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Peter Christensen
science forum Guru Wannabe

Joined: 02 Jan 2006
Posts: 130

Posted: Tue Jul 18, 2006 1:42 pm    Post subject: Re: P = m*V = (E/c,p) ?

 Quote: there are no problems with 3 vectors, yo foken moron

Can't even spell...
The TimeLord
science forum Guru Wannabe

Joined: 12 Jun 2005
Posts: 182

Posted: Tue Jul 18, 2006 6:17 am    Post subject: Re: P = m*V = (E/c,p) ?

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>

 Quote: P [4-vector] = m*V [4-vector] = (E/c,p [3-vector] ) Does this always apply?

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.

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!
Peter Christensen
science forum Guru Wannabe

Joined: 02 Jan 2006
Posts: 130

Posted: Sun Jul 16, 2006 1:11 pm    Post subject: Re: P = m*V = (E/c,p) ?

 Quote: crap

PC
tendon
science forum beginner

Joined: 13 Jul 2006
Posts: 18

Posted: Sat Jul 15, 2006 3:55 pm    Post subject: Re: P = m*V = (E/c,p) ?

Peter Christensen wrote:
 Quote: "Tom Roberts" 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. crap  Quote: Thanks for the reply. PC tendon science forum beginner Joined: 13 Jul 2006 Posts: 18 Posted: Sat Jul 15, 2006 3:52 pm Post subject: Re: P = m*V = (E/c,p) ? 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. 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... there are no problems with 3 vectors, yo foken moron > PC Peter Christensen science forum Guru Wannabe Joined: 02 Jan 2006 Posts: 130 Posted: Sat Jul 15, 2006 12:17 pm Post subject: Re: P = m*V = (E/c,p) ? "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).

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.

PC
Tom Roberts
science forum Guru

Joined: 24 Mar 2005
Posts: 1399

Posted: Fri Jul 14, 2006 1:56 am    Post subject: Re: P = m*V = (E/c,p) ?

Peter Christensen wrote:
 Quote: 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).

Tom Roberts
Peter Christensen
science forum Guru Wannabe

Joined: 02 Jan 2006
Posts: 130

Posted: Thu Jul 13, 2006 10:38 pm    Post subject: Re: P = m*V = (E/c,p) ?

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

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
tendon
science forum beginner

Joined: 13 Jul 2006
Posts: 18

Posted: Thu Jul 13, 2006 6:15 pm    Post subject: Re: P = m*V = (E/c,p) ?

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

what is a 4 vector, a function a 4 parameters?

draw a 4 vector
Peter Christensen
science forum Guru Wannabe

Joined: 02 Jan 2006
Posts: 130

 Posted: Thu Jul 13, 2006 9:36 am    Post subject: P = m*V = (E/c,p) ? 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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