STR Notes

Special Theory of Relativity

Frame of Reference

A coordinate system relative to which the measurements are taken is known as frame of reference. A coordinate system in which the law of inertial is valid is known as inertial frame of reference, while the accelerated frame is known as non-inertial frame of reference.

Special Theory of Relativity

Postulates of Special Theory of Relativity:

In 1905, Albert Einstein formulated his special theory of relativity in terms of two postulates:

1. Principle of Relativity

The laws of Physics have the same form in all frames of reference moving with constant velocities with respect to one another. It can also be stated as

“The laws of Physics are invariant o a transformation between all inertial frames”

2. Principle of Constancy of Speed of Light

The speed of light in free space has the same value for all observers regardless of their state of motion. It can also be stated as:

The speed of light in free space has the same value “c” in all inertial frames of references.

Relativistic Effects:

Theory of Relativity shows the effects of relative motion on physical quantities. These effects are observed at relativistic speed. (The speed $\geq \frac{c}{10}$ is called relativistic speed, where c is speed of light.)

Discussion:

The first postulate shows that laws of Physics are absolute and universal and are same for all inertial observers. So the laws of Physics that hold for one inertial observer can’t be violated for any other inertial observer.

To understand 2^nd postulate, consider three observers A, B and C at rest in three different inertial frames.

• A flash of light emitted by observer A is observed by him to travel at speed c.

• If the frame of B is moving away from A at a speed of $\frac{c}{4}$ then according to Galilean Kinematics, B measure the speed of flash emitted by A: $c - \frac{c}{4} = \frac{3c}{4}$.

• If the frame C is moving towards A with a speed of $\frac{c}{4}$ then according to Galilean Transformation, C measures the value $c + \frac{c}{4} = \frac{5c}{4}$ for the speed of flash emitted by A.

However according to 2^nd postulate all the three observers measure the same speed of flash of light. However ordinary objects don’t obey 2^nd postulate e.g.,

$\begin{pmatrix} Velocity\ of\ a\ projectile\ \\ fired\ from\ a\ moving\ car \\ \ with\ respect\ to\ ground \\ \end{pmatrix} = \begin{pmatrix} Velocity\ of\ \ \\ projectile \\ w.\ r.t\ car \\ \end{pmatrix} + \begin{pmatrix} Velocity\ of\ \ \\ car \\ w.\ r.t\ ground \\ \end{pmatrix}$

But velocities of waves and particles moving at speeds close to c behave in this wave. When Einstein put forward these postulates, there was no experimental test for the verification of these postulates. However, in 1964 a proton accelerator produced a beam of neutral pions (π^∘) which rapidly decay into γ − rays:

π^∘ → γ + γ

Now γ − rays are electromagnetic waves and move with the speed of light. The speed of moving Pions was measured equal to 0.99995c.

According to Galileo, the γ − rays emitted in the direction of motion of Pions should have a speed equal to c + 0.99995c But the measured speed of γ − rays was equal to c. This is consistent with 2^nd postulate.

Galilean Transformation

Consider two observers in two different inertial frames S and S′. Frame S is at rest and S′ is moving with uniform velocity v along x-axis with respect to frame S. Suppose at t = 0, the origins of two frames coincide.

Both the observers observe the same event. The position and time of event observed by S is denoted by (x,y,z,t) and position and time of the event observed by S′ is denoted by (x′,y′,z′,t′). According to Galilean Transformation:

$$\begin{Bmatrix} x’ = x - vt\ \\ y’ = y\ \\ z’ = z \\ t^{'} = t \\ \end{Bmatrix}$$

Fundamental Equation of Special Theory of Relativity

Consider two observers in two different inertial frames S and S′. Frame S is at rest and S′ is moving with uniform velocity v along x-axis with respect to frame S. Suppose at t = 0, the origins of two frames coincide.

Both the observers observe the same event. The position and time of event observed by S is denoted by (x,y,z,t) and position and time of the event observed by S′ is denoted by (x′,y′,z′,t′). Consider a wave of light starts from O and O^′ at t = 0 with speed c. Let the wave reaches a point P after time t from O and takes the time t^′ to reach at P from point O^′. Then the distance covered by light ray from point O to point P:

|OP| = ct

$$\Longrightarrow \sqrt{x^{2} + y^{2} + z^{2}} = ct$$

 ⇒ x² + y² + z² = c²t²

 ⇒ x² + y² + z² − c²t² = 0   −  −  −  − −    (1)

And the distance covered by light ray from point O′ to point P:

|O′P| = ct′

$$\Longrightarrow \sqrt{{x'}^{2} + {y'}^{2} + {z'}^{2}} = ct'$$

 ⇒ x′² + y′² + z′² = c²t′²

 ⇒ x^′2 + y^′2 + z^′2 − c²t^′2 = 0    −  −  −  − −    (2)

Comparing these equations, we get:

x² + y² + z² − c²t² = x^′2 + y^′2 + z^′2 − c²t^′2   −  −  −  − −   (3)

This is the fundamental equation of special theory of relativity given by Einstein in 1905.

Galilean Transformations Doesn’t Satisfy the Fundamental Equation of Relativity

Applying the values of x′, y′, z′, t′ from Galilean Transformation in Fundamental Equation of Relativity

x² + y² + z² − c²t² = (x−vt)² + y² + z² − c²t²

 ⇒ x² = (x−vt)²

This is clearly impossible until t = 0. Hence Galilean Transformation fail to satisfy Fundamental Equation of Relativity.

Lorentz Transformations Satisfy the Fundamental Equation of Relativity

Hence we need such transformations which satisfy Fundamental Equation of Special Theory of Relativity. Such transformations are called Lorentz Transformation. These are given below:

$$\begin{Bmatrix} x’ = \gamma\ (x - vt)\ \\ y’ = y\ \\ z’ = z \\ t^{'} = \ \gamma\left( t - \frac{vx}{c^{2}} \right) \\ \end{Bmatrix}$$

Derivation of Lorentz Transformation

Consider two inertial frames of reference S and S’. Frame S is at rest and S’ is moving with velocity v in the direction of increasing x.

As relative velocity v is along x − axis. Moreover, x − axis and x’ − axis coincides at t = 0, so: y’ = y and z’ = z. So the fundamental equation of special theory of relativity x² + y² + z² − c²t² = x′² + y′² + z′² − c²t′² becomes:

x² − c²t² = x^′2 − c²t^′2    −  −  − −  (A)

• As O’ move along x − axis with relative velocity v with respect to ‘S’, so distance covered by O’ with respect to S after time t will be vt.

• As O also appear to move along negative x’ − axis with relative velocity  − v. So the distance covered by O with respect to S’ after time t is equal to  − vt′.

These two requirements can be satisfied by putting:

x’ = a(x−vt)   −  −  − −   (1)

and

x = b(x’+vt′)   −  −  − −   (2)

In equation (1) and (2), if we know values of ‘a’ and ‘b’, then we can find the relations between (x,y,z,t) and (x′,y′,z′,t′), which satisfy equation (A).

To find ‘a’ and ‘b’, we put the value of x’ from equation (1) in equation (2), we get:

x = b[x’ + vt′]

 ⇒ x = b[a(x−vt)+vt′]

 ⇒ x = ab(x−vt) + bvt′

 ⇒ x = abx − abvt + bvt′

 ⇒ bvt^′ = abvt − abx + x

$$\Longrightarrow t^{'} = at - \frac{ax}{v} + \frac{x}{bv}$$

$$\Longrightarrow t^{'} = at - \frac{ax}{v}\left( 1 - \frac{1}{ab} \right)$$

$$\Longrightarrow t^{'} = a\left\lbrack t - \frac{x}{v}\left( 1 - \frac{1}{ab} \right) \right\rbrack\ \ - - - - \ \ \ (3)$$

Putting the values of x’ and t^′ from equations (1) and (3) in equation (A):

x² − c²t² = x^′2 − c²t^′2

$$\Longrightarrow x^{2} - c^{2}t^{2} = \left\lbrack a(x - vt) \right\rbrack^{2} - c^{2}a^{2}\left\lbrack t - \frac{x}{v}\left( 1 - \frac{1}{ab} \right) \right\rbrack^{2}$$

$$\Longrightarrow x^{2} - c^{2}t^{2} = a^{2}\left( x^{2} + v^{2}t^{2} - 2xvt \right) - c^{2}a^{2}\left\lbrack t^{2} - 2\frac{xt}{v}\left( 1 - \frac{1}{ab} \right) + \frac{x^{2}}{v^{2}}\left( 1 - \frac{1}{ab} \right)^{2} \right\rbrack$$

$$\Longrightarrow x^{2} - c^{2}t^{2} = a^{2}x^{2} + {a^{2}v}^{2}t^{2} - 2a^{2}xvt - c^{2}a^{2}t^{2} + \frac{2c^{2}a^{2}xt}{v}\left( 1 - \frac{1}{ab} \right) - c^{2}a^{2}\frac{x^{2}}{v^{2}}\left( 1 - \frac{1}{ab} \right)^{2}$$

$$\Longrightarrow x^{2} - c^{2}t^{2} - a^{2}x^{2} - {a^{2}v}^{2}t^{2} + 2a^{2}xvt + c^{2}a^{2}t^{2} - \frac{2c^{2}a^{2}xt}{v}\left( 1 - \frac{1}{ab} \right) + c^{2}a^{2}\frac{x^{2}}{v^{2}}\left( 1 - \frac{1}{ab} \right)^{2} = 0$$

$$\Longrightarrow x^{2}\left\lbrack 1 - a^{2} + \frac{c^{2}a^{2}}{v^{2}}\left( 1 - \frac{1}{ab} \right)^{2} \right\rbrack + xt\left\lbrack 2a^{2}v - \frac{2c^{2}a^{2}}{v}\left( 1 - \frac{1}{ab} \right) \right\rbrack + t^{2}\left\lbrack - c^{2} - {a^{2}v}^{2} + c^{2}a^{2} \right\rbrack = 0$$

This relation must hold for all values of x and t. So the coefficients of x², xt and t² must be zero separately. So we get three equations:

$$1 - a^{2} + \frac{c^{2}a^{2}}{v^{2}}\left( 1 - \frac{1}{ab} \right)^{2} = 0\ \ - - - - \ \ \ (4)$$

$$2a^{2}v - \frac{2c^{2}a^{2}}{v}\left( 1 - \frac{1}{ab} \right) = 0\ \ - - - - \ \ \ (5)$$

 − c² − a²v² + c²a² = 0   −  −  − −   (6)

From equation (6), we have:

 − c² − a²v² + c²a² = 0

 ⇒ c² = c²a² − a²v²

$$\Longrightarrow c^{2} = a^{2}c^{2}\left( 1 - \frac{v^{2}}{c^{2}} \right)$$

$$\Longrightarrow 1 = a^{2}\left( 1 - \frac{v^{2}}{c^{2}} \right)$$

$$\Longrightarrow a^{2} = \frac{1}{\left( 1 - \frac{v^{2}}{c^{2}} \right)}$$

$$\Longrightarrow a = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$$

Now consider equation (5):

$$2a^{2}v - \frac{2c^{2}a^{2}}{v}\left( 1 - \frac{1}{ab} \right) = 0$$

$$\Longrightarrow 2a^{2}v = \frac{2c^{2}a^{2}}{v}\left( 1 - \frac{1}{ab} \right)$$

$$\Longrightarrow v^{2} = c^{2}\left( 1 - \frac{1}{ab} \right)$$

$$\Longrightarrow \frac{v^{2}}{c^{2}} = 1 - \frac{1}{ab}$$

$$\Longrightarrow \frac{1}{ab} = 1 - \frac{v^{2}}{c^{2}}$$

$$\Longrightarrow \frac{1}{ab} = \frac{1}{a^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \because a = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} \Longrightarrow \frac{1}{a} = \sqrt{1 - \frac{v^{2}}{c^{2}}}$$

$$\Longrightarrow \frac{1}{b} = \frac{1}{a}\ \ \ \ $$

 ⇒ b = a    

$$\Longrightarrow b = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$$

Putting values of ‘a’ and ‘b’ in equation (1) and (2), we get:

$$x’ = \frac{x - vt}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} = \gamma\ (x - vt)\ \ \ - - - - \ \ \ (7)$$

$$x = \frac{x’ + vt}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}\ = \gamma\ (x’ + vt)\ \ \ - - - - \ \ \ (8)$$

Where $\gamma = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$ is called Lorentz’s factor.

Now putting $a = b = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$ in equation (3), we get:

$$t^{'} = a\left\lbrack t - \frac{x}{v}\left( 1 - \frac{1}{ab} \right) \right\rbrack$$

$${\Longrightarrow t}^{'} = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}\left\lbrack t - \frac{x}{v}\left( 1 - \left\{ 1 - \frac{v^{2}}{c^{2}} \right\} \right) \right\rbrack$$

$${\Longrightarrow t}^{'} = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}\left\lbrack t - \frac{x}{v}\left( 1 - 1 + \frac{v^{2}}{c^{2}} \right) \right\rbrack$$

$${\Longrightarrow t}^{'} = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}\left\lbrack t - \frac{x}{v}\left( \frac{v^{2}}{c^{2}} \right) \right\rbrack$$

$${\Longrightarrow t}^{'} = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}\left( t - \frac{vx}{c^{2}} \right)$$

$${\Longrightarrow t}^{'} = \ \gamma\left( t - \frac{vx}{c^{2}} \right)\ \ \ \ \ \ - - - - \ \ \ (9)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \because\gamma = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$$

Conclusion:

The set of transformation equations

$$\begin{Bmatrix} x’ = \gamma\ (x - vt)\ \\ y’ = y\ \\ z’ = z \\ t^{'} = \ \gamma\left( t - \frac{vx}{c^{2}} \right) \\ \end{Bmatrix}$$

is called Lorentz’s Transformation. When v ≪ c i.e., $\frac{v}{c} \approx 0$, the Lorentz Transformations transform into Galilean transformation. Thus the Galilean Transformations is special case of Lorentz Transformation.

Inverse Lorentz Transformation

The Lorentz Transformation equations are as follows:

$$\begin{Bmatrix} x’ = \gamma\ (x - vt)\ \ \ \ - - - - \ \ \ (1) \\ y’ = y\ \ \ \ \ - - - - \ \ \ (2) \\ z’ = z\ \ \ \ - - - - \ \ \ (3) \\ t^{'} = \ \gamma\left( t - \frac{vx}{c^{2}} \right)\ \ \ \ - - - - \ \ \ (4) \\ \end{Bmatrix}$$

From equation (4):

$$\frac{t^{'}}{\gamma} = \ t - \frac{vx}{c^{2}}$$

$$\Longrightarrow t = \ \frac{vx}{c^{2}} + \frac{t^{'}}{\gamma}$$

Putting the value of t in equation (1):

$$x’ = \gamma\ \left\lbrack x - v\left( \ \frac{vx}{c^{2}} + \frac{t^{'}}{\gamma} \right) \right\rbrack$$

$$\Longrightarrow \frac{x’}{\gamma} = \ x - \frac{v^{2}x}{c^{2}} - \frac{vt^{'}}{\gamma}$$

$$\Longrightarrow \frac{x’}{\gamma} = \ x\left( 1 - \frac{v^{2}}{c^{2}} \right) - \frac{vt^{'}}{\gamma}$$

$$\Longrightarrow x\left( 1 - \frac{v^{2}}{c^{2}} \right) = \ \frac{x’}{\gamma} + \frac{vt^{'}}{\gamma}$$

$$\Longrightarrow \frac{x}{\gamma^{2}} = \ \frac{1}{\gamma}\left( x’ + vt^{'} \right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \because\gamma = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} \Longrightarrow \frac{1}{\gamma^{2}} = 1 - \frac{v^{2}}{c^{2}}$$

 ⇒ x = γ (x’+vt^′)   −  −  −  − −    (5)

Now consider equation (1):

x’ = γ (x−vt)

$$\Longrightarrow \frac{x’}{\gamma} = x - vt$$

$$\Longrightarrow x = \frac{x’}{\gamma} + vt$$

Putting value of x in equation (4), we get:

$$t^{'} = \ \gamma\left\lbrack t - \frac{v}{c^{2}}\left( \frac{x’}{\gamma} + vt \right) \right\rbrack$$

$$\Longrightarrow \frac{t^{'}}{\gamma} = \ t - \frac{vx’}{\gamma c^{2}} - \frac{v^{2}}{c^{2}}t$$

$$\Longrightarrow \frac{t^{'}}{\gamma} = \ t\left( 1 - \frac{v^{2}}{c^{2}} \right) - \frac{vx’}{\gamma c^{2}}$$

$$\Longrightarrow \frac{t^{'}}{\gamma} = \ \frac{t}{\gamma^{2}} - \frac{vx’}{\gamma c^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \because\gamma = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} \Longrightarrow \frac{1}{\gamma^{2}} = 1 - \frac{v^{2}}{c^{2}}$$

$$\Longrightarrow \frac{t}{\gamma^{2}} = \frac{t^{'}}{\gamma} + \frac{vx’}{\gamma c^{2}}\ $$

$$\Longrightarrow \frac{t}{\gamma^{2}} = \frac{1}{\gamma}\left( t^{'} + \frac{vx’}{c^{2}} \right)\ $$

$$\Longrightarrow t = \gamma\left( t^{'} + \frac{vx’}{c^{2}} \right)\ \ \ - - - - - \ \ \ \ (6)$$

Hence the Inverse Lorentz Transformations are:

$$\begin{Bmatrix} x = \gamma\ (x’ + vt) \\ y = y’ \\ z = z’ \\ t = \ \gamma\left( t^{'} + \frac{vx}{c^{2}} \right) \\ \end{Bmatrix}$$

Important Note:

We can obtain the inverse Lorentz Transformations just by interchanging primed and unprimed coordinates and replacing v by –v.

Transformation of Velocities

The equations of Lorentz Transformations can be used to get a relation between velocity u of a particle measured by an observer in S frame and velocity u′ of the same particle measured by an observer in S′ frame who is moving with velocity v with respect to S.

Suppose according to SFOR, particle moves from (x₁, y₁, z₁, t₁) to (x₂, y₂, z₂, t₂) and according to S′FOR, the particle moves from (x′₁, y′₁, z′₁, t′₁) to (x′₂, y′₂, z′₂, t′₂).

The x-component of velocity u′ measured by S′FOR will be:

$${u'}_{x} = \frac{\mathrm{\Delta}x'}{\mathrm{\Delta}t'}\ \ - - - - - \ \ \ (1)$$

By Lorentz Transformations, we have:

x’ = γ (x−vt)

 ⇒ Δx’ = γ (Δx−vΔt)

Also,

$$t^{'} = \ \gamma\left( t - \frac{vx}{c^{2}} \right)$$

$$\Longrightarrow {\mathrm{\Delta}t}^{'} = \ \gamma\left( \mathrm{\Delta}t - \frac{v\mathrm{\Delta}x}{c^{2}} \right)$$

Putting values in equation (1):

$${u'}_{x} = \frac{\mathrm{\Delta}x'}{\mathrm{\Delta}t'}$$

$$\Longrightarrow {u'}_{x} = \frac{\gamma\ (\mathrm{\Delta}x - v\mathrm{\Delta}t)}{\gamma\left( \mathrm{\Delta}t - \frac{v\mathrm{\Delta}x}{c^{2}} \right)}$$

$$\Longrightarrow {u'}_{x} = \frac{\mathrm{\Delta}x - v\mathrm{\Delta}t}{\mathrm{\Delta}t - \frac{v\mathrm{\Delta}x}{c^{2}}}$$

$$\Longrightarrow {u'}_{x} = \frac{\mathrm{\Delta}t\left( \frac{\mathrm{\Delta}x}{\mathrm{\Delta}t} - v \right)}{\mathrm{\Delta}t\left( 1 - \frac{v}{c^{2}}\frac{\mathrm{\Delta}x}{\mathrm{\Delta}t} \right)}$$

$$\Longrightarrow {u^{'}}_{x} = \frac{\frac{\mathrm{\Delta}x}{\mathrm{\Delta}t} - v}{1 - \frac{v}{c^{2}}\frac{\mathrm{\Delta}x}{\mathrm{\Delta}t}}$$

$$\Longrightarrow {u^{'}}_{x} = \frac{u_{x} - v}{1 - \frac{vu_{x}}{c^{2}}}\ \ \ \ \ \ \ \because\frac{\mathrm{\Delta}x}{\mathrm{\Delta}t} = u_{x}$$

The y-component of velocity u′ measured by S′FOR will be:

$${u'}_{y} = \frac{\mathrm{\Delta}y'}{\mathrm{\Delta}t'}\ \ - - - - - \ \ \ (2)$$

As y^′ = y ⇒ Δy^′ = Δy

Putting values in equation (2), we get:

$${u'}_{y} = \frac{\mathrm{\Delta}y}{\gamma\left( \mathrm{\Delta}t - \frac{v\mathrm{\Delta}x}{c^{2}} \right)}$$

$$\Longrightarrow {u'}_{y} = \frac{\mathrm{\Delta}y}{\gamma\mathrm{\Delta}t\left( 1 - \frac{v}{c^{2}}\frac{\mathrm{\Delta}x}{\mathrm{\Delta}t} \right)}$$

$$\Longrightarrow {u'}_{y} = \frac{\left( \frac{\mathrm{\Delta}y}{\mathrm{\Delta}t} \right)}{\gamma\left( 1 - \frac{v}{c^{2}}\frac{\mathrm{\Delta}x}{\mathrm{\Delta}t} \right)}$$

$$\Longrightarrow {u^{'}}_{y} = \frac{u_{y}}{\gamma\left( 1 - \frac{vu_{x}}{c^{2}} \right)}\ \ \ \ \ \because\frac{\mathrm{\Delta}x}{\mathrm{\Delta}t} = u_{x}\ \&\ \frac{\mathrm{\Delta}y}{\mathrm{\Delta}t} = u_{y}\ $$

Similarly,

$${u^{'}}_{z} = \frac{u_{z}}{\gamma\left( 1 - \frac{vu_{x}}{c^{2}} \right)}$$

It should be noted that u^′_y ≠ u_y and u^′_z = u_z even though Δy^′ = Δy and Δz^′ = Δz. This is another difference between Lorentz and Galilean transformations.

Inverse Velocity Transformations

We can obtain inverse velocity transformation from equations of velocity transformation simply by changing v by –v replacing primed coordinates with unprimed coordinates and vice versa. So inverse velocity transformations are:

$$u_{x} = \frac{{u^{'}}_{x} + v}{1 + \frac{v{u^{'}}_{x}}{c^{2}}}$$

$$u_{y} = \frac{{u^{'}}_{y}}{\gamma\left( 1 + \frac{v{u^{'}}_{y}}{c^{2}} \right)}$$

$$u_{z} = \frac{{u^{'}}_{z}}{\gamma\left( 1 + \frac{v{u^{'}}_{z}}{c^{2}} \right)}$$

Lorentz Velocity Transformations under Non-Relativistic Limit

Under non-relativistic limit (i.e., for u ≪ c, we put $\frac{v}{c^{2}} = 0$ ) the equations of velocity transformations takes the form:

u^′_x = u_x − v

u^′_y = u_y

u^′_z = u_z

This set of equation is called Galilean Velocity Transformation. So under non-relativistic limits, the Lorentz Velocity Transformation change into Galilean Velocity Transformation.

The Lorentz Velocity Transformation and Einstein’s 2^nd Postulate

We can derive the result of Einstein’s 2^nd postulate from Lorentz Velocity Transformations. According to Einstein’s 2^nd postulate, the speed of light is constant for all observers. So speed c measured by an observer must also be measured to be c by any other observer.

Suppose the two observers observe a common event of passage of light beam along x-axis in frame S and S’. According to observer in S, the velocity of light beam along x-axis u_x = c and u_y = u_z = 0. So, according to Lorentz Velocity Transformation, the velocity measured by S’:

$${u^{'}}_{x} = \frac{u_{x} - v}{1 - \frac{vu_{x}}{c^{2}}} = \frac{c - v}{1 - \frac{vc}{c^{2}}}$$

$$\Longrightarrow {u^{'}}_{x} = \frac{c - v}{\left( 1 - \frac{v}{c} \right)} = \frac{c - v}{\left( \frac{c - v}{c} \right)}$$

 ⇒ u^′_x = c

And

$${u^{'}}_{y} = \frac{u_{y}}{\gamma\left( 1 - \frac{vu_{x}}{c^{2}} \right)} = \frac{0}{\gamma\left( 1 - \frac{v(0)}{c^{2}} \right)} = 0\ \ $$

$$\&\ {u^{'}}_{z} = \frac{u_{z}}{\gamma\left( 1 - \frac{vu_{x}}{c^{2}} \right)} = \frac{0}{\gamma\left( 1 - \frac{v(0)}{c^{2}} \right)} = 0$$

Therefore, velocities measured by S’FOR are u^′_x = c, u^′_y = u^′_z = 0. So the observer S’ also measures the same speed. Hence the speed of light is same for all observers.

CONSEQUENCE OF SPECIAL THEORY OF RELATIVITY

Relativity of Time

Consider two frames of references S and S′. S is at rest and S′ is moving with uniform velocity v with respect to S. Suppose an event occurs at one and same place ‘x’ in frame S.

• The duration of event measured by the observer in frame S is Δt = t₂ − t₁.

• The duration of same event measured by the observer in frame S′ is Δt′ = t′₂ − t′₁.

By using Lorentz Transformation:

$$t_{1}^{'} = \ \gamma\left( t_{1} - \frac{vx_{1}}{c^{2}} \right)$$

$$and\ t_{2}^{'} = \ \gamma\left( t_{2} - \frac{vx_{2}}{c^{2}} \right)$$

Now

Δt′ = t′₂ − t′₁

$$\Longrightarrow \mathrm{\Delta}t' = \ \gamma\left( t_{2} - \frac{vx_{2}}{c^{2}} \right) - \ \gamma\left( t_{1} - \frac{vx_{1}}{c^{2}} \right)$$

$$\Longrightarrow \mathrm{\Delta}t' = \ \gamma\left( t_{2} - \frac{vx_{2}}{c^{2}} - t_{1} + \frac{vx_{1}}{c^{2}} \right)$$

$$\Longrightarrow \mathrm{\Delta}t^{'} = \ \gamma\left\lbrack \left( t_{2} - t_{1} \right) - \frac{v}{c^{2}}\left( x_{2} - x_{1} \right) \right\rbrack\ \ \ - - - - \ \ \ \ (1)$$

Because the event occurs at the same place, therefore: x₁ = x₂. Equation (1) takes the form:

$$\mathrm{\Delta}t^{'} = \ \gamma\left\lbrack \mathrm{\Delta}t - \frac{v}{c^{2}}(0) \right\rbrack$$

 ⇒ Δt^′= γΔt

$$\Longrightarrow \mathrm{\Delta}t^{'} = \frac{\mathrm{\Delta}t}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}\ \ \ \ \ \ \ \ \ \ \because\gamma = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$$

Since $\frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}\ > 1$

So, Δt^′ > Δt

Hence the observer in frame S′ will conclude that the clock in frame S is slowed down i.e., time is dilated.

Relativity of Length (Length Contraction)

Consider a rod lying at rest along x-axis in stationary frame S. Let the coordinates of its ends in this frame be x₁ and x₂. Then, length l of the rod is called proper length and described as:

l = x₂ − x₁

Let the length of the rod seen in moving frame S′ moving with velocity v be l′. Let the coordinates of ends of the rod in S^′ frame of reference (FOR) are x₁^′ and x₂^′. Then the length l of the rod observed with respect to S′FOR will be:

l′ = x₂^′ − x₁^′

It should be noted that the measurements are made simultaneously in both frames.

By Lorentz Transformation, we have

1. x₁ = γ(x₁^′+vt₁^′)   −  −  −  −  − (1)

2. x₂ = γ(x₂^′+vt₂^′)   −  −  −  −  − (2)

Subtracting (1) and (2), we have

x₂ − x₁ = γ[x₂^′−x₁^′+vt₂^′−vt₁^′]

 ⇒ x₂ − x₁ = γ[x₂^′−x₁^′+v(t₂^′−t₁^′)]

 ⇒ l = γ[l^′+vΔt^′]     ∵Δt^′ = t₂^′ − t₁^′

Putting Δt^′ = 0, because measurements are made simultaneously:

l = γl^′

$$\Longrightarrow l^{'} = \frac{l}{\gamma}$$

$$\Longrightarrow l^{'} = l\sqrt{1 - \frac{v^{2}}{c^{2}}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \because\frac{1}{\gamma} = \sqrt{1 - \frac{v^{2}}{c^{2}}}$$

$$Since\ \sqrt{1 - \frac{v^{2}}{c^{2}}} < 1$$

Therefore,

l^′ < l

i.e., the length l of the rod appears to be reduced in moving frame S′. This effect is called Length Contraction.

Relativity of Mass

The Einstein’s Energy-Mass relationship is described as:

E = mc²

Here E, m & c represent energy, mass and speed of light respectively.

Differentiating both sides, we have:

dE = c²dm   −  −  −  −  − (1)                         ∵In STR, c (speed of light) is constant

Let the work dW is done on the object by the force F which displaced it through distance dr. As the work done dW on an object appears in the form of change in its energy dE, therefore, we can write:

dE = dW

 ⇒ dE = F dr               ∵dW = F dr

By Newton’s second law of motion, the applied force on an object is equal to time rate of change of linear momentum p i.e., $F = \frac{dp}{dt}$. Therefore,

$$dE = \frac{dp}{dt}\ dr$$

$$\Longrightarrow dE = \left\lbrack \frac{d}{dt}(mv) \right\rbrack\ dr\ \ \ \ \ \ \ \ \ \ \ \ \because p = mv$$

$$\Longrightarrow dE = \left\lbrack m\frac{dv}{dt} + v\frac{dm}{dt} \right\rbrack\ dr$$

$$\Longrightarrow dE = \left\lbrack m\frac{dv}{dt} + v\frac{dm}{dt} \right\rbrack\ v\ dt\ \ \ \ \ \ \ \ \ \ \because dr = v\ dt$$

 ⇒ dE = mvdv + v²dm   −  −  −  − −   (2)

Comparing eq. (1) and (2), we get:

c²dm   = mvdv + v²dm  

 ⇒ c²dm − v²dm = mvdv  

 ⇒ (c²−v²)dm = mvdv  

$$\Longrightarrow \frac{dm}{m} = \frac{vdv}{\left( c^{2} - v^{2} \right)}\ \ $$

$$\Longrightarrow \frac{dm}{m} = \frac{vdv}{c^{2}\left( 1 - \frac{v^{2}}{c^{2}} \right)}\ \ $$

$$\Longrightarrow \frac{dm}{m} = \frac{\frac{v}{c^{2}}dv}{\left( 1 - \frac{v^{2}}{c^{2}} \right)}\ \ $$

Integrating both sides, we have:

$$\int_{m_{o}}^{m}\frac{dm}{m} = \int_{0}^{v}\frac{\frac{v}{c^{2}}dv}{\left( 1 - \frac{v^{2}}{c^{2}} \right)}\ \ $$

$$\Longrightarrow \int_{m_{o}}^{m}\frac{dm}{m} = \frac{1}{( - 2)}\int_{0}^{v}\frac{( - 2)\frac{v}{c^{2}}dv}{\left( 1 - \frac{v^{2}}{c^{2}} \right)}\ \ $$

$$\Longrightarrow \left| \ln m \right|_{m_{o}}^{m} = \left| - \frac{1}{2}\ln\left( 1 - \frac{v^{2}}{c^{2}} \right) \right|_{0}^{v}\ \ $$

$$\Longrightarrow \left| \ln m \right|_{m_{o}}^{m} = \left| \ln\left( 1 - \frac{v^{2}}{c^{2}} \right)^{- \frac{1}{2}} \right|_{0}^{v}$$

$$\Longrightarrow \ln m - \ln m_{o} = \ln\left( 1 - \frac{v^{2}}{c^{2}} \right)^{- \frac{1}{2}} - \ln{(1)}$$

$$\Longrightarrow \ln\frac{m}{m_{o}} = \ln\left( 1 - \frac{v^{2}}{c^{2}} \right)^{- \frac{1}{2}}\ \ \ \ \ \ \ \ \ \ \ \because\ln(1) = 0$$

$$\Longrightarrow \frac{m}{m_{o}} = \left( 1 - \frac{v^{2}}{c^{2}} \right)^{- \frac{1}{2}}$$

$$\Longrightarrow m = m_{o}\left( 1 - \frac{v^{2}}{c^{2}} \right)^{- \frac{1}{2}}$$

$$\Longrightarrow m = \frac{m_{o}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$$

This is the expression relativistic mass that describe mass variation with respect to velocity.

Equivalence of Mass and Energy (or Proof of E = mc²)

Suppose a force F acts on a body and as the result of this force, the body covers a distance dx in direction of force. The work done by this force is:

$$dW = \overrightarrow{F}.\overrightarrow{dx}$$

 ⇒ dW = F dx   −  −  − −  (1)   ∵θ = 0^∘ 

By Work-Energy Theorem, the work done dW on a body result in increase of its kinetic energy dK:

dW = dK   −  −  − −   (2)  

Equating (1) and (2), we have:

dK = F dx  −  −  − −   (3)

By Newton’s 2^nd Law of Motion, the time rate of change of linear momentum of body is equal applied force:

$$F = \frac{dp}{dt}$$

$$\Longrightarrow F = \frac{d}{dt}(mv)$$

$$\Longrightarrow F = m\ \frac{dv}{dt} + v\ \frac{dm}{dt}$$

Equation (3) becomes:

$$dK = \left( m\ \frac{dv}{dt} + v\ \frac{dm}{dt} \right)\ dx$$

$$\Longrightarrow dK = m\ \frac{dv}{dt}\ dx\ + v\ \frac{dm}{dt}dx$$

$$\Longrightarrow dK = m\ dv\frac{dx}{dt}\ \ + v\ dm\frac{dx}{dt}$$

 ⇒ dK = mv dv   + v² dm    −  −  − −    (4)

From relativistic mechanics,

$$m = \frac{m_{o}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$$

$$\Longrightarrow m = {m_{o}\left( 1 - \frac{v^{2}}{c^{2}} \right)}^{\frac{- 1}{2}}$$

$$\Longrightarrow dm = {m_{o}\left( \frac{- 1}{2} \right)\left( 1 - \frac{v^{2}}{c^{2}} \right)}^{\frac{- 3}{2}}\left( \frac{- 2v}{c^{2}} \right)dv$$

$$\Longrightarrow dm = \frac{m_{o}}{\left( 1 - \frac{v^{2}}{c^{2}} \right)^{\frac{3}{2}}}\ \ \ \frac{v}{c^{2}}dv$$

$$\Longrightarrow dm = \frac{m_{o}}{\left( 1 - \frac{v^{2}}{c^{2}} \right)^{\frac{1}{2}}}\ \ \ \frac{1}{\left( 1 - \frac{v^{2}}{c^{2}} \right)}\ \ \ \frac{v}{c^{2}}dv$$

$$\Longrightarrow dm = m\ \frac{1}{\left( \frac{c^{2} - v^{2}}{c^{2}} \right)}\ \ \ \frac{v}{c^{2}}dv$$

$$\Longrightarrow dm = \frac{mv\ dv}{c^{2} - v^{2}}$$

 ⇒ mv dv = (c²−v²)dm

Putting values in equation (4), we get:

dK = (c²−v²)dm   + v² dm

 ⇒ dK = (c²−v²)dm   + v² dm

 ⇒ dK = (c²−v²+v²)dm

 ⇒ dK = c² dm

Integrating bother sides:

K = c² m + A   −  −  − −   (5)

where A is constant of integration.

At t = 0, m = m_o, and K = 0, equation (5) ⇒ A =  − m_oc² 

Now the equation (5) becomes:

K = mc² − m_oc²

 ⇒ K + m_oc² = mc²   −  −  − −   (6)

This equation shows that when K = 0, the body still possess some energy equal to m_oc², called rest mass energy. Here K + m_oc² = E is called total energy. Equation (6) takes the form:

E = mc²

This equation is called Einstein’s Energy-Mass Relationship.

Relativistic Energy

From Einstein’s Energy-Mass Relationship:

E = mc²

$$\Longrightarrow E = {\frac{m_{o}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}c}^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \because m = \frac{m_{o}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$$

$$\Longrightarrow E^{2} = \frac{m_{o}^{2}c^{4}}{\left( 1 - \frac{v^{2}}{c^{2}} \right)}$$

$$\Longrightarrow \frac{E^{2}}{c^{2}} = \frac{m_{o}^{2}c^{2}}{\left( 1 - \frac{v^{2}}{c^{2}} \right)}\ \ \ \ - - - - \ \ \ \ (1)$$

The linear momentum p a particle having mass m moving with velocity v is described as:

p = mv

$$\Longrightarrow p = \frac{m_{o}v}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$$

$$\Longrightarrow p^{2} = \frac{m_{o}^{2}v^{2}}{\left( 1 - \frac{v^{2}}{c^{2}} \right)}\ \ - - - - \ \ \ \ (2)\ $$

Subtracting equation (1) and (2):

$$\frac{E^{2}}{c^{2}} - p^{2} = \frac{m_{o}^{2}c^{2}}{\left( 1 - \frac{v^{2}}{c^{2}} \right)} - \frac{m_{o}^{2}v^{2}}{\left( 1 - \frac{v^{2}}{c^{2}} \right)}$$

$$\Longrightarrow \frac{E^{2}}{c^{2}} - p^{2} = \frac{m_{o}^{2}}{\left( 1 - \frac{v^{2}}{c^{2}} \right)}\left( c^{2} - v^{2} \right)$$

$$\Longrightarrow \frac{E^{2}}{c^{2}} - p^{2} = \frac{m_{o}^{2}}{\left( \frac{c^{2} - v^{2}}{c^{2}} \right)}\left( c^{2} - v^{2} \right)$$

$$\Longrightarrow \frac{E^{2}}{c^{2}} - p^{2} = m_{o}^{2}c^{2}$$

 ⇒ E² = p²c² + m_o²c⁴

$$\Longrightarrow E = \sqrt{p^{2}c^{2} + m_{o}^{2}c^{4}}$$

This is the expression of relativistic energy.

Sample Problem 6: What is momentum of a proton moving with speed of v = 0.86 c?

Solution:

Rest mass of proton m = 1.67 × 10⁻²⁷ kg

Speed of Proton v = 0.86 c

Momentum p = ?

As p = mv

$$\Longrightarrow p = \frac{m_{o}v}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$$

$$\Longrightarrow p = \frac{1.67 \times 10^{- 27} \times 0.86\ c}{\sqrt{1 - \frac{{(0.86)^{2}\ c}^{2}}{c^{2}}}}$$

$$\Longrightarrow p = \frac{1.67 \times 10^{- 27} \times 0.86 \times 3 \times 10^{8}}{\sqrt{1 - (0.86)^{2}}}$$

 ⇒ p = 8.44 × 10⁻¹⁹ kg m s⁻¹