(1) Radiation in the Relativistic Case

(1.1) Calculating the Power Received per Unit Solid Angle Using the Relativistic Doppler Effect

In the rest frame of the blackbody sphere ($\Sigma'$), the radiation is isotropic, and the power emitted per unit solid angle is:

where $P'$ is the total power emitted by the blackbody sphere in its rest frame.

When observed from the frame $\Sigma$ (observer's frame), due to the relativistic Doppler effect and aberration of light, both the frequency and angular distribution of the radiation change.

The relativistic Doppler factor $\delta$ is given by:

where:

• $\beta = \frac{v}{c}$ is the dimensionless velocity,
• $\gamma = \frac{1}{\sqrt{1 - \beta^2}}$ is the Lorentz factor,
• $\theta$ is the angle between the direction of motion and the observation direction in the $\Sigma$ frame.

The observed power per unit solid angle transforms according to:

This transformation accounts for the change in energy, time intervals, and solid angles due to relativistic effects. Therefore, substituting $\frac{dP'}{d\Omega'}$:

This is the expression for the radiated power received per unit solid angle in the $\Sigma$ frame.

(1.2) Calculating the Angular Distribution of Received Power and the Total Power

When integrating $\frac{dP}{d\Omega}$ over all solid angles, we find that it does not yield the total power $P'$ emitted by the blackbody sphere. This discrepancy arises because, in the observer's frame $\Sigma$, the time intervals required to receive energy emitted in different directions are not the same. Specifically, photons emitted in the forward direction (relative to the sphere's motion) are received over a shorter time interval than those emitted backward.

To account for this, we consider the time transformation between the frames. The time interval $dt$ in the observer's frame relates to $dt'$ in the rest frame by:

This implies that the observer receives energy over a time interval $dt$ that depends on $\theta$. Therefore, the power received per unit solid angle during the same time interval $dt$ is:

Simplifying using $\delta = \frac{1}{\gamma (1 - \beta \cos \theta)}$:

However, recognizing that $\delta^5 / \gamma = \delta^4$, we see that:

This suggests that, despite the varying time intervals, the expression for power per unit solid angle remains consistent.

To find the total power $P_{\text{obs}}$ received by the observer, we integrate over all solid angles:

This integral does not equal $P'$ because $\delta^4$ does not integrate to $\gamma^{-4}$ over the sphere. The correct total power received is related to the total emitted power by:

This means the observer measures a total power that is $\gamma^2$ times greater than the power emitted in the rest frame. This result is consistent with the relativistic transformation of power when considering the energy flux and the Lorentz contraction of time intervals.

(2) Re-analysis from the Electromagnetic Field Perspective

(2.1) Deriving the Energy Density Transformation Using Electromagnetic Field Transformations

In the rest frame $\Sigma'$, consider a plane electromagnetic wave propagating at an angle $\theta'$. The electric and magnetic fields are related by $E' = c B'$ for electromagnetic waves in vacuum.

Using the Lorentz transformation for electromagnetic fields (Equation 6.1), we can express the fields in the observer's frame $\Sigma$:

Assuming the fields in $\Sigma'$ are:

Transforming to the $\Sigma$ frame and calculating the energy density $w = \frac{1}{2} \varepsilon_0 E^2 + \frac{B^2}{2\mu_0}$, we find that:

This shows that the energy density transforms with $\delta^4$, consistent with the earlier result obtained using the relativistic Doppler effect.

(2.2) Analyzing Energy Transformation in a Volume Element

Consider a small rectangular volume $V'$ in $\Sigma'$, with sides $dx', dy', dz'$. When transforming to $\Sigma$, due to the relativity of simultaneity, the volume does not remain synchronized, and there is an energy flow into or out of the volume during the time difference.

The energy $W$ in $V$ (the transformed volume in $\Sigma$) is related to $W'$ in $V'$ by:

where $S'$ is the Poynting vector in $\Sigma'$, $\Delta A'$ is the area of one face of the volume, and $\cos \theta'$ accounts for the direction of energy flow relative to the motion.

Calculating $W$ and considering the energy flow due to the Poynting vector, we find:

Using the relation between $\theta$ and $\theta'$ from aberration of light:

we can express $W$ entirely in terms of quantities in the $\Sigma$ frame.

(2.3) Reproducing the Results from (1.1) and (1.2)

By applying the electromagnetic field transformations and accounting for the energy flow into or out of the volume due to the lack of simultaneity, we obtain the same transformation for the energy density as before:

This confirms that the observed power per unit solid angle is:

Repeating the integration over all solid angles and considering the varying time intervals as in (1.2), we again find that the total power received by the observer is:

This consistency between the two approaches (relativistic Doppler effect and electromagnetic field transformations) verifies the correctness of the results obtained in (1.1) and (1.2).

Conclusion

Through both the relativistic Doppler effect and electromagnetic field transformations, we have shown that the power per unit solid angle received by an observer from a moving blackbody sphere is enhanced by a factor of $\delta^4$ compared to the isotropic emission in the rest frame. Additionally, the total power received is greater than the emitted power due to relativistic effects, specifically time dilation and the beaming effect, resulting in $P_{\text{obs}} = \gamma^2 P'$.

Problem: Four-Stroke Diesel Engine Modeling

(1) Quantitative Description of Pressure and Temperature Changes During the Four Strokes

In a simplified model of a four-stroke diesel engine, we analyze the pressure and temperature variations of the gas within the cylinder during each stroke. The four strokes are:

1. Intake Stroke:

   • Process: The intake valve $A$ opens, and the piston moves downward from the top dead center (TDC) to the bottom dead center (BDC), drawing in a fresh charge of diesel-air mixture from the infinite reservoir at pressure $p_1$ and temperature $T_1$.
   • Assumptions: The intake process is isobaric (constant pressure) and isothermal (constant temperature) since $p_1 \approx p_0$, and the gas enters from an infinite reservoir.
   • Equations and Trends:
     • Pressure remains constant: $p = p_1$.
     • Temperature remains constant: $T = T_1$.
     • Volume increases from $V_{\text{min}}$ to $V_{\text{max}} = V_0$.

2. Compression Stroke:

   • Process: Both valves are closed, and the piston moves upward from BDC to TDC, compressing the gas adiabatically (no heat exchange).
   • Assumptions: The process is adiabatic and reversible, so it follows the adiabatic equation for ideal gases.
   • Equations and Trends:
     • Adiabatic condition: $p V^\gamma = \text{constant}$, where $\gamma = \frac{C_p}{C_v}$ for the diesel-air mixture.
     • Pressure increases: $p \uparrow$ as volume decreases.
     • Temperature increases: $T \uparrow$ due to compression work.
     • Volume decreases from $V_0$ to $V_{\text{min}} = V_0 - V$.

3. Combustion (Power) Stroke:

   • Process: At the end of the compression stroke, the compressed mixture reaches critical pressure and temperature, causing spontaneous combustion (explosion). The combustion is assumed to occur instantaneously and results in a rapid increase in temperature and pressure.
   • Assumptions: The combustion process is approximated as an adiabatic and isochoric (constant volume) process because it happens rapidly and the piston does not move significantly during combustion.
   • Equations and Trends:
     • Isochoric condition: $V = \text{constant}$.
     • Pressure spikes: $p \uparrow$ sharply due to heat release from combustion.
     • Temperature spikes: $T \uparrow$ significantly due to the exothermic reaction.
     • Energy added: $\Delta U$ per mole of diesel fuel combusted.

4. Expansion Stroke (Power Stroke Continuation):

   • Process: After combustion, the high-pressure and high-temperature gases push the piston downward from TDC to BDC, doing work on the piston. Both valves remain closed, and the gas expands adiabatically.
   • Assumptions: The expansion is adiabatic and reversible.
   • Equations and Trends:
     • Adiabatic condition: $p V^\gamma = \text{constant}$.
     • Pressure decreases: $p \downarrow$ as volume increases.
     • Temperature decreases: $T \downarrow$ due to expansion work.
     • Volume increases from $V_{\text{min}}$ to $V_0$.

5. Exhaust Stroke:

   • Process: The exhaust valve $B$ opens, and the piston moves upward from BDC to TDC, pushing out the combustion products to the atmosphere at pressure $p_0$ and temperature $T_0$.
   • Assumptions: The exhaust process is isobaric and isothermal with the environment.
   • Equations and Trends:
     • Pressure remains constant: $p = p_0$.
     • Temperature adjusts to $T \approx T_0$ as gases mix with the atmosphere.
     • Volume decreases from $V_0$ to $V_{\text{min}}$.

Summary of Processes:

• Intake and Exhaust Strokes: Isobaric and isothermal processes at $p = p_1 \approx p_0$ and $T = T_1 \approx T_0$.
• Compression and Expansion Strokes: Adiabatic processes following $p V^\gamma = \text{constant}$.
• Combustion Event: Isochoric and adiabatic process with rapid increase in $p$ and $T$.

(2) Analysis Under Steady Operation on Flat Terrain

Given:

• Flywheel rotates at constant angular velocity $\omega$.
• Vehicle moves at constant speed on a flat road.
• Transmission ratio between wheel and flywheel: $\alpha$.
• Flywheel connected to wheels without energy loss.

(2.1) Calculating Temperature $T$ and Pressure $p$ Before Combustion

At the end of the compression stroke, just before combustion, the gas has been compressed adiabatically from volume $V_0$ to $V_{\text{min}} = V_0 - V$.

Using Adiabatic Compression Equations:

For an ideal gas undergoing adiabatic compression:

Where:

• $p_{\text{comp}}$, $T_{\text{comp}}$: Pressure and temperature after compression.
• $p_{\text{intake}} = p_1$, $T_{\text{intake}} = T_1$: Intake pressure and temperature.
• $V_{\text{comp}} = V_{\text{min}} = V_0 - V$.
• $V_{\text{intake}} = V_0$.

Calculations:

1. Pressure After Compression:

2. Temperature After Compression:

(2.2) Calculating Temperature and Pressure After Combustion and Average Power

Combustion Process:

• Occurs at constant volume $V = V_{\text{min}}$.
• Fuel combusts completely with the oxygen in the mixture.
• Energy released per mole of diesel: $\Delta U$.

Total Moles Before Combustion:

Let $n_{\text{total}}$ be the total number of moles before combustion.

• Moles of air (mostly nitrogen and oxygen): $n_{\text{air}}$.
• Moles of diesel fuel: $n_{\text{fuel}}$.

Assuming ideal gas behavior, the total number of moles is proportional to the volume and inversely proportional to temperature:

Temperature After Combustion:

The heat released $Q$ increases the internal energy of the gas:

For an isochoric process, the change in internal energy $\Delta U_{\text{total}}$ is:

Where:

• $C_V$ is the molar heat capacity at constant volume.
• $\Delta T = T_{\text{combustion}} - T_{\text{comp}}$.

Assuming Diatomic Gas with Modified Heat Capacity:

Since combustion products include triatomic gases (e.g., $\text{CO}_2$, $\text{H}_2\text{O}$), we need to account for different heat capacities.

Let’s denote:

• $C_{V,\text{before}}$: Heat capacity before combustion.
• $C_{V,\text{after}}$: Heat capacity after combustion.

For simplification, we can approximate $C_{V,\text{before}} \approx \frac{5}{2} R$ (diatomic gases), and $C_{V,\text{after}} \approx \frac{7}{2} R$ (due to presence of polyatomic gases).

Calculating Temperature Increase:

Where $C_{V,\text{avg}}$ is an average heat capacity.

For an approximate calculation:

Pressure After Combustion:

Since volume remains constant during combustion:

Where $n_{\text{after}}$ accounts for the change in moles due to combustion (number of moles may change due to the reaction stoichiometry).

Expansion Stroke (Adiabatic Expansion):

After combustion, the gas expands adiabatically from $V_{\text{min}}$ back to $V_0$.

Using the adiabatic condition:

Work Done During Expansion:

The work done by the gas during the expansion stroke is:

Simplifying:

Average Power Output:

Since the engine operates in cycles, the average power $P$ is:

Where $\tau$ is the time per cycle, related to the engine speed.

(2.3) Relationship Between Flywheel Angular Velocity $\omega$ and Diesel Volume Fraction $a$

Given:

• Resistive torque on front wheels: $\tau = \beta \omega$.
• Flywheel provides torque to overcome this resistance.
• Transmission ratio: $\alpha$ (wheel angular velocity $\omega_{\text{wheel}} = \alpha \omega$).

Assumptions:

• $a$ is small; higher-order terms in $a$ can be neglected.
• Exponential terms involving $a$ can be approximated by expanding and neglecting $a$ in exponents.

Using the Hint and Defining Dimensionless Parameter $A$:

Given the parameter $A$ defined as:

Deriving the Relationship:

1. Net Work per Cycle:

The net work done per cycle is equal to the work during expansion minus the work during compression (since intake and exhaust strokes do no work).

2. Relating Work to Torque:

The net work per cycle is also equal to the torque times the angular displacement per cycle:

3. Steady-State Condition:

In steady operation, the net work produced by the engine per cycle balances the resistive work:

But $\Delta \theta_{\text{wheel}} = \alpha \Delta \theta_{\text{flywheel}}$, so:

4. Relating Work to Fuel Fraction $a$:

The net work $W_{\text{net}}$ depends on the amount of fuel combusted, which is proportional to $a$. Therefore, we can express $W_{\text{net}}$ in terms of $a$ and $A$.