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I have seen it written that for a continuum undergoing deformation, if we ignore body forces and heat transfer, the work done is equal to stress power:

$\cfrac{dW}{dt}=\sigma_{ij}D_{ij}$, where $D_{ij}$ is the velocity gradient.

Then we have $\cfrac{dW}{dt}=\sigma_{ij}D_{ij}=\sigma_{ij}v_{i,j}=\sigma_{ij}\cfrac{dF_{iA}}{dt}F^{-1}_{Aj}=\cfrac{1}{J}P_{Bi}x_{j,B}\cfrac{dF_{iA}}{dt}F^{-1}_{Aj}$, where $P_{Bi}$ is the first Piola-Kirchhoff tensor and we have used the fact that $J\sigma_{ij}=P_{Ai}\cfrac{dF_{iA}}{dt}$.

This result simplifies to $\cfrac{dW}{dt}=\cfrac{1}{J}P_{Ai}\cfrac{dF_{iA}}{dt}$.

However, in other sources I see it written that $\cfrac{dW}{dt}=P_{Ai}\cfrac{dF_{iA}}{dt}$, and I see no reason to assume that the Jacobian should be equal to 1. Could you please tell me where I went wrong? Thank you.

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    $\begingroup$ Aren't the "other sources" writing the power in the vicinity of the reference configuration? $\endgroup$ – Joce Apr 25 '16 at 13:36
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I think @Joce has the correct answer in his comment, but I'm going to expand their answer a little bit to provide some more detail.

The stress power is specified per unit volume, but you have to specify which unit of volume. The stress power integrated over an entire body in the current configuration $\Omega_t$ is given by:

\begin{align} \int_{\Omega_t} \mathbf{\sigma}:\mathbf{D} \; dv_t &= \int_{\Omega_t} \mathbf{\sigma}:\mathbf{L} \; dv_t \\ &= \int_{\Omega_t} \sigma : (\dot{\mathbf{F}}\mathbf{F}^{-1}) \; dv_t \\ &= \int_{\Omega_t} \frac{1}{J}(\mathbf{P}\mathbf{F}^T):(\dot{\mathbf{F}}\mathbf{F}^{-1}) \; dv_t \\ &= \int_{\Omega_t} \frac{1}{J}\mathbf{P}:\dot{\mathbf{F}}\; dv_t \end{align}

Now, taking $dv_t$ to be a differential volume in the current configuration and $dV$ to be a differential volume in the reference configuration, we use the relation $dv_t = J dV$ to change to integration over the reference body $\Omega_0$:

\begin{align} \int_{\Omega_t} \frac{1}{J}\mathbf{P}:\dot{\mathbf{F}}\; dv_t &= \int_{\Omega_0} \frac{1}{J}\mathbf{P}:\dot{\mathbf{F}}\; JdV \\ &= \int_{\Omega_0} \mathbf{P}:\dot{\mathbf{F}}\; dV \end{align}

So, you see, both expressions are correct, but you have to be precise about the units of volume that you use. Unfortunately, this isn't always specified clearly in papers, so you have to do a little work to figure out exactly what they're doing.

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