Can we see light if we move our eyes at speed of light?

  • #1
adityakiran18
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Hello,
We know we can not see light.It is only the effect of light that we actually see .
Is it because of the fact that our eyes cant move at a speed of light? or its just basically the nature of light that we can never see it..?
If I move my eyeballs at 'c', would I be able to see light?


Remember: We theoretically assume that movement of eye-balls at the speed of light is possible. The discussion is not about achieving speed of light.,etc.
 
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Answers and Replies

  • #2
joemmonster
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actually, i saw a video on youtube. It was about a camera that can capture the movement of a light through a coke bottle,

so, if you have that kind of eyes, then you might able to see the light just like a glowing object..
 
  • #3
russ_watters
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Welcome to PF!

Of course we can see light! Light is what we see -- that's basically a tautology. We see light that hits our eyes and can't see light that doesn't hit our eyes, so moving doesn't have any impact on whether we can see light that doesn't hit our eyes.

This is like asking if one were to walk next to a river at the same speed as the river flows, would you get wet?
 
  • #4
Dale
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Remember: We theoretically assume that movement of eye-balls at the speed of light is possible.
Under which theory can we theoretically assume that? We cannot assume it under special relativity.
 
  • #5
HallsofIvy
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Hello,
We know we can not see light.It is only the effect of light that we actually see .
Is it because of the fact that our eyes cant move at a speed of light? or its just basically the nature of light that we can never see it..?
What, exactly, do you mean by "see" here? Many would argue that light is the only thing we see because it is the only thing that affects the retinas of or eyes. When we say we see objects we really mean we see images of them in the light- we see the light coming from the objects.

If I move my eyeballs at 'c', would I be able to see light?


Remember: We theoretically assume that movement of eye-balls at the speed of light is possible. The discussion is not about achieving speed of light.,etc.
While we can "theoretically assume" anything we want, even that relativity is NOT true, it makes no sense to assume that relativity is NOT true and then ask "what would relativity say about this situation". And that is what you doing here. What you could do is ask "suppose we are moving at 99.9% the speed of light (relative to some frame of reference), what would we see? And the answer to that is that we would exactly what we would see at any speed relative to that frame of reference- we would see light coming toward us at "c". The speed of light is the one constant speed in relativity.
 
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  • #6
ghwellsjr
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Hello,
We know we can not see light.It is only the effect of light that we actually see .
Is it because of the fact that our eyes cant move at a speed of light? or its just basically the nature of light that we can never see it..?
If I move my eyeballs at 'c', would I be able to see light?


Remember: We theoretically assume that movement of eye-balls at the speed of light is possible. The discussion is not about achieving speed of light.,etc.
It's true that when we emit a flash of light, we cannot see the light as it is traveling away from us. Is that what you are talking about?

If so, then we can put a reflector at some distant point and wait for the light to bounce off of it and return back to us. The net result is that we can see that the light reached the reflector but because of the additional time that it took to return, we can't really say when it hit the reflector. However, Einstein came up with the idea that we can assume that the light hit the reflector half way during its round trip. Is this the issue that you are concerned with?
 
  • #7
someGorilla
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It's true that when we emit a flash of light, we cannot see the light as it is traveling away from us.

Though if light is moving in air, so at <c, you could catch up with it if you run fast enough.
 
  • #8
HallsofIvy
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Though if light is moving in air, so at <c, you could catch up with it if you run fast enough.
And if you can run fast enough, be sure to enter the Olympics!
 
  • #9
ghwellsjr
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Though if light is moving in air, so at <c, you could catch up with it if you run fast enough.
Long before you can run that fast in air, you will burn up.
 
  • #10
OMFG Clown
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actually, i saw a video on youtube. It was about a camera that can capture the movement of a light through a coke bottle,

so, if you have that kind of eyes, then you might able to see the light just like a glowing object..

Sorry could you link that vid?
 
  • #11
adityakiran18
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actually, i saw a video on youtube. It was about a camera that can capture the movement of a light through a coke bottle,

so, if you have that kind of eyes, then you might able to see the light just like a glowing object..

Hello joemmonster, can you send me the link of that video please
 
  • #12
adityakiran18
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Welcome to PF!

Of course we can see light! Light is what we see -- that's basically a tautology. We see light that hits our eyes and can't see light that doesn't hit our eyes, so moving doesn't have any impact on whether we can see light that doesn't hit our eyes.

This is like asking if one were to walk next to a river at the same speed as the river flows, would you get wet?
Hi russ_watters,
Well I partly agree with that..
Now consider the situation below:
I am standing at (0,-2) and facing X-axis....
A light ray is moving along positive X-axis direction and is at the origin(0,0) at present
(i mean its nearby to me)...
....There is no object anywhere...Its just the single ray of light...
Can I still see that..?

Assume linear propogation as I said single light ray...Dont consider huygen's wave or imaginary spheres.....
 
  • #13
adityakiran18
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It's true that when we emit a flash of light, we cannot see the light as it is traveling away from us. Is that what you are talking about?

If so, then we can put a reflector at some distant point and wait for the light to bounce off of it and return back to us. The net result is that we can see that the light reached the reflector but because of the additional time that it took to return, we can't really say when it hit the reflector. However, Einstein came up with the idea that we can assume that the light hit the reflector half way during its round trip. Is this the issue that you are concerned with?


hi ghwellsjr,
Thanx 4 the reply...
Actually I am afraid its not about that....This is the situation I was talking of:
I am standing at (0,-2) and facing X-axis....
A light ray is moving along positive X-axis direction and is at the origin(0,0) at present
(i mean its nearby to me)...
....There is no object anywhere...Its just the single ray of light...
Can I still see that..?

Assume linear propogation as I said single light ray(like a LASER source,,etc)...Dont consider huygen's wave or imaginary spheres.....
 
  • #14
adityakiran18
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What, exactly, do you mean by "see" here? Many would argue that light is the only thing we see because it is the only thing that affects the retinas .............................................that we would exactly what we would see at any speed relative to that frame of reference- we would see light coming toward us at "c". The speed of light is the one constant speed in relativity.



HI HallsofIvy,
Actually thats not exactly my doubt..Yeah,ofcourse wen i say 'assuming speed of 'c', imean 0.99c or somthing like that......

This is the situation I was talking of:
Consider the co-ordinate system...
.I am standing at (0,-2) and facing X-axis....
A light ray is moving along positive X-axis direction and is at the origin(0,0) at present
(i mean its nearby to me)...
....There is no object anywhere to reflect it.etc...Its just the single ray of light...
Can I still see that..?

Assume linear propogation as I said single light ray(like a LASER source,,etc)...Dont consider huygen's wave or imaginary spheres.....
 
  • #15
adityakiran18
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Under which theory can we theoretically assume that? We cannot assume it under special relativity.
When I say moving at 'c',obviously i mean 0.99c...NOt breaking any relativity laws..
 
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  • #16
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Cola bottle + light video together with the corresponding TED talk.

I am standing at (0,-2) and facing X-axis....
A light ray is moving along positive X-axis direction and is at the origin(0,0) at present
(i mean its nearby to me)...
....There is no object anywhere...Its just the single ray of light...
Can I still see that..?
Light is moving away from you? No, you cannot see it.
Why did you ask this in 3 posts in a row?
When I say moving at 'c',obviously i mean 0.99c...NOt breaking any relativity laws..
That is not obvious at all, as it is a completely different situation.
 
  • #17
adityakiran18
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Cola bottle + light video together with the corresponding TED talk.


Light is moving away from you? No, you cannot see it.
Why did you ask this in 3 posts in a row?

That is not obvious at all, as it is a completely different situation.

--------------------------------------------
Well I had 2 reply differently to different answers.So asked in 3 posts..Could hav written one instead...Anyways,....so u say we cant see that light ray...

(yup ..by 'obviously' , I meant it is understood that you cant move your eyeballs at more than 'c'.So it shud b a bit less than 'c'..Thats wat i meant...)

Anyways, thanx 4 d answer bro..
 
  • #18
Dale
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I am standing at (0,-2) and facing X-axis....
A light ray is moving along positive X-axis direction and is at the origin(0,0) at present
(i mean its nearby to me)...
....There is no object anywhere...Its just the single ray of light...
Can I still see that..?
No. You only ever see light that is moving towards you, you can never see light which is moving away from you.
 
  • #19
HallsofIvy
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HI HallsofIvy,
Actually thats not exactly my doubt..Yeah,ofcourse wen i say 'assuming speed of 'c', imean 0.99c or somthing like that......
Then say "near c" !

But you say "0.99c or something like that" relative to what frame of reference. In any case, as several people have said, the speed of light, relative to any frame of reference is "c". How you are moving, relative to anything, is irrelevant to light.

This is the situation I was talking of:
Consider the co-ordinate system...
.I am standing at (0,-2) and facing X-axis....
A light ray is moving along positive X-axis direction and is at the origin(0,0) at present
(i mean its nearby to me)...
....There is no object anywhere to reflect it.etc...Its just the single ray of light...
Can I still see that..?

Assume linear propogation as I said single light ray(like a LASER source,,etc)...Dont consider huygen's wave or imaginary spheres.....
Once again the ONLY light you "see" is light that goes into your eye and to your retina. If the light was " moving along positive X-axis direction" and you are at (-2, 0), it is moving away from you and cannot go into your eye. You cannot see it at all.

You may be thinking of a situation in which a person in front of you is shining a flash light and you see a beam of light going out from the flash light. What you are seeing is light that has reflected of water globules, dust, etc. in the air and has been reflected back to your eyes. If this were done in a perfect vacuum, you would see nothing at all.
 
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  • #20
ghwellsjr
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It's true that when we emit a flash of light, we cannot see the light as it is traveling away from us. Is that what you are talking about?

If so, then we can put a reflector at some distant point and wait for the light to bounce off of it and return back to us. The net result is that we can see that the light reached the reflector but because of the additional time that it took to return, we can't really say when it hit the reflector. However, Einstein came up with the idea that we can assume that the light hit the reflector half way during its round trip. Is this the issue that you are concerned with?

hi ghwellsjr,
Thanx 4 the reply...
Actually I am afraid its not about that....This is the situation I was talking of:
I am standing at (0,-2) and facing X-axis....
A light ray is moving along positive X-axis direction and is at the origin(0,0) at present
(i mean its nearby to me)...
....There is no object anywhere...Its just the single ray of light...
Can I still see that..?

Assume linear propogation as I said single light ray(like a LASER source,,etc)...Dont consider huygen's wave or imaginary spheres.....
Your explanation is very confusing. What does (0,-2) stand for?

What does it mean that you are facing the X-axis?

But as I answered before: "we cannot see the light as it is traveling away from us" and then you said that's not what you are asking about and then you asked if you can see a single ray of light which I presume is either traveling away from you or traveling across your field of view, and it is not hitting any object so that it cannot reflect off of anything, but even if it is traveling toward you, you will not see it unless it enters your eye.

Somehow, I don't think I have answered your question but that is because I don't understand your question. You have to be very precise when you ask a question like this. Assume I'm really stupid and have no idea what you are talking about. Don't take anything for granted.

Please note in the Coke bottle video that they showed a ray of light coming from a laser and they pointed out that the only reason you could see it is because it is reflecting off of particles in the air. At first this was a constant ray of light so you could see it as a steady beam. But then they changed it to a very short pulse so that you could see its progress as if it were a bullet. That's why I used the word "flash of light". You need to do something similar in your explanation. Are you talking about a constant ray of light or a short burst of light?
 
  • #21
Dale
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Whether or not you see something is pretty straight forward: if the light hits your retina you see it. (Assuming no health issues)
 
  • #22
nitsuj
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lol,

All else is "elsewhere"
 
  • #23
adityakiran18
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Your explanation is very confusing. What does (0,-2) stand for?

What does it mean that you are facing the X-axis?

But as I answered before: "we cannot see the light as it is traveling away from us" and then you said that's not what you are asking about and then you asked if you can see a single ray of light which I presume is either traveling away from you or so that it cannot reflect off of anything, but even if it is traveling toward you, you will not see it unless it enters your eye.

Somehow, I don't think I have answered your question but that is because I don't understand your question. You have to be very precise when you ask a question like this. Assume I'm really stupid and have no idea what you are talking about. Don't take anything for granted.

Please note in the Coke bottle video that they showed a ray of light coming from a laser and they pointed out that the only reason you could see it is because it is reflecting off of particles in the air. At first this was a constant ray of light so you could see it as a steady beam. But then they changed it to a very short pulse so that you could see its progress as if it were a bullet. That's why I used the word "flash of light". You need to do something similar in your explanation. Are you talking about a constant ray of light or a short burst of light?

hello george,
As u hav assumed ,the light is traveling across my field of view, and it is not hitting any object ,in my question .And it is a flash of light.
By giving those numbers,I meant the relative positions on the XY-coordinates.

Anyway, I agree that it is difficult to explain the question. The point is that in physics, a simple illustrative diagram makes a perfect sense and explains a whole lot .

I got the answer and my doubt got cleared....
Thanks for your answer man...
-Aditya
 
  • #24
adityakiran18
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Whether or not you see something is pretty straight forward: if the light hits your retina you see it. (Assuming no health issues)

health issues...haha..lol!!

nyway thanx 4 d reply...!!
 
  • #25
Jeronimus
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hi ghwellsjr,
Thanx 4 the reply...
Actually I am afraid its not about that....This is the situation I was talking of:
I am standing at (0,-2) and facing X-axis....
A light ray is moving along positive X-axis direction and is at the origin(0,0) at present
(i mean its nearby to me)...
....There is no object anywhere...Its just the single ray of light...
Can I still see that..?

Assume linear propogation as I said single light ray(like a LASER source,,etc)...Dont consider huygen's wave or imaginary spheres.....

Because i do not know about all the special cases how light behaves when it is not within a vacuum away of gravitational fields i will answer the question only for that special situation in which we assume you are in a perfect vacuum and there is no gravity field present.

So what we have here is the first postulate (some like to call it axiom) of special relativity.
Light will ALWAYS travel at C in a vacuum absent of any gravity.

Therefore, if you hold a flashlight towards any direction. The photons it emits will travel at C away of you. It would be impossible to see those photons unless they hit an object and bounce back. At least that is what one has to assume when he wants to arrive at the formulas Einstein arrived at for special relativity.
My knowledge about QM is too minimal to even attempt a guess on which scale it would change the formulas if photons would not travel as Einstein has imagined them to travel or as he deliberately postulated to "keep things simple enough" for his first shot.

Anyway, if we want to arrive at the formulas Einstein arrived at we HAVE TO consider those idealistic photons all traveling at C or at least "lightbeams" traveling at C within a vacuum away of gravity.

l*gamma = l' with
gamma = sqrt(1 - (v^2/c^2))
as an example of one of the formulas you will arrive at if you follow the two postulates for special relativity strictly.
(There are actually more postulates required but then it gets almost philosophical if we dive that deep)

And here is the clue. YOU yourself, seen from your own reference system DO NOT MOVE at all. You have to consider yourself standing still, and everything else is moving around you.
Your eyeballs move at exactly zero m/s within your reference frame. Remember it is called relativity for a reason. Everything moves relative to each other.

You could see someone ELSE's eyeballs move at CLOSE TO but never reach C, following a photon closely and getting distance to it very very slow.
The mindbending part here is that the eyeballs of the other guy you see traveling close to C, just a tiny bit slower than the photon they follow, will, when seen from that other guy's perspective(reference system) see that photon traveling at C (and so will everyone else in any inertial reference system - i repeat ALL WILL SEE THAT PHOTON TRAVEL AT C no matter which inertial reference system they are observing it from), and this guy or his eyeballs will have to consider himself NOT moving.
If your mind did not blow up yet, continue reading...

This is not easy at all to get your mind around. It took me months. But after drawing this

http://img204.imageshack.us/img204/2425/89825284.jpg [Broken]

representing two observers traveling at v= 0.5c RELATIVE to each other (Both consider themselves not moving, but see each other passing at 0.5c) i finally managed to arrive myself at the formulas, using only the two postulates(or axioms if you prefer).

I now KNOW that Einstein was right first hand, FINALLY!! (As long as the two postulates are valid. It would be "easy" to adjust the formulas if QM told us that photons don't behave as idealistic as assumed in the postulates but nevertheless we should be very close to same formulas still).


Those are the two postulates required to build the formulas for special relativity. This, maths and maybe some geometry will get you the formulas.

1. Light(photons? lightbeams?) in a vacuum away of any gravity field will ALWAYS travel at C in all inertial frames of reference.

2. The laws of physics are the same in all inertial reference systems.



Or as i would say.. if two reference systems A and B are equal in quality, then there is no reason to consider any of the two special.

Whatever an observer a within a ref system A sees in ref system B under certain conditions, an observer b in system B will see the same if same conditions in A are met.

If for example observer a sees a ruler which is 1 meter within b's reference system (not moving relative to b) shrink by 80% to 0.8m, then b will see a different ruler which is 1 meter within a's reference system (not moving relative to a) also shrink by 80% to 0.8m.

Why? Because neither a's reference system nor b's reference system are special.
The laws of physics are the same in both reference systems(2nd postulate).
a sees b moving and considers himself motionless. b sees a moving and considers himself motionless. They move relative to each other at a certain speed. Their relative speed to each other is a common factor which cannot be used to assume either reference system A or B being special in any way.

Start drawing two diagrams representing two reference systems, and follow strictly the postulates like i did. The yellow lines are lightbeams. As you can see, the diagrams are drawn in a way, so llightbeams moving at C are at 45° always (see Minkowski diagrams)




After you understood this, your next stop will be the twin paradoxon most likely. The key to this is that acceleration changes the reference system the accelerating twin is in which creates an asymmetrical situation. One twin being in the same reference system at all times, while the other twin is passing through several reference systems by accelerating at times.
 
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  • #26
ghwellsjr
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Jeronimus, I'm just curious--everyone else orders Einstein's two postulates the other way around, why do you call 1 the "light constancy" postulate and 2 the "laws of physics are the same" postulate?

Also, I don't know how you can get Einstein's special relativity correct with your definition of gamma, both in your text and in your drawing, you have it inverted. Why is that?
 
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  • #27
Jeronimus
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Jeronimus, I'm just curious--everyone else orders Einstein's two postulates the other way around, why do you call 1 the "light constancy" postulate and 2 the "laws of physics are the same" postulate?

Also, I don't know how you can get Einstein's special relativity correct with your definition of gamma, both in your text and in your drawing, you have it inverted. Why is that?

Why was it so important to you to even ask why i switched the two postulates in position like it matters in any imaginable way. I guess it is because my memory is not the best and i just remembered the postulates rather than in which order they were numbered. Really strange question there. So the answer to this would be, i did it just because i happened to remember them in this order.

As for getting gamma wrong. I will have to look at it again, because this is a drawing i did several years ago, using a scanned page for the cubes, open office and some other open office draw thingy for the formulas i don't remember the name for. Yes, i do have all the originals here in case you are suggesting that i copied this of anywhere.
As for getting gamma wrong... I did this long ago as a result of my school teacher (shortly before i was supposed to enter university) asked me doing a project on Einstein's relativity. I broke my head on it, trying to understand it, and never could.
So i asked my physics teacher if he did, and he said no. He only understood how to apply the formulas perfectly well and use the maths.
I never finished the project because it took me several months to come up with this.

This was years ago, and i am only reading occasionally on physics topics out of interest. Since i think to understand relativity quite well, i thought i would answer this question. A pity if i got gamma wrong after so much work, but it should be some stupid minor mistake because the thought process at how i arrived at it is right or so i think. I am looking into it now to see where the supposed error is...
 
  • #28
Jeronimus
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Jeronimus, I'm just curious--everyone else orders Einstein's two postulates the other way around, why do you call 1 the "light constancy" postulate and 2 the "laws of physics are the same" postulate?

Also, I don't know how you can get Einstein's special relativity correct with your definition of gamma, both in your text and in your drawing, you have it inverted. Why is that?

l is the green "rocket" or "ruler" if you want, in system A being at REST

gamma being sqrt(1-(v^2/c^2)) at 0.5c relative speed, it gives me if i calculated correctly, .86... which is the expected factor of the length contraction a ruler or rocket within a reference frame B would be subject to, given that the ruler or rocket was at rest in reference frame A it moves at a speed of 0.5c relative to.

So going by the second postulate i wrote down, that if anything happens with the size or time being "shrinked" or "increased" by some factor, which had to be in order to maintain the light beams traveling at C for all observers seen from an arbitrary inertial reference frame,

l*gamma = l'

...the same thing would happen the other way around in a similar manner with a length l2 (the blue line) which is at rest in system B. So i wrote down,

l2*gamma = l2'

Which again goes in accord with the theory when you enter the values. l2' in system A SHOULD be shrinked by a factor of about 0.86... (gamma) since it has the size of l2 in B when motionless.

This allowed me to solve for gamma and was the hardest part of the thinking process. Writing down in formulas what the second postulate implies.

Maybe they use l and l' differently in the "official" versions of deriving the formula where you read it, or maybe i am completely deluded, But unless you can point me at the exact mistake i made, i am blind at the moment about it. I could check every formula i created but i rather have you point me at what you think is most definitely wrong.
 
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  • #29
ghwellsjr
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I don't want to study your argument. I'm just pointing out that you have taken upon yourself to redefine terms that already have standard definitions in the context of Special Relativity. This can only lead to confusion. Why would you persist along this direction?

Look up Length Contraction in wikipedia. You will see that your formula is backwards from the commonly accepted formula because your definition of gamma is the reciprocal of the commonly accepted definition. It wouldn't be so bad if you used a different English or Greek letter to define what you are calling gamma but to use the same letter in the same context where it already has an accepted meaning is not wise.

The same thing applies regarding the first and second postulates. Why would you want to continue to number them differently than everyone else in the world, starting with Einstein?
 
  • #30
Jeronimus
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I don't want to study your argument. I'm just pointing out that you have taken upon yourself to redefine terms that already have standard definitions in the context of Special Relativity. This can only lead to confusion. Why would you persist along this direction?

Look up Length Contraction in wikipedia. You will see that your formula is backwards from the commonly accepted formula because your definition of gamma is the reciprocal of the commonly accepted definition. It wouldn't be so bad if you used a different English or Greek letter to define what you are calling gamma but to use the same letter in the same context where it already has an accepted meaning is not wise.

The same thing applies regarding the first and second postulates. Why would you want to continue to number them differently than everyone else in the world, starting with Einstein?

You do see that the only difference wikipedia has is naming l, L0, and l' L. I want to believe that standards do not go that far as in telling you what kind of letters to use for the size of lengths as long as you remain within "some" boundaries.
Gamma is not really different from what i arrived at. It's not "inverse" as you say. You thought it is because i use different letters for the length in system A than what wiki did.

Furthermore, we are talking about just two postulates here. TWO. About SR. I sincerely doubt that there is a "standard" for which to name first and which to name second. I would even place a high bet that somewhere out there, there is SOME book by some known scientist who dared to switch the numbering on those two postulates. God forbid.
 
  • #31
ghwellsjr
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You do see that the only difference wikipedia has is naming l, L0, and l' L. I want to believe that standards do not go that far as in telling you what kind of letters to use for the size of lengths as long as you remain within "some" boundaries.
Gamma is not really different from what i arrived at. It's not "inverse" as you say. You thought it is because i use different letters for the length in system A than what wiki did.
Let's see if that's the only difference.

Your equation is:

l*gamma = l'

Now we change l to L0 and l' to L:

L0*gamma = L

Wiki says:

L = L0/gamma

Can't you see that your gamma is the inverse of the wiki gamma?

Then you also defined your gamma as:

gamma = sqrt(1 - (v^2/c^2))

But wiki says the inverse of that.
Furthermore, we are talking about just two postulates here. TWO. About SR. I sincerely doubt that there is a "standard" for which to name first and which to name second. I would even place a high bet that somewhere out there, there is SOME book by some known scientist who dared to switch the numbering on those two postulates. God forbid.
I wouldn't say it's a standard, it's just conventional. I'll take that bet. But you have to do the search. If you can find a book within 24 hours that switches the numbering of Einstein's postulates, I'll give you a dollar. No help from anybody else, please.
 
  • #32
nitsuj
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...that switches the numbering of Einstein's postulates...

They're numbered:bugeye:
 
  • #33
Jeronimus
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7
My derivation is based solely on thought processes trying to extract the right formulas out of the two postulates. I am sorry that it did not turn out an EXACT carbon copy of wikipedia. Maybe wikipedia should change theirs because i explain every thought process. Not like i care, but it seems that SOME people care about every little detail.

Wikipedia also has Lo*sqrt(1-(v^2/c^2))= L which is exactly what my formulas describe. My gamma is better than theirs because it's fully explained and derived. It's THEM which have it inverse. But again i do not care, because it takes just a few steps and we both end up with the same formulas.

This is not an attempt to have the last word. But we are only derailing the thread with this nonsense. I will not reply anymore unless you can point out an error in my derivation. Not being a carbon copy of wikipedia is NOT an error.
 
  • #34
ghwellsjr
Science Advisor
Gold Member
5,122
150
This thread, according to the OP, was answered to his satisfaction before your first post. In your post, you quoted from post #13 where I offered him exactly the same issue you raised in your post and I asked him if that was the issue he was concerned with and he said "no". So why did you bring it up again? Wouldn't that be considered derailing the thread? There was no need to add anything more to this thread. The OP was quite happy when his question finally got answered.

Look, Jeronimus, I'm trying to help you. When I saw that you had inverted gamma, I didn't trust anything else you had to offer with respect to Special Relativity. That's what other people are going to conclude too. Do you want people to discount and reject what you have to offer, just because you derived the formula on your own and couldn't remember the well-established form? That's quite impressive that you did that, and even if your version of gamma is better (which I agree with), we're not going to change the rest of the world. It is, of course, quite arbitrary but trying to change it or hang-on to your definition is, as I said before, not wise.
 
  • #35
joemmonster
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0
 
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