Can you help me with this equality?

[Abrain]

I'd like to understand why $$\int{\rho}dV = \frac{1}{c} \int{j^0}dV = \frac{1}{c} \int{j^i}dS_i$$ (the second equality), where

$$j^i = \rho \frac{dx^i}{dt}$$ is the current density 4-vector

$$\mathbf{j} = \rho \mathbf{v}$$ is the current density 3-vector

$$j^i = (c\rho, \mathbf{j})$$

$$\rho$$ is the charge density

Are you able to explain me this equality?

Thank you very much!

 

[Fredrik]

The tags need to be "tex" or "itex" instead of "latex". You also need to know that there's a bug that makes the wrong images appear in the previews most of the time. The workaround is to refresh and resend after each preview. You can edit your posts for 24 hours, so there's still time to fix the one above.

 

[Abrain]

The tags need to be "tex" or "itex" instead of "latex". You also need to know that there's a bug that makes the wrong images appear in the previews most of the time. The workaround is to refresh and resend after each preview. You can edit your posts for 24 hours, so there's still time to fix the one above.

It doesn't seem to work neither with [tex] or [itex]

 

[JesseM]

I think your problem is that you've got the wrong kind of slash in the closing tags, should be /tex not \tex (and don't use capital TEX). Let's see if changing those works:I'd like to understand why $$\int{\rho}dV = \frac{1}{c} \int{j^0}dV = \frac{1}{c} \int{j^i}dS_i$$ (the second equality), where

$$j^i = \rho \frac{dx^i}{dt}$$ is the current density 4-vector

$$\mathbf{j} = \rho \mathbf{v}$$ is the current density 3-vector

$$j^i = (c\rho, \mathbf{j})$$

$$\rho$$ is the charge density

$$dS_i$$ is the element $$-dx-dy-dz+cdt$$

Are you able to explain me this equality?

Thank you very much!

 

[Abrain]

I think your problem is that you've got the wrong kind of slash in the closing tags, should be /tex not \tex (and don't use capital TEX).

Oh thank you! I feel so stupid about this...

 

[Phrak]

$$\int{\rho}dV = \frac{1}{c} \int{j^0}dV = \frac{1}{c} \int{j^i}dS_i$$

Your units don't seem to match-up. I take V to be in units of volume. In such case, S should also have units of volume.

Rather instead, I think you have mistakenly taken an element of S to be it's surface normal vector, and S should be the boundary of the volume V. Does this sound close?

 

[Abrain]

Your units don't seem to match-up. I take V to be in units of volume. In such case, S should also have units of volume.

Rather instead, I think you have mistakenly taken an element of S to be it's surface normal vector, and S should be the boundary of the volume V. Does this sound close?

This could be... the book does'n say exactly what dS is.
However $$dS = -dx-dy-dz+cdt$$ is a 4-volume, isn't it?

But,if you are right, the equality is holding only because of the Gauss law? This sound pretty correct

 

[Phrak]

OK, apparently this really is supposed to be the charge continuity equation; the time rate change of charge contained in a volume is equal to the charge entering.

If so, the left hand side needs a d/dt operator and S would be the surface normal vector having units of area pointing inward.