I do not have a problem with the concept of the constant speed of light as it has no mass and therefore no inertia and therefore no relationship to any IFR. However it seems to be expressed as constant in all IFR's which I do not understand. This seems to say that if I am traveling at 1/2c and I shine a torch forward the light moved away from me at c and if I shine the light backwards it also travels away from me at c.
This seems to say that, in the first case, the light is traveling away from the point in space that it is created (independent of my IFR), at 1.5c.
To clarify this for me, my question is:
If I am traveling towards a light source at .5c and I have two light detectors with a set space between them to measure the time that the light takes to transition from one to the other, will the time taken correspond to a speed of c? And, if I decrease my speed to .25c and do the same measurement will the measurement also indicate a speed of c?
This is an experimental fact and so is not something you should set out to understand using your regular Galilei transformation. In fact, as you have discovered, it directly violates Galilei addition of velocities. The conclusion you should draw from this is that Galilei addition of velocities does not work in this extreme. Other conclusions such as the relativity of simultaneity follow in a relatively straight forward fashion.
#4
Quandry
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So you are saying that regardless of my speed, the measurement between to two detectors is always the same?
This seems to say that, in the first case, the light is traveling away from the point in space that it is created (independent of my IFR), at 1.5c.
The point in space is a frame-dependent concept. Say your flashlights are on a table on a train. To someone on the train, the point in space where the light was emitted is a point just above the table. To someone standing beside the track, the point in space where the light was emitted is some point above the track where the flashlights happened to be when they emitted.
The emission event is frame-independent, but you can't measure speed with respect to that.
#6
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Orodruin said:
The conclusion you should draw from this is that Galilei addition of velocities does not work in this extreme
I do not set out to understand using Galilei transformation. In fact if you refer to my post I am clear that this does not happen (although it seems a logical, but incorrect, conclusion to the speed of light being a constant in all frames of reference, which by logic says that it is different between two frames of reference). But if I am approaching a source of light which is traveling at c and I am traveling at .5c the relationship between my detector and the source of light is changing at 1.5c, even although the light is traveling at c. There are many incidences where relationships change faster than the speed of light - e.g. phase relationships between traveling waves can move down a waveguide faster than the speed of light.
There is no implication of adding velocities in this process.
#7
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Ibix said:
The point in space is a frame-dependent concept.
It is not the point in space that is frame dependent. It is the observers definition of that point in space relative to other factors.
Ibix said:
The emission event is frame-independent, but you can't measure speed with respect to that
Which I guess is what I am saying - but if you are in the same frame of reference as the photons you can measure speed - but since you are not...
It is not the point in space that is frame dependent. It is the observers definition of that point in space relative to other factors.
How do you define a point without reference to other factors?
Which I guess is what I am saying
An event is a point in spacetime. It doesn't have a velocity for you to compare an object's velocity to.
but if you are in the same frame of reference as the photons you can measure speed - but since you are not...
Light doesn't have a reference frame for you to be in, so measuring speed if you were in it doesn't make sense.
#10
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A.T. said:
How did you calculate that 1.5c then?
I am saying that that is the implication of the light traveling away from me at c, when it is traveling away from the point in space that it was created at c. These two things are incompatible and would require Galilei transformation to make it work.
#11
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Ibix said:
How do you define a point without reference to other factors?
The event happens at a point whether or not you can define it
Ibix said:
An event is a point in spacetime. It doesn't have a velocity for you to compare an object's velocity to.
That is correct, but if the event is defined as the emission of photons you can assume a velocity= c
Ibix said:
Light doesn't have a reference frame for you to be in, so measuring speed if you were in it doesn't make sense.
I guess that's still what I am saying.
#12
Quandry
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Just going back to my question, it seems that y'all are saying the answer is yes?
Quandry said:
To clarify this for me, my question is:
If I am traveling towards a light source at .5c and I have two light detectors with a set space between them to measure the time that the light takes to transition from one to the other, will the time taken correspond to a speed of c? And, if I decrease my speed to .25c and do the same measurement will the measurement also indicate a speed of c?
The event happens at a point whether or not you can define it
An event is a point in spacetime. There will come a time when you say that that event is in the past - it is not in the slice of spacetime you call "now". So you cannot measure a spatial distance to the event "now". You can only decide that some point in "now" is the same as the (spatial part of) the event. That choice is frame dependent - in fact, it's one definition of a choice of frame (up to spatial rotation).
Quandry said:
That is correct, but if the event is defined as the emission of photons you can assume a velocity= c
This is pure nonsense. An event is a point in spacetime. Velocity is the slope of a line in spacetime. A point does not have a slope and there is no way to define one for it. Simply "assuming" an undefined quantity has a particular value is meaningless.
Quandry said:
I guess that's still what I am saying.
Then you are talking nonsense. There is no frame of reference for light - it's a contradiction in terms.
Just going back to my question, it seems that y'all are saying the answer is yes?
Quandry said:
To clarify this for me, my question is:
If I am traveling towards a light source at .5c and I have two light detectors with a set space between them to measure the time that the light takes to transition from one to the other, will the time taken correspond to a speed of c? And, if I decrease my speed to .25c and do the same measurement will the measurement also indicate a speed of c?
Everyone will always measure a time between the reception events consistent with the motion of the detectors (as measured in their frame) and the constant speed of light. Between length contraction, time dilation the relativity of simultaneity, and the different motion of the detectors in any other frame, everyone will always be able to come up with a consistent explanation for why everyone else also comes up with the same invariant speed of light.
It's a numerical value. How did you calculate it exactly?
#16
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Ibix said:
This is pure nonsense. An event is a point in spacetime. Velocity is the slope of a line in spacetime. A point does not have a slope and there is no way to define one for it. Simply "assuming" an undefined quantity has a particular value is meaningless.
Your response is not relevant. If the event is the result of turning on a torch (as defined) it is reasonable to assume that the event results in photon emission.
Ibix said:
Then you are talking nonsense. There is no frame of reference for light - it's a contradiction in terms.
I m sorry that you think this is nonsense. I was agreeing with you.
'nuff said!
#17
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A.T. said:
It's a numerical value. How did you calculate it exactly?
In addition to what has been said already, one thing that confuses many people is the difference between separation speed and relative speed.
If you are moving to the left with 0.5c relative to an observer A and a light signal travels to the right, then A will see the distance between you and the light growing with 1.5c. This is separation speed.
The above does not mean that you will see the light moving at 1.5c! To draw that conclusion you must assume absolute space and time, which directly contradict SR. In fact, they are basic assumptions behind the Galilei transformation! The basic assumption in SR is that the light will have speed c in your inertial frame too. This is the relative speed.
Also note that there is no way you can objectively say "I am moving at 0.5c" as velocities are relative and change between inertial frames. You need to specify relative to what you move at 0.5c.
Your response is not relevant. If the event is the result of turning on a torch (as defined) it is reasonable to assume that the event results in photon emission.I m sorry that you think this is nonsense. I was agreeing with you.
In that case your writing is imprecise, because I'm having trouble reading you as agreeing with me even when you say you are. I think I shall duck out of this conversation for now.
There is no implication of adding velocities in this process.
Quandry said:
I added 1 and .5 and I got 1.5.
Hmm...
#22
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Orodruin said:
If you are moving to the left with 0.5c relative to an observer A and a light signal travels to the right, then A will see the distance between you and the light growing with 1.5c. This is separation speed.
Thanks for the term separation speed. It clarifies the confusion caused by my use of relative speed to mean the same thing.
Can I therefore conclude that, if I am the observer, I see observer A moving right at .5 and, since light belongs to no IFR, I see the same light signal moving to the right and a separation speed between them of .5c
Thanks for the term separation speed. It clarifies the confusion caused by my use of relative speed to mean the same thing.
Can I therefore conclude that, if I am the observer, I see observer A moving right at .5 and, since light belongs to no IFR, I see the same light signal moving to the right and a separation speed between them of .5c
Yes, and there's nothing special about light in this specific case. If you see a spaceship approaching from the left at speed ##v### with ##v<c## and another spaceship approaching from the right with speed ##u## (also less than ##c##), you will find that the distance between them is shrinking by ##(u+v)<2c## - that's a separation velocity and it's not the speed of anything in any frame. The right-moving spaceship on the left will consider himself to be at rest while you are moving at speed ##v## and the left-mover is moving at speed ##(u+v)/(1+uv/c^2)## which will always be less than ##c##. Likewise, the left-moving ship on the right will consider himself to be at rest while you are moving at speed ##u## and the right-mover on the left is moving at the same speed ##w##.
#24
Stephanus
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@Quandry welcome to PF Forum,
Perhaps I can help you to understand it.
Quandry said:
I do not have a problem with the concept of the constant speed of light as it has no mass and therefore no inertia and therefore no relationship to any IFR. However it seems to be expressed as constant in all IFR's which I do not understand.
It's hard to quote your questions and answer them at the same time.
So, I'll answer in my own format.
You: This seems to say that if I am traveling at 1/2c and I shine a torch forward the light moved away from me at c and if I shine the light backwards it also travels away from me at c.
You: that if I am traveling at 1/2c.
Answer: With respect to whom? You always have to say in SR that, "I'm traveling at some velocity with respect to something"
So let's just say that you're traveling 1/2c wrt Alice.
You: and I shine a torch forward the light moved away from me at c and if I shine the light backwards it also travels away from me at c.
Answer: Yes!, From your ifr you are at rest. From Alice ifr it's you who is traveling at 1/2c. You are always at rest from your ifr!
You: This seems to say that, in the first case, the light is traveling away from the point in space that it is created (independent of my IFR), at 1.5c.
Answer: No! The light travels at 1c.
You: the light is traveling away from the point in space that it is created (independent of my IFR)
Answer: Yes.
This might be a little confusing. This is where the velocity addition formula comes to mind.
##w = \frac{u+v}{1+uv}##
Let's say you (u) travels at 0.5c and fires a missile 0.4c (v), so from Alice (w) ifr the speed of the missile is
##w = \frac{0.5+0.4}{1+0.5*0.4} = \frac{0.9}{1.2} = 0.75c##
So the speed of the missile in Alice ifr is 0.75c
From yours is (of course) 0.4c
Let's say you (u) travels at 0.5c and shines a light at, of course c or a.
##w = \frac{0.5+1}{1+0.5*1} = \frac{1.5}{1.5} = 1c##
From Alice ifr your light travels at c
from yours, it travels at c.
I see that you already understand that from your responses above.
You: To clarify this for me, my question is:
If I am traveling towards a light source at .5c and I have two light detectors with a set space between them to measure the time that the light takes to transition from one to the other, will the time taken correspond to a speed of c?
Answer: a speed of c? Sure, why not. But it's blue-shifted.
You: And, if I decrease my speed to .25c and do the same measurement will the measurement also indicate a speed of c?
Answer: at a speed of c? Yes, But be careful here, how you do the measurement!
Let's have a detailed thought experiment here.
You (YY) are moving 0.5c to the east, and Alice(AA) from your east is at rest. Y2 is your friend far away from you at east and is comoving with you, that is he travels 0.5c to the east also.
Or Alice will say that it's you who are at rest, and she travels to the west at 0.5c
A2 is Alice friend's who is comoving with her. That is, A2 is traveling to the west at 0.5c
Let's say the distance between YY and Y2 is 1000 light seconds or 300 millions km wrt you!
The distance between A2 and AA is 1000 light seconds wrt you!
Wrt Alice A2 and AA is ##1000/\sqrt{0.75} = 1154.7 \text{ seconds or } 346.4 \text{ millions km}##
But YY and Y2 wrt Alice is ##1000*\sqrt{0.75} = 866.03 \text{ seconds or } 259.8 \text{ millions km}##
Be careful, there will be length contraction and relative simultaneity of event that will come later.
BB is Bob which we will use in later case.
Okay, so here's the picture.
....BB........B2
....YY........Y2
...<------AA........A2
Surely, you will see Alice who is moving toward you at 0.5c to the west and when A2 meets Y2 then A2 shines a light. This light will reach YY 1000 seconds later wrt you, blue shifted. If you don't move 0.25c.
But what if you move 0.25c to the west? There will be two of you here. One who was at rest, and the one who is traveling 0.25c to the west.
So, let's add Bob (BB) who is at rest in your first frame. And now you moves 0.25c to the west wrt Bob. Let's see when YY receives the signal
And let's put B1 at some distance where when YY receives the signal YY will meet B1. ...<---YY.......Y2 ... ......B1...BB.......B3...B2
...<------AA........A2.....
This can be solved by simple algebra.
##0.25T + 1000 = T, T = 1333##
So when YY receives the signal the clock show will 1333. B1 clock not YY clock!
What does YY clock show?
Time dilation.
##\gamma = \frac{1}{\sqrt{1-0.25^2}} = 1.033##
##1333/1.033 = 1291##
YY clock will show 1291 not 1000 not 1333. And the light still travels at c wrt you (YY).
How come that be?
So the only explanation for this situation is this.
Facts that you and Bob both agree
1. When BB meets YY and B2 meets Y2, all of you synchronize your clock.
2. Then A2 meets Y2 and (of course B2) and A2 sends a signal to the west and at that time YY and Y2 moves to the west at 0.25c
3. Now this is tricky. You change frame.
In your new frame, it seems that the light comes further back from Y2.
I think this is rather advanced, although for some people this is very, very bacic. Perhaps you ask about read time dilation, length contraction and relative simultaneity of events.
And the answer about your two detector should come after that. There's a relative simultaneity of events there.
I hope I give the correct answer and the right answer, the answer that you're looking for.
Sincerely
#25
Quandry
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Stephanus said:
Perhaps I can help you to understand it.
Stephanus, thank you for taking the time to provide such a detailed answer. I was going to say complete, but clearly there is more. However, you have answered my question very well.
Now my challenge is to understand why an observer is required, and why the observer appears to require intelligence.
Thank you for getting me past the barrier,
Bill
Now my challenge is to understand why an observer is required, and why the observer appears to require intelligence.
No observer is required and no intelligence is involved. Putting an observer into the picture is just a convenient way of describing a thought experiment, or of visualizing the origin of a reference frame.
#29
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Quandry said:
Stephanus, thank you for taking the time to provide such a detailed answer. I was going to say complete, but clearly there is more. However, you have answered my question very well.
Now my challenge is to understand why an observer is required, and why the observer appears to require intelligence.
Thank you for getting me past the barrier,
Bill
Hm..., slow down buddy. You are trying to understand SR in an instant.
A.T. , Orodruin and Nugatory have given you a very good answers about this so called "observer"
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#30
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I seem to have a turn of phrase that causes confusion.
I do not say that observers are required, or that intelligence is required. I have seen no explanation which does not use a description of this nature and I wonder how this can be described without observers or intelligence or with just a single IFR (which is a bit of an oxymoron) which is the whole universe.
I am not asking this question in this forum (yet), as I need to understand for myself what this means.
Bill
#31
Stephanus
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Perhaps I can help you...
You are moving 0.1c.
So you are moving from whom? To whom? what?
Can you explain what "you are moving 0.1c" mean?
I wonder how this can be described without observers or intelligence
Just replace "observer" with "reference frame", "see" with "is measured in that reference frame" etc.
Quandry said:
or with just a single IFR (which is a bit of an oxymoron) which is the whole universe
In SR there are infinitely many IRFs, and they all contain the whole universe (extend to infinity).
#33
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A.T. said:
In SR there are infinitely many IRFs, and they all contain the whole universe (extend to infinity).
I presume you are referring to Multi Universe, Many Word Interpretation. Not sure that meta-theories should be expressed as assertions, however, I need to come to grips with one universe before tackling infinity.
Many thanks.
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Quandry said:
I presume you are referring to Multi Universe, Many Word Interpretation. Not sure that meta-theories should be expressed as assertions, however, I need to come to grips with one universe before tackling infinity.
Many thanks.
I think not. What A.T. means by many IFRs is not multiverse. They are all in THIS universe. Even though not OBSERVABLE, but they are in THIS universe.
If you travels 100 km / hour to the west, and I travel 100 km/hour to the east. You have your own IFR and I have mine. That's just two of them.
Someone moves 50km/hour to the north would make three.
#35
Quandry
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OK, I thought he was saying that every IFR contained the whole universe - multiple IFR's means multiple universes. So single universe infinite(?) IFR's is good :)
I thought he was saying that every IFR contained the whole universe
Yes, in SR and Newtonian physics. Not in GR.
Quandry said:
multiple IFR's means multiple universes.
No. it doesn't. It just means multiple ways to assign space-time coordinates to events in the same universe.
#37
Quandry
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Thanks guys.
#38
Stephanus
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TIME DILATION
Stephanus said:
..They are all in THIS universe. Even though not OBSERVABLE...
"THIS universe". I do like the idea of multiverse.
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 experiment 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.
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.
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.
#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
#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.
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.
##\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##
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
1
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.
#42
Stephanus
1,316
104
sheshank said:
I think experimental facts can't be manipulated, our reasoning and understanding can.
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.
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.
sheshank said:
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.
for General Relativity, of course, Eddington proved it. 1 year after the release of GR I think.
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
1,316
104
PeterDonis 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.
Thanks for the correction. I will check my source before I post.
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
241
11
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
1,316
104
exmarine said:
Wouldn't these discussions be far more fruitful if they were about "the velocity of a photon" instead of about "the velocity of light"?
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.
Wouldn't these discussions be far more fruitful if they were about "the velocity of a photon" instead of about "the velocity of light"?
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.
Are you sure that the speed of photon is not equal to the speed of light?
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.
Wouldn't these discussions be far more fruitful if they were about "the velocity of a photon" instead of about "the velocity of light"?
Certainly not. Stay away from photons unless you wish to discuss them properly in a QFT setting. Photons are not "little balls of light".
Edit: I see I got beat to it by Peter. The point still stands.
#50
exmarine
241
11
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.
