- #1

- 212

- 4

ds

^{2}= [1- (2GM/rc

^{2})]c

^{2}dt

^{2}- dr

^{2}/[1- (2GM/rc

^{2})] - r

^{2}(dθ

^{2}+ sin

^{2}(θ)d∅

^{2})

Now, here is the first example problem that I created:

Let's say that there is a perfectly spherical moon in a vaccum. The moon has mass M = 100 kg, and it has radius r = 6.58765432 * 10

^{-34}Ls (Ls stands for light seconds). Now, we have two massless observers (I made them massless just for simplicity of the problem). We will call these observers Mario and Luigi. Now let's say that Mario stands on top of this tiny moon, so his distance from the moon's center of mass is r = 6.58765432 * 10

^{-34}Ls. His angular coordinates with respect to the center of the moon are θ = 0, and ∅ = π/2.

Now, Luigi floats somewhere off in space. His distance from the moon's center of mass is r = 4 Ls. His angular coordinates with respect to the moon's center are the exact same as Mario's.

If Luigi's watch reads that 20 seconds have passed, then how much time has passed from Mario's perspective?

Now here is my answering process:

Note: I used units where c = 1 Ls/s, G = 2.47037037 * 10

^{-36}Ls

^{3}/ (Kg * s

^{2})

Using this line element:

ds

^{2}= [1- (2GM/rc

^{2})]c

^{2}dt

^{2}- dr

^{2}/[1- (2GM/rc

^{2})] - r

^{2}(dθ

^{2}+ sin

^{2}(θ)d∅

^{2})

I just plugged in r = 6.58765432 * 10

^{-34}(Mario's distance from the moon's center) , dt =20 (the amount of time Luigi counted) and dr = the difference between Luigi's r coordinate and Mario's r coordinate, which ≈ 4. θ, dθ,and d∅ should all be 0 since Mario and Luigi are at the exact same angular coordinates with respect to the center of the moon.

The final result comes out to be ds

^{2}= dτ

^{2}≈ 36

The square root of 36 is 6, so this means that Mario should see a time of about 6 seconds when Luigi counts 20 seconds.

Now I have two questions about this:

a) Is this result correct? I just want to make sure I have an accurate understanding of things here.

b) I noticed that this result differs from what I'd get using the gravitational time dilation formula. If I were to manipulate the formula t = τ /sqrt(1 - (2GM/rc

^{2})) where t is the time that an external observer sees while τ is the proper time that the person in the gravitational field (Mario in this case) would see, then I could plug in t = 20 seconds and arrive at the result τ = 10 seconds (meaning Mario would see 10 seconds). However, upon plugging dt = 20 seconds into the line element, I noticed that 10 seconds is what I would get only if dr, dθ, and d∅ were set to 0. Would I be correct in surmising that the gravitational time dilation formula assumes that dr, dθ, and d∅ are all 0?

If a) is not correct, then please tell me where I went wrong.

I actually have some more questions pertaining to space-like intervals (I made sure the above example had a time-like interval), but this thread is getting kind of long and I am getting tired of typing, so I will just make a part 2 to this thread at a later time.

Thank you in advance.