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[Edit] switch to consistent geometric units and get rid of factors of 'c'.

I've been thinking about the relativistic hoop/disk again, and as a preliminary step I decided to look at what happens to the stress-energy tensor of a wire when we apply a load to it.

Suppose we have a wire, of mass m, length L, and cross sectional area A.

Then if the wire is not under stress, T^00 will (edit: in geometric units) just be m/(LA). Other components of the stress-energy tensor will be zero.

Suppose we apply a tension, T, gradually, and that the wire is within its elastic limit.

Then the wire will elongate from L to (L+d). This will require some amount of work W. To find the exact amount of work required would mean knowing the relationship between stress and strain, but if we use Hooke's law, it will be just

W = .5 K d^2

where K is the spring constant.

The wire may change its area when put under load, depending on it's Poisson's ratio http://en.wikipedia.org/wiki/Poisson's_ratio

Call the new area AA

What I'm interested in is the value for T^00 of the wire under load. This should be, by the conservation of energy, edit (in geometric units)

[tex]\frac{m + W}{(L+d)(AA)}[/tex]

If we take the ratio of T^00 under load to the initial value, we get

T^00 (loaded) / T^00(initial) = [tex]\frac{1+W}{(1+d/L)(AA/A)}[/tex]

The other component of the stress energy tensor will be the strain T in the wire.

[add]

T^11 = -T

[add]

I think this analysis should be OK even if the stresses exceed the elastic limit, as long as any temperature rise the wire may experience due to the non-reversible stretching (which will increase entropy) isn't allowed to radiate away and is uniformly distributed.

I've been thinking about the relativistic hoop/disk again, and as a preliminary step I decided to look at what happens to the stress-energy tensor of a wire when we apply a load to it.

Suppose we have a wire, of mass m, length L, and cross sectional area A.

Then if the wire is not under stress, T^00 will (edit: in geometric units) just be m/(LA). Other components of the stress-energy tensor will be zero.

Suppose we apply a tension, T, gradually, and that the wire is within its elastic limit.

Then the wire will elongate from L to (L+d). This will require some amount of work W. To find the exact amount of work required would mean knowing the relationship between stress and strain, but if we use Hooke's law, it will be just

W = .5 K d^2

where K is the spring constant.

The wire may change its area when put under load, depending on it's Poisson's ratio http://en.wikipedia.org/wiki/Poisson's_ratio

Call the new area AA

What I'm interested in is the value for T^00 of the wire under load. This should be, by the conservation of energy, edit (in geometric units)

[tex]\frac{m + W}{(L+d)(AA)}[/tex]

If we take the ratio of T^00 under load to the initial value, we get

T^00 (loaded) / T^00(initial) = [tex]\frac{1+W}{(1+d/L)(AA/A)}[/tex]

The other component of the stress energy tensor will be the strain T in the wire.

[add]

T^11 = -T

[add]

I think this analysis should be OK even if the stresses exceed the elastic limit, as long as any temperature rise the wire may experience due to the non-reversible stretching (which will increase entropy) isn't allowed to radiate away and is uniformly distributed.

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