# 1ve broken the 1st law of thermodynamics, ive created energy

i think

mirrors reflect light and light source 1 with x energy is reflected by the mirror which uses some of that energy to create a double of that light source and thus creating light source y. is light being created?

PHYSICS IS BROKEN. LOOK TOWARDS THE SKY IF U CAN SPOT THE TEAR IN SPACETIME

oh and i know the question sounds like a homework question but IT IS NOT. its a personal question of something i was thinking about yesterday

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Born2bwire
Gold Member
Lisa, in this house we obey the laws of thermodynamics!

Doc Al
Mentor
mirrors reflect light and light source 1 with x energy is reflected by the mirror which uses some of that energy to create a double of that light source and thus creating light source y. is light being created?
No. Think of the incoming beam as sending a certain amount of energy towards the mirror per second; the reflection has less energy--you're losing energy, not creating it!

PHYSICS IS BROKEN. LOOK TOWARDS THE SKY IF U CAN SPOT THE TEAR IN SPACETIME
Those are tears of laughter, not tears in spacetime. :rofl:

russ_watters
Mentor
Yes the two beams have more energy than the one beam would if the mirror were replaced with an opaque surface. When pointing a laser into space you can create a beam with an arbitrarily large energy content.... But it may be useful to learn the difference between energy and power ;)

Those are tears of laughter, not tears in spacetime. :rofl:
Great joke, Doc! I must appreciate your sense of humour.

nice but there is one minor problem....u haven't actually "Created Energy"..

DaveC426913
Gold Member
I must be missing something. You've got 1000 lumens coming in and 900 lumens going out. Where is this extra energy you claim?

Are you suggesting that the sun and the sun's reflection in the mirror are both light sources for a total of 1900 lumens?

FredGarvin
Lisa, in this house we obey the laws of thermodynamics!
A classic line.

I enjoy how light curves in your second diagram.

Danger
Gold Member
I enjoy how light curves in your second diagram.
At least she's not ignoring gravity, although I don't see a source for it...

If you bounce a ball of the wall a ball bounces back. Doesn't mean you have doubled anything. If you bounce a light packet off a mirror it bounces back.

DaveC426913
Gold Member
I think what you're failing to realize is that, whether or not you've set up a mirror, the sun is putting out X lumens of light in all directions all the time. Except for the light that impinges on your detector, all the rest of that light just gets absorbed or otherwise dissipated.

What you are considering your "system" (the elements you consider important to your experiment) competely ignores this sunlight.

Inserting a mirror into the area where sunlight is streaming merely redirects existing sunlight, causing you to now consider it as part of your "system".

All you've done is make the light ray equivalent of a funnel, gathering it from a wider area and redirecting it to a smaller area.

This does bring up an interesting point now though.

A single photon reflects off of a surface in free space. The photon transfers some momentum to the object. To conserve energy, the photon must loose energy. Must the photon become blue-shifted?

This does bring up an interesting point now though.

A single photon reflects off of a surface in free space. The photon transfers some momentum to the object. To conserve energy, the photon must loose energy. Must the photon become blue-shifted?
No, it would be red shifted because it lost energy in the interaction. Although I'm thinking (I may be wrong) that you probably wouldn't really call it red shift, because the original photon was absorbed, the interaction happened, and another (different) photon was emitted with the energy not absorbed. One normally thinks of red shift as something that happens to the original photon regardless of any interaction (or I do, anyway).

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Those are tears of laughter, not tears in spacetime. :rofl:
Good one!

No, it would be red shifted because it lost energy in the interaction. Although I'm thinking (I may be wrong) that you probably wouldn't really call it red shift, because the original photon was absorbed, the interaction happened, and another (different) photon was emitted with the energy not absorbed. One normally thinks of red shift as something that happens to the original photon regardless of any interaction (or I do, anyway).
Shifts are due to relative velocity bet source and observer, you receive more no. of photons in the front than at rest so frequency increases and the spectrum blueshifts. The opposite on the back
Here it does not shift(you do not see color change after reflcn)
Waves are either reflected, absorbed or transmitted on interaction with matter
If more than one of the three happens its intensity(amplitude )decreases