Gravitational time dilation reason

[rogerk8]

WannabeNewton,

I got stuck early in your explanation i.e at:

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

Please explain this like I have no clue about physics at all.

If something is free-falling, does the length it falls equal 1/2gt^2. If so, why?

And why is it gt^2 for O?

Roger

 

[WannabeNewton]

Hi roger! For an object under the force $mg$, the path in general is given by $z(t) = z_0 + v_0 t + \frac{1}{2}gt^2$ where $z_0$ is the initial position and $v_0$ is the initial velocity. This is gotten simply by solving Newton's 2nd law for $m\ddot{x} = F = mg$. $O'$s initial position is at the bottom so he has $z_0 = 0$ and he starts out at rest so $v_0 = 0$. $O$ is at the top so his initial position is $h$ (the height of the rocket) and he also starts out at rest so $v_0 = 0$.

 

[rogerk8]

Hi roger! For an object under the force $mg$, the path in general is given by $z(t) = z_0 + v_0 t + \frac{1}{2}gt^2$ where $z_0$ is the initial position and $v_0$ is the initial velocity. This is gotten simply by solving Newton's 2nd law for $m\ddot{x} = F = mg$. $O'$s initial position is at the bottom so he has $z_0 = 0$ and he starts out at rest so $v_0 = 0$. $O$ is at the top so his initial position is $h$ (the height of the rocket) and he also starts out at rest so $v_0 = 0$.

Hi WannabeNewton!

Would you mind solving the differential equation for me also?

I think I can see that if you integrate x'' once you'll get vt and if you integrate it twice you'll get 1/2gt^2.

But it was such a long time since I worked with differential equations so I do not remember and I cannot see it clearly.

Hope you don't think I am lazy for asking this.

Roger

 

[rogerk8]

Hi WannabeNewton!

Solving...

$$mx''=F=mg$$


$$x''=F/m=g$$

$$x'=F/m(t+A)$$

$$x=F/m(t^2/2+At+B)$$

$$x(0)=F/m*B=x_0$$

$$x'(o)=F/m*A=v_0$$

Which leads to

$$x(t)=x_0+v_0t+1/2gt^2$$

Thank you for not solving it for me!

I will not be that lazy again :)

Roger

 

[WannabeNewton]

Yep that works fine :)

 

[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