Question on Spatial Curvature: Laser Experiment on a Spaceship

  • Thread starter kuahji
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In summary: For a photon fired in the x direction, the x component of it's 4 momentum will be constant (but non-zero). In summary, the path of a light beam in a spaceship accelerating to the right and hitting a mirror on the left will follow a curved path, reflecting off the mirror at equal angles of incidence and reflection. This can be viewed in both an inertial frame outside the spaceship and in the accelerated frame of the spaceship. Additionally, the choice of coordinate system can affect the path of the light beam.
  • #1
kuahji
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Ok, I'll try & lay the picture as best as possible. I'm on a spaceship & have a laser on the floor pointing towards the ceiling. When I'm not accelerating the laser shoots up & hits the ceiling. No problem there. If I'm accelerating at a fast enough pace the laser wouldn't go straight to the ceiling. We'll say it shoots out & hits the wall to my left while the ship is accelerating straight ahead. If the wall to my left was a mirror, where would the light then reflect to? Here is a basic picture http://img155.imageshack.us/img155/740/spaceship3hp.jpg [Broken] (red box = laser, blue line = light beam).
I don't understand the math behind all this just yet but I'm just trying to get a basic understand for which direction the light would travel. Would it bounce straight back in the same direction? Head towards the ceiling? Curve again & hit the wall? or something totally different?
 
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  • #2
I'am not expert at this but I think it wouldn't look too differn't as long your traveling not at the speed of light it would still apper to be pointing at the celling beacuse the information your getting it form is light.

I'am not sure if this has anything to do with spatial curvature I think this is more of a SR question then a GR question.
 
  • #3
scott1 said:
I'am not expert at this but I think it wouldn't look too differn't as long your traveling not at the speed of light it would still apper to be pointing at the celling beacuse the information your getting it form is light.

I'am not sure if this has anything to do with spatial curvature I think this is more of a SR question then a GR question.

I believe it has to do with GR because of "acceleration." SR doesn't deal with acceleration does it?
 
  • #4
I don't get a clear sense from your diagram which way the spaceship is accelerating.

Is the spaceship supposed to be accelerating from left to right?
 
  • #5
The ship is suppose to be accelerating to the right which cases the beam of light to curve to the left.
 
  • #6
kuahji said:
I believe it has to do with GR because of "acceleration." SR doesn't deal with acceleration does it?

Special relativity most certainly does - see https://www.physicsforums.com/showthread.php?t=119201".

This problem could involve either special relativity or general relativity, depending on whether the background spacetime is curved or not. Let's assume that the spaceship is deep in interstellar space, ie., far away from all massive bodies, so that the analysis can proceed in special relativity.

I hope to get back to this thread later today.

Regards,
George
 
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  • #7
The light will travel in a mirror image path, very roughly something like this (using your initial diagram as a guide).

[tex]
\[
\unitlength 1mm
\begin{picture}(50.00,80.00)(0,0)

\linethickness{0.15mm}
\qbezier(0.00,50.00)(50.00,45.00)(50.00,20.00)

\linethickness{0.15mm}
\qbezier(0.00,50.00)(50.00,55.00)(50.00,80.00)
\end{picture}
\]
[/tex]

If the light kept going, it would curve downwards again, and continue to "bounce" back and forth between the mirror, and a maximum "height".
 
  • #8
Assuming a flat spacetime, the light-ray (viewed from an inertial frame outside the accelerating ship) would still travel along a straight worldline on the light cone of the source event, then along a straight worldline on the light cone of the reflection event... in accordance with the law of reflection. One would have to transform into the accelerating ship's frame to obtain the light-ray's trajectory as viewed in that frame.
 
  • #9
robphy said:
Assuming a flat spacetime, the light-ray (viewed from an inertial frame outside the accelerating ship) would still travel along a straight worldline on the light cone of the source event, then along a straight worldline on the light cone of the reflection event... in accordance with the law of reflection. One would have to transform into the accelerating ship's frame to obtain the light-ray's trajectory as viewed in that frame.

I think there is an easier way. Let the x-axis point to the right (the direction of the acceleration of the ship), and the y-axis point up.

Then the y-component of the momentum of the photon should be the same before and after the collision with the mirror.

Some inertial obsevers will see the x-component of the momentum of the photon change, and hence they will see its energy E^2 = px^2 + py^2 will change.

However, the accelerated observer should not see the energy of the photon change , i.e. at any given height (x-coordinate of the accelerated observer), the photon should have the same energy (same frequency, no gravitational red or blue shift).The only way to keep the y component of the momentum unchanged and the energy unchanged is to have the angle of incidence equal the angle of reflection, i.e. py = constant and px changes sign.
 
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  • #10
George Jones said:
Special relativity most certainly does - see https://www.physicsforums.com/showthread.php?t=119201".

This problem could involve either special relativity or general relativity, depending on whether the background spacetime is curved or not. Let's assume that the spaceship is deep in interstellar space, ie., far away from all massive bodies, so that the analysis can proceed in special relativity.

I hope to get back to this thread later today.

Regards,
George

Thanks for the information. Glad I came here because it's clear the book I have has it wrong or at least a rather incomplete definition of the two. Thanks for all the other replies. It's all been pretty informative.
 
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  • #11
There are several possible choices for "the" coordinate system of an accelerated observer, but one specific choice is fairly popular (MTW, "Gravitation", pg 173 for instance)

[tex]t=x^0 = (1/g + \chi^1) \mathrm{sinh}(g \chi^0)[/tex]
[tex]x=x^1 = (1/g + \chi^1) \mathrm{cosh} (g \chi^0)[/tex]
[tex]y=x^2 = \chi^2[/tex]
[tex]z=x^3 = \chi^3[/tex]

where [itex]x^i[/itex] are the inertial coordinates, and [itex]\chi^i[/itex] are the accelerated coordinates. I've given the inertial coordinates a second, non-indexed description as well.

Anyway, you can transform any specific lightbeam by specifying it's path in inertial coordinates, then doing some algebra.

For instance

[itex]x^2 = c x^0, x^1,x^3 \mathrm{constant}[/itex] defines the curve for one light beam in inertial coordinates (one moving in the y direction), parameterized by time ([itex]x^0[/itex]) this appears as

[tex]\chi^1 = \frac{1}{g \mathrm{cosh}(g \chi^0)} - \frac{1}{g}[/tex]
[tex]\chi^2 = \frac{c}{g} \mathrm{tanh} (g \chi^0) [/tex]

parameterized by 'time' ([itex]\chi^0[/itex], a different time than before)

Something that isn't obvious about the coordinate system used is that [itex]x^1 = -1/g[/itex] is the surface of an "event horizon" called the Rindler horizon.

[add]I've attached a graph of the actual path for c=g=1.
 

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1. What is spatial curvature?

Spatial curvature is a measure of the degree to which the geometry of space deviates from the flat, Euclidean geometry we typically think of. It is a fundamental concept in the field of cosmology and is closely related to the curvature of space-time as described by Einstein's theory of general relativity.

2. How is spatial curvature measured?

Spatial curvature is typically measured using astronomical observations, particularly of the cosmic microwave background radiation. Scientists also use mathematical models and equations to calculate the curvature based on other observable phenomena, such as the distribution of matter in the universe.

3. What are the implications of a non-zero spatial curvature?

A non-zero spatial curvature would indicate that the universe is not perfectly flat, and could have significant implications for our understanding of the universe's overall structure and evolution. It could also impact predictions about the ultimate fate of the universe and the behavior of large-scale structures like galaxies and galaxy clusters.

4. Can spatial curvature change over time?

According to the currently accepted model of the universe, spatial curvature is relatively constant and does not change over time. However, some alternative theories and models suggest that it could vary over time, potentially affecting the expansion rate of the universe and other key cosmological parameters.

5. How does spatial curvature relate to the concept of dark energy?

Spatial curvature is closely linked to the concept of dark energy, which is thought to be responsible for the accelerated expansion of the universe. A non-zero spatial curvature could indicate that the effects of dark energy are not uniform throughout the universe, potentially leading to new insights into the nature of this mysterious force.

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