Energy - momentum relationship

[jjson775]

Homework Statement

Show that the energy momentum relationship E²=p²c² + (mc²)² follows from the expressions E=ymc² and p = ymu

Relevant Equations

y = √ (1 - v²/c²) ^-1/2

The textbook says that by squaring and subtracting the expressions you can eliminate u.
E² - p² = y² (mc²)² - y²m²u²

 

[Steve4Physics]

Homework Statement:: Show that the energy momentum relationship E²=p²c² + (mc²)² follows from the expressions E=ymc² and p = ymu
Relevant Equations:: y = √ (1 - v²/c²) ^-1/2

The textbook says that by squaring and subtracting the expressions you can eliminate u.
E² - p² = y² (mc²)² - y²m²u²

Are you sure you mean "√ (1 - v²/c²) ^-1/2" - square rooting AND raising to the power "-1/2"?
What is 'u'(as opposed to 'v')?

 

[jjson775]

My mistake. There shouldn’t be a square root symbol. U and v are the same thing, velocity.

 

[Steve4Physics]

The textbook says that by squaring and subtracting the expressions you can eliminate u.
E² - p² = y² (mc²)² - y²m²u²

Well, you did what the book suggested but (in my opinion) the book's suggestion ("squaring and subtracting") isn't very helpful.

There's no physics needed - just some algebraic 'juggling'. So be prepared to spend time playing around with the equations.

Hints:
Replace $\gamma^2$ by $\frac {1}{1 - (v/c)^2}$
Rather than use $p = \gamma mv$, use $pc = \gamma mvc$

 

[jjson775]

Well, you did what the book suggested but (in my opinion) the book's suggestion ("squaring and subtracting") isn't very helpful.

There's no physics needed - just some algebraic 'juggling'. So be prepared to spend time playing around with the equations.

Hints:
Replace $\gamma^2$ by $\frac {1}{1 - (v/c)^2}$
Rather than use $p = \gamma mv$, use $pc = \gamma mvc$

Thanks for the help Steve. I never would have got this myself and had to look at a video of the derivation online where the trick is to insert the term (v^2 - v^2)

 

[robphy]

Using rapidities, these identities are more apparent and less mysterious (assuming you are comfortable with trig identities and their hyperbolic analogues).

With (v/c)=\tanh\theta, then \gamma=\cosh\theta since
1 \equiv \cosh^2\theta -\sinh^2\theta = \cosh^2\theta(1 -\tanh^2\theta).

 

[Steve4Physics]

The solution looks OK but a bit clumsy because of the recurring expression $1- \frac{v^2}{c^2}$.

This suggests a neater/more economical approach, leaving $\gamma$ in the working till near the end. Since you’ve already posted the solution, I hope I’m not out-of-order showing an improved version:

$E = \gamma mc^2 ⇒ E^2 = \gamma^2 m^2 c^4$

$p = \gamma mu ⇒ p^2 c^2 = \gamma^2 m^2 u^2 c^2$

$E^2 - p^2 c^2 = \gamma^2 m^2 c^4 - \gamma^2 m^2 u^2 c^2 = \gamma^2 m^2 c^2 (c^2 - u^2)$

Now we can get rid of $\gamma$

$\gamma^2 = \frac{1}{1 - \frac {u^2}{c^2}} = \frac {c^2 }{c^2 – u^2}$

Substituting for $\gamma^2$ will then give the required answer immediately (since the $(c^2 – u^2)$ terms cancel).

However the suggestion from @robphy is much more elegant mathematically.

 

[jjson775]

Very good. Yours is the solution I was trying for but didn't see that y² = 1/(1-u²/c²). I'm not familiar with rapidities. I am a retired engineer 78 years old, trying to learn something about modern physics through self study. I love it.

 

[PeroK]

What about: $$p^2c^2 + m^2c^4 = \gamma^2 m^2 v^2 c^2 + m^2c^4 = \gamma^2m^2c^4(\frac{v^2}{c^2} + \frac{1}{\gamma^2}) = E^2(\frac{v^2}{c^2} + 1 - \frac{v^2}{c^2}) = E^2$$

 

[PeroK]

Or: $$p = \gamma mv = \gamma m c^2 (\frac v {c^2}) = E\frac v {c^2}$$ $$E^2 - p^2c^2 = E^2 - E^2\frac {v^2} {c^2} = E^2(1 - \frac {v^2} {c^2}) = \frac{E^2}{\gamma^2} = m^2c^4$$

 

[PeroK]

Very good. Yours is the solution I was trying for but didn't see that y² = 1/(1-u²/c²). I'm not familiar with rapidities. I am a retired engineer 78 years old, trying to learn something about modern physics through self study. I love it.

Here's a really useful identity that you don't see much, but I find very handy: $$\gamma \frac{v}{c} = \sqrt{\gamma^2 - 1}$$ It also means that you can express $p$ without $v$: $$p = mc \sqrt{\gamma^2 - 1}$$ In general, I find it useful to get rid of $v$ and use the gamma factor - things seems to work out better once you have eliminated $v$ from your equations. E.g. the above identity falls out of this: $$p^2c^2 = m^2c^4(\gamma^2 - 1) = E^2 - m^2c^4$$

 

[neilparker62]

Very good. Yours is the solution I was trying for but didn't see that y² = 1/(1-u²/c²). I'm not familiar with rapidities. I am a retired engineer 78 years old, trying to learn something about modern physics through self study. I love it.

Enterprising of you - all the best!

 

[robphy]

Here's a really useful identity that you don't see much, but I find very handy: $$\gamma \frac{v}{c} = \sqrt{\gamma^2 - 1}$$ It also means that you can express $p$ without $v$: $$p = mc \sqrt{\gamma^2 - 1}$$

Trigonometrically, that reads
\cosh\theta \tanh\theta=\sqrt{\cosh^2\theta -1}=\sinh\theta
So that
pc=mc^2\sinh\theta
(the spatial component of the energy-momentum 4-vector)
and
E=mc^2\cosh\theta
(the temporal component of the energy-momentum 4-vector).
...
So the square-magnitude is
E^2-(pc)^2=(mc^2)^2(\cosh^2\theta-\sinh^2\theta)=(mc^2)^2

 

[neilparker62]

If you let: $$\frac{v}{c}=\sin\theta$$,

you'll find you can also prove as required using the trig identity: $$\sec^2(\theta)=1+\tan^2(\theta)$$

In this case referring to Perok's earlier post:

$$\gamma \frac{v}{c} = \sqrt{\gamma^2 - 1}$$ can be written as $$\sec\theta \sin\theta=\sqrt{\sec^2\theta-1}$$

 

[robphy]

If you let: $$\frac{v}{c}=\sin\theta$$,
...

Yes, interesting,...
but be aware that the use of Euclidean trigonometry like this for special relativity (also found in works by Loedel, Brehme, Shadowitz, and those derived from them) has limited value compared to the use of hyperbolic trigonometry using rapidities.

For the simplest problems in special relativity or for using it as a mathematical trick to derive an identity, it works.
But for more complicated problems, it likely won't work as well.

I should note that the angle measures between the two types are different.
They are actually related by https://en.wikipedia.org/wiki/Gudermannian_function (See also https://mathworld.wolfram.com/Gudermannian.html ).
\sin({\rm gd\ }\theta) =\tanh \theta

Let \phi={\rm gd\ }\theta. So that v= \sin\phi =\tanh \theta .

In particular, the Doppler factor, which is eigenvalue of the Lorentz transformation, is \exp\theta =\cosh\theta + \sinh\theta using rapidities, whereas it is \displaystyle \frac{1+\sin\phi}{\cos\phi}=\sec\phi+\sec\phi\sin\phi using the Euclidean method.

(See also my reply at https://physics.stackexchange.com/q...-u-c-sin-theta-equation-in-special-relativity )

 

[kuruman]

Using rapidities, these identities are more apparent and less mysterious (assuming you are comfortable with trig identities and their hyperbolic analogues).

With (v/c)=\tanh\theta, then \gamma=\cosh\theta since
1 \equiv \cosh^2\theta -\sinh^2\theta = \cosh^2\theta(1 -\tanh^2\theta).

A rapidity I stumbled upon in grad school relies on geometry rather than trigonometry. From the definition of $\gamma$ and using trivial algebra, one gets $$\gamma^2v^2=\gamma^2c^2-c^2$$ This relation suggests a right triangle as shown below

Then if one rescales the triangle by a factor of $mc$ and applies the Pythagorean theorem (the old meets the new), one gets $$(mc)^2\gamma^2c^2=(mc)^2c^2+(mc)^2\gamma^2v^2.$$With $E=\gamma mc^2$ and $pc=\gamma m v c$, the energy relation is proven.

 

[robphy]

A rapidity I stumbled upon in grad school relies on geometry rather than trigonometry. From the definition of $\gamma$ and using trivial algebra, one gets $$\gamma^2v^2=\gamma^2c^2-c^2$$ This relation suggests a right triangle as shown below

View attachment 277258
Then if one rescales the triangle by a factor of $mc$ and applies the Pythagorean theorem (the old meets the new), one gets $$(mc)^2\gamma^2c^2=(mc)^2c^2+(mc)^2\gamma^2v^2.$$With $E=\gamma mc^2$ and $pc=\gamma m v c$, the energy relation is proven.

This approach is related to the Euclidean methods by Loedel, Brehme, and others mentioned above.
The algebraic relations are correctly represented in this Euclidean right-triangle.
But, as above, they are of limited use.

Please note that
on a position-vs-time graph (a.k.a. the Minkowski spacetime diagram)
with the geometry of Minkowski spacetime,

• the hypotenuse is along the worldline and along the 4-velocity of the particle
• and the "legs" [the spatial component \gamma v
  and the temporal component \gamma c of the 4-velocity]
  are Minkowski-orthogonal--- not the legs with magnitude "c" and \gamma v.

From the symmetries of special relativity,
it is the Minkowski version of the Pythagorean theorem that is fundamental c^2=(\gamma c)^2 - (\gamma v)^2 on a position-vs-time graph---not the Euclidean version.

Here's an energy-momentum diagram from my answer on
https://physics.stackexchange.com/q...al-in-a-spacetime-diagram-relat/551250#551250

To see this geometry applied to a more complicated situation,
visit https://physics.stackexchange.com/a/510201/148184
and https://physics.stackexchange.com/questions/594212/momentum-diagram-for-two-colliding-particles

 

[PeroK]

A rapidity I stumbled upon in grad school relies on geometry rather than trigonometry. From the definition of $\gamma$ and using trivial algebra, one gets $$\gamma^2v^2=\gamma^2c^2-c^2$$

That is the algebraic trick to express $\gamma v$ in terms of $\gamma$ only. The geometry, I suggest, doesn't go very deep.

The hyperbolic geometry of the rapidity, however, carries over into the generators of the Lorentz Group, where it provides a foundation for more advanced physics, such as QFT.

 

[neilparker62]

Yes, interesting,...
but be aware that the use of Euclidean trigonometry like this for special relativity (also found in works by Loedel, Brehme, Shadowitz, and those derived from them) has limited value compared to the use of hyperbolic trigonometry using rapidities.

For the simplest problems in special relativity or for using it as a mathematical trick to derive an identity, it works.
But for more complicated problems, it likely won't work as well.

I should note that the angle measures between the two types are different.
They are actually related by https://en.wikipedia.org/wiki/Gudermannian_function (See also https://mathworld.wolfram.com/Gudermannian.html ).
\sin({\rm gd\ }\theta) =\tanh \theta

Let \phi={\rm gd\ }\theta. So that v= \sin\phi =\tanh \theta .

In particular, the Doppler factor, which is eigenvalue of the Lorentz transformation, is \exp\theta =\cosh\theta + \sinh\theta using rapidities, whereas it is \displaystyle \frac{1+\sin\phi}{\cos\phi}=\sec\phi+\sec\phi\sin\phi using the Euclidean method.

(See also my reply at https://physics.stackexchange.com/q...-u-c-sin-theta-equation-in-special-relativity )

Thanks for the very interesting references on the "Gundermannian function". A lot of transcendental type functions seem to have originated in that particular epoch (18th Century). Per your Wikipedia reference, the inventor of this function was Johann Heinrich Lambert who also invented the Lambert-W function which I ran into trying to solve for interest rate in an equation for a theoretical "continuous repayment mortgage'. Seems to have several different applications in various areas of Physics if you browse through the Wiki reference.

Angle relationships fascinating but a bit beyond me unfortunately.