^2. Is this true?
Example: P = (E/c,p) = m*[gamma*c, gamma*v] = m*V
I know, that (E/c,p) makes sence for photons with m = 0, while m*V
really doesn't.
PC
what is a 4 vector, a function a 4 parameters?
draw a 4 vector
Something very fundamental to know in relativity. In the name of the
newsgroup, maybe you've noticed.
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yo fool
more fundamental than anythin is a vector
draw or explain a foken 4 vector projectef in
a 4d coordinat system
remember fool, orthogonal relationship
| Quote: |
Please try to draw an ordinary 3-vector on a piece of paper...
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there are no problems with 3 vectors, yo foken moron
> PC |
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tendon
science forum beginner
Joined: 13 Jul 2006
Posts: 18
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| Posted: Sat Jul 15, 2006 3:55 pm Post subject:
Re: P = m*V = (E/c,p) ?
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Peter Christensen wrote:
| Quote: | "Tom Roberts" <tjroberts137@sbcglobal.net> skrev i en meddelelse
news:IRCtg.64739$fb2.11731@newssvr27.news.prodigy.net...
Peter Christensen wrote:
P [4-vector] = m*V [4-vector] = (E/c,p [3-vector] )
Does this always apply?
Yes, in SR. Note that your "p" is shorthand for 3 components, and there is
no gamma anywhere.
The 4-velocity V has components relative to coordinates {x^i} defined by:
V^i = (d/d\tau) x^i where \tau is the proper time of
the object.
The sometimes-seen equation p = m*gamma*v, for 3-vectors p and v, is not
an invariant equation and must be written in a specific inertial frame. In
particular, both p and v use _coordinate_time_ in their definition, not
the object's proper time (the change from coordinate to proper time
involves a factor of gamma in this case).
Yes I agree, on spacetime, it's known that:
c^2*(d\tau)^2 = c^2*(dt)^2 - (dx)^2
When using v=dx/dt, it's easy to show that dt = gamma*d\tau. -But it's hard
to understand the difference between t (coordinate time) and \tau (proper
time) intuitively, I think. The advantage with proper time (\tau) is that it
is an invariant. That's why the R 4-vector (position) and the P 4-vector
(momentum) can be differentiated with respect to \tau and still be
4-vectors: V (velocity), A (acceleration) and F (force) respectively.
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crap
| Quote: |
Thanks for the reply.
PC |
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Peter Christensen
science forum Guru Wannabe
Joined: 02 Jan 2006
Posts: 130
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| Posted: Sun Jul 16, 2006 1:11 pm Post subject:
Re: P = m*V = (E/c,p) ?
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That's what your posting is.
PC |
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The TimeLord
science forum Guru Wannabe
Joined: 12 Jun 2005
Posts: 182
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| Posted: Tue Jul 18, 2006 6:17 am Post subject:
Re: P = m*V = (E/c,p) ?
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I've read a couple of responses and most are very good.
However, I thought I'd add a couple of my thoughts for
completeness.
On Thu, 13 Jul 2006 02:36:30 -0700, "Peter Christensen" <PeCh@MailAPS.org>
wrote in <1152783389.984891.267220@m73g2000cwd.googlegroups.com>:
| Quote: | P [4-vector] = m*V [4-vector] = (E/c,p [3-vector] )
Does this always apply?
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Mostly when m/=0. For the photon, since m=0, there is a
little bit of a different approach. We know the 4-wave
vector is
k^i = [2*Pi/lambda; k (3-vector)]
Then p^i = h/(2*Pi) * k^i. This is then consistent with
the observed relation E=p*c with E=h*c/lambda and p=h/lambda.
| Quote: |
I just wonder, because I've seen the P (momentum 4-vector) defined in both
different ways. Maybe the relation m*V = (E/c,p) always holds, if the
momentum [3-vector] is defined as p = gamma*m*v [3-vector] and E =
gamma*m*c^2. Is this true?
|
Firstly it is only true in flat space, because how it is
derived. Secondly, you have to be careful how you use
P=m*V, since m=0 and p/=0 for photons.
| Quote: |
Example: P = (E/c,p) = m*[gamma*c, gamma*v] = m*V
I know, that (E/c,p) makes sence for photons with m = 0, while m*V really
doesn't.
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Bingo! Note that since v=c for photons, gamma is undefined
for photons in free space (vacuum) as a quantity.
I usually define the above as Vproper = d/dtau * X
then Vproper = gamma * V = dt/dtau * d/dt * X
thus P = m * Vproper = m * gamma * V
This way things tend to remind me that I need to be more
careful since m*gamma is like zero*infinity for photons.
This is where a lot of older books introduce m*gamma as
relativistic mass. However, today it is more common to
think of Vproper=gamma*V and P=m*Vproper instead of
mrel=gamma*m and P=mrel*V. The modern way tends to not
introduce all the logical errors that the older way did.
--
// The TimeLord says:
// Pogo 2.0 = We have met the aliens, and they are us! |
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Peter Christensen
science forum Guru Wannabe
Joined: 02 Jan 2006
Posts: 130
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| Posted: Tue Jul 18, 2006 1:42 pm Post subject:
Re: P = m*V = (E/c,p) ?
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| Quote: | there are no problems with 3 vectors, yo foken moron
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Can't even spell... |
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