- #1

- 24

- 1

- 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

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