Chapter 2: Q109E (page 70)

From $p = γ_{u} m u$ (i.e., $p_{x} = γ_{u} m u_{x}, p_{y} = γ_{u} m u_{y}$ and $p_{z} = γ_{u} m u_{z}$ ), the relativistic velocity transformation $(2 - 20)$ , and the identity $γ_{u^{'}} = (1 - u_{x} v / c^{2}) γ_{v} γ_{u}$ show that $p_{y}^{'} = p_{y}$ and $p_{z}^{'} = p_{z}$ .

Short Answer

Expert verified

It is shown that both frames’ y and z component momentum are equal by using the relativistic velocity and the given identity for $γ_{w^{'}}$ .

Step by step solution

State the relativistic velocity transformation equations.

The relativistic velocity transformation between two frames $S & S^{'}$ is expressed as,

$u_{x}^{'} = \frac{u_{x} - v}{1 - \frac{u_{x} v}{c^{2}}}$

$u_{y}^{'} = \frac{u_{y}}{γ_{r} (1 - \frac{u_{x} η^{3}}{c^{2}})}$

$u_{z}^{'} = \frac{u_{z}}{y_{v} (1 - \begin{matrix} u_{x} v \\ c^{2} \end{matrix})}$

Insert the above equations in the corresponding momentum equation.

The relativistic momentum in $S^{'}$ frame for the y-component is, $p_{y}^{'} = γ_{u} m u_{y}^{'}$

Inserting the expression for $u_{y}^{'}$ and the given identity for $γ_{w^{'}}$ in the above equation,

$p_{y}^{'} = [(1 - \frac{u_{x} v}{c^{2}}) γ_{v} γ_{n}] (m) [\frac{u_{y}}{γ_{v} (1 - \frac{u_{x} v}{c^{2}})}]$

$= γ_{u} m u_{y}$

$p_{y}^{'} = p_{y}$

Similarly, for the z-component in the frame $S^{'}$ ,

$p_{z}^{'} = γ_{u^{m}} m m u_{z}^{'}$

$= [(1 - \frac{u_{c} v}{c^{2}}) γ_{v} γ_{i u}] (m) [\frac{u_{z}}{γ_{v} (1 - \frac{u_{x} v}{c^{2}})}]$

$= γ_{u} m u_{z}$

$p_{z}^{'} = p_{z}$

Thus, unlike relativistic x- component momentum, the y and z component momentum of both frames is equal.

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From $p = γ_{u} m u$ (i.e., $p_{x} = γ_{u} m u_{x}, p_{y} = γ_{u} m u_{y}$ and $p_{z} = γ_{u} m u_{z}$ ), the relativistic velocity transformation $(2 - 20)$ , and the identity $γ_{u^{'}} = (1 - u_{x} v / c^{2}) γ_{v} γ_{u}$ show that $p_{y}^{'} = p_{y}$ and $p_{z}^{'} = p_{z}$ .

Short Answer

Step by step solution

State the relativistic velocity transformation equations.

Insert the above equations in the corresponding momentum equation.

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From p=γumu (i.e., px=γumux,py=γumuy and pz=γumuz ), the relativistic velocity transformation (2-20), and the identity γu'=(1-uxv/c2)γvγu show that py'=py and pz'=pz.

Short Answer

Step by step solution

State the relativistic velocity transformation equations.

Insert the above equations in the corresponding momentum equation.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

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From $p = γ_{u} m u$ (i.e., $p_{x} = γ_{u} m u_{x}, p_{y} = γ_{u} m u_{y}$ and $p_{z} = γ_{u} m u_{z}$ ), the relativistic velocity transformation $(2 - 20)$ , and the identity $γ_{u^{'}} = (1 - u_{x} v / c^{2}) γ_{v} γ_{u}$ show that $p_{y}^{'} = p_{y}$ and $p_{z}^{'} = p_{z}$ .