Time derivative of a time-dependent vector and scalar

  • #1

Homework Statement



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

Homework Equations



See above

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}}}) ##
 

Answers and Replies

  • #2
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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)} ##
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
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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 [tex] \eta \equiv m [/tex]

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

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

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

Thus with little work we have to show that

[tex] W = \int \frac{dp}{dt} . u dt = \int \frac{dE}{dt} dt [/tex]

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

[tex] \frac{ mu }{ (1-u^2/c^2)^{3/2} } . \frac{du}{dt} \left( (1-u^2/c^2)^3 + u^2/c^2 \right) [/tex]
 
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  • #4
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.
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
PeroK
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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
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## (\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}}} ##
Pull out the common factor
[tex] \frac{m \vec {u} \cdot \frac{d \vec{u}}{dt}}{(1-u^2/c^2)^{3/2}}[/tex]
 
  • #7
Pull out the common factor
[tex] \frac{m \vec {u} \cdot \frac{d \vec{u}}{dt}}{(1-u^2/c^2)^{3/2}}[/tex]
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
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You end up with this conundrum because you are not differentiating correctly.
[tex]\frac{d}{dt} u^2 = \frac{d}{dt} \left(\vec{u} \cdot \vec{u} \right) = 2 \vec{u} \cdot \frac{d \vec{u}}{dt}[/tex]
 
  • #9
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(m⃗uududtc2(1−u2c2)3/2+md⃗udt√1−u2c2)⋅⃗u=mu3dudtc2(1−u2c2)3/2+md⃗udt⋅⃗u√1−u2c2 (\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}}}
Be careful of the [tex] u^3 [/tex] 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 [tex] u.u u [/tex]

The fact that the one of the time derivatives .....
The easiest way to think about it is by writing it by components using Einstein summation convection with upper and lower indexes. Thus: [tex] 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} [/tex]

Unfortunately no idea how the get rid of the extra term in brackets, maybe some series....
 
  • #11
PeroK
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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}##
 
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  • #12
You end up with this conundrum because you are not differentiating correctly.
[tex]\frac{d}{dt} u^2 = \frac{d}{dt} \left(\vec{u} \cdot \vec{u} \right) = 2 \vec{u} \cdot \frac{d \vec{u}}{dt}[/tex]
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
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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} ##
It comes from the derivative of the denominator.
 
  • #14
It comes from the derivative of the denominator.
Of course; I should have noticed that sooner, I was a bit too fixated on the parentheses there. Thanks for the assistance.
 

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