 A spaceship with mass $m = 25:kg$ is moving toward a red giant with $M = M_{sun}$ and $L = 1000:L_{sun}$ (L: Luminosity). It is moving on a line (so this is not an orbital question). When it gets too close to the red giant, it's sensor open its solar sails (they get and then reflect the light completely), hence the radiation pressure help it to escape from the red giant (on the same line). Find the area of solar sail. (It is from National Olympiad of an Asian Country)

My problem is with the radiation pressure formula. for solving this problem, somewhere said the radiation pressure is: ($c$: light speed, $sigma$: Stefan-Boltzman Constant)

$$P= frac{4sigma T^4}{3c} = frac{4}{3}frac{L}{4pi D^2}$$ Where $T$ is the equivalent Temprature in distance $D$.

Somewhere else said:

$$P= frac{4sigma T^4}{c} = 4 frac{L}{4pi D^2}$$

In another book it is:

$$P=2 frac{4sigma T^4}{c} = 2 imes 4 frac{L}{4pi D^2}$$

2 is because it gets the light and then reflecting it.

At the final:

$$P imes A = frac{GmM}{D^2}$$

I want to know what is the right solution for calculating radiation pressure?

I must say that the below picture is also with the question but in the books, it was used for another part of the question which is not my problem and I didn't mentioned it above. I might venture into some math which is slightly above what you need, but I wanted to show you the reason behind the correct equation to use. To start with, the pressure due to radiation, in the most general sense, is given by

$$P_{mathrm{absorption}} = frac{langle S angle}{c} cos( heta)$$ $$P_{mathrm{reflection}} = 2frac{langle S angle}{c} cos^2( heta)$$

The top equation represents the case where some light hits your object and the light gets absorbed, and the bottom equation represents the case where light hits and bounces off your object (really it's re-emitted). Your case is the bottom equation so I'll work with that from now on and drop the reflection subscript. Without needing to explain exactly what it is, just accept that $langle S angle$ is the average energy of the photon being absorbed or emitted. Now, that's for a single photon. If we have a lot of photons, generally the energy distribution of all the photons together is said to be described by some function, such as $I_lambda$ (the subscript refers to wavelength). We can define the full pressure then by not only integrating over all wavelengths of all the photons, but importantly all the directions that the photons may hit or be emitted.

$$P = frac{2}{c}int_0^inftyint_0^{2pi}int_0^{pi/2} I_lambda(lambda) cos^2( heta)sin( heta):d heta:dphi:dlambda$$

Note now that we've replaced $langle S angle$ with $I_lambda(lambda)$. Also, since we're assuming this is pressure on a flat surface, we're not integrating over every angle, but only half of them $-$ any photon, in order to hit the solar sail, has to come from the direction the sail is facing, and can only reflect in the direction it came from, i.e., we're discluding photons "behind" the sail. For that reason, the $d heta$ integral only goes to $pi/2$.

If you don't understand the math at this point, that's fine but here's the important takeaway: If we only integrated over all wavelengths, you'd get the equation without the three that you listed. What that means is that the equation without the three doesn't take into account all the possible angles and orientations of the incoming photons. It simply assumes all your photons hit your surface perpendicularly, and then bounce off perpendicularly. This is of course not physically accurate as the photons can come in from a variety of angles.

If you do the angle integrals out, you find that you get the following:

$$P = frac{2}{c}int_0^{2pi}:dphi int_0^{pi/2} cos^2( heta)sin( heta):d hetaint_0^infty I_lambda(lambda):dlambda$$

$$P = frac{2}{c}frac{2pi}{3}int_0^infty I_lambda(lambda):dlambda$$

You can see there in the last line that the three popped out of those integrals over all the possible angles (specifically the $d heta$ integral). Now, you get the rest of this by assuming some sort of function for $I_lambda(lambda)$ which describes the energy distribution of the photons as a function of wavelength. Since your light comes from a star, you know that's just going to be the Planck distribution (specifically the equation for $B_lambda(lambda,T)$). This integral is non-trivial, but if you again run through the math, you find

$$P = frac{2}{c}frac{2pi}{3}int_0^infty frac{2hc^2}{lambda^5}frac{1}{e^{hc/lambda k_BT}-1}:dlambda = frac{2}{c}frac{2pi}{3}frac{sigma T^4}{pi}$$

$$oxed{P = frac{4sigma T^4}{3c}}$$

This is the true equation for radiation pressure on the solar sail, assuming reflection is occurring. This includes the three because we need to account for the photons hitting and bouncing off at all possible angles (for the solar sail). Note one important differences here is that my equation already has the extra "2" due to reflection factored in. Your last equation differs from mine in that it throws in that extra "2" for reflection. So either it erroneously included the factor of "2" twice, or (more likely) its account for reflection at every angle, not just all possible angles. Remember above, we only integrated the $d heta$ angle up to $pi/2$ for the reason explained there. If we integrated up to $pi$ to include every angle, then we'd get an extra "2" as well and my equation would match your last one. Your solar sails though, don't receive or re-emit photons at every angle, and so my equation differs by that factor of "2".

I actually think you are supposed to assume the solar sail is unfurled while the spaceship is still quite far from the red giant (despite "too close"), because otherwise the problem is extremely difficult. At least if you are far from the red giant, you can assume all the light is incident along the normal to the sail, and if it's a mirror, it will also reflect back the same way. In that situation (only), you can simply take the radiative flux density (L / 4pi D^2), divide by c (to get a momentum flux), multiply by the area A, and double it (to account for straight-back reflection), resulting in an expression that is like zephyr's except without the 1/3 (which comes from assuming we are only talking about one instant and it is right at the surface of a blackbody with no limb darkening, none of which we should expect from a red giant), and my expression also has a 1/D^2 which is actually important in what follows.

The reason I think this is what is intended is because only then is D not essential to treat in detail. It then becomes easy to figure out what the area of the sail needs to be, we can do it with energy considerations. At large D, the radiative force will be a constant multiple of the gravity, and that multiple needs to be "2" in order to get the spaceship to go back to infinity where it came from (assuming either the original speed at infinity was small, or that your goal is to return so whatever that original speed was, as these are the only ways you don't need to know that original speed). The gravitational force is GMm/D^2, so P times A needs to be twice that, where P is given above. The answer comes out A = 4pi G M m c / L, which is independent of the D where the sail is unfurled.

The reason it's much harder if you don't make D large is that the incident flux is not along the normal any more, so you need an r-dependent factor that ranges from 1/3 right at the surface, to 1 farther out. You'd have to integrate the role of that factor to get the necessary A, whereas my answer doesn't need any of that. What's more, if you unfurl the sail right at the surface, the spacecraft will enter the star, and the radiative environment becomes even more complicated, not to mention possibly destroying the spacecraft.

So I don't think any of those expressions are what you want, because you can't just get the force at the surface, you need to know the force that would succeed in getting the spacecraft to turn around, so it must exceed gravity by a factor that would be very hard to compute unless you take D large.

• • Contributed by Jeremy Tatum
• Emeritus Professor (Physics & Astronomy) at University of Victoria

Radiation Pressure. Recall equation 1.18.5 and the conditions for which it is valid. It was derived for isotropic radiation. In the atmosphere, radiation is not isotropic there is a net flux of radiation outwards. Therefore the radiation density must go inside the integral sign. We can also write the equation in terms of specific intensity, making use of equations 1.15.3 and 1.17.1. The equation for the radiation pressure then becomes

[P = frac<1> int_ <4pi>I cos^2 heta domega , label<4.6.1>]

where by now we are used to the abbreviated notation.

If the radiation is isotropic, this is not zero it is (4pi/(3c)). In the expressions for (J) and for (P), the power of (cos heta) is even (0 and 2 respectively) and one can see both physically and mathematically that neither of them is zero for isotropic radiation. One the other hand, the expression for (F) has an odd power of (cos heta), and it is therefore zero for isotropic radiation, as expected.

• • Contributed by OpenStax
• General Physics at OpenStax CNX

By the end of this section, you will be able to:

• Describe the relationship of the radiation pressure and the energy density of an electromagnetic wave
• Explain how the radiation pressure of light, while small, can produce observable astronomical effects

Material objects consist of charged particles. An electromagnetic wave incident on the object exerts forces on the charged particles, in accordance with the Lorentz force. These forces do work on the particles of the object, increasing its energy, as discussed in the previous section. The energy that sunlight carries is a familiar part of every warm sunny day. A much less familiar feature of electromagnetic radiation is the extremely weak pressure that electromagnetic radiation produces by exerting a force in the direction of the wave. This force occurs because electromagnetic waves contain and transport momentum.

To understand the direction of the force for a very specific case, consider a plane electromagnetic wave incident on a metal in which electron motion, as part of a current, is damped by the resistance of the metal, so that the average electron motion is in phase with the force causing it. This is comparable to an object moving against friction and stopping as soon as the force pushing it stops (Figure (PageIndex<1>)). When the electric field is in the direction of the positive y-axis, electrons move in the negative y-direction, with the magnetic field in the direction of the positive z-axis. By applying the right-hand rule, and accounting for the negative charge of the electron, we can see that the force on the electron from the magnetic field is in the direction of the positive x-axis, which is the direction of wave propagation. When the (vec) field reverses, the (vec) field does too, and the force is again in the same direction. Maxwell&rsquos equations together with the Lorentz force equation imply the existence of radiation pressure much more generally than this specific example, however. Figure (PageIndex<1>): Electric and magnetic fields of an electromagnetic wave can combine to produce a force in the direction of propagation, as illustrated for the special case of electrons whose motion is highly damped by the resistance of a metal.

Maxwell predicted that an electromagnetic wave carries momentum. An object absorbing an electromagnetic wave would experience a force in the direction of propagation of the wave. The force corresponds to radiation pressure exerted on the object by the wave. The force would be twice as great if the radiation were reflected rather than absorbed.

Maxwell&rsquos prediction was confirmed in 1903 by Nichols and Hull by precisely measuring radiation pressures with a torsion balance. The schematic arrangement is shown in Figure (PageIndex<2>). The mirrors suspended from a fiber were housed inside a glass container. Nichols and Hull were able to obtain a small measurable deflection of the mirrors from shining light on one of them. From the measured deflection, they could calculate the unbalanced force on the mirror, and obtained agreement with the predicted value of the force. Figure (PageIndex<2>): Simplified diagram of the central part of the apparatus Nichols and Hull used to precisely measure radiation pressure and confirm Maxwell&rsquos prediction.

The radiation pressure (p_) applied by an electromagnetic wave on a perfectly absorbing surface turns out to be equal to the energy density of the wave:

If the material is perfectly reflecting, such as a metal surface, and if the incidence is along the normal to the surface, then the pressure exerted is twice as much because the momentum direction reverses upon reflection:

We can confirm that the units are right:

[[u] = dfrac = dfrac = dfrac = units , of , pressure.]

Equations ef and ef give the instantaneous pressure, but because the energy density oscillates rapidly, we are usually interested in the time-averaged radiation pressure, which can be written in terms of intensity:

Radiation pressure plays a role in explaining many observed astronomical phenomena, including the appearance of comets . Comets are basically chunks of icy material in which frozen gases and particles of rock and dust are embedded. When a comet approaches the Sun, it warms up and its surface begins to evaporate. The coma of the comet is the hazy area around it from the gases and dust. Some of the gases and dust form tails when they leave the comet. Notice in Figure (PageIndex<3>) that a comet has two tails. The ion tail (or gas tail) is composed mainly of ionized gases. These ions interact electromagnetically with the solar wind, which is a continuous stream of charged particles emitted by the Sun. The force of the solar wind on the ionized gases is strong enough that the ion tail almost always points directly away from the Sun. The second tail is composed of dust particles. Because the dust tail is electrically neutral, it does not interact with the solar wind. However, this tail is affected by the radiation pressure produced by the light from the Sun. Although quite small, this pressure is strong enough to cause the dust tail to be displaced from the path of the comet. Figure (PageIndex<3>): Evaporation of material being warmed by the Sun forms two tails, as shown in this photo of Comet Ison. (credit: modification of work by E. Slawik&mdashESO)

Example (PageIndex<1>): Halley&rsquos Comet

On February 9, 1986, Comet Halley was at its closest point to the Sun, about (9.0 imes 10^ <10>m) from the center of the Sun. The average power output of the Sun is (3.8 imes 10^ <26>, W).

1. Calculate the radiation pressure on the comet at this point in its orbit. Assume that the comet reflects all the incident light.
2. Suppose that a 10-kg chunk of material of cross-sectional area (4.0 imes 10^ <-2>m^2) breaks loose from the comet. Calculate the force on this chunk due to the solar radiation. Compare this force with the gravitational force of the Sun.

Calculate the intensity of solar radiation at the given distance from the Sun and use that to calculate the radiation pressure. From the pressure and area, calculate the force.

a. The intensity of the solar radiation is the average solar power per unit area. Hence, at (9.0 imes 10^ <10>m) from the center of the Sun, we have

Assuming the comet reflects all the incident radiation, we obtain from Equation ef

b. The force on the chunk due to the radiation is

[eginF &= pA onumber [4pt] &= (2.5 imes 10^ <-5>N/m^2)(4.0 imes 10^ <-2>m^2) onumber [4pt] &= 1.0 imes 10^ <-6>, N, onumber end onumber]

whereas the gravitational force of the Sun is

[egin F_g &= dfrac onumber [4pt] &= dfrac <(6.67 imes 10^<-11>, N cdot m^2 /kg^2)(2.0 imes 10^ <30>kg)(10 , kg)> <(9.0 imes 10^<10>m)^2> onumber [4pt] &= 0.16 , N. onumber end onumber]

Significance

The gravitational force of the Sun on the chunk is therefore much greater than the force of the radiation.

After Maxwell showed that light carried momentum as well as energy, a novel idea eventually emerged, initially only as science fiction. Perhaps a spacecraft with a large reflecting light sail could use radiation pressure for propulsion. Such a vehicle would not have to carry fuel. It would experience a constant but small force from solar radiation, instead of the short bursts from rocket propulsion. It would accelerate slowly, but by being accelerated continuously, it would eventually reach great speeds. A spacecraft with small total mass and a sail with a large area would be necessary to obtain a usable acceleration.

When the space program began in the 1960s, the idea started to receive serious attention from NASA. The most recent development in light propelled spacecraft has come from a citizen-funded group, the Planetary Society. It is currently testing the use of light sails to propel a small vehicle built from CubeSats, tiny satellites that NASA places in orbit for various research projects during space launches intended mainly for other purposes.

The LightSail spacecraft shown below (Figure (PageIndex<4>)) consists of three CubeSats bundled together. It has a total mass of only about 5 kg and is about the size as a loaf of bread. Its sails are made of very thin Mylar and open after launch to have a surface area of (32 , m^2). Figure (PageIndex<3>): Two small CubeSat satellites deployed from the International Space Station in May, 2016. The solar sails open out when the CubeSats are far enough away from the Station.

Example (PageIndex<2>): LightSail Acceleration

The first LightSail spacecraft was launched in 2015 to test the sail deployment system. It was placed in low-earth orbit in 2015 by hitching a ride on an Atlas 5 rocket launched for an unrelated mission. The test was successful, but the low-earth orbit allowed too much drag on the spacecraft to accelerate it by sunlight. Eventually, it burned in the atmosphere, as expected. The next Planetary Society&rsquos LightSail solar sailing spacecraft is scheduled for 2018. The Lightsail is based on the on NASA's NanoSail-D project. (Public domain NASA).

LightSail Acceleration

The intensity of energy from sunlight at a distance of 1 AU from the Sun is (1370 , W/m^2). The LightSail spacecraft has sails with total area of (32 , m^2) and a total mass of 5.0 kg. Calculate the maximum acceleration LightSail spacecraft could achieve from radiation pressure when it is about 1 AU from the Sun.

The maximum acceleration can be expected when the sail is opened directly facing the Sun. Use the light intensity to calculate the radiation pressure and from it, the force on the sails. Then use Newton&rsquos second law to calculate the acceleration.

The resulting acceleration is

Significance

If this small acceleration continued for a year, the craft would attain a speed of 1829 m/s, or 6600 km/h.

How would the speed and acceleration of a radiation-propelled spacecraft be affected as it moved farther from the Sun on an interplanetary space flight?

Its acceleration would decrease because the radiation force is proportional to the intensity of light from the Sun, which decreases with distance. Its speed, however, would not change except for the effects of gravity from the Sun and planets.

The Formation of Massive Stars

Stellar physics is a broad field that touches on a range of phenomena from magnetic fields to radiative processes and thermonuclear fusion to plasmas. Stars form through the gravitational collapse of cold, dense, dusty proto-stellar cores, themselves embedded in thick molecular clouds or filaments. Massive stars, defined as those with a mass greater than 8 solar masses, are of key interest in star formation. Although they are extremely rare, comprising less than 1% of the total stellar population, they make their presence known by dominating the surrounding interstellar medium (ISM) with their powerful stellar winds as well as shocks from their eventual supernovae. Their formation is known to be impeded by several feedback mechanisms, including outflows, radiation pressure and magnetic fields. Today’s paper uses a series of radiative magnetohydrodynamic (RMHD) simulations to understand the overall impact that these combined mechanisms have on star formation.

Pushing the Boundaries

The fact that massive stars are so rare is reflective of a more general problem with star formation: its inefficiency. Estimates of star formation efficiencies (SFEs) are as low as 33%. As massive stars begin to form, they launch powerful molecular outflows from their poles. These jets can interact with the surrounding molecular cloud and eject large quantities of material. This, in combination with other feedback mechanisms, limits the star’s ability to accrete material, ultimately limiting its final mass. Knowing the upper limit of just how massive a star can be is incredibly valuable, for it allows us to set the upper boundary of the initial mass function (IMF). An IMF is a model of the initial distribution of stellar masses for a given population of stars. It is impossible to simulate the evolution of a stellar population without one. This is where massive stars are important, for they are the dominant source of radiative feedback and energy injection into the ISM through supernovae. So, to help determine these upper mass limits, we must simulate the processes that inhibit star formation in as much detail as possible.

Massively Magnetic Outflows Radiation Pressure (Games) Figure 1 (Figure 2 in the paper): Density plots for the three simulations, with the most massive star in the center of each panel.

In Figure 1, after the stellar mass of the proto-stellar core exceeds 30 solar masses we see several pressure-dominated bubbles expanding away from the star (this is most noticeable in the middle row TurbRad+OF simulation). This process is known as the “flashlight effect”, where thick material is beamed away from the poles, causing low-density bubbles to expand outwards.

Go With The Flow

Over time, strong entrained outflows begin to break through the proto-stellar core and eject large quantities of material, as can be seen in Figure 2. Figure 2 (Figure 10 in the paper): Projected yz densities of the entrained outflows.

The outflows in Figure 2 become steadier and more directed over time. Although the proto-stellar core is initially highly turbulent, as it accretes material its rotational axis stabilises over time. One of the key results of these simulations is that the momentum feedback from these outflows is the dominant feedback mechanism (compared to radiation pressure) and helps to eject significant fractions of material, reducing the star formation efficiency. Outflows also help to act as conduit through which radiation can escape, weakening the feedback effects from radiation pressure.

Don’t Forget the B Field Figure 3 (Figure 15 in the paper): The star formation efficiencies for the total stellar population (top) and the primary, most massive star (bottom) as functions of simulation time for the three different simulations.

Magnetic fields are known to affect star formation. Indeed, Figure 3 shows that the SFE is further reduced by the presence of magnetic fields (compare the purple dashed line to the pink dashed line). Overall, the simulations that contained outflows resulted in smaller SFEs. So in order to reconcile observations that place overall star formation efficiency at around 33%, this work shows that it is necessary to account for the effects of outflows.

In such an involved phenomena like star formation, there are many nuances. Magnetic fields slow the growth rate of stars by helping to prevent the core from fragmenting, however there are several non-ideal effects (such as the Hall effect) that could theoretically impact the star formation process. These non-ideal effects were not considered, although it is unknown whether such effects have any noticeable impact on SFEs.

A Joint Effort

This comprehensive series of simulations, one of the first to account for so many factors, demonstrates the role of outflows, magnetic fields and radiation pressure in limiting the formation of massive stars and reducing the overall SFE. This study shows that feedback from outflows dominates the feedback from radiation pressure, and that magnetic fields further inhibit star formation. Importantly, both outflows and magnetic fields are needed to reproduce the low efficiencies obtained from observations.

• • Contributed by Jeremy Tatum
• Emeritus Professor (Physics & Astronomy) at University of Victoria

Photons carry momentum (h/lambda) and hence exert pressure. Pressure is rate of change of momentum (i.e. force) per unit area.

The pressure (P) exerted by radiation (in ( ext^<-2>), or ( ext)) is related to the energy density (u) of radiation (in ( ext^<-3>)) by

depending on the circumstances!

First, we may imagine a parallel beam of photons that have come a long way from their original source. For example, they might be photons that have arrived at a comet from the Sun, and they are about to push material out from the comet to form the tail of the comet. Each of them is travelling with speed (c). We suppose that there are (n) of them per unit volume, and therefore the number of them per unit area arriving per unit time is (nc). Each of them carries momentum (h/lambda). [As in section 1.17 they need not all carry the same momentum. The total momentum is the sum of each.] The rate of arrival of momentum per unit area is (nhc/lambda = nh u). But (h u) is the energy of each photon, so the rate of arrival of momentum per unit area is equal to the energy density. (Verify that these are dimensionally similar.) If all the photons stick (i.e. if they are absorbed), the rate of change of momentum per unit area (i.e. the pressure) is just equal to the energy density (Equation ( ef<1.18.2>)) but if they are reflected elastically, the rate of change of momentum per unit area is twice the energy density (Equation ( ef<1.18.1>)).

If the radiation is isotropic, the situation is different. The radiation may be approximately isotropic deep in the atmosphere of a star, though I fancy not completely isotropic, because there is sure to be a temperature gradient in the atmosphere. I suppose for the radiation to be truly isotropic, you'd have to go to the very center of the star.

We'll start from Equation 1.17.4, which gives the rate at which photons arrive at a point per unit area. ("at a point per unit area"? This makes sense only if you bear in mind the meaning of "per unit"!) If the energy of each photon is ( ext), the momentum of each is ( ext/c). (This is the relation, from special relativity, between the energy and momentum of a particle of zero rest mass.) However, it is the normal component of the momentum which contributes to the pressure, and the normal component of each photon is (( ext cos heta)/c). The rate at which this normal component of momentum arrives per unit area is found by multiplying the integrand in Equation 1.17.4 by this. Bearing in mind that (n ext) is the energy density (u), we obtain

[frac <4pi>int_0^ <2pi>int_0^ sin heta cos^2 heta d heta dphi. ag <1.18.5>label<1.18.5>]

The pressure on the surface is the rate at which the normal component of this momentum is changing. If the photons stick, this is

[frac <4pi>int_0^ <2pi>int_0^ sin heta cos^2 heta d heta dphi = u/6. ag <1.18.6>label<1.18.6>]

Stars along the main sequence (MS) all have one thing in common: they are all burning hydrogen into helium in their cores. This process is called fusion. Fusion creates heavier elements from lighter elements (for example, helium from hydrogen). Fusion also releases energy (photons) which travel outward from the core of the star (where the fusion is taking place) and eventually escape from the surface of the star (this is the light we see from the star). These photons exert a pressure on the stellar material as they travel outward from the core this pressure is called Radiation Pressure.

In the Main Sequence Phase of a star's evolution, radiation pressure pushing outward exactly balances the gravitational pressure pulling inward (this balance is called Hydrostatic Equilibrium). Because these two forces are exactly in balance, the star is stable (this means it neither shrinks nor expands). A star will spend almost 90% of its lifetime on the MS. The lifetime of a star on the MS is determined by its mass. It turns out that the lifetime of stars on the MS is proportional to their mass to the -2.5 power. Therefore, more massive stars spend less time on the MS. For example, a 40 solar mass star will spend about 1 million years on the MS while a 0.5 solar mass star will spend 56 billion years on the MS. A star like our Sun will spend about 10 billion years as a MS star (and since it is already about 5 billion years old, the Sun is half-way through its MS life).

Stars along the MS also obey a mass - luminosity relation. This relation says that the more mass a star has, the greater the pressure needed to hold that mass up, and therefore the faster the fusion reactions need to occur. Therefore, higher mass stars on the MS fuse their hydrogen into helium more quickly than lower mass stars and have hotter cores that release more energy, which makes these stars more luminous. (This also explains why massive stars live shorter lives they burn through their fuel more quickly.) So, as you move up the MS, you find stars of higher temperature and greater luminosity. The luminosity - radius - temperature relation also tells us that the stars higher up on the MS will have larger radii.

The more massive a star is, the stronger its gravity. The stronger the gravity of a star, the more compressed its core becomes and therefore the higher the temperature rises in the star's core. This means that it is easier for more massive stars to reach temperatures at which fusion can begin. (For hydrogen fusion to occur, core temperatures of 10 million K are needed!) This also means that if the mass of a protostar is below some limit, gravity will never raise core temperatures high enough for fusion to occur. In fact, stars with masses less than 0.08 solar masses cannot raise their core temperatures high enough to ignite hydrogen fusion. These stars are known as Brown Dwarfs. Brown Dwarfs are not in balance (hydrostatic equilibrium) because they have no radiation pressure to counter their own gravity. These stars contract slowly and never become Main Sequence stars.

Department of Astronomy, University of Maryland
College Park, MD 20742-2421
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A question from my kid.

It's really not a bad question at all. Really, you're getting at Fourier uncertainty (very similar to Heisenberg uncertainty). Essentially, time and frequency are convoluted, meaning you can express functions of time in terms of frequency instead. But, there is some uncertainty. If you have a single frequency of light, then that wave must stretch on forever (cos(w*t) will never end, where w is the frequency). If you have an infinitely short pulse of light, it must contain infinite frequencies.

Additionally, if you have a pulse of light that is too short for a certain wavelength, then that wavelength can't exist. 532-nm light takes 1.77 fs (femtosecond, or 10^-15 second) to fully oscillate. When within that 1.77 fs does this wave begin to exist? Impossible to tell. Additionally, it's not accelerating within this time.

It seems to me questions like this would be easier to investigate in the long-wave radio part of the spectrum.

#27 bcgilbert

It's really not a bad question at all. Really, you're getting at Fourier uncertainty (very similar to Heisenberg uncertainty). Essentially, time and frequency are convoluted, meaning you can express functions of time in terms of frequency instead. But, there is some uncertainty. If you have a single frequency of light, then that wave must stretch on forever (cos(w*t) will never end, where w is the frequency). If you have an infinitely short pulse of light, it must contain infinite frequencies.

Additionally, if you have a pulse of light that is too short for a certain wavelength, then that wavelength can't exist. 532-nm light takes 1.77 fs (femtosecond, or 10^-15 second) to fully oscillate. When within that 1.77 fs does this wave begin to exist? Impossible to tell. Additionally, it's not accelerating within this time.

I am unaware of Fourier uncertainty, do tell? Pulses of light no matter how short will produce equivalent frequencies, they may be x rays or even gamma rays. There is no radiation if there is no acceleration?

My little bit of doubt comes from this line of thinking.

A wave, if I am not mistaken, has three parts: A crest, a trough and another crest. The question I have is when the photon is created is the wave created all at once or does it start as a crest followed by the trough then the second crest thus completing the initial wave. If its the latter then it would seem logical some non zero time has to pass as the wave parts are created. Perhaps some other physics is happening in the brief time the wave is being built. Once fully formed and on its way established physics takes over the modeling.

The electron is the main game in town, the detailed motion of the electron determines the wavelike behaviour of the radiated fields, just like the paper cone in your speakers do. If the electron wriggles back and forth then the wave follows, if the the electron moves in an abrupt discontinuous motion then the field follows, and there will be no recognizable waves radiated in the field, however by Fourier analysis abstract mathematical waves can extracted from the time domain waveform, the same for your speakers. Classically this is simple quantum mechanically it's a mess, I'll challenge QMs to work it out for us, for sound waves too while you're at it?

All of the paradoxes and confusion stem from the absurd notion of the "photon".

I hope your kid has equal time learning EM theory derived from Maxwell's equations.

Also some history, including the debate between Max Planck and Albert Einstein on the notion of the packets of light we now call the "photon".

There is a great deal of misconception and misinformation surrounding the topic of radiation pressure in stars. I have found there to be a widespread idea that the cores of stars are held up primarily by radiation pressure, yet this is very rarely true. To see how bad the misinformation there is, consider these quotes:

"Radiation pressure counterbalances the gravitational forces due to the star’s mass
which tend to make it contract. When the star’s energy production ceases and the radiation
pressure is removed, the star will start to collapse. "

The above comes from the very definition of radiation pressure in the online astronomy dictionary, no less, yet it is completely wrong. Not wrong as in some technically detailed sense, but wrong as in the opposite of true. Radiation pressure throughout most stars (especially our Sun) is well known to those who understand stellar interiors to be fairly negligible, even in the core. What's more, the end of fusion most certainly does not bring about the "removal" of the radiation pressure-- the radiation pressure generally increases when fusion ends in a stellar core, but it doesn't matter because it wasn't doing anything important anyway. The only time radiation pressure matters in the core is for stars much more massive than the Sun, like as massive as stars get.

The misinformation is not limited to the online dictionary. Here is a NASA site, no less:
http://imagine.gsfc.nasa.gov/docs/science/know_l2/stars.html
"To add more nucleons to the iron nucleus requires an input of energy,
and so, once the center of the star consists of iron, no more
energy can be extracted. The star's core then has no resistance to the force of gravity,
and once it starts to contract a very rapid collapse will take place."

You can google lots of astronomy course websites and answer-all websites to see this wrong information repeated over and over, I'll just give a few examples of hammering this false theme that the end of fusion causes a loss of radiation pressure which leads to dramatic changes in the core:

marquard/astronomy/sunlike.htm
At some point in time the hydrogen nuclei in the core will run out.
When this occurs there will no longer be a balance between gravitational
pressure inward and photon pressure outward. This is due to the fact that
the photon production ceases when the fusion process runs out of fuel.
At this point the core (almost pure helium) will begin to collapse again.

The main sequence ends when the star's core runs out of hydrogen nuclei. Without the
radiation pressure generated by hydrogen fusion, equilibrium is lost. The star's core,
made up almost entirely of helium now, begins to collapse.

Quite remarkable how consistent is this incorrect theme, once the mainstream understanding of stellar interiors is actually mastered.

A question from my kid.

My son is taking high school physics and they are studying light and its properties. His lesson the other day covered how photons have a rest mass of zero and an inertial (relativistic) mass of some number once they get going.

Why can't you fell the kick of the photons leaving a flashlight when its turned on?

His thought is when the photons are created and accelerated they gain mass. The mass gain should be felt as a kick. Obviously no kick is felt so we talked about it but I'm sure I left him more confused then when we started.

Do the photons actually gain mass?

If so is there a lag time between the creation, acceleration and mass gaining of the photon?

Is the flashlight just not powerful enough to feel the effect?

How come in Star Wars the laser guns kick like a shotgun?

Do I need to go back to physics class

#2 Tangerman

Photons indeed have momentum. You should try calculating the momentum from the intensity of a flashlight. It's a trivial amount, much less than you would ever feel. Even from high-powered lasers it's minimal. Also, photons don't really "get going." The speed of light is always the same, in vacuum or not, even though it can appear different due to the refractive index. See https://www.feynmanl. h.edu/I_31.html for more about that.

#3 photoracer18

Maybe. Photons gain mass when they gain energy (E=mc 2 ) but in the scheme of things that mass is very very small. Light exhibits the properties of both particles and waves depending on what property you are looking at. Some stars, the extremely hot ones (Type O and Wolf-Rayets), have enough light pressure to push matter away from the star at a particular distance. High energy photons of any type have mass and therefore momentum and will transfer the momentum to anything they strike. There are equations to calculate it and it has to be factored in for things like spacecraft solar panels because the light over time can change the orbit. But under the gravity field of a planet the effect is so infinitesimal its not detectable without instrumentation.

#4 JamesMStephens

The "kick" is too small to perceive. Radiation does exert a force on a material body, this causes the gas tail of a comet to point away from the Sun--radiation pressure pushes the ions. (The dust tail is driven mainly by solar wind.)

Any kick you feel when operating a flashlight is (in principle) the same as the kick you feel when firing a shotgun or rifle, except the shell has much more momentum than a pulse of radiation carrying the same energy. Fire a 1.9 gram bullet from a 0.22 rifle at, say, 340 m/s and the momentum is (m*v) 0.65 kg-m/s, and the kinetic energy ((1/2)mv 2 ) is 110 Joules. For radiation the momentum of the pulse is energy divided by the speed of light. For a 110 J radiation pulse the momentum would be 3.7*10 −7 kg-m/s. The bullet would have 1.8 million times the momentum of a radiation pulse of equal energy! For a pulse of radiation to carry the momentum of the bullet its energy would have to be 198 megajoules.

The bullet would leave the barrel in about three milliseconds. If a flashlight emitted 198 megajoules in 3 ms it would give the same kick as the 0.22 rifle. It would have to have a power output of 66 billion Watts. (You would have to toggle it on and then back off in 3 ms.)

If you find a flashlight that powerful keep away from it!

Edited by JamesMStephens, 02 October 2020 - 04:06 PM.

#5 AstroBrett

If you think everything in Star Wars (or any Hollywood movie) obeys the laws of physics, then yes, you should go back to Physics Class !

#6 Tangerman

Maybe. Photons gain mass when they gain energy (E=mc 2 ) but in the scheme of things that mass is very very small. Light exhibits the properties of both particles and waves depending on what property you are looking at. Some stars, the extremely hot ones (Type O and Wolf-Rayets), have enough light pressure to push matter away from the star at a particular distance. High energy photons of any type have mass and therefore momentum and will transfer the momentum to anything they strike. There are equations to calculate it and it has to be factored in for things like spacecraft solar panels because the light over time can change the orbit. But under the gravity field of a planet the effect is so infinitesimal its not detectable without instrumentation.

We still haven't found any rest mass of photons. Also, for particles without a rest mass, E =mc^2 should be modified to E^2 = p^2*c^2 + m^2*c^4, where p is momentum. Thus, for photons where rest mass m is 0, E = pc. Well, for light, the energy of a photon is also equal to h*c/lambda, where h is Planck's constant and lambda is the wavelength of the light. You can then solve for the momentum of a photon.

#7 Keith Rivich

If you think everything in Star Wars (or any Hollywood movie) obeys the laws of physics, then yes, you should go back to Physics Class !

I don't. I was joking on that one!

#8 Keith Rivich

Photons indeed have momentum. You should try calculating the momentum from the intensity of a flashlight. It's a trivial amount, much less than you would ever feel. Even from high-powered lasers it's minimal. Also, photons don't really "get going." The speed of light is always the same, in vacuum or not, even though it can appear different due to the refractive index. See https://www.feynmanl. h.edu/I_31.html for more about that.

So, is the acceleration of the light from creation to its final speed instantaneous? Or does it take an incredibly small but real amount of time to go from zero to c?

#9 EJN

This is how the momentum of a photon is derived:

Momentum of a photon is: p = hf/c where h is Planck's constant, f is the frequency, c = speed of light.

This is derived from 3 relationships, E = hf (often written E = hv, where v is the greek letter nu),
E = mc 2 , and p = mv

E = mc 2 can be rearranged to solve for mass, m = E/c 2

Since a photon has no rest mass but has energy, and v = c, by substitution p = E/c 2 * c = E/c
By substitution again of E = hf,
p = hf/c

#10 DaveC2042

So, is the acceleration of the light from creation to its final speed instantaneous? Or does it take an incredibly small but real amount of time to go from zero to c?

The photon cannot travel at anything other than c. When the photon is created it is already travelling at c. This is a very fundamental aspect of relativity. And applies to any particle with zero rest mass.

#11 Tangerman

Light is a very, very interesting thing. As much as it's been studied and as fundamental as it is, most do not understand it very well (I have much to learn about it yet).

Newton actually believed that light was made of particles (what we may today call photons). Maxwell showed mathematically that light is a wave, and it had also been observed using diffraction gratings. Then along came a famous double-slit experiment which showed that light truly does come in discrete packets (like a particle) but if you let enough light through, you see the wave behavior. Wave-particle duality is really cool, but can also be quite confusing, especially as those who try to explain it switch between describing light as a wave and light as a particle, depending on what they're trying to explain.

#12 Tony Flanders

So, is the acceleration of the light from creation to its final speed instantaneous?

It is incorrect to call it acceleration. At one moment, the photon does not exist at all. An immeasurably short time later, it is traveling at the speed of light. Photons cannot travel slower than c, and objects with rest mass cannot travel at c.

#13 Keith Rivich

It is incorrect to call it acceleration. At one moment, the photon does not exist at all. An immeasurably short time later, it is traveling at the speed of light. Photons cannot travel slower than c, and objects with rest mass cannot travel at c.

It's what is happening between existence and the "immeasurably short time later" I am curious about. Is it shorter then one unit of Planck time? During that brief interlude is light (by this I mean waves or particles) ramping up to c or is the acceleration time truly 0?

If one uses 0 in this formula "m=f/a" you get nonsense as division by 0 is undefined. I thought the universe doesn't like this. Am I missing something?

#14 JamesMStephens

If one uses 0 in this formula "m=f/a" you get nonsense as division by 0 is undefined. I thought the universe doesn't like this. Am I missing something?

You're looking at a relativistic problem for a particle with zero mass and wring Newton's second law as a low speed (v<<C) for a particle with non-zero mass. A better statement of Newtons's second law is (dp/dt)= F, with p = gamma*m*v, m being rest mass. There is no "m", per se, for light, the momentum is energy divided by speed of light. If we write F = dp/dt for light it's OK to write F = dp/dt = d((E/c)/dt) = (1/c)*P, where P is power (and E is energy). This is EJN pointed this out in his post and I used this in my earlier post to estimate the power output your flashlight would need to give a kick comparable to a 0.22 rifle.

A concise way to look at your question is that for a given energy lighy carries a miniscule amount of momentum as compared to matter. As far as the acceleration issue is concerned the photon doesn't undergo acceleration, it eith exists (and is in motion--at c) or it doesn't!

#15 Keith Rivich

I found this article "How does a photon accelerate to light speed so quickly?" that explains the phenomena in pretty plain English.

I still have a little doubt that there is not an incredibly tiny but for now unmeasurable gap between 0 and c. But I can accept the conclusions and finish my discussion with my son armed a little bit better.

Edited by Keith Rivich, 03 October 2020 - 02:13 PM.

#16 BillP

The photon cannot travel at anything other than c. When the photon is created it is already travelling at c. This is a very fundamental aspect of relativity. And applies to any particle with zero rest mass.

c being its fixed speed, which however is different in different mediums. So c changes depending on the medium it is in at the time.

#17 DaveC2042

I found this article "How does a photon accelerate to light speed so quickly?" that explains the phenomena in pretty plain English.

I still have a little doubt that there is not an incredibly tiny but for now unmeasurable gap between 0 and c. But I can accept the conclusions and finish my discussion with my son armed a little bit better.

The 'tiny gap' idea isn't entirely stupid, but you need to appreciate it would have profound consequences, overturning a lot of very well established physics.

First, Maxwell's equations say that any periodic disturbance in the EM field travels at c - no ifs or buts. So if we find light travelling at less than c, even for the tiniest instant, Maxwell's equations need an overhaul. That's a big deal.

Secondly, the whole basis of relativity collapses. The initial motivation for it was Einstein's observation that if you could find a frame of reference in which you could travel alongside light, you would not see anything that made sense - you'd see a propagating wave that wasn't propagating, a contradiction in terms. The strict requirement that light always travels at c for all observers, is what prevents this, and leads to special relativity.

Finally, general relativity is based on the idea that the geometry of space-time is defined by the paths that light follows. If light can travel at a speed other than c, it cannot serve that function.

Now, of course, if someone were to demonstrate that light did 'accelerate' to c, then that's what we'd have to deal with. But it's extremely unlikely, given how much of our science is based on it not being possible.

#18 Tangerman

c being its fixed speed, which however is different in different mediums. So c changes depending on the medium it is in at the time.

See the article I referenced. c is the same for any medium, although it effectively becomes different because what we see is the time for molecules to re-emit light in a constructively interfering pattern.

#19 Keith Rivich

The 'tiny gap' idea isn't entirely stupid, but you need to appreciate it would have profound consequences, overturning a lot of very well established physics.

First, Maxwell's equations say that any periodic disturbance in the EM field travels at c - no ifs or buts. So if we find light travelling at less than c, even for the tiniest instant, Maxwell's equations need an overhaul. That's a big deal.

Secondly, the whole basis of relativity collapses. The initial motivation for it was Einstein's observation that if you could find a frame of reference in which you could travel alongside light, you would not see anything that made sense - you'd see a propagating wave that wasn't propagating, a contradiction in terms. The strict requirement that light always travels at c for all observers, is what prevents this, and leads to special relativity.

Finally, general relativity is based on the idea that the geometry of space-time is defined by the paths that light follows. If light can travel at a speed other than c, it cannot serve that function.

Now, of course, if someone were to demonstrate that light did 'accelerate' to c, then that's what we'd have to deal with. But it's extremely unlikely, given how much of our science is based on it not being possible.

My little bit of doubt comes from this line of thinking.

A wave, if I am not mistaken, has three parts: A crest, a trough and another crest. The question I have is when the photon is created is the wave created all at once or does it start as a crest followed by the trough then the second crest thus completing the initial wave. If its the latter then it would seem logical some non zero time has to pass as the wave parts are created. Perhaps some other physics is happening in the brief time the wave is being built. Once fully formed and on its way established physics takes over the modeling.

We define the energy and momentum of a photon as follows, where h is Planck's constant, c is the speed of light, is the frequency of the light, and s is a unit vector which points in the direction of propagation.

We begin by considering the macroscopic properties of the radiation field. In the absence of diffraction effects, radiation can be treated as traveling in straight lines (rays). We consider the amount of energy dE which crosses through an area s . dA, where is the angle between the normal to the surface area and s, into a solid angle ds in time dt with frequency range d.

and the specific intensity I has units of erg/s/cm 2 /str/Hz. Although I is expressed as a scalar quantity, it is defined as the intensity in a particular direction (along the normal ray in the above diagram). The exception to this rule is the case of an isotropic radiation field, for which the specific intensity is the same in all directions. For a blackbody,

The solid angle ds is defined so that, integrated over a sphere, it is equal to 4.

The mean intensity J is the zeroth moment of the radiation field. This is the average of I over all solid angles. For isotropic radiation, J = I .

The energy density u is the amount of radiant energy per unit volume at frequency . Visualize a cylinder of side area dA and length c × dt, drawn around a set of light rays traveling through ds. There is a certain amount of energy within the cylinder. As the speed of propagation of light (in a vacuum) is c,

The net flux density F in the direction z at frequency is the integral of I cos over all solid angles. This is equivalent to the net energy flow through the area dA into the solid angle ds. For an isotropic radiation field, the net energy flow is zero.

We define H as the first moment of the intensity I .

The second moment K is then

which is related to the radiation pressure pr as follows. The momentum flux, integrated over all frequencies along the direction s is equal to dF/c. The radiation pressure is simply the component of the momentum flux in the z direction.