Using light to see the world requires speed of light to be maximum?

[pliu123123]

"Using light to see the world requires speed of light to be maximum?"

As in some posts, one may argue that massive body can move faster than light has nothing at all to do with whether or not we "observe" the universe through light. Or a blind person would have a maximum speed of sound..

But the blind person can simply do experiment himself or know from other people that something traveling at higher speed than sound

And in many illustration in explaining relativity, for example the train with a person in the centre, and the calibration of A ,B ,C clocks, etc..
they require the use of light rays.

But one may argue that after we do experiment through light, we can be confident to describe physical law the same way as we don't use light.

However it sounds not very natural .

Is it a common sense that if we use light to calibrate something, we must set light to have the maximum speed?

 

[russ_watters]

Or a blind person would have a maximum speed of sound..

So you're saying that a blind person would not be able to ride in or hear the Concorde? Why?

 

[pliu123123]

So you're saying that a blind person would not be able to ride in or hear the Concorde? Why?

i follow the idea by this post:
https://www.physicsforums.com/showpost.php?p=2488068&postcount=21"

 

[Dale]

The premise of the OP doesn't make a lot of sense to me. You don't measure the speed of light by more light.

 

[russ_watters]

i follow the idea by this post:
https://www.physicsforums.com/showpost.php?p=2488068&postcount=21"

Ok...you seem to be disagreeing with that poster. Why?

Let me word the issues a little clearer:
1. Just because a blind person uses sound to observe things, it does not mean he/she can't observe things moving faster than the speed of sound.
2. Just because we use light to observe the universe, it does not mean we can't observe things moving faster than the speed of light*.

You seem to disagree with this...if so, why?

*In fact, we do observe some objects "moving" faster than the speed of light due to the expansion of the universe.

 

[pliu123123]

Ok...you seem to be disagreeing with that poster. Why?

Let me word the issues a little clearer:
1. Just because a blind person uses sound to observe things, it does not mean he/she can't observe things moving faster than the speed of sound.
2. Just because we use light to observe the universe, it does not mean we can't observe things moving faster than the speed of light*.

You seem to disagree with this...if so, why?

*In fact, we do observe some objects "moving" faster than the speed of light due to the expansion of the universe.

But as you say, we observe some objects "moving" faster than the speed of light due to the expansion of the universe. will this violate the principle of relativity?

 

[HallsofIvy]

No, it doesn't.

 

[pliu123123]

No, it doesn't.

could you explain more?

 

[George Jones]

could you explain more?

Special relativity is a local, not global, approximation to general relativity.

There is a distinction between coordinate speed and physical speed. Physical speed can only be measured locally. Let any observer measure the speed of a photon that whizzes by in his local neighbourhood. The answer is the same (the speed of light) for all observers. Also, the measured speed of any matter in a local neighbourhood will be less than this (light) speed.

 

[Dale]

Is it a common sense that if we use light to calibrate something, we must set light to have the maximum speed?

Again, we do not use more light to measure the speed of light. Here is a brief description of some of the mechanisms used to measure the speed of light:

http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/measure_c.html

None of them were limited to 299792458 m/s.

 

[pliu123123]

Special relativity is a local, not global, approximation to general relativity.

There is a distinction between coordinate speed and physical speed. Physical speed can only be measured locally. Let any observer measure the speed of a photon that whizzes by in his local neighbourhood. The answer is the same (the speed of light) for all observers. Also, the measured speed of any matter in a local neighbourhood will be less than this (light) speed.

Yes, according to relativity,for any observer, when they measure the speed of light (photon) or photon (light), they get the same result.

One can understand like this:
A and B are moving relative to each other. So they have their own description to natural laws. But there is a natural law that whenever they describe ,they would get the same result.
That is SPEED OF LIGHT.

But here comes my question:

If ,<- just if ,there were something moving faster than speed of light , light cannot be reflected from that object to us, this would violate our daily experience in doing experiment.
On the other hand, when we "see" the thing, it must let light to reflect away from it to enter our reference frame,i.e. it has lower speed than that of light
.
So does it make sense that the reason why speed of light is constant to any obeserver is that we must have all observer the " right " to see and describe all kind of physical phenomena. And there should not be one observer see the object and the other can't see the object?

 

[kcdodd]

Say that we had two forms of "light", instead of one, and each one propagated at different speeds c_1 and c_2, where c_2 > c_1 or vice versa. The second form of light is what our mysterious super-luminal matter would use to travel.

The first principle of relativity states that physics are the same in every inertial reference frame. This means that c_1 and c_2 are measured to be the same in every inertial reference frame. The question is this: can we reconcile two different velocities being the same in every reference frame?

 

[HallsofIvy]

Yes, according to relativity,for any observer, when they measure the speed of light (photon) or photon (light), they get the same result.

One can understand like this:
A and B are moving relative to each other. So they have their own description to natural laws. But there is a natural law that whenever they describe ,they would get the same result.
That is SPEED OF LIGHT.

But here comes my question:

If ,<- just if ,there were something moving faster than speed of light , light cannot be reflected from that object to us, this would violate our daily experience in doing experiment.
On the other hand, when we "see" the thing, it must let light to reflect away from it to enter our reference frame,i.e. it has lower speed than that of light.

How do you arrive at that? If something were coming toward us at greater than the speed of light, any light it emitted would still be seen by us- we wouldn't "see" it until it was past us but we would still see it!

So does it make sense that the reason why speed of light is constant to any obeserver is that we must have all observer the " right " to see and describe all kind of physical phenomena. And there should not be one observer see the object and the other can't see the object?

 

[Dale]

Good point HallsOfIvy, the claim is like saying that we can't hear a gunshot because a bullet goes faster than the speed of sound.

 

[kcdodd]

Ok, so we already have the case of a light-pulse expanding sphere in two reference frames.

$$c^2 t^2 - x^2 = 0$$
$$c^2 t'^{2} - x'^{2} = 0$$

We have two restraints and two unknowns, given we know t and x in one frame. So we already know the solution to keep this invariant:

$$ct' = \gamma (ct - \beta x)$$
$$x' = \gamma (x - \beta ct)$$

which is just a lorentz transformation. Now, suppose we add a second sphere of a second type of light expanding at a different speed. Now we have two additional equations:

$$c_2^2 t^2 - x_2^2 = 0$$
$$c_2^2 t'^{2} - x_2'^{2} = 0$$

We also want this invariant, so we need a new solution. But notice we can just set c_2 = b*c, as a constant multiple. Then this becomes:

$$b^2 c^2 t^2 - x_2^2 = 0$$
$$b^2 c^2 t'^{2} - x_2'^{2} = 0$$

There is the obvious solution where x_2 = b*x. Then we can use the lorentz transformation we already have to keep this invariant.

Now, looking at the differential form just setting x -> dx and t -> dt we have

$$c dt' = \gamma (c dt - \beta dx)$$
$$dx' = \gamma (dx - \beta c dt)$$

and so

$$\frac{dx'}{dt'} = \frac{dx/dt - \beta c}{1 - \frac{\beta}{c}dx/dt}$$

Now, we want to make sure that our second light speed is also invariant, so we need c_2' = c_2, or in other words, b'c = bc. So, pluging this into the velocity equation:

$$b' c = \frac{b c - \beta c}{1 - \beta b c / c} = b c \frac{1 - \beta c/{b c}}{1 - \beta b}$$

So, for b'c = bc we have to have $$\beta c/{b c} = \beta b$$, or $$1/b = b$$. The only way that can happen is for b = 1, which means c_2 = c. There can only be one invariant speed.

Ok, so why does that mean it is a maximum speed, since we could just have non-invariant super-luminal travel. Well, suppose we make the stipulation that any non-fundamental physical action (such as a non-invariant one) has to be made up of fundamental physical actions (invariant ones). This is the first postulate stated again, because we say that the "fundamental laws of physics are invariant to reference frame". Obviously you can not add together two luminal actions and get a super-luminal one. But nothing stops you from adding luminal actions together and get a sub-luminal one.

 

[pliu123123]

How do you arrive at that? If something were coming toward us at greater than the speed of light, any light it emitted would still be seen by us- we wouldn't "see" it until it was past us but we would still see it!

if we think about A and B are separated by a distance and we are A. An object moves towards us(A) at speed greater than the speed of light. we could still see the object but B cannot. This would violate the principle of relativity that all the observer should describe physical laws in the same way or expression. And it seems that it also violates our daily life experience...

How can we remedy this?Thx very much.

 

[russ_watters]

if we think about A and B are separated by a distance and we are A. An object moves towards us(A) at speed greater than the speed of light. we could still see the object but B cannot.

The object is moving from B to A? An object moving from B to A would not be able to be detected by someone at B by shining a light on it, yes.

But that's only one specific scenario - there are many observers who could shine a light on the object and see it (such as one at a location perpendicular to the line between A and B. And, of course, if the object emits light, someone at B could see it.

This would violate the principle of relativity that all the observer should describe physical laws in the same way or expression. And it seems that it also violates our daily life experience... How can we remedy this?Thx very much.

Except that we've never observed such an object, there is nothing particularly profound about this. I see no violations of Relativity here and nothing to "remedy".

 

[pliu123123]

Good point HallsOfIvy, the claim is like saying that we can't hear a gunshot because a bullet goes faster than the speed of sound.

Yes,me thought is that if we are given 3 devices that can emit sound wave at a speed of F,then

think about this game:
The game is played in infinitely dark environment and enough silent room.

but before the referees turn off the light, you are confident that there are three tunnels, situated around you at angle 0,120 and 240 degree, for the referee to give a shot of bullet,in fact not necessarily a bullet , an object moving at V<F is also OK.

When the game starts,the light is turned off,the environment becomes infinitely dark.
And once you are signalled the game started
the referee requires you to detect in which tunnel he has shot out the bullet or object.

Assume the 3 devices are so concentrated to shoot out a beam of sound at speed of F.So that you cannot hear directly the sound given out by the devices, but you can detect the reflection of the sound wave by your ear and whatever ideal conditions

And of course,assume you want to win the game as fast as possible, you place the 3 devices at 0 120 and 240 degree.

Your strategy is to find out in which tunnel you detect a sound , then that's the answer.

But the question is that if the referee wrongly give you the device with sound speed of E, in which E<V. Can we still have a chance to win the game?

 

[russ_watters]

That scenario makes little sense and I don't see a good reason to complicate a straightforward issue by using a longwinded narrative to describe it. In fact, I think by doing so, you are mostly just confusing yourself.

 

[Bararontok]

The speed of light is not maximum in the atmosphere of the Earth because the air's optical resistance decreases the velocity of light. Light is not instantaneous and therefore the light traveling from a distant object will be conveying information from past events and not events occurring in real time. This case is true for all sensory information delivered to the human body because the information must first be converted into electrical energy by the senses and then that electricity will travel through the highly resistive neural network at a speed of 90km/h towards the brain. After the information reaches the brain, the brain must process and record the incoming data. All of these processes take time and thus it is impossible for the human body to experience real time data recording however the time delay caused by the brain, senses and the energy that carries the information is very minimal, usually in the milli to micro seconds range and that is why people do not notice this effect.

 

[Dale]

Yes,me thought is that if we are given 3 devices that can emit sound wave at a speed of F,then ... Can we still have a chance to win the game?

First of all, how is any of this at all relevant to the topic? The point is that the fact that c is the "universal speed limit" has nothing to do with using light to see things. I would encourage you to look at kcdodd's posts which show that there can mathematically only be one invariant speed. Then it is a simple matter of experiment to determine what that invariant speed is, whether it is infinite (corresponding to Galilean relativity) or whether it is finite (corresponding to the Lorentz transform). It has nothing to do with using light in our experiments, and in fact, if the photon were eventually discovered to have some incredibly small mass then the overwhelming experimental evidence for the Lorentz transform would still remain intact and the modifications required for special relativity would be very minor.

However, to answer your question, if an object travels at some velocity which is greater than the speed of the waves used to detect the object then:
1) if the object is emitting the waves then they can be detected regardless of the direction of travel
2) if the object is not emitting the waves but only reflecting the waves then question depends on the geometry of the emitter, detector, and the reflector. In general for some position of emitter and detector the reflector will be detectable regardless of velocity.

 

[pliu123123]

That scenario makes little sense and I don't see a good reason to complicate a straightforward issue by using a longwinded narrative to describe it. In fact, I think by doing so, you are mostly just confusing yourself.

Dear russ_watters, actually i want to learn to use type in the mathematical formula but i don't know how to use Latex. may i know where can i learn it,thx very much

 

[russ_watters]

I googled and found a tutorial: http://www.maths.tcd.ie/~dwilkins/LaTeXPrimer/

I don't see how there is much math complicated math to this issue, though. Really, it's more a diagram that I think you need. Or just to think about what (and when!) a supersonic airplane sounds like.

 

[pliu123123]

I googled and found a tutorial: http://www.maths.tcd.ie/~dwilkins/LaTeXPrimer/

I don't see how there is much math complicated math to this issue, though. Really, it's more a diagram that I think you need. Or just to think about what (and when!) a supersonic airplane sounds like.

thanks , russ_watters.i hope i can express idea more clearly in the future using LATEX
THX~