| Topic: |
Science > Physics |
| User: |
"GR_GR" |
| Date: |
22 Feb 2005 05:41:02 AM |
| Object: |
Another four-vector problem |
First of all, thanks to those of you who helped me with my last question.
I have another question related to four-vectors and the associated
four-velocity and four-acceleration.
The problem statement give a non-parameterized world line as:
x[t] = a t + b Sin[w t];
y[t] = b Cos[w t];
z[t] = 0
Compute the components of the four-velocity and four-acceleration.
For what values of the parameters b, w is this motion possible?
----
Here is my solution so far.
----
First I write the world line as a vector, r.
r = {a t + b Sin[w t], b Cos[w t],0}
Therefor the velocity is simply:
V = D[r,t]
V = {a + b w Cos[t w],-b w Sin[t w],0}
Now, the components of the four-velocity as related to the 3-velocity
are of the form in Hartle,
u^alpha = {gamma, (gamma) V}
where;
gamma = (1-V.V)^(-1/2)
Needless to say, I can then compute each component of the four-velocity,
and by differentiating each of those components with respect to time I
can obtain the components of the four-acceleration.
However, my main question has to do with the question:
"For what values of the parameters b, w is this motion possible?"
I assume that (V.V)^(1/2) < 1, or less than the speed of light. (Note
that c is suppressed as c = 1 in this discussion)
V.V = a^2 + b^2 w^2 + 2 a b w Cos[t w]
I also note that Cos[t w] ranges from 1, ... , -1
Is it correct to determine the answer by solving v.v = 1 for b and plug
in the values of Cos[wt] = 1 or -1 ?
With the answer:
-(1+a)/w < b < (1+a)/w
For:
-1 < Cos[wt] < 1
Does this make any sense, or is lack of sleep reducing my thought
process to rubbish?
Thanks for the help.
.
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| User: "Androcles Androcles@ MyPlace.org" |
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| Title: Re: Another four-vector problem |
22 Feb 2005 06:04:24 AM |
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"GR_GR" <nyb@colorado.edu> wrote in message
news:cvf5oh$mrr$1@peabody.colorado.edu...
First of all, thanks to those of you who helped me with my last
question.
I have another question related to four-vectors and the associated
four-velocity and four-acceleration.
Time is not a vector quantity. There is no (-t) + t = 0.
1: v + w X V
2: (v + w) + x = v + (w + x)
3: there exist the element 0 X V such that: v + 0 = 0 + v = v
4: for all v there exists -v such that v + (-v) = 0
5: v + w = w + v
These first 5 axioms imply that the binary operation of addition +: V x
V -> V forms an abelian group -
1: implies the mapping is closed (it wouldn't be a mapping if it
wasn't!)
2: implies addition is associative
3: there exists an identity
4: says every element has an inverse
5: addition is commutative (the parallelogram rule) - this fith axiom is
what makes the group abelian.
http://members.tripod.com/~Paul_Kirby/Linear/linear.html
A set V is a vector space over a field F (for example, the field of real
or of complex numbers) if, given
a.. an operation vector addition defined in V, denoted v + w (where v,
w ? V), and
b.. an operation scalar multiplication in V, denoted a * v (where v ?
V and a ? F),
the following ten properties hold for all a, b ? F and u, v, and w ? V:
1.. v + w belongs to V.
(Closure of V under vector addition.)
2.. u + (v + w) = (u + v) + w.
(Associativity of vector addition in V.)
3.. There exists a neutral element 0 in V, such that for all elements
v in V, v + 0 = v.
(Existence of an additive identity element in V.)
4.. For all v in V, there exists an element w in V, such that v + w =
0.
(Existence of additive inverses in V.)
5.. v + w = w + v.
(Commutativity of vector addition in V.)
6.. a * v belongs to V.
(Closure of V under scalar multiplication.)
7.. a * (b * v) = (ab) * v.
(Associativity of scalar multiplication in V.)
8.. If 1 denotes the multiplicative identity of the field F, then 1 *
v = v.
(Neutrality of one.)
9.. a * (v + w) = a * v + a * w.
(Distributivity with respect to vector addition.)
10.. (a + b) * v = a * v + b * v.
(Distributivity with respect to field addition.)
Properties 1 through 5 indicate that V is an abelian group under vector
addition. The rest, properties 6 through 10, apply to scalar
multiplication of a vector v ? V by a scalar a ? F. Note that property 5
actually follows from the other 9.
http://en.wikipedia.org/wiki/Vector_space
10 Axioms of a Vector Space
1. u + v is in V (closure under addition)
2. u + v = v + u (commutativity)
3. (u + v) + w = u + (v + w) (associativity)
4. There exists an element 0 in V , called a zero vector, such that
u+0=u.
5. For each u in V , there is an element ?u in V such that u + (?u)=0.
6. cu is in V (closure under scalar multiplication)
7. c(u + v) = cu + cv (distributivity)
8. (c + d)u = cu +du (distributivity)
9. c(du) = (cd)u
10. 1u = u
http://pirate.shu.edu/~wachsmut/Teaching/MATH3912/Projects/papers/zybura_vector_spaces.pdf
Androcles.
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| User: "Dirk Van de moortel" |
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| Title: Re: Another four-vector problem |
22 Feb 2005 12:48:27 PM |
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"Androcles" <Androcles@ MyPlace.org> wrote in message news:cdFSd.117291$68.73070@fe1.news.blueyonder.co.uk...
"GR_GR" <nyb@colorado.edu> wrote in message
news:cvf5oh$mrr$1@peabody.colorado.edu...
First of all, thanks to those of you who helped me with my last
question.
I have another question related to four-vectors and the associated
four-velocity and four-acceleration.
Time is not a vector quantity. There is no (-t) + t = 0.
Right, when we set our clock to zero, all the events
that happened before that cannot be described anymore,
since Androshit goes into panic mode with the concept
of negative time.
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/NegTime.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/NegTime2.html
Dirk Vdm
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| User: "GR_GR" |
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| Title: Re: Another four-vector problem |
22 Feb 2005 01:17:58 PM |
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Androcles wrote:
--plonk--
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| User: "Dirk Van de moortel" |
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| Title: Re: Another four-vector problem |
22 Feb 2005 10:46:01 AM |
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"GR_GR" <nyb@colorado.edu> wrote in message news:cvf5oh$mrr$1@peabody.colorado.edu...
First of all, thanks to those of you who helped me with my last question.
I have another question related to four-vectors and the associated
four-velocity and four-acceleration.
The problem statement give a non-parameterized world line as:
x[t] = a t + b Sin[w t];
y[t] = b Cos[w t];
z[t] = 0
Compute the components of the four-velocity and four-acceleration.
For what values of the parameters b, w is this motion possible?
----
Here is my solution so far.
----
First I write the world line as a vector, r.
r = {a t + b Sin[w t], b Cos[w t],0}
Therefor the velocity is simply:
V = D[r,t]
V = {a + b w Cos[t w],-b w Sin[t w],0}
Now, the components of the four-velocity as related to the 3-velocity
are of the form in Hartle,
u^alpha = {gamma, (gamma) V}
where;
gamma = (1-V.V)^(-1/2)
Needless to say, I can then compute each component of the four-velocity,
and by differentiating each of those components with respect to time I
can obtain the components of the four-acceleration.
However, my main question has to do with the question:
"For what values of the parameters b, w is this motion possible?"
I assume that (V.V)^(1/2) < 1, or less than the speed of light. (Note
that c is suppressed as c = 1 in this discussion)
V.V = a^2 + b^2 w^2 + 2 a b w Cos[t w]
I also note that Cos[t w] ranges from 1, ... , -1
Is it correct to determine the answer by solving v.v = 1 for b and plug
in the values of Cos[wt] = 1 or -1 ?
With the answer:
-(1+a)/w < b < (1+a)/w
For:
-1 < Cos[wt] < 1
Does this make any sense, or is lack of sleep reducing my thought
process to rubbish?
Whatever x(t), y(t), z(t), you will *always* have the four-vector
square v.v = 1 as an identity, by design. Likewise, you'll have
the four-vector product a.v = 0 as another identity. You can't
get any information about x(t), y(t), z(t) out of these identities,
so you won't get wiser about a, b or w by using them.
From the condition
V.V = a^2 + b^2 w^2 + 2 a b w cos(w t) < 1 (for any t),
You get
-1 < a + b w < 1 if a b w > 0
and
-1 < a - b w < 1 if a b w < 0
The first gives
-1-a < bw < 1-a if a b w > 0
The second gives
-1+a < bw < 1+a if a b w < 0
Supposing both 0 < a < 1 and 0 < b < 1 the first condition gives
w < (1-a)/b if w > 0
and the second gives
-(1-a)/b < w if w < 0
which gives the combined
-(1-a)/b < w < (1-a)/b
which seems to differ from your solution in the sign of a,
unless I made a mistake somewhere...
Dirk Vdm
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| User: "Tom Roberts" |
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| Title: Re: Another four-vector problem |
23 Feb 2005 09:49:58 AM |
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GR_GR wrote:
The problem statement give a non-parameterized world line as:
x[t] = a t + b Sin[w t];
y[t] = b Cos[w t];
z[t] = 0
NOTE: you must assume that these coordinates are inertial, because you
(implicitly) need the metric to solve this problem. You probably want to assume
a, b, and w are all constants, too.
With those assumptions this is circular motion in an inertial frame moving with
velocity a wrt the frame with the coordinates {x,y,z,t}.
Compute the components of the four-velocity and four-acceleration.
For what values of the parameters b, w is this motion possible?
First I write the world line as a vector, r.
r = {a t + b Sin[w t], b Cos[w t],0}
Therefor the velocity is simply:
V = D[r,t]
That is clearly a 3-velocity in this specific inertial frame. I assume
Mathematica notation.
V = {a + b w Cos[t w],-b w Sin[t w],0}
Now, the components of the four-velocity as related to the 3-velocity
are of the form in Hartle,
u^alpha = {gamma, (gamma) V}
where;
gamma = (1-V.V)^(-1/2)
Note this implicitly assumes V is referenced to inertial coordinates (because
computing the norm of u uses the metric and this formula gives a norm of 1 only
for metric components in an inertial frame).
Needless to say, I can then compute each component of the four-velocity,
and by differentiating each of those components with respect to time I
can obtain the components of the four-acceleration.
You must differentiate u wrt PROPER time of the object, not coordinate time t.
That's another factor of gamma.
However, my main question has to do with the question:
"For what values of the parameters b, w is this motion possible?"
I assume that (V.V)^(1/2) < 1, or less than the speed of light. (Note
that c is suppressed as c = 1 in this discussion)
[...]
Is it correct to determine the answer by solving v.v = 1 for b and plug
in the values of Cos[wt] = 1 or -1 ?
This approach looks OK, except the limits will involve a, b, and w; not just b.
I have not checked the algebra.
Tom Roberts tjroberts@lucent.com
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| User: "Dirk Van de moortel" |
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| Title: Re: Another four-vector problem |
23 Feb 2005 09:59:13 AM |
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"Tom Roberts" <tjroberts@lucent.com> wrote in message news:cvi8n6$4ee@netnews.proxy.lucent.com...
GR_GR wrote:
The problem statement give a non-parameterized world line as:
x[t] = a t + b Sin[w t];
y[t] = b Cos[w t];
z[t] = 0
NOTE: you must assume that these coordinates are inertial, because you
(implicitly) need the metric to solve this problem. You probably want to assume
a, b, and w are all constants, too.
With those assumptions this is circular motion in an inertial frame moving with
velocity a wrt the frame with the coordinates {x,y,z,t}.
Compute the components of the four-velocity and four-acceleration.
For what values of the parameters b, w is this motion possible?
First I write the world line as a vector, r.
r = {a t + b Sin[w t], b Cos[w t],0}
Therefor the velocity is simply:
V = D[r,t]
That is clearly a 3-velocity in this specific inertial frame. I assume
Mathematica notation.
V = {a + b w Cos[t w],-b w Sin[t w],0}
Now, the components of the four-velocity as related to the 3-velocity
are of the form in Hartle,
u^alpha = {gamma, (gamma) V}
where;
gamma = (1-V.V)^(-1/2)
Note this implicitly assumes V is referenced to inertial coordinates (because
computing the norm of u uses the metric and this formula gives a norm of 1 only
for metric components in an inertial frame).
Needless to say, I can then compute each component of the four-velocity,
and by differentiating each of those components with respect to time I
can obtain the components of the four-acceleration.
You must differentiate u wrt PROPER time of the object, not coordinate time t.
That's another factor of gamma.
However, my main question has to do with the question:
"For what values of the parameters b, w is this motion possible?"
I assume that (V.V)^(1/2) < 1, or less than the speed of light. (Note
that c is suppressed as c = 1 in this discussion)
[...]
Is it correct to determine the answer by solving v.v = 1 for b and plug
in the values of Cos[wt] = 1 or -1 ?
This approach looks OK, except the limits will involve a, b, and w; not just b.
I have not checked the algebra.
Don't you agree that v.v (both as 4-vectors) give v.v = 1,
whatever the values of a, b and w are, so you can't get
any information out of it?
After all, the components are
( g , g v_z , g v_y , g v_z ) with g = 1/sqrt(1-V^2)
which gives -by design- for the square:
g^2 - (g v_z)^2 - (g v_y)^2 - (g v_z)^2
= g^2 ( 1 - V^2 )
= 1
I checked the algebra and found
-(1-a)/b < w < (1-a)/b
assuming 0 < a < 1 and 0 < b
Dirk Vdm
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| User: "GR_GR" |
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| Title: Re: Another four-vector problem |
23 Feb 2005 01:24:54 PM |
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Dirk Van de moortel wrote:
"Tom Roberts" <tjroberts@lucent.com> wrote in message news:cvi8n6$4ee@netnews.proxy.lucent.com...
GR_GR wrote:
The problem statement give a non-parameterized world line as:
x[t] = a t + b Sin[w t];
y[t] = b Cos[w t];
z[t] = 0
NOTE: you must assume that these coordinates are inertial, because you
(implicitly) need the metric to solve this problem. You probably want to assume
a, b, and w are all constants, too.
With those assumptions this is circular motion in an inertial frame moving with
velocity a wrt the frame with the coordinates {x,y,z,t}.
Compute the components of the four-velocity and four-acceleration.
For what values of the parameters b, w is this motion possible?
First I write the world line as a vector, r.
r = {a t + b Sin[w t], b Cos[w t],0}
Therefor the velocity is simply:
V = D[r,t]
That is clearly a 3-velocity in this specific inertial frame. I assume
Mathematica notation.
V = {a + b w Cos[t w],-b w Sin[t w],0}
Now, the components of the four-velocity as related to the 3-velocity
are of the form in Hartle,
u^alpha = {gamma, (gamma) V}
where;
gamma = (1-V.V)^(-1/2)
Note this implicitly assumes V is referenced to inertial coordinates (because
computing the norm of u uses the metric and this formula gives a norm of 1 only
for metric components in an inertial frame).
Needless to say, I can then compute each component of the four-velocity,
and by differentiating each of those components with respect to time I
can obtain the components of the four-acceleration.
You must differentiate u wrt PROPER time of the object, not coordinate time t.
That's another factor of gamma.
However, my main question has to do with the question:
"For what values of the parameters b, w is this motion possible?"
I assume that (V.V)^(1/2) < 1, or less than the speed of light. (Note
that c is suppressed as c = 1 in this discussion)
[...]
Is it correct to determine the answer by solving v.v = 1 for b and plug
in the values of Cos[wt] = 1 or -1 ?
This approach looks OK, except the limits will involve a, b, and w; not just b.
I have not checked the algebra.
Don't you agree that v.v (both as 4-vectors) give v.v = 1,
v.v = -1 and is timelike. Question, is this true regardless of the sign
of the metric we choose? I notice that many people have been stating
that v.v = 1, but in Hartle v.v > 0 is space like, v.v = -1 is timelike,
and v.v = 0 is null.
(Some time passes)
Well, just answered my own question.
If I choose the metric to be (+ - - -) the norm is positive.
Why can't these people standardize things?
Wolfram uses the (+ - - -) convention, while Hartle uses the (- + + +)
convention.
I can see how this causes confusion.
whatever the values of a, b and w are, so you can't get
any information out of it?
I agree.
The only things we have to work with is the ansatz of the 3-vector never
exceeding the speed of light.
After all, the components are
( g , g v_z , g v_y , g v_z ) with g = 1/sqrt(1-V^2)
which gives -by design- for the square:
g^2 - (g v_z)^2 - (g v_y)^2 - (g v_z)^2
= g^2 ( 1 - V^2 )
= 1
I checked the algebra and found
-(1-a)/b < w < (1-a)/b
assuming 0 < a < 1 and 0 < b
Dirk Vdm
I agree with you here.
Note for gamma:
dt/dtau = gamma = (1-V.V)^(-1/2)
Where V is the three-vector.
This would indicate that V.V < 1 or gamma is imaginary.
V.V = a^2 + b^2 w^2 - 2 a b Cos[w t] < 1
Cos ranges: -1 < Cos[w t] < 1
Also, we can use a series approximation for Cos[w t] ~ 1
Therefore:
a^2 + b^2 w^2 - 2 a b < 1
This has solutions (Thanks Mathematica, although I can do by hand! :-))
(w < 0 && (1 - a)/w < b < (-1 - a)/w) || (w == 0 && -1 < a < 1) || (w >
0 && (-1 - a)/w < b < (1 - a)/w)
.
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| User: "Dirk Van de moortel" |
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| Title: Re: Another four-vector problem |
23 Feb 2005 01:57:57 PM |
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"GR_GR" <nyb@colorado.edu> wrote in message news:cvila8$et3$1@peabody.colorado.edu...
Dirk Van de moortel wrote:
"Tom Roberts" <tjroberts@lucent.com> wrote in message news:cvi8n6$4ee@netnews.proxy.lucent.com...
GR_GR wrote:
[snip]
However, my main question has to do with the question:
"For what values of the parameters b, w is this motion possible?"
I assume that (V.V)^(1/2) < 1, or less than the speed of light. (Note
that c is suppressed as c = 1 in this discussion)
[...]
Is it correct to determine the answer by solving v.v = 1 for b and plug
in the values of Cos[wt] = 1 or -1 ?
This approach looks OK, except the limits will involve a, b, and w; not just b.
I have not checked the algebra.
Don't you agree that v.v (both as 4-vectors) give v.v = 1,
v.v = -1 and is timelike. Question, is this true regardless of the sign
of the metric we choose? I notice that many people have been stating
that v.v = 1, but in Hartle v.v > 0 is space like, v.v = -1 is timelike,
and v.v = 0 is null.
Ah yes, I had assumed the other signature because you
explicitly talked about v.v = 1.
No problem, it doesn't matter.
(Some time passes)
Well, just answered my own question.
If I choose the metric to be (+ - - -) the norm is positive.
Why can't these people standardize things?
History, and sometimes convenience I guess.
Wolfram uses the (+ - - -) convention, while Hartle uses the (- + + +)
convention.
I can see how this causes confusion.
It's a bit annoying but one gets used to it - eventually :-)
whatever the values of a, b and w are, so you can't get
any information out of it?
I agree.
The only things we have to work with is the ansatz of the 3-vector never
exceeding the speed of light.
After all, the components are
( g , g v_z , g v_y , g v_z ) with g = 1/sqrt(1-V^2)
which gives -by design- for the square:
g^2 - (g v_z)^2 - (g v_y)^2 - (g v_z)^2
= g^2 ( 1 - V^2 )
= 1
I checked the algebra and found
-(1-a)/b < w < (1-a)/b
assuming 0 < a < 1 and 0 < b
Dirk Vdm
I agree with you here.
Note for gamma:
dt/dtau = gamma = (1-V.V)^(-1/2)
Where V is the three-vector.
This would indicate that V.V < 1 or gamma is imaginary.
V.V = a^2 + b^2 w^2 - 2 a b Cos[w t] < 1
Cos ranges: -1 < Cos[w t] < 1
Also, we can use a series approximation for Cos[w t] ~ 1
Therefore:
a^2 + b^2 w^2 - 2 a b < 1
This has solutions (Thanks Mathematica, although I can do by hand! :-))
(w < 0 && (1 - a)/w < b < (-1 - a)/w) || (w == 0 && -1 < a < 1) || (w >
0 && (-1 - a)/w < b < (1 - a)/w)
There is a w-factor missing in the Cos term of your
V.V = a^2 + b^2 w^2 - 2 a b Cos[w t] < 1
but I guess it's a typo.
Perhaps you couldn't find my yesterday's calculation, so I'll
reproduce it here (but I will remove the b < 1 assumption
since it's not needed, and I will add a few lines)
From the condition
V.V = a^2 + b^2 w^2 + 2 a b w cos(w t) < 1 (for any t),
You get
a^2 + b^2 w^2 + 2 a b w < 1 (if a b w > 0)
giving
-1 < a + b w < 1 (if a b w > 0)
and
a^2 + b^2 w^2 - 2 a b w < 1 (if a b w < 0)
giving
-1 < a - b w < 1 (if a b w < 0)
The first gives
-1-a < bw < 1-a (if a b w > 0)
The second gives
-1+a < bw < 1+a (if a b w < 0)
Supposing both 0 < a < 1 and 0 < b the first condition gives
w < (1-a)/b (if w > 0)
and the second gives
-(1-a)/b < w (if w < 0)
which gives the combined
-(1-a)/b < w < (1-a)/b
which seems to differ from your solution in the sign of a,
unless I made a mistake somewhere...
Anyway I checked (with NuCalc) the parametrized curves
of the speed
v_x(t) = a + b w cos(w t)
v_y(t) = -b w sin(w t)
and found that for
-(1-a)/b < w < (1-a)/b
the speed circle lies inside the unit circle, whereas when
I take w just outside the interval, the velocity circle just
crosses the unit circle.
hm... algebra :-)
Cheers,
Dirk Vdm
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| User: "GR_GR" |
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| Title: Re: Another four-vector problem |
23 Feb 2005 03:33:48 PM |
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Dirk Van de moortel wrote:
"GR_GR" <nyb@colorado.edu> wrote in message news:cvila8$et3$1@peabody.colorado.edu...
Dirk Van de moortel wrote:
"Tom Roberts" <tjroberts@lucent.com> wrote in message news:cvi8n6$4ee@netnews.proxy.lucent.com...
GR_GR wrote:
<CHOP>
I agree with you here.
Note for gamma:
dt/dtau = gamma = (1-V.V)^(-1/2)
Where V is the three-vector.
This would indicate that V.V < 1 or gamma is imaginary.
V.V = a^2 + b^2 w^2 - 2 a b Cos[w t] < 1
Cos ranges: -1 < Cos[w t] < 1
Also, we can use a series approximation for Cos[w t] ~ 1
Therefore:
a^2 + b^2 w^2 - 2 a b < 1
This has solutions (Thanks Mathematica, although I can do by hand! :-))
(w < 0 && (1 - a)/w < b < (-1 - a)/w) || (w == 0 && -1 < a < 1) || (w >
0 && (-1 - a)/w < b < (1 - a)/w)
There is a w-factor missing in the Cos term of your
V.V = a^2 + b^2 w^2 - 2 a b Cos[w t] < 1
but I guess it's a typo.
Yes! Thanks for noticing.
Perhaps you couldn't find my yesterday's calculation, so I'll
reproduce it here (but I will remove the b < 1 assumption
since it's not needed, and I will add a few lines)
From the condition
V.V = a^2 + b^2 w^2 + 2 a b w cos(w t) < 1 (for any t),
You get
a^2 + b^2 w^2 + 2 a b w < 1 (if a b w > 0)
giving
-1 < a + b w < 1 (if a b w > 0)
and
a^2 + b^2 w^2 - 2 a b w < 1 (if a b w < 0)
giving
-1 < a - b w < 1 (if a b w < 0)
The first gives
-1-a < bw < 1-a (if a b w > 0)
The second gives
-1+a < bw < 1+a (if a b w < 0)
Supposing both 0 < a < 1 and 0 < b the first condition gives
w < (1-a)/b (if w > 0)
and the second gives
-(1-a)/b < w (if w < 0)
which gives the combined
-(1-a)/b < w < (1-a)/b
which seems to differ from your solution in the sign of a,
unless I made a mistake somewhere...
Anyway I checked (with NuCalc) the parametrized curves
of the speed
v_x(t) = a + b w cos(w t)
v_y(t) = -b w sin(w t)
and found that for
-(1-a)/b < w < (1-a)/b
the speed circle lies inside the unit circle, whereas when
I take w just outside the interval, the velocity circle just
crosses the unit circle.
hm... algebra :-)
:-)
I was finally able to corner the prof and he agrees with our method of
determining a, b and w.
BTW, there is a grad student here who started giving me some help in
this subject. He has been very helpful and his office is on the same
floor as mine.
He used to post on Usenet and blew a gasket when I mentioned that I
posted a question here.
He said he left Usenet in disgust last summer and that it is an
addictive waste of time, filled with "cranks" and "crack pots"
interspersed with helpful people! :-)
I assume that you are one of the helpful people he was talking about.
Thanks again.
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| User: "Dirk Van de moortel" |
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| Title: Re: Another four-vector problem |
23 Feb 2005 04:08:38 PM |
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"GR_GR" <nyb@colorado.edu> wrote in message news:cvisrv$lnl$1@peabody.colorado.edu...
[snip]
I was finally able to corner the prof and he agrees with our method of
determining a, b and w.
BTW, there is a grad student here who started giving me some help in
this subject. He has been very helpful and his office is on the same
floor as mine.
He used to post on Usenet and blew a gasket when I mentioned that I
posted a question here.
He said he left Usenet in disgust last summer and that it is an
addictive waste of time, filled with "cranks" and "crack pots"
interspersed with helpful people! :-)
Yes, we both know who you are talking about.
Don't forget to say hello from me ;-)
I assume that you are one of the helpful people he was talking about.
Thanks again.
Welcome.
And... let's be careful out there...
Dirk Vdm
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| User: "Eric Gisse" |
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| Title: Re: Another four-vector problem |
23 Feb 2005 08:35:01 PM |
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GR_GR wrote:
[snip]
BTW, there is a grad student here who started giving me some help in
this subject. He has been very helpful and his office is on the same
floor as mine.
He used to post on Usenet and blew a gasket when I mentioned that I
posted a question here.
He said he left Usenet in disgust last summer and that it is an
addictive waste of time, filled with "cranks" and "crack pots"
interspersed with helpful people! :-)
It is that time of year again, isn't it?
I have a friend, who is *not myself* [I still don't know what im doing
yet], who really wants to go to U-O-Colorado for study of Bose-Einstein
condensates. I was wondering if you had any useful advice for him.
BTW, has the second generation oscillator been built yet?
I assume that you are one of the helpful people he was talking about.
Dirk is exceedingly helpful.
Thanks again.
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| User: "GR_GR" |
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| Title: Re: Another four-vector problem |
23 Feb 2005 09:07:25 PM |
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Eric Gisse wrote:
GR_GR wrote:
[snip]
BTW, there is a grad student here who started giving me some help in
this subject. He has been very helpful and his office is on the same
floor as mine.
He used to post on Usenet and blew a gasket when I mentioned that I
posted a question here.
He said he left Usenet in disgust last summer and that it is an
addictive waste of time, filled with "cranks" and "crack pots"
interspersed with helpful people! :-)
It is that time of year again, isn't it?
I have a friend, who is *not myself* [I still don't know what im doing
yet], who really wants to go to U-O-Colorado for study of Bose-Einstein
condensates. I was wondering if you had any useful advice for him.
Get in line? :-)
There has been a glut of graduate students since the Nobel prizes, about
45 + per year. Everybody and their mother wants into JILA and the AMO
groups.
Debby Jin is a possibility.
Eric Cornell is still recovering from his battle with the flesh eating
bacteria, and is still not back in the lab.
BTW, has the second generation oscillator been built yet?
?
I assume that you are one of the helpful people he was talking about.
Dirk is exceedingly helpful.
Thanks again.
.
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| User: "Eric Gisse" |
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| Title: Re: Another four-vector problem |
23 Feb 2005 09:59:44 PM |
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GR_GR wrote:
Eric Gisse wrote:
GR_GR wrote:
[snip]
BTW, there is a grad student here who started giving me some help
in
this subject. He has been very helpful and his office is on the
same
floor as mine.
He used to post on Usenet and blew a gasket when I mentioned that I
posted a question here.
He said he left Usenet in disgust last summer and that it is an
addictive waste of time, filled with "cranks" and "crack pots"
interspersed with helpful people! :-)
It is that time of year again, isn't it?
I have a friend, who is *not myself* [I still don't know what im
doing
yet], who really wants to go to U-O-Colorado for study of
Bose-Einstein
condensates. I was wondering if you had any useful advice for him.
Get in line? :-)
Seriously, not me. I don't make a habit of asking "for a friend
<wink>".
I am yet to find out where my interests settle, though "gravity" is the
overriding theme. I'm finding that I like the abstract more than I
thought I would.
There has been a glut of graduate students since the Nobel prizes,
about
45 + per year. Everybody and their mother wants into JILA and the AMO
groups.
JIMO isn't his thing. AMO sounds like the right group, but as I said,
not me. You are the only person I know, even tangentially, who is
schooling in Colorado - so I ask.
The word will be passed on the next time grad school comes up in
conversation.
Debby Jin is a possibility.
Eric Cornell is still recovering from his battle with the flesh
eating
bacteria, and is still not back in the lab.
What the *****? There has got to be a story behind that one.
BTW, has the second generation oscillator been built yet?
?
http://www.phys.lsu.edu/mog/mog22/node9.html
I forgot what I asked you originally, but the idea was implanted in my
mind that the device was being upgraded [Better cooling, IIRC] to be
more sensitive. Did nothing come of that? Or am I totally off the mark?
[snippage]
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| User: "Y.Porat" |
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| Title: Re: Another four-vector problem |
25 Feb 2005 03:18:23 AM |
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ok now Gisee disclosed his real identity !!!! (:-)
just for those who are interested....]
i was wondering long ago who is that zero scintist mumbler
anyway if he is doing some useful experiments
that is a nice surprise for me .
Y.P
------------------------------------------------
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| User: "Eric Gisse" |
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| Title: Re: Another four-vector problem |
25 Feb 2005 03:40:41 AM |
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Y.Porat wrote:
ok now Gisee disclosed his real identity !!!! (:-)
Idiot.
just for those who are interested....]
i was wondering long ago who is that zero scintist mumbler
anyway if he is doing some useful experiments
that is a nice surprise for me .
Y.P
------------------------------------------------
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| User: "Y.Porat" |
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| Title: Re: Another four-vector problem |
25 Feb 2005 04:11:21 AM |
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Hi varney!!
you are angree eh?
i like it
bye imposter zero
--------------------
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| User: "Dirk Van de moortel" |
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| Title: Re: Another four-vector problem |
25 Feb 2005 05:59:51 AM |
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"Y.Porat" <maporat@012.net.il> wrote in message news:1109326281.949390.270540@o13g2000cwo.googlegroups.com...
Hi varney!!
you are angree eh?
i like it
bye imposter zero
--------------------
;-)
Dirk Vdm
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| User: "Y.Porat" |
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| Title: Re: Another four-vector problem |
25 Feb 2005 06:16:15 AM |
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which side you are Dirk?
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| User: "Dirk Van de moortel" |
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| Title: Re: Another four-vector problem |
25 Feb 2005 06:57:32 AM |
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"Y.Porat" <maporat@012.net.il> wrote in message news:1109333775.423849.22210@f14g2000cwb.googlegroups.com...
which side you are Dirk?
For me to know and for you to guess ;-)
Dirk Vdm
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| User: "GR_GR" |
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| Title: Re: Another four-vector problem |
25 Feb 2005 07:40:20 AM |
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Dirk Van de moortel wrote:
"Y.Porat" <maporat@012.net.il> wrote in message news:1109333775.423849.22210@f14g2000cwb.googlegroups.com...
which side you are Dirk?
For me to know and for you to guess ;-)
Dirk Vdm
Porat must be one of the "cranks and crack pots" mentioned.
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| User: "Dirk Van de moortel" |
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| Title: Re: Another four-vector problem |
25 Feb 2005 09:21:21 AM |
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"GR_GR" <nyb@colorado.edu> wrote in message news:cvn9s7$i49$1@peabody.colorado.edu...
Dirk Van de moortel wrote:
"Y.Porat" <maporat@012.net.il> wrote in message news:1109333775.423849.22210@f14g2000cwb.googlegroups.com...
which side you are Dirk?
For me to know and for you to guess ;-)
Dirk Vdm
Porat must be one of the "cranks and crack pots" mentioned.
meta - ;-)
Dirk Vdm
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| User: "Y.Porat" |
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| Title: Re: Another four-vector problem |
25 Feb 2005 07:43:40 AM |
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ok enough with it
we are supposed (supposed !) to be serious people.
Bye
Y.Porat
---------------------------
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| User: "Sam Wormley" |
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| Title: Re: Another four-vector problem |
25 Feb 2005 09:00:39 PM |
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Y.Porat wrote:
ok enough with it
we are supposed (supposed !) to be serious people.
Bye
Y.Porat
---------------------------
*smirk*
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| User: "Dirk Van de moortel" |
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| Title: Re: Another four-vector problem |
26 Feb 2005 03:58:59 AM |
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"Sam Wormley" <swormley1@mchsi.com> wrote in message news:rDRTd.55510$4q6.14175@attbi_s01...
Y.Porat wrote:
ok enough with it
we are supposed (supposed !) to be serious people.
Bye
Y.Porat
---------------------------
*smirk*
That reminds me of a little anecdote my father experienced
a long time ago. My dad and two of my uncles took the
car to go for a fishing trip when we were having a family
holidays at the coast.
When they were leaving the village, their was a small
scale traffic jam... Suddenly a very angry and aggressive,
clearly overstressed policeman summoned them to open
the window and explain what the hell they were doing
coming out of that one way street! My uncle calmly
explained that he must have missed the sign. The cop
shouted that a car driver is supposed to use his bloody
eyes! My other uncle calmly objected that they were on
a holiday and that they weren't familiar with the layout
of the village.... The cop turned purple by now and
ordered everyone to leave the car and hand over their
identity cards. He took the ID-cards, glanced at them
for a while and then suddenly amazed the entire
company: "Hm... ah... a doctor... and an engineer... well...
since we're all intellectuals here, there's no reason to
make a fuzz about this, is there? Have a nice day,
gentlemen!"
Dirk Vdm
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| User: "Ken S. Tucker" |
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| Title: Re: Another four-vector problem |
26 Feb 2005 04:27:08 AM |
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Dirk Van de moortel wrote:
"Sam Wormley" <swormley1@mchsi.com> wrote in message
news:rDRTd.55510$4q6.14175@attbi_s01...
Y.Porat wrote:
ok enough with it
we are supposed (supposed !) to be serious people.
Bye
Y.Porat
---------------------------
*smirk*
That reminds me of a little anecdote my father experienced
a long time ago. My dad and two of my uncles took the
car to go for a fishing trip when we were having a family
holidays at the coast.
When they were leaving the village, their was a small
scale traffic jam... Suddenly a very angry and aggressive,
clearly overstressed policeman summoned them to open
the window and explain what the hell they were doing
coming out of that one way street! My uncle calmly
explained that he must have missed the sign. The cop
shouted that a car driver is supposed to use his bloody
eyes! My other uncle calmly objected that they were on
a holiday and that they weren't familiar with the layout
of the village.... The cop turned purple by now and
ordered everyone to leave the car and hand over their
identity cards. He took the ID-cards, glanced at them
for a while and then suddenly amazed the entire
company: "Hm... ah... a doctor... and an engineer... well...
since we're all intellectuals here, there's no reason to
make a fuzz about this, is there? Have a nice day,
gentlemen!"
Dirk Vdm
Back in the 90's we were going from
Ontario (yuk) canada in a van, and at
the border, the US lady decided to
inspect us.
We explained we're headed to the MENSA
convention in Boston, showed ID,
Same thing ...have a nice day.
The moral of the story Dirky
doctors and engineers are NOT
intellects, they're programmable.
Dirky, you have a rather queer idea of
what intellect is, did your mom ever
let you wear knickers???
Know wonder Dirky is an OCD fruit cake.
Regards Psycho, get help...
Ken S. Tucker
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| User: "Eric Gisse" |
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| Title: Re: Another four-vector problem |
25 Feb 2005 05:57:27 AM |
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Y.Porat wrote:
Hi varney!!
you are angree eh?
i like it
bye imposter zero
--------------------
Idiot.
.
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| User: "Y.Porat" |
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| Title: Re: Another four-vector problem |
25 Feb 2005 06:03:17 AM |
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sure you are what i sayed now we have another proove
i could smell your pigg smell from thousand of miles
but could not believe how low even you can go....
------------------
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| User: "Eric Gisse" |
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| Title: Re: Another four-vector problem |
25 Feb 2005 03:03:55 PM |
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Y.Porat wrote:
sure you are what i sayed now we have another proove
i could smell your pigg smell from thousand of miles
but could not believe how low even you can go....
------------------
Idiot.
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| User: "GR_GR" |
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| Title: Re: Another four-vector problem |
23 Feb 2005 10:45:05 AM |
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Tom Roberts wrote:
GR_GR wrote:
The problem statement give a non-parameterized world line as:
x[t] = a t + b Sin[w t];
y[t] = b Cos[w t];
z[t] = 0
NOTE: you must assume that these coordinates are inertial, because you
(implicitly) need the metric to solve this problem. You probably want to
assume a, b, and w are all constants, too.
Right. This problem assumes a Minkowsky metric, and a, b and w are all
constants.
With those assumptions this is circular motion in an inertial frame
moving with velocity a wrt the frame with the coordinates {x,y,z,t}.
Compute the components of the four-velocity and four-acceleration.
For what values of the parameters b, w is this motion possible?
First I write the world line as a vector, r.
r = {a t + b Sin[w t], b Cos[w t],0}
Therefor the velocity is simply:
V = D[r,t]
That is clearly a 3-velocity
Correct. I could parameterize the motion with respect to proper time,
but if I do not want to bother I can determine the four-velocity with
respect to the three-velocity, which I simply call the velocity.
in this specific inertial frame. I assume
Mathematica notation.
Yes. It is as universal as I know.
V = {a + b w Cos[t w],-b w Sin[t w],0}
Now, the components of the four-velocity as related to the 3-velocity
are of the form in Hartle,
u^alpha = {gamma, (gamma) V}
where;
gamma = (1-V.V)^(-1/2)
Note this implicitly assumes V is referenced to inertial coordinates
(because computing the norm of u uses the metric and this formula gives
a norm of 1 only for metric components in an inertial frame).
I noted that V was the three-velocity. Also, the norm in the (-,+,+,+) I
am using should be time like, and thus is negative 1, not 1.
Needless to say, I can then compute each component of the
four-velocity, and by differentiating each of those components with
respect to time I can obtain the components of the four-acceleration.
You must differentiate u wrt PROPER time of the object, not coordinate
time t.
Correct, or without converting to proper time:
a = gamma D[u,t]
That's another factor of gamma.
Yes.
However, my main question has to do with the question:
"For what values of the parameters b, w is this motion possible?"
I assume that (V.V)^(1/2) < 1, or less than the speed of light. (Note
that c is suppressed as c = 1 in this discussion)
[...]
Is it correct to determine the answer by solving v.v = 1 for b and
plug in the values of Cos[wt] = 1 or -1 ?
This approach looks OK, except the limits will involve a, b, and w; not
just b. I have not checked the algebra.
My professor gave the semi-cryptic hint in response to an email:
First Response:
"You need the condition that the four velocity is a time like vector."
My follow up question:
"You mean that u.u = -1 < 0 => Time like
I have verified this relation directly from my solution, but I am not
sure how that will help."
His last response:
"Yes but to ensure that you need the condition that d\tau^2 = -ds^2 >0-
this gives a constraint on the parameter space."
After a nap I will crunch on this more.
Thanks for your help.
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| User: "Bruce Scott TOK" |
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| Title: Re: Another four-vector problem |
22 Feb 2005 08:30:05 AM |
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|> x[t] = a t + b Sin[w t];
|> y[t] = b Cos[w t];
|> z[t] = 0
|>
|> Compute the components of the four-velocity and four-acceleration.
|> For what values of the parameters b, w is this motion possible?
The easiest way to do this is simply differentiate with respect to t and
then normalise the four vector to -1 (it is a "unit vector"):
u dot u = g_ab u^a u^b = -1
This incidentally gives you the gamma factor.
Then find a = du/dtau = gamma du/dt and ensure that as a four vector a
is perpendicular to u
a dot u = 0
--
cu,
Bruce
drift wave turbulence: http://www.rzg.mpg.de/~bds/
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