SR Question

  • #1
220
0
An atom is at rest in an inertial reference frame. Suddenly, two photons are emitted from it, at right angles to one another.

What is the relative speed of the photons?

P.S. Assume they depart from the atom at 299792458 meters per second in the atomic frame, as has been found true by experiment.
 
Last edited:

Answers and Replies

  • #2
617
1
SR Answer: c
 
  • #3
220
0
Prove that the relative speed of the two photons is c if you can.

Regards,

StarThrower

P.S.

By Euclidean geometry, if the relative speed of the photons is c, then the two photons form an equilateral triangle with the atom, whose area is increasing in time. The interior angles of an equilateral triangle are 60 degrees. However, the emission angle was stipulated to be 90 degrees. This is a contradiction.
 
Last edited:
  • #4
977
1
What exactly allows you to apply Euclidean geometry to spacetime in this case? (I am sincerely asking this because I don't know the answer.)
 
  • #5
220
0
Chen said:
What exactly allows you to apply Euclidean geometry to spacetime in this case? (I am sincerely asking this because I don't know the answer.)
Answer: The Pythagorean Theorem

Keep in mind that the time dilation formula of the special theory of relativity is derived by assuming that the pythagorean theorem is true in the kind of reference frame we have in my question here.

By stipulation, the photons are departing from the atom at right angles to one another in the rest frame of the atom, which is an inertial reference frame by stipulation. The distance between them at any moment in time is found using the pythagorean theorem. We actually have an isosceles triangle here. The hypotenuse of which is greater than either leg.

However, if the relative speed of the photons is c, as was suggested by Jdavel, then you can prove that the triangle is equilateral. Thence, the emission angle of the photons is 60 degrees.

But it was stipulated that the emission angle is 90 degrees.

Kind regards,

StarThrower

P.S. The very first theorem of Euclid, was to prove that an equilateral triangle could be constructed. Here is a link to Euclid's first theorem:

Euclid 's First Theorem

In Euclid's eleventh proposition, he proves that a right angle can be constructed:

Euclid Book One Theorem Eleven
 
Last edited:
  • #6
286
0
First Issue:
I posed this question to StarThrower in post #92 on this page of the following thread:
https://www.physicsforums.com/showthread.php?t=17024&page=7&pp=15

I proposed that the photons would never observe each other and therefore could never determine how fast they were traveling relative to each other. An observer (in this case the atom) at point C could certainly propose how fast they were traveling with respect to himself, or how fast they were traveling with respect to each other with respect to himself. I'm intrigued about the legitimate answer.

Second issue:
Flat space is only one kind of space. As you say, "By stipulation, the photons are departing from the atom at right angles to one another in the rest frame of the atom." But you cannot necessarily assume that the photon is restricted to Euclidian 3-space. The Pythagorean theorem applies to flat space. For curved spaces you must use a modified version of the theorem that takes into account the metric of the curvature of space.

http://mathworld.wolfram.com/Distance.html

The metric is a (sometimes complex) matrix that tells you HOW to measure the distance between any two points in a specific locale of the space.
 
  • #7
220
0
Severian596 said:
First Issue:
I posed this question to StarThrower in post #92 on this page of the following thread:
https://www.physicsforums.com/showthread.php?t=17024&page=7&pp=15

I proposed that the photons would never observe each other and therefore could never determine how fast they were traveling relative to each other. An observer (in this case the atom) at point C could certainly propose how fast they were traveling with respect to himself, or how fast they were traveling with respect to each other with respect to himself. I'm intrigued about the legitimate answer.

Second issue:
Flat space is only one kind of space. As you say, "By stipulation, the photons are departing from the atom at right angles to one another in the rest frame of the atom." But you cannot necessarily assume that the photon is restricted to Euclidian 3-space. The Pythagorean theorem applies to flat space. For curved spaces you must use a modified version of the theorem that takes into account the metric of the curvature of space.

http://mathworld.wolfram.com/Distance.html

The metric is a (sometimes complex) matrix that tells you HOW to measure the distance between any two points in a specific locale of the space.
Let me reiterate:

Keep in mind that the time dilation formula of the special theory of relativity is derived by assuming that the pythagorean theorem is true in the kind of reference frame we have in my question here.

All kinds of mathematical nonsense is resting on the incorrect assumption that speed of light is c in any inertial reference frame. Space is three dimensional Euclidean.

From the site quoted we have:

In Euclidean three-space, the distance between points A=(x1,y1,z1) and B=(x2,y2,z2) is:

[tex] D(A,B) = \sqrt{ (x2-x1)^2 + (y2-y1)^2+(z2-z1)^2 } [/tex]

You cannot bend the vacuum.

Kind regards,

StarThrower
 
Last edited:
  • #8
977
1
StarThrower said:
All of that kind of nonsense is predicated on the incorrect assumption that speed of light is c in any inertial reference frame. Space is three dimensional Euclidean.

You cannot bend the vacuum.
You're in the wrong profession, mate. You should have become a comedian.

How come light, which travels at straight lines, can and does bend around the sun, allowing us to see objects which are actually behind it? Or are you going to suggest that light does not travel at a straight line now?
 
  • #9
286
0
StarThrower said:
Space is three dimensional Euclidean. You cannot bend the vacuum.
How do you define gravity in 3-space?
 
  • #10
220
0
Chen said:
You're in the wrong profession, mate. You should have become a comedian.

How come light, which travels at straight lines, can and does bend around the sun, allowing us to see objects which are actually behind it? Or are you going to suggest that light does not travel at a straight line now?
For light to travel in a curved path, it must experience a force. Otherwise, by Newton's First Law of Motion (Which is roughly Aristotle's law of Inertia), the photon would continue to move in a straight line at a constant speed.

BUT YOU ARE TELLING ME THE PHOTON IS IN A GRAVITATIONAL FIELD. Therefore, something is interacting with the photon to exert a force on the photon. Hence the photon isn't in an inertial reference frame anymore.

Kind regards,

StarThrower
 
Last edited:
  • #11
286
0
StarThrower said:
For light to travel in a curved path, it must experience a force. Otherwise, by Newton's First Law of Motion (Which is roughly Aristotle's law of Inertia), the photon would continue to move in a straight line at a constant speed.

BUT YOU ARE TELLING ME THE PHOTON IS IN A GRAVITATIONAL FIELD. Therefore, something is interacting with the photon to exert a force on the photon. Hence the photon isn't in an inertial reference frame anymore.

Kind regards,

StarThrower
Like I said define the force of gravity in FLAT 3-space.
 
  • #12
220
0
Severian596 said:
How do you define gravity in 3-space?

[tex] \vec{F} =- GM_1 M_2 \frac{\vec{R}}{|\vec{R}|^3} [/tex]
 
  • #13
286
0
StarThrower said:
[tex] \vec{F} =- \frac{GM_1 M_2}{|vec{R}|^2} [/tex]
Indeed it was. According to this equation when you wave your hand from left to right (specifically when you modify R in the equation above), you instantaneously influence atoms on jupiter, the moon...etc, because your hand exerts a gravitational force on them. Are you comfortable with this?
 
  • #14
turin
Homework Helper
2,323
3
Space-time does NOT have a Euclidean geometry; it has a Euclidean topology. Since the pathagorean theorem is a geometrical theorem for Euclidean geometry, it does not apply to space-time.
 
  • #15
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
19
What is the relative speed of the photons?
Undefined, since, as we've pointed out to you before, photons don't have rest frames.


By Euclidean geometry, if the relative speed of the photons is c, then the two photons form an equilateral triangle with the atom, whose area is increasing in time. The interior angles of an equilateral triangle are 60 degrees. However, the emission angle was stipulated to be 90 degrees. This is a contradiction.
Allow me to remind you that the difference of two velocity vectors in a given reference frame is not, in general, give you the relative velocity of the said objects in SR.


BUT YOU ARE TELLING ME THE PHOTON IS IN A GRAVITATIONAL FIELD. Therefore, something is interacting with the photon to exert a force on the photon. Hence the photon isn't in an inertial reference frame anymore.
According to GR, the photon is travelling in a straight line, period. And, of course, photons never have inertial rest frames.
 
  • #16
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
8
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.

StarThrower said:
By Euclidean geometry, if the relative speed of the photons is c, then the two photons form an equilateral triangle with the atom, whose area is increasing in time. The interior angles of an equilateral triangle are 60 degrees. However, the emission angle was stipulated to be 90 degrees. This is a contradiction.
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.
 
  • #17
286
0
Hurkyl said:
Undefined, since, as we've pointed out to you before, photons don't have rest frames.
Hurkyl, is this because in order to determine the relative velocity of two objects they must be in inertial rest frame(s) (IRF)?

Let me pose a sub-light question (god forbit)...two objects A and B speed away from point C. At time [tex]t_{0}= 0[/tex] the distance between any two of these three points (A, B, and C) can be considered zero. A and B are traveling at 0.7c wrt point C.

A
^
|
|
|
|
C---------->B

How do we determine how fast A and B are travelling with respect to each other? Usually in SR this kind of thing is stated up front (S' is traveling at speed v with respect to S).

StarThrower asserts (as he usually does) that we can state how fast A and B are traveling with respect to each other if we just use our observations from our own frame of reference and apply [you know who's] theorem. I don't believe this is correct. If we concentrate on sub-light speeds for this problem perhaps we can get "real" SR answers instead of having to listen to StarThrower assert something else about absolute time and space...
 
Last edited:
  • #18
617
1
StarThrower said:
Prove that the relative speed of the two photons is c if you can.

I can prove it. But I can't prove it to you, because your mind is closed to the possibility that I can. So I'm not going to waste my time.

A few days ago I showed you the source of all your confusion here: you don't understand what the variables used in SR mean. You think you do. But you don't.
 
Last edited:
  • #19
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
19
Hurkyl, is this because in order to determine the relative velocity of two objects they must be in inertial rest frame(s) (IRF)?
Yes, I do interpret "relative velocity" to mean the velocity of one object in the rest frame of the other object.
 
  • #20
russ_watters
Mentor
19,932
6,406
StarThrower said:
All kinds of mathematical nonsense is resting on the incorrect assumption that speed of light is c in any inertial reference frame.
On what do you base this assertion? And don't say the Pythagorean theorem - I want emperical evidence, for example, give me just one (confirmed) test of the speed of light that shows a variation from C.
 
  • #21
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
19
Hurkyl, is this because in order to determine the relative velocity of two objects they must be in inertial rest frame(s) (IRF)?

Let me pose a sub-light question (god forbit)...two objects A and B speed away from point C. At time the distance between any two of these three points (A, B, and C) can be considered zero. A and B are traveling at 0.7c wrt point C.

A
^
|
|
|
|
C---------->B

How do we determine how fast A and B are travelling with respect to each other? Usually in SR this kind of thing is stated up front (S' is traveling at speed v with respect to S).

StarThrower asserts (as he usually does) that we can state how fast A and B are traveling with respect to each other if we just use our observations from our own frame of reference and apply [you know who's] theorem. I don't believe this is correct. If we concentrate on sub-light speeds for this problem perhaps we can get "real" SR answers instead of having to listen to StarThrower assert something else about absolute time and space...


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.
 
Last edited:
  • #22
617
1
Severian596 said:
If we concentrate on sub-light speeds for this problem perhaps we can get "real" SR answers instead of having to listen to StarThrower assert something else about absolute time and space...
Ok, Hurkyl just did it. But StarThrower isn't going read that post. He wouldn't have clue what Hurkyl was talking about, because he doesn't understand what the variables mean.

You can even do a little better than what Hurkyl did. Start the problem using an unspecified sub-light speed v. Then once you find the relative speed, let v go to c as a limiting value, and watch the relative speed go to c. But StarThrower wouldn't understand that either!
 
  • #23
31
0
StarThrower said:
An atom is at rest in an inertial reference frame. Suddenly, two photons are emitted from it, at right angles to one another.

What is the relative speed of the photons?

P.S. Assume they depart from the atom at 299792458 meters per second in the atomic frame, as has been found true by experiment.

If the part of the universe the two photons are passing though is at rest with respect to the source atom then neither photon will have suffered the effects of SR. Detectors at a known distance from the source atom would register simultaneous arrivals. Thus a logical person would deduce that the relative velocity of the two photons were square root of 2 times 299792458 meters per second or approx: 423970560 meters per second.

SR is irrelevent to the calculation as you are not trying to observe a photon travelling between the two.
 
  • #24
286
0
Thank you very much, Hurkyl! This is my first full-fledged 4-space Lorentz transformation. Usually they factor out the y and z to make it more friendly, so I worked along side you to see if I got the same result.

My work agrees up until the transformation of C to A. I'm working out of Hogg's publication and am using his Lorentz Transformation notation. I cut it out of the PDF and posted it http://copperplug.no-ip.org/homesite/Spacetime_files/Lorentz.gif [Broken]...it's probably important to explain my discrepancy.

To transform from y to y', I did the following:

[tex]t = t[/tex]
[tex]y = 0.7ct[/tex]
[tex]\beta_y = 0.7[/tex]
[tex]\beta^{2} = 0.7^{2}[/tex]

Therefore the transformation should go:

[tex]y' = t(-\gamma \beta_{y}) + 0 + y(1+\frac{(\gamma -1)(\beta^{2}_{y})}{\beta^{2}}) + 0[/tex]
[tex]y' = (t)(-\gamma 0.7) + (0.7ct)(1+\frac{(\gamma -1)(0.7^{2})}{0.7^{2}})[/tex]

Eventually I simplify this to the following (I resubstituted y to compare it to yours):

Mine was: [tex]y' = \gamma(y-0.7t)[/tex]
Yours was: [tex]y' = \gamma(y-0.7ct)[/tex]

I don't have a c term in the final result where you do...where did I go wrong?
 
Last edited by a moderator:
  • #25
31
0
Hurkyl said:


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.
Spacetime is only a representation of the interaface between frames of reference so the obove analysis is also irrelevent.
 

Related Threads on SR Question

  • Last Post
2
Replies
30
Views
3K
  • Last Post
11
Replies
268
Views
20K
  • Last Post
Replies
20
Views
4K
  • Last Post
Replies
14
Views
868
  • Last Post
Replies
4
Views
2K
Replies
8
Views
2K
Replies
159
Views
11K
Replies
6
Views
2K
Replies
2
Views
4K
  • Last Post
5
Replies
100
Views
12K
Top