- #1

Eugene Shubert

- 22

- 0

1/ds^2 = 1/dt^2 – 1/(dx^2 + dy^2 + dz^2)

Eugene Shubert

http://www.everythingimportant.org/relativity

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- Mathematica
- Thread starter Eugene Shubert
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- #1

Eugene Shubert

- 22

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1/ds^2 = 1/dt^2 – 1/(dx^2 + dy^2 + dz^2)

Eugene Shubert

http://www.everythingimportant.org/relativity

- #2

isn't that just

ds^{2} = dt^{2}- (dx^{2}+dy^{2}+dz^{2})

?

ds

?

- #3

Eugene Shubert

- 22

- 0

Eugene Shubert

http://www.everythingimportant.org/relativity

- #4

Brad_Ad23

- 502

- 1

- #5

Yes I do remember elementary algebra, good for me. you can invert that equation.

- #6

plus

- 178

- 1

- #7

Considering

- 75

- 0

I thought flipping a fraction such as 1/3^2 would result in 3^-2.

Doesn't it change the exponent?

Doesn't it change the exponent?

- #8

C'mon guys....

1/ds^{2} = 1/dt^{2} – 1/(dx^{2} + dy^{2} + dz^{2})

1/ds^{2} = (dx^{2} + dy^{2} + dz^{2} - dt^{2})/[ (dx^{2} + dy^{2} + dz^{2})(dt^{2}) ]

ds^{2} = [ (dx^{2} + dy^{2} + dz^{2})(dt^{2}) ]/(dx^{2} + dy^{2} + dz^{2} - dt^{2})

Which just doesn't look any cleaner.

edit: changed to using integrated superscript.

1/ds

1/ds

ds

Which just doesn't look any cleaner.

edit: changed to using integrated superscript.

Last edited by a moderator:

- #9

plus

- 178

- 1

Originally posted by suffian

C'mon guys....

1/ds^{2}= 1/dt^{2}– 1/(dx^{2}+ dy^{2}+ dz^{2})

1/ds^{2}= (dx^{2}+ dy^{2}+ dz^{2}- dt^{2})/[ (dx^{2}+ dy^{2}+ dz^{2})(dt^{2}) ]

ds^{2}= [ (dx^{2}+ dy^{2}+ dz^{2})(dt^{2}) ]/(dx^{2}+ dy^{2}+ dz^{2}- dt^{2})

Which just doesn't look any cleaner.

edit: changed to using integrated superscript.

That was my point.

- #10

Eugene Shubert

- 22

- 0

I’m thinking of ds as an invariant that represents a differential increment of proper time. That would imply that the total amount of elapsed proper time t' would equal t/sqrt (1-1/V^2) where V^2 = (dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2. I would interpret V^2 > 1 to be a superluminal velocity.

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