Time dilation for a clock thrown vertically

[gnieddu]

Homework Statement

Two clocks are placed on the earth's surface. One is kept still while the other is thrown upwards with an initial small velocity $v_0$. How will the time measurements differ between the two clocks according to SR only and to GR only?

Relevant Equations

Equation describing the motion of a body in earth's gravitational field following Newton's laws: $z=v_0t-\frac{1}{2}gt^2$

Proper time under Minkowski metric (with (+, -, -, -) signature): $c^2d{\tau}^2=-ds^2=c^2dt^2-dz^2$

The non-moving clock will see the other one move upwards and land as predicted by Newton's laws, so using the equation $z=v_0t-\frac{1}{2}gt^2$, and assuming the moving clock starts at $t=0$, it will land at $t=\frac{2v_0}{g}$.

Now, using SR only, and the Minkowski metric (with signature (-,+,+,+)) the time $\tau$ measured by the moving clock is obtained by using the equation:

$$c^2dτ^2=−ds^2=c^2dt^2−dz^2$$

where dz can be obtained from the equation of motion by differentiating:

$$dz=d[v_0t−\frac {1} {2} gt^2]=(v_0−gt)dt$$

In the end, we need to evaluate:

$$τ=\frac {1} {c} \int_0^{\frac {2v_0}{g}}[c^2−(v_0−gt)^2]dt $$

If we consider GR only, the solution follows the same line, but the metric would be the one for the weak field approximation:

$$ds^2=−(1+\frac{2gh}{c^2})c^2dt^2+dz^2$$

with the added complexity of expressing h as $v_0t-\frac{1}{2}gt^2$.

In both cases, the integrals are not nice, but can possibly be simplified by using the approximation $\sqrt{1+x}=1+\frac{x}{2}$. All of this, of course, if my plan of action makes sense and I'm not missing anything.

Thanks in advance

Gianni

 

[TSny]

Now, using SR only, and the Minkowski metric (with signature (-,+,+,+)) the time $\tau$ measured by the moving clock is obtained by using the equation:

$$c^2dτ^2=−ds^2=c^2dt^2−dz^2$$

where dz can be obtained from the equation of motion by differentiating:

$$dz=d[v_0t−\frac {1} {2} gt^2]=(v_0−gt)dt$$

In the end, we need to evaluate:

$$τ=\frac {1} {c} \int_0^{\frac {2v_0}{g}}[c^2−(v_0−gt)^2]dt $$

There is an important typographical error in the last equation. Note that the right-hand side as written does not have the dimensions of time. Otherwise, I think this is correct for the "SR only" calculation.

If we consider GR only, the solution follows the same line, but the metric would be the one for the weak field approximation:

$$ds^2=−(1+\frac{2gh}{c^2})c^2dt^2+dz^2$$

with the added complexity of expressing h as $v_0t-\frac{1}{2}gt^2$.

OK. It's not clear to me what they mean by "GR only". In the above expression, the $dz^2$ term will reproduce the "SR only" contribution. By "GR only" they might mean the contribution due to only the $(1+\frac{2gh}{c^2})c^2dt^2$ part of the expression above. I don't know.

In both cases, the integrals are not nice, but can possibly be simplified by using the approximation $\sqrt{1+x}=1+\frac{x}{2}$.

Yes.

All of this, of course, if my plan of action makes sense and I'm not missing anything.

I think you are on the right track.

 

[gnieddu]

There is an important typographical error in the last equation. Note that the right-hand side as written does not have the dimensions of time. Otherwise, I think this is correct for the "SR only" calculation.OK. It's not clear to me what they mean by "GR only". In the above expression, the $dz^2$ term will reproduce the "SR only" contribution. By "GR only" they might mean the contribution due to only the $(1+\frac{2gh}{c^2})c^2dt^2$ part of the expression above. I don't know.Yes.

I think you are on the right track.

Hi TSny, thanks for your feedback (and apologies for my late reply). You're right about the typo in my formula: the expression under the integral should be square-root-ed. Regarding the meaning of "GR only", my assumption is that I should compute the time dilation using only the GR metric vs. also considering special relativistic effects connected with the clock's speed (i.e. the $\sqrt{1-\frac{v^2}{c^2}}$ factor normally found in SR). But perhaps I'm only unnecessarily messing up things here...

 

[TSny]

Regarding the meaning of "GR only", my assumption is that I should compute the time dilation using only the GR metric vs. also considering special relativistic effects connected with the clock's speed (i.e. the $\sqrt{1-\frac{v^2}{c^2}}$ factor normally found in SR). But perhaps I'm only unnecessarily messing up things here...

The GR metric $d\tau^2=(1+\frac{2gh}{c^2})dt^2-dz^2/c^2$ includes the SR effect of "moving clocks run slow". For example, imagine letting the gravitational acceleration $g$ go to zero in the metric. Then the metric reduces to the SR Minkowski metric $d\tau^2=dt^2-dz^2/c^2$. Thus $d\tau= dt\sqrt{1 - (v_z/c)^2}$ which is the SR time dilation.

So, in the GR metric, the $dz^2/c^2$ accounts for "moving clocks run slow". The $\frac{2gh}{c^2}$ part takes care of "clocks at higher gravitational potential run faster". GR handles everything. We wouldn't add the SR effect to the result of the GR metric.

 

[gnieddu]

The GR metric $d\tau^2=(1+\frac{2gh}{c^2})dt^2-dz^2/c^2$ includes the SR effect of "moving clocks run slow". For example, imagine letting the gravitational acceleration $g$ go to zero in the metric. Then the metric reduces to the SR Minkowski metric $d\tau^2=dt^2-dz^2/c^2$. Thus $d\tau= dt\sqrt{1 - (v_z/c)^2}$ which is the SR time dilation.

So, in the GR metric, the $dz^2/c^2$ accounts for "moving clocks run slow". The $\frac{2gh}{c^2}$ part takes care of "clocks at higher gravitational potential run faster". GR handles everything. We wouldn't add the SR effect to the result of the GR metric.

Thanks. I thought I'd have to have some expression for $dz^2$ in order to account for the fact that the clock's z coordinate is changing in time (and not uniformly due to the accelerated motion), but I not I get yor point.