# Expression for the total energy of a particle of rest mass

1. May 2, 2008

### snoopies622

I am looking for an expression for the total energy of a particle of rest mass $$m_0$$ that includes kinetic and gravitational potential, if there is such a thing. If I take the product of the time-components of the four-velocity and four-momentum vectors, I get

$$m_0 c^2 \gamma^2$$

where
$$\gamma^2 = (dt/d\tau)^2 = \frac{dt^2 c^2}{g_a_b dx^a dx^b}$$

whereas if I take the dot product of the two vectors I simply get $$m_0 c^2$$.

Are either of these expressions correct?

2. May 2, 2008

### olgranpappy

The kinetic energy of the particle is
$$\sqrt{p^2c^2+m_0^2c^4}-m_0c^2\;.$$
If the particle is in a potential field $V(\vec r)$ the total energy (i.e., kinetic plus potential) is
$$\sqrt{p^2c^2+m_0^2c^4}-m_0c^2 + V(\vec r)\;.$$
of course, you can drop the -m_0c^2 if you want since it's a constant

3. May 3, 2008

### snoopies622

And that's with $$p=m_0 v \gamma$$?

4. May 3, 2008

### olgranpappy

plug in p=m_0vgamma and find out...
$$\sqrt{m_0^2v^2c^2\gamma^2+m_0^2c^4}-m_0c^2+V(r)$$
$$=m_0 c^2 \gamma - m_0c^2 + V(\vec r)=m_0c^2(\gamma -1) + V(\vec r)$$

5. May 4, 2008

### snoopies622

Thanks, olgranpappy; I appreciate that.

And now for the big money question: Does anyone know how to quantify gravitational potential energy using GR? Sometimes the energy of a particle is expressed with$$E=\gamma m_0 c^2$$ where
$$\gamma = \frac{dt}{d\tau}$$

but this implies that as an object is moved deeper (lower) into a gravitational field and $$dt/d\tau$$ grows, the potential energy grows as well, which is the opposite of what one would expect, of course. The other expressions I quoted above either repeat this problem (the one with $$\gamma^2$$) or remain constant at any height ($$m_0 c^2$$), ignoring gravitational potential energy altogether.

So I am stumped.

Last edited: May 4, 2008
6. May 4, 2008

### LHS1

The total energy of a particle = relativistic mass times c^2. As many physcists do not like the term relativistic mass, the equation may be written as E = gamma times rest mass times c^2. Here the total energy means the so called ' rest mass energy ', which includes energy from chemical bonding, nuclear energy which be released during nuclear reaction etc. plus the kinetic energy of the body.The gravitational potential energy is indeed stored in the gravitational field, not on the particle. Therefore it is not included in the equation.

7. May 4, 2008

### snoopies622

So there is no GR equivalent to $$mgh$$ or $$\frac{-Gm_1 m_2}{r}$$ from classical mechanics?

8. May 6, 2008

### pmb_phy

The first thing you need to know is that E does not generally equal mc2 when the object is in a gravitational field (m = relativistic mass). The energy is given by E = P0 rather than E = P0 as it is in SR. In the presence of a gravtitational field E is a function of both the position of the object and its velocity. It seems quite reasonable to me to think of "energy as a virtue of position" as potential energy. In the weak field limit the energy of the object is a sum of rest energy, kinetic energy and potential energy just as one might assume.

Pete

9. May 6, 2008

### DrGreg

I am still very much a beginner in GR.

I find here section 12.2 pages 54-55, the equation

$$E = mc^2 \frac{\sqrt{1 - 2GM / (r c^2)}}{\sqrt{1 - v^2 / c^2}}$$​

for the total energy (rest energy + kinetic energy + potential energy) of a small mass m near a large mass M. (Here "mass" means "rest mass".)

For small v and large r this approximates to

$$E = mc^2 + \frac{1}{2}mv^2 - \frac{GMm}{r}$$​

I don't pretend to fully understand this, but that's what it says. (Actually it doesn't quite say that because it assumes m = c = G = 1 and I've reexpressed the full equation without those assumptions.)

(The website I quoted contains the lecture notes on which the book General Relativity by NMJ Woodhouse was based.)

Perhaps some GR experts could comment on this.

10. May 6, 2008

### olgranpappy

... or for large r but arbitrary v it approximates to
$$E=m c^2\gamma - \frac{GMm\gamma}{r}=\sqrt{p^2c^2+m^2c^4}-\frac{GMm\gamma}{r}$$
where, again, m is rest mass.

Thanks for sharing the formula DrGreg.

11. May 7, 2008

### snoopies622

Thanks, mes amis. For some reason DrGreg's link isn't working on my computer, but I now have an idea of where that first equation may have come from, at least a version I was looking for.

If, by definition, $$p^0=m_0 c \frac{dt}{d\tau}$$ which equals -- in the case of a particle at rest and using the Schwarzschild metric --

$$\frac {m_0 c}{\sqrt{1-\frac{2GM}{rc^2}}}$$, and $$p_0 = g_0_x p^x$$ which (again for a particle at rest and using the Schwarzschild metric)

$$=g_0_0 p^0 = (1-\frac{2GM}{rc^2})p^0 = mc_0 \sqrt{1-\frac{2GM}{rc^2}}$$, and the energy of a particle is $$cp_0$$ ,then we get

$$E=m_0c^2 \sqrt{1-\frac{2GM}{rc^2}}$$

which looks like DrGreg's equation with $$v=0$$.

Oui?

Last edited: May 7, 2008
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