# Measuring the thermoelectric effect of a simple metal

1. Dec 6, 2004

### SmD

By using the following formulation of the thermoelectric effect:

$$\mathbf{j} = \mathbf{\L}_{11}\mathbf{\xi} + \mathbf{\L}_{12}(-\mathbf{\nabla}T)$$
$$\mathbf{j} = \mathbf{\L}_{21}\mathbf{\xi} + \mathbf{\L}_{22}(-\mathbf{\nabla}T)$$

where $$\mathbf{\L}_{ij}$$ are tensors that charaterizes a given material.

What would be the easiest conceivable experiment to measure $$\mathbf{\L}_{ij}$$ of a certain material. One cannot exploit the Seebeck effect to get the ABSOLUTE VALUES of the coefficients of ONE METAL, since when you connect the leads of a voltmeter on the metal, it creates a temperature gradient inside of the meter itself (which would lead to false data). On the other hand, the use of the Peltier and Thomson effect leads to certain problems. I'm pretty sure that I must use one these effects to measure these coefficients, but I'm not sure how to do it .

Last edited: Dec 6, 2004
2. Dec 6, 2004

### Gonzolo

What if you use two metals, as in a thermocouple, with one of them a known reference? I think you can then figure out the unknown Seebeck coefficient from this.

If you really want to stick to one metal, I think you can use a reference temperature (passing one of the leads through ice or maintaining a point along the line at constant temperature.)

Either way, I think you can simply use the Seebeck effect.