Photons have no time

[Boerseun]

However, I do believe in a spiritual force that can INFLUENCE the physical. So these tiny corrections can be spiritually manipulated.

This may surprise you about what I write here, but my LIFE experiences have proven this to me.

Well, there we go. There's no arguing with this.

 

New Science, I think you're at the wrong forum. Hypo is about empirical science. Your personal experiences are nothing we can discuss with you, or test or experiment upon. And I don't care what you take as valid or invalid, if it's not repeatable, it's not science. Nobody will have the same experiences as you, so nobody will know what the hell you're on about.

 

Cheers.

[InfiniteNow]

These thought experiments are a departure from reality IMO.

 

However, I do believe in a spiritual force that can INFLUENCE the physical. So these tiny corrections can be spiritually manipulated.

This may surprise you about what I write here, but my LIFE experiences have proven this to me.

I'm not even going to touch the juxtiposition of your comments about thought experiements departing from reality, but spiritual forices being okay... I think I understand your stance, but it's more a semantic issue now. You seem to be confusing your own subjectively biased experiences as empirical evidence itself. You appear to be treating them as some greater form of proof which out weighs predictable evidence... evidence which is objectively reproducable by others. In the empirical minds of most of us here, your own subjective interpretations hold little merit unless supported by anything more than your own faith, hence your continued sense that you must defend yourself from attack.

 

 

So, I kinda ignore these deparures from our realistic environment and prefer to stick with the Earth, Moon and the Sun clocks as our standards.

The other trivialities are ignored by me,

Of course, I can accept the current 'cesium' clocks as the current standad.

What you cite above has no relavence to by brand of Cosmology.

I suppose we should also ignore your continued departures from relevant posts to this thread. You're simply preaching, not teaching.

 

I am a 'free thinker' and refuse to be regimented by the 'power' science establishment.

That's good. The problem is, your freedom frequently comes across as inaccuracy and lack of understanding. Thinking of new concepts and ideas is not the problem. Doing so in the face of conflicting evidence is.

 

 

If Copernicus didn't correct the churches power science in the past, our modern science today would not exist.

We would still be accepting the 'geocentric' theory that the current BBU is based on with some major modifications.

 

Two things. One: I don't think Copernicus had the bloated uninformed ego you do. Two: I don't think any of his work had direct relevance to the concept of a photon having no time from it's own perspective, which is the topic of this thread.

 

 

:lol:

 

So, the idea that photons have no time due to the effects of time dilation appears to be the consensus.

 

As an object approaches the speed of light, it experiences increasing dilation of time relative to outside observers. Per our current understanding of this process, when an object reaches the speed of light its time dilation is infinite. This tends to be interpreted that photons do not experience passage of time, and hence is everywhere at once (per our limited human interpretations where we try to understand things from our own perspectives). Further, no object except a massless particle can travel at the speed of light. Anything with mass would take larger and larger amounts of energy to continually accelerate, and as speeds approach the speed of light, the amount of energy required to continue accelerating becomes greater than the amount of energy available in the universe. How am I doing?

 

This is fun. :cup:

[Mike C]

As an object approaches the speed of light, it experiences increasing dilation of time relative to outside observers. Per our current understanding of this process, when an object reaches the speed of light its time dilation is infinite. This tends to be interpreted that photons do not experience passage of time, and hence is everywhere at once (per our limited human interpretations where we try to understand things from our own perspectives). Further, no object except a massless particle can travel at the speed of light. Anything with mass would take larger and larger amounts of energy to continually accelerate, and as speeds approach the speed of light, the amount of energy required to continue accelerating becomes greater than the amount of energy available in the universe. How am I doing?

 

Your description above has raised an interesting question?

Since photons have zero rest mass, what is their mass at the 'velocity of light'?

Sorry about the departure from the format.

 

NS

[CraigD]

Your description above has raised an interesting question?

Since photons have zero rest mass, what is their mass at the 'velocity of light'?

This is indeed an interesting question. In my college days (ca. 1980) most people didn’t get the chance to encounter it until their first college course in Modern Physics, typically in their 3rd or 4th undergraduate year. It’s not a complicated derivation, however, so, via the wonder of the internet in general and hypography in particular, here it is:

 

The momentum of a photon is given by [math]p = \frac{h f}{c}[/math]

where [math]h[/math] is Planck’s constant, about [math]6.6 \times 10^{-34} \, \mbox{kg m^2 / s^2}[/math],

[math]f[/math] is the frequency of the photon - about [math]6 \times 10^{14} \, \mbox{cycles / s}[/math] for one of green visible light,

and [math]c[/math] is the speed of light, about [math]3\times 10^8\, \mbox{m / s}[/math].

 

Momentum is defined as [math]p = m v[/math]

where [math]v[/math] is velocity – in the case of the photon, [math]v=c[/math]

 

So [math]m = \frac{p}{v} = \frac{hf}{v c} = \frac{hf}{c^2}[/math].

 

Using the values above, then the mass of a single photon of visible green light when it is, as always, traveling at [math]c[/math], is about [math]4 \times 10^{-36} \, \mbox{kg}[/math] (if I haven’t made a mistake in my hurried work).

[Mike C]

This is indeed an interesting question. In my college days (ca. 1980) most people didn’t get the chance to encounter it until their first college course in Modern Physics, typically in their 3rd or 4th undergraduate year. It’s not a complicated derivation, however, so, via the wonder of the internet in general and hypography in particular, here it is:

 

The momentum of a photon is given by [math]p = \frac{h f}{c}[/math]

where [math]h[/math] is Planck’s constant, about [math]6.6 \times 10^{-34} \, \mbox{kg m^2 / s^2}[/math],

[math]f[/math] is the frequency of the photon - about [math]6 \times 10^{14} \, \mbox{cycles / s}[/math] for one of green visible light,

and [math]c[/math] is the speed of light, about [math]3\times 10^8\, \mbox{m / s}[/math].

 

Momentum is defined as [math]p = m v[/math]

where [math]v[/math] is velocity – in the case of the photon, [math]v=c[/math]

 

So [math]m = \frac{p}{v} = \frac{hf}{v c} = \frac{hf}{c^2}[/math].

 

Using the values above, then the mass of a single photon of visible green light when it is, as always, traveling at [math]c[/math], is about [math]4 \times 10^{-36} \, \mbox{kg}[/math] (if I haven’t made a mistake in my hurried work).

 

Your data above has a major flaw that is easily noted.

 

h = 6.6 ^ - 34 is 'Joules per second' and does not involve mass and seconds is NOT squared.

 

NS

[Qfwfq]

One joule is one metre times one newton.

 

One newton is one kg times one metre/(second squared).

[Jay-qu]

Um NS how about you check your own data before accusing other people that theirs it wrong.

 

The units of h are [math]m^2.kgs^{-1}[/math]

 

so when h is multiplied by frequency units [math]s^{-1}[/math] and divided by c squared units [math]m^2.s^{-2}[/math] it all cancels nicely leaving kg :)

[InfiniteNow]

Your data above has a major flaw that is easily noted.

 

h = 6.6 ^ - 34 is 'Joules per second' and does not involve mass and seconds is NOT squared.

In addition to the fallibility in your attack regarding units addressed above by Q & J, you also have again written the value of [math]h[/math] incorrectly.

 

What you meant was [math]h = 6.6 \times 10^{-34}[/math].

 

 

The value you posted of [math]6.6^{-34}[/math] equates to [math]1.366 \times 10^{-28}[/math], and we know that is not the proper value of [math]h[/math]

[EStein]

The problem we get into is imagining that a photon has a point of view. It hasn't. Before it's detected, it doesn't even exist.

[TheFaithfulStone]

God.

 

TFS

[exasperation or explanation? make of it what you will.]

[EStein]

God.

 

TFS

[exasperation or explanation? make of it what you will.]

 

We may know a photon's mass yet not know it's position...and visa versa. It may more properly be said to have 'mass equivilance'. All subatomic "particles" have weirdness in common. We, at no time, are allowed to know all of their attributes. In fact, knowing one attribute implies an automatic error on the other. Thank You, Jay-qu for correctly stating the math.

[TheFaithfulStone]

I suppose that's what I get for attempting subtlety on the internet.

 

tfs

[EStein]

Using the values above, then the mass of a single photon of visible green light when it is, as always, traveling at [math]c[/math], is about [math]4 \times 10^{-36} \, \mbox{kg}[/math] (if I haven’t made a mistake in my hurried work).

 

The following is from NASA's "Ask A Physicist"

Mass of the Photon

 

 

No, photons do not have mass, but they do have momentum. The proper, general equation to use is E^2 = m^2c^4 + p^2c^2 So in the case of a photon, m=0 so E = pc or p = E/c. On the other hand, for a particle with mass m at rest (i.e., p = 0), you get back the famous E = mc^2.

[arkain101]

Through these conversions, I derived a geometric fact and equation that demonstrates Energy (in the form of EMR) has 2-vector momentum.

 

 

Now, you will have to bare with me as it is iffy mathamatics.

 

However, the focus point is to contemplate circular motion / energy / momentum converted into square geometry.

 

What the findings were is that at each of the 4 turns (we assume they are instantanious) there is and must be an equal opposite force supplied from the outside of the (square) orbiting mass or particle.

 

As such each corner of this system has 2 times the total energy of the moving object in order to sustain the orbiting system.

 

The end result is an 8 times system.

 

A statement that can be made is that, the same force or energy is involved to turn an object 90 degrees as it is to stop the object.

 

Thus the square orbiting system thus contains a two vector unit. Which is to say, energy always has two momentum vectors.

 

E^2 = m^2c^4 + p^2c^2

 

In this equation, I point out that p^2c^2 gives support to this thought that a photon is a 2 dimensional and/or 2-vector unit.

[arkain101]

These are some old pictures of some works I had been doing at home.

 

Hopefully the image can be seen clearly enough to help demonstrate my points.

 

I am prepared to hear my amateur follies! :omg:

[EStein]

Now, you will have to bare with me.....

 

You mean we have to get naked?

[arkain101]

Why would you ask such a silly question?

 

Who does physics with cloaths on?