# Photon energy and gravitation

I don't understand what you are asking me, are you questioning the validity of e=mc^2 or just that the Earth doesn't have a uniform mass?
You said "in general relativity mass and energy are the same thing essentially."
I am not trying to say anything but am trying to understand.
Are you saying:
Earth mass = 5.9742 × 1024 kilograms
and
Earth energy = 5.9742 × 1024 kilograms
if not,
Earth energy = ???

You said "in general relativity mass and energy are the same thing essentially."
I am not trying to say anything but am trying to understand.
Are you saying:
Earth mass = 5.9742 × 1024 kilograms
and
Earth energy = 5.9742 × 1024 kilograms
if not,
Earth energy = ???
If you could take all the matter on Earth and convert it over to energy entirely efficiently then yes. But lets just say that a proton and an anti proton if they collided and taking account of their energy in total in theory have equivalent energy to mass according to e=mc^2.

Put it this way if you restricted the energy in a box and didn't let it escape and you could convert energy to mass then it would be equivalent to e=mc^2.

The answer you got was really lame. He just told you what everyone knows. I would find that annoying too...
... So it definitely takes more than a single university course in quantum mechanics (or a quick look at a Wikipedia article ) to really know what a photon is.
I am new to these forums and came here to learn. Twice in my post I indicated I am a layman when it comes to physics. I usually do not try to answer questions unless they are basic and within my understanding.

Lightarrow indicated he didn't know that light is a photon and I gave him very basic reply and pointed him to a link he could get a better understanding. You on the other hand told him

"So it definitely takes more than a single university course in quantum mechanics (or a quick look at a Wikipedia article ) to really know what a photon is."

While you claim my post was lame, yours post was a lot of words that said NOTHING to answer the question.

I didn't come here to be insulted and if a degree in physics is needed to fit in I will leave.

I am new to these forums and came here to learn. Twice in my post I indicated I am a layman when it comes to physics. I usually do not try to answer questions unless they are basic and within my understanding.

Lightarrow indicated he didn't know that light is a photon and I gave him very basic reply and pointed him to a link he could get a better understanding. You on the other hand told him

"So it definitely takes more than a single university course in quantum mechanics (or a quick look at a Wikipedia article ) to really know what a photon is."

While you claim my post was lame, yours post was a lot of words that said NOTHING to answer the question.

I didn't come here to be insulted and if a degree in physics is needed to fit in I will leave.
It takes an interest in physics, I don't have a degree in physics either. Perhaps he was having a bad day?

If you could take all the matter on Earth and convert it over to energy entirely efficiently then yes. But lets just say that a proton and an anti proton if they collided and taking account of their energy in total in theory have equivalent energy to mass according to e=mc^2.

Put it this way if you restricted the energy in a box and didn't let it escape and you could convert energy to mass then it would be equivalent to e=mc^2.
Thank you, I have been trying to use google calculator to calculate E=MC^2 with different arbitrary #'s with out luck. Could you give me an example?

I understand that a photon could be the length of the known universe with a frequency from 1 through gamma rays. It has 0 mass and exhibits both wave and particle properties. In flat space (vacuum) it travels at the speed of light and that speed is relative to it's current location (frame). The link I provided has a lot of information.
The frequency cannot be less than 1? Why?
The speed is c even in a curved spacetime.

However, we were discussing if a photon bends spacetime, and I wrote that I don't even know what a photon is. Let's say that you find that a photon is "a quantum of excitation of the electromagnetic field". Does it help you to understand if it bends spacetime? To me, it doesn't.

Ok take the mass of proton and anti-proton:

both = 2x1.6726 x 10-27, so you have a combined mass of 2 protons.

Plug them in to e=mc^2.

Sorry I don't know the level of people here, so I mistakenly thought you were trying to make a subtle point I wasn't getting.

Interestingly masses of particles this small are often given in units known as e/c^2, because m=e/c^2.

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The answer you got was really lame. He just told you what everyone knows. I would find that annoying too. But if you meant to suggest that only a person who can tell you the location, size and shape of a photon can claim to know what a photon is, I have to disagree. The answers to 1-3 are all "not well defined" (which was probably your point), and I'm not sure that question 4 even makes sense.
I agree with you. However, if we could know where a photon is and how much is its volume (and I'm NOT saying that such answers can be found) we could try to compute if, how and how much a photon bends space. The volume would be required to have the energy density (the parameter in the stress-energy tensor).

I agree with you. However, if we could know where a photon is and how much is its volume (and I'm NOT saying that such answers can be found) we could try to compute if, how and how much a photon bends space. The volume would be required to have the energy density (the parameter in the stress-energy tensor).
Since the curvature experienced by the lensing by the moons gravitation can be precisely modelled in an eclipse, if light did curve space to any measurable degree we would be able to pick it up as a discrepancy in what we measured, or we could just do it in the lab. If it does bend space it logically is so small that we could not measure it with the technology we have now.

The frequency cannot be less than 1? Why?
The speed is c even in a curved spacetime.

However, we were discussing if a photon bends spacetime, and I wrote that I don't even know what a photon is. Let's say that you find that a photon is "a quantum of excitation of the electromagnetic field". Does it help you to understand if it bends spacetime? To me, it doesn't.
Frequency is often measured in cycles per second. You can drop down to a straight line, 1 cycle per second frequency.

"a quantum of excitation of the electromagnetic field" means nothing to me as far as proving light bends spacetime.

Frequency is often measured in cycles per second. You can drop down to a straight line, 1 cycle per second frequency.

"a quantum of excitation of the electromagnetic field" means nothing to me as far as proving light bends spacetime.
Since the whole thing is more or less arm waving, nothing we can define light as can be held to account in a theory of light bending space. If it did though then there should in theory be a discrepancy in the http://en.wikipedia.org/wiki/Michelson-Morley_experiment" [Broken] experiment also, but just how you'd go about setting up something to measure a "particle" like a photons effect on space-time is beyond anyone, even dare I say it the finest minds in physics, so it's a moot point.

Generally we just assume light speed is constant in a vacuum and leave it at that. For the same reason we assume the photon has no mass, because to say otherwise is a matter for philosophy not science.

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jtbell
Mentor
nothing we can define light as can be held to account in a theory of light bending space.
As far as I know, classical electrodynamics adapts OK to curved spacetime, and that's what people use when calculating the gravitational effects of light. I've seen references to such calculations here.

It's photons that we have problems with, because we don't have a generally accepted quantum theory of gravity yet.

Since the whole thing is more or less arm waving, nothing we can define light as can be held to account in a theory of light bending space. If it did though then there should in theory be a discrepancy in the http://en.wikipedia.org/wiki/Michelson-Morley_experiment" [Broken] experiment also, but just how you'd go about setting up something to measure a "particle" like a photons effect on space-time is beyond anyone, even dare I say it the finest minds in physics, so it's a moot point.

Generally we just assume light speed is constant in a vacuum and leave it at that. For the same reason we assume the photon has no mass, because to say otherwise is a matter for philosophy not science.
Well said and I second the motion.

What I was looking for earlier, and I get a real DUH for not catching my own mistake. I need to work on making myself clear.
E=MC^2
mass of Earth * (the speed of light^2) = 5.36934319 × 10^41 joules

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Frequency is often measured in cycles per second. You can drop down to a straight line, 1 cycle per second frequency.
It's better if you study again the concept of frequency.
"a quantum of excitation of the electromagnetic field" means nothing to me as far as proving light bends spacetime.
Exactly. So, if you post here what wikipedia or anything else says about photons, doesn't mean anything for what concern that question...

Well said and I second the motion.

What I was looking for earlier, and I get a real DUH for not catching my own mistake. I need to work on making myself clear.
E=MC^2
mass of Earth * (the speed of light^2) = 5.36934319 × 10^41 joules
To be accurate you'd probably need to account for all the different masses, and since it's a massive object it will no doubt be extremely complicated, with various energy interactions not to mention the system is not closed. But in theory I suppose that is correct. Of course this is all hypothetical until I finish constructing my dooms day device, but your calculations are appreciated.

It's better if you study again the concept of frequency....
I said "Frequency is often measured in cycles per second. You can drop down to a straight line, 1 cycle per second frequency." You may be right though, an oscilloscope measures "0" frequency so 1 may be wrong. This of course is not in the visible light spectrum. Maybe think DC current. If you have a link that can teach me something please post it.

To be accurate you'd probably need to account for all the different masses, and since it's a massive object it will no doubt be extremely complicated, with various energy interactions not to mention the system is not closed. But in theory I suppose that is correct. Of course this is all hypothetical until I finish constructing my dooms day device, but your calculations are appreciated.
Actually I let google do all my math since it knows all about the mass of Earth and the speed of light. You probably already know this but anyone who doesn't, just copy and paste (mass of earth)c^2 into google search and hit enter. There are more meaningful and more complicated equations it can work with.

I am glad that I am not the only one building a dooms day device. Most of my friends think I am nuts.

Actually I let google do all my math since it knows all about the mass of Earth and the speed of light. You probably already know this but anyone who doesn't, just copy and paste (mass of earth)c^2 into google search and hit enter. There are more meaningful and more complicated equations it can work with.

I am glad that I am not the only one building a dooms day device. Most of my friends think I am nuts.
Well that's not a problem for me as I have gotten rid of my friends, if you know what I mean, they knew too much!

I said "Frequency is often measured in cycles per second. You can drop down to a straight line, 1 cycle per second frequency." You may be right though, an oscilloscope measures "0" frequency so 1 may be wrong. This of course is not in the visible light spectrum. Maybe think DC current. If you have a link that can teach me something please post it.
There is no need of links, just think that frequency is 1/T (T = period of oscillation) so if something, included EM fields, takes more than 1 second to complete the oscillation, the frequency is lower than 1 Hz. Take a point charge and put it on the border of a disk and rotate the disk at constant speed so that it takes 2 seconds to make one rev. Then the EM field generated has a frequency of 0.5Hz.

There is no need of links, just think that frequency is 1/T (T = period of oscillation) so if something, included EM fields, takes more than 1 second to complete the oscillation, the frequency is lower than 1 Hz. Take a point charge and put it on the border of a disk and rotate the disk at constant speed so that it takes 2 seconds to make one rev. Then the EM field generated has a frequency of 0.5Hz.
Didn't I correct myself in my last post when I said "You may be right though, an oscilloscope measures "0" frequency so 1 may be wrong."? So I believe a straight line would be 0 hz.

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Didn't I correct myself in my last post when I said "You may be right though, an oscilloscope measures "0" frequency so 1 may be wrong."? So I believe a straight line would be 0 hz.
Ok, I didn't write that to show you again that you are incorrect I just meant to show you how I reasoned about it (then you can use the information as you prefer...)