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)

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?

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.

Have a look at this triangle
http://en.wikipedia.org/wiki/Image:T...y_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)

*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).

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:T...y_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.

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:T...y_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:T...y_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).

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.

In article .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*.



--
Self control is what keeps us from being rapist.

Observations of Bernard - No 100

"BernardZ" wrote in message
news:MPG.1ee80d1ada138f659899d1@news...
| In article .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
|
|

Bilge wrote:

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.