User:AndreRD/Analogues in Physics

In Physics, there are many cases where two different phenomena that might seem unrelated share some similar underlying structure. When this happens, there is often a correspondence between the units and formulas of the two systems.

Translational Motion vs Rotational Motion

Translational Analog	Symbol	Unit	Rotational Analog	Symbol	Unit
class="wikitable"
Position	$\mathbf{x}$	m	AngleIf an axis of rotation is defined, then the angular position can be treated like a pseudoscalar quantity. Otherwise, it is best represented as a three-dimensional rotational matrix.	$\theta$	radAngles are technically unitless, the radian being equal to 1, but it's often useful to include it in some rotational quantities to better clarify what the quantity represents. For example, a measurement of 6 rad s^-1 is a lot clearer than just 6 s^-1, which is potentially ambiguous as it could be understood as 6 Hz by mistake, or not even understood to be referring to rotation.
Time	$t$	s	Time	$t$	s
Linear VelocityThe magnitude of the velocity has a special name, the speed. Similarly, the magnitude of the angular velocity is called the angular speed. Confusingly though, another name for the angular speed is the angular frequency.	$\mathbf{v} = \frac{\operatorname{d}\mathbf{x}}{\operatorname{d}t}$	m s^-1	Angular Velocity	$\boldsymbol{\omega} = \frac{\operatorname{d}\theta}{\operatorname{d}t}\cdot\hat{\mathbf{u}}$ Where $\hat{\mathbf{u}}$ is the unit vector along the axis of rotation as given by the right hand rule.	rad s^-1
Linear Acceleration	$\mathbf{a} = \frac{\operatorname{d}\mathbf{v}}{\operatorname{d}t}$	m s^-2	Angular Acceleration	$\boldsymbol{\alpha} = \frac{\operatorname{d}\boldsymbol{\omega}}{\operatorname{d}t}$	rad s^-2
Mass	$m$	kg	Moment of Inertia	$I$	kg m²
Linear Momentum	$\mathbf{p} = m\mathbf{v}$	kg m s^-1 or N s	Angular Momentum	$\mathbf{L}=I\boldsymbol{\omega}$	kg m² rad s^-1 or N m s
Force	$\mathbf{F} = \frac{\operatorname{d}\mathbf{p}}{\operatorname{d}t} = m \frac{\operatorname{d}\mathbf{v}}{\operatorname{d}t} = m\mathbf{a}$	N = kg m s^-2	Torque	$\boldsymbol{\tau} = \frac{\operatorname{d}\mathbf{L}}{\operatorname{d}t} = I\frac{\operatorname{d}\boldsymbol{\omega}}{\operatorname{d}t} = I \boldsymbol{\alpha}$	kg m² rad s^-2 or N m
Translational Kinetic Energy	$E_k = \frac{1}{2}mv^2 = \frac{p^2}{2m}$	J = kg m² s^-2	Rotational Kinetic Energy	$E_r = \frac{1}{2}I\omega^2 = \frac{L^2}{2I}$	J = kg m² s^-2
Translational Work	$W = \int\mathbf{F}\cdot\operatorname{d}\mathbf{x}$	J = N m = kg m² s^-2	Rotational Work	$W = \int \mathbf{\tau} \ \cdot\operatorname{d}\theta\$	J = N m = kg m² s^-2
Laws of rectilinear motion under constant acceleration	$\begin{align}\\& v_f = v_i+at \\& \Delta x = \tfrac{1}{2}(v_i+v_f)t \\& \Delta x = v_i t + \tfrac{1}{2} at^2 \\& \Delta x = v_f t - \tfrac{1}{2} at^2 \\& v_f^2 = u_f^2 + 2a\Delta x \end{align}$		Laws of circular motion under constant torque	$\begin{align}\\& \omega_f = \omega_i+\alpha t \\& \Delta\theta = \tfrac{1}{2}(\omega_i+\omega_f)t \\& \Delta\theta = \omega_i t + \tfrac{1}{2} \alpha t^2 \\& \Delta\theta = \omega_f t - \tfrac{1}{2} \alpha t^2 \\& \omega_f^2 = \omega_i^2 + 2\alpha\Delta\theta\end{align}$

Electric Fields and Gravitational Fields

Electric Analog	Symbol	Unit	Gravitational Analog	Symbol	Unit
class="wikitable"
Electric charge	$q$	C (= A s)	Gravitational mass	$m$	kg
Coulomb's constant	$k_e = \frac{1}{4 \pi \varepsilon_0} = 8.987 \times 10^9 \ \mathrm{N \ m^2 \ C^{-2}}$	N m² C^-2	Gravitational constant	$G = 6.67 \times 10^{-11} \ \mathrm{N \ m^2 \ kg^{-2}}$	N m² kg^-2
Electric field of a point charge at a given position	$\mathbf{E} = \frac{k_e q}{r^2} \hat \mathbf{r}$ Where ${\mathbf{r}}$ is the vector from the point charge/mass to the given position, hence $\hat {\mathbf{r}}$ being its unit vector and $r^2$ being its magnitude squared	N C^-1	Gravitational field of a point mass at a given position	$\mathbf{g} = \frac{G m}{r^2} \hat \mathbf{r}$	N kg^-1 = m s^-2
Electric force on point charge 2 from point charge 1	$\mathbf{F} = \mathbf{E}_1 q_2 = \frac{k_e q_1 q_2}{r^2} \hat \mathbf{r}$ Where ${\mathbf{r}}$ is the vector from point charge/mass 1 to point charge/mass 2, hence $\hat {\mathbf{r}}$ beings its unit vector and $r^2$ being its magnitude squared	N	Gravitational force on point mass 2 from point mass 1	$\mathbf{F} = \mathbf{g}_1 m_2 = \frac{G m_1 m_2}{r^2} \hat \mathbf{r}$	N
Work done across an electric field	$W = \int \mathbf{F} \cdot \operatorname{d} \mathbf{x} = q \int \mathbf{E} \cdot \operatorname{d} \mathbf{x} = qV$	J = N m = C V	Work done across gravitational field	$W = \int \mathbf{F} \cdot \operatorname{d} \mathbf{x} = m \int \mathbf{g} \cdot \operatorname{d} \mathbf{x} \ ( = mg\Delta x )$ This last equality only holds when $\mathbf{g}$ is constant along the path taken and parallel to that path at every point. However, this is a common form of the formula seen in high-school textbooks where $\mathbf{g}$ is Earth's local gravity which is taken to be constant and down everywhere.	J = N m = kg m² s^-2
potential	Example	Example	Example	Example	Example
flux	Example	Example	Example	Example	Example
gauss's law	Example	Example	Example	Example	Example
Example	Example	Example	Example	Example	Example
Example	Example	Example	Example	Example	Example

AC Circuits and Masses on springs

Electricity like water in a pipe

File:Circuit analogy water pipes big.jpg

Since electrons cannot be seen and are intangible, a very popular model of explaining how electric circuit work is by the Water-in-Pipes model. This is a useful model to use because water flowing through pipes is a mechanical system that works in uch the same way as an electrical circuit, but with different units.

The main ideas are that:

The pipe is like the wire in the electric circuit
The pump is like the battery.
The pressure generated by the pump drives water through the pipe; that pressure is like the voltage generated by the battery which drives electrons through the circuit.
The seashells plug up the pipe and slow the flow of water, creating a pressure difference from one end to the other. In a similar way the resistance in the electric circuit resists the flow of electricity and creates a voltage drop from one end to the other. Energy is lost across the resistor and shows up as heat.

The power in the circuit equals the voltage times the current. The same power can be carried by a high voltage and a low current as is carried by a low voltage and a high current. The higher the current flow, however, the more energy is lost as heating of the wires. That's why high voltage and low current is used when transporting electrical energy along power lines. http://www.windows2universe.org/image_linking.html

User:AndreRD/Analogues in Physics

Translational Motion vs Rotational Motion

Electric Fields and Gravitational Fields

AC Circuits and Masses on springs

Electricity like water in a pipe

Mechanical waves, sound waves and light waves

E = ½xy² laws

Correspondence Principle

See Also

References and Notes

External Links