# I Integral of the momentum (with respect to time)

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1. Jan 4, 2017

### Aaron Bex

Hi everybody !

Can anyone help me with this problem:

Which is the (indefinite) integral with respect to time of the momentum of a particle of rest mass $m_0$?

$\int \dfrac{m_0\;\mathbf{v}}{{\sqrt{1-\dfrac{\mathbf{v}\cdot\mathbf{v}}{c^2}}}}\;dt$

where $m_0$ is invariant with respect to time.

Thanks !

Last edited: Jan 4, 2017
2. Jan 4, 2017

### Staff: Mentor

Without knowledge about v(t) it is impossible to simplify this integral.

Is this part of a homework problem?

3. Jan 4, 2017

### Aaron Bex

Then it is impossible to simplify this integral in a general way.

In this case, it is a personal interest.

Thank you !

4. Jan 5, 2017

### Battlemage!

This is kind of odd. I've seen people take the derivative of that with respect to time, and then integrate over distance to get the relativistic energy equation.

If you looked at it in the limit as v/c -> 0, this integral should represent a displacement through space of the mass, I think.

Since $v = \frac{dx}{dt}$, then $m_0 vdt = m_0 dx$. So integrate the right: $\int m_0 dx = m_0 x + C$ or if you use x(a) and x(b) as your end points on the integral, you get a displacement of the mass, i.e. $m_0 [x(b) - x(a)]$, which definitely looks like mass times a displacement.

As for not doing it when v/c is close to zero, you have the problem of v being a function of time. This isn't a problem if you do what I spoke about in my first sentence (taking the derivative or momentum with respect to time and then integrating over x) since you end up with nice dummy variables that turn the integral into a single variable one (only the speed remains and the integral is done over speed rather than distance). With this, though, you're left integrating over distance after canceling a dt, but your variable still has dt in it (assuming v is a function of time).

Also I can't imagine the physical interpretation of this being as simple as the case involving the low speed version, since time and space can be mixed coordinates in special relativity (and usually are, per the Lorentz transformation equations).

5. Jan 5, 2017

### Aaron Bex

Therefore, the main equation is equal to the following equation:

$\int \dfrac{m_0}{{\sqrt{1-\dfrac{\mathbf{v}\cdot\mathbf{v}}{c^2}}}}\;d\mathbf{r}$

But it is also impossible to simplify this integral in a general way.

Thank you !