- #36
A.T.
Science Advisor
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Yes, in SR and Newtonian physics. Not in GR.Quandry said:
No. it doesn't. It just means multiple ways to assign space-time coordinates to events in the same universe.Quandry said:
Can I get clarification on the constant speed of light
- B
- Thread starter Quandry
- Start date
In summary: The implication is that the speed of the light is different depending on the frame of reference you are in.
- #37
Quandry
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Thanks guys.
- #38
Stephanus
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TIME DILATION
I'm not in the office now. Perhaps I can spend my time by talking SR with you.
I'd like to talk about the proof of time dilation.
The hard proof is the muonexperiment observation as many people know about it.
Perhaps I can give you a rather soft proof about time dilation.
This is the two postulate of SR.
1. The speed of light is always, ever and forever constant for every observer.
2. Different observers observes different law of physics.
By law of physics, it means length, time, simultaneity, mass, momentum etc.
Now no matter what (thought) experiments that you do. You do not violate those two postulate!The proof of time dilation.
Supposed you are standing in a box in space. You box' height is 300 km. There's a mirror on the ceiling.
You shine a light vertically upward and the light bounce back to you. What will your stop watch show when you do this experiment? 0.2ms, right?
You can do this experiments one hundreds times, and your stopwatch will always show 0.2ms. Then, someone moves toward you. Or we can say, there are a crowd of people are moving toward/away from you.
You can do your experiments as many as you like, and still 0.2ms. Regardless how many people are moving.
You are D and your mirror is E
Now, there are three people (A,B and C) who move 0.6c toward you. Separated each other so that the first time you shine your light. You'll meet A. When the light reach the ceiling, your mirror meets B. And when you receive back your light you'll meet C.
Just ignore the distance in the picture, I'll explain them later.
This is the fact that both of the 2 participants agree.
The 2 participants are
1. You
2. A,B,C. Because they are comoving, which is "Move at the same speed and direction".
The facts are:
1. When you shine the light you meet A.
2. When your mirror bounce the light it meets B.
3. When you receive your light back you meet C.
What the two participants won't agree is the clock!
Explanations.
if ABC move 0.6c, that is 180,000 km/s then ABC will see that it's you who moves 0.6c toward them!.
All ABC know are the 3 facts above. A meet you when A see you turn on the torch, B meets the mirror when B see the light, C meets you when you receive your light back.
Math question:
What A-C distance so that ABC knows that you move at 0.6c? Speed is invariant.
Now if we trace back the experiment: AC distances is 450 km.
ABC synchronize their clocks before.
When A meets you, it writes down its clock, say at 12:00:10.00
When C meets you, it writes down its clock: 12:00:12.50
So what is your speed according to ABC? 0.6c. As written above. But in this experiment, we already know the answer and we set the situation to prove it.
So your clock shows 2ms and ABC clock will show 2.5ms. This is my thought experiment to prove time dilation.
2.5/2 is 1.25
It conforms Lorentz Factor for 0.6c speed.
##\gamma = \frac{1}{\sqrt{1-0.6^2}} = \frac{1}{0.8} = 1.25##
And in the end it will cause length contraction and relative simultaneity of events.
Now we can continue to Length contraction.
"THIS universe". I do like the idea of multiverse.Stephanus said:
I'm not in the office now. Perhaps I can spend my time by talking SR with you.
I'd like to talk about the proof of time dilation.
The hard proof is the muon
Perhaps I can give you a rather soft proof about time dilation.
This is the two postulate of SR.
1. The speed of light is always, ever and forever constant for every observer.
2. Different observers observes different law of physics.
By law of physics, it means length, time, simultaneity, mass, momentum etc.
Now no matter what (thought) experiments that you do. You do not violate those two postulate!The proof of time dilation.
Supposed you are standing in a box in space. You box' height is 300 km. There's a mirror on the ceiling.
You shine a light vertically upward and the light bounce back to you. What will your stop watch show when you do this experiment? 0.2ms, right?
You can do this experiments one hundreds times, and your stopwatch will always show 0.2ms. Then, someone moves toward you. Or we can say, there are a crowd of people are moving toward/away from you.
You can do your experiments as many as you like, and still 0.2ms. Regardless how many people are moving.
You are D and your mirror is E
Now, there are three people (A,B and C) who move 0.6c toward you. Separated each other so that the first time you shine your light. You'll meet A. When the light reach the ceiling, your mirror meets B. And when you receive back your light you'll meet C.
Just ignore the distance in the picture, I'll explain them later.
This is the fact that both of the 2 participants agree.
The 2 participants are
1. You
2. A,B,C. Because they are comoving, which is "Move at the same speed and direction".
The facts are:
1. When you shine the light you meet A.
2. When your mirror bounce the light it meets B.
3. When you receive your light back you meet C.
What the two participants won't agree is the clock!
Explanations.
if ABC move 0.6c, that is 180,000 km/s then ABC will see that it's you who moves 0.6c toward them!.
All ABC know are the 3 facts above. A meet you when A see you turn on the torch, B meets the mirror when B see the light, C meets you when you receive your light back.
Math question:
What A-C distance so that ABC knows that you move at 0.6c? Speed is invariant.
This is a simple pythagoras theorem.
So, we know the vertical line is 300 km.
if ABC considers that AB diagonal distance is 1 unit (light moves at 1c), and AB horizontal distance is 0.6 unit (your speed) then ABC will consider that DE distance is ##DE = \sqrt{1^2-0.6^2} = 0.4 \text{ unit}## You will see ##x = \sqrt{1^2-0.6^2} = 0.4 z## pattern in Lorentz Factor
So, if 300 km is 0.4 unit, then AB diagonal distance is 375 km and AB horizontal distance is 225 km that makes AC distance is 450km.
So from ABC frame, the light travels from A to B and to C = 375 + 375 km = 750 km. Because the speed of light is invariant, so ABC stopwatch will show 2.5 ms as opposed to your 2ms!
Now close the spoiler back, why don't we setup the distances first then calculate again.
So, we know the vertical line is 300 km.
if ABC considers that AB diagonal distance is 1 unit (light moves at 1c), and AB horizontal distance is 0.6 unit (your speed) then ABC will consider that DE distance is ##DE = \sqrt{1^2-0.6^2} = 0.4 \text{ unit}## You will see ##x = \sqrt{1^2-0.6^2} = 0.4 z## pattern in Lorentz Factor
So, if 300 km is 0.4 unit, then AB diagonal distance is 375 km and AB horizontal distance is 225 km that makes AC distance is 450km.
So from ABC frame, the light travels from A to B and to C = 375 + 375 km = 750 km. Because the speed of light is invariant, so ABC stopwatch will show 2.5 ms as opposed to your 2ms!
Now close the spoiler back, why don't we setup the distances first then calculate again.
ABC synchronize their clocks before.
When A meets you, it writes down its clock, say at 12:00:10.00
When C meets you, it writes down its clock: 12:00:12.50
So what is your speed according to ABC? 0.6c. As written above. But in this experiment, we already know the answer and we set the situation to prove it.
So your clock shows 2ms and ABC clock will show 2.5ms. This is my thought experiment to prove time dilation.
2.5/2 is 1.25
It conforms Lorentz Factor for 0.6c speed.
##\gamma = \frac{1}{\sqrt{1-0.6^2}} = \frac{1}{0.8} = 1.25##
And in the end it will cause length contraction and relative simultaneity of events.
Now we can continue to Length contraction.
- #39
Stephanus
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LENGTH CONTRACTION
So we already know that AC distance is 450 km in ABC frame!.
But what is AC distance in your frame?
Again remember those three facts before:
1. When you shine the light you meet A.
2. When your mirror bounce the light it meets B.
3. When you receive your light back you meet C.
But those don't proof what ABC distance according to your frame.
So why don't we set another observer in your frame that when you meet C, your observer meets A.
Four facts:
1. When you shine the light you meet A.
2. When your mirror bounce the light it meets B.
3. When you receive your light back you meet C.
4: When D meets C then at the same time in your frame (not the same time for ABC!) F meets A.
Now, how far that you have to place F, so when you(D) meet C then F will meet A?
Now, we know that the time it takes the light from D to E back to D is 2ms.
For 2ms, A will travel 360 km. So DF is 360 km.
But that is cheating.
Perhaps we should do the problem backward?
You have a friend some 360 km away.
Something meets you then 2 ms later it meets your friend. So its speed must be 0.6c
So from previous experiments: AC distance from ABC frame is 450 km, while
FD distance from your frame is 360 km.
This is the thought experiment to proof length contraction.
While 450km/360 km = 1.25. Again it matches Lorentz Factor
So we already know that AC distance is 450 km in ABC frame!.
But what is AC distance in your frame?
Again remember those three facts before:
1. When you shine the light you meet A.
2. When your mirror bounce the light it meets B.
3. When you receive your light back you meet C.
But those don't proof what ABC distance according to your frame.
So why don't we set another observer in your frame that when you meet C, your observer meets A.
Four facts:
1. When you shine the light you meet A.
2. When your mirror bounce the light it meets B.
3. When you receive your light back you meet C.
4: When D meets C then at the same time in your frame (not the same time for ABC!) F meets A.
Now, how far that you have to place F, so when you(D) meet C then F will meet A?
Now, we know that the time it takes the light from D to E back to D is 2ms.
For 2ms, A will travel 360 km. So DF is 360 km.
But that is cheating.
Perhaps we should do the problem backward?
You have a friend some 360 km away.
Something meets you then 2 ms later it meets your friend. So its speed must be 0.6c
So from previous experiments: AC distance from ABC frame is 450 km, while
FD distance from your frame is 360 km.
This is the thought experiment to proof length contraction.
While 450km/360 km = 1.25. Again it matches Lorentz Factor
- #40
Stephanus
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RELATIVE SIMULTANEITY OF EVENTS
So, from previous experiments Time dilation and Length contraction, we'll get relative simultaneity of events as the consequence.
Why there is relative simultaneity of events?
Consider this.
Now you have a friend: G. Right in the middle between you and F.
When you meet C and A meets F then there's two light sources turned on.
One near F (and A) and the other one near you (and C).
Because G is in the middle between you and F then the light will arrive at G at the same time. To makes matter straight. We put some detector in G. That IF G receives both light at the same time. G will raise up its Flag. If the lights don't arrive at the same time. G keeps its flag down.
So after the experiment is done. Both you and ABC agree that the light arrive at G at the same time. Both of you will see G raises up its flag.
Now these are the facts that are agreed by both sides. (You and ABC).
1. When you shine the light you meet A. (we don't need the fact anymore in our next thought experiment)
2. When your mirror bounce the light it meets B. (we don't need the fact anymore in our next thought experiment)
3. When you receive your light back you meet C. (we don't need the fact anymore in our next thought experiment)
4: When D meets C at the same time in your frame (not the same time for ABC!) F meets A. (we don't need the fact anymore in our next thought experiment)
5: When D meets C there is a light source near DC turned on
6: When F meets A there is a light sources near AF turned on.
7. Both lights arrive at G at the same time in your frame and ABC frame.
So, here is the ilustration.
In your frame
It's you who are at rest. You'll see AC travels 0.6c to the west.
When F meets A and D meets C,
In ABC frame
Now what will ABC sees?
in ABC frame, it's them who are at rest. It's you who are moving. Now what happens if ABC sees that DF length is equal to their AC?
Now this is wrong.
If FD = AC then from ABC frame the lights won't reach G at the same time. See pic 4
Remember. The lights reach G at the same time is the fact that both observers agree. Both will see that G raises its flag.
So in AC frame, A has to meet F first, before C meets D.
So the solution of the problem. AC has to be longter then DF.
Let's see the agreed facts again.
5: When D meets C there is a light source near DC turned on, see pic 5
6: When F meets A there is a light sources near AF turned on, see pic 6
5 and 6 happens simulaneously in DF frame. See pic 1 but not in AC frame.
7. Both lights arrive at G at the same time in your frame and ABC frame. see pic 7.
So this is the thought experiment proof of Simultaneity of events.
So, from previous experiments Time dilation and Length contraction, we'll get relative simultaneity of events as the consequence.
Why there is relative simultaneity of events?
Consider this.
Now you have a friend: G. Right in the middle between you and F.
When you meet C and A meets F then there's two light sources turned on.
One near F (and A) and the other one near you (and C).
Because G is in the middle between you and F then the light will arrive at G at the same time. To makes matter straight. We put some detector in G. That IF G receives both light at the same time. G will raise up its Flag. If the lights don't arrive at the same time. G keeps its flag down.
So after the experiment is done. Both you and ABC agree that the light arrive at G at the same time. Both of you will see G raises up its flag.
Now these are the facts that are agreed by both sides. (You and ABC).
5: When D meets C there is a light source near DC turned on
6: When F meets A there is a light sources near AF turned on.
7. Both lights arrive at G at the same time in your frame and ABC frame.
So, here is the ilustration.
In your frame
It's you who are at rest. You'll see AC travels 0.6c to the west.
When F meets A and D meets C,
In ABC frame
Now what will ABC sees?
in ABC frame, it's them who are at rest. It's you who are moving. Now what happens if ABC sees that DF length is equal to their AC?
Now this is wrong.
If FD = AC then from ABC frame the lights won't reach G at the same time. See pic 4
Remember. The lights reach G at the same time is the fact that both observers agree. Both will see that G raises its flag.
So in AC frame, A has to meet F first, before C meets D.
So the solution of the problem. AC has to be longter then DF.
Let's see the agreed facts again.
5: When D meets C there is a light source near DC turned on, see pic 5
6: When F meets A there is a light sources near AF turned on, see pic 6
5 and 6 happens simulaneously in DF frame. See pic 1 but not in AC frame.
7. Both lights arrive at G at the same time in your frame and ABC frame. see pic 7.
So this is the thought experiment proof of Simultaneity of events.
Facts:
v = 0.6 in both frame.
In DF frame
DF = 360, AC = 360 (of course)
DG, FG = 180
in AC frame
AC = 450
DF = ?? now this what we are going to find.
Let's call DF length = r and DG = 0.5r.
When A meets F, G has already in 0.5r distance from A and travels 0.6c away.
So the light will reach G from A at ##t1 = \frac{0.5r}{1-0.6} = 1.25r## light second distance.
When C meets D, G is 0.5r distance from C and travels 0.6c toward.
So the light will reach G from C at ##t2 = \frac{0.5r}{1+0.6} = 0.3125## light second distance.
So in r unit. AC length is 1.5625 r light second distance.
We know that AC length is 450 km = 1.5ms
So ##1.5 ms = 1.5625r; r = 0.96ms## That makes DF length from AC frame is 288 km.
The simultaneity of events?
We know that when A meets C, 1.25 seconds later the light will meet G.
And we also know that when the light meet G, it's already 0.3125 seconds later after C meets D.
So from DF frame
A meets F and C meets D happens at the same time.
From AC frame.
A meets F happens 0.9375 seconds before C meets D
v = 0.6 in both frame.
In DF frame
DF = 360, AC = 360 (of course)
DG, FG = 180
in AC frame
AC = 450
DF = ?? now this what we are going to find.
Let's call DF length = r and DG = 0.5r.
When A meets F, G has already in 0.5r distance from A and travels 0.6c away.
So the light will reach G from A at ##t1 = \frac{0.5r}{1-0.6} = 1.25r## light second distance.
When C meets D, G is 0.5r distance from C and travels 0.6c toward.
So the light will reach G from C at ##t2 = \frac{0.5r}{1+0.6} = 0.3125## light second distance.
So in r unit. AC length is 1.5625 r light second distance.
We know that AC length is 450 km = 1.5ms
So ##1.5 ms = 1.5625r; r = 0.96ms## That makes DF length from AC frame is 288 km.
##\gamma = \frac{1}{\sqrt{1-0.6^2}} = 1.25##
DF length in DF frame = 360.
DF length in AC frame = ##360 / \gamma = 360/1.25 = 288##
AC length in AC frame = 450
AC length in DF frame = ##450 / \gamma = 450/1.25 = 360##
DF length in DF frame = 360.
DF length in AC frame = ##360 / \gamma = 360/1.25 = 288##
AC length in AC frame = 450
AC length in DF frame = ##450 / \gamma = 450/1.25 = 360##
The simultaneity of events?
We know that when A meets C, 1.25 seconds later the light will meet G.
And we also know that when the light meet G, it's already 0.3125 seconds later after C meets D.
So from DF frame
A meets F and C meets D happens at the same time.
From AC frame.
A meets F happens 0.9375 seconds before C meets D
- #41
sheshank
- 27
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I think experimental facts can't be manipulated, our reasoning and understanding can. Speed of light is constant irrespective of the Frame of Reference. Make that as an axiom, just like how Einstein did, and start from there. Make it as a basis and try to think, explain what you are 'thought experimenting'. That's it. There is no way around, as everything in it is counter intuitive and beats our common sense notion of space and time.
On a side note, I think one needs to go more deeper into classical mechanics, into the Hamiltonian Dynamics to understand the core principle of motion viz. principle of least action, to understand why Laws of physics are independent of frame of reference. Then, in addition to that adding all the other experimental facts like speed of light constant, or Quantization of Energy, we will make our starting points (a.k.a axioms) from where we can resume our reasoning again. It is more math less 'science' (in the sense of being a direct evidence) when it comes to extremities like this. Why? I have no answer for that. It is just given and that's it.
On a side note, I think one needs to go more deeper into classical mechanics, into the Hamiltonian Dynamics to understand the core principle of motion viz. principle of least action, to understand why Laws of physics are independent of frame of reference. Then, in addition to that adding all the other experimental facts like speed of light constant, or Quantization of Energy, we will make our starting points (a.k.a axioms) from where we can resume our reasoning again. It is more math less 'science' (in the sense of being a direct evidence) when it comes to extremities like this. Why? I have no answer for that. It is just given and that's it.
- #42
Stephanus
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May I differ? The first time Einstein formulate his theory of Special Relativity, he did that with pencil and paper (and a typewrite). The https://en.wikipedia.org/wiki/Time_dilation_of_moving_particles#Experiments was conducted (as I suspect it) in 1940. Long after Einstein released his SR theory, even long after General Relativity.sheshank said:
While for General Relativity, of course, Eddington proved it. 1 year after the release of GR I think.
Of course there are some off course theory, just as Aristotle's 4 elements. Fire, Water, Air and Earth.
Thanks for the answer.sheshank said:
- #43
PeterDonis
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Stephanus said:
Eddington's solar eclipse observations were done in 1919, four years after Einstein's publication of GR. Eddington claimed that his observations confirmed the GR prediction of the bending of light by the Sun (which in itself doesn't "prove GR", it just confirms one particular prediction), but later analysis has made it pretty clear that his data was not accurate enough to actually confirm the GR prediction. Later, more accurate experiments have confirmed it.
- #44
Stephanus
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Thanks for the correction. I will check my source before I post.PeterDonis said:
Nevertheless, according to this link, this was the experiment that made Einstein famous overnight.
I didn't know that this experiment was wrong. I've read long ago that there's a solar eclipse done not long after GR was released that made Einstein famous overnight.
Now, I'm afraid that I post some incorrect information in this thread. So I check again.
- #45
exmarine
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Wouldn't these discussions be far more fruitful if they were about "the velocity of a photon" instead of about "the velocity of light"?
- #46
Stephanus
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Are you sure that the speed of photon is not equal to the speed of light? I don't know. I'm new in physics anyway.exmarine said:
- #47
PeterDonis
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exmarine said:
Probably not, since the concept of a "photon" is problematic and likely to lead to misunderstandings in a discussion of this sort. Briefly, a "photon" is not a point particle that moves at the speed of light; it is more complicated than that. Using the term "light" instead of "photon" avoids the complications.
- #48
PeterDonis
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Stephanus said:
In QM the concept of the "speed" of a photon is not really well-defined. That's one of the complications I mentioned in my response to exmarine just now. Better to just avoid all that for this discussion.
- #49
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Certainly not. Stay away from photons unless you wish to discuss them properly in a QFT setting. Photons are not "little balls of light".exmarine said:
Edit: I see I got beat to it by Peter. The point still stands.
- #50
exmarine
- 241
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Well, isn’t “light” the plural of “photon”? If so, does it make any sense to discuss “THE speed of photons”? How about the “the SPEEDS of photons”, or “the SPEEDS of light”?
And if QFT allows meaningful discussions of “the speeds of photons”, why does it prohibit one about “the speed of A photon”? I have seen many discussions about double-slit experiments where it is claimed that “one photon at a time” still creates the fringe patterns, etc. So it appears to be an empirical fact that one photon at a time can exist and be tested. It just seems to me that referring to a single photon’s behavior when explaining things would significantly reduce the confusion.
Thanks.
And if QFT allows meaningful discussions of “the speeds of photons”, why does it prohibit one about “the speed of A photon”? I have seen many discussions about double-slit experiments where it is claimed that “one photon at a time” still creates the fringe patterns, etc. So it appears to be an empirical fact that one photon at a time can exist and be tested. It just seems to me that referring to a single photon’s behavior when explaining things would significantly reduce the confusion.
Thanks.
- #51
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No.exmarine said:
These discussions are usually very heuristic.exmarine said:
Truth with modification that needs qualifications and should not be taken too litterally.exmarine said:
Certainly not. You cannot discuss photons properly without discussing QFT and quantification of gauge fields. Photons are among the most complicated objects in QFT and I suggest taking any heuristic discussion speaking about single photons with a huge grain of salt unless you understand the underlying theory properly.exmarine said:
- #52
exmarine
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Quote: Oroduin, post #51
Would you care to explain your answers a bit more?
So light is not comprised of photons?
They have not done double-slit experiments “one photon at a time”? Or they have, but…. what?
As for studying QFT, I can do the math in Feynman’s All Paths version for a photon. Is that what you mean, or something else? Yes a photon is a complicated deal, but I don’t see why one can’t study, and measure – at least in theory, the speed of a single one. It covered some distance and took some component time, so…
Thanks.
Would you care to explain your answers a bit more?
So light is not comprised of photons?
They have not done double-slit experiments “one photon at a time”? Or they have, but…. what?
As for studying QFT, I can do the math in Feynman’s All Paths version for a photon. Is that what you mean, or something else? Yes a photon is a complicated deal, but I don’t see why one can’t study, and measure – at least in theory, the speed of a single one. It covered some distance and took some component time, so…
Thanks.
- #53
PeterDonis
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exmarine said:
Not in the sense you are using the term "photon". Light is not comprised of little massless point particles moving at ##c##.
exmarine said:
That's how it is often described in pop science articles, but it's not a good description of the actual physics involved.
exmarine said:
You may think that's what you're doing, but it isn't.
exmarine said:
Look up "Newton-Wigner localization" and how it does not work for massless quantum fields.
- #54
PeterDonis
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exmarine said:
If you want an entertaining introduction to the complications, try John Baez's "Photons, Schmotons" series:
http://math.ucr.edu/home/baez/photon/schmoton.htm
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