- #1

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[tex]

m_0 c^2 \gamma^2

[/tex]

where

[tex]

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

[/tex]

whereas if I take the dot product of the two vectors I simply get [tex]m_0 c^2[/tex].

Are either of these expressions correct?

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- Thread starter snoopies622
- Start date

- #1

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[tex]

m_0 c^2 \gamma^2

[/tex]

where

[tex]

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

[/tex]

whereas if I take the dot product of the two vectors I simply get [tex]m_0 c^2[/tex].

Are either of these expressions correct?

- #2

Homework Helper

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[tex]

\sqrt{p^2c^2+m_0^2c^4}-m_0c^2\;.

[/tex]

If the particle is in a potential field [itex]V(\vec r)[/itex] the total energy (i.e., kinetic plus potential) is

[tex]

\sqrt{p^2c^2+m_0^2c^4}-m_0c^2 + V(\vec r)\;.

[/tex]

of course, you can drop the -m_0c^2 if you want since it's a constant

- #3

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And that's with [tex]p=m_0 v \gamma[/tex]?

- #4

Homework Helper

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[tex]

\sqrt{m_0^2v^2c^2\gamma^2+m_0^2c^4}-m_0c^2+V(r)

[/tex]

[tex]

=m_0 c^2 \gamma - m_0c^2 + V(\vec r)=m_0c^2(\gamma -1) + V(\vec r)

[/tex]

- #5

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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[tex] E=\gamma m_0 c^2 [/tex] where

[tex] \gamma = \frac{dt}{d\tau}[/tex]

but this implies that as an object is moved deeper (lower) into a gravitational field and [tex]dt/d\tau[/tex] 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 [tex]\gamma^2[/tex]) or remain constant at any height ([tex]m_0 c^2[/tex]), ignoring gravitational potential energy altogether.

So I am stumped.

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[tex] E=\gamma m_0 c^2 [/tex] where

[tex] \gamma = \frac{dt}{d\tau}[/tex]

but this implies that as an object is moved deeper (lower) into a gravitational field and [tex]dt/d\tau[/tex] 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 [tex]\gamma^2[/tex]) or remain constant at any height ([tex]m_0 c^2[/tex]), ignoring gravitational potential energy altogether.

So I am stumped.

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- #7

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- #8

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The first thing you need to know is that E does not generally equal mc

Pete

- #9

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I am still very much a beginner in GR.

I find http://www2.maths.ox.ac.uk/~nwoodh/gr/gr03.pdf [Broken] section 12.2 pages 54-55, the equation

[tex]E = mc^2 \frac{\sqrt{1 - 2GM / (r c^2)}}{\sqrt{1 - v^2 / c^2}}[/tex]

for the total energy (rest energy + kinetic energy + potential energy) of a small mass

For small

[tex]E = mc^2 + \frac{1}{2}mv^2 - \frac{GMm}{r}[/tex]

I don't pretend to fully understand this, but that's what it says. (Actually it doesn't quite say that because it assumes

(The website I quoted contains the lecture notes on which the book

Perhaps some GR experts could comment on this.

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- #10

Homework Helper

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I am still very much a beginner in GR.

I find http://www2.maths.ox.ac.uk/~nwoodh/gr/gr03.pdf [Broken] section 12.2 pages 54-55, the equation

[tex]E = mc^2 \frac{\sqrt{1 - 2GM / (r c^2)}}{\sqrt{1 - v^2 / c^2}}[/tex]

for the total energy (rest energy + kinetic energy + potential energy) of a small massmnear a large massM. (Here "mass" means "rest mass".)

For smallvand largerthis approximates to

... or for large r but arbitrary v it approximates to

[tex]

E=m c^2\gamma - \frac{GMm\gamma}{r}=\sqrt{p^2c^2+m^2c^4}-\frac{GMm\gamma}{r}

[/tex]

where, again, m is rest mass.

Thanks for sharing the formula DrGreg.

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- #11

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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, [tex]p^0=m_0 c \frac{dt}{d\tau}[/tex] which equals -- in the case of a particle at rest and using the Schwarzschild metric --

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

[tex]=g_0_0 p^0 = (1-\frac{2GM}{rc^2})p^0

= mc_0 \sqrt{1-\frac{2GM}{rc^2}}[/tex], and the energy of a particle is [tex]cp_0[/tex] ,then we get

[tex]E=m_0c^2 \sqrt{1-\frac{2GM}{rc^2}}[/tex]

which looks like DrGreg's equation with [tex]v=0[/tex].

*Oui?*

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

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

[tex]=g_0_0 p^0 = (1-\frac{2GM}{rc^2})p^0

= mc_0 \sqrt{1-\frac{2GM}{rc^2}}[/tex], and the energy of a particle is [tex]cp_0[/tex] ,then we get

[tex]E=m_0c^2 \sqrt{1-\frac{2GM}{rc^2}}[/tex]

which looks like DrGreg's equation with [tex]v=0[/tex].

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