- #1

bernhard.rothenstein

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In summary, the simplest way to find out if a physical quantity is the component of a four vector is to use equations (1) and (2).

- #1

bernhard.rothenstein

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- #2

bernhard.rothenstein

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bernhard.rothenstein said:

I add some details.

Consider that observers from I detect simultaneously the space coordinates of the ends of a moving rod, at rest in the I', located along the x' axis. The result is that the proper length of the rod L

The space coordinates of the ends measured in the stationary frame do not depend on the time when they are measured. Consider that observers from I' detect simultaneously the space coordinates of the stationary rod, the measurements being associated with the space-time coordinates (x'

L=[tex]\gamma[/tex]L

which accounts for the Lorentz expansion (dilation?).

Equation (1) is derived by radar detection of the space coordinates of the moving rod.

Virtues of (1) are mentioned.

Which is the simplest way to show that (1) is the component of a four vector the length contracted length being not?

- #3

DrGreg

Science Advisor

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[tex]\textbf{X'} = \Lambda \, \textbf{X}[/tex]

where

[tex]\Lambda = \left[ \begin{array}{cccc}

\gamma & -\gamma v & 0 & 0 \\

-\gamma v & \gamma & 0 & 0 \\

0 & 0 & 1 & 0 \\

0 & 0 & 0 & 1 \\

\end{array} \right][/tex]

\gamma & -\gamma v & 0 & 0 \\

-\gamma v & \gamma & 0 & 0 \\

0 & 0 & 1 & 0 \\

0 & 0 & 0 & 1 \\

\end{array} \right][/tex]

"Lorentz-contracted length" doesn't satisfy this because the length is measured between two events that are simultaneous in the frame where the measurement is made. When you change frames, you also change events. The Lorentz transform applies only when you measure the

Other examples of 4-vectors are

Energy-momentum... [tex]\textbf{P} = (\frac{E}{c}, p_1, p_2, p_3)^T = m\frac{d\textbf{X}}{d\tau}[/tex]

4-force...... [tex]\textbf{F} = \frac{d\textbf{P}}{d\tau}[/tex]

4-current..... [tex]\textbf{J} = (\rho c, j_1, j_2, j_3)^T[/tex]

electromagnetic 4-potential [tex]\Psi = (\phi, a_1 c, a_2 c, a_3 c)^T[/tex]

4-force...... [tex]\textbf{F} = \frac{d\textbf{P}}{d\tau}[/tex]

4-current..... [tex]\textbf{J} = (\rho c, j_1, j_2, j_3)^T[/tex]

electromagnetic 4-potential [tex]\Psi = (\phi, a_1 c, a_2 c, a_3 c)^T[/tex]

(See four-vector on Wikipedia.)

- #4

bernhard.rothenstein

- 991

- 1

Thank you. Please have a look at the following lines.DrGreg said:xaxis), we can say this is a 4-vector because

[tex]\textbf{X'} = \Lambda \, \textbf{X}[/tex]

where

[tex]\Lambda = \left[ \begin{array}{cccc}

\gamma & -\gamma v & 0 & 0 \\

-\gamma v & \gamma & 0 & 0 \\

0 & 0 & 1 & 0 \\

0 & 0 & 0 & 1 \\

\end{array} \right][/tex]

"Lorentz-contracted length" doesn't satisfy this because the length is measured between two events that are simultaneous in the frame where the measurement is made. When you change frames, you also change events. The Lorentz transform applies only when you measure thesamepair of events in two different frames.

Other examples of 4-vectors are

Energy-momentum... [tex]\textbf{P} = (\frac{E}{c}, p_1, p_2, p_3)^T = m\frac{d\textbf{X}}{d\tau}[/tex]

4-force...... [tex]\textbf{F} = \frac{d\textbf{P}}{d\tau}[/tex]

4-current..... [tex]\textbf{J} = (\rho c, j_1, j_2, j_3)^T[/tex]

electromagnetic 4-potential [tex]\Psi = (\phi, a_1 c, a_2 c, a_3 c)^T[/tex]

(See four-vector on Wikipedia.)

Consider that in a 2D approach A is the scalar component of a "two vector" and let A

A=[tex]\gamma[/tex](u)A

Let

(1) and (2) ensuring the invariance of the relativistic interval.

As we see the Lorentz contracted length is not the component of a four vector

whereas the dilated length L=L

The dilated length could be obtained simultaneously detecting the ends of the rod in its rest frame. Did Einstein take into account that measurement variant?

Would you accept such an approach?

A four vector component is a mathematical concept used in the field of physics and special relativity. It is a set of four numbers that represent the position and direction of an object in space and time.

A four vector component is different from a regular vector in that it includes time as a fourth component, while a regular vector only has three components representing space. This allows four vector components to accurately describe the position and motion of an object in four-dimensional space-time.

A four vector component is usually denoted by a symbol with an arrow above it, such as *A* → or *p* →. It will also have four components, typically labeled as *x*, *y*, *z*, and *t* for space and time, respectively.

Four vector components are important in physics because they provide a way to accurately describe the position and motion of objects in space and time. They are used in special relativity, electromagnetism, and other areas of physics to calculate and describe physical phenomena.

Yes, four vector components can be used to describe any type of motion, as long as it occurs in four-dimensional space-time. This includes both linear and rotational motion, as well as motion at any speed, including the speed of light.

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