Jump to content

Recommended Posts

Posted

I have a new theory

 

My theory is that I will never reach the intellectual heights of people like Newton or Einstein so all I can do is ask questions to try and understand as best I can the theories that these people intellectually superior to me come up with.

 

So I was just wondering if there is an upper and lower limit to how much energy a photon can have.

 

I was thinking... if you travel away from a source of light, the faster you move away the more the light appears red shifted, it has a lower frequency and a longer wave length and so less energy. It still travells at the speed of light relative to you though. So if you were to travel ever faster you would reach a point where the photon, in your frame of reference, has zero energy however you would be travelling at c, which you cannot do. This is like saying that in a frame of reference in which a photon is at rest, that photon has zero energy and so does not exist therefore a photon cannot have a frame which it is at rest with respect to.

 

That was kind of my train of thought, if you move away from a photon fast enough it has no energy but if you slow down it has energy again. Forget the relativity part, what, if any, are the upper and lower limits of energy a photon can have?

Posted

In the simplest case there would be a lower limit of zero in the limit of an infinite wavelength and no upper bound ("upper limit of infinite energy") in the limit of zero wavelength.

 

... and now it's showtime for the people telling you about pair production and Planck lengths ...

Posted

 

if you move away from a photon fast enough

 

All light has the same identical velocity, c, in every inertial frame to every observer.

You need to be careful identifying the photon as a particle that travels at c though.

Posted (edited)

Showtime !!

 

There is no predicted lower limit, you can go to zero.

 

Energy gravitates and as such produces a gravitational field. A high enough concentration of energy in a given volume is predicted to form a black hole. See J. A. Wheeler's work with gravitational collapse of geons ( just going by memory).

Is that an upper limit ? I don't know, you can still add more energy to a black hole.

Edited by MigL
Posted

 

All light has the same identical velocity, c, in every inertial frame to every observer.

You need to be careful identifying the photon as a particle that travels at c though.

 

Bad english, I meant something more along the lines of... moving away from the source of the light fast enough

Showtime !!

 

There is no predicted lower limit, you can go to zero.

 

Energy gravitates and as such produces a gravitational field. A high enough concentration of energy in a given volume is predicted to form a black hole. See J. A. Wheeler's work with gravitational collapse of geons ( just going by memory).

Is that an upper limit ? I don't know, you can still add more energy to a black hole.

 

I heard somewhere that a photon with enough energy would be a black hole, a black hole travelling at c though gives one pause for thought. Wouldn't Hawking radiation evapourate it away though and would a black hole travelling at c have any gravitational influence upon its surroundings?

 

Also, does a photon falling into a black hole add any energy to that black hole i.e. as it enters it red shifts so much it now has zero energy?

Posted (edited)

 

I heard somewhere that a photon with enough energy would be a black hole, a black hole travelling at c though gives one pause for thought.

Interestingly, in that case you kind of gave a reasoning against that statement yourself: Assuming a flat gravitational background: If you move in the direction of the photon fast enough, you can get the photon energy as low as you want. And if you move towards it you can get its energy as high as you want. If the statement about high-energy photons being black holes was correct, every free photon would be a black hole and not a black hole at the same time. Before you get excited about having found a new groundbreaking result in physics, the black hole duality: The simple solution is that photons are not black holes.

Edited by timo
Posted (edited)

Interestingly, in that case you kind of gave a reasoning against that statement yourself: Assuming a flat gravitational background: If you move into the direction of the photon fast enough, you can get the photon energy as low as you want. And if you move towards it you can get its energy as high as you want. If the statement about high-energy photons being black holes was correct, every free photon would be a black hole and not a black hole at the same time. Before you get excited about having found a new groundbreaking result in physics, the black hole duality: The simple solution is that photons are not black holes.

 

Bold Mine.

 

The "I have a new theory" title to this thread was pure sarcasm, I am not under the delusion that I have altered the understanding of science. I am just struggling with the logic of some things which I know is due to my lack of knowledge/understanding.

 

I'm aware this humble approach is rare in this forum which is a quite deliberate approach after reading some of the nonsense here.

 

What I'm struggling with is - if when you are travelling in one direction the photon has zero energy how can it have energy again if you start travelling in the opposite direction. The photon can only have zero energy if you are travelling at the speed of light, which you cannot do, so the photon can never have zero energy answers this logical paradox.

 

However if a photon can have zero energy then I'm back in my mental whirlpool and seeking further knowledge/understanding.

 

Edit for grammer

Edited by between3and26characterslon
Posted (edited)

You cannot boost to a frame in which the photon has zero energy (EDIT to be precise: at least that is what I would imagine from physical intuition, if you want to verify it then look up redshift formulas explicitly). Maybe I should have phrased "energy as low as you want" explicitly as "non-zero energy as low as you want". Also note that in my initial post I was explicitly referring to a limit process. This implies that for any positive "minimum energy" you would impose you can find a case in which the photon energy would be lower than this minimum. It is common practice to refer to this as the minimum being zero, even if no such state actually exists.

 

EDIT: Since I just ranted about "new theories" somewhere else, I feel that a clarification is in order for this post: The bold part was not written as a reference to the thread title, and technically you did not claim something like being up to revolutionize physics anywhere. The reason I wrote it was much simpler: While writing I realized that going into more details would get me on a slippery slope where one could get lost in a lot of, in my opinion ultimately irrelevant, details. So I wrote this to jump off the slope ;)

Edited by timo
Posted

"non-zero energy as low as you want".

 

Thanks timo, that's starting to make more sense.

 

So is it true to say that zero energy of a photon is a limit which can never be achieved similar to 0K is a limt of temperature which can never be achieved similar to c is a speed limit of a massive body which can never be achieved.

Posted (edited)

I beg to differ.

Any photon emerging from the event horizon of a black hole has exactly zero energy.

 

Hah-Hah !

Edited by MigL
Posted

Practically, no, its 'black' and has no energy to do any work on any detector.

Theoretically, yes, but it would take forever to measure its infinite wavelength.

Posted

I beg to differ.

Any photon emerging from the event horizon of a black hole has exactly zero energy.

 

Hah-Hah !

 

 

Practically, no, its 'black' and has no energy to do any work on any detector.

Theoretically, yes, but it would take forever to measure its infinite wavelength.

 

Why would a photon leave the event horizon? There's Hawking radiation but that is not "zero energy".

Posted

What about 2 photons travelling towards each other for a head on collision,relative to each other they are not blue shifted,otherwise all photon collisions would create electron/positron pairs.

Posted

What about 2 photons travelling towards each other for a head on collision,relative to each other they are not blue shifted,otherwise all photon collisions would create electron/positron pairs.

 

The photon's frame is not a valid one, so it doesn't make sense to talk about a blue shift. (if a photon had a rest frame, wouldn't its frequency be zero?) Pick an inertial frame, and do the analysis in that frame.

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.