# Relative motion problem

1. Aug 2, 2008

### mtworkowski@o

I have a problem with relitive motion. If a light source is moving toward me a some high percent of c, the light will not reach me any sooner than if the light were stationary. The intantanious position of the light is the only thing that matters. Light is bound by a maximum propigation rate that is not determined by v. And if I am the one that is moving, position is still the only determining factor. I don't see how to determine the speed exclusive of position. I know this probably sounds dopey, but how does all this work?

2. Aug 2, 2008

### JesseM

Re: confused

Each inertial observer can use rulers and clocks at rest relative to themselves. For example, say I have a ruler exactly 5 light-seconds long at rest relative to me, and I attach clocks to either end, with the clocks synchronized according to what is called the "Einstein synchronization convention", based on the assumption that light's speed is uniform in all directions in my frame (so I could synchronize the two clocks in my frame by setting off a flash at the midpoint between them, and setting both to read the same time when the light from the flash reaches them). In this case, if I place a light emitter next to one clock and a light detector next to the other, and note the time T1 on the clock next to the emitter when it is turned on, and note the time T2 on the clock next to the detector when it detects light, then T2 - T1 will be 5 seconds, and since the distance between the clocks as measured by the ruler is 5 light-seconds, I have measured a distance/time of (5 light seconds)/(5 seconds) = 1 light second/second, or exactly c. If each inertial observer uses rulers and synchronized clocks at rest relative to themselves to measure the time and distance between light being emitted and light being detected, they will all get the answer of c, even though the values for the distance and time may be different.

3. Aug 2, 2008

### mtworkowski@o

Re: confused

A light second is the distance light travels in one second? So this is a distance? How does v relate to this? Please?

4. Aug 2, 2008

### JesseM

Re: confused

Yes, just like a light-year is a measure of distance. But you're also free to define it as 299792458 meters.
The v of what object, as viewed in what frame? Are you talking about the speed of the light's source as measured in the observer's rest frame? The observer can use the same system of rulers and clocks at rest with respect to herself to measure the speed of a sublight object like a light emitter (for example, if one clock reads T1=0 seconds as the emitter passes it, and another clock 5 light-seconds away on the ruler reads T2=20 seconds as the emitter passes it, then the emitter took 20 seconds to cross a distance of 5 light-seconds in this frame, so its speed must be 0.25c). And she'll find that the speed of light that comes out of the emitter (as measured by her own ruler/clock system) is always c regardless of the speed of the emitter (again relative to her own ruler/clock system).

5. Aug 2, 2008

### mtworkowski@o

Re: confused

I guess I'm remembering the formula for adding velocities at and around c. Two ref frames moving near c. I think I don't believe it's exactly true that v1 and v2 = u = (v + w) /(1+ vw /c^2). I don't even think I remember that right, but I don't think c is pegged to v in any way at all. Is it nuts to just watch the distance between two ref frames diminish and always use those distances for calcs.

6. Aug 2, 2008

### JesseM

Re: confused

Again, what does v refer to? In the velocity addition equation the two velocities v and w represent the speed of an object A in the rest frame of some inertial observer B, and the speed of B in the rest frame of some other inertial observer C, with u being the speed of A in the frame of C.
How do you physically measure the distance, and how do you measure the rate it's diminishing, if not by using something like the ruler/clock system I describe above? Remember, rulers in motion relative to one another measure distance differently, and clocks in motion relative to one another measure time differently. There is no "true" measure of the spatial distance between objects in relativity.

7. Aug 2, 2008

### mtworkowski@o

Re: confused

For the purpose of calculations, can't we define the distance ourselves and stipulate that this is what happens when.....? Can't we say that when the distance is half of what it was, then the light will take half the time to travel the distance? Can't we reduce the motion to a snap shot photo and measure the d with a ruler and say that at 1/4d then the time will be 1/4? Doesn't this get v out of the problem?

8. Aug 3, 2008

9. Aug 3, 2008

### mtworkowski@o

Re: confused

That's alot to swallow. I have to sit down with that. Thank you. JesseM, are you standing by?

10. Aug 3, 2008

### MeJennifer

Re: confused

Good question!

While it is in principle possible to determine the speed of an approaching object by applying rulers and clocks it would in most situations be completely impractical. Imagine someone placing hundreds of clocks and rulers around us with a radius of say one light year. :) And beyond this impracticality, the gravitational field will deviate their inertial paths with respect to each other, e.g. the clocks would drift with respect to each other.

In a practical sense we would measure the frequency of the object coming at us and compare this frequency to the frequency we expect. If it is a spaceship we would expect a prior agreed on frequency, if it is a naturally radiating object, for instance a particular type of star, we would compare the expected light spectrum with what we actually measure.

Then it is just a matter of applying a relatively simple formula given to us by Doppler and Einstein.

11. Aug 3, 2008

### mtworkowski@o

Re: confused

The doppler effect has always been one of my favorites. My understanding of it is incomplete. The shift is from the source moving fast enough to stack waves tighter, or the waves hit the sensor faster because the entire wave formation has a higher v.(ie. delivering a higher f.)

12. Aug 3, 2008

### JesseM

Re: confused

I asked you twice before what you meant by "v", could you please clarify? And I also don't understand what you mean by "the time will be 1/4"--1/4 of what? 1/4 of the time it took to travel the distance d? That'd be true of course if the object is moving inertially, but it'd be true regardless of whether the object was moving at 1c, 0.5c, 0.001c, etc...it tells you nothing about the actual speed.

13. Aug 3, 2008

### JesseM

Re: confused

For light it has nothing to do with the waves moving at a different speed, it's just that if the source is moving towards you, each successive peak has a shorter distance to travel to reach you (because the source has gotten closer since it emitted the last one) and this reduces the time between peaks hitting you (beyond the actual time between peaks being emitted from the source in your frame).

14. Aug 3, 2008

### mtworkowski@o

Re: confused

Sorry, I was using d for distance and v for the velocity of the light sources.

15. Aug 4, 2008

### cryptic

Sagnac, Michelson-Gale,...

If it is so why http://gallica.bnf.fr/ark:/12148/bpt6k31103/f1410.table" measured different c?

Last edited by a moderator: Apr 23, 2017
16. Aug 4, 2008

### JesseM

Re: Sagnac, Michelson-Gale,...

Where did you get the idea that the Sagnac effect is inconsistent with the idea that light moves at c in all inertial frames? This page says:

Last edited by a moderator: Apr 23, 2017
17. Aug 4, 2008

### cryptic

Re: Sagnac, Michelson-Gale,...

It is not important what the page says, as you can see they measured c'=c-v and c'=c+v.

18. Aug 4, 2008

### JesseM

Re: Sagnac, Michelson-Gale,...

No, the page says that 'the sum of the speeds of the wave front and the receiver at the "end" point is c-v in the co-rotating direction and c+v in the counter-rotating direction.' The speed of the wave front itself is still c (at least in inertial frames; in the non-inertial frame of an observer rotating with the ring the speed would be different in opposite directions).

Last edited: Aug 4, 2008
19. Aug 4, 2008

### cryptic

Re: Sagnac, Michelson-Gale,...

I don't speak of rotating ring but of moving mirrors. Relative speed of moving mirrors is zero and absolute, tangential non accelerated speed in earth's frame is v. Light rays are of course not rotating. Light rays are only reflected at different places in earth's frame. The speed of the wave front is c only in this frame, in moving frame it is c+v and c-v as measured by those experiments.

20. Aug 4, 2008

### JesseM

Re: Sagnac, Michelson-Gale,...

How are they moving, exactly? The wikipedia article on the Sagnac effect has a setup with 3 or 4 mirrors, but placed on a rotating platform. In the History of the Sagnac effect section it also mentions that the Michelson-Gale experiment was a large interferometer that detected the rotation of the Earth, similar to how the smaller-scale version detects the rotation of the platform it's mounted on. Of course, any rotating frame is not an inertial one! This includes any frame where any fixed point on the surface of the Earth has a fixed position-coordinate.
What do you mean by "tangential non accelerated speed"? Do you understand that an object rotating in a circle is experiencing constant acceleration, since acceleration is a change in the magnitude or direction of an object's velocity vector?
What do you mean by "Earth's frame" and "moving frame", exactly? Does Earth's frame mean a frame where the center of the Earth has a fixed position-coordinate, or one where points on the surface have fixed position-coordinates? If it's just based on a fixed center but with the surface rotating, then is the "moving frame" one which is moving with constant speed and direction relative to the center, or is it rotating relative to it?

21. Aug 4, 2008

### cryptic

Re: Sagnac, Michelson-Gale,...

Rotation has nothing to do with light rays. Light rays move after reflection with c, relative to position of reflection, independent of any rotation and without any acceleration. They do not move in a rotating system but in an IS between two rotating mirrors. We can use translating or rotating mirrors the result would be the same.
Light rays are not accelerated. If the mirrors are accelerated is absolutely inconsequential.
Yes.

22. Aug 4, 2008

### JesseM

Re: Sagnac, Michelson-Gale,...

Obviously. The point is that if the mirrors are themselves rotating, the round trip time for the light to get back to the mirror it started from is different depending on which direction the light went, since in any inertial frame the mirrors will themselves be moving.
If you use translating mirrors, you can find an inertial frame where the mirrors are at rest, and in this frame the round-trip time is the same in both the clockwise and counterclockwise direction, meaning that if you place a clock at rest with respect to the mirror it will measure the same time for light to return from both directions. The same is not true if the mirrors are rotating.

23. Aug 5, 2008

### cryptic

Re: Sagnac, Michelson-Gale,...

It is true only if mirrors not move. Translating and rotating mirrors will produce the same result with only difference that the relevant mirror speed, in case of rotation is v'=v/√2 because the mirrors move under an angle of 45° with respect to light.

If mirrors move in any direction with respect to the center of the earth the time difference for both directions of light would always be:

Δt=4L/(c-v) - 4L/(c+v), as Sagnac and Michelson-Gale measured.

L is the distance between mirrors and v is the relevant velocity of mirrors.

24. Aug 5, 2008

### JesseM

Re: Sagnac, Michelson-Gale,...

You mean you think that even if the mirrors are moving inertially (translating but not rotating or accelerating in any other way), the round-trip time for light will be different in different directions? That is definitely incorrect, and I'm sure no version of the Sagnac or Michelson-Gale experiment has shown this since it would be a simple disproof of relativity.
In the equation v refers to the tangential velocity in the frame of the center of the axis of rotation (the center of the Earth in the case of the Michelson-Gale experiment). Are you claiming the equation would work in other contexts besides rotation, like if the mirrors were moving in a purely inertial way relative to the center of the Earth? If so you are mistaken.

25. Aug 5, 2008

### cryptic

Re: Sagnac, Michelson-Gale,...

Both experiments actually disproved relativity. Michelson wrote in his paper: "The calculated value of the displacement on the assumption of a stationary ether es well as in accordance with relativity is...". But relativity can not be applied because whole apparatus move with the same speed and the time needed to make a measurement is less then 10 µs. During this time the surface of the earth moved less then 2mm and the rotation can be neglected.

I'm not mistaken. http://www.aip.org/history/gap/PDF/michelson.pdf" measured also a displacement of interference fringes and found (page 341) that “the relative velocity of the earth and the ether is probably less than one sixth the earth's orbital velocity (5 km/s), and certainly less than one fourth (7.5 km/s).”

This displacement is due to earth's rotation and was too overestimated by Michelson (v≈250m/s).

Last edited by a moderator: Apr 23, 2017