Will the Earth Observers See a Burning Spaceship?

[BTBlueSkies]

I had this though back in collage and asked my professor and he gave me a very short attempt to answer the question then said he was never quite sure about all this relativistic stuff... I didn't get my answer though but I have not forgot my question.

Say a spaceship was headed toward the sun at 0.9 c. Now to the guys on the spaceship should see the Earth and sun look very oblique as they would have contracted their length in the direction of their motion and would also see the sun way closer than the guys on the earth. The guys on the Earth would see the spaceship looking very short in length, and it would be the normal 1AU from the sun just as it passes by the earth.

My question was, would the guys on Earth see the spaceship to be on fire, as in the space ships frame the ship would be very close to the sun. Also, the spaceship guys would see the sun equaly close to the them and the Earth so they should maybe see the Earth on fire on the sunward side.

My teacher said something about intensity of the radiation and then quickly dropped the conversation and said, I have never quite signed onto this theory... but in any case of his opinion, I didn't get an answer.

So, would the guys on Earth see the spaceship burning up as it passed the earth?

Thanks.

 

[Vanadium 50]

You have many ideas entangled there.

If you have an object traveling such that its clocks are slowed down by a factor x, the rate at which it absorbs energy from the surroundings (like the sun) appears to the traveler to be x times larger. However, the time it spends absorbing is x times smaller. Depending on the details, including how fast it can shed energy, it might or might not catch fire, but no matter which one happens, all observers would agree on that fact.

 

[jartsa]

Let's forget distance to sun. Instead we consider distance between photons. I mean those photons that are moving from the sun to the space ship. That distance is is contracted by factor of 4.4 because of the velocity of the ship. So the photon stream becomes 4.4 times denser when velocity increases by 0.9 c.

Now we consider energy of each photon. Wavelength of photon is contracted by factor of 4.4. This means 4.4 times more energy per photon.

Now we multiply the two factors: 4.4*4.4 = 19.4

Power of sunlight is 19.4 times the normal, according to spaceship passengers.

 

[PeterDonis]

Power of sunlight is 19.4 times the normal, according to spaceship passengers.

I'm not sure this is right. If we view the Sun's radiation as black-body radiation (which to a good approximation it is), then its temperature should be multiplied by the Doppler factor (in this case, 4.4), as described, for example, here. But radiation intensity goes like the fourth power of the temperature, so the radiation intensity (watts per square meter) seen by the spaceship should be 4.4 to the fourth power or about 375 times the intensity seen by an observer at rest relative to the Sun.

 

[jartsa]

I'm not sure this is right. If we view the Sun's radiation as black-body radiation (which to a good approximation it is), then its temperature should be multiplied by the Doppler factor (in this case, 4.4), as described, for example, here. But radiation intensity goes like the fourth power of the temperature, so the radiation intensity (watts per square meter) seen by the spaceship should be 4.4 to the fourth power or about 375 times the intensity seen by an observer at rest relative to the Sun.

I see. My photon density was too low. It should be 375/4.4 = 85 times the density seen by an observer at rest relative to the Sun. Maybe average distance between photons is always isotropic in black-body radiation. Then my calculation would succeed as I would say that to obey this law the same contraction must occur in the two other dimensions too. 4.4 * 4.4 * 4.4 = 85.

 

[pervect]

I believe jarsta gets basically the right answerr (as long as you use the square of the doppler shift factor). The formal argument that it is right is that we approximate the solar radiation by a null dust, and ask how the energy density scales , which we compute via the stress energy tensor. The answer is the square of the doppler shift factor.

For a paper on the topic, I'd suggest "In search of the ’’starbow’’: The appearance of the starfield from a relativistic spaceship" by Mckinley and Doeherty, the abstract is at http://dx.doi.org/10.1119/1.11834, I'm not sure if the full text version is online (it was at one time).

 

[PeterDonis]

we approximate the solar radiation by a null dust, and ask how the energy density scales , which we compute via the stress energy tensor. The answer is the square of the doppler shift factor.

I agree that this is how the energy density, viewed as the 0-0 component of the SET, scales. However, I'm not sure how this fits in with the radiation intensity varying as the fourth power of the Doppler factor. Here's a partial heuristic guess at how the two results could be consistent:

The total radiation intensity (when integrated over all frequencies) is a power per unit solid angle, which can be converted to a power per unit area at a given distance from the source by computing the total area of a 2-sphere at that distance. Power is energy per unit time, so an observer moving towards the source should see power increased by one extra gamma factor relative to energy density, because of time dilation; and the conversion of solid angle to area will pick up another factor of gamma for an observer moving towards the source, because the distance to the source will be length contracted.

This gets us two factors of gamma, but we still need two factors of $1 + v$ to get two more Doppler factors. So I'm still not sure how all this fits together.

 

[pervect]

I agree that this is how the energy density, viewed as the 0-0 component of the SET, scales. However, I'm not sure how this fits in with the radiation intensity varying as the fourth power of the Doppler factor. Here's a partial heuristic guess at how the two results could be consistent:

The total radiation intensity (when integrated over all frequencies) is a power per unit solid angle, which can be converted to a power per unit area at a given distance from the source by computing the total area of a 2-sphere at that distance. Power is energy per unit time, so an observer moving towards the source should see power increased by one extra gamma factor relative to energy density, because of time dilation; and the conversion of solid angle to area will pick up another factor of gamma for an observer moving towards the source, because the distance to the source will be length contracted.

This gets us two factors of gamma, but we still need two factors of $1 + v$ to get two more Doppler factors. So I'm still not sure how all this fits together.

There was a long PF thread on this before, but I'm not sure anymore what conclusion I drew at that time. I think I concluded that you did wind up with a blackbody spectrum, but the emission constant was different. I know I made at least one mistake along the way, which doesn't help me remember the right result :(.

https://www.physicsforums.com/threa...elativistic-speeds.696108/page-3#post-4411342
https://www.physicsforums.com/threa...elativistic-speeds.696108/page-5#post-4412774

 

[m4r35n357]

I found this paper: http://people.physics.anu.edu.au/~cms130/vrproject/resources/Notes_Relativity_2008.pdf when doing my twin paradox raytracing stuff. The last part of section 10 gives the POV of the Real-Time Relativity authors on this matter. They appear to specify either the third or fourth power of doppler, depending on what you are looking for. I never bothered to fully understand it as I had no intention of implementing it accurately ;)

 

[PeterDonis]

There was a long PF thread on this before

I think pervect summed it up pretty well (and resolved the question I had posed in post #7) in this post; his summary is basically this (he is using r to denote the Doppler factor):

The total photon arrival from the source, integrated over the entire view, scales as r. So the photon arrival rate increases if you're moving towards the object.

The total energy delivered scales as r^2, because $E = h \nu$ and $\nu$ gets doppler shifted. The usual definition of intensity is via delivered energy, so the intensity scales as r^2.

Relativistic aberation makes the angle subtended by the object smaller - this causes the object's apparent area to shrink by a factor of r^2. This would make the surface brightness scale as r^4 (the same energy is delivered in a smaller area).

So energy density does scale like the square of the Doppler factor, while intensity--power per unit area--does scale like the fourth power of the Doppler factor. Which is the "correct" answer depends on which of the two--energy density or intensity--is driving the effect you're interested in.