Hyperbolic relationship of time and distance in relativity

[stevmg]

Homework Statement

In space-time when one "travels" a distance x from a point (within the light cone) this, in effect, comes off the time it takes. This is a hyperbolic relationship

Homework Equations

tau = SQRT[t² - x²]

We'll stay with one dimension (x)

The Attempt at a Solution

tau² = t²-x² - Now that IS a hyperbola

The question is, where did tau = SQRT[t² - x²] come from. I've looked at the Lorentz transforms

x' = \gamma(x - vt)
t' = \gamma[t - v²x/c²]

I cannot make that happen but it is because of the Lorentz transforms that this time-distance relationship is true

 

[Cyosis]

Use the time dilation formula.

 

[vela]

Try calculating (ct')²-x'². Your Lorentz transformation equations seem a bit messed up unitwise. Are you using units where c=1? If so, c shouldn't appear in your equations.

 

[stevmg]

Try calculating (ct')²-x'². Your Lorentz transformation equations seem a bit messed up unitwise. Are you using units where c=1? If so, c shouldn't appear in your equations.

You are right

tau = SQRT[t² - (x/c)²]

Now what? Where do I go from here? There is a step between the original Lorentz transformation equation and this that I am missing and I don't see it, no matter how obvious it should be.

 

[stevmg]

Use the time dilation formula.

Cyosis - much help in the past but this site was absolutely no help to me.

I know that \Deltat = \gamma\Deltat' where \Deltat' is measured in S', the "moving" frame relative to S.

To wit, if \gamma = 1.25 and elapsed time for a spaceship is 8 years in S', then the elapsed time for the Earth is 10 light years, assuming we use the worldline of Earth as going straight up.

This is fine and dandy but it doesn't get me from the \Deltat's to the \Deltax's.

There is something obvious that I am missing and that's what I need to be pointed out.

Steve G

 

[vela]

Did you calculate (ct')²-x'²?

 

[Cyosis]

If I am not mistaken you're wondering where \tau=\sqrt{t^2-x^2/c^2} comes from. Here \tau is the proper timer. You could read this directly off the metric or compare the proper time expression \tau=t/\gamma and see that they are the same.

 

[Cyosis]

We don't just hand out the answers. Secondly you've been asked to calculate (ct')^2-x'^2 by vela or compare the two proper time expressions by me yet have not showed a single attempt.

 

[stevmg]

If I am not mistaken you're wondering where \tau=\sqrt{t^2-x^2/c^2} comes from. Here \tau is the proper timer. You could read this directly off the metric or compare the proper time expression \tau=t/\gamma and see that they are the same.

Now, we are getting somewhere. Exactly what is proper time. Is it some mythical frame of reference in which objects do not move but just "get older" and that all other frames of reference move at a velocity in relation to this mythical frame?

In that case it is easy to uderstand that \tau=t/\gamma in which t is the elapsed time in the "moving" frame. If that were so, then the hyperbolic relationship becomes obvious.

For given \tau 1/\gamma = \sqrt{(1-v^2/c^2)}

thus \tau=t\sqrt{(1-v^2/c^2)}
\tau=\sqrt{(t^2-v^2t^2/c^2)} but vt = x
\tau=\sqrt{(t^2-x^2/c^2)} and
\tau^2=t^2-x^2/c^2 for all x and t relating to a given \tau and the hyperbolic relationship is obvious.
But, now, where does your original definition or derivation of \tau come from? It does appear on page 100 of AP French's Special Relativity in an obtuse way. It appears that the elapsed time in the moving frame t is greater than the elapsed time in the resting frame \tau by his equations.

Am I getting somewhere?

 

[stevmg]

We don't just hand out the answers. Secondly you've been asked to calculate (ct')^2-x'^2 by vela or compare the two proper time expressions by me yet have not showed a single attempt.

I was editing my answer when you answered - look above (post #9) to see what I ultimately came up with...

Steve G

 

[Cyosis]

Now, we are getting somewhere. Exactly what is proper time. Is it some mythical frame of reference in which objects do not move but just "get older" and that all other frames of reference move at a velocity in relation to this mythical frame?

Proper time is the time interval measured between two events that occur at the same point in a particular frame.

But, now, where does your original definition or derivation of LaTeX Code: \\tau come from? It does appear on page 100 of AP French's Special Relativity in an obtuse way.

I don't get why you chose to use 'my' method that relies heavily on the proper time formula derived from the metric over velas method which relies on Lorentz transformations, transformations that you seem to be familiar with (post #1).

Using the Lorentz transformation in combination with the definition of proper time I have given you will yield you the time dilation formula.

It appears that the elapsed time in the moving frame t is greater than the elapsed time in the resting frame LaTeX Code: \\tau by his equations.

There is no moving and or resting frame in an absolute sense. In this case there are two frames that move with respect to each other with a relativity velocity v. Whether t' runs slower than t or t runs slower than t' depends entirely from which frame the measurement is made.

 

[stevmg]

I don't get why you chose to use 'my' method that relies heavily on the proper time formula derived from the metric over velas method which relies on Lorentz transformations, transformations that you seem to be familiar with (post #1).

Actually I was thinking along those lines before we were exchanging posts. It seemed easier.

IS IT RIGHT?

That's where your expertise comes in.

 

[Cyosis]

Actually I was thinking along those lines before we were exchanging posts. It seemed easier.

IS IT RIGHT?

That's where your expertise comes in.

The mathematics in post #9 is correct.

 

[stevmg]

Proper time is the time interval measured between two events that occur at the same point in a particular frame.



I don't get why you chose to use 'my' method that relies heavily on the proper time formula derived from the metric over velas method which relies on Lorentz transformations, transformations that you seem to be familiar with (post #1).

Using the Lorentz transformation in combination with the definition of proper time I have given you will yield you the time dilation formula.



There is no moving and or resting frame in an absolute sense. In this case there are two frames that move with respect to each other with a relativity velocity v. Whether t' runs slower than t or t runs slower than t' depends entirely from which frame the measurement is made.

Actually, there is an excellent mathematical demonstration of this in AP French's Special Relativity which uses vela's equation (post #7) and the Lorentz equations to arrive at an invarant s² which firmly establishes the equation of a hyperbola. had I seen that earlier, I would not have even bothered with establishing this thread.

Much thanks for your time.