- #1
tsuwal
- 105
- 0
I am a bit confused the concept of energy density. I was taught that, in vacum:
<br /> \frac{1}{2}\int_{V}^{ } \rho P dV = \frac{1}{2}\varepsilon _0 \int_{V}^{ } E^{2} dV<br />
where P is the eletric potencial.
This is true because:
<br /> \rho P=\bigtriangledown \cdot (PE)+E^{2}<br />
where the first term can be ignored because if we integrate in the whole space and use the divergence theorem we find that this term goes to zero.
This expression only makes sense if we integrate in the whole space right? (because we ignored a term in the dedution on this basis). Similarly, can we do the same to gravitic fields? This is:
<br /> \frac{1}{2}\int_{V}^{ } \rho P dV = \frac{1}{2}\varepsilon _g \int_{V}^{ } F_g^{2} dV<br />
where \rho is the mass density \epsilon_g=\frac{1}{4\pi G} is the "gravitic permitivity" and F_g is gravitacional force?
If this is true, then if we consider a sphere with a mass m and compute its energy by this formula should we get mc^{2}? (I tried it but didn't get this)
<br /> \frac{1}{2}\int_{V}^{ } \rho P dV = \frac{1}{2}\varepsilon _0 \int_{V}^{ } E^{2} dV<br />
where P is the eletric potencial.
This is true because:
<br /> \rho P=\bigtriangledown \cdot (PE)+E^{2}<br />
where the first term can be ignored because if we integrate in the whole space and use the divergence theorem we find that this term goes to zero.
This expression only makes sense if we integrate in the whole space right? (because we ignored a term in the dedution on this basis). Similarly, can we do the same to gravitic fields? This is:
<br /> \frac{1}{2}\int_{V}^{ } \rho P dV = \frac{1}{2}\varepsilon _g \int_{V}^{ } F_g^{2} dV<br />
where \rho is the mass density \epsilon_g=\frac{1}{4\pi G} is the "gravitic permitivity" and F_g is gravitacional force?
If this is true, then if we consider a sphere with a mass m and compute its energy by this formula should we get mc^{2}? (I tried it but didn't get this)