Equivalence Principle: Photon Fall Near Earth's Surface

k3r0
Messages
15
Reaction score
0

Homework Statement


A photon near the surface of the Earth travels a horizontal distance of 3 km. How far (in meters) does the photon 'fall' in this time? (Hint: think equivalence principle).

Homework Equations


N/A


The Attempt at a Solution


My understanding of the equivalence principle is that experiments carried out in a 'free fall' frame will have the same results as an experiment done floating in space (far from a large body). I'm confused because the photon near the surface of Earth is in neither of these situations -- so would the answer be that it doesn't 'fall' at all? I would expect it to travel horizontally only.

Thanks.
 
Physics news on Phys.org
k3r0 said:

Homework Statement


A photon near the surface of the Earth travels a horizontal distance of 3 km. How far (in meters) does the photon 'fall' in this time? (Hint: think equivalence principle).

Homework Equations


N/A


The Attempt at a Solution


My understanding of the equivalence principle is that experiments carried out in a 'free fall' frame will have the same results as an experiment done floating in space (far from a large body). I'm confused because the photon near the surface of Earth is in neither of these situations -- so would the answer be that it doesn't 'fall' at all? I would expect it to travel horizontally only.

Thanks.

The usual picture is to imagine an experiment carried out in an elevator (3km wide), where you shoot a beam of photons straight across the elevator. If you carry this out in space it will travel straight across. If you are in the same elevator at the Earth's surface in free fall, accelerating towards the center of the earth, from the view point of someone in the elevator it will also travel straight across. You are supposed to think of how it looks from the view point of someone who is not in the elevator and standing on the Earth's surface.
 
Thanks, I've got it now!
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top