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Golden Boar
science forum Guru
Joined: 17 May 2005
Posts: 651
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 Posted: Wed May 31, 2006 3:38 am Post subject:
Another look at the Lorentz factor.
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The Lorentz factor is usually given by the equation:
gamma = 1/sqrt(1-v^2/c^2)
or by
beta = v/c
gamma = 1/sqrt(1-beta^2)
The equation can be written in a more intuitive (for me at least) way
as shown below:
gamma = c / sqrt(c^2 - v^2) |
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Eric Gisse
science forum Guru
Joined: 04 May 2005
Posts: 1999
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| Posted: Wed May 31, 2006 4:50 am Post subject:
Re: Another look at the Lorentz factor.
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Golden Boar wrote:
| Quote: | The Lorentz factor is usually given by the equation:
gamma = 1/sqrt(1-v^2/c^2)
or by
beta = v/c
gamma = 1/sqrt(1-beta^2)
The equation can be written in a more intuitive (for me at least) way
as shown below:
gamma = c / sqrt(c^2 - v^2)
|
Excellent - you can do algebra. More than can be said for a fair bit of
folks here, though.
Now why are you doing this? |
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Golden Boar
science forum Guru
Joined: 17 May 2005
Posts: 651
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| Posted: Wed May 31, 2006 5:27 am Post subject:
Re: Another look at the Lorentz factor.
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Eric Gisse wrote:
| Quote: | Golden Boar wrote:
The Lorentz factor is usually given by the equation:
gamma = 1/sqrt(1-v^2/c^2)
or by
beta = v/c
gamma = 1/sqrt(1-beta^2)
The equation can be written in a more intuitive (for me at least) way
as shown below:
gamma = c / sqrt(c^2 - v^2)
Excellent - you can do algebra. More than can be said for a fair bit of
folks here, though.
Now why are you doing this?
|
No reason really, it's just a neater more intuitive equation, which I
came up with, and you never know it might help someone.
Have a look at this triangle
http://en.wikipedia.org/wiki/Image:Trigonometry_triangle.svg
The speed of light is side h.
The velocity is side a.
So side b = sqrt(c^2 - a^2)
The Lorentz factor is then c / b = sec(A) = 1/cos(A) |
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Golden Boar
science forum Guru
Joined: 17 May 2005
Posts: 651
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| Posted: Wed May 31, 2006 5:35 am Post subject:
Re: Another look at the Lorentz factor.
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*correction*
The Lorentz factor is then h / b = sec(A) = 1/cos(A).
Also, gamma * beta is then a / b = tan(A) = sin(A)/cos(A). |
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Eric Gisse
science forum Guru
Joined: 04 May 2005
Posts: 1999
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| Posted: Wed May 31, 2006 7:06 am Post subject:
Re: Another look at the Lorentz factor.
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Golden Boar wrote:
| Quote: | Eric Gisse wrote:
Golden Boar wrote:
The Lorentz factor is usually given by the equation:
gamma = 1/sqrt(1-v^2/c^2)
or by
beta = v/c
gamma = 1/sqrt(1-beta^2)
The equation can be written in a more intuitive (for me at least) way
as shown below:
gamma = c / sqrt(c^2 - v^2)
Excellent - you can do algebra. More than can be said for a fair bit of
folks here, though.
Now why are you doing this?
No reason really, it's just a neater more intuitive equation, which I
came up with, and you never know it might help someone.
|
....and what makes you think it is more intuitive? The form of the
equation as usually presented is the result of the derivation. Adding
another step of algebra doesn't add any insight.
Uhhh...no.
All you did, whether you realise it or not, is multiply it by a form of
1 while bringing the extra c term inside the square root. |
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Golden Boar
science forum Guru
Joined: 17 May 2005
Posts: 651
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| Posted: Wed May 31, 2006 8:30 am Post subject:
Re: Another look at the Lorentz factor.
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Eric Gisse wrote:
| Quote: | Golden Boar wrote:
Eric Gisse wrote:
Golden Boar wrote:
The Lorentz factor is usually given by the equation:
gamma = 1/sqrt(1-v^2/c^2)
or by
beta = v/c
gamma = 1/sqrt(1-beta^2)
The equation can be written in a more intuitive (for me at least) way
as shown below:
gamma = c / sqrt(c^2 - v^2)
Excellent - you can do algebra. More than can be said for a fair bit of
folks here, though.
Now why are you doing this?
No reason really, it's just a neater more intuitive equation, which I
came up with, and you never know it might help someone.
...and what makes you think it is more intuitive? The form of the
equation as usually presented is the result of the derivation. Adding
another step of algebra doesn't add any insight.
Have a look at this triangle
http://en.wikipedia.org/wiki/Image:Trigonometry_triangle.svg
The speed of light is side h.
The velocity is side a.
So side b = sqrt(c^2 - a^2)
The Lorentz factor is then c / b = sec(A) = 1/cos(A)
Uhhh...no.
All you did, whether you realise it or not, is multiply it by a form of
1 while bringing the extra c term inside the square root.
|
That c was meant to be h.
I made a few mistakes in the above posts becuase at first I was using a
triangle with sides a,b and c. So I will satrt again.
Have a look at this triangle,
http://en.wikipedia.org/wiki/Image:Trigonometry_triangle.svg
The speed of light is side h.
The velocity is side a.
So side b = sqrt(h^2 - a^2)
The Lorentz factor is then h / b = sec(A) = 1/cos(A).
Also, gamma * beta is then a / b = tan(A) = sin(A)/cos(A). |
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Bilge
science forum Guru
Joined: 30 Apr 2005
Posts: 2816
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| Posted: Wed May 31, 2006 8:55 am Post subject:
Re: Another look at the Lorentz factor.
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Golden Boar:
| Quote: | The Lorentz factor is usually given by the equation:
gamma = 1/sqrt(1-v^2/c^2)
or by
beta = v/c
gamma = 1/sqrt(1-beta^2)
The equation can be written in a more intuitive (for me at least) way
as shown below:
gamma = c / sqrt(c^2 - v^2)
|
OK, your next assignment is to begin with an arbitrary lagrangian
density, L[q(x^u), d_v q(x^u)] and find the conserved currents and
charges if L is invariant under an infinitesimal spacetime displacement,
x^u -> x^u + a^u. |
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Eric Gisse
science forum Guru
Joined: 04 May 2005
Posts: 1999
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| Posted: Wed May 31, 2006 9:15 am Post subject:
Re: Another look at the Lorentz factor.
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Golden Boar wrote:
| Quote: | Eric Gisse wrote:
Golden Boar wrote:
Eric Gisse wrote:
Golden Boar wrote:
The Lorentz factor is usually given by the equation:
gamma = 1/sqrt(1-v^2/c^2)
or by
beta = v/c
gamma = 1/sqrt(1-beta^2)
The equation can be written in a more intuitive (for me at least) way
as shown below:
gamma = c / sqrt(c^2 - v^2)
Excellent - you can do algebra. More than can be said for a fair bit of
folks here, though.
Now why are you doing this?
No reason really, it's just a neater more intuitive equation, which I
came up with, and you never know it might help someone.
...and what makes you think it is more intuitive? The form of the
equation as usually presented is the result of the derivation. Adding
another step of algebra doesn't add any insight.
Have a look at this triangle
http://en.wikipedia.org/wiki/Image:Trigonometry_triangle.svg
The speed of light is side h.
The velocity is side a.
So side b = sqrt(c^2 - a^2)
The Lorentz factor is then c / b = sec(A) = 1/cos(A)
Uhhh...no.
All you did, whether you realise it or not, is multiply it by a form of
1 while bringing the extra c term inside the square root.
That c was meant to be h.
I made a few mistakes in the above posts becuase at first I was using a
triangle with sides a,b and c. So I will satrt again.
Have a look at this triangle,
http://en.wikipedia.org/wiki/Image:Trigonometry_triangle.svg
The speed of light is side h.
The velocity is side a.
So side b = sqrt(h^2 - a^2)
|
Since you fail to grasp my point while beliving your method has merit
in of itself, why don't you do the same thing but for a velocity in an
arbitrary direction [not just along one direction on a line connecting
to the observer] and see what you get.
To make it simpler, consider a stationary observer with an object
moving in the x-y plane with velocity u, x velocity v, and y velocity
w.
| Quote: |
The Lorentz factor is then h / b = sec(A) = 1/cos(A).
Also, gamma * beta is then a / b = tan(A) = sin(A)/cos(A).
|
If you look hard enough, you can find "derivations" for many
expressions in such a manner. I gurantee you that you won't be able to
reconstruct the general lorentz transformation matrix. In case you
aren't sure what it is, I'll explain. The general transformation matrix
allows for arbitrary boosts and/or rotations while having a unit
determinant while preserving the length between events in Minkowski
space.
Have fun. |
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BernardZ
science forum beginner
Joined: 14 Jul 2005
Posts: 32
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| Posted: Wed May 31, 2006 9:29 am Post subject:
Re: Another look at the Lorentz factor.
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In article <1149046688.508095.182340@i40g2000cwc.googlegroups.com>,
goldenboar@hotmail.com says...
| Quote: | The Lorentz factor is usually given by the equation:
gamma = 1/sqrt(1-v^2/c^2)
or by
beta = v/c
gamma = 1/sqrt(1-beta^2)
The equation can be written in a more intuitive (for me at least) way
as shown below:
gamma = c / sqrt(c^2 - v^2)
|
Reminds me of a discussion I had with my engineering professor. I said
that some units could be better written in a different form. He reckoned
*so what*.
--
Self control is what keeps us from being rapist.
Observations of Bernard - No 100 |
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The Sorcerer
science forum Guru
Joined: 22 May 2006
Posts: 363
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| Posted: Wed May 31, 2006 10:03 am Post subject:
Re: Another look at the Lorentz factor.
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"BernardZ" <bernardZ@Nospam.com> wrote in message
news:MPG.1ee80d1ada138f659899d1@news...
| In article <1149046688.508095.182340@i40g2000cwc.googlegroups.com>,
| goldenboar@hotmail.com says...
| > The Lorentz factor is usually given by the equation:
| >
| > gamma = 1/sqrt(1-v^2/c^2)
| >
| > or by
| >
| > beta = v/c
| > gamma = 1/sqrt(1-beta^2)
| >
| > The equation can be written in a more intuitive (for me at least) way
| > as shown below:
| >
| > gamma = c / sqrt(c^2 - v^2)
| >
| >
|
| Reminds me of a discussion I had with my engineering professor. I said
| that some units could be better written in a different form. He reckoned
| *so what*.
So what?
Androcles.
|
| --
| Self control is what keeps us from being rapist.
|
| Observations of Bernard - No 100
|
| |
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Golden Boar
science forum Guru
Joined: 17 May 2005
Posts: 651
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| Posted: Wed May 31, 2006 10:17 am Post subject:
Re: Another look at the Lorentz factor.
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Bilge wrote:
| Quote: | Golden Boar:
The Lorentz factor is usually given by the equation:
gamma = 1/sqrt(1-v^2/c^2)
or by
beta = v/c
gamma = 1/sqrt(1-beta^2)
The equation can be written in a more intuitive (for me at least) way
as shown below:
gamma = c / sqrt(c^2 - v^2)
OK, your next assignment is to begin with an arbitrary lagrangian
density, L[q(x^u), d_v q(x^u)] and find the conserved currents and
charges if L is invariant under an infinitesimal spacetime displacement,
x^u -> x^u + a^u.
|
Nah, I don't think I will. |
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Golden Boar
science forum Guru
Joined: 17 May 2005
Posts: 651
|
| Posted: Wed May 31, 2006 10:30 am Post subject:
Re: Another look at the Lorentz factor.
|
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|
Eric Gisse wrote:
| Quote: | Golden Boar wrote:
Eric Gisse wrote:
Golden Boar wrote:
Eric Gisse wrote:
Golden Boar wrote:
The Lorentz factor is usually given by the equation:
gamma = 1/sqrt(1-v^2/c^2)
or by
beta = v/c
gamma = 1/sqrt(1-beta^2)
The equation can be written in a more intuitive (for me at least) way
as shown below:
gamma = c / sqrt(c^2 - v^2)
Excellent - you can do algebra. More than can be said for a fair bit of
folks here, though.
Now why are you doing this?
No reason really, it's just a neater more intuitive equation, which I
came up with, and you never know it might help someone.
...and what makes you think it is more intuitive? The form of the
equation as usually presented is the result of the derivation. Adding
another step of algebra doesn't add any insight.
Have a look at this triangle
http://en.wikipedia.org/wiki/Image:Trigonometry_triangle.svg
The speed of light is side h.
The velocity is side a.
So side b = sqrt(c^2 - a^2)
The Lorentz factor is then c / b = sec(A) = 1/cos(A)
Uhhh...no.
All you did, whether you realise it or not, is multiply it by a form of
1 while bringing the extra c term inside the square root.
That c was meant to be h.
I made a few mistakes in the above posts becuase at first I was using a
triangle with sides a,b and c. So I will satrt again.
Have a look at this triangle,
http://en.wikipedia.org/wiki/Image:Trigonometry_triangle.svg
The speed of light is side h.
The velocity is side a.
So side b = sqrt(h^2 - a^2)
Since you fail to grasp my point while beliving your method has merit
in of itself, why don't you do the same thing but for a velocity in an
arbitrary direction [not just along one direction on a line connecting
to the observer] and see what you get.
|
First of all, it may not be velocity we're working with here, but
distance.
h is the distance a photon travels in 1 second.
a is the distance an electron travels in 1 second.
Secondly, I can now visualise the Lorentz factor, i'ts a right-angled
triangle.
Thirdly, I don't care whether you find it helpful or not.
| Quote: |
To make it simpler, consider a stationary observer with an object
moving in the x-y plane with velocity u, x velocity v, and y velocity
w.
The Lorentz factor is then h / b = sec(A) = 1/cos(A).
Also, gamma * beta is then a / b = tan(A) = sin(A)/cos(A).
If you look hard enough, you can find "derivations" for many
expressions in such a manner. I gurantee you that you won't be able to
reconstruct the general lorentz transformation matrix. In case you
aren't sure what it is, I'll explain. The general transformation matrix
allows for arbitrary boosts and/or rotations while having a unit
determinant while preserving the length between events in Minkowski
space.
Have fun.
|
I couldn't care less about Lorentz transformations. |
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Eric Gisse
science forum Guru
Joined: 04 May 2005
Posts: 1999
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| Posted: Wed May 31, 2006 10:49 am Post subject:
Re: Another look at the Lorentz factor.
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Golden Boar wrote:
[snip]
Exactly. |
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Spoonfed (www.spoonfedrel
science forum Guru Wannabe
Joined: 28 Apr 2006
Posts: 144
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| Posted: Wed May 31, 2006 3:55 pm Post subject:
Re: Another look at the Lorentz factor.
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Eric Gisse wrote:
| Quote: | Golden Boar wrote:
Eric Gisse wrote:
Golden Boar wrote:
Eric Gisse wrote:
Golden Boar wrote:
The Lorentz factor is usually given by the equation:
gamma = 1/sqrt(1-v^2/c^2)
or by
beta = v/c
gamma = 1/sqrt(1-beta^2)
The equation can be written in a more intuitive (for me at least) way
as shown below:
gamma = c / sqrt(c^2 - v^2)
Excellent - you can do algebra. More than can be said for a fair bit of
folks here, though.
Now why are you doing this?
No reason really, it's just a neater more intuitive equation, which I
came up with, and you never know it might help someone.
...and what makes you think it is more intuitive? The form of the
equation as usually presented is the result of the derivation. Adding
another step of algebra doesn't add any insight.
Have a look at this triangle
http://en.wikipedia.org/wiki/Image:Trigonometry_triangle.svg
The speed of light is side h.
The velocity is side a.
So side b = sqrt(c^2 - a^2)
The Lorentz factor is then c / b = sec(A) = 1/cos(A)
Uhhh...no.
All you did, whether you realise it or not, is multiply it by a form of
1 while bringing the extra c term inside the square root.
That c was meant to be h.
I made a few mistakes in the above posts becuase at first I was using a
triangle with sides a,b and c. So I will satrt again.
Have a look at this triangle,
http://en.wikipedia.org/wiki/Image:Trigonometry_triangle.svg
The speed of light is side h.
The velocity is side a.
So side b = sqrt(h^2 - a^2)
Since you fail to grasp my point while beliving your method has merit
in of itself, why don't you do the same thing but for a velocity in an
arbitrary direction [not just along one direction on a line connecting
to the observer] and see what you get.
To make it simpler, consider a stationary observer with an object
moving in the x-y plane with velocity u, x velocity v, and y velocity
w.
The Lorentz factor is then h / b = sec(A) = 1/cos(A).
Also, gamma * beta is then a / b = tan(A) = sin(A)/cos(A).
If you look hard enough, you can find "derivations" for many
expressions in such a manner. I gurantee you that you won't be able to
reconstruct the general lorentz transformation matrix. In case you
aren't sure what it is, I'll explain. The general transformation matrix
allows for arbitrary boosts and/or rotations while having a unit
determinant while preserving the length between events
|
....where "lengths" between events in Minkowski space have the awkward
definition of sqrt(c^2 t^2 - x^2).
| Quote: | in Minkowski
space.
Have fun. |
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PD
science forum Guru
Joined: 03 May 2005
Posts: 4363
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| Posted: Wed May 31, 2006 5:40 pm Post subject:
Re: Another look at the Lorentz factor.
|
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Golden Boar wrote:
| Quote: | The Lorentz factor is usually given by the equation:
gamma = 1/sqrt(1-v^2/c^2)
or by
beta = v/c
gamma = 1/sqrt(1-beta^2)
The equation can be written in a more intuitive (for me at least) way
as shown below:
gamma = c / sqrt(c^2 - v^2)
|
I think it would benefit you to see the form of the Lorentz
transformations in terms of hyperbolic transcendental functions, where
gamma = cosh(rapidity),
about as simple as you can get.
Moreover, the rapidity variable is exceeding useful as well as being
more physically fundamental.
See, for example, http://www.everything2.com/index.pl?node_id=312518
The important thing to note is that, rather than reading up on things
that have already been done a long long time ago, you are spending your
time futzing with equations to stumble on the work yourself. This in
itself is not a bad thing. Thinking that you've done something original
or of interest to others --- that you'll find people have a lesser
opinion of.
PD |
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