Relativistic Energy Derivation

[Calu]

Whilst reading following a derivation of the Relativistic Energy equation I came across the following:

d/dt[mu/(1-u²/c²)^1/2] = [m/(1-u²/c²)^3/2] du/dt.

I was wondering how that step was done.

 

[mfb]

This is just the chain rule and then some simplifications.

 

[Calu]

This is just the chain rule and then some simplifications.

I see it now, thanks.

 

[Calu]

Okay, so I now have:

d/dt[mu/(1-u²/c²)^-1/2]

= m((1-u²/c²)^-1/2) + mu(-u/c²).(1-u²/c²)^-3/2

I'm not sure how to simplify from here, any hints?

 

[vanhees71]

I have no clue, how you came to this derivative. First write the whole thing a bit more clearly with LaTeX:
$$E=\frac{m c^2}{\sqrt{1-u^2/c^2}}.$$
Here, $m$ is the invariant mass of the particle and $u=|\vec{u}|$ the usual (non-covariant) velocity with respect to the computational (inertial) frame.

Now taking the derivative with respect to time, you indeed just have to use the chain rule:
$$\frac{\mathrm{d} E}{\mathrm{d} t}=\frac{\mathrm{d} \vec{u}}{\mathrm{d} t} \cdot \vec{\nabla}_{u} E.$$

 

[vela]

Calu is calculating dp/dt using the product rule, but there's a typo, an extra negative sign in the first line. There's another sign error in the second line.

 

[Calu]

Derivation of Relative Energy Equation

I have no idea how to write in Latex, sorry, I'll type out the entire thing:

Assume m and c are constants.

W = ∫^x2_x1 F . dx = ∫^x2_x1 dp/dt . dx

dp/dt = d/dt [mu . (1-u²/c²)^-1/2]

by the product rule:

dp/dt = m(1-u²/c)^-1/2) + mu.(1-u²/c)^-3/2 . (-u/c²)

Then as γ = (1-u²/c)^1/2

dp/dt = m (γ - u/c²(γ³))

I'm not sure how to simplify to get the intended answer of m(1-u²/c)^-3/2 . du/dt

 

[vela]

I forgot to point out another error. You've differentiated with respect to u, not with respect to t.

 

[mfb]

You've differentiated with respect to u, not with respect to t.

Well that is easy to fix as m and c do not depend on t, but it is important to get it right.

You can combine both fractions to a single one if you expand the first one (the one that gives γ). Then simplify and you get the right result.

 

[Calu]

What do you mean by expand? Sorry I'm very confused.

 

[vela]

Not quite. You're still have an algebra mistake. The factor of 2 in the second term shouldn't be there.

 

[mfb]

@vela: Which factor of 2?

What do you mean by expand? Sorry I'm very confused.

Expand the fraction

Combining two fractions to a single one is taught several years before special relativity, you should know how to do that.

 

[vela]

@vela: Which factor of 2?

Calu had posted another attempt but edited it out or deleted the post after I replied.

 

[Calu]

@vela: Which factor of 2?

Expand the fraction

Combining two fractions to a single one is taught several years before special relativity, you should know how to do that.

Oh I see, we don't call that "expanding" a fraction, sorry for the confusion.

Also I edited the post because I thought it easier just to continue with what we had.

EDIT: I'm sorry but could you help me out? I'm not sure what I'm meant to do. And as far as I know we have been taught little about "expanding" fractions as we are usually told to simplify (reduce) them. I can't think of a time I've been asked to do the opposite. Usually to combine 2 fractions I would cross multiply to get a common denominator.

Let me just step away from this a second, because I have done the above (cross multiplied and found a common denominator) and have arrived at an answer similar to one I arrived at earlier but decided I could go no further.

The answer I have is as follows:

dp/dt = m [γ - u²/c²(γ³)]

Is there a way to take out a factor of (1-u²/c²) (i.e. γ²) from this? Or am I completely wrong?

And I'll post my result from combining the fractions:

[m(1-u²/c²)^3/2) - m(u²/c^c)(1-u²/c²)^1/2)] / (1-u²/c²)²

Which when we replace with γ is:

m(γ³ - (u²/c²)γ)

which as I said is similar to the answer I got earlier.

 

[Fredrik]

The answer I have is as follows:

dp/dt = m [γ - u²/c²(γ³)]

Is there a way to take out a factor of (1-u²/c²) (i.e. γ²) from this?

First take a factor of $\gamma$ outside. Then rewrite the $\gamma$ that remains on the inside in terms of u. Then simplify and think about what the simplified result looks like.

Edit: I don't think that minus sign is correct.

 

[Calu]

Right, so I finally did this after going back and differentiating wrt t as I was supposed to.

d/dt [mu(1-u²/c²)^-1/2]

= m.du/dt.(1-u²/c²)^-1/2 + (-1/2)mu((1-u²/c²)^-3/2.2u/c².du/dt

=m.du/dt.(1-u²/c²)^-1/2 + u²/c².m.((1-u²/c²)^-3/2.du/dt

= m(1-u²/c² + u²/c²)(1-u²/c²)^-3/2).du/dt

=m(1-u²/c²)^-3/2).du/dt as required.

Also, how would I solve:

m ∫^u_o [u(1-u²/c²)^-3/2] . du

Would it be integration by parts?

 

[Fredrik]

Looks good except that in the second line, 2u/c^2 should be (-2u/c^2), and there's also a left parenthesis too many.

No need to use integration by parts. You're supposed to write the integrand in a form that enables you to just use the formula $\int_a^b f'(x)dx=f(b)-f(a)$.

The work can be expressed as $\int_0^u mu\gamma^3\mathrm du$ or as $\int_0^t mu\gamma^3\dot u\mathrm dt$. Does either of the integrands look like a derivative?

 

[Calu]

I'm not sure how you would decide whether something "looks like" a derivative.

 

[Fredrik]

What I meant is that you will have to see that at least one of them is a derivative. Is either of the integrands equal to a derivative that you calculated earlier?

 

[Calu]

I managed to perform the integration by using a substitution of z = 1 - u²/c².

I still don't see how either is equivalent to a derivative I calculated earlier though.

 

[Fredrik]

OK, I see that it's more difficult to see in your approach than in the one I've been using (because mine involves a calculation of $\dot\gamma$):
\begin{align}
&\dot\gamma =\gamma^3u\dot u\\
&\dot p=\frac{d}{dt}(\gamma m u) =\dot\gamma m u+\gamma m\dot u =\dots=\gamma^3m\dot u\\
&W=\int F\mathrm dx =\int \frac{dp}{dt}u\mathrm dt =\int\gamma^3 m\dot u u\mathrm dt =\int\frac{d}{dt}(m\gamma)\mathrm dt.
\end{align} I have done this calculation a few times before, but it has actually never occurred to me that you can finish it with a simple variable substitution.

Edit: Here's another way:
$$W=\int \frac{dp}{dt}u\mathrm du =\int\gamma^3 m u\mathrm du =\int\frac{d}{du}(m\gamma)\mathrm du.$$

 

[Calu]

Oh yes that makes a lot more sense now. Thanks very much to all of you!