E=pc vs v=E/p: Exploring the Scale Difference

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In summary: 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...In summary, velocity is energy divided by momentum. The difference in scale between them is v, but the equation ##E=pc## suggests the difference in scale is C not v. Why is this?
  • #1
JohnH
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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?
 
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  • #2
JohnH said:
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##.
 
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  • #3
JohnH said:
If velocity is energy divided by momentum
I think that is backwards. I think it is ##\vec v=\vec p/E##
 
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  • #4
Ah I see. This makes sense. Thank you Nugatory.
 
  • #5
Dale said:
I think that is backwards. I think it is ##\vec v=\vec p/E##
The units don't match, do they?
 
  • #6
nasu said:
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.
 
  • #7
jbriggs444 said:
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.
jbriggs444 said:
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).
 
  • #8
nasu said:
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.
JohnH said:
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.
 
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  • #9
nasu said:
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.
 
  • #10
@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.
 
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  • #11
nasu said:
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
$$v=\frac{p}{E}$$
 
  • #12
nasu said:
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}##).
 
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  • #13
@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.
 
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  • #14
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.
 
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  • #15
Dale said:
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.)
 
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  • #16
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.
 
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FAQ: E=pc vs v=E/p: Exploring the Scale Difference

What is the difference between E=pc and v=E/p?

E=pc and v=E/p are two different equations that represent the relationship between energy (E), momentum (p), and velocity (v). In E=pc, the energy is equal to the product of momentum and the speed of light (c). In v=E/p, the velocity is equal to the energy divided by the momentum. Essentially, these equations show the different ways in which energy, momentum, and velocity are related.

Which equation is more accurate?

Both equations are equally accurate in their respective contexts. E=pc is used to describe the energy of a particle with mass, while v=E/p is used to describe the velocity of a massless particle such as a photon. Therefore, the accuracy of each equation depends on the specific scenario being studied.

Can these equations be used interchangeably?

No, these equations cannot be used interchangeably. As mentioned before, E=pc is used for particles with mass, while v=E/p is used for massless particles. Additionally, the units for each equation are different. E=pc has units of energy (Joules), while v=E/p has units of velocity (meters per second).

How do these equations relate to Einstein's theory of relativity?

E=pc and v=E/p are both derived from Einstein's theory of relativity. They are used to describe the behavior of particles at different scales, and they demonstrate the concept of mass-energy equivalence, which is a fundamental aspect of relativity.

What are some real-world applications of these equations?

E=pc and v=E/p have a wide range of applications in fields such as particle physics, astrophysics, and engineering. They are used to understand the behavior of particles in high-energy collisions, the movement of objects in space, and the design of advanced technologies such as particle accelerators and spacecraft. These equations have also played a crucial role in the development of modern physics and our understanding of the universe.

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