What is the relative speed of photons?

[Chen]

Ever heard of the Michelson-Morley experiment?

 

[Severian596]

It is interesting to postulate that Newton's law [...] neither here nor there.

The reason I ask the initial question was because I wanted you to consider contemplating it. If you won't be swayed by emperical evidence, and you won't be swayed by accepted theory, and you sure won't be swayed by our input, put current theory on your list of things to contemplate. There's a reason we've progressed further than Newton.

Namely this is experimental evidence (which led to the development of theories which then sufficiently predicted and/or explained the data), but your most frustrating trait is you take the stance "I can think it, therefore it is." I believe the Physical sciences differ from Philosophy because physical sciences seek evidencial proof. "Common sense," and "what any logical person would conclude," come crashing down in the face of valid, correct experimental data.

 

[StarThrower]

After all the time people have spent showing you that a reference frame moving at c is ill-defined in SR, I can't believe we're seeing another "what does motion look like from the photon's point of view?" thread.



So what??

The angle is stipulated to be 90 degrees in the lab frame. Even in Galilean relativity, the angle would not be 90 degrees in any other frame. In fact, it is always possible to transform to a frame in which both photons are moving along the same line! It's called the "center of momentum" frame.

Tom, first of all not a single physicist on this planet has proved to me that a reference frame traveling along with a photon isn't an inertial reference frame. The reason being of course, that it is (provided the photon isn't being subjected to a force).

That being said, the burden now falls squarely on the shoulders of the relativists to prove the impossible. Namely that a frame at which a photon is at rest, is non-inertial, in the case where the speed of this particular photon is constant in some IRF.

It's not that the reference frame is ill-defined in SR, that involves a bit of logico/deductive confusion on your part, or handwaiving, I'm not sure. Also, I am a bit confused as to why you wrote that in Galilean relativity it wouldn't be 90 degrees in any other frame. Perhaps you can elaborate on that?

Kind regards,

StarThrower

 

[StarThrower]

Ever heard of the Michelson-Morley experiment?

Yes, indeed I have. The first thing to say, is when I did a derivation for the 'interferometer formula' using the S and S` frames found in a textbook of mine, if I do recall, I spotted an error. Yes I do recall that.

At any rate, if a photon must leave a source at speed c=299792458, regardless of what kind of reference frame the photon is in, then the Michelson Morely results are explained without using SR.

Regards,

StarThrower

 

[Severian596]

At any rate, if a photon must leave a source at speed c=299792458, regardless of what kind of reference frame the photon is in, then the Michelson Morely results are explained without using SR.

SR is the theory that asserted 'a photon must leave its source at speed c regardless of its reference frame'. Michelson/Morely exhibited this behavior of the assertion experimentally. Therefore the Michelson/Morely results are explained by special relativity, not without using them.

The fact is you need to educate yourself on the topic. Period. Or you will never learn the answers to the questions you ask.

 

[StarThrower]

Ok, let us put coordinates on this system in order to solve it. (Things are much clearer with coordinates, which is probably one of the reasons why anti-SR crackpots tend to dislike them)


In the rest frame of C, let us put the event where A, B, and C are all at the same place at the origin of the coordinate system. Let the coordinate axes x, y, and z be right, up, and out of the page respectively, and let t be the coordinate time.

The worldlines of A and B are given parametrically by (being sloppy and parametrizing by t):

A: (t, 0, 0.7ct, 0)
B: (t, 0.7ct, 0, 0)

Letting t', x', y', and z' be the coordinates in A's reference frame, the Lorentz transform from C's frame to A's frame gives the relations:

x' = x
y' = γ(y - 0.7ct)
z' = z
t' = γ(t - 0.7y/c)

where γ := 1/√(1-0.7²) = 1/√0.51

So, the worldlines are given parametrically in A's frame by (yes, I mean to parametrize by t and not t'):

A: (γ(t - 0.7 (0.7ct)/c), 0, γ(0.7ct - 0.7ct), 0)
A: (γ 0.51 t, 0, 0, 0)
B: (γ(t - 0.7 0/c), 0.7ct, γ(0-0.7ct), 0)
B: (γt, 0.7ct, -γ0.7ct, 0)

(I only did A as a sanity check)

So, we have that the coordinate velocity of B in A's frame is given by
v = (0.7c √0.51, -0.7c, 0)

So, unless I've made a mistake along the way, B's speed relative to A is 0.7 c √1.51.
[/size]

No, you didn't make a visible mistake along the way, and the reason you didn't is because the initial speed was taken as .7c. Let it be c, and redo the calculation.

Kind regards,

StarThrower

Here, I will start you off:

GOAL: Compute the relative speed of B and A.

As Hurkyl did, start off viewing the motion of A,B in the rest frame of C.

Let there be a clock at rest (anywhere in this frame, it doesn't matter), whose readings are denoted by the letter t. At the beginning of this event, the clock reads zero. As time goes foward, this clock's readings increase numerically. And let us presume that the clock ticks so fast, that whenever point A or point B changes coordinates in this frame, this clock has changed readings. Furthermore, let the tick rate of this clock be constant. This having been said...

At any moment in time t in this frame, the y coordinate of point A is ct on the y axis, x coordinate = 0, and z coordinate =0. Simultaneously in this frame, the x coordinate of B in this frame is ct, y coordinate =0, and z coordinate =0.

Or using coordinates so as to not be accused of being an anti-SR crackpot, we have:

A: (t, 0, ct, 0)
B: (t, ct, 0, 0)

Now, we are interested in viewing the motion of B in the rest frame of A. Let a coordinate system have been set up, whose origin is the point A. Let the coordinate axes x`, y`, and z` be right, up, and out of the page respectively, and let t` be the coordinate time. The motion of B is going to be diagonally down, and to the right, placing B in the fourth quadrant of this frame FA, once t`>0.

There is a clock at rest in this frame, whose rest rate matches that of the other clock. In fact, this clock is constructed identically to the other clock. When the event begins, this clock reads zero too. Hence, if t=0 then t`=0.

The Lorentz transformations from rest frame C, to rest frame A give:

x' = x
y' = γ(y - ct)
z' = z
t' = γ(t - y/c)

where γ := 1/√(1-1) = 1/0

Oops, I guess I can't use the Lorentz tranformations. Let me try the Galilean transformations instead... etc. etc.

Lets check out the location of B in FA, when t` = 1.

We have a clock at rest in frame A, which we shall call clock A. In the time it took clock A to go from zero, to 1, the point C moved downwards, onto the negative y` axis. This distance down it moved (in this rest frame FA), is given by ct`. Now, the y coordinate of A in reference frame C, at any moment in time according to clock C is ct, and the y` coordinate of C in reference frame A according to clock A is -ct`. Separation distance must be equivalent, hence we must have |-ct`| = |ct| which implies t=t`. In other words, the current coordinates of point C in frame A are: (1, 0, -c, 0), and clock C also reads 1.

Using the galilean tranformation we have t = t`, hence t = 1.

In this time, the point B has moved onto the positive x-axis of frame C, at an x coordinate of ct or ct` (take your pick). This places point B in the fourth quadrant of the x`,y` plane, at a location with the following coordinates in frame A:

(t`,x`,y`,z`) = (1,c,c,0)

At this point in the argument c is in meters, because c = 299792458 m/s, but t` = one second in frame A= one second in frame C = t, hence the distance in meters is 299792458 meters.

The hypotenuse of this triangle can be computed using the pythagorean theorem. It corresponds to the distance traveled by point B in frame A, in the time it took clock A to go from zero to one. Using the pythagorean theorem we have:

D^2 = c^2 + c^2 = 2c^2

From which we infer that the distance traveled is:

D = (\sqrt{2}) c = (\sqrt{2}) (299792458) meters

Hence, the speed of B relative to A, in the rest frame of A is:

|\vec{V}| = (\sqrt{2}) (299792458) meters/second

Because it did so in one tick of clock A, and I recently stipulated that the unit of clock A is the second.

Let objects A,B be photons. Thus, the relative speed of these two photons exceeds the speed of light in the rest frame of A. Thus, if the fundamental postulate of the special theory of relativity is true, then FA is not an inertial reference frame. FC is an inertial reference frame by stipulation, and FA is moving relative to FC at a constant speed. Therefore, FA is an inertial reference frame. Therefore, FA is and isn't an inertial reference frame, which is impossible. Therefore, the fundamental postulate of SR is false.

Kind regards,

StarThrower

 

[StarThrower]

SR is the theory that asserted 'a photon must leave its source at speed c regardless of its reference frame'. Michelson/Morely exhibited this behavior of the assertion experimentally. Therefore the Michelson/Morely results are explained by special relativity, not without using them.

The fact is you need to educate yourself on the topic. Period. Or you will never learn the answers to the questions you ask.

This is incorrect.

Proof:

(forthcoming)


Kind regards,

StarThrower

 

[jdavel]

StarThrower said: "We are viewing the motion of A,B in the rest frame of C."

That's your problem!

With Galileo's relativity, you can calculate the relative speed of A and B in the rest frame of C or B or A and always get the same result, namely sqrt(2)*c

With Einstein's relativity, you can calculate the relative speed of A and B in the rest frame of A or B and always get the same result, namely c. But you'll get a different result in the rest frame of C, namely sqrt(2)*c.

But the postulate that says photons travel at c in all inertial frames means just that. Any measurement of a photon's speed made in an inertial frame, will always give the result c. In your setup there are two photons (A and B) moving at c in the rest frame of C, just like the postulate says there should be. If it were physically possible (but it's not) to get a measuring apparatus into the rest frame of B, then B would vanish (photons don't exist at rest) and you would measrue the speed of A as c.

By the way, I don't understand why you made the photons go at a right angle to each other. Why not simplify the problem and have the atom emit them in opposite directions. That way, showing that each photon is moving through the other's frame at speed c (in other words that their relative speed is c) might be easy enough for you to understand.

 

[jdavel]

StarThrower said: "There is a clock at rest in this frame, whose rest rate matches that of the other clock."

Relativity says this is impossible! You can't disprove a theory by DESCRIBING an experiment that would disprove the theory, if the theory says the experiment is impossible. You have to DO the experiment. In fact, it wouldn't even matter what the result of doing this experiment was. Just being able to do it would disprove relativity.

 

[StarThrower]

StarThrower said: "We are viewing the motion of A,B in the rest frame of C."

That's your problem!

With Galileo's relativity, you can calculate the relative speed of A and B in the rest frame of C or B or A and always get the same result, namely sqrt(2)*c

With Einstein's relativity, you can calculate the relative speed of A and B in the rest frame of A or B and always get the same result, namely c. But you'll get a different result in the rest frame of C, namely sqrt(2)*c.

But the postulate that says photons travel at c in all inertial frames means just that. Any measurement of a photon's speed made in an inertial frame, will always give the result c. In your setup there are two photons (A and B) moving at c in the rest frame of C, just like the postulate says there should be. If it were physically possible (but it's not) to get a measuring apparatus into the rest frame of B, then B would vanish (photons don't exist at rest) and you would measrue the speed of A as c.

By the way, I don't understand why you made the photons go at a right angle to each other. Why not simplify the problem and have the atom emit them in opposite directions. That way, showing that each photon is moving through the other's frame at speed c (in other words that their relative speed is c) might be easy enough for you to understand.

You cannot divide by zero, therefore you cannot use the Lorentz transformations to make a coordinate transformation here. Go back and read more carefully.

Kind regards,

StarThrower

 

[StarThrower]

StarThrower said: "There is a clock at rest in this frame, whose rest rate matches that of the other clock."

Relativity says this is impossible! You can't disprove a theory by DESCRIBING an experiment that would disprove the theory, if the theory says the experiment is impossible. You have to DO the experiment. In fact, it wouldn't even matter what the result of doing this experiment was. Just being able to do it would disprove relativity.

You cannot divide by zero, therefore you cannot use the Lorentz transformations here.

Kind regards,

StarThrower

 

[Chen]

You cannot divide by zero, therefore you cannot use the Lorentz transformations here.

You also cannot perform this in reality (something you cannot seem to grasp for some reason), so what seems to be the problem? Again, why should physical equational yield sensible answers for non-physical situations? Just to cater to your whim?

 

[StarThrower]

You also cannot perform this in reality (something you cannot seem to grasp for some reason), so what seems to be the problem? Again, why should physical equational yield sensible answers for non-physical situations? Just to cater to your whim?

My whole point, is that you can perform this experiment in reality. You can fire two photons at right angles to one another simultaneously. According to the theory of relativity, these two photons cannot move relative to each other. That is impossible. Hence, SR contradicts.

Kind regards,

StarThrower

 

[Chen]

My whole point, is that you can perform this experiment in reality. You can fire two photons at right angles to one another simultaneously. According to the theory of relativity, these two photons cannot move relative to each other. That is impossible. Hence, SR contradicts.

By saying this you assume that "a reference frame traveling along with a photon isn't an inertial reference frame". You have still not proven this. You told us you think it's right, you told us no one has proven otherwise, but you have never actually proved that it's correct. And until you do, nothing we can tell you will make any difference.

Now you can either frown upon us and say that "a reference frame traveling along with a photon is inertial until proven otherwise", in which case we can end the discussion right here, or you can actually attempt to prove this statement.

It's like I'd come to some mathematicians and tell them "I believe 1 + 1 = 3, and therefore (1 + 1) + (1 + 1) = 6 and your math contradicts itself."

 

[StarThrower]

By saying this you assume that "a reference frame traveling along with a photon isn't an inertial reference frame". You have still not proven this. You told us you think it's right, you told us no one has proven otherwise, but you have never actually proved that it's correct. And until you do, nothing we can tell you will make any difference.

Now you can either frown upon us and say that "a reference frame traveling along with a photon is inertial until proven otherwise", in which case we can end the discussion right here, or you can actually attempt to prove this statement.

It's like I'd come to some mathematicians and tell them "I believe 1 + 1 = 3, and therefore (1 + 1) + (1 + 1) = 6 and your math contradicts itself."

You have what I am saying backwards. I am saying the following:

If a photon isn't experiencing a force then it is in an inertial reference frame.
From this it follows that if the speed of a photon is constant in some inertial reference frame, then the rest frame of the photon is an inertial reference frame.

By the fundamental ASSUMPTION of SR, the speed of a photon in any inertial reference frame is c>0.

From that assumption the following contradiction can be reached using perfect binary logic:

In inertial reference frame X, c=0 and not (c=0), where c denotes the speed of light in frame X.

Therefore, the fundamental postulate of SR must be false, by deduction.

Kind regards,

StarThrower

 

[Chen]

You have what I am saying backwards. I am saying the following:

If a photon isn't experiencing a force then it is in an inertial reference frame.
From this it follows that if the speed of a photon is constant in some inertial reference frame, then the rest frame of the photon is an inertial reference frame.

That is not allowed by SR. You are basically saying this:

Let us assume something that contradicts a specific theory. Let us attempt to analyze the assumed situation. We reach a contradiction, therefore the specific theory is wrong.

 

[StarThrower]

That is not allowed by SR. You are basically saying this:

Let us assume something that contradicts a specific theory. Let us attempt to analyze the assumed situation. We reach a contradiction, therefore the specific theory is wrong.

No, that's not what I am saying. I've repeatedly said what I am saying, but I can say it again, no harm done. I am saying this:



If the fundamental postulate of SR is true, then there is at least one statement X, such that (X and not X). Therefore, the fundamental postulate of SR is false.

The statement X being reached in this example problem is this: c denotes the speed of light in inertial reference frame F, and c=0 and not (c=0).

Kind regards,

StarThrower

 

[Chen]

If the fundamental postulate of SR is true, then there is at least one statement X, such that (X and not X). Therefore, the fundamental postulate of SR is false.

The statement X being reached in this example problem is this: c denotes the speed of light in inertial reference frame F, and c=0 and not (c=0).

Except that X does not conform with the theory of SR. Which is what we have been trying to tell you for the past... what, two weeks now?

 

[StarThrower]

Except that X does not conform with the theory of SR. Which is what we have been trying to tell you for the past... what, two weeks now?

You don't understand the reductio absurdum.

What do you mean by, the word 'conform' in your sentence, and I quote

"statement X does not conform with the theory of SR."

What's that mean?

Kind regards,

StarThrower

 

[Severian596]

No, that's not what I am saying. I've repeatedly said what I am saying, but I can say it again, no harm done. I am saying this:



If the fundamental postulate of SR is true, then there is at least one statement X, such that (X and not X). Therefore, the fundamental postulate of SR is false.

The statement X being reached in this example problem is this: c denotes the speed of light in inertial reference frame F, and c=0 and not (c=0).

Kind regards,

StarThrower

Let's cut through the rather poor wording. You should not begin this kind of argument by saying "if something is true, then [by some argument that depends on that something being true] that something is false." Instead:

(1) Special Relativity depends on the condition that the speed of light through a vacuum c is constant (remains unchanged) in any inertial frame of reference.
(2) There is at least one inertial frame of reference where the speed of light through a vacuum does NOT equal c.
(3) Therefore, based on Special Relativity's dependence on condition (1) and the existence of condition (2), Special Relativity is not consistent.

Now all you have to do (as was stated before) is prove that condition (2) does actually exist. If it does, it makes Special Relativity inconsistent. If it does not exist then Special Relativity is not proven inconsistent.

Now...how do you plan on proving condition (2)?

 

[Severian596]

In essence, StarThrower, you have constructed a theory. How do you plan on proving your theory?

..:: EDIT ::..

Thus far you have only been able to illustrate that if your theory is true SR is inconsistent. Perhaps we should just end this discussion and conclude that your theory, if it is ever proven, could bring the downfall of SR?

 

[jdavel]

StarThrower said: "You cannot divide by zero, therefore you cannot use the Lorentz transformations to make a coordinate transformation here."

Do you have to divide by zero to find the derivative of a function? No, you find it by calculating the limit of a ratio as both the numerator and denominator approach zero.

So try this? Start with two particles traveling at speed v in the rest frame of C, and calculate the speed of A in the rest frame of B. That's the relative speed of A and B, right?

Now see if that relative speed has a finite limit as v approaches c.

It will, and it will be c.

 

[Severian596]

I guess I should also say that you should be prepared for the consequences of your theory, too. You'll receive a lot of fame and attention if your theory proves true, but much of the fame and attention will be negative. People will expect you to present a lot of proof that your theory is true.

And I doubt you'll be able to wow them with, "Okay. Suppose the following scenario...[insert the scenario here]...SEE! Bam! Inconsistent."

*StarThrower dusts his hands and is quite pleased with himself*

 

[quantumdude]

Tom, first of all not a single physicist on this planet has proved to me that a reference frame traveling along with a photon isn't an inertial reference frame. The reason being of course, that it is (provided the photon isn't being subjected to a force).

It has already been shown that motion is ill-defined if we take the speed of light to be the same in every frame. If you assume that, then it is not possible to put the a photon at the origin of a frame that is at rest. Jeez, instead of waiting for a physicist to prove it for you, just work it out yourself. It's not that hard.

That being said, the burden now falls squarely on the shoulders of the relativists to prove the impossible. Namely that a frame at which a photon is at rest, is non-inertial, in the case where the speed of this particular photon is constant in some IRF.

There is no frame in which a photon is at rest.

What you are doing here is no different that asking me to prove that all mermaids have tails. Since the class of mermaids is empty, there is nothing for me to prove.

Similarly, the class of reference frames in which the photon is at rest is empty, by assumption.

It's not that the reference frame is ill-defined in SR, that involves a bit of logico/deductive confusion on your part, or handwaiving, I'm not sure.

Actually, it is ill defined in SR. Starting from any frame that is moving with a speed that is less than light, a Lorentz boost to v=c results in the Lorentz factor g to having a zero denominator, which makes it *drumroll, please* undefined[/color].

Come on, man, do a little work for yourself. Your insistence on having others prove everything for you when it is so easy to derive gives me the impression that you are here only to stir things up, rather than having any genuine interest in learning.

Also, I am a bit confused as to why you wrote that in Galilean relativity it wouldn't be 90 degrees in any other frame. Perhaps you can elaborate on that?

Again: Try doing a little work for yourself. It won't kill you.

Assume Galilean relativity. Let S be stationary and S' be moving with a velocity vi with respect to S (v>0). Let two objects, U₁ and U₂, be moving at the same speed, but right angles in S. Give them velocities u₁=ui and u₂uj, respectively.

Now transform them to frame S' according to Galilean relativity:

u₁'=u₁-v
u₂'=u₂-v

Substituting the velocities given above yields:

u₁'=(u-v)i
u₂'=-vi+uj

Take the dot product:

u₁^.u₂=v²-uv,

which in general does not vanish.

Kind regards,

Right back atcha.

 

[Severian596]

FINALLY, to sum things up, look at it this way:

Postulates of the Theory Special Relativity (SR):

(1) All laws of physics apply uniformly to all inertial frames of reference
(2) All inertial frames of reference measure the speed of light in a vacuum c to be the same value

Now consider StarThrower's Theory (S) that depends on the following conditions:
(S1) A photon is in an inertial reference frame
(S2) A photon measures c differently than at least one other inertial frame of reference

StarThrower's Theory asserts the following:
((S1) \bigwedge (S2)) \rightarrow (S) \rightarrow (\sim 2) \rightarrow (\sim SR)

So there you go! All you have to prove is (S1) \bigwedge (S2) and you've got the theory of SR right where you want it.

 

[jdavel]

StarThrower said: "By the fundamental ASSUMPTION of SR, the speed of a photon in any inertial reference frame is c>0."

Suppose Einstein had said this:

For any inertial reference frame in which photons in empty space and matter can be observed, all the photons will be observed to be moving at a speed c, and all the matter will be observed to be moving at speeds less than c. Since all observers are made of matter, nothing can be observed in a frame of reference where only photons might (or might not) exist. Thus, such a frame of reference (if it exists) is irrelevant to all of physics, and is not considered in this theory.

That wording would lead just as inevitably to the Lorentz transforms and the other predictions of SR. But isn't Einstein's wording more elegant? And really, everyone who understands the theory knows what he meant.

 

[ramcg1]

"Thus a logical person would deduce that the relative velocity of the two photons were square root of 2 times 299792458 meters per second..."

Logical? Not if they were aware of SR.

"SR is irrelevant to the calculation as you are not trying to observe a photon traveling between the two."

Not true. SR is relevant to any calculation involving motion.

You only believe this because you were taught this.

It does not make it true.

Try this thought:

A photon arriving at an atom can either join that atom or continue on its journey. At each atom there is an interaction. Each time the photon moves on it moves on with the velocity c with respect to the atom it just interacted with. If that atom was traveling at a different velocity to the previous atom then SR is applied to its frequency and wavelength (it may also have to adjust its amplitude).

SR and GR are only descriptions of these interactions not the macro universe.

Have a good easter.

 

[quantumdude]

A photon arriving at an atom can either join that atom or continue on its journey. At each atom there is an interaction. Each time the photn moves on it moves on with the velocity c with respect to the atom it just interacted with. If that atom was traveling at a different velocity to the previous atom then SR is applied to its frequency and wavelength (it may also have to adjust its amplitude).

SR and GR are only descriptions of these interactions not the macro universe.

?

The paragraph above does not in any way imply the conclusion.

 

[jdavel]

StarThrower, how about this?

Define: io to be a frame of reference "in which the equations of mechanics hold good"

Let I be the set of io and all other frames of reference in (n>0) that move at constant speed with respect to io

Let C be the subset of I in which all photons travel at c

Let V be the subset of I in which at least one photon travels at a speed v, not equal to c.

Let P be the subset of I which are possible in the physical world

Would you agree that I is the union of C and V?

Would you agree that the intersection of C and V is the null set?

If so, can you use this formalism to explain what you think the postulate of relativity means?

I think it means this: (C is not the null set) AND (P=C)

 

[Chen]

You don't understand the reductio absurdum.

What do you mean by, the word 'conform' in your sentence, and I quote

"statement X does not conform with the theory of SR."

What's that mean?

I'll make it easier to understand.

Let's say you have a theory which includes two statements, X and Y. These statements must be true for the theory to be correct. Now let's take a statement Z and prove that it is self-contradictory. What does it have to do with the theory at all? How exactly does it prove the theory to be wrong?!