A general discussion on the relationship between Photons and Time.

[Alias Moniker]

My understanding in a nutshell is that the photon travels so fast that it moves at the relative speed of time itself. This causes the Photon to not experience the passage of time, as demonstrated by the Lorentz equation saying when Velocity is the Speed of Light, Duration or Time = 0. Using the Train analogy, the Photon and Time are each trains moving at equal speeds, so relative to the Photon, Time seems to be standing still. In the local, special relativity sense, Time is the other frame of reference moving at the Speed of Light. The Photon is created, already existing at all points of its existence simultaneously, but for no amount or passing or length of time to any "observer" except time itself. Meanwhile, since the Photon travels with time, it measures itself in length, since it does not experience change relative to Time.

To observe or measure a Photon, first you would need to stop time at the exact moment when the Photon was created and destroyed, and if you were able to discern it from all other light created at that moment, and if you were able to somehow focus on it, it would appear as a solid line touching all points of its path including its origin and destination. The Photon does not have a "speed" because it does not experience the passage of time for its locations to be compared against each other from one moment to the next, its locations could only be compared from one moment to the same moment, because the Photon only gets the one moment to exist. And actually "moment" even implies the passing of time so it's more like the Photon gets 0 moment to exist, whether it travels across a room or across all of existence it only takes, literally, no time at all.

The Photon is the particle of light, and is distinct from the electromagnetic wave of light, which is caused by the Photon disrupting either space from time, or time from itself. So as an analogy, the Photon is lightning and the electromagnetic wave, "light", is thunder, a direct but delayed result of the Photon's passing. Or the Photon is a boat and "light" is the wake of the boat as it passes through water. Since the Photon is instantaneous, it can't be the Photon that takes 8 observable human minutes to travel from the Sun to the Earth. 8 minutes is longer than instantaneous. Instead, the Photon travels this distance at its own unknown pace while everything else existing appears to be paused, and after the Photon is absorbed, Time continues rattling on and the electromagnetic wave follows the path of the already extinct Photon through space and time; where as the Photon only traveled "through" space, and in relationship to time, it traveled "with".

And also, theoretically it isn't necessary to "accelerate a particle" to the speed of light, theoretically you could "reduce a particle's mass to 0".

Edited July 18, 2014 by Alias Moniker

[Klaynos]

Speeds are all relative. You must give them related to something else. You are making two common mistakes, changing your reference frame but not changing the required observations and using a photon frame which is not valid.

 

The invalid frame of a photon has bug repercussions for your whole thread as statements like there being no time are meaningless and unfounded as it is not a valid frame and cannot be discussed in that way.

 

The photon is the electromagnetic wave (assuming a single photon emission). Distinguishing them will probably only lead to confusion.

 

Photons do not accelerate, they travel at c.

 

I don't see how you could change any particles mass to such an extent. It is true that all massless particles travel at c.

[Strange]

 

My understanding in a nutshell is that the photon travels so fast that it moves at the relative speed of time itself.

 

There is no "speed of time". Photons move at the speed of light.

 

 

This causes the Photon to not experience the passage of time, as demonstrated by the Lorentz equation saying when Velocity is the Speed of Light, Duration or Time = 0.

 

The Lorentz transform cannot be applied in this case because you end up dividing by zero. (So much of the rest of your post becomes irrelevant.)

 

 

To observe or measure a Photon, first you would need to stop time at the exact moment when the Photon was created and destroyed

 

We observe and measure photons all the time without stopping time, so I guess this must be wrong.

 

 

The Photon is the particle of light, and is distinct from the electromagnetic wave of light

 

They are different models for exactly the same thing.

 

 

And also, theoretically it isn't necessary to "accelerate a particle" to the speed of light, theoretically you could "reduce a particle's mass to 0".

 

An interesting idea, but it is just as impossible to reduce an object's mass to zero as it is to accelerate it to the speed of light.

[swansont]

I think you may be causing yourself trouble if you try to think of photons and EM waves as separate things. Light is an EM wave. It also has a curious feature that the energy is quantized, which has certain ramifications. As the previous posters have explained, the frame of a photon is not valid — we have no physics that works in that frame. We can't talk about what happens from a photon's perspective.

[Alias Moniker]

Just because you can't talk about what happens from a Photon's perspective doesn't mean that the Photon doesn't exist and travel at the speed of light. Math is just a representation of the natural world and all you're saying is that you haven't yet figured out how to mathematically represent the Photon's point of view. Just think for a second about how much Newtonian Physics changed when it got to Mercury.

Special Relativity: The laws of physics are the same for all observers in uniform motion relative to one another.
We are never able to travel at the speed of light and the Photon is never able to not travel at the speed of light so the laws of physics can't be the same for a Photon as they are for "anything that isn't a Photon".

If the laws of physics require that its participants have a valid frame of reference, and if light does not have a valid frame of reference, how can you use a system that requires its participants to have a valid frame of reference to understand and explain something that exists without having a valid frame of reference?

Also here's a link to an article from 2012 from the Cornell University Library talking about theoretical ways to detect a Photon, can you show me something that says a Photon has been observed and measured and what it's, whatever, dimensions are?
http://arxiv.org/abs/1211.4182

Here's a quote from Wiki on the Photon saying that we still haven't verified it's "mass-less" property: "The photon is currently understood to be strictly massless, but this is an experimental question."

It just doesn't sound like anyone has ever seen one.

Because if something's existence is "instantaneous", or if it's "T=0", it isn't just that you wouldn't be able to represent this with current math, it's that you would never experience the "space without time", in order to exist in the Photon's reality. The Photon is "under dimensional" and instead of existing in 4d "space space space time" (or 5d space, space, space, time, gravity) etc... the Photon only exists in 3d "space space space" or 4d (space, space, space, gravity) etc (the Photon exists in a reality that does not recognize the existence of time). As the observer, you would never "notice" that time stopped, because for you it didn't. You existed in your reality where time is continually ticking by and the Photon existed in its reality where there is no such thing as time, and the Photon's existence... "interrupts" our existence, but for 0 of our time. Theoretically there could be one single Photon providing all of the infinite amount of light in all of OUR existence, streaking through our reality whenever it's needed, even if it needed to be two places at the same of our time(s).

Watch this Neil deGrasse Tyson video:

Edited July 19, 2014 by Alias Moniker

[ajb]

Math is just a representation of the natural world and all you're saying is that you haven't yet figured out how to mathematically represent the Photon's point of view.

What we are saying is that usually by "point of view" one is making an explicit reference to an inertial rest frame of some observer. Mathematically this is picking a natural set of coordinates within the framework of special relativity. The point is that there is no inertial frame for which a photon can be considered at rest.

 

As special relativity is well tested and is inherent in Maxwells equations which describe light classically, we have no reason to think that we can have a rest frame for a photon. This is tied to the fact that photons are massless. (I'll comment more on this in a moment)

 

 

Here's a quote from Wiki on the Photon saying that we still haven't verified it's "mass-less" property: "The photon is currently understood to be strictly massless, but this is an experimental question."

Theoretically, photons are massless. All evidence collected todate is consitent with the photon being massless. However, in any experiment you can never show that something is zero exactly, there are always experimental as well as systematic errors. The best you can do is show that the mass of a photon is very small. The measured mass is small enough to be consistent with zero mass as far as all out observations are conserned. [math]10^{-14} - 10^{-18}eV/c^{2} [/math] are the upper bounds of the mass of a photon as derived from several experiments.

 

It just doesn't sound like anyone has ever seen one.

We have devices that can detect single photons.

 

Because if something's existence is "instantaneous", or if it's "T=0", it isn't just that you wouldn't be able to represent this with current math, it's that you would never experience the "space without time", in order to exist in the Photon's reality. The Photon is "under dimensional" and instead of existing in 4d "space space space time", the Photon only exists in 3d "space space space". As the observer, you would never "notice" that time stopped, because for you it didn't. You existed in your reality where time is continually ticking by and the Photon existed in its reality where there is no such thing as time, and the Photon's existence... "interrupts" our existence, but for 0 of our time. Theoretically there could be one single Photon providing all of the infinite amount of light in all of OUR existence, streaking through our reality whenever it's needed, even if it needed to be two places at the same of our time(s).

We use 4d space-time when describing photons, so I am at a loss as to what you really mean here. It sounds like you are again trying to make sense of the "photons point of view", which is a delacate matter as explained above,

[Alias Moniker]

I'm saying, that the laws of physics are the same for all "observers", and light is not a "valid observer", so why do you assume that the laws of physics are the same for light? Or, why do you assume that the laws of physics are the same for an observer as they are for a non observer? They can be very similar without being the same.

Edited July 19, 2014 by Alias Moniker

[Strange]

I'm saying, that the laws of physics are the same for all "observers", and light is not a "valid observer", so why do you assume that the laws of physics are the same for light? .

 

Because we are the observer of light, not the other way round.

 

 

Or, why do you assume that the laws of physics are the same for an observer as they are for a non observer?

 

What is a non-observer?

 

 

They can be very similar without being the same.

 

What are you trying to say. I assumed at first that you were trying to clarify your understanding of photons and related terminology.

 

Now you seems to be saying that current science is wrong ... or something?

[ajb]

I'm saying, that the laws of physics are the same for all "observers", and light is not a "valid observer", so why do you assume that the laws of physics are the same for light? They can be very similar without being the same.

Light is not considered an inertial observer. More carefully, an inertial observer is really just a special choice of coordinates, that is we pick what we call an inertial frame. This is really just employing the standard coordinates (ct, x, y z) and all such frames are related by a Lorentz transformation. If you have some massive body then you can always find such a frame with the body at the origin. (I assume no gravitational fields here, otherwise you can only do this locally and not set up a global coordinate system). However, you cannot find an inertial frame for which a photon can be considered at rest. Thus a photon is not a valid inertial observer.

 

In relativity you don't really need to treat massive and massless particles differently, but you have to be careful trying to physically interepret things. I am not sure what mathematical understanding you have of special relativity, so I am reluctant to say anything technical other than you should look up the notion of null, space-like and time-like.

[Alias Moniker]

There's a logical contradiction outside of physics:
The laws of physics are the same for all "observers"
Light is "not considered an inertial observer" - Light, is not, an observer.
So the laws of physics are not necessarily the same for an "observer" as they are for "not considered an inertial observer".
The laws of physics are only the same for all "observers"...
Since light is not an observer.
The laws of physics are not necessarily the same for light as they are for you.
Because you're an observer.
But light is "not considered an inertial observer".

What are you trying to say, that current science is omniscient?

Edited July 19, 2014 by Alias Moniker

[Strange]

Are the laws of physics the same if you're not an observer?

 

What do you mean by "not an observer"?

 

But this sounds like that old philosophy joke, "if a tree falls in a forest and there is no one there ..."

[Klaynos]

The laws of physics are the same in all inertial reference frames. The laws include the behaviour of light (Maxwell's equations). If the laws were not the same we would make different measurements than we do, the statement is experimentally consistent.

[ajb]

Are the laws of physics the same if you're not an observer?

Lets stick to special relativity to avoid confusion. If you employ non-inertial coordinates then the laws of physics will look different. You will typically have fictitious forces and the speed of light will almost never be c. But with care you can still calculate what every you want and this will be the same as employing inertial coordinates, just it will get messy and maybe things are hard to interpret, especially if you want to change frames. The physics will not really depend on what coordinates you use. It just happens that for special relativity we have a prefered class of frames with some nice properties, the inertial frames. The statement is something like "the (non-gravitational) laws of physics take the same form in all inertial coordinates".

 

Now, if you employ non-inertial coordinates you are most of the way towards understanding general relativity, but the space-time is still flat. In general relativty we have a stronger statment than in special relativity "all the laws of physics are the same in any frame". Meaning that the physics really does not care what coordinates you use and that there are really no prefered classes of frames.

[Alias Moniker]

"All the laws of physics are the same in any frame"
But I understand that light is not a frame.
So there's no reason to assume that "all the laws of physics are the same in (not a frame)".

Physics wouldn't exist where there is not a frame.
Light is not a frame.

Edited July 19, 2014 by Alias Moniker

[ajb]

"All the laws of physics are the same in any frame"

Yes, nothing meaningful should depend on your choice of coordinates.

 

But I understand that light is not a frame.

You mean that we cannot find an inertial frame of reference such that a photon can be considered at rest.

 

So there's no reason to assume that "all the laws of physics are the same in (not a frame)".

What is "not a frame"?

 

A frame is really a choice of coordinates and that is it.

[Klaynos]

Physics includes a description of how light behaves in any inertial reference frame.

[Alias Moniker]

To whom do the laws of physics apply?
"All observers"
Is light an observer?

"No"
The laws of physics apply to all "observers", but this does not mean that the laws of physics apply to light, which is not an observer.

You apply physics to light because physics is how you understand particles and waves and light is perfectly both of those, a particle and a wave, so it makes sense that physics should apply.
But

Being an "observer" overrules "being a particle and a wave". Since light is not an observer and since the laws of physics only apply to observers you can't accurately apply physics to light even though it is both a particle and a wave because it is still not an observer and physics (as far as you understand it) only applies to all observers. And light is not an observer.

Why do you keep asking me what something that, "isn't an observer" or "doesn't have a frame" is? You are the one telling me that light is not an observer and doesn't have a frame.

Edited July 19, 2014 by Alias Moniker

[Klaynos]

The laws of physics include how light behaves. What is not true is that everything acts the same. Electrons do not behave in the same way to protons which do not behave in the same way to photons. Whilst the laws are the same everywhere they are complicated.

 

This has been experimental observed.

[ajb]

Q: Is light considered to be an "observer?"

A: "No"

That is correct, given what we have already discussed.

 

Q: Do the laws of physics apply to something if it is NOT an "observer"?

A: We only know that the laws of physics apply to something if it IS an observer.

No, we know the physics of photons and these cannot be used to set-up an inertial frame of reference. For a massive particle we can find an inertial frame for which this particle is considered to be at rest, this is the so-called rest frame. An observer (in this context) is really just a choice of coordiates.

 

Reason: The laws of physics apply to all "observers", but this does not mean that the laws of physics apply to light, which is not an observer.

But again, we know the physics of light in all inertial frames and we can extend this to non-inertial frames.

 

Why do you keep asking me what something that, "isn't an observer" or "doesn't have a frame" is? You are the one telling me that light is not an observer and doesn't have a frame.

Once again, a photon is not considered to be an inertial observer in the sense that you cannot find an inertial frame of reference for which the photon can be considered to be at rest. There is no rest frame for a photon and usually by an "observer" we mean a choice of inertial coordinates. Typically you set this up with respect to something "the observer" which could be for example your lab. You pick an inertial frame for which your lab is considered to be at rest.

 

You may pick some other inertial coordinates, for example it is not uncommon in calculations of relativistic scatterings to pick the centre of momentum frame. There is no actual "observer" there, it is a choice of coordinates for your calculations.

 

In a more general setting one could take "observer" to simply mean any choice ot coordinates, but in special relativity we have a special class of "observers".

 

To whom do the laws of physics apply?

"All observers"

The laws of physics apply to all physical entities in the Universe. I will confess I am not really sure what is a physical entity as there are objects in physics that we don't directly see. Anyway, buy observer you should have in mind picking some coordinate system in order to describe the physics.

[Alias Moniker]

Do the laws of physics apply to something if it is not an observer?

Edited July 19, 2014 by Alias Moniker

[ajb]

Do the laws of physics apply to something if it is not an observer?

I think this is an ill-posed question.

 

If you are asking "do the laws of physics apply to photons" then yes. Photons and light are described in physics. Classically we have Maxwell's equations and we know how to describe photons. They do not violate anything in special or general relativity, if that is what you are asking.

[swansont]

We don't have any physics that works in the frame of a photon, i.e. we can't describe physics from the point of view of a photon. But that's not a problem, because we can never be in that frame. We have a pretty good description of how a photon behaves from any frame of reference we are in.

[Alias Moniker]

But that IS a problem, because even though we are never in that frame, the photon is in that frame. That frame of light, that physics can't describe, does exist, and the photon may behave radically different in that frame than it "appears" to, in our frame. The laws of physics may be radically different in that "non-observers frame", than they are in any "observers" frame. The speed of light is on the other side of a horizon where time doesn't exist. It can't be accurate to apply physics that depend on time (an observer's physics) to a particle that doesn't depend on time (light/a non observer).

Edited July 19, 2014 by Alias Moniker

[ajb]

But that IS a problem, because even though we are never in that frame, the photon is in that frame. That frame of light, that physics can't describe, does exist, and the photon may behave radically different in that frame than it "appears" to, in our frame. The laws of physics may be radically different in that "non-observers" frame, than they are in any "observers" frame.

We have not defined what we mean by the photon frame. For sure it is not an inertial coordinate frame. We do however have light-cone coordinates which can be set up so that a light ray moving in a given direction has light-cone velocity zero. A light ray moving in the opposite direction will have light-cone velocity infinity. The key point is that this is not an inertial frame and is not connected to inertial frames via a Lorentz transformation. However we are free to use this coordinate system if we like and people do when considering relativistic collisions and string theory.

[Klaynos]

If nothing can interact with that frame from the valid frames without following the laws we understand then how can we ever test it? If we cannot test out it isn't science of we can test it then it falls within the "laws of physics".