# Differentiating the relativistic charge wrt velocity

• spaghetti3451
In summary, to show that ##\frac{dQ'}{dv} = \frac{1}{c^{2}} \int d^{3}x \bigg[ x \big( \frac{\partial \rho}{\partial t} - \frac{v}{c^{2}} \frac{\partial j^{x}}{\partial t} \big) - j^{x} \bigg] \bigg|_{t = \frac{vx}{c^{2}}, x, y, z}##, you need to use the chain rule and product rule to differentiate the first and second terms of ##Q'## with respect to ##v##. This will result in ##\frac{\partial
spaghetti3451

## Homework Statement

Given that ##Q' = \int d^{3}x \bigg[ \rho \big( \frac{vx}{c^{2}},x,y,z \big) - \frac{v}{c^{2}} j^{x} \big( \frac{vx}{c^{2}},x,y,z \big) \bigg]##, show the following:

##\frac{dQ'}{dv} = \frac{1}{c^{2}} \int d^{3}x \bigg[ x \big( \frac{\partial \rho}{\partial t} - \frac{v}{c^{2}} \frac{\partial j^{x}}{\partial t} \big) - j^{x} \bigg] \bigg|_{t = \frac{vx}{c^{2}}, x, y, z}##

## Homework Equations

3. The Attempt at a Solution [/B]

I understand that the second term can be easily obtained by differentiating ##- \frac{v}{c^{2}} j^{x}## with respect to ##v##. However, I am having trouble differentiating the first term with respect to ##v##.

In essence, I need to show that ##\frac{\partial \rho}{\partial v} = \frac{1}{c^{2}} x \bigg( \frac{\partial \rho}{\partial t} - \frac{v}{c^{2}} \frac{\partial j^{x}}{\partial t} \bigg)##.

Any hints?

You have ##\frac{\partial }{\partial v}\left( p(t(v),x,y,z) - \frac {v}{c^2}j^x(t(v),x,y,z)\right) ##
Using the chain rule and product rule, you should be able to work this out.
##\frac{\partial }{\partial v} p(t(v),x,y,z) = \frac{\partial p }{\partial t}\frac{\partial t }{\partial v} ##
##\frac{\partial }{\partial v}\frac {v}{c^2} j^x(t(v),x,y,z)=\left( \frac{\partial }{\partial v}\frac {v}{c^2}\right)j^x+\frac {v}{c^2}\frac{\partial j^x}{\partial v} ##
Notice that you will need the chain rule one more time for that last derivative.

Thanks! Got it!

## 1. What is the relativistic charge?

The relativistic charge refers to the electric charge of a particle as described by Einstein's theory of relativity. It takes into account the effects of time and space dilation at high speeds.

## 2. How is the relativistic charge different from the classical charge?

The classical charge, described by Coulomb's law, is a constant value for a given particle. However, the relativistic charge takes into account the changes in the particle's mass and length at high speeds, leading to a variable charge.

## 3. What is meant by "wrt velocity" in differentiating the relativistic charge?

"wrt velocity" stands for "with respect to velocity." In this context, it means calculating the rate of change of the relativistic charge with respect to the particle's velocity.

## 4. Why is it important to differentiate the relativistic charge wrt velocity?

By differentiating the relativistic charge wrt velocity, we can understand how the charge changes as the particle's velocity changes. This is crucial in understanding the behavior of charged particles at high speeds and in the presence of electromagnetic fields.

## 5. How is the relativistic charge differentiated wrt velocity?

The relativistic charge is differentiated using the Lorentz transformation equations, which describe how measurements of time, length, and mass change with velocity in accordance with the theory of relativity. These equations are used to find the rate of change of the relativistic charge with respect to velocity.

Replies
1
Views
1K
Replies
3
Views
984
Replies
1
Views
1K
Replies
10
Views
1K
Replies
1
Views
1K
Replies
5
Views
1K
Replies
32
Views
880
Replies
1
Views
1K
Replies
2
Views
2K
Replies
9
Views
2K