# Time derivative of a time-dependent vector and scalar

1. Jun 29, 2016

### cwbullivant

1. The problem statement, all variables and given/known data

$\frac{d}{dt}\gamma(t)\vec{u(t)}$

2. Relevant equations

See above

3. The attempt at a solution

This comes from trying to verify a claim in Chapter 12 of Griffiths Electrodynamics, 4th. edition (specifically Eq. 12.62 -> Eq. 12.63, if anyone has it on hand).

I would have expected that the usual product rule applies in this case, and that:

$\frac{d}{dt}\gamma(t)\vec{u(t)} = \frac{d\gamma(t)}{dt}\vec{u(t)} + \gamma(t)\frac{d\vec{u(t)}}{dt}$

And that all vector operations on the result would proceed as normal (i.e. $\frac{d}{dt}(\gamma(t)\vec{u(t)}) \cdot \vec{u(t)} = \frac{d\gamma(t)}{dt}\vec{u(t)} \cdot \vec{u(t)} + \gamma(t)\frac{d\vec{u(t)}}{dt} \cdot \vec{u(t)}$).

But the text of the book appears to imply that:

$\frac{d}{dt}(\gamma(t)\vec{u(t)}) \cdot \vec{u(t)} = \frac{d\gamma(t)}{dt}\frac{d\vec{u(t)}}{dt} \cdot \vec{u(t)}$

What am I doing wrong here? Does the product rule not apply when one of the quantities to be differentiated is a vector and the other a scalar? That just doesn't sound right.

Edit: The exact chain of equations from the textbook is:

$\frac{d\vec{p}}{dt} = \frac{d}{dt}(\frac{m\vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}) \cdot \vec{u} = \frac{m\vec{u}}{(1-\frac{u^2}{c^2})^{\frac{3}{2}}} \cdot \frac{d\vec{u}}{dt} = \frac{d}{dt}(\frac{mc^2}{\sqrt{1-\frac{u^2}{c^2}}})$

2. Jun 29, 2016

### Fightfish

It doesn't; work it out carefully and you'll see that that isn't true. The normal product rule of course applies - all Griffiths did was to show the simplified expression.

3. Jun 30, 2016

### dpopchev

EDIT: poor replay skills

I followed Griffits and I am stuck in the calculation.

The scalar field is not time dependent as actually it is the mass of a test particle, so his original γ becomes $$\eta \equiv m$$

In the calculation we want to derive the work-energy theorem thus $$W = \int \frac{dp}{dt} dl = \int \frac{dp}{dt} . u dt$$

Of course p is a vector and is define with the proper velocity η

$$p = m\eta = m\frac{u}{\sqrt{1-u.u/c^2}}$$

Thus with little work we have to show that

$$W = \int \frac{dp}{dt} . u dt = \int \frac{dE}{dt} dt$$

Making the calculation on the left leads me to the following expression, where the right thingy in the brackets is missing in the book:

$$\frac{ mu }{ (1-u^2/c^2)^{3/2} } . \frac{du}{dt} \left( (1-u^2/c^2)^3 + u^2/c^2 \right)$$

Last edited: Jun 30, 2016
4. Jun 30, 2016

### cwbullivant

I get a two term expression when attempting this. Namely:

$\frac{d}{dt}(\frac{m\vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}) = \frac{m\vec{u}u\frac{du}{dt}}{c^2(1-\frac{u^2}{c^2})^{3/2}} + \frac{m\frac{d\vec{u}}{dt}}{\sqrt{1-\frac{u^2}{c^2}}}$

Take the dot product with u on both terms and it becomes:

$(\frac{m\vec{u}u\frac{du}{dt}}{c^2(1-\frac{u^2}{c^2})^{3/2}} + \frac{m\frac{d\vec{u}}{dt}}{\sqrt{1-\frac{u^2}{c^2}}}) \cdot \vec{u} = \frac{mu^3\frac{du}{dt}}{c^2(1-\frac{u^2}{c^2})^{3/2}} + \frac{m\frac{d\vec{u}}{dt} \cdot \vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}$

The derivative of relativistic energy, on the other hand, is:

$\frac{d}{dt}(\frac{mc^2}{\sqrt{1-\frac{u^2}{c^2}}}) = \frac{mu\frac{du}{dt}}{(1-\frac{u^2}{c^2})^{3/2}}$

I'm at a loss for how to simplify the two term derivative into that last line.

5. Jun 30, 2016

### PeroK

Why not try a specific pair of simple functions and see whether your solution and/or Griffiths' are correct?

Sorry, I see now that you are dealing with specific functions. I missed your edit to the OP.

6. Jun 30, 2016

### Fightfish

Pull out the common factor
$$\frac{m \vec {u} \cdot \frac{d \vec{u}}{dt}}{(1-u^2/c^2)^{3/2}}$$

7. Jun 30, 2016

### cwbullivant

The fact that the one of the time derivatives is $\frac{du}{dt}$, and the other is $\frac{d\vec{u}}{dt}$ doesn't make a difference with regard to pulling them out? I'd think the unit vector in the u direction wouldn't be time independent in general (so $\frac{d\vec{u}}{dt} = \frac{du}{dt}\hat{u} + u\frac{d\hat{u}}{dt}$), so shouldn't that add another product rule expansion?

I see that this does, in fact, get the right answer if you assume that $\frac{d\hat{u}}{dt} = \vec{0}$, but is that assertion always valid?

8. Jun 30, 2016

### Fightfish

You end up with this conundrum because you are not differentiating correctly.
$$\frac{d}{dt} u^2 = \frac{d}{dt} \left(\vec{u} \cdot \vec{u} \right) = 2 \vec{u} \cdot \frac{d \vec{u}}{dt}$$

9. Jun 30, 2016

### dpopchev

Be careful of the $$u^3$$ as despite not explicitly labelled with a arrow it is a vector, thus always respect the dot and write it. In this case do it as $$u.u u$$

The easiest way to think about it is by writing it by components using Einstein summation convection with upper and lower indexes. Thus: $$p_i = \left( \frac{d}{dt}\left( \frac{mu}{\sqrt{...}} \right) . u \right)_i = \left( \frac{d}{dt} \frac{mu}{\sqrt{...}} \right)_i u^i = \frac{m}{\sqrt{...}} \frac{du^i}{dt} u_i + m u_i u^i \frac{d}{dt} \frac{1}{\sqrt{....}} = \frac{m}{\sqrt{...}}u.\frac{du}{dt} + \frac{m}{(...)^{3/2}}u.\frac{du}{dt} \frac{u.u}{c^2}$$

Unfortunately no idea how the get rid of the extra term in brackets, maybe some series....

10. Jun 30, 2016

### dpopchev

11. Jun 30, 2016

### PeroK

You can break this down logically as follows. First, you need to show (by differentiating carefully) that:

$\frac{d\gamma}{dt} = \frac{\gamma^3}{c^2} \vec{u} \cdot \frac{d\vec{u}}{dt}$

Hence, with $E = \gamma mc^2$ you have:

$\frac{dE}{dt}= \gamma^3 m \vec{u} \cdot \frac{d\vec{u}}{dt}$

And, with $\vec{p} = \gamma m \vec{u}$ and using the product rule, you have:

$\frac{d\vec{p}}{dt} \cdot \vec{u} = \frac{d\gamma}{dt} mu^2 + \gamma m \vec{u} \cdot \frac{d\vec{u}}{dt} = \gamma^3 m (\frac{u^2}{c^2} + \frac{1}{\gamma^2}) \vec{u} \cdot \frac{d\vec{u}}{dt}$

Finally, as $\frac{u^2}{c^2} + \frac{1}{\gamma^2} = 1$ we have:

$\frac{dE}{dt} = \frac{d\vec{p}}{dt} \cdot \vec{u}$

Last edited: Jul 8, 2016
12. Jul 6, 2016

### cwbullivant

I don't see where the $\frac{d}{dt} u^2$ term is in the initial expression. The initial one as written is:

$\frac{d}{dt}(\frac{m\vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}) \cdot \vec{u} = (\frac{m\vec{u}u\frac{du}{dt}}{c^2(1-\frac{u^2}{c^2})^{3/2}} + \frac{m\frac{d\vec{u}}{dt}}{\sqrt{1-\frac{u^2}{c^2}}}) \cdot \vec{u}$

And only one of the vectors is inside the time derivative on the left (in your expression, both u vectors are inside the time derivative: $\frac{d}{dt} \left(\vec{u} \cdot \vec{u} \right)$; I know that expression is correct, I just don't see how it applies here). Am I supposed to bring the dot and vector on the right inside the time derivative before applying it, so that it looks like this?

$\frac{d}{dt}(\frac{m\vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}) \cdot \vec{u} = \frac{d}{dt}(\frac{m\vec{u}}{\sqrt{1-\frac{u^2}{c^2}}} \cdot \vec{u})$

13. Jul 6, 2016

### Fightfish

It comes from the derivative of the denominator.

14. Jul 7, 2016

### cwbullivant

Of course; I should have noticed that sooner, I was a bit too fixated on the parentheses there. Thanks for the assistance.