I Is the Scale Difference Between E=pc and v=E/p Explained by Special Relativity?

[JohnH]

If velocity is energy divided by momentum it seems like the difference in scale between them is v and yet E=pc suggests the difference in scale is C not v. Why is this?

 

[Nugatory]

If velocity is energy divided by momentum it seems like the difference in scale between them is v and yet E=pc suggests the difference in scale is C not v. Why is this?

Because $E=pc$ only applies to massless particles, and these always move with speed $c$. The general relationship that you want is $E^2=(mc^2)^2+(pc)^2$, which reduces to $E=pc$ for massless particles and to the famous $E=mc^2$ when $p$ is zero (massive particle at rest).

There’s also the classical formula for the kinetic energy of something moving at speeds that are small compared with the speed of light: $E_k=mv^2/2=p^2/2m$.

 

[Dale]

If velocity is energy divided by momentum

I think that is backwards. I think it is $\vec v=\vec p/E$

 

[JohnH]

Ah I see. This makes sense. Thank you Nugatory.

 

[nasu]

I think that is backwards. I think it is $\vec v=\vec p/E$

The units don't match, do they?

 

[jbriggs444]

The units don't match, do they?

In units where c=1, they do.

A good sanity check is the fact that the left hand side is a vector quantity and the right hand side has a vector quantity in the numerator and a scalar in the denominator. The competing formula, $\vec{v}=E/\vec{p}$ fails that sanity check -- can't divide a scalar by a vector.

Edit: To be fair you could multiply through by $\vec{p}$ and cast the competing equation as $\vec{v} \cdot \vec{p} = E$, thereby getting something dimensionally consistent.

 

[nasu]

In units where c=1, they do.

A good sanity check is the fact that the left hand side is a vector quantity and the right hand side has a vector quantity in the numerator and a scalar in the denominator. The competing formula, $\vec{v}=E/\vec{p}$ fails that sanity check -- can't divide a scalar by a vector.

In units where c=1, they do.

A good sanity check is the fact that the left hand side is a vector quantity and the right hand side has a vector quantity in the numerator and a scalar in the denominator. The competing formula, $\vec{v}=E/\vec{p}$ fails that sanity check -- can't divide a scalar by a vector.

Have you seen this formula with vectors in some book? I never said it should be written this way (with vectors, and vector p in the denominator).

 

[jbriggs444]

Have you seen this formula with vectors in some book? I never said it should be written this way (with vectors, and vector p in the denominator).

I saw it in this thread.

If velocity is energy divided by momentum

Velocity is a vector. Energy is a scalar. Momentum is a vector. It is worthwhile emphasizing that fact if you are one is planning to divide a scalar by a vector.

Edit: Rephrased to avoid implicit accusation which had apparently been taken badly.

 

[Dale]

The units don't match, do they?

You are right, but you can always throw in factors of c to fix that. In units where c=1 it is fine.

The problem I have with the other way is (in units where c=1) $E>|\vec p|$ so $E/\vec p > 1$ which would mean $v>c$. At least $\vec p/E$ is a vector and is 0 when an object is at rest and is 1 for a massless object.

I am not sure $\vec v = \vec p/E$ is right, but it seems more plausible than the other way around.

 

[nasu]

@jbriggs44
No, I am not planning do do anything like this. When did I say that $\vec{v}=\frac{E}{\vec{p}}$ is the correct formula? But this formula being wrong does not make $\vec{v}=\frac{\vec{p}}{E}$ correct by default. Definitely does not work in classical mechanics. And in relativistic mechanics, I don't see how making c=1 will make this work. With c=1 we have $E=\sqrt{m2+p^2}$ so $\frac{p}{E}=\frac{p}{\sqrt{m2+p^2}}$. Can you manipulate this to give the velocity in the end?
Being consistent in terms of vector quantities does not make the equation physically right. This is all I was trying to say.
Nugatory already showed the OP that his equation is not right for massive particles. You cannot make it right by just switching the numerator and denominator.

@Dale
I did not say that the formula proposed by the OP is right. It's simply wrong and you cannot make it right just by fixing the vector part.

 

[Dale]

Nugatory already showed the OP that his equation is not right for massive particles. You cannot make it right by just switching the numerator and denominator.

Actually, I just worked it out for massive particles and you can make it right by switching the numerator and denominator. For a massive object in units where c=1 we have:$$m^2=E^2-p^2$$ $$p= \frac{m v}{\sqrt{1-v^2}}$$ Which you can solve for $v$ and eliminate $m$ to get
$$v=\frac{p}{E}$$

 

[SiennaTheGr8]

With c=1 we have $E=\sqrt{m^2 + p^2}$ so $\frac{\vec{p}}{E}=\frac{\vec{p}}{\sqrt{m^2 + p^2}}$. Can you manipulate this to give the velocity in the end?

Yes:

$\dfrac{\vec p}{\sqrt{m^2 + p ^2}} = \dfrac{\gamma m \vec v}{\sqrt{m^2 + \gamma^2 m^2 v ^2}} = \dfrac{\gamma}{\sqrt{1 + (\gamma v)^2}} \, \vec v = \dfrac{\gamma}{\sqrt{\gamma^2}} \, \vec v = \vec v$

(where $\gamma = 1 / \sqrt{1 - v^2}$).

 

[nasu]

@Dale
Nice. :) You are right.
I was too lazy to try it. It did not seem possible to simplify the formula to this.
So it works for relativistic mechanics.

@SiennaTheGr8
Yes, I tried it too after Dale said it works.
You are right.

 

[robphy]

Concerning $\displaystyle v=\frac{p}{E}$, it's helpful to think geometrically, with rapidities $\theta$,
the Minkowski-angle between inertial worldlines, with $v=c\tanh\theta$.
I've thrown in the factors of $c$ and
$\cosh\theta=\gamma=\frac{1}{\sqrt{1-(v/c)^2}}=\frac{1}{\sqrt{1-\tanh^2\theta}}$ to help in the translation.

$p$ is the spatial component of the 4-momentum: $p=m[c]\sinh\theta$.
$E$ is the temporal component of the 4-momentum: $E=m[c^2]\cosh\theta$.
So, $$\displaystyle \frac{v}{[c]}=\frac{p[c]}{E}=\frac{m[c] \gamma \frac{v}{c} [c]}{m[c^2]\gamma}=
\frac{m[c]\sinh\theta [c]}{m[c^2]\cosh\theta}=\tanh\theta,$$
the slope of the 4-momentum vector on an energy-momentum diagram.

The space-time analogue (or should that be time-space?) is
$$\displaystyle \frac{v}{[c]}=\frac{\Delta x/[c]}{\Delta t}
=\frac{\tau \gamma \frac{v}{c}}{\tau \gamma}
=\frac{\tau \sinh\theta}{\tau \cosh\theta}=\tanh\theta,$$
the slope of the 4-velocity vector on a spacetime diagram.

 

[PeterDonis]

I am not sure $\vec v = \vec p/E$ is right

It is. There might be a slicker way of showing it, but the brute force way of just checking each of the two possible cases (massive and massless) is straightforward:

Massive particle: $E = m \gamma$, $\vec{p} = m \gamma \vec{v}$, so obviously $\vec{v} = \vec{p} / E$. (Edit: I see others have already gotten this.)

Massless particle: $E = | \vec{p} |$, so $\vec{p} / E = \vec{p} / | \vec{p} | = \vec{v}$, since $\vec{v}$ has a magnitude of $1$ (in units where $c = 1$) for this case and $\vec{v}$ points in the same direction as $\vec{p}$. (Edit: It does not appear that anyone has considered this case.)

 

[Herman Trivilino]

Lev Okun calls $\vec{p}=\big(\frac{E}{c^2}\big) \vec{v}$ one of the fundamental equations of special relativity. It does indeed apply to both massless and massive particles.