# Question on Spatial Curvature

1. May 4, 2006

### kuahji

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 (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?

2. May 4, 2006

### scott1

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. May 4, 2006

### kuahji

I believe it has to do with GR because of "acceleration." SR doesn't deal with acceleration does it?

4. May 4, 2006

### pervect

Staff Emeritus
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. May 5, 2006

### kuahji

The ship is suppose to be accelerating to the right which cases the beam of light to curve to the left.

6. May 5, 2006

### George Jones

Staff Emeritus
Special relativity most certainly does - see this thread.

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

7. May 5, 2006

### pervect

Staff Emeritus
The light will travel in a mirror image path, very roughly something like this (using your initial diagram as a guide).

$$$\unitlength 1mm \begin{picture}(50.00,80.00)(0,0) \linethickness{0.15mm} \qbezier(0.00,50.00)(50.00,45.00)(50.00,20.00) \linethickness{0.15mm} \qbezier(0.00,50.00)(50.00,55.00)(50.00,80.00) \end{picture}$$$

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. May 5, 2006

### robphy

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. May 5, 2006

### pervect

Staff Emeritus
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.

Last edited: May 5, 2006
10. May 5, 2006

### kuahji

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.

11. May 5, 2006

### pervect

Staff Emeritus
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)

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

where $x^i$ are the inertial coordinates, and $\chi^i$ 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

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

$$\chi^1 = \frac{1}{g \mathrm{cosh}(g \chi^0)} - \frac{1}{g}$$
$$\chi^2 = \frac{c}{g} \mathrm{tanh} (g \chi^0)$$

parameterized by 'time' ($\chi^0$, a different time than before)

Something that isn't obvious about the coordinate system used is that $x^1 = -1/g$ 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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Last edited: May 5, 2006