Does the single speed of light mean an absolute frame of reference?

[xyzt]

Are the photons emitted simultaneously in all frames?

You know the answer to this question: no. Yet, as I explained, RoS is irrelevant in this case. The relative speed of the photons is zero in all frames.

[Delta1212]

You know the answer to this question: no. Yet, as I explained, RoS is irrelevant in this case. The relative speed of the photons is zero in all frames.

I don't think md is saying that the relative speed will be different, but the relative position will be if they are not emitted simultaneously. That relative position won't change within any single frame, but it won't necessarily be the same in every frame, and so you could conceivably construct a frame where the relative distance between the photons is greater than the distance between the photons and their destination, leading to one photon arriving before the other is emitted in that frame. The relative speed will, obviously, be the same but the order of events would not be.

 

Of course, this only make sense as long as they have both different start and end points.

[xyzt]

I don't think md is saying that the relative speed will be different, but the relative position will be if they are not emitted simultaneously. That relative position won't change within any single frame, but it won't necessarily be the same in every frame, and so you could conceivably construct a frame where the relative distance between the photons is greater than the distance between the photons and their destination, leading to one photon arriving before the other is emitted in that frame. The relative speed will, obviously, be the same but the order of events would not be.

 

Of course, this only make sense as long as they have both different start and end points.

Any two particles that have a comoving frame, are comoving in all frames. This is the point that both of you are missing.

The fact that the distance between the particles is different (due to length contraction) is utterly irrelevant.

Edited July 4, 2014 by xyzt

[Delta1212]

Any two particles that have a comoving frame, are comoving in all frames. This is the point that both of you are missing.

The fact that the distance between the particles is different (due to length contraction) is utterly irrelevant.

No one said they wouldn't be co-moving in every frame. But although co-moving objects will be co-moving in every frame, they may not simultaneously exist in every frame.

 

Take the pole moving at relativistic speed through a barn. There are four events: The front of the pole enters the barn, the front of the pole exists the back of the barn, the back of the pole enters the barn, and the back of the pole exists the back of the barn.

 

Now, in the frame of the barn with the length contracted pole, the order is "front of the pole enters, back of the pole enters, front of the pole exists, back of the pole exists."

 

In the frame of the pole, with the length contracted barn, the order is "front of the pole enters, front of the pole exits, back of the pole enters, back of the pole exists."

 

In the first instance, both the front of the pole and back of the pole exist within barn at the same time. In the latter, they do not. In all frames, the front and back of the pole are co-moving.

 

Now, you could set up a situation where you have the events "first photon is emitted, second photon is emitted, first photon is absorbed, second photon is absorbed" where the photons travel parallel to each other and so are co-moving.

 

If the events are all separated in space, you could conceivably find a frame where the order is "first photon is emitted, first photon is absorbed, second photon is emitted, second photon is absorbed." The photons will be co-moving in the sense that their relative velocities will be zero, but they will not exist simultaneously in that frame.

[xyzt]

No one said they wouldn't be co-moving in every frame. But although co-moving objects will be co-moving in every frame, they may not simultaneously exist in every frame.

 

Huh? There is no relative motion between the two particles . In any frame.

There are no "events", just two particles with no relative motion.

Since the language of physics is math, let me explain this to you mathematically:

 

In a frame S, there are two photons being emitted simultaneously, [math]x_0[/math] apart. The equations of motion (not "events") are:

 

[math]x_1=ct[/math]

[math]x_2=x_0+ct[/math]

 

Their separation is [math]x_2-x_1=x_0[/math] and their relative speed is [math]\frac{dx_2}{dt}-\frac{dx_1}{dt}=0[/math].

 

In ANY other frame S' moving with speed V wrt S:

 

 

 

[math]x'_1=\gamma(V)(x_1-Vt)[/math]

[math]x'_2=\gamma(V)(x_2-Vt)[/math]

 

The photon separation in frame S' is [math]x'_2-x'_1=\gamma(V)x_0[/math] and their relative speed is [math]\frac{dx'_2}{dt}-\frac{dx'_1}{dt}=0[/math].

Edited July 4, 2014 by xyzt

[md65536]

Any two particles that have a comoving frame, are comoving in all frames. This is the point that both of you are missing.

The fact that the distance between the particles is different (due to length contraction) is utterly irrelevant.

Do the photons have a comoving frame?

 

Delta1212 understands what I was getting at. I was speaking only of OP's premise that the photons are "always" side-by-side. I agree it's not an essential part of the answer. If OP's thought experiment requires the photons to be separated and side-by-side in all frames, then it's a problem, but the experiment can be restated without that requirement.

[xyzt]

Do the photons have a comoving frame?

 

Delta1212 understands what I was getting at. I was speaking only of OP's premise that the photons are "always" side-by-side. I agree it's not an essential part of the answer. If OP's thought experiment requires the photons to be separated and side-by-side in all frames, then it's a problem, but the experiment can be restated without that requirement.

The problem with you two is that you write prose instead of math. The language of physics is math. Why don't you try to put the prose above into math?

Edited July 4, 2014 by xyzt

[Dekan]

The problem with you two is that you write prose instead of math. The language of physics is math.

 

Could "Math" be the last refuge of Physics. When an explanation of Physics, written in plain prose, exposes blatant absurdity, "Math" comes to the rescue, and explains it all away.

Edited July 4, 2014 by Dekan

[md65536]

The photon separation in frame S' is [math]x'_2-x'_1=\gamma(V)x_0[/math] and their relative speed is [math]\frac{dx'_2}{dt}-\frac{dx'_1}{dt}=0[/math].

In post #1 the photons are described as moving parallel to each other. The only separation is along the y-axis, not in the direction of travel.

 

Good for you for using math, I admit once again that I'm deficient. However the math is useless if it describes something entirely different from the prose. If you wish to continue discussing your variation instead of OP's, why not start a new thread?

[xyzt]

In post #1 the photons are described as moving parallel to each other. The only separation is along the y-axis, not in the direction of travel.

 

Good for you for using math, I admit once again that I'm deficient. However the math is useless if it describes something entirely different from the prose. If you wish to continue discussing your variation instead of OP's, why not start a new thread?

Then you can make [math]x_0=0[/math] in my post. If you know how to do basic algebra you will find that the "photons are side by side" in ALL frames. I gave out the GENERAL solution, you can derive all the particular solutions from it. At one point, "Delta1212" was asking about photons that were separated in the x direction, so I posted the general solution. .

Huh? There is no relative motion between the two particles . In any frame.

There are no "events", just two particles with no relative motion.

Since the language of physics is math, let me explain this to you mathematically:

 

In a frame S, there are two photons being emitted simultaneously, [math]x_0[/math] apart. The equations of motion (not "events") are:

 

[math]x_1=ct[/math]

[math]x_2=x_0+ct[/math]

 

Their separation is [math]x_2-x_1=x_0[/math] and their relative speed is [math]\frac{dx_2}{dt}-\frac{dx_1}{dt}=0[/math].

 

In ANY other frame S' moving with speed V wrt S:

 

 

 

[math]x'_1=\gamma(V)(x_1-Vt)[/math]

[math]x'_2=\gamma(V)(x_2-Vt)[/math]

 

The photon separation in frame S' is [math]x'_2-x'_1=\gamma(V)x_0[/math] and their relative speed is [math]\frac{dx'_2}{dt}-\frac{dx'_1}{dt}=0[/math].

What if the photons are not emitted simultaneously in frame S? Then , one needs to change the equations as follows:

 

In a frame S, there are two photons being emitted simultaneously, [math]x_0[/math] apart in space and [math]t_0[/math] apart in timr. The equations of motion (not "events") are:

 

[math]x_1=ct[/math]

[math]x_2=x_0+c(t+t_0)[/math]

 

Their separation is [math]x_2-x_1=x_0+ct_0[/math] and their relative speed is [math]\frac{dx_2}{dt}-\frac{dx_1}{dt}=0[/math].

 

In ANY other frame S' moving with speed V wrt S:

 

 

 

[math]x'_1=\gamma(V)(x_1-Vt)[/math]

[math]x'_2=\gamma(V)(x_2-Vt)[/math]

 

The photon separation in frame S' is [math]x'_2-x'_1=\gamma(V)(x_0+ct_0)[/math] and their relative speed is [math]\frac{dx'_2}{dt}-\frac{dx'_1}{dt}=0[/math].

Edited July 4, 2014 by xyzt

[Strange]

 

When an explanation of Physics, written in plain prose, exposes blatant absurdity, "Math" comes to the rescue, and explains it all away.

 

When natural language is vague and ambiguous, math comes to the rescue and fills in the missing details.

[Sensei]

Math in science is to predict future, and calculate past state.

 

f.e. you detect apple to be at time t0 at position x0,

later at t1, at position x1,

later at t2, at position x2,

you can calculate velocity v=(x1-x0)/(t1-t0)

then you can calculate acceleration

a=(v1-v0)/(t1-t0)

and knowing equation d=1/2*a*t^2,

you can calculate when apple will hit ground,

at which branch of tree apple grew,

whether it will smash, or land safely etc. etc.

[xyzt]

 

When natural language is vague and ambiguous, math comes to the rescue and fills in the missing details.

Right. And when you don't have the math tools, you can't claim to be doing physics.

[Sensei]

Right. And when you don't have the math tools, you can't claim to be doing physics.

 

Not quite. You can make experiment without knowing math describing it. Equations will appear later after gathering enough data and deep analyze. First must be experiment or observation.

Edited July 5, 2014 by Sensei

[xyzt]

 

Not quite. You can make experiment without knowing math describing it. Equations will appear later after gathering enough data and deep analyze. First must be experiment or observation.

Not necessarily, I write the theory of my experiments before I perform the experiments. It can go either way.

Bottom line, in both cases you need to be able to write down the math.

[DimaMazin]

 

You can make experiment without knowing math describing it.

The experiment can be dangerous for you and exactly will be useless for science.

[md65536]

In a frame S, there are two photons being emitted simultaneously, [math]x_0[/math] apart in space and [math]t_0[/math] apart in timr. The equations of motion (not "events") are:

[...]

The photon separation in frame S' is [math]x'_2-x'_1=\gamma(V)(x_0+ct_0)[/math] and their relative speed is [math]\frac{dx'_2}{dt}-\frac{dx'_1}{dt}=0[/math].

But what if the separation in space is not in the direction of V, such as in OP's setup? Edited July 6, 2014 by md65536

[xyzt]

But what if the separation in space is not in the direction of V, such as in OP's setup?

I already explained, make [math]x_0=0[/math] in my formulas.Do you think that you can do that or I need to do it for you?

Edited July 6, 2014 by xyzt

[Phi for All]

!

Moderator Note

Let's save the side-discussion about math vs prose for another thread, folks.

[md65536]

I already explained, make [math]x_0=0[/math] in my formulas.Do you think that you can do that or I need to do it for you?

The latter, because I'm having trouble following your math. I suspect that you're answering a different question but I might be wrong. If there is a y-axis separation, and x_0 = 0, and letting t_0 = 0 (the photons are emitted simultaneously in S frame), then the x separation is always 0 in your formula. Is that true for all observers, even ones that aren't limited to motion in the x direction?

 

I think this is relevant, because it is fairly common for people to claim that there is an absolute frame of reference by ignoring any frames where their premises don't hold. I believe that the premise implied by OP that the x-separation is 0 in all frames is wrong, but your math seems to support it.

[xyzt]

The latter, because I'm having trouble following your math. I suspect that you're answering a different question but I might be wrong. If there is a y-axis separation, and x_0 = 0, and letting t_0 = 0 (the photons are emitted simultaneously in S frame), then the x separation is always 0 in your formula.

 

 

Yes, you finally got it.

 

 

 

Is that true for all observers, even ones that aren't limited to motion in the x direction?

 

Your question doesn't make sense, try re-posting it.

 

 

 

I believe that the premise implied by OP that the x-separation is 0 in all frames is wrong, but your math seems to support it.

 

What gives you this idea? If [math]x_0=0[/math] then basic algebra teaches you that [math]\gamma(V)x_0=0[/math].

Edited July 6, 2014 by xyzt

[md65536]

Your question doesn't make sense, try re-posting it.

Is the x-axis separation of the photons equal to 0 for all observers, including those whose velocity relative to the S frame is not in the same direction as V?

 

What gives you this idea? If [math]x_0=0[/math] then basic algebra teaches you that [math]\gamma(V)x_0=0[/math].

Yes, if your equations are complete then I'm wrong.

But don't your equations also hold that [math]\gamma(V)t_0=0[/math] in the case that x_0 = 0 and t_0 = 0? Which implies that if the photon emissions are simultaneous in frame S they're simultaneous in all frames?

Edited July 6, 2014 by md65536

[xyzt]

Is the x-axis separation of the photons equal to 0 for all observers, including those whose velocity relative to the S frame is not in the same direction as V?

 

Yes, if your equations are complete then I'm wrong.

 

Yes, you are wrong.

 

 

 

But don't your equations also hold that [math]\gamma(V)t_0=0[/math] in the case that x_0 = 0 and t_0 = 0? Which implies that if the photon emissions are simultaneous in frame S they're simultaneous in all frames?

 

Two events that are simultaneous and co-located in one frame are simultaneous and co-located in ALL frames. This is why [math](x_0,t_0)=(0,0)[/math] makes the vents co-located in all frames. Proof:

 

[math]t'=\gamma(V)(t-vx/c^2)[/math]

 

Therefore the temporal separation between events, [math]\Delta t'[/math] satisfies:

 

[math]\Delta t'=\gamma(V)(\Delta t-v \Delta x/c^2)[/math]

 

If [math](\Delta t, \Delta x)=(0,0)[/math] then [math]\Delta t'=0[/math] in ALL frames.

Edited July 6, 2014 by xyzt

[md65536]

Two events that are simultaneous and co-located in one frame are simultaneous and co-located in ALL frames. This is why [math](x_0,t_0)=(0,0)[/math] makes the vents co-located in all frames. Proof:

 

[math]t'=\gamma(V)(t-vx/c^2)[/math]

 

Therefore the temporal separation between events, [math]\Delta t'[/math] satisfies:

 

[math]\Delta t'=\gamma(V)(\Delta t-v \Delta x/c^2)[/math]

 

If [math](\Delta t, \Delta x)=(0,0)[/math] then [math]\Delta t'=0[/math] in ALL frames.

I think I'm starting to get it. And you say this works if the separation is in any direction, it doesn't matter if it's in the direction of V? Say, as per OP, a separation only in the y-axis, that's the same as a separation only in the x-axis?

 

Or in other words, your math with one spatial dimension works the same as with 2 or 3, and it doesn't matter how the separation vector and the velocity vector are oriented relative to the other?

[xyzt]

I think I'm starting to get it. And you say this works if the separation is in any direction, it doesn't matter if it's in the direction of V? Say, as per OP, a separation only in the y-axis, that's the same as a separation only in the x-axis?

 

Or in other words, your math with one spatial dimension works the same as with 2 or 3, and it doesn't matter how the separation vector and the velocity vector are oriented relative to the other?

Yep