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Charge and existence?

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Can it be said that energy and matter interact by means of charge?

This question reflects a fundamental misconception of the concept "energy". Energy is a property, specifically a capacity. It is not an entity, a thing, a material, etc. Energy does not "interact" with matter, matter HAS energy.

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This question reflects a fundamental misconception of the concept "energy". Energy is a property, specifically a capacity. It is not an entity, a thing, a material, etc. Energy does not "interact" with matter, matter HAS energy.

That is the Newtonian scheme but it is not true in the widest context known today.

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  • 2 weeks later...

Energy must be manifested in matter or radiation.

Energy and mass aren't separate quantities in the modern sense. In the same way you can relate the sides of a triangle by a^2 + b^2 = c^2 in space, relativity gives an equivalent relation E^2 = m^2 + p^2 (in units where we do not need to put in the speed of light c). That is to say a particle has both mass 'm' and momentum 'p' and it can exchange momentum with other particles via different interactions (changing the energy 'E' of the particle). Light has only momentum (E = p) and a 'resting' piece of matter has only mass energy (E = m).

However, if you were trying to ask, does electromagnetic radiation (light) interact with matter via charge, then yes. We attach our field concept of the electron to the field concept of light by a mathematical charge term. This charge term facilitates the interaction between the electromagnetic field and the field of electrons.

Edited by Q.E.D.
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  • 1 month later...
Energy and mass aren't separate quantities in the modern sense. In the same way you can relate the sides of a triangle by a^2 + b^2 = c^2 in space, relativity gives an equivalent relation E^2 = m^2 + p^2 (in units where we do not need to put in the speed of light c). That is to say a particle has both mass 'm' and momentum 'p' and it can exchange momentum with other particles via different interactions (changing the energy 'E' of the particle). Light has only momentum (E = p) and a 'resting' piece of matter has only mass energy (E = m).

That is incorrect. The proper equations are E^2=m^2*c^4 + p^2*c^2; for a particle without mass E=pc and the rest mass of a matter particle is E=mc^2.

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Always the same sad story with these discussions.

They start by asking some fundamental question, then people realize that there is confusion because noone understands the basic physics concepts, then the dubject is dropped and it turns into a contest of algebra skills.

I agree

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Always the same sad story with these discussions.

They start by asking some fundamental question, then people realize that there is confusion because noone understands the basic physics concepts, then the dubject is dropped and it turns into a contest of algebra skills.

My very humble opinion is this: If you don't know physics then you should shut the hell up. :lol:

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That is the Newtonian scheme but it is not true in the widest context known today.

Indeed, energy can change into matter. If you let two protons collide with enough kinetic energy, this energy is converted into new particles that didn't exist before. In other words, a "property" (kinetic energy) becomes a "thing" (matter).

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Q.E.D. is correct, he gives the formulas in units that are used in theoretical physics, in which c = 1.

Yes, you're right. It depends, sometimes setting c = 1 is done to simplify the equations. I'm used to using c = speed of light and I over looked that he said c = 1.

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  • 2 weeks later...

I noticed some people might be confused about setting c=1 so I thought I'd draw an example from economics. Say you had a piggy bank containing quarters and a bank account in England containing Euros and you wanted a formula for how much money you have. If P is the amount of pennies in your piggy bank and E is the amount of money in your bank account, your total money M in dollars will be something like this:

M = kE + 0.01P

where k is the constant determining the conversion rate from Euros to Dollars. If we change units so that we count our piggy bank in dollars and our English bank account in dollars our equation becomes

M = E + P (much simpler)

Even if 'k' were truly constant in nature (euros to dollars never changed) it would still be possible to eliminate it from equations by choosing units where k=1. I hope that will make it clearer to anyone who reads this in the future.

Edited by Q.E.D.
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