# Power-Energy Relationship

1. Oct 29, 2012

### danielu13

I was looking at the mass-energy relationship equation, which by taking the derivative I would think that you you should get a power-energy relationship. What I get is:

$P = \frac{mc^2}{2\sqrt{1-\frac{v^2}{c^2}}}*2a$

If this were to be true, a particle without acceleration would have no power, and thus release no energy. Or, it could be that all of the energy is instantaneously released and shows no power. Can someone help me where I went wrong in this?

2. Oct 29, 2012

### Staff: Mentor

You figured correctly: no power is expended to keep an object moving at constant speed in a lossless environment.

3. Oct 29, 2012

### danielu13

But is there any way of figuring the power produced by the destruction of mass, as in the mass-energy relationship?

4. Oct 29, 2012

### Staff: Mentor

No, it depends on how fast it is produced.

5. Oct 29, 2012

### danielu13

So the conversion between matter and energy is not instantaneous then?

6. Oct 30, 2012

### JustinRyan

If it was, there would be infinite power for an infinitely short time.

Conversion of mass to energy means the mass would no longer exist.

When a photon is emitted, it does not accelerate, it moves at c instantaneously.

My guess is that the consideration of power in this case is not meaningful.

7. Oct 30, 2012

### JustinRyan

A power energy relationship is time.

The derivative would be a power vs mass relationship.

I do find it an interesting thought.

8. Oct 30, 2012

### Staff: Mentor

On a molecular or atomic level I think it is -- therefore power is meaningless because as said it would be infinite or undefined.

I have, however seen people calculate power using a propagation rate or reaction rate of even a very fast reaction such as in a bomb. It provides interesting hyperbole but I don't think it is really all that useful.

Last edited: Oct 30, 2012