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How can I show that $$ \int_V\mathbf x\rho(\mathbf x)\,\mathrm dV=0, $$ where we've defined $\mathbf r=\mathbf R+\mathbf x$, with $\mathbf R$ being the center of mass.

The definition of center of mass is $$ \mathbf R=\frac{\int_V\mathbf r\rho(\mathbf r)\,\mathrm dV}{\int_V\mathbf \rho(\mathbf r)\,\mathrm dV}=\frac{1}{M}\int_V\mathbf r\rho(\mathbf r)\,\mathrm dV. $$ The situation is illustrated here: enter image description here I started as follows: $$ \int_V\mathbf x\rho(\mathbf x)\,\mathrm dV=\int_V\mathbf r-\mathbf R\rho(\mathbf x)\,\mathrm dV=\int_V\mathbf r\rho(\mathbf x)\,\mathrm dV-\int_V\mathbf R\rho(\mathbf x)\,\mathrm dV. $$ However, I don't really know how to continue from here, because I get $$ \int_V \mathbf r\rho(\mathbf x)\,\mathrm dV-\int_V\mathbf R\rho(\mathbf x)\,\mathrm dV=\int_V \mathbf r\rho(\mathbf x)\,\mathrm dV-\frac{1}{M}\int_V\mathbf r\rho(\mathbf r)\,\mathrm dV\cdot\int_V\rho(\mathbf x)\,\mathrm dV. $$ My problem is that $\int_V\rho(\mathbf x)\,\mathrm dV\neq\mathbf M$. Otherwise it would have been easy.

EDIT I am being told that $\rho(\mathbf r)=\rho(\mathbf x)$, but then I don't understand how this integration works...

enter image description here Because I would think that integrating over $\mathbf x$ gives a different mass than integrating over $\mathbf r$, because the position is different.

EDIT2 I made the mistake to think that vector $\mathbf x$ originated at the origin, but he originates at $\mathbf R$ obviously.

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You manipulated the expression and eventually got to: $$ \int_V \mathbf r\rho(\mathbf x)\,\mathrm dV-\int_V\mathbf R\rho(\mathbf x)\,\mathrm dV. $$

Now, using your equation for $\mathbf R$, we have $\int_V\mathbf r \rho(\mathbf x) \,\mathrm dV= \mathbf R\cdot M$. Also, $$\int_V \mathbf R \rho(\mathbf x)\,\mathrm dV=\mathbf R \int_V \rho(\mathbf x),\mathrm dV =\mathbf R \cdot M.$$ Thus subtracting the two gives zero.

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  • $\begingroup$ I don't understand how $\int_V \rho(\mathbf x)\,\mathrm dV=M$. See the edit in my post. $\endgroup$ Commented Feb 11, 2017 at 13:55
  • $\begingroup$ Oh wait, I get it. I have to think of $\mathbf x$ as being a vector with a fixed position! I thought that $\mathbf x$ was centered at the origin, but that is not true, apparently. $\endgroup$ Commented Feb 11, 2017 at 13:57
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    $\begingroup$ In the integral $\int_V \rho(\mathbf x)\, \mathrm dV$, $\mathbf x$ is just a variable. The integral is over the region $V$, so you are free to change the "dummy variable" to whatever you want. $\endgroup$ Commented Feb 11, 2017 at 14:01
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The trick is that R is a constant. Thus, $$\mathrm{d}\textbf{r}=\mathrm{d}\textbf{x}$$ and also $$\mathrm{d}V(\textbf{x})=\mathrm{d}V(\textbf{r})$$ The fact, that it is constant can be seen from it's $$ \mathbf R=\frac{\int_V\mathbf r\rho(\mathbf r)\,\mathrm dV}{\int_V\mathbf \rho(\mathbf r)\,\mathrm dV}=\frac{1}{M}\int_V\mathbf r\rho(\mathbf r)\,\mathrm dV. $$ in here, $M$ is constant and definite integration also gives you a constant, hence the defined value is also a constant. Therefore, substituing for r in the definition of mass, gives you the same value for x.

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