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# Do all photons have equal energy?

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Do all photons have equal amounts of energy in 4 dimensional space-time?

They all travel at the speed of light.

What matters to us is how frequently they cross through our bit of 3 dimensional space?

Edited by derek w

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A photon's energy depends on its wavelength. So two photons can have different energy.

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A photon's energy depends on its wavelength. So two photons can have different energy.

I understand that a photon's energy is related to it's wavelength.

But what I am asking is it's energy related to wavelength because we are only looking at a 3 dimensional cross section of 4 dimensions.

In 4 dimensions photons do not have different energies?

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I understand that a photon's energy is related to it's wavelength.

But what I am asking is it's energy related to wavelength because we are only looking at a 3 dimensional cross section of 4 dimensions.

No. There is no time dependence, so it doesn't matter if you included a fourth dimension.

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I thought all photons were the same. Like electrons are all the same.

Are there different kinds of photons?

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Are there different kinds of photons?

They're all alike, only difference is their frequency. A "blue" photon is the exact same as a "red" one, except for its frequency (or wavelength/energy, but those 3 are all related to eachother).

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For those whose idol is Sheldon Cooper (poor souls): Being really pedantic one could argue that the mere concept of "a photon's energy" doesn't make sense in the first place in "4 dimensional space-time". The answer to the original question would probably still qualify as "no", though.

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I understand that a photon's energy is related to it's wavelength.

But what I am asking is it's energy related to wavelength because we are only looking at a 3 dimensional cross section of 4 dimensions.

In 4 dimensions photons do not have different energies?

I'm not sure what you mean by "in 4 dimensional space-time." Are you talking about four-vectors? If so, then in a sense yes, they all have a magnitude zero four-momentum:

$p^\mu=(E,\mathbf{p})$

with $p^\mu p_\mu = 0$. They also have an associated wave-vector which contains the information about the frequency and wavelength of the light: $k^\mu = (\omega, \mathbf{k})$. Since $p^\mu = \hbar k^\mu$, the magnitude of the wave-vector is also zero: $k^\mu k_\mu =0$.

EDIT: I'm not sure what's going on with the LaTeX, but that </span> thing shouldn't be there.

Edited by ydoaPs
fixed LaTeX
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All photons have an angular momentum of (h),but are massless.

constant speed/circumference of circle=frequency

h x f = energy of photon

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All photons have an angular momentum of (h),but are massless.

constant speed/circumference of circle=frequency

h x f = energy of photon

Pardon?

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The angular momentum is spin, aka intrinsic angular momentum. There's no circle, or anything speeding around it.

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I think he's referencing the idea to unify the wave-particle theory... How do particles move in a 2 dimensional wave in three dimensional space? If they're traveling in a helical path (spinning while moving). Thus, a higher wavelength could be considered to have more energy even though the particle is massless.

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I think he's referencing the idea to unify the wave-particle theory... How do particles move in a 2 dimensional wave in three dimensional space? If they're traveling in a helical path (spinning while moving). Thus, a higher wavelength could be considered to have more energy even though the particle is massless.

There's no helical path. As I said, it's intrinsic angular momentum. There is no physical motion associated with it.

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What gets called "spin" isn't actually spin.

If you calculate the tangential velocity for an electron based on its "spin" it turns out to be faster than light (or so I was told, I never checked the maths).

It's a problem the particle physicists subsequently avoided by calling these properties things like "charm", "colour" and "flavour".

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It's a problem the particle physicists subsequently avoided by calling these properties things like "charm", "colour" and "flavour".

OTOH, not really. The color interaction has nothing to do with actual color (which has no meaning at that scale), and quarks really don't have flavors. It's book-keeping. For charge there's - and +, which can add to zero for a neutral system. For a ternary system, it's more complicated. In the system that was devised, colors have to add up to white, and that model reflects the actual behavior that we see.

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If you have 4 dimensions,then it's possible to create a model of an object/wave that can deposit energy at different wave lengths in 3 dimensional space?

t2 = r2 - x2

There are 2 points when t=0.

(x-r,y=0,z=0,t=0) and (x+r,y=0,z=0,t=0)

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If you have 4 dimensions,then it's possible to create a model of an object/wave that can deposit energy at different wave lengths in 3 dimensional space?

t2 = r2 - x2

There are 2 points when t=0.

(x-r,y=0,z=0,t=0) and (x+r,y=0,z=0,t=0)

Possible, probably, but EM waves don't do that. E x B = S where S is the Poynting vector

IOW, the direction of energy propagation, E and B fields are all mutually perpendicular.

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At point (x=0,y=0,z=0,t-r) would a virtual electron appear?

And at point (x=0,y=0,z=0,t+r) would a virtual positron appear?

Do virtual electrons and positrons appear at anti-nodes of EM waves?

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Do all photons have equal amounts of energy in 4 dimensional space-time?

They all travel at the speed of light.

What matters to us is how frequently they cross through our bit of 3 dimensional space?

The energy of a photon is given by $E=pc$, thus photons with different momentum will have different energy. This is similar to what happen with 1000 kg cars. Their energy is given by $E=(1/2000) p^2$, thus cars with different momentum will have different energy.

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The energy of a photon is given by $E=pc$, thus photons with different momentum will have different energy. This is similar to what happen with 1000 kg cars. Their energy is given by $E=(1/2000) p^2$, thus cars with different momentum will have different energy.

In which way would a photon change it's momentum,it can't change it's velocity and it can't change it,s mass.

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In which way would a photon change it's momentum,it can't change it's velocity and it can't change it,s mass.

A change in frequency, which implies a change in energy and thus momentum. (It's not actually that a photon can change its energy, but that two photons can have different energy)

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A change in frequency, which implies a change in energy and thus momentum. (It's not actually that a photon can change its energy, but that two photons can have different energy)

But you can change a photons frequency without destroying it.

Edited by derek w
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But you can change a photons frequency without destroying it.

No, I don't think you can.

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No, I don't think you can.

You can move towards or away from a source of light.

And a photons frequency would change if it was headed towards or away from a mass(the greater the mass the greater the change).

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You can move towards or away from a source of light.

And a photons frequency would change if it was headed towards or away from a mass(the greater the mass the greater the change).

No, that doesn't count. You have changed your reference frame when you do that; energy is not an invariant quantity.

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