Infinite Energy: Is Constant Exchange of Photons Possible?

[Lasha]

In gravitational field its a constant exchange of photons right? Then in ideal circumstances, if one object orbits the other one forever, then it means we get the exchange of photons forever right?which means infinite energy, I get that it can't be observed but that's possible for infinite energy to exist right?

 

[ZapperZ]

In gravitational field its a constant exchange of photons right?

This appears to be your starting premise, and it is imperative that you get this right (which it isn't) FIRST and foremost. So can you please cite a proper source to back this up?

If not, the rest of what you have written is moot, because it is based on a faulty concept.

Zz.

 

[Lasha]

I read a thread on this forum :D

 

[StevieTNZ]

Best to post the link to that thread.

 

[Drakkith]

Not only is gravity not mediated by photons, but theoretically by gravitons, these gravitons would be virtual gravitons and you would not be able to get infinite energy.

 

[Naty1]

Then in ideal circumstances, if one object orbits the other one forever, then it means we get the exchange of photons forever right?which means infinite energy

But things like planets, stars, even galaxies, are not ideal and do not 'orbit forever'...the orbits slowly decay.

A short bit here:
http://en.wikipedia.org/wiki/Orbital_decay#By_gravitational_radiation

edit: just remembered this discussion is in "quantum physics'...
so better to check here:
http://en.wikipedia.org/wiki/Atomic_orbital

For subatomic particles, no gravity has yet been included in the Standard Model of Particle physics

 

[ZapperZ]

I read a thread on this forum :D

This is lazy and unacceptable. Cite the link to the exact thread!

https://www.physicsforums.com/blog.php?b=2703

Zz.

 

[Drakkith]

But things like planets, stars, even galaxies, are not ideal and do not 'orbit forever'...the orbits slowly decay.

While true, even if they did not lose energy to gravitational waves, it would still not be infinite energy. I just want to make sure the OP understands that.

 

[Lasha]

This is lazy and unacceptable. Cite the link to the exact thread!

https://www.physicsforums.com/blog.php?b=2703

Zz.

I am really sorry for that, it was a long ago.I assume that it was about the electrical field and I somehow understood it that way

While true, even if they did not lose energy to gravitational waves, it would still not be infinite energy. I just want to make sure the OP understands that.

Wait so what does "losing energy to gravitational waves" mean? By Newtonian mechanics it would have been easily explained, in ideal circumstances it would move evenly, which means no force,no energy.But it takes speed of light for gravitational field to travel right? So there's no quants of energy traveling?

 

[Drakkith]

Wait so what does "losing energy to gravitational waves" mean?

In general relativity two orbiting bodies lose orbital energy by emitting waves in the metric, aka gravitational waves.

By Newtonian mechanics it would have been easily explained, in ideal circumstances it would move evenly, which means no force,no energy.But it takes speed of light for gravitational field to travel right? So there's no quants of energy traveling?

In Newtonian mechanics there is no such thing as a gravitational wave because there is no metric. Also, it is important to understand that a static field does not propagate. It is only changes in that field that propagate.

Finally, don't confuse regular photons and other force carrier particles with their virtual counterparts. It is these virtual particles that mediate forces.

 

[ZapperZ]

I am really sorry for that, it was a long ago.I assume that it was about the electrical field and I somehow understood it that way

Then either your memory got corrupted, or you understood the wrong thing. This is why we require that you cite the source completely. After all, none of us here want to defend and explain something that is wrong.

Gravity isn't mediated by photons. Photons (virtual ones) are responsible for all electromagnetic interactions as described using quantum field theory.

So the starting premise of your thread is false.

Zz.

 

[Akaisora]

I read a thread on this forum :D

Use the search function and search for keywords to find the thread.