Invariance of Energy Momentum Relativistic

[agnimusayoti]

Homework Statement

Suppose an observer O measures a particle of mass m moving in the x direction to have speed v, energy E, and momentum p. Observer O', moving at speed u in the x direction; measures v', E', and p' for the same object. (a) Use the Lorentz velocity transformation to find E' and p' in terms of m, u, and v. (b) Reduce (E')^2 - (p'c)^2 to its simplest form.

Relevant Equations

Lorentz velocity transformation:
$$v'=(v-u)/(1-(uv/c^2))$$

Relationship between energy-momentum:
$$E^2=(pc)^2 +(mc^2)^2$$

I try to use relativistic energy equation:
$$E'=\gamma mc^2$$
But, I use
$$\gamma=\frac{1}{\sqrt{(1-(\frac{v'}{c})^2}}$$
then I use Lorentz velocity transformation.
$$v'=\frac{v-u}{1-\frac{uv}{c^2}}$$
At the end, I end up with messy equation for E' but still have light speed c in the terms. How should I do to get E' just in terms of m, u, and v?

 

[PeroK]

It's hard to help without seeing what you've done. Part a) is going to get messy doing it the way the question asks.

 

[agnimusayoti]

This is my work

 

[PeroK]

What I would do is use the Energy-Momentum transformation to calculate $E', p'$ the sensible way. Then, at least you know what you are aiming for.

 

[agnimusayoti]

Energy momentum Transformation is:
$$E^2 = (PC)^2 + (m c^2)^2$$?
Or what equation do you refer to?

Is my early step is right?

 

[PeroK]

Energy momentum Transformation is:
$$E^2 = (PC)^2 + (m c^2)^2$$?
Or what equation do you refer to?

Is my early step is right?

The Lorentz Transformation applies to all four-vectors, including the energy-momentum four vector. If you haven't seen that yet, then don't worry.

The problem you have is that it's not easy to see how to simplify what you have unless you know what expression you are aiming for.

Hint: try to find factors of $\gamma_u$ and $\gamma_v$ in what you have.

 

[vela]

At the end, I end up with messy equation for E' but still have light speed c in the terms. How should I do to get E' just in terms of m, u, and v?

The answer is going to have factors of $c$ in it.

One suggestion to avoid algebra errors: Keep a factor of $c$ with each $u$ and $v$.
$$\frac {v'}{c} = \frac{\frac vc - \frac uc}{1- \left(\frac vc\right)\left(\frac uc\right)}$$

 

[agnimusayoti]

The Lorentz Transformation applies to all four-vectors, including the energy-momentum four vector. If you haven't seen that yet, then don't worry.

The problem you have is that it's not easy to see how to simplify what you have unless you know what expression you are aiming for.

Hint: try to find factors of $\gamma_u$ and $\gamma_v$ in what you have.

Hmm unfortunately, I do not know what is four vector. So, could it be done with another method?

 

[agnimusayoti]

The answer is going to have factors of $c$ in it.

One suggestion to avoid algebra errors: Keep a factor of $c$ with each $u$ and $v$.
$$\frac {v'}{c} = \frac{\frac vc - \frac uc}{1- \left(\frac vc\right)\left(\frac uc\right)}$$

But the books said without c?

 

[vela]

The book didn't explicitly mention $c$ because it's understood it will appear. It's asking you to find $E'$ and $p'$ in terms of the unprimed variables as opposed to, say, $v'$.

 

[agnimusayoti]

Ok I see.
Well, I tried to solve the problem again and got this:
$$E'=\gamma_{v} \gamma_{u}(1-\frac{v}{c} \frac{u}{c}) (mc^2)$$
$$p'=\gamma_{v} \gamma_{u}m(v-u)$$

Then, I got
$${E'}^2 - {p'c}^2={mc^2}^2$$

Is it right?
Thankss for the suggestion @vela and @PeroK!

 

[agnimusayoti]

Oh, here I defined:
$$\gamma_{v}=\frac{1}{\sqrt{1-{\frac{v}{c}}^2}}$$
$$\gamma_{u}=\frac{1}{\sqrt{1-{\frac{u}{c}}^2}}$$

 

[PeroK]

hen, I got
$${E'}^2 - {p'c}^2={mc^2}^2$$

Is it right?

You're better writing that as: $$E'^2 - p'c^2=(mc^2)^2$$ or $$E'^2 - p'c^2=m^2c^4$$

 

[Guillermo Navas]

Homework Statement:: Suppose an observer O measures a particle of mass m moving in the x direction to have speed v, energy E, and momentum p. Observer O', moving at speed u in the x direction; measures v', E', and p' for the same object. (a) Use the Lorentz velocity transformation to find E' and p' in terms of m, u, and v. (b) Reduce (E')^2 - (p'c)^2 to its simplest form.
Relevant Equations:: Lorentz velocity transformation:
$$v'=(v-u)/(1-(uv/c^2))$$

Relationship between energy-momentum:
$$E^2=(pc)^2 +(mc^2)^2$$

I try to use relativistic energy equation:
$$E'=\gamma mc^2$$
But, I use
$$\gamma=\frac{1}{\sqrt{(1-(\frac{v'}{c})^2}}$$
then I use Lorentz velocity transformation.
$$v'=\frac{v-u}{1-\frac{uv}{c^2}}$$
At the end, I end up with messy equation for E' but still have light speed c in the terms. How should I do to get E' just in terms of m, u, and v?

Homework Statement:: Suppose an observer O measures a particle of mass m moving in the x direction to have speed v, energy E, and momentum p. Observer O', moving at speed u in the x direction; measures v', E', and p' for the same object. (a) Use the Lorentz velocity transformation to find E' and p' in terms of m, u, and v. (b) Reduce (E')^2 - (p'c)^2 to its simplest form.
Relevant Equations:: Lorentz velocity transformation:
$$v'=(v-u)/(1-(uv/c^2))$$

Relationship between energy-momentum:
$$E^2=(pc)^2 +(mc^2)^2$$

I try to use relativistic energy equation:
$$E'=\gamma mc^2$$
But, I use
$$\gamma=\frac{1}{\sqrt{(1-(\frac{v'}{c})^2}}$$
then I use Lorentz velocity transformation.
$$v'=\frac{v-u}{1-\frac{uv}{c^2}}$$
At the end, I end up with messy equation for E' but still have light speed c in the terms. How should I do to get E' just in terms of m, u, and v?

 

[Guillermo Navas]

Here is the pdf file.
In the image, where it says "we know that (7) is equal to mc^2" it should say "we know that (7) is equal to (mc)^2".

 

[PeroK]

What about? $$\frac{1}{\gamma'^2} = 1 - (v'/c)^2 = 1- \frac{v^2 + u^2 -2uv}{c^2(1- \frac{uv}{c^2})^2} = \frac{c^2(1 - \frac{v^2}{c^2} - \frac{u^2}{c^2} + \frac{u^2v^2}{c^4})}{c^2(1- \frac{uv}{c^2})^2} = \frac{(1- \frac{v^2}{c^2})(1 - \frac{u^2}{c^2})}{(1- \frac{uv}{c^2})^2}$$ Hence:
$$\gamma' = \gamma_u \gamma_v (1- \frac{uv}{c^2})$$

 

[robphy]

What about? $$\frac{1}{\gamma'^2} = 1 - (v'/c)^2 = 1- \frac{v^2 + u^2 -2uv}{c^2(1- \frac{uv}{c^2})^2} = \frac{c^2(1 - \frac{v^2}{c^2} - \frac{u^2}{c^2} + \frac{u^2v^2}{c^4})}{c^2(1- \frac{uv}{c^2})^2} = \frac{(1- \frac{v^2}{c^2})(1 - \frac{u^2}{c^2})}{(1- \frac{uv}{c^2})^2}$$ Hence:
$$\gamma' = \gamma_u \gamma_v (1- \frac{uv}{c^2})$$

Here's a trigonometric translation (using (u/c)=\tanh U and (v/c)=\tanh V, so \gamma_u=\cosh U and \gamma_v=\cosh V).
of the result.

$$\gamma' = \gamma_u \gamma_v (1- \frac{uv}{c^2})$$
is
$$
\begin{eqnarray*}
\cosh(U-V)
& =&\cosh U \cosh V (1-\tanh U \tanh V)\\
& =&\cosh U \cosh V -\sinh U \sinh V\\
\end{eqnarray*}
$$

The relative-velocity formula is trigonometrically:
$$
\begin{eqnarray*}
\tanh(U-V)
& =&\frac{\tanh U - \tanh V}{1-\tanh U \tanh V}\\
\end{eqnarray*}
$$

Sometimes a trigonometric ["angle" ; "rapidity"] intuition
can guide and interpret a calculation involving ["slopes" ; "velocities"].