Inaccuracy in the speed of light

[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.

Thanks in advance,
Ryan

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

 

[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.

 

[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.

 

[jtbell]

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.

 

[parkner]

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

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.

 

[pervect]

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.

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.

 

[phinds]

It adds normally: c - v.

As pervect pointed out, you misunderstand what this thread is about.

 

[sophiecentaur]

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.

You would need to specify who is measuring this time interval, in what frame.
Read this link about a dream Einstein claimed to have had.

 

[parkner]

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.

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.

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.[

]

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

 

[Dale]

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'.

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

 

[phinds]

I can't agree with this interpretation.

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

 

[Maxila]

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).

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).

 

[pervect]

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).

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.

 

[Maxila]

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

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

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 observer to the coordinates for the second.

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).

 

[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!

 

[A.T.]

relative velocity is a coordinate independent thing.

relative = coordinate dependent

 

[Dale]

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!

You are thinking of relative speed, not relative velocity.

 

[parkner]

You are thinking of relative speed, not relative velocity.

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

 

[PeterDonis]

In SR this is also preserved: there is the same v in the both frames: in the stationary and in the moving.

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.)

In a one dimensional case: v > 0 means a distance grows, and v < 0 it decreases

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

 

[Dale]

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

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

Please provide a reference.

 

[sophiecentaur]

It seems to me that some contributors to this thread may have forgotten what SR tells us. Michelson and Morley sorted this out years ago, didn't they?

 

[parkner]

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 $- v$.

Nope. The relative quantities in physics are always frame independent, because these are just the elementary relations between the bodies.

The reference system is irrelevant here, because the relation is just an operator: a op b, and it must be, and it's always a symmetric operator, due to the obvious fact: we always observe other bodies from our point of view!
[/QUOTE]

v-ab = v-ba, identically, as the distance is equal on both sides: a-b = b-a.

A position vector: r-b = A-B, this is for body B (he is always in 0, means in his own place - we can't go out of himself), but this is r-a = B-A for body A, this is always reversed, because A is in our own place: me <> he, others.

 

[PeterDonis]

Nope.

You need to review the basic math of SR. Whatever concepts you are talking about are not the correct concepts for the topic under discussion.

The reference system is irrelevant here

Not for the topic under discussion.

the relation is just an operator: a op b, and it must be, and it's always a symmetric operator, due to the obvious fact: we always observe other bodies from our point of view!

But we don't all observe each other in the same direction. If you are in the positive $x$ direction relative to me, then I am in the negative $x$ direction relative to you. Position is not just a magnitude; it's a magnitude and a direction. At least, that's the concept of "position" that's relevant for the topic under discussion.

Similarly, if you are moving in the positive $x$ direction relative to me, then I am moving in the negative $x$ direction relative to you. Velocity is not just a magnitude; it's a magnitude and a direction. At least, that's the concept of "velocity" that's relevant for the topic under discussion.

Perhaps you have other concepts in mind that, to you, go by the names "position" and "velocity". It certainly seems that way from the things you are saying. I can't comment on whether those concepts, in and of themselves, are consistent or not; but I can say that they are not the correct ones for the topic under discussion. The correct ones for the topic under discussion are the ones I described above. So continuing to talk about your concepts is pointless; for this discussion, you need to talk about the concepts that are relevant for this discussion.

 

[pervect]

You are thinking of relative speed, not relative velocity.

I agree, though I had to think about it for a bit, even though it's a simple issue.

Velocity has a magnitude and a direction. (As you say, speed doesn't).

Restricting ourselves now to 1 dimensional problems, there becomes only two possible directions. Representing the direction with a plus or minus sign is a common convention. The choice of which direction is positive and which is negative is arbitrary and human (basically a coordinate choice), but the existence of two possible directions is not.

When we consider the case of swapping two observers, we do indeed see that the direction changes - i.e. if from A's perspective B is approaching from the east from B's perspective A is approaching from the west. (To pick out east and west as our two possible 1d directions).

It might or might not be clearer to say that there's a change in direction rather than a change in sign - however the velocity addition formulas I quoted earlier DO require that one direction be represented by positive numbers and the other direction by negative numbers in order to be consistent, so it is probably a good idea to refer to the change in direction as a sign change.

You need these conventions even in non-relativity theory. If you add a velocity of 20 miles per hour in the east direction, and 10 miles in the west, you have to subtract (20-10) to get 10 miles per hour net velocity in the east direction. This is equivalent to saying that positive velocities are in the east direction, negative velocities in the west direction, and that the total velocity is +20 -10 = +10

[add]
Ryan already seems familiar with the notion of signed numbers representing velocities that I described above, as evidenced by his comfortableness in adding velocities, so I think this has become a bit of a digression. I.e parkner's issues have to some extent derailed the thread started by Ryan who had different issues.

 

[Dale]

v-ab = v-ba, identically, as the distance is equal on both sides: a-b = b-a.

This is not difficult to calculate. Using notation $r_{ij}$ to denote the position of object i in the reference frame of object j and similarly using $v_{ij}=dr_{ij}/dt_j$ to denote the velocity of object i in the reference frame of object j. Then, consider objects A and B with their corresponding reference frames. For frame A we can write:
$r_{AA}=(0,0,0)$
$v_{AA}=dr_{AA}/dt_A=(0,0,0)$
$r_{BA}=(vt_A,0,0)$
$v_{BA}=dr_{BA}/dt_A=(v,0,0)$

Then by the Lorentz transform:
$r_{AB}=(-\gamma v t_A,0,0)=(-v t_B,0,0)$
$v_{AB}=dr_{AB}/t_B=(-v,0,0)=-v_{BA}$
$r_{BB}=(0,0,0)$
$v_{BB}=(0,0,0)$

 

[A.T.]

The relative quantities in physics are always frame independent,

The other way around.

 

[A.T.]

If you see a body then you directly know whether it moves away or closer to you.

That's just the sign of the distance change.

This is independent of any reference system,

But the rate at which the distance change is frame dependent.

 

[pervect]

That's just the sign of the distance change.


But the rate at which the distance change is frame dependent.

Yes, this is one of the areas where relativistic physics / kinematics differs from pre-relativistic physics. It may be worth a more detailed example, and I'll try to use terminology that I think everyone in the thread understands and agrees with for the sake of clarity. (Not that I think there was anything wrong about what I said before, but I'd rather sidestep the semantic issues.)

If you have a frame of reference in which one object is moving to the right at .4c, and another object is moving to the left at .4c, their rate of closure / the rate at which the distance changes between the two objects is .8c in that frame of reference.

However, if you switch to a frame of reference co-moving with either object, the relativisitic velocity additon formula gives the rate of closure as (.4 + .4)/(1+.4 * .4) $\approx$ .68965 c.

Thus we can see that when we include the effects of relativity, the rate of closure depends on the frame of reference, it is dependent on the observer. This didn't used to be true, before relativity the rate of closure was independent of the frame of reference.

The rest of the discussion is, I think, a consequence of this basic fact.

 

[universal_101]

I think the point parkner is making, is that relative properties between two inertial frames is frame independent, in a sense that differential ageing is frame invariant, similarly relative velocity of one frame w.r.t a particular frame is also frame invariant.

Basically, 'motion' is relative, but 'relative motion' is undoubtedly absolute.

 

[ghwellsjr]

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.

You are proposing that this scheme to measure the speed of the propagation of light in one direction will detect a difference under some inertial circumstances even when the measurement of the round-trip speed of light is the same in those different inertial circumstances, correct? I hope that is what you are suggesting because a round-trip measurement of the speed of light is known to always yield the same value under inertial circumstances.

I'm going to change your scenario in a way that remains faithful to your original scenario but will reveal a flaw in your conclusion that the results can be skewed when everything is not stationary.

Instead of a console at the center point between the laser and the receiver and detectors at both the laser and the receiver to start the laser or start and stop the timer at the receiver, I'm going to do it all with a couple mirrors and with the the laser at the receiver/timer.

Consider this:

We have a laser pointing at a half-silvered mirror two feet away and a regular mirror four feet away. When the laser emits a pulse of light, it travels to the half-silvered mirror which reflects half of the pulse of light back to the laser where there is a receiver that starts a timer. The other half of the pulse of light travels through the half-silvered mirror to the regular mirror and back to the receiver to stop the timer. This is fundamentally the same measurement that you proposed that would allegedly measure the speed of light going from the regular mirror (or laser, in you scenario) to the receiver, correct?

Here is a spacetime diagram depicting this scenario:

The laser/receiver are located at the Coordinate Distance of 4 feet and are depicted in red. The laser emits the thin red pulse of light toward the mirrors. The half-silvered mirror is depicted in black at the Coordinate Distance of 2 feet and reflects a thin black pulse of light back to the red laser/receiver to start the timer at the Coordinate Time of 4 nanoseconds. Meanwhile, the thin red laser pulse continues on through the black half-silvered mirror to the regular mirror depicted in blue at the Coordinate Distance of 0 which reflects the thin blue pulse of light back to the red laser/receiver to stop the timer at the Coordinate Time of 8 nanoseconds.

The timer has measured a difference in the Coordinate Times of 4 nanoseconds and since the distance from the blue regular mirror to the red receiver is 4 feet, the measurement of the speed of light is 4 feet in 4 nanoseconds or 1 foot per nanosecond, agreed?

Can you see how this is identical to your scenario if we treat the two light pulses coming out of the black half-silvered mirror as functionally equivalent to your console?

However, now imagine a situation where the whole system was moving in one direction at a speed. This would skew the results.

What you are suggesting is that the speed of light going in one direction would be different than the speed of light going in the other direction and so the timer would measure a different value than for the stationary case, correct?

Well, let's see if that is in fact true. Let's make the speed of light faster going to the left than going to the right such that the round trip time is the same, which it is known to be. Here is another spacetime diagram depicting what happens in that case:

As you can see, the timer starts at the Coordinate Time of 4 nanoseconds and stops at the Coordinate Time of 8 nanoseconds for a difference of 4 nanosecond, just as in the stationary case.

Let's try the opposite case where the propagation speeds in the two directions are interchanged:

Again, we get the exact same result.

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.

Thanks in advance,
Ryan

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

As I have shown, since the round-trip measurement of the speed of light comes out the same value under all inertial circumstances, we cannot jump to the conclusion that your proposed measurement scheme is capable of measuring the speed of the propagation of light in one direction because it always yields the same answer, even in circumstances that you say should reveal a skew.

 

[bloby]

My two cents for the OP:

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

Yes, yes, yes.

The true speed of light would be the calculated speed plus the speed of the system.

No.
To avoid belives, you would have to set an experiment that measures the speed of light of the moving system, and you would find the astonishing fact that the result you obtain and the one obtained by your friend moving with the system agree.

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.

Yes.

If things behaved like classical mechanics predict it to, we would not speak of the speed of light.(evenless the true speed)