Can Humans See Bodies Traveling at the Speed of Light?

[Sana]

Dear Sir,
Can anyone inform me and discuss about it that"If a body travels or comes with speed of light so can it'll be visible for human, can human see that body"? or simple "Can human see the body if it come with speed of light"?
All comments r welcomed

Regards

Sana

 

[Reflector]

Can't you speak english properly? I couldn't understand half of what you asked. And who are you addressing 'Dear Sir to?'...

 

[Nenad]

I think I understang your question (barelly). This is a paradox that Einstein had when he was thinking about light. If someone travels at the speed of light, and they hold out a mirror in front of them, can they see their image in the mirror? The answer is YES. The speed of light in a tricky subject. If you do the relativistic addition of welocity, you will get the ruslt that the light emmited out towards the mirror by your body is c, or the speed of light, so the person travellong at the speed of light in a spaceship wouldent even know it. This following is a calculation of the result.

u' = v + u / (1 + vu/c^2)
u' = 1c + 1c / (1 + (1c)(1c)/c^2)
u' = 2c / (1 + 1)
u' = 1c

 

[quantumdude]

Can't you speak english properly? I couldn't understand half of what you asked. And who are you addressing 'Dear Sir to?'...

If you don't understand the question, then don't bother with it.

Can anyone inform me and discuss about it that"If a body travels or comes with speed of light so can it'll be visible for human, can human see that body"? or simple "Can human see the body if it come with speed of light"?
All comments r welcomed

Sana,

I've moved your post from the Quantum forum to the Relativity forum, because that's what this is really about. For those who don't understand, what Sana is asking about is this.

Consider a body that radiates in a part of the EM spectrum that is not visible (say, beyond either infrared or ultraviolet). The question is: Is it possible for the body to move at such a speed that the radiation is Doppler shifted to the visible spectrum? That is, is it possible that a body which is not visible to the naked eye can move so fast as to become visible?

The answer is: Yes.

Check out the following page from hyperphysics:

Relativistic Doppler Shift Calculator

It states the mathematical formulae, and has a neat little calculator that will spit out the Doppler-shifted results for you.

So, for instance, if you want to see how fast you would have to go for a red object (l_source=700*10^-9nm) to disappear into the infrared (l_observed>700*10^-9nm), then you just have to input those two parameters into the appropriate places and left-click outside the fields. The calculator will fill in all the other fields for you.

 

[Sana]

Question is simply:
Can we see a body which travels with speed of light?

Sana

 

[quantumdude]

Question is simply:
Can we see a body which travels with speed of light?

Oh, I guess Nenad was right.

The answer is: Sure you can. They're called photons!

 

[selfAdjoint]

Oh, I guess Nenad was right.

The answer is: Sure you can. They're called photons!

Up in Mkaku forum I answered Sana's question no. Our eyes interact with scattered photons to enable us to see other things, but we don't see the photons themselves.

 

[j8hart]

When a photon arrives at your "eye" you "see" it.

However nothing can tell you it's coming, since nothing can out pace it.

If there was such a thing as a body that shines and also travels towards us at the speed of light we would not be able to see it coming.

Is that what you were asking?

 

[quantumdude]

Up in Mkaku forum I answered Sana's question no. Our eyes interact with scattered photons to enable us to see other things, but we don't see the photons themselves.

That's one way to look at it. Another way to look at it would be to say that photons are the only thing that anyone ever really sees. But I don't think that's the issue so much as whether or not an object moving at c is detectable. And photons clearly are, else you wouldn't see the things that they scatter off of.

 

[reilly]

Question is simply:
Can we see a body which travels with speed of light?

Sana

It all depends. If the body is traveling toward you, the energy of the radiation/photons will burn you to a crisp - "blue shift",the relativistic doppler shift will give frequency -> infinity as v->c. A body going away from the observer at c will be invisible, due to the red shift.

Yes, we see via photons, but only ones in a limited part of the frequency spectrum, one that does not include 0 or abitrarily large frequencies.
Regards,
Reilly Atkinson

 

[Reflector]

Question is simply:
Can we see a body which travels with speed of light?

Sana

Oh I see. Sure you can. Photons travel at the speed of light and you can see them...

 

[sal]

This is a paradox that Einstein had when he was thinking about light. If someone travels at the speed of light, and they hold out a mirror in front of them, can they see their image in the mirror? The answer is YES.

There's a problem, though, which is anything traveling at C is moving on a null geodesic, and time doesn't pass for that object. Therefore, if you were traveling at C, you could not see yourself in a mirror, nor do anything else, because your "clock" would stop.

If you do the relativistic addition of welocity, you will get the ruslt that the light emmited out towards the mirror by your body is c, or the speed of light, so the person travellong at the speed of light in a spaceship wouldn't even know it. This following is a calculation of the result.

But your calculation used the composition of velocities rule, which is derived from the Lorentz transforms, which are not valid for an observer moving at C. There's a divide by zero in there someplace if v=c, so you can't really conclude anything about how fast a photon would appear to travel from the PoV of another photon using CoV -- and that's really what you're trying to do here.

To put it another way, if one photon is following another one, how fast does the one behind see the one in front pulling away? It'll be something like

dx/dtau

where 'x' is the distance between them, and 'tau' is the photon's proper time. But dx is zero (the distance between them doesn't change) and dtau is zero, because "proper time" doesn't pass for a photon -- so it's 0/0.

In other words, the answer, according to SR, is undefined.

 

[Nenad]

Im preety shure it is defined. Most physics books will tell you that you will see your image when traveling at the speed of light. And in my calculation, there is no division by 0. Ill do it again for you in latex form.

$$u' = \frac {v + u} {1 + vu/c^2}$$

$$u' = \frac {1c + 1c} {1 + (1c)(1c)/c^2}$$

$$u' = \frac {2c} {1 + 1}$$

$$u' = 1c$$

 

[Tom McCurdy]

Are photons pure energy? Well I supose everything is pure energy if your a string person, but I mean anything with mass can't go the speed of light so if you were continuingly pushing something it turns into energy so what determins what it would be if it were to travel at c?

 

[sal]

Im preety shure it is defined.

Show me a quote from any physics text indicating that velocity composition is well-defined for a frame of reference moving at C.

Most physics books will tell you that you will see your image when traveling at the speed of light. And in my calculation, there is no division by 0. Ill do it again for you in latex form.

$$u' = \frac {v + u} {1 + vu/c^2}$$

$$u' = \frac {1c + 1c} {1 + (1c)(1c)/c^2}$$

$$u' = \frac {2c} {1 + 1}$$

$$u' = 1c$$

Perhaps I was not clear.

The formula you used, composition of velocities, is a consequence of the Lorentz transform.

The Lorentz transformation is not valid at C -- you cannot use it to shift your point of view to an FoR traveling at C. The transform incorporates gamma explicitly, and gamma->infinity as v->c.

In consequence, nothing derived from the Lorentz transform -- including the CoV rule -- is valid for v=c. The derivation of the composition of velocities law involves a divide-by-zero if one frame is moving at C, so the law itself is not valid for that case.

As to "most" physics books claiming you can see yourself in a mirror while traveling at C -- could you give an example of one? Since physics texts which treat relativity universally conclude that you can't travel at C, it would seem that any claim based on relativity theory about what you could do if you did travel at C must be vacuous, wouldn't you think?

In other words, you can't use a theory that says "X is impossible" to determine the behavior of "X" since, by definition, it's outside the domain of applicability of the theory.

The one thing SR does predict about travel at the speed of light is that proper time doesn't pass in that case. This prediction has consequences. In particular, massless particles cannot decay. Another consequence is that something traveling at C can't "do" anything -- photons are "frozen" except when they interact with other particles.

As an example of a consequence of the statement "time doesn't pass when you're moving at C", neutrino oscillations imply that nuetrinos have mass, because otherwise they would necessarily travel at C, and in that case they couldn't oscillate because something traveling at C can't "do" anything.

So, if you were a massless particle traveling at C, you could be hit by a mirror. But you couldn't look at the mirror, because that act of "looking" would imply some time had passed for you. Acts imply duration, and there isn't any when traveling at C.

 

[pmb_phy]

Dear Sir,
Can anyone inform me and discuss about it that"If a body travels or comes with speed of light so can it'll be visible for human, can human see that body"? or simple "Can human see the body if it come with speed of light"?
All comments r welcomed

Regards

Sana

There sure were a lot of answers to this very simple question. The problem is that you've assumed that a body, i.e. something which can emit or reflect photons, can travel at the speed of light. Since relativity says that is impossible the the question itself has no meaning. That was one of the more important conclusions Einstein's theory predicts.

Pete

 

[Neo]

Hi Sal

Does this imply that photons, presuming they were created in the process of
Big Bang, are as old as the universe? "Immortalized photons?" Would this imply, do you think, that the usable energy of the universe would never run out and that photons cannot undergo entropy? Does this idea of "eternal photons" dismiss the possibility of "heat death?"

 

[sal]

Hi Sal

Does this imply that photons, presuming they were created in the process of
Big Bang, are as old as the universe? "Immortalized photons?" Would this imply, do you think, that the usable energy of the universe would never run out and that photons cannot undergo entropy? Does this idea of "eternal photons" dismiss the possibility of "heat death?"

No, not at all -- photons can interact with other particles, and they can be absorbed; they just can't do anything spontaneously.

All it means is photons can't decay without the mediation of another particle, and nobody ever suggested they did that anyway, as far as I know.

The neutrino issue, on the other hand, is real and significant -- neutrinos were suspected of being massless and traveling at C, like photons. But determining that they oscillate -- or, in other words, they decay into other types of neutrinos -- shows that they can't be massless. If they were massless, then the neutrino half life would necessarily be infinite.

 

[Neo]

No, not at all -- photons can interact with other particles, and they can be absorbed; they just can't do anything spontaneously.

Interesting...spontaneous processes require free energy. So does that imply that photons have no free energy? Or that they cannot use their free energy?

 

[sal]

No, not at all -- photons can interact with other particles, and they can be absorbed; they just can't do anything spontaneously.

Interesting...spontaneous processes require free energy. So does that imply that photons have no free energy? Or that they cannot use their free energy?

Dunno. I'm out of my depth once we step off the pier into the waters of thermodynamics.

 

[Neo]

Dunno. I'm out of my depth once we step off the pier into the waters of thermodynamics.

(delta G) = H - T(deltaS)

where deltaG is the free energy change, H is the enthalpy of reaction, T is temp, and deltaS is the change in entropy.

So you are saying that photons cannot do anything spontaneous, which means:

deltaG is nonnegative for photonic reactions?
deltaG > 0

For photon reaction, H - T(deltaS) > 0

What could we say about entropy for such a reaction? Theoretically, if a photon does not interact with any other particles, then the entropy change would be zero.

So in this case:

H - 0 > 0

So H must be positive, which means that the reaction must be endothermic. What does this tell us--anything??

Does this have any relation to photoelectric effect? Basically the photoionization of metallic atoms...but what exactly happens in terms of enthalpy of reaction during photoelectric effect?




By the way Sal, I'm new here...where are you a student and what year? I'm studying biology right now at Harvard and I'm in my second year. I'm an aspiring pre-med and conduct cancer research but have deep interest in theoretical physics, as well.

 

[sal]

By the way Sal, I'm new here...where are you a student and what year? I'm studying biology right now at Harvard and I'm in my second year. I'm an aspiring pre-med and conduct cancer research but have deep interest in theoretical physics, as well.

When I took thermo, the text used had some problems. Part way into it, I ran into a proof of one of the fundamental theorems that I just could not fathom -- as far as I could see the proof was invalid, and I didn't see why the theorem should be true. At the next lecture, I stuck my hand up, and asked the prof about it.

That was when I found out he was almost as deaf as a rock.

So as I sat there in the front row of a hall filled with about 300 students, the prof cupped his ear, and walked slowly over to my seat, cupping his ear the whole way, and leaned over so I could shout in his ear. I did that, and he went back to the board, and answered ... the wrong question. He still hadn't heard me after all.

Some time after that I dropped the course, and never did get around to learning the subject.

That was, oh, about ... 30 years ago, I guess.

(delta G) = H - T(deltaS)

where deltaG is the free energy change, H is the enthalpy of reaction, T is temp, and deltaS is the change in entropy.

So you are saying that photons cannot do anything spontaneous, which means:

deltaG is nonnegative for photonic reactions?
deltaG > 0

I don't see that...?

First, since thermo is intrinsically statistical, I'm not sure how to apply it to a single particle. But more specifically, why should a reaction that must involve more than one particle necessarily result in an increase in free energy?

Temperature also would seem to be somewhat ill-defined for a photon, or maybe my ignorance is just showing.

For photon reaction, H - T(deltaS) > 0

What could we say about entropy for such a reaction? Theoretically, if a photon does not interact with any other particles, then the entropy change would be zero.

Well, in any situation where nothing interacts with anything, the entropy change is likely to be zero, isn't it?

So in this case:

H - 0 > 0

So H must be positive, which means that the reaction must be endothermic. What does this tell us--anything??

Sure doesn't tell me anything.

Does this have any relation to photoelectric effect? Basically the photoionization of metallic atoms...but what exactly happens in terms of enthalpy of reaction during photoelectric effect?

 

[Neo]

increase[/I] in free energy?

Temperature also would seem to be somewhat ill-defined for a photon, or maybe my ignorance is just showing.

Here is the slightly revised version:

(delta G) = H - T(deltaS)
where deltaG is the free energy change, H is the enthalpy of reaction, T is temp, and deltaS is the change in entropy.
No spontaneity:
deltaG is negative for photonic reactions; deltaG < 0
For photon reaction, H - T(deltaS) < 0
Entropy change is zero; So in this case: H - 0 < 0
Therefore: H<0, which means that all reactions of photons are exothermic insofar as they don't interact with other particles?

 

[Neo]

Perhaps you could compare the situation with the speed of sound, for example. If you're traveling at sonic speed, you will not hear sound waves traveling "behind" you as they are trying to "catch up" with you. They have to "wash over" you in order for you to be able to hear them, so-to-speak.

Perhaps due to the consequences of luminal speed in vacuo, it is impossible to feel the effects of time because you're traveling "as fast as" or "faster" than time itself.

The closer something is moving to light speed, the "slower" time is moving for it. So, theoretically, if you're moving at light speed, time is infinitesimally slow, correct?

 

[sal]

The closer something is moving to light speed, the "slower" time is moving for it. So, theoretically, if you're moving at light speed, time is infinitesimally slow, correct?

No. At lightspeed, time's rate is identically zero. It's not infinitesimal. This isn't just a hypothetical, "what if" statement; massless particles actually move at C, and for them, dtau/dt = 0, where t is time in any Lorentz frame.

This is why photons have no 4-velocity, and have no "frame of reference" of their own. The "timelike" component in the 4-velocity is dt/dtau and it's undefined, since dtau is zero.

Infinitesimal values aren't well defined, in any case, except in hyperreal analysis.
In other situations they're just shorthand for a limit process.

 

[Neo]

No. At lightspeed, time's rate is identically zero. It's not infinitesimal. This isn't just a hypothetical, "what if" statement; massless particles actually move at C, and for them, dtau/dt = 0, where t is time in any Lorentz frame.

So, during the Big Bang, when space was expanding explosively (probably at light speed?) -- what was time doing? Was time at a standstill during the explosive growth of the three spatial dimensions? Or was time expanding explosively since it's in a continuum with space?

 

[Neo]

Are electron clouds formed because the electrons are traveling near the speed of light?

If so, why is a "cloud" formed --- a colloid, that is. If it is possible to form a more uncommon type of matter like a colloid by traveling near the speed of light, could you possibly form dark matter by traveling near the speed of light?

Normally, you think of "normal," baryonic matter as being formed when one travels near light speed (arbitrarily say 99.99% C). But what if it were possible that dark matter formed as a result of this acceleration toward C?