Gravitational time dilation reason

[rogerk8]

Hi WannabeNewton!

The last part of this statement:

The position of $O'$ will be given by $z_{O'} = \frac{1}{2}gt^{2}$ and the position of $O$ will be given by $z_{O} = h + gt^{2}$.

has to be wrong.

Because the change in z should be the same for both O and O', right?

By the way, shouldn't g be substituted for the more general a while the rocket is moving upwards?

Roger

 

[WannabeNewton]

No roger $z_{O}$ and $z_{O'}$ don't represent changes in $z$ in the manner of which you speak. They simply track the positions of $O$ and $O'$ respectively relative to the origin of the inertial frame (which is at $z = 0$).

$g$ is just a label here. It was chosen as the label because in this example the rocket is at rest on the Earth so via the equivalence principle it's acceleration is $g$.

 

[rogerk8]

Hi WannabeNewton!

So zo-zo' should not equal h?

Do you mind explaining why?

Roger

 

[WannabeNewton]

What? Yes $z_{O} - z_{O'} = h$ but how does that relate to your statement in post #36? This represents the distance between the two observers, not the position of the individual observers themselves relative to the inertial frame.

 

[rogerk8]

Post #36:

$$z_0=h+gt^2$$

and

$$z_0'=\frac{1}{2}gt^2$$

then

$$z_0-z_0'=h+\frac{1}{2}gt^2$$

which has to be wrong, right?

Roger

 

[WannabeNewton]

Oh sorry, there should be a $\frac{1}{2}$ in front of the $gt^2$ for $z_{O}$. My bad, I totally missed that! Sorry I wrote that original post up in a haste, so I made some careless errors.

 

[rogerk8]

I knew it!

I knew I was right!

But don't feel sorry, I will get back to you with more questions to be sorry about ;)

Roger