# Inaccuracy in the speed of light

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1. Jul 9, 2014

### RyanXXVI

Imagine a system with a laser and a receiver with the ability to detect when light from the laser reaches it. There is also a console equidistant from both the receiver and the laser which sends a signal to each instrument, making the laser turn on and the receiver start a timer. The distance between the receiver and the laser is known and everything is stationary. When the receiver receives the light, the timer stops, then does a calculation to discover the speed of light.

In that situation, the result would be completely accurate. However, now imagine a situation where the whole system was moving in one direction at a speed. This would skew the results. The true speed of light would be the calculated speed plus the speed of the system. Of course, this would be un-calculable if the speed of the system was unknown. Also, to any observer in this system, the system would be stationary.

My point is, everyone on this planet moves at the speed at which the speed at which the Earth move as well as our own individual speeds. Would this not mean that our measurement of the speed of light is inaccurate? Unless one knew the true velocity of the Earth, it would be. Could someone please explain to me how I am wrong. I imagine I am because physicists much more intelligent than me have determined the speed of light.

Ryan

P.S. The alleged inaccuracy extends farther than just Earth and includes any measurements taken in our galaxy for the same reason.

2. Jul 9, 2014

### ModusPwnd

If you have the light return the way it came that problem would not exist. Note there is no "true velocity of the earth". Speed and velocity are relative. To to us on the earth the speed of the earth is zero, to a craft whizzing by its non-zero. In each case the speed of light would be observed to be the same speed. That is one of the wondrous behaviors of the universe.

3. Jul 9, 2014

### phinds

Light speed does not add to other speeds the way you are assuming it does.

If I'm in a plane going 1000 mph and I fire a bullet at you at 100 mph, you see the bullet going 1,100 mph if you are in front of the plane and 900 mph if you are behind the plane.

If I'm in the same plane and shoot a light beam at you at c, you see the beam arrive at c and that's true whether you are in front of the plane or behind it (or off to the side or whatever).

That is one of the fundamental postulates of SR. Light travels at c in all inertial frames of reference.

Last edited: Jul 9, 2014
4. Jul 9, 2014

### Staff: Mentor

For the general rule for "adding" velocities, see

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel.html

In the formula given on that page, let u' = c (for the velocity of the projectile e.g. a light pulse with respect to the "moving observer" B), and you will always get u = c (for the velocity of the projectile relative to the "stationary observer" A), regardless of the relative velocity v of A and B.

5. Jul 9, 2014

### parkner

It adds normally: c - v.

For example: we send a signal to the space probe at distance d, and moving away with a speed v.
What is a time of the signal journey to reach the probe?

Of course, this is not d/c, but just d/(c-v)
Thus the signal speed wrt the probe is c-v exactly, not c, what is very easy to verify.

6. Jul 9, 2014

### pervect

Staff Emeritus
This is true, but it's not related to the velocity addition formula. The velocities you are talking about are not actual velocities, but rates of closure in a particular coordinate system.

To relate rates of closure to actual velocities, if you have two objects, the relative velocity between them is equal to the rate of closure between the two objects as measured in a coordinate system where one of the objects is at rest.

If you specify the relative velocity between A and B as u, and the relative velocity between B and c as v, the relative velocity between A and C is giving by the velocity addition formula.

Let A be a light beam, and B and C be two observers

Note that in this case A doesn't have a frame of reference in relativity, but B and C do.

Then the relative velocity between A and B is c (we measure this in a frame where B is at rest). The relative velocity between B and C is v (we can measure this either in a frame where B is at rest or C is at rest. The two results will be equal except for the necessary sign inversion). Finally, the relative velocity between A and C is still equal to c ( we measure this in a frame where C is at rest) because the velocity addition formula in special relativity is not linear.

The exact nature of the velocity addition formula in relativity is a consequence of the Lorentz transform. The Lorentz transform describes the mathematical details of how one changes frames of referernce in relativity (i.e. from B to C in this example). It is different from the pre-relativity Gallilean transform. The exact mathematical details can be found in textbooks and the web, at this point I am only trying to say that relativity is different from pre-relativity in this respect (in how one changes frames), and that the OP has made assumptions that are not true in relativity, presumably due to unfamiliarity with the theory. The solution is to become familar with the theory.

Because "changing frames of reference" is an abstract concept, I have chose to focus on velocity addition as a less-abstract consequence, hopefully it's easier to understand the issue this way.

Last edited: Jul 9, 2014
7. Jul 9, 2014

8. Jul 9, 2014

### sophiecentaur

You would need to specify who is measuring this time interval, in what frame.

9. Jul 10, 2014

### parkner

I can't agree with this interpretation.
The relative velocity is just the 'rate of closure': v = dx/dt, ie. it's a relative distance hange.

And in a general case: v = v_r + v_t, where: v_t = wr = df/dt r
so there is possible some additional tangential displacement: an angular position change x distance.

]

You are talking rather about something what can be called as a 'relativistic velocity', ie. some model-dependend abstract term, not an 'actual vielocity'.

10. Jul 10, 2014

### Staff: Mentor

No, pervect's terminology is the common one. Relative velocity is specifically the velocity of one object in the rest frame of the other.

http://en.wikipedia.org/wiki/Relative_velocity

11. Jul 10, 2014

### phinds

It is not an interpretation, it if a fundamental fact of Special Relativity and has been demonstrated empirically.

12. Jul 10, 2014

### Maxila

First to be clear of my meaning below, I understand this, empirical evidence supports this, and I fully agree with it. With that said, we know B has a relative perspective of length and time compared to C, therefore the light beam (A) and the constant c must also be proportional and relative to each view of x/t (where x is length and t is time). This is not questioning c as a constant for each frame, rather it is a observation of the constant being relative to each view of space and time, just as their space (Euclidean) and time are relative to each other. In other words as views of space and time change so must c exactly, in order to be constant for the change in space and time.

For example this problem done by GSU physics dept. to explain the Muon experiment and relativity http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html the comparison of frames shows what we already know, x does not equal x' nor does t = t'. We also know x/t = c = x'/t', but common to both is a ratio of their x/t as a constant, not the specific and relative values. Again to be clear, this is not an argument against any evidence or facts shown, merely an observation that in order for the constant c to remain constant for all, it is also relative to all frames view of space and time (x and t).

13. Jul 10, 2014

### pervect

Staff Emeritus
I won't argue overmuch with any approach that gets identical answers , but I'll suggest that philosophically it's simpler to think of space and time as not changing, but that the description of them via coordinates changes when one chooses a frame of reference.

This goes along with the idea that coordinates are labels without direct physical significance that I've mentioned in other recent threads.

Aside from being (IMO) much simpler, this is usually the manner in which relativity is usually explained. With this viewpoint, while space and time as abstract entities don't change when you change observers. Thee coordinates DO change, but the underlying entities don't. The manner in which the coordinates change in relativity is given by the Lorentz transform, which relates the coordinates for one oberver to the coordinates for the second.

14. Jul 10, 2014

### Maxila

Personally I'd modify that a bit in saying coordinates are labels that only have direct physical significance to the frame they are assigned.

The thought I brought up came from working with and trying to see if Lorentz transformations and Lorentz covariance (ds^2 = c^2dt^2 - dx^2 -dy^2 -dz^2) relate with the empirical existence and evidence of time (it always describes a change in distance indistinguishable from speed, if one looks to where the direct reference to its change is assigned).

15. Jul 10, 2014

### parkner

I wish to remind only:
relative velocity is a coordinate independent thing.

This is just a relation of the two bodies - a distance derivative.
In SR this is also preserved: there is the same v in the both frames: in the stationary and in the moving.
In a one dimensional case: v > 0 means a distance grows, and v < 0 it decreases - exactly the same on both sides, ie. never v on one frame and -v on the other side!

16. Jul 10, 2014

### A.T.

relative = coordinate dependent

17. Jul 10, 2014

### Staff: Mentor

You are thinking of relative speed, not relative velocity.

18. Jul 11, 2014

### parkner

No, because a speed is just |v|, what can't be a negative number.

19. Jul 11, 2014

### Staff: Mentor

No, there isn't. If the velocity of frame B relative to frame A is $v$, then the velocity of frame A relative to frame B is $- v$. Look at the math for Lorentz transformations. The transformation that goes from frame B to frame A is the inverse of the transformation that goes from frame A to frame B, and the transformation with $- v$ is the inverse of the transformation with $v$. (We're talking about what you call the "one dimensional case" here--pure boosts along a single direction, with no spatial rotation.)

Not when you're doing Lorentz transformations. See above.

20. Jul 11, 2014

### Staff: Mentor

If you were not thinking of relative speed then your claims were wrong. That is not how relative velocity transforms.