- #1

- 47

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter QuArK21343
- Start date

- #1

- 47

- 0

- #2

- 36,426

- 6,956

You could specify a threshold temperature rise; or you could define the 'wavefront' as being the location at which the temperature gradient is a maximum.

- #3

- 47

- 0

Also, is there some simple way to estimate the velocity of the wavefront along which the temperature gradient is maximum?

Thanks

- #4

- 36,426

- 6,956

In the set up you state, there is zero length of wire at T_0, so I think you'll find the temperature will become T_1 everywhere immediately. You need more of a heat reservoir, or some external source maintaining one end at T_0.

Once you've clarified the set up you can try solving the equation and then seeing if some reasonable definition of wavefront provides an algebraic solution, but I don't hold out much hope.

- #5

- 18,054

- 9,004

- #6

- 36,426

- 6,956

More thoughts...

At http://www.mth.pdx.edu/~marek/mth510pde/notes%202.pdf [Broken] there's a useful equation (2.4.12) for the infinite uniform bar case. For t > 0, substituting x+2y√t for y inside the integral gives u(x, t) = (1/√π)∫e^(-y^2).g(x+2y√t).dy.

That can be read as indicating that the influence of an initial temperature g(x) at point x spreads out as the square root of time.

At http://www.mth.pdx.edu/~marek/mth510pde/notes%202.pdf [Broken] there's a useful equation (2.4.12) for the infinite uniform bar case. For t > 0, substituting x+2y√t for y inside the integral gives u(x, t) = (1/√π)∫e^(-y^2).g(x+2y√t).dy.

That can be read as indicating that the influence of an initial temperature g(x) at point x spreads out as the square root of time.

Last edited by a moderator:

- #7

- 47

- 0

- #8

- 36,426

- 6,956

The integral gives a positive result for all x.

I don't think you need to invoke relativity to explain why this doesn't happen in practice. It's probably limited by the velocity of sound in the object.

Share:

- Replies
- 12

- Views
- 2K

- Replies
- 10

- Views
- 2K