A thought experiment of the relativity of light

[CClyde]

This makes no sense. Coordinates don't "move".

All coordinates move relative to an appropriately chosen frame of reference.

My point is none of the coordinates of light emissions are appropriate.

 

[PeterDonis]

All coordinates move relative to an appropriately chosen frame of reference.

No, they don't. This is nonsense. Objects move.

My point is none of the coordinates of light emissions are appropriate.

What coordinates? Your own? What are you talking about?

 

[Ibix]

All coordinates move relative to an appropriately chosen frame of reference.

I think you are trying to say that a thing having "constant spatial coordinates" in one frame is moving in another.

My point is none of the coordinates of light emissions are appropriate.

Appropriate for what?

 

[Dale]

I cannot think of a more clear way to state this.

I can: With math!

1. The spatial coordinates of a light emission (t0, x0, y0, z0) remain at the center of the sphere that is the propagating wavefront.

1. In an inertial frame and using units where $c=1$, the equation for a light wave emitted at $\mathrm{R}=(t_0,x_0,y_0,z_0)$ is given by $(t-t_0)^2=(x-x_0)^2+(y-y_0)^2+(z-z_0)^2$ for $t\ge t_0$. For any given $t$ this is the equation of a sphere centered at $(x_0,y_0,z_0)$ with radius $t-t_0$.

2. The constancy of c holds the wavefront symmetrical for all observers.

2. The invariance of $c$ means that the equation is also $(t’-t’_0)^2=(x’-x’_0)^2+(y’-y’_0)^2+(z’-z’_0)^2$ in any other (primed) inertial frame.

3. All such light emission coordinates are, by the definition of lights independence of the motion of the source incapable of motion relative to each other.

This one I don’t know what you are trying to say. Please try to express it clearly using math as demonstrated for 1. and 2.

 

[CClyde]

No, they don't. This is nonsense. Objects move.

If you are moving toward an object and the object disappears, are you still moving, or did the coordinates of its position only exist because the object existed?

How do you define yourself in motion relative to the object, but not in motion relative to the coordinates of the object?
The principle of relativity says such motion is not a property of you, or the object but a measure by the observer.

If there were some reference, some coordinate you could actually prove cannot move, not just state it as you are, but prove the existence of, and by law that such coordinates cannot move, well, then you could measure your position over time relative to such coordinates and know if it is you or the object that is moving.
I think Dale has just offered a mathematic definition.

 

[Dale]

@CClyde you really need to start expressing things mathematically as much as possible. Your communication in English is hopelessly confusing.

The problem here is that you are using the word “coordinates”, which means things like $(t_0,x_0,y_0,z_0)$. @PeterDonis rightly objects to describing those as “moving”.

Whereas what you mean by that word are things like $(t,x_0,y_0,z_0)$. These are not coordinates, but the integral curves of the timelike tetrad field for a given reference frame. They are worldlines so they can be said to move in the sense that they have a timelike tangent.

That said, I still cannot parse your point 3. Please try to express it mathematically as much as possible.

 

[hutchphd]

If you are moving toward an object and the object disappears, are you still moving, or did the coordinates of its position only exist because the object existed?

The continuing spatial co-incidence of the object (flashbulb) and the center of the resultant spherical (to all observers) light pulse is true only in the rest frame of the flashbulb. These positions coincide for all observers only at t=0 when the flashbulb fires.

For all other times everthing depends upon reference frame motion (as described by Lorentz transformations)

 

[CClyde]

I think you are trying to say that a thing having "constant spatial coordinates" in one frame is moving in another.

The principle of relativity says motion is not a property of some “thing”, it is a measure made by an observer. A thing you measure at rest is in motion relative to a frame not at rest with you. This reciprocal symmetry of kinematics allows any inertial frame to be at rest, or in uniform motion by simply defining it relative to the appropriate frame.

The frame of a light emission can be at rest or in motion just as mentioned above, when such a measure is made relative to an appropriate “inertial” frame of reference, either at rest or in motion relative to the light emission frame.

But the frame of a light emission cannot be set in motion by choosing any other frame of light emission.

So I said, none of the coordinates of light emissions are appropriate.

 

[Dale]

So I said, none of the coordinates of light emissions are appropriate.

You should stop saying this because it is wrong. We are here to try to understand what you are saying (which currently is unclear) so that we can help you see where your reasoning fails. But to be clear your reasoning must fail somewhere because your conclusion here is false. Do not repeat this conclusion

 

[Dale]

The frame of a light emission

What is the frame of a light emission?

A light emission is a single event $(t_0,x_0,y_0,z_0)$. It is in all frames at some coordinates obtained by the Lorentz transform, but it does not define any frame.

I suspect this may be your key misconception

 

[CClyde]

The continuing spatial co-incidence of the object (flashbulb) and the center of the resultant spherical (to all observers) light pulse is true only in the rest frame of the flashbulb. These positions coincide for all observers only at t=0 when the flashbulb fires.

For all other times everthing depends upon reference frame motion (as described by Lorentz transformations)

Yes, I agree but I am not concerned with the whereabouts of the flashbulb after emission. I am concerned with the center of the spherical light pulse which remains at the center of the sphere it defines relative to the center of all other such centers, none which can move relative to each other without violating the constancy of the speed of light

 

[hutchphd]

So I said, none of the coordinates of light emissions are appropriate.

Nobody understands what this means. What is "a coordinate of light emission" The center of the light sphere is not a good marker.
My brother is a land surveyer in Maine. One of his least favorite historical deeds located his clients property to "the hole in the ice on the pond" This seems a similar exercise.

For each observers in uniform relative motion we can choose coordinate systems such that the emission event occurs at (t,x,y,z)=(0,0,0,0). The rest is as given by Lorentz transformation and does not comport with the picture in the head of the OP. Physics is hard.

 

[PeroK]

Yes, I agree but I am not concerned with the whereabouts of the flashbulb after emission. I am concerned with the center of the spherical light pulse which remains at the center of the sphere it defines relative to the center of all other such centers, none which can move relative to each other without violating the constancy of the speed of light

So, where do you go from here? I would claim you are new to the concept of having the same events described in two different reference frames and, in particular, cannot reconcile the invariance of $c$ with your mental picture of the motion of light. What do you do next? Accept you are wrong? Persist in your error? The choice is yours. No one can force you to understand SR.

 

[CClyde]

For each observers in uniform relative motion we can choose coordinate systems such that the emission event occurs at (t,x,y,z)=(0,0,0,0). The rest is as given by Lorentz transformation and does not comport with the picture in the head of the OP. Physics is hard.

Please explain how two light spheres from two emission events (0,0,0,0) and (0,1, 1, 1) move away from each other such that at a time > 0 the centers of these spheres, are no longer at (0,0,0,0) and (0,1, 1, 1).

 

[Vanadium 50]

@CClyde I reread this thread and confess I am confused as to your goial.

Is is:
1. To clear up your misconceptions?
2. To convince us that relativity is internally inconsistent and you are the first person in more tnan a century who noticed?

 

[CClyde]

@CClyde I reread this thread and confess I am confused as to your goial.

Is is:
1. To clear up your misconceptions?
2. To convince us that re;ativity is internally inconsistent and you are the first person in more tnan a century who notic4d?

My goal is to find out if I have misconceptions, or if I have discovered an internal inconsistency in the principle of relativity.

 

[Dale]

My goal is to find out if I have misconceptions, or if I have discovered an internal inconsistency in the principle of relativity.

Relativity is self-consistent. If any chain of reasoning leads you to believe otherwise then that chain of reasoning is incorrect.

Several problems with your reasoning have been pointed out as have several points where your communication is unclear. So the goal has been resolved as the first one:

you have misconceptions (some already identified and some remaining to be clarified).

 

[Dale]

Please explain how two light spheres from two emission events (0,0,0,0) and (0,1, 1, 1) move away from each other such that at a time > 0 the centers of these spheres, are no longer at (0,0,0,0) and (0,1, 1, 1).

What do you mean by this?

 

[Nugatory]

If there were some reference, some coordinate you could actually prove cannot move, not just state it as you are, but prove the existence of, and by law that such coordinates cannot move, well, then you could measure your position over time relative to such coordinates

There is such a reference. If you are moving inertially (you determine this by the absence of proper acceleration - an accelerometer you are holding reads zero) you can choose your position in three-dimensional space to be at rest with coordinates (0,0,0) and your position in four-dimensional spacetime to have coordinates (t,0,0,0) where t is your wristwatch time. Once you’ve done that, you can determine the coordinates of any other point in space, and it is a simple calculation to convert these values to other coordinate systems.

However, you may be hung up on an earlier misunderstanding:

and know if it is you or the object that is moving.

There is no such distinction. When we say “THIS is moving and THAT is not” we are actually saying “We have chosen to use a coordinate system in which THIS is moving and THAT is not”. Thus we are free to consider either to be at rest, just by choosing our coordinates.

In fact, we choose coordinates for our convenience and change them so routinely that we often aren’t aware of it. When driving a car I nearly always consider the surface of the earth to be at rest - but I instantly and effortlessly switch to coordinates in which I and the car are at rest when I reach for the cup of coffee in the armrest cup holder.

 

[Frabjous]

My goal is to find out if I have misconceptions, or if I have discovered an internal inconsistency in the principle of relativity.

Here’s a rule of thumb: If you do not do the math, you have not found an inconsistency. Words are not precise enough to perform physics with.

 

[Nugatory]

Please explain how two light spheres from two emission events (0,0,0,0) and (0,1, 1, 1) move away from each other such that at a time > 0 the centers of these spheres, are no longer at (0,0,0,0) and (0,1, 1, 1).

Ahhhh…. That would be #2 of this thread.
But it would be a good exercise to actually calculate the position of the spreading flash of light twice, using the two coordinates systems in which each of two observers moving relative to one another are at rest.

 

[Vanadium 50]

Great, we can work with that.

It's probably worth pointing out that if you options are "I misunderstand something" and "I discovered something major that Einstein and everybody who followed him missed" it'd a good bet to go with the former.

As @Dale points out, relativity is known to be mathematically consistent. So that's a non-starter.

If you wanted to go and prove something mathematically inconsistent, the way you are going about it is not the way to do tt. You would provide two calculations of the same thing and show they get different results: i.e. you would use 10x as many numbers as you are using and one-tenth as many words.

You want "if I calculate x thsi way [numbers numbers numbers] I get 7 and when i calculate x thsi oter way [numbers numbers numbers] I get 11," You don't want increasingly complex scenarios with lots of words,

But as mentioned, this will be futile. as it is known that SR is internally consistent.

 

[vanhees71]

So, where do you go from here? I would claim you are new to the concept of having the same events described in two different reference frames and, in particular, cannot reconcile the invariance of $c$ with your mental picture of the motion of light. What do you do next? Accept you are wrong? Persist in your error? The choice is yours. No one can force you to understand SR.

Let's see, whether we can make the entire trouble concrete. Since it's only about kinematics, let's just consider a massless scalar field. Wo so we can simply look at the spherically symmetric solution of the wave equation,
$$\Box \Phi(x)=0, \quad \Box=1/c^2 \partial_t^2-\vec{\nabla}^2.$$
Then to have a nice model for a "light source", just consider a spherical harmonic wave, i.e.,
$$\Phi(x)=\phi(r) \exp(-\mathrm{i} \omega t), \quad r=\vec{x}.$$
The solution "outgoing" solution reads
$$\phi(r)=\frac{1}{r} \exp(\mathrm{i} k r), \quad k=\omega/c.$$
This is the solution for a light source sitting at rest at $r=0$ (the singularity of the wave).

Now it's convenient to write this in a manifestly covariant way, using the four-velocity of the light source $(u^{\mu})=(1,0,0,0)$. Then you get
$$\Phi(x)=\frac{1}{\sqrt{(u \cdot x)^2-x \cdot x}} \exp[-\mathrm{i} k (u \cdot x-\sqrt{(u \cdot x)-x \cdot x}].$$
In a frame of reference $\Sigma'$, where the light-source moves with velocity $v= c \beta$ in the positive 1-direction you have
$$u'=\gamma (1,\beta,0,0).$$
Then in this frame the wave is represented by
$$\Phi'(x')=\frac{1}{\sqrt{\gamma^2(x'-v t')^2+y^{\prime 2}+ z^{\prime 2}}} \exp[\mathrm{i} k [\gamma (ct'-\beta x')-\sqrt{\gamma^2(x'-v t')^2+y^{\prime 2}+ z^{\prime 2}}],$$
where I made use of $\gamma=1/\sqrt{1-\beta^2}$ to get
$$r=\sqrt{(u' \cdot x')-x' \cdot x'}=\sqrt{\gamma^2(x'-v t')^2+y^{\prime 2}+ z^{\prime 2}}.$$
As you see for an observer in this frame the singularity is at
$$\gamma^2 (x'-v t')^2+y^{\prime 2}+z^{\prime 2}=0 \; \Rightarrow \; x'=y'=0, \quad x'=v t'$$
i.e., the singularity moves with velocity $v$ in the positive $x'$-direction along the $x'$ axis, as constructed.

At the same time the motion of the surfaces of constant phase $0$ is given by
$$\gamma (ct'-\beta x')-\sqrt{\gamma^2(x'-v t')^2+y^{\prime 2}+ z^{\prime 2}}=0,$$
and indeed after some algebra you find that it's given by
$$c^2 t^{\prime 2}=\vec{x}^{\prime 2},$$
i.e., the surface of constant phase is moving with the speed of light out from the origin.

That must be so because of the properties of the Lorentz transformation, i.e., because $x \cdot x=x' \cdot x'$.

 

[Nugatory]

Let's see, whether we can make the entire trouble concrete….

Although there is nothing wrong here, there are easier ways of approaching the math and OP may want to start with those.

 

[hutchphd]

Please explain how two light spheres from two emission events (0,0,0,0) and (0,1, 1, 1) move away from each other such that at a time > 0 the centers of these spheres, are no longer at (0,0,0,0) and (0,1, 1, 1).

If there are two emission events, different observers will not agree as to either their distance apart or their simultaneity. You cannot intuit the details (nor can I)

 

[PeroK]

Ahhhh…. That would be #2 of this thread.
But it would be a good exercise to actually calculate the position of the spreading flash of light twice, using the two coordinates systems in which each of two observers moving relative to one another are at rest.

I suspect calculations are anathema to the OP.

 

[vanhees71]

Although there is nothing wrong here, there are easier ways of approaching the math and OP may want to start with those.

If you have a more simple model for a "light source", let's see it. Perhaps it's my limited imagination that I couldn't find a simpler one ;-)).

 

[vanhees71]

I suspect calculations are anathema to the OP.

Then, s/he has no chance to ever talk about physics in such a way that s/he can understand it.

 

[Vanadium 50]

Well, B-level probably does preclude a D'Alembartian (and there should be a squared there! Just like the Oxford Comma!")

But the general point is valid - one cannot prove a mathematical inconsistency withoutm you know, mathematics,

 

[vanhees71]

Where should be a "squared"?

If you are not allowed to use the wave equation to discuss waves, then the question cannot be answered adequately at any level.

 

[Vanadium 50]

$\Box^2$ over $\Box$

 

[vanhees71]

That's a very strange notation. Does anybody use this?

 

[Vanadium 50]

Everybody.

It emphasizes that it;s a second derivative.,

 

[Dale]

Please explain how two light spheres from two emission events (0,0,0,0) and (0,1, 1, 1) move away from each other such that at a time > 0 the centers of these spheres, are no longer at (0,0,0,0) and (0,1, 1, 1).

So, instead of talking about light spheres, for a moment let's talk about light cones. In this case in some given (unprimed) inertial frame with coordinates $(t,x,y,z)$ we have two light cones: $$\mathrm{A}=(t,x,y,z) : \ t^2=x^2+y^2+z^2$$$$\mathrm{B}=(t,x,y,z) : \ t^2= (x-1)^2+(y-1)^2+(z-1)^2$$ So here we have two light cones, each of which is a full 4D object. These 4D objects represent right circular cones with the axis along the $t$ direction and with the apex at $(0,0,0,0)$ and $(0,1,1,1)$ respectively.

Now for any $t=t_1>0$ we can take the full 4D objects $\mathrm{A}$ and $\mathrm{B}$ we can obtain the 3D objects that are 3D "slices" of the light cones $\mathrm{A}$ and $\mathrm{B}$ at the specific time $t_1$. These 3D objects are $$\left. \mathrm{A} \right|_{t=t_1} = \mathrm{A}(t_1) = (x,y,z) : \ t_1^2=x^2+y^2+z^2$$$$\left. \mathrm{B} \right|_{t=t_1} = \mathrm{B}(t_1) = (x,y,z) : \ t_1^2=(x-1)^2+(y-1)^2+(z-1)^2$$ These are 3D objects at the time $t=t_1$, and each represents a sphere of radius $t_1$ with the center at $(0,0,0)$ and $(1,1,1)$ respectively.

Now, with that clarification and notation. Please write what you mean about light spheres moving away from each other.

 

[PeroK]

My guess is that the confusion stems from implicity using the Galilean transformation with universal simultaneity and intuitively noting that this is incompatible with an invariant speed. And, since the Lorentz transformation is "mathematics", it's not an intuitive alternative.